Quantum Field Theory for the Gifted Amateurfiore/libricorsoptr/QuantumFieldTheory...Quantum Field Theory for the Gifted Amateur

Quantum Field Theory for the Gifted Amateur

Quantum Field Theoryfor the Gifted Amateur

Tom LancasterDepartment of Physics, University of Durham

Stephen J. BlundellDepartment of Physics, University of Oxford

3

3

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Preface

BRICK: Well, they say nature hates a vacuum, Big Daddy.BIG DADDY: That’s what they say, but sometimes I thinkthat a vacuum is a hell of a lot better than some of the stuffthat nature replaces it with.Tennessee Williams (1911–1983) Cat on a Hot Tin Roof

Quantum field theory is arguably the most far-reaching and beautifulphysical theory ever constructed. It describes not only the quantum vac-uum, but also the stuff that nature replaces it with. Aspects of quantumfield theory are also more stringently tested, as well as verified to greaterprecision, than any other theory in physics. The subject nevertheless hasa reputation for difficulty which is perhaps well-deserved; its practition-ers not only manipulate formidable equations but also depict physicalprocesses using a strange diagrammatic language consisting of bubbles,wiggly lines, vertices, and other geometrical structures, each of whichhas a well defined quantitative significance. Learning this mathematicaland geometrical language is an important initiation rite for any aspir-ing theoretical physicist, and a quantum field theory graduate course isfound in most universities, aided by a large number of weighty quantumfield theory textbooks. These books are written by professional quan-tum field theorists and are designed for those who aspire to join themin that profession. Consequently they are frequently thorough, seriousminded and demand a high level of mathematical sophistication.

The motivation for our book is the idea that quantum field theory istoo important, too beautiful and too engaging to be restricted to theprofessionals. Experimental physicists, or theoretical physicists in otherfields, would benefit greatly from knowing some quantum field theory,both to understand research papers that use these ideas and also tocomprehend and appreciate the important insights that quantum fieldtheory has to offer. Quantum field theory has given us such a radicallydifferent and revolutionary view of the physical world that we thinkthat more physicists should have the opportunity to engage with it.The problem is that the existing texts require far too much in the wayof advanced mathematical facility and provide too little in the way ofphysical motivation to assist those who want to learn quantum fieldtheory but not to be professional quantum field theorists. The gapbetween an undergraduate course on quantum mechanics and a graduatelevel quantum field theory textbook is a wide and deep chasm, and oneof the aims of this book is to provide a bridge to cross it. That beingsaid, we are not assuming the readers of this are simple-minded folk who

vi Preface

can be fobbed off with a trite analogy as a substitute for mathematicalargument. We aim to introduce all the maths but, by using numerousworked examples and carefully worded motivations, to smooth the pathfor understanding in a manner we have not found in the existing books.

We have chosen this book’s title with great care.1 Our imagined reader1After all, with the number of chapterswe ended up including, we could havecalled it ‘Fifty shades of quantum fieldtheory’.

is an amateur, wanting to learn quantum field theory without (at leastinitially) joining the ranks of professional quantum field theorists; but(s)he is gifted, possessing a curious and adaptable mind and willing toembark on a significant intellectual challenge; (s)he has abundant cu-riosity about the physical world, a basic grounding in undergraduatephysics, and a desire to be told an entertaining and intellectually stim-ulating story, but will not feel patronized if a few mathematical nicetiesare spelled out in detail. In fact, we suspect and hope that our book willfind wide readership amongst the graduate trainee quantum field theo-rists who will want to use it in conjunction with one of the traditionaltexts (for learning most hard subjects, one usually needs at least twobooks in order to get a more rounded picture).

One feature of our book is the large number of worked examples, whichare set in slightly smaller type. They are integral to the story, and fleshout the details of calculations, but for the more casual reader the gutsof the argument of each chapter is played out in the main text. Toreally get to grips with the subject, the many examples should providetransparent demonstrations of the key ideas and understanding can beconfirmed by tackling the exercises at the end of each chapter. Thechapters are reasonably short, so that the development of ideas is keptat a steady pace and each chapter ends with a summary of the key ideasintroduced.

Though the vacuum plays a big part in the story of quantum field the-ory, we have not been writing in one. In many ways the present volumerepresents a compilation of some of the best ideas from the literatureand, as a result, we are indebted to these other books for providingthe raw material for many of our arguments. There is an extensive listof further reading in Appendix A where we acknowledge our sources,but we note here, in particular, the books by Zee and by Peskin andSchroeder and their legendary antecedent: the lectures in quantum fieldtheory by Sidney Coleman. The latter are currently available online asstreamed videos and come highly recommended. Also deserving of spe-cial mention is the text by Weinberg which is ‘a book to which we areall indebted, and from which none of us can escape.’22T.S. Eliot on Ulysses.

It is a pleasure to acknowledge the help we have received from var-ious sources in writing this book. Particular mention is due to SonkeAdlung at Oxford University Press who has helped steer this project tocompletion. No authors could wish for a more supportive editor andwe thank him, Jessica White and the OUP team, particularly Mike Nu-gent, our eagle-eyed copy editor. We are very grateful for the commentsand corrections we received from a number of friends and colleagues whokindly gave up their time to read drafts of various chapters: Peter Byrne,Claudio Castelnovo, John Chalker, Martin Galpin, Chris Maxwell, Tom

Preface vii

McLeish, Johannes Moller, Paul Tulip and Rob Williams. They deservemuch credit for saving us from various embarrassing errors, but any thatremain are due to us; those that we find post-publication will be postedon the book’s website:

http://www.dur.ac.uk/physics/qftgabook

For various bits of helpful information, we thank Hideo Aoki, Nikitas Gi-dopoulos, Paul Goddard and John Singleton. Our thanks are also due tovarious graduate students at Durham and Oxford who have unwittinglyserved as guinea pigs as we tried out various ways of presenting thismaterial in graduate lectures. Finally we thank Cally and Katherine fortheir love and support.

TL & SJBDurham & Oxford

January 2, 2014

Contents

0 Overture 1

0.1 What is quantum field theory? 10.2 What is a field? 20.3 Who is this book for? 20.4 Special relativity 30.5 Fourier transforms 60.6 Electromagnetism 7

I The Universe as a set of harmonic oscillators 9

1 Lagrangians 10

1.1 Fermat’s principle 101.2 Newton’s laws 101.3 Functionals 111.4 Lagrangians and least action 141.5 Why does it work? 16Exercises 17

2 Simple harmonic oscillators 19

2.1 Introduction 192.2 Mass on a spring 192.3 A trivial generalization 232.4 Phonons 25Exercises 27

3 Occupation number representation 28

3.1 A particle in a box 283.2 Changing the notation 293.3 Replace state labels with operators 313.4 Indistinguishability and symmetry 313.5 The continuum limit 35Exercises 36

4 Making second quantization work 37

4.1 Field operators 374.2 How to second quantize an operator 394.3 The kinetic energy and the tight-binding Hamiltonian 434.4 Two particles 44

x Contents

4.5 The Hubbard model 46Exercises 48

II Writing down Lagrangians 49

5 Continuous systems 50

5.1 Lagrangians and Hamiltonians 505.2 A charged particle in an electromagnetic field 525.3 Classical fields 545.4 Lagrangian and Hamiltonian density 55Exercises 58

6 A first stab at relativistic quantum mechanics 59

6.1 The Klein–Gordon equation 596.2 Probability currents and densities 616.3 Feynman’s interpretation of the negative energy states 616.4 No conclusions 63Exercises 63

7 Examples of Lagrangians, or how to write down a theory 64

7.1 A massless scalar field 647.2 A massive scalar field 657.3 An external source 667.4 The φ4 theory 677.5 Two scalar fields 677.6 The complex scalar field 68Exercises 69

III The need for quantum fields 71

8 The passage of time 72

8.1 Schrodinger’s picture and the time-evolution operator 728.2 The Heisenberg picture 748.3 The death of single-particle quantum mechanics 758.4 Old quantum theory is dead; long live fields! 76Exercises 78

9 Quantum mechanical transformations 79

9.1 Translations in spacetime 799.2 Rotations 829.3 Representations of transformations 839.4 Transformations of quantum fields 859.5 Lorentz transformations 86Exercises 88

10 Symmetry 90

10.1 Invariance and conservation 90

Contents xi

10.2 Noether’s theorem 9210.3 Spacetime translation 9410.4 Other symmetries 96Exercises 97

11 Canonical quantization of fields 98

11.1 The canonical quantization machine 9811.2 Normalizing factors 10111.3 What becomes of the Hamiltonian? 10211.4 Normal ordering 10411.5 The meaning of the mode expansion 106Exercises 108

12 Examples of canonical quantization 109

12.1 Complex scalar field theory 10912.2 Noether’s current for complex scalar field theory 11112.3 Complex scalar field theory in the non-relativistic limit 112Exercises 116

13 Fields with many components and

massive electromagnetism 117

13.1 Internal symmetries 11713.2 Massive electromagnetism 12013.3 Polarizations and projections 123Exercises 125

14 Gauge fields and gauge theory 126

14.1 What is a gauge field? 12614.2 Electromagnetism is the simplest gauge theory 12914.3 Canonical quantization of the electromagnetic field 131Exercises 134

15 Discrete transformations 135

15.1 Charge conjugation 13515.2 Parity 13615.3 Time reversal 13715.4 Combinations of discrete and continuous transformations 139Exercises 142

IV Propagators and perturbations 143

16 Propagators and Green’s functions 144

16.1 What is a Green’s function? 14416.2 Propagators in quantum mechanics 14616.3 Turning it around: quantum mechanics from the

propagator and a first look at perturbation theory 14916.4 The many faces of the propagator 151Exercises 152

xii Contents

17 Propagators and fields 154

17.1 The field propagator in outline 15517.2 The Feynman propagator 15617.3 Finding the free propagator for scalar field theory 15817.4 Yukawa’s force-carrying particles 15917.5 Anatomy of the propagator 162Exercises 163

18 The S-matrix 165

18.1 The S-matrix: a hero for our times 16618.2 Some new machinery: the interaction representation 16718.3 The interaction picture applied to scattering 16818.4 Perturbation expansion of the S-matrix 16918.5 Wick’s theorem 171Exercises 174

19 Expanding the S-matrix: Feynman diagrams 175

19.1 Meet some interactions 17619.2 The example of φ4 theory 17719.3 Anatomy of a diagram 18119.4 Symmetry factors 18219.5 Calculations in p-space 18319.6 A first look at scattering 186Exercises 187

20 Scattering theory 188

20.1 Another theory: Yukawa’s ψ†ψφ interactions 18820.2 Scattering in the ψ†ψφ theory 19020.3 The transition matrix and the invariant amplitude 19220.4 The scattering cross-section 193Exercises 194

V Interlude: wisdom from statistical physics 195

21 Statistical physics: a crash course 196

21.1 Statistical mechanics in a nutshell 19621.2 Sources in statistical physics 19721.3 A look ahead 198Exercises 199

22 The generating functional for fields 201

22.1 How to find Green’s functions 20122.2 Linking things up with the Gell-Mann–Low theorem 20322.3 How to calculate Green’s functions with diagrams 20422.4 More facts about diagrams 206Exercises 208

Contents xiii

VI Path integrals 209

23 Path integrals: I said to him, ‘You’re crazy’ 210

23.1 How to do quantum mechanics using path integrals 21023.2 The Gaussian integral 21323.3 The propagator for the simple harmonic oscillator 217Exercises 220

24 Field integrals 221

24.1 The functional integral for fields 22124.2 Which field integrals should you do? 22224.3 The generating functional for scalar fields 223Exercises 226

25 Statistical field theory 228

25.1 Wick rotation and Euclidean space 22925.2 The partition function 23125.3 Perturbation theory and Feynman rules 233Exercises 236

26 Broken symmetry 237

26.1 Landau theory 23726.2 Breaking symmetry with a Lagrangian 23926.3 Breaking a continuous symmetry: Goldstone modes 24026.4 Breaking a symmetry in a gauge theory 24226.5 Order in reduced dimensions 244Exercises 245

27 Coherent states 247

27.1 Coherent states of the harmonic oscillator 24727.2 What do coherent states look like? 24927.3 Number, phase and the phase operator 25027.4 Examples of coherent states 252Exercises 253

28 Grassmann numbers: coherent states

and the path integral for fermions 255

28.1 Grassmann numbers 25528.2 Coherent states for fermions 25728.3 The path integral for fermions 257Exercises 258

VII Topological ideas 259

29 Topological objects 260

29.1 What is topology? 26029.2 Kinks 26229.3 Vortices 264Exercises 266

xiv Contents

30 Topological field theory 267

30.1 Fractional statistics a la Wilczek:the strange case of anyons 267

30.2 Chern–Simons theory 26930.3 Fractional statistics from Chern–Simons theory 271Exercises 272

VIII Renormalization: taming the infinite 273

31 Renormalization, quasiparticles and the Fermi surface 274

31.1 Recap: interacting and non-interacting theories 27431.2 Quasiparticles 27631.3 The propagator for a dressed particle 27731.4 Elementary quasiparticles in a metal 27931.5 The Landau Fermi liquid 280Exercises 284

32 Renormalization: the problem and its solution 285

32.1 The problem is divergences 28532.2 The solution is counterterms 28732.3 How to tame an integral 28832.4 What counterterms mean 29032.5 Making renormalization even simpler 29232.6 Which theories are renormalizable? 293Exercises 294

33 Renormalization in action:

propagators and Feynman diagrams 295

33.1 How interactions change the propagator in perturbationtheory 295

33.2 The role of counterterms: renormalization conditions 29733.3 The vertex function 298Exercises 300

34 The renormalization group 302

34.1 The problem 30234.2 Flows in parameter space 30434.3 The renormalization group method 30534.4 Application 1: asymptotic freedom 30734.5 Application 2: Anderson localization 30834.6 Application 3: the Kosterlitz–Thouless transition 309Exercises 312

35 Ferromagnetism: a renormalization group tutorial 313

35.1 Background: critical phenomena and scaling 31335.2 The ferromagnetic transition and critical phenomena 315Exercises 320

Contents xv

IX Putting a spin on QFT 321

36 The Dirac equation 322

36.1 The Dirac equation 32236.2 Massless particles: left- and right-handed wave functions 32336.3 Dirac and Weyl spinors 32736.4 Basis states for superpositions 33036.5 The non-relativistic limit of the Dirac equation 332Exercises 334

37 How to transform a spinor 336

37.1 Spinors aren’t vectors 33637.2 Rotating spinors 33737.3 Boosting spinors 33737.4 Why are there four components in the Dirac equation? 339Exercises 340

38 The quantum Dirac field 341

38.1 Canonical quantization and Noether current 34138.2 The fermion propagator 34338.3 Feynman rules and scattering 34538.4 Local symmetry and a gauge theory for fermions 346Exercises 347

39 A rough guide to quantum electrodynamics 348

39.1 Quantum light and the photon propagator 34839.2 Feynman rules and a first QED process 34939.3 Gauge invariance in QED 351Exercises 353

40 QED scattering: three famous cross-sections 355

40.1 Example 1: Rutherford scattering 35540.2 Example 2: Spin sums and the Mott formula 35640.3 Example 3: Compton scattering 35740.4 Crossing symmetry 358Exercises 359

41 The renormalization of QED and two great results 360

41.1 Renormalizing the photon propagator: dielectric vacuum 36141.2 The renormalization group and the electric charge 36441.3 Vertex corrections and the electron g-factor 365Exercises 368

X Some applications from the worldof condensed matter 369

42 Superfluids 370

42.1 Bogoliubov’s hunting license 370

xvi Contents

42.2 Bogoliubov’s transformation 37242.3 Superfluids and fields 37442.4 The current in a superfluid 377Exercises 379

43 The many-body problem and the metal 380

43.1 Mean-field theory 38043.2 The Hartree–Fock ground state energy of a metal 38343.3 Excitations in the mean-field approximation 38643.4 Electrons and holes 38843.5 Finding the excitations with propagators 38943.6 Ground states and excitations 39043.7 The random phase approximation 393Exercises 398

44 Superconductors 400

44.1 A model of a superconductor 40044.2 The ground state is made of Cooper pairs 40244.3 Ground state energy 40344.4 The quasiparticles are bogolons 40544.5 Broken symmetry 40644.6 Field theory of a charged superfluid 407Exercises 409

45 The fractional quantum Hall fluid 411

45.1 Magnetic translations 41145.2 Landau Levels 41345.3 The integer quantum Hall effect 41545.4 The fractional quantum Hall effect 417Exercises 421

XI Some applications from the worldof particle physics 423

46 Non-abelian gauge theory 424

46.1 Abelian gauge theory revisited 42446.2 Yang–Mills theory 42546.3 Interactions and dynamics of Wµ 42846.4 Breaking symmetry with a non-abelian gauge theory 430Exercises 432

47 The Weinberg–Salam model 433

47.1 The symmetries of Nature before symmetry breaking 43447.2 Introducing the Higgs field 43747.3 Symmetry breaking the Higgs field 43847.4 The origin of electron mass 43947.5 The photon and the gauge bosons 440Exercises 443

Contents xvii

48 Majorana fermions 444

48.1 The Majorana solution 44448.2 Field operators 44648.3 Majorana mass and charge 447Exercises 450

49 Magnetic monopoles 451

49.1 Dirac’s monopole and the Dirac string 45149.2 The ’t Hooft–Polyakov monopole 453Exercises 456

50 Instantons, tunnelling and the end of the world 457

50.1 Instantons in quantum particle mechanics 45850.2 A particle in a potential well 45950.3 A particle in a double well 46050.4 The fate of the false vacuum 463Exercises 466

A Further reading 467

B Useful complex analysis 473

B.1 What is an analytic function? 473B.2 What is a pole? 474B.3 How to find a residue 474B.4 Three rules of contour integrals 475B.5 What is a branch cut? 477B.6 The principal value of an integral 478

Index 480

0Overture

0.1 What is quantum field the-ory? 1

0.2 What is a field? 2

0.3 Who is this book for? 2

0.4 Special relativity 3

0.5 Fourier transforms 6

0.6 Electromagnetism 7

To begin at the beginningDylan Thomas (1914–1953)

Beginnings are always troublesomeGeorge Eliot (1819–1880)

0.1 What is quantum field theory?

Every particle and every wave in the Universe is simply an excitationof a quantum field that is defined over all space and time.

That remarkable assertion is at the heart of quantum field theory. Itmeans that any attempt to understand the fundamental physical lawsgoverning elementary particles has to first grapple with the fundamentalsof quantum field theory. It also means that any description of compli-cated interacting systems, such as are encountered in the many-bodyproblem and in condensed matter physics, will involve quantum fieldtheory to properly describe the interactions. It may even mean, thoughat the time of writing no-one knows if this is true, that a full theoryof quantum gravity will be some kind of quantum upgrade of generalrelativity (which is a classical field theory). In any case, quantum fieldtheory is the best theory currently available to describe the world aroundus and, in a particular incarnation known as quantum electrodynamics(QED), is the most accurately tested physical theory. For example, themagnetic dipole moment of the electron has been tested to ten significantfigures.

The ideas making up quantum field theory have profound conse-quences. They explain why all electrons are identical (the same ar-gument works for all photons, all quarks, etc.) because each electron isan excitation of the same electron quantum field and therefore it’s notsurprising that they all have the same properties. Quantum field theoryalso constrains the symmetry of the representations of the permutationsymmetry group of any class of identical particles so that some classesobey Fermi–Dirac statistics and others Bose–Einstein statistics. Inter-actions in quantum field theory involve products of operators which arefound to create and annihilate particles and so interactions correspondto processes in which particles are created or annihilated; hence there

2 Overture

is also the possibility of creating and destroying virtual particles whichmediate forces.

0.2 What is a field?

This is all very well, but what is a field? We will think of a field as somekind of machine that takes a position in spacetime and outputs an ob-ject representing the amplitude of something at that point in spacetime(Fig. 1). The amplitude could be a scalar, a vector, a complex number,a spinor or a tensor. This concept of a field, an unseen entity whichpervades space and time, can be traced back to the study of gravity dueto Kepler and ultimately Newton, though neither used the term andthe idea of action-at-a-distance between two gravitationally attractingbodies seemed successful but nevertheless utterly mysterious. Euler’sfluid dynamics got closer to the matter by considering what we wouldnow think of as a velocity field which modelled the movement of fluid atevery point in space and hence its capacity to do work on a test parti-cle imagined at some particular location. Faraday, despite (or perhapsbecause of) an absence of mathematical schooling, grasped intuitivelythe idea of an electric or magnetic field that permeates all space andtime, and although he first considered this a convenient mental picturehe began to become increasingly convinced that his lines of force had anindependent physical existence. Maxwell codified Faraday’s idea and theelectromagnetic field, together with all the paraphernalia of field theory,was born.

xµ

φ(xµ)

Fig. 1 A field is some kind of ma-chine that takes a position in space-time, given by the coordinates xµ, andoutputs an object representing the am-plitude of something at that point inspacetime. Here the output is thescalar φ(xµ) but it could be, for ex-ample, a vector, a complex number, aspinor or a tensor.

Thus in classical physics we understand that gravity is a field, elec-tromagnetism is a field, and each can be described by a set of equationswhich governs their behaviour. The field can oscillate in space and timeand thus wave-like excitations of the field can be found (electromagneticwaves are well-known, but gravity waves are still to be observed). Theadvent of quantum mechanics removed the distinction between what hadbeen thought of as wave-like objects and particle-like objects. Thereforeeven matter itself is an excitation of a quantum field and quantum fieldsbecome the fundamental objects which describe reality.

0.3 Who is this book for?

Quantum field theory is undoubtedly important, but it is also noto-riously difficult. Forbidding-looking integrals and a plethora of funnysquiggly Feynman diagrams are enough to strike fear in many a heartand stomach. The situation is not helped by the fact that the many ex-cellent existing books are written by exceedingly clever practioners whostructure their explanations with the aspiring professional in mind. Thisbook is designed to be different. It is written by experimental physicistsand aimed at the interested amateur. Quantum field theory is too in-teresting and too important to be reserved for professional theorists.However, though our imagined reader is not necessarily an aspiring pro-


fessional (though we hope quite a few will be) we will assume that (s)heis enthusiastic and has some familiarity with non-relativistic quantummechanics, special relativity and Fourier transforms at an undergradu-ate physics level. In the remainder of this chapter we will review a fewbasic concepts that will serve to establish some conventions of notation.

0.4 Special relativity

Quantum fields are defined over space and time and so we need a properdescription of spacetime, and so we will need to use Einstein’s specialtheory of relativity which asserts that the speed c of light is the samein every inertial frame. This theory implies that the coordinates of anevent in a frame S and a frame S (moving relative to frame S at speedv along the x-axis) are related by the Lorentz transformation

t = γ(t− vx

c2

),

x = γ(x− vt),

y = y,

z = z, (1)

where γ = (1 − β2)−1/2 and β = v/c. Because the speed of light setsthe scale for all speeds, we will choose units such that c = 1. For similarreasons1 we will also set ~ = 1. 1In some of the early chapters we will

include the factors of ~ and c so thatthe reader can make better contact withwhat they already know, and will givenotice when we are going to removethem.

A good physical theory is said to be covariant if it transforms sen-sibly under coordinate transformations.2 In particular, we require that

2A good counterexample is the two-component ‘shopping vector’ that con-tains the price of fish and the price ofbread in each component. If you ap-proach the supermarket checkout withthe trolley at 45 to the vertical, youwill soon discover that the prices ofyour shopping will not transform ap-propriately.

quantities should be Lorentz covariant if they are to transform ap-propriately under the elements of the Lorentz group (which include theLorentz transformations of special relativity, such as eqn 1). This willrequire us to write our theory in terms of certain well-defined mathe-matical objects, such as scalars, vectors and tensors.3

3We will postpone discussion of tensorsuntil the end of this section.

• Scalars: A scalar is a number, and will take the same value inevery inertial frame. It is thus said to be Lorentz invariant.Examples of scalars include the electric charge and rest mass of aparticle.

• Vectors: A vector can be thought of as an arrow. In a particularbasis it can be described by a set of components. If the basisis rotated, then the components will change, but the length ofthe arrow will be unchanged (the length of a vector is a scalar).In spacetime, vectors have four components and are called four-

vectors. A four-vector is an object which has a single time-likecomponent and three space-like components. Three-vectors will bedisplayed in bold italics, such as x or p for position and momentumrespectively. The components of three-vectors are listed with aRoman index taken from the middle of the alphabet: e.g. xi, withi = 1, 2, 3. Four-vectors are made from a time-like part and aspace-like part and are displayed in italic script, so position in

4 Overture

spacetime is written x where x = (t,x). Components for four-vectors will be given a Greek index, so for example xµ where µ =0, 1, 2, 3. We say that the zeroth component, x0, is time-like.

Example 0.1

Some other examples of four-vectors4 are:4These are written with c = 1.

• the energy-momentum four-vector p = (E,p),

• the current density four-vector j = (ρ, j),

• the vector potential four-vector A = (V,A).

The four-dimensional derivative operator ∂µ is also a combination of a time-like partand a space-like part, and is defined5 by

5Though strictly ∂µ refers only to theµth component of the four-vector oper-ator ∂, rather than to the whole thing,we will sometimes write a subscript (orsuperscript) in expressions like this toindicate whether coordinates are listedwith the indices lowered or with themraised.

∂µ ≡ ∂

∂xµ=

„∂

∂t,∇

«

=

„∂

∂t,∂

∂x,∂

∂y,∂

∂z

«

. (2)

Note the lower index written in ∂µ, contrasting with the upper index on four-vectorslike xµ, which means that the four-dimensional derivative is ‘naturally lowered’. Thisis significant, as we will now describe.

A general coordinate transformation from one inertial frame to anothermaps xµ → xµ, and the vector aµ transforms as

aµ =

(∂xµ

∂xν

)aν . (3)

Here we have used the Einstein summation convention, by whichtwice repeated indices are assumed to be summed.6 Certain other

6In full, this equation would be

aµ =X

ν

„∂xµ

∂xν

«

aν .

vectors will transform differently. For example, the gradient vector∂µφ ≡ ∂φ/∂xµ transforms as

∂φ

∂xµ=

(∂xν

∂xµ

)∂φ

∂xν. (4)

The jargon is that aµ transforms like a contravariant vector and∂φ/∂xµ ≡ ∂µφ transforms like a covariant vector,7 though we will

7These unfortunate terms are dueto the English mathematician J. J.Sylvester (1814–1897). Both types ofvectors transform covariantly, in thesense of ‘properly’, and we wish to re-tain this sense of the word ‘covariant’rather than using it to simply label onetype of object that transforms properly.Thus we will usually specify whetherthe indices on a particular object are‘upstairs’ (like aµ) or ‘downstairs’ (like∂µφ) and their transformation proper-ties can then be deduced accordingly.

avoid these terms and just note that aµ has its indices ‘upstairs’ and∂µφ has them ‘downstairs’ and they will then transform accordingly.

The Lorentz transformation (eqn 1) can be rewritten in matrix formas

txyz

=

γ −βγ 0 0−βγ γ 0 0

0 0 1 00 0 0 1

txyz

, (5)

or for short asxµ = Λµν x

ν , (6)

where Λµν ≡ (∂xµ/∂xν) is the Lorentz transformation matrix. In thesame way, the energy-momentum four-vector transforms as

pµ = Λµν pν . (7)


A downstairs vector aµ transforms as

aµ = Λµν aν , (8)

where Λµν ≡ (∂xν/∂xµ) is the inverse of the Lorentz transformation

matrix Λµν .8

8 Note that in the equation xµ = Λµνxνwe have

Λµν =

0

BB@

γ βγ 0 0βγ γ 0 00 0 1 00 0 0 1

1

CCA. (9)

In fact

Λµν Λµρ =

„∂xµ

∂xν

«„∂xρ

∂xµ

«

= δρν ,

where the Kronecker delta δji is de-fined by

δji =

1 i = j0 i 6= j.

The δji symbol is named after LeopoldKronecker (1823–1891).

The Lorentz transformation changes components but leaves the lengthof the four-vector x unchanged. This length is given by the square root9

9Of course |x| can be negative in specialrelativity, so it is better to deal with|x|2, the square of the length.

of|x|2 = x · x = (x0)2 − (x1)2 − (x2)2 − (x3)2. (10)

In general, the four-vector inner product10 is

10Note that other conventions are pos-sible and some books write a · b =−a0b0 + a · b and define their metrictensor differently. This is an entirely le-gitimate alternative lifestyle choice, butit’s best to stick to one convention in asingle book.

a · b = a0b0 − a · b, (11)

which we can writea · b = gµν a

µbν , (12)

where the metric tensor gµν is given by

gµν =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

. (13)

Upstairs and downstairs vectors are related by the metric tensor via

aµ = gµν aν , (14)

so that we can lower or raise an index by inserting the metric tensor.The form of the metric tensor in eqn 13 allows us to write

a0 = a0 ai = −ai, (15)

and hencea · b = gµν a

µbν = aµbµ = aµbµ. (16)

Note also that a · b = gµν aµbν and gµν = gµν .

Exercise: You can check that

gµν gνρ = δρµ

and also that

Λµν = gµκ Λκρ g

ρν .

Example 0.2

(i) An example of an inner product is

p · p = pµpµ = (E,p) · (E,p) = E2 − p2 = m2, (17)

where m is the rest mass of the particle.

(ii) The combination ∂µxν = ∂xν

∂xµ = δνµ and hence the inner product ∂µxµ = 4(remember the Einstein summation convention).

(iii) The d’Alembertian operator ∂2 is given by a product of two derivative op- Named in honour of the French math-ematician Jean le Rond d’Alembert(1717–1783). In some texts thed’Alembertian is written as and insome as

2. Because of this confusion,we will avoid the symbol altogether.

erators (and is the four-dimensional generalization of the Laplacian operator).It is written as

∂2 = ∂µ∂µ =∂2

∂t2− ∂2

∂x2− ∂2

∂y2− ∂2

∂z2(18)

=∂2

∂t2− ∇

2. (19)

6 Overture

To complete this discussion, we can also define a general tensor T i···kℓ···nwith an arbitrary set of upstairs and downstairs indices. This transformsas

T i′···k′ℓ′···n′ =

∂xi′

∂xi· · · ∂x

k′

∂xk∂xℓ

∂xℓ′· · · ∂x

n

∂xn′ Ti···kℓ···n . (20)

Example 0.3

(i) The Kronecker delta δij is a ‘mixed tensor of second rank’,11 and one can check11It has two indices (hence secondrank) and one is upstairs, one is down-stairs (hence mixed).

that it transforms correctly as follows:

δji =∂xi

∂xk∂xℓ

∂xjδℓk =

∂xi

∂xk∂xk

∂xj= δij . (21)

Note that whenever δij or δij are written, they are not tensors and are simply ashorthand for the scalar 1, in the case when i = j, or 0 when i 6= j.

(ii) The antisymmetric symbol or Levi-Civita symbol12 εijkℓ is defined in four12This is named after Italian physi-cist Tullio Levi-Civita (1873–1941).Useful relationships with the Levi-Civita symbol include results for three-dimensional vectors:

(b× c)i = εijkbjck,

a · (b× c) = εijkaibjck,

and matrix algebra:

detA = εi1i2···inA1i1A2i2 · · ·Anin ,

where A is a n×n matrix with compo-nents Aij .

dimensions by (i) all even permutations ijkℓ of 0123 (such as ijkℓ = 2301) haveεijkℓ = 1; (ii) all odd permutations ijkℓ of 0123 (such as ijkℓ = 0213) have εijkℓ =−1; (iii) all other terms are zero (e.g. ε0012 = 0). The Levi-Civita symbol can bedefined in other dimensions.13 We will not treat this symbol as a tensor, so the

13The version in two dimensions israther simple:

ε01 = −ε10 = 1,

ε00 = ε11 = 0.

In three dimensions, the nonzero com-ponents are:

ε012 = ε201 = ε120 = 1,

ε021 = ε210 = ε102 = −1.

version with downstairs indices εijk is identical to εijk.

0.5 Fourier transforms

We will constantly be needing to swap between representations of anobject in spacetime and in the corresponding frequency variables, thatis spatial and temporal frequency. Spatial frequency k and temporalfrequency ω also form a four-vector (ω,k) and using E = ~ω and p =~k we see that this is the energy-momentum four-vector (E,p). (Infact, with our convention ~ = 1 the two objects are identical!) Toswap between representations, we define the four-dimensional Fouriertransform f(k) of a function f(x) of spacetime x as

Jean Baptiste Joseph Fourier (1768–1830)

f(k) =

∫d4x eik·xf(x), (22)

where four-dimensional integration is defined by∫

d4x =

∫dx0dx1dx2dx3. (23)

The inverse transform is

f(x) =

∫d4k

(2π)4e−ik·xf(k), (24)

and contains four factors of 2π that are needed for each of the fourintegrations. Another way of writing eqn 22 is

f(ω,k) =

∫d3xdt ei(ωt−k·x)f(t,x). (25)

In the spirit of this definition, we will try to formulate our equations sothat every factor of dk comes with a (2π), hopefully eliminating one ofthe major causes of insanity in the subject, the annoying factors14 of 2π.

14Getting these right is actually impor-

tant: if you have (2π)4 on the top of anequation and not the bottom, your an-swer will be out by a factor of well overtwo million.

0.6 Electromagnetism 7

Example 0.4

The Dirac delta function δ(x) is a function localized at the origin and whichhas integral unity. It is the perfect model of a localized particle. The integral of ad-dimensional Dirac delta function δ(d)(x) is given by

Z

ddx δ(d)(x) = 1. (26)

It is defined by Z

ddx f(x)δ(d)(x) = f(0). (27)

Consequently, its Fourier transform is given by

δ(d)(k) =

Z

ddx eik·xδ(d)(x) = 1. (28)

Hence, the inverse Fourier transform in four-dimensions isZ

d4k

(2π)4e−ik·x = δ(4)(x). (29)

0.6 Electromagnetism

In SI units Maxwell’s equations in free space can be written: James Clerk Maxwell (1831–1879)

∇ · E = ρǫ0, ∇ × E = −∂B

∂t ,

∇ · B = 0, ∇ × B = µ0J + 1c2∂E∂t .

(30)

In this book we will choose15 the Heaviside–Lorentz16 system of units 15Although SI units are preferable formany applications in physics, the desireto make our (admittedly often compli-cated) equations as simple as possiblemotivates a different choice of units forthe discussion of electromagnetism inquantum field theory. Almost all bookson quantum field theory use Heaviside–Lorentz units, though the famous text-books on electrodynamics by Landauand Lifshitz and by Jackson do not.

16These units are named after the En-glish electrical engineer O. Heaviside(1850–1925) and the Dutch physicistH. A. Lorentz (1853–1928).

(also known as the ‘rationalized Gaussian CGS’ system) which can beobtained from SI by setting ǫ0 = µ0 = 1. Thus the electrostatic potentialV (x) = q/4πǫ0|x| of SI becomes

V (x) =q

4π|x| , (31)

in Heaviside–Lorentz units, and Maxwell’s equations can be written

∇ · E = ρ, ∇ × E = − 1c∂B∂t ,

∇ · B = 0, ∇ × B = 1c (J + ∂E

∂t ).(32)

Using our other choice of c = ~ = 1 obviously removes the factors ofc from these equations. In addition, the fine structure constant α =e2/4π~c ≈ 1

137 simplifies to

α =e2

4π. (33)

Note that we will give electromagnetic charge q in units of the electroncharge e by writing q = Q|e|. The charge on the electron correspondsto Q = −1.

Part I

The Universe as a set of

harmonic oscillators

In this introductory part of the book, we trace the development of thepicture of the Universe which underpins quantum field theory. Thispicture is one in which harmonic oscillators are ubiquitous, and suchoscillators form a central paradigm of quantum systems. We show thatnon-interacting harmonic oscillators have eigenfunctions which look andbehave like particles and find some elegant methods for describing sys-tems of coupled oscillators.

• In Chapter 1 we provide a formulation of classical mechanics whichis suitable for a quantum upgrade. This allows us to talk aboutfunctionals and Lagrangians.

• The simple harmonic oscillator, presented in Chapter 2, is well-known from basic quantum physics as an elementary model of anoscillating system. We solve this simple model using creation andannihilation operators and show that the solutions have the charac-teristics of particles. Linking masses by springs into a chain allowsus to generalize this problem and the solutions are phonons.

• The next step is to change our viewpoint and get rid of wave func-tions entirely and develop the occupation number representationwhich we do in Chapter 3. We show that bosons are described bycommuting operators and fermions are described by anticommut-ing operators.

• Already we have many of the building blocks of a useful theory.In Chapter 4 we consider how to build single-particle operatorsout of creation and annihilation operators, and this already givesus enough information to discuss the tight-binding model of solidstate physics and the Hubbard model.

1 Lagrangians

1.1 Fermat’s principle 10

1.2 Newton’s laws 10

1.3 Functionals 11

1.4 Lagrangians and least action14

1.5 Why does it work? 16

Chapter summary 17

Exercises 17

In this chapter, we give an introduction to Lagrangians in classical me-chanics and explain why the Lagrangian formulation is a sensible wayto describe quantum fields.

1.1 Fermat’s principle

Fig. 1.1 Refraction of a light raythrough a slab of glass. The ray findsthe path of least travel time from A toB.

We begin with an example from the study of optics. Consider the passageof a light ray through a slab of glass as shown in Fig. 1.1. The bendingof the light ray near the air/glass interface can be calculated using theknown refractive indices of air and glass using the famous Snell’s law ofrefraction, named after Willebrord Snellius, who described it in 1621,but discovered first by the Arabian mathematician Ibn Sahl in 984. In1662, Pierre de Fermat produced an ingenious method of deriving Snell’slaw on the basis of his principle of least time. This states that thepath taken between two points A and B by a ray of light is the paththat can be traversed by the light in the least time. Because light travelsmore slowly in glass than in air, the ray of light crosses the glass at asteeper angle so it doesn’t have so much path length in the glass. If thelight ray were to take a straight line path from A to B this would takelonger. This was all very elegant, but it didn’t really explain why lightwould choose to do this; why should light take the path which took theleast time? Why is light in such a hurry?

Fermat’s principle of least time is cute, and seems like it is telling ussomething, but at first sight it looks unhelpful. It attempts to replacea simple formula (Snell’s law), into which you can plug numbers andcalculate trajectories, with a principle characterizing the solution of theproblem but for which you need the whole apparatus of the calculus ofvariations to solve any problems. Fermat’s principle is however the keyto understanding quantum fields, as we shall see.

1.2 Newton’s laws

t

x

0 τ

Fig. 1.2 A particle moves from A to Bin time τ and its path is described byNewton’s laws of motion.

A somewhat similar problem is found in the study of dynamics. Considera particle of mass m subject to a force F which moves in one spatialdimension x from point A to B, as shown in the spacetime diagram inFig. 1.2. Here time is on the horizontal axis and space is on the verticalaxis. The exact path x(t) that the particle takes is given by Newton’s

1.3 Functionals 11

laws of motion, i.e.

F = mx. (1.1)

This equation can be integrated to find x(t). However, when you stopand think about it, this is a very quantum-unfriendly approach. Thesolution gives you the position of the particle at every time t from t = 0to t = τ . Quantum mechanics tells us that you might measure theparticle’s position at t = 0 and find it at A, and you might measure itagain at t = τ and find it at B, but you’d not be able to know preciselywhat it did in between. Having a method which lets you calculate x(t)from a differential equation is not a good starting point.

Dynamics need to be formulated in a completely different way if thesubject is to be generalized to quantum mechanics. This is exactly whatJoseph-Louis Lagrange and William Rowan Hamilton did, although they J.-L. Lagrange (1736–1813) was an

Italian-born French mathematician andphysicist.

W. R. Hamilton (1805–1865) was anIrish mathematician and physicist.

had no idea that what they were doing would make dynamics morequantum-friendly. We will take a slightly different approach to theirsand arrive at the final answer by asking ourselves how kinetic energy Tand potential energy V vary during the trajectory of the particle. Weknow that they must sum to the total energy E = T+V which must be aconstant of the motion. But during the trajectory, the balance betweenkinetic and potential energy might change.

It is simple enough to write down the average kinetic energy T duringthe trajectory, which is given by

T =1

τ

∫ τ

0

1

2m[x(t)]2 dt. (1.2)

The average potential energy V during the trajectory is given by

V =1

τ

∫ τ

0

V [x(t)] dt. (1.3)

These two quantities must sum to give the total energy E = E = T + V .However, what we want to do is to consider how T and V vary as youalter the trajectory. To do this, we need a little bit of mathematics,which we cover in the next section.

1.3 Functionals

NUMBER NUMBER

FUNCTION

NUMBER

3 27

e−xx2

3

19

function

functional

Fig. 1.3 A function turns a numberinto another number. A functional op-erates on an entire function and pro-duces a number.

The expressions in eqns 1.2 and 1.3 are functionals of the trajectory x(t).What does this mean?

Let us recall that a function is a machine (see Fig. 1.3) which turns anumber into another number. For example, the function f(x) = 3x2 willturn the number 1 into the number 3, or the number 3 into the number27. Give a function a number and it will return another number.

A functional is a machine which turns a function into a number. Youfeed the machine with an entire function, like f(x) = x2 or f(x) = sinxand it returns a number.

12 Lagrangians

Example 1.1

Here are some examples of functionals.

• The functional F [f ] operates on the function f as follows:

F [f ] =

Z 1

0f(x) dx. (1.4)

Hence, given the function f(x) = x2, the functional returns the number

F [f ] =

Z 1

0x2 dx =

1

3. (1.5)

• The functional G[f ] operates on the function f as follows:

G[f ] =

Z a

−a5[f(x)]2 dx. (1.6)

Hence, given the function f(x) = x2, the functional returns the number

G[f ] =

Z a

−a5x4 dx = 2a5. (1.7)

• A function can be thought of as a trivial functional. For example, the func-tional Fx[f ] given by

Fx[f ] =

Z ∞

−∞f(y)δ(y − x)dy = f(x), (1.8)

returns the value of the function evaluated at x.

We now want to see how a functional changes as you adjust the functionwhich is fed into it. The key concept here is functional differentiation.Recall that a derivative of a function is defined as follows:

df

dx= limǫ→0

f(x+ ǫ) − f(x)

ǫ. (1.9)

The derivative of the function tells you how the number returned bythe function f(x) changes as you slightly change the number x that youfeed into the ‘machine’. In the same way, we can define a functional

derivative of a functional F [f ] as follows:

δF

δf(x)= limǫ→0

F [f(x′) + ǫδ(x− x′)] − F [f(x′)]

ǫ. (1.10)

The functional derivative tells you how the number returned by thefunctional F [f(x)] changes as you slightly change the function f(x) thatyou feed into the machine.

Example 1.2

Here are some examples of calculations of functional derivatives. You can workthrough these if you want to acquire the skill of functional differentiation, or skip tothe next bit of text if you are happy to take the results on trust.

1.3 Functionals 13

• The functional I[f ] =R 1−1 f(x) dx has a functional derivative given by

δI[f ]

δf(x0)= lim

ǫ→0

1

ǫ

»Z 1

−1[f(x) + ǫδ(x− x0)] dx−

Z 1

−1f(x) dx

–

=

Z 1

−1δ(x− x0) dx

=

(

1 −1 ≤ x0 ≤ 1

0 otherwise.(1.11)

• The functional J [f ] =R[f(y)]pφ(y) dy has a functional derivative with respect

to f(x) given by

δJ [f ]

δf(x)= lim

ǫ→0

1

ǫ

»Z

[f(y) + ǫδ(y − x)]pφ(y) dy −Z

[f(y)]pφ(y) dy

–

= p[f(x)]p−1φ(x). (1.12)

• The functional H[f ] =R ba g[f(x)] dx, where g is a function whose derivative is

g′ = dg/dx, has a functional derivative given by

δH[f ]

δf(x0)= lim

ǫ→0

1

ǫ

»Z

g[f(x) + ǫδ(x− x0)] dx−Z

g[f(x)] dx

–

= limǫ→0

1

ǫ

»Z

(g[f(x)] + ǫδ(x− x0)g′[f(x)]) dx−Z

g[f(x)] dx

–

=

Z

δ(x− x0) g′[f(x)] dx

= g′[f(x0)]. (1.13)

• Using the result of the previous example, the functional V [x] = 1τ

R τ0 V [x(t)] dt

so thatδV [x]

δx(t)=

1

τV ′[x(t)]. (1.14)

• The functional J [f ] is defined by J [f ] =Rg(f ′) dy where f ′ = df/dy. Hence

δJ [f ]

δf(x)= limǫ→0

1

ǫ

»Z

dy g

„∂

∂y[f(y) + ǫδ(y − x)]

«

−Z

dy g

„∂f

∂y

«–

, (1.15)

and using

g

„∂

∂y[f(y) + ǫδ(y − x)]

«

= g`f ′ + ǫδ′(y − x)

´≈ g(f ′) + ǫδ′(y − x)

dg(f ′)

df ′,

(1.16)then the calculation can be reduced to an integral by parts

δJ [f ]

δf(x)=

Z

dy δ′(y−x) dg(f ′)

df ′=

»

δ(y − x)dg(f ′)

df ′

–

−Z

δ(y−x) d

dy

„dg(f ′)

df ′

«

.

(1.17)The term in square brackets vanishes assuming x is inside the limits of theintegral, and we have simply

δJ [f ]

δf(x)= − d

dx

„dg(f ′)

df ′

«

. (1.18)

• An example of the previous result is that for F [φ] =R “ ∂φ

∂y

”2dy,

δF [φ]

δφ(x)= −2

∂2φ

∂x2. (1.19)

Equation 1.19 can be easily generalizedto three dimensions and leads to a veryuseful result which is worth memoriz-ing, namely that if

I =

Z

(∇φ)2 d3x,

thenδI

δφ= −2∇

2φ.

• Another example is for T = 1τ

R τ0

12m[x(t)]2 dt,

δT [x]

δx(t)= −mx

τ. (1.20)

14 Lagrangians

1.4 Lagrangians and least action

With these mathematical results under our belt, we are now ready toreturn to the main narrative. How does the average kinetic energy andthe average potential energy vary as we adjust the particle trajectory?We now have the main results from eqns 1.14 and 1.20 which are:

δV [x]

δx(t)=V ′[x(t)]

τ,

δT [x]

δx(t)= −mx

τ. (1.21)

In fact, Newton’s laws tell us that for the classical trajectory of theparticle we have that mx = −dV/dx. This means that for variationsabout the classical trajectory

δV [x]

δx(t)=δT [x]

δx(t). (1.22)

This means that if you slightly deviate from the classical trajectory,then both the average kinetic energy and the average potential energywill increase1 and by the same amount. This equation can be rewritten1They might both decrease by the same

amount if the classical trajectory max-imizes (rather than minimizes) the ac-tion, see below, but this case is not theone usually encountered.

asδ

δx(t)(T [x] − V [x]) = 0, (1.23)

i.e. that the difference between the average kinetic energy and the av-erage potential energy is stationary about the classical trajectory. Thisshows that there is something rather special about the difference be-tween kinetic energy and potential energy, and motivates us to define aquantity known as the Lagrangian L as

L = T − V. (1.24)

The integral of the Lagrangian over time is known as the action S

S =

∫ τ

0

Ldt, (1.25)

and so the action has dimensions of energy×time, and hence is measuredin Joule-seconds. This is the same units as Planck’s constant h, and wewill see later in this chapter why it is often appropriate to think ofmeasuring S in units of Planck’s constant. Our variational principle(eqn 1.23) connecting variations of average kinetic energy and averagepotential energy can now be rewritten in a rather appealing way sincefor this problem S =

∫ τ0

(T − V ) dt = τ(T [x] − V [x]), so that

t

x

0 τ

Fig. 1.4 Small adjustments to the pathof a particle from its classical trajectorylead to an increase in the action S.

δS

δx(t)= 0, (1.26)

and this is known as Hamilton’s principle of least action.2 It states

2To be pedantic, the principle onlyshows the action is stationary. It couldbe a maximum, or a saddle point, justas easily as a minimum. ‘Stationaryaction’ would be better than ‘least ac-tion’. But we are stuck with the name.

that the classical trajectory taken by a particle is such that the action isstationary; small adjustments to the path taken by the particle (Fig. 1.4)only increase the action (in the same way that small adjustments to thepath taken by a ray of light from the one determined by Snell’s lawlengthen the time taken by the light ray).

1.4 Lagrangians and least action 15

Example 1.3

The Lagrangian L can be written as a function of both position and velocity. Quitegenerally, one can think of it as depending on a generalized position coordinate x(t)and its derivative x(t), called the velocity. Then the variation of S with x(t) isδS/δx(t) and can be written as

δS

δx(t)=

Z

du

»δL

δx(u)

δx(u)

δx(t)+

δL

δx(u)

δx(u)

δx(t)

–

=

Z

du

»δL

δx(u)δ(u− t) +

δL

δx(u)

d

dtδ(u− t)

–

=δL

δx(t)+

»

δ(u− t)δL

δx(u)

–tf

ti

−Z

du δ(u− t)d

dt

δL

δx(u)

=δL

δx(t)− d

dt

δL

δx(t), (1.27)

and hence the principle of least action (eqn 1.26) yields

δL

δx(t)− d

dt

δL

δx(t)= 0, (1.28)

which is known as the Euler–Lagrange equation.

Leonhard Euler (1707–1783). In thewords of Pierre-Simon Laplace (1749–1827): Read Euler, read Euler, he is

the master of us all.

The Lagrangian L is related to the Lagrangian density L by

L =

∫dxL, (1.29)

so that the action S is given by

S =

∫dt L =

∫dtdxL. (1.30)

The following example introduces the idea of a Lagrangian density, aconcept we will come back to frequently, but also provides a nice way toderive the classical wave equation.

ψ(x, t)

x

dx

ρ dx

Fig. 1.5 Waves on a string. The dis-placement from equilibrium is ψ(x, t)and the equation of motion can be de-rived by considering an element of thestring of length dx and mass ρ dx. Thefigure shows a short section in the mid-dle of the string which is assumed to betethered at either end so that ψ(0, t) =ψ(ℓ, t).

Example 1.4

Consider waves on a string of mass m and length ℓ. Let us define the mass densityρ = m/ℓ, tension T and displacement from the equilibrium ψ(x, t) (see Fig. 1.5). The

kinetic energy T can then be written as T = 12

R ℓ0 dx ρ(∂ψ/∂t)2 and the potential

energy V = 12

R ℓ0 dx T (∂ψ/∂x)2. The action is then

S[ψ(x, t)] =

Z

dt (T − V ) =

Z

dt dxL„

ψ,∂ψ

∂t,∂ψ

∂x

«

, (1.31)

where

L„

ψ,∂ψ

∂t,∂ψ

∂x

«

=ρ

2

„∂ψ

∂t

«2

− T2

„∂ψ

∂x

«2

(1.32)

is the Lagrangian density. We then have immediately

0 =δS

δψ=

∂L∂ψ

− d

dx

∂L∂(∂ψ/∂x)

− d

dt

∂L∂(∂ψ/∂t)

= 0 + T ∂2ψ

∂x2− ρ

∂2ψ

∂t2, (1.33)

and so the wave equation (∂2ψ/∂x2) = (1/v2)(∂2ψ/∂t2) with v =p

T /ρ emergesalmost effortlessly.

16 Lagrangians

As a final trick, let us put the Euler–Lagrange equation on a fully rela-tivistic footing, bracing ourselves to use some four-vector notation andthe Einstein summation convention (see Section 0.4). If the Lagrangiandensity L depends on a function φ(x) (where x is a point in spacetime)and its derivative3 ∂µφ, then the action S is given by3Recall from the argument in Sec-

tion 0.4 that the index µ in ∂µφ is nat-urally lowered; see eqn 0.2.

S =

∫d4xL(φ, ∂µφ). (1.34)

By analogy with eqn 1.27, the action principle gives44Remember that we are using the Ein-stein summation convention, by whichtwice repeated indices are assumedsummed. Equation 1.35 is simplythe four-dimensional generalisation ofeqns 1.27 and 1.28.

δS

δφ=∂L∂φ

− ∂µ

(∂L

∂(∂µφ)

)= 0, (1.35)

the four-vector version of the Euler–Lagrange equation.

Example 1.5

As a simple example of this, consider the case when L is given by

L =1

2(∂µφ)2 − 1

2m2φ2, (1.36)

where (∂µφ)2 = (∂µφ)(∂µφ). Simple differentiation then gives

∂L∂φ

= −m2φ, and∂L

∂(∂µφ)= ∂µφ. (1.37)

Use of the action principle (eqn 1.35) gives

δS

δφ= −m2φ− ∂µ∂

µφ = 0 and hence (∂2 +m2)φ = 0. (1.38)

1.5 Why does it work?

In this chapter, we have considered two variational principles: Fermat’sprinciple of least time and Hamilton’s principle of least action. Onedescribes the path taken by a ray of light, the other describes the pathtaken by a classical particle. They are very elegant, but why not stickto using Snell’s law and Newton’s law? And why do they both work?

The answer to both of these questions is quantum mechanics. We willtalk about this in more detail later in the book, but the motion of aparticle (photon or billiard ball) going from A to B involves all possiblepaths, the sensible classical ones and completely insane ones. You haveto sum them all up, but they are each associated with a phase factor, andfor most sets of paths the different phases mean that the contributionscancel out. It is only when the phase is stationary that nearby paths allgive a non-cancelling contribution. The wave function for a particle hasa phase factor5 given by

5Fermat’s principle of least time tells ussomething about optical path length ofa ray, that is the difference between thephase at the beginning and end of a ray.By analogy with Hamilton’s principleof least action of a classical mechanicalsystem, one can posit that the action Sis given by a constant multiplied by thephase of a wave function. This definesthe constant which is given the symbol~. Thus we take the phase to be S/~.

eiS/~, (1.39)

Exercises 17

where S =∫Ldt is the action, so a stationary phase equates to a sta-

tionary action (eqn 1.26). (Running the argument in reverse, the phasefactor for a photon of energy E is e−iEt/~, and so stationary phaseequates to stationary time, which is Fermat’s principle.) We will seehow this approach leads naturally to Feynman’s path-integral approachlater in the book (Chapter 23). But for now, notice simply that if theaction is stationary then the classical path which minimizes the actionis the one that is observed and everything else cancels.

Snell’s law and Newton’s law are enough to solve classical systems.But neither allow the generalization to quantum systems to be performedwith ease. Thus, to formulate quantum field theory (the grand task ofthis book), we have to start with a Lagrangian picture. Our next step isto gain some insight into what happens in non-classical systems, and soin the next chapter we will turn our attention to an archetypal quantumsystem: the simple harmonic oscillator.

Chapter summary

• Fermat’s principle of least time states that light takes a path whichtakes the least time.

• The Lagrangian for a classical particle is given by L = T − V .

• Classical mechanics can be formulated using Hamilton’s principleof least action.

δS

δx(t)= 0, (1.40)

where the action S =∫Ldt.

• Both Fermat’s and Hamilton’s principles show how the classicalpaths taken by a photon or massive particle are ones in which thephase of the corresponding wave function is stationary.

Exercises

(1.1) Use Fermat’s principle of least time to derive Snell’slaw.

(1.2) Consider the functionals H[f ] =R

G(x, y)f(y) dy,

I[f ] =R 1

−1f(x) dx and J [f ] =

Z „

∂f

∂y

«2

dy of the

function f . Find the functional derivativesδH[f ]

δf(z),

δ2I[f3]

δf(x0)δf(x1)and

δJ [f ]

δf(x).

(1.3) Consider the functional G[f ] =R

g(y, f) dy. Showthat

δG[f ]

δf(x)=∂g(x, f)

∂f. (1.41)

Now consider the functional H[f ] =R

g(y, f, f ′) dyand show that

δH[f ]

δf(x)=∂g

∂f− d

dx

∂g

∂f ′ , (1.42)

18 Lagrangians

where f ′ = ∂f/∂y. For the functional J [f ] =R

g(y, f, f ′, f ′′) dy show that

δJ [f ]

δf(x)=∂g

∂f− d

dx

∂g

∂f ′ +d2

dx2

∂g

∂f ′′ , (1.43)

where f ′′ = ∂2f/∂y2.

(1.4) Show that

δφ(x)

δφ(y)= δ(x− y), (1.44)

andδφ(t)

δφ(t0)=

d

dtδ(t− t0). (1.45)

(1.5) For a three-dimensional elastic medium, the poten-tial energy is

V =T2

Z

d3x (∇ψ)2, (1.46)

and the kinetic energy is

T =ρ

2

Z

d3x

„

∂ψ

∂t

«2

. (1.47)

Use these results, and the functional derivative ap-proach, to show that ψ obeys the wave equation

∇2ψ =

1

v2

∂2ψ

∂t2, (1.48)

where v is the velocity of the wave.

(1.6) Show that if Z0[J ] is given by

Z0[J ] = exp

„

−1

2

Z

d4xd4y J(x)∆(x− y)J(y)

«

,

(1.49)where ∆(x) = ∆(−x) then

δZ0[J ]

δJ(z1)= −

»Z

d4y∆(z1 − y)J(y)

–

Z0[J ]. (1.50)

2Simple harmonic oscillators

2.1 Introduction 19

2.2 Mass on a spring 19

2.3 A trivial generalization 23

2.4 Phonons 25

Chapter summary 27

Exercises 27

2.1 Introduction

The advent of quantum mechanics convinced people that things that hadpreviously been thought of as particles were in fact waves. For example,it was found that electrons and neutrons were not simply little rigidbits of matter but they obey a wave equation known as the Schrodingerequation. This idea is known as first quantization. To summarize:

First quantization: Particles behave like waves. (2.1)

However, this is not the end of the story. It was also realized thatthings that had been previously thought of as waves were in fact par-ticles. For example, electromagnetic waves and lattice waves were notsimply periodic undulations of some medium but they could actuallybehave like particles. These phenomena were given particle-like names:photons and phonons. This idea is known as second quantization. Tosummarize:

Second quantization: Waves behave like particles. (2.2)

Quite how these ideas link up is one of main themes in quantum fieldtheory which sees the underlying reality behind both waves and particlesas a quantum field. However, before we get to this point, it is worthspending some time reviewing second quantization in a bit more detailas it is the less familiar idea. In this chapter, we focus on the mostfamous example of a wave phenomenon in physics: the oscillations of amass on a spring.

2.2 Mass on a spring

m

K

Fig. 2.1 A mass m suspended on aspring, of spring constant K.

Consider a mass on a spring (as shown in Fig. 2.1), one of the simplestphysical problems there is. Assume we have mass m, spring constantK, the displacement of the mass from equilibrium is given by x and themomentum of the mass is given by p = mx. The total energy E is thesum of the kinetic energy p2/2m and the potential energy 1

2Kx2.

In quantum mechanics, we replace p by the operator −i~∂/∂x, andwe then have the Schrodinger equation for a harmonic oscillator

(− ~2

2m

∂2

∂x2+

1

2Kx2

)ψ = Eψ. (2.3)

20 Simple harmonic oscillators

The solutions to this equation can be obtained by a somewhat involvedseries-solution method and are given by

ψn(ξ) =1√2nn!

(mωπ~

)1/4

Hn(ξ)e−ξ2/2, (2.4)

where Hn(ξ) is a Hermite polynomial and ξ =√mω/~x. As shown

in Fig. 2.2, these eigenfunctions look very wave-like. However, theydo have a ‘particle-like’ quality which is apparent from the eigenvalues,which turn out to be

En =

(n+

1

2

)~ω, (2.5)

where ω =√K/m. This is a ladder of energy levels (see Fig. 2.3), though

note when n = 0 the energy is not zero but ~ω/2. This is called thezero-point energy. Because the energy levels form a ladder, you mustadd a quantum of energy ~ω to move up the ladder by one rung. It is asif you are adding a lump, or particle, of energy. Can we make this vaguefeeling any more concrete? Yes we can. Moreover, we can do it quiteelegantly and without dirtying our hands with Hermite polynomials!

Fig. 2.2 Eigenfunctions of the simpleharmonic oscillator.

E = 0

E

n = 0

n = 1

n = 2

n = 3

n = 4

n = 5

···

hω

Fig. 2.3 The ladder of energy levels forthe simple harmonic oscillator.

To accomplish this, we start with the Hamiltonian for the simple har-monic oscillator written out as follows:

H =p2

2m+

1

2mω2x2, (2.6)

where we have expressed the spring constant as K = mω2. One feelsthe seductive temptation to factorize this Hamiltonian and write it as

1

2mω2

(x− i

mωp

)(x+

i

mωp

), (2.7)

but we hit a problem: in quantum mechanics, the operators x and p donot commute. Hence, multiplying out the brackets in eqn 2.7 doesn’tquite give us back H but instead produces

1

2mω2

(x− i

mωp

)(x+

i

mωp

)=

1

2mω2x2 +

p2

2m+

iω

2[x, p], (2.8)

where [x, p] ≡ xp − px = i~ is the commutator of x and p. The right-hand side of eqn 2.8 is then just H − ~ω

2 , which is nearly H but hasthe correction −~ω

2 due to the zero-point energy being subtracted. Thefactorization can therefore be made to work and we realize that theoperators x− i

mω p and x+ imω p are clearly going to be useful. Since p

and x are Hermitian, the operators x− imω p and x+ i

mω p are adjointsof each other (and therefore are not themselves Hermitian, so cannotcorrespond to any observable). We will give them the special names a†

and a, although we will include a multiplicative constant√mω/2~ in

our definition so that they have particularly nice properties. Hence wewill write

a =

√mω

2~

(x+

i

mωp

), (2.9)

a† =

√mω

2~

(x− i

mωp

). (2.10)

2.2 Mass on a spring 21

Example 2.1

The commutator [a, a†] can be evaluated as follows (remembering that [x, p] = i~):

[a, a†] =mω

2~

„

− i

mω[x, p] +

i

mω[p, x]

«

=mω

2~

„~

mω+

~

mω

«

= 1. (2.11)

The definitions of a and a† can be inverted to give

x =

√~

2mω(a+ a†), (2.12)

p = −i

√~mω

2(a− a†). (2.13)

Putting our new definitions of a and a† into our equation for the Hamil-tonian yields

H = ~ω

(a†a+

1

2

). (2.14)

The active ingredient in this Hamiltonian is the combination a†a. Ifa†a has an eigenstate |n〉 with eigenvalue n, then H will also have aneigenstate |n〉 with eigenvalue ~ω(n+ 1

2 ), so that we have recovered theeigenvalues of a simple harmonic oscillator in eqn 2.5. However, we needto prove that n takes the values 0, 1, 2, . . .. The first step is to show thatn ≥ 0. We can do that by noting that

n = 〈n|a†a|n〉 = |a|n〉|2 ≥ 0. (2.15)

Next, we have to show that n takes only integer values, and we willdo that below but beforehand let us introduce a bit of notation to savesome writing. We define the number operator n by

n = a†a, (2.16)

and hence writen|n〉 = n|n〉. (2.17)

Number of what? The quantity n labels the energy level on the ladder(see Fig. 2.3) that the system has reached, or equivalently the numberof quanta (each of energy ~ω) that must have been added to the systemwhen it was in its ground state. We can rewrite the Hamiltonian as

H = ~ω

(n+

1

2

), (2.18)

and therefore

H|n〉 =

(n+

1

2

)~ω|n〉, (2.19)

so that |n〉 is also an eigenstate of the Hamiltonian. Thus |n〉 is aconvenient shorthand for the more complicated form of ψn(ξ) shown ineqn 2.4. The next examples show that the eigenvalue n indeed takesinteger values.


Example 2.2

This example looks at the property of the state defined by a†|n〉. One of the thingswe can do is to operate on this with the number operator:

na†|n〉 = a†aa†|n〉. (2.20)

Using the commutator in eqn 2.11 gives aa† = 1 + a†a and hence

na†|n〉 = (n+ 1)a†|n〉. (2.21)

The above example shows that the state a†|n〉 is an eigenstate of H butwith an eigenvalue one higher than the state |n〉. In other words, theoperator a† has the effect of adding one quantum of energy. For thisreason a† is called a raising operator.

Example 2.3

This example looks at the property of the state defined by a|n〉. We can operate onthis with the number operator:

na|n〉 = a†aa|n〉. (2.22)

Using the commutator in eqn 2.11 gives a†a = aa† − 1 and hence

na|n〉 = (n− 1)a|n〉. (2.23)

The above example shows that the state a|n〉 is an eigenstate of H butwith an eigenvalue one lower than the state |n〉. In other words, theoperator a has the effect of subtracting one quantum of energy. For thisreason a is called a lowering operator.

Example 2.4

Question: Normalize the operators a and a†.Solution: We have shown that a|n〉 = k|n− 1〉, where k is a constant. Hence, takingthe norm of this state (i.e. premultiplying it by its adjoint) gives

|a|n〉|2 = 〈n|a†a|n〉 = |k|2〈n− 1|n− 1〉 = |k|2, (2.24)

where the last equality is because the simple harmonic oscillator states are normalized(so that 〈n− 1|n− 1〉 = 1). However, we notice that a†a = n is the number operatorand hence

〈n|a†a|n〉 = 〈n|n|n〉 = n. (2.25)

Equations 2.24 and 2.25 give k =√n. (This assumes k to be real, but any state

contains an arbitrary phase factor, so we are free to choose k to be real.)In the same way, we have shown that a†|n〉 = c|n + 1〉, where c is a constant.

Hence,|a†|n〉|2 = 〈n|aa†|n〉 = |c|2〈n+ 1|n+ 1〉 = |c|2, (2.26)

and using aa† = 1 + a†a = 1 + n, we have that

〈n|aa†|n〉 = 〈n|1 + n|n〉 = n+ 1, (2.27)

and hence c =√n+ 1 (choosing c to be real). In summary, our results are:

a|n〉 =√n|n− 1〉, (2.28)

a†|n〉 =√n+ 1|n+ 1〉. (2.29)

2.3 A trivial generalization 23

If we keep hitting the state |n〉 with a, we eventually get to |0〉, theground state of the simple harmonic oscillator. At this point, we willannihilate the state completely with a further application of a becausea|0〉 = 0. Notice that

H|0〉 = ~ω

(n+

1

2

)|0〉 =

1

2~ω|0〉, (2.30)

so this really is the ground state and we see that the energy is 12~ω, the

zero-point energy.1

1To recap, we can write

|a|n〉|2 = 〈n|a†a|n〉

= 〈n| H~ω

− 1

2|n〉

=En

~ω− 1

2,

where En = (n+ 12)~ω. Thus the con-

dition a|0〉 = 0 implies E0~ω

− 12

= 0, andhence the energy of the ground state isgiven by E0 = 1

2~ω.

It is also possible to go the other way, operating on |0〉 with the raisingoperator a†. This leads to

a†|0〉 = |1〉, (2.31)

a†|1〉 =√

2|2〉 =⇒ |2〉 =(a†)2√

2|0〉, (2.32)

a†|2〉 =√

3|3〉 =⇒ |3〉 =(a†)3√3 × 2

|0〉, (2.33)

and in general

|n〉 =(a†)n√n!

|0〉, (2.34)

so the state |n〉 can be obtained by repeated application of the operatora†. This leads to a completely different way of thinking about thesenew operators: we call a† a creation operator and a an annihilation

operator. We imagine a† acting to create a quantum of energy ~ω andmove the oscillator up one rung of the ladder. Its adjoint, a, acts toannihilate a quantum of energy ~ω and move the oscillator down onerung of the ladder (see Fig. 2.4). These quanta of energy behave likeparticles; we are adding and subtracting particles by the application ofthese operators and we have realized the dream of second quantization.A wave problem has spontaneously produced particles!

E = 0E

|0〉

|1〉

|2〉

|3〉

|4〉

|5〉

···

a†

a†

a†

a†

a†

a

a

a

a

a

a

Fig. 2.4 The ladder of energy levels forthe simple harmonic oscillator. The op-erator a† moves the oscillator up onerung of the ladder. Its adjoint, a, actsto move the oscillator down one rungof the ladder. Note that a|0〉 = 0, sodropping off the bottom of the ladderresults in zero.2.3 A trivial generalization

The next thing we can do is a completely trivial generalization of whatwe have done before, but it is worth thinking about because certaincomplicated problems reduce to it. Consider a set of N uncoupled simpleharmonic oscillators. They are uncoupled, so one could be on your desk,another sitting in your bathroom, another one out in the park. Theydon’t talk to each other, influence each other or affect each other inany way. Nevertheless, just for fun, we are going to consider their jointHamiltonian H which is simply the sum of individual Hamiltonians Hk

where k runs from 1 to N . Hence

H =

N∑

k=1

Hk, (2.35)


and

Hk =p2k

2mk+

1

2mkω

2kx

2k. (2.36)

So far, nothing complicated. We now define the operator a†k which cre-ates a quantum of energy in the kth oscillator (and leaves all the othersunaffected). We also define the operator ak which annihilates a quan-tum of energy in the kth oscillator (and leaves all the others unaffected).Acting with an operator ak only affects the number of quanta for thekth oscillator. We can therefore write

a†k|n1, n2, ..., nr, ...〉 ∝ |n1, n2, ..., nk + 1, ...〉, (2.37)

ak|n1, n2, ..., nk, ...〉 ∝ |n1, n2, ..., nk − 1, ...〉.

The operators will have the commutation rules:

[ak, aq] = 0, (2.38)[a†k, a

†q

]= 0, (2.39)

[ak, a

†q

]= δkq. (2.40)

In eqn 2.40, the point is that operators acting on different oscillatorscommute (and therefore do not affect each other). Hence we can followthe results of the previous section and write the Hamiltonian in eqn 2.35as

H =

N∑

k=1

~ωk

(a†kak +

1

2

). (2.41)

Again we have to define a vacuum state |0, 0, 0, 0, 0, ...〉 (usually justcalled |0〉), such that

ak|0〉 = 0, (2.42)

for all k. This is the state in which every one of the oscillators is in itsground state.

A general state of the system, written as |n1, n2, · · · , nN 〉 is known asthe occupation number representation. Using the same techniquesas in Section 2.2, we can write this general state as

|n1, n2, · · · , nN 〉 =1√

n1!n2! · · ·nN !(a†1)

n1(a†2)n2 · · · (a†N )nN |0, 0, · · · 0〉.

(2.43)The idea here is that we are operating on the vacuum state with aproduct of creation operators necessary to put n1 quanta of energy intooscillator number 1, n2 quanta of energy into oscillator number 2, etc.We can express this even more succinctly using the following notation

|nk〉 =∏

k

1√nk

(a†k)nk |0〉, (2.44)

where |0〉 is the vacuum state, as explained above.In this model, all of the oscillators have been completely indepen-

dent. It is now time to tackle a more challenging problem, coupling theoscillators together, and that is covered in the next section.

2.4 Phonons 25

2.4 Phonons

Consider a linear chain of N atoms (see Fig. 2.5), each of mass m, andconnected by springs of unstretched length a and which have springconstant K. The masses are normally at position Rj = ja, but can bedisplaced slightly by an amount xj . The momentum of the jth mass ispj . The Hamiltonian for this system is

m m m

j − 1 j j + 1

Kxj

Rj = ja

Fig. 2.5 A linear chain of atoms.

H =∑

j

p2j

2m+

1

2K(xj+1 − xj)

2. (2.45)

In contrast to the model in the previous section, we are now dealingwith a coupled problem. Each mass is strongly coupled to its neighbourby the springs and there is no way in which we can consider them asindependent. However, we will show that the excitations in this systembehave exactly as a set of totally independent oscillators. This worksbecause we can Fourier transform the problem, so that even thoughthe masses are coupled in real space, the excitations are uncoupled inreciprocal space. How this is done is covered in the following example.

Example 2.5

We begin by Fourier transforming both xj and pj , by writing

For the time being, we will drop the‘hats’ on xj and pj to save clutteringthe algebra. We note that there is an-other way of solving this problem whichinvolves writing the operators xj andpj in terms of creation and annihila-tion operators of particular modes. Itis arguably a more elegant route to theanswer and will generalize more easilyto additional cases, but as it is a littlemore abstract we will save it for laterin the book.

xj =1√N

X

k

xkeikja, (2.46)

pj =1√N

X

k

pkeikja, (2.47)

and equivalently of course

xk =1√N

X

j

xje−ikja, (2.48)

pk =1√N

X

j

pje−ikja. (2.49)

We impose periodic boundary conditions2 forcing eikja = eik(j+N)a The wave vector

2Periodic boundary conditions are de-scribed in detail in the next chapter.

k therefore takes the values 2πm/Na, where m is an integer in the range −N/2 <m ≤ N/2. Note that X

j

eikja = Nδk,0. (2.50)

The commutation relations are

The second quantization trickto evaluate sums of the form

1

N

X

j

X

kq

pkpqei(k+q)ja : (2.51)

Step 1: Perform the spatialsum, making use of the identityP

j ei(k−k′)ja = Nδkk′ . This gives

X

pq

pkpqδk,−q . (2.52)

Step 2: Use the Kronecker delta to doone of the momentum sums. This hasthe effect of setting q = −k, leaving uswith a sum over a single index:

X

k

pkp−k. (2.53)

ˆxj , pj′

˜= i~δjj′ , (2.54)

[xk, pk′ ] =1

N

X

j

X

j′

e−ikjae−ik′j′aˆxj , pj′

˜

=i~

N

X

j

e−i(k+k′)ja

= i~δk,−k′ . (2.55)

We can now work out terms in the Hamiltonian, so that

X

j

p2j =X

j

1√N

X

k

pkeikja

!

1√N

X

k′

pk′eik′ja

!

=1

N

X

j

X

k

X

k′

pkpk′ei(k+k′)ja. (2.56)


This combination of sums over one spatial and two momentum indices will occurfrequently and is easily dealt with using the trick shown in the box. The result is

X

j

p2j =X

k

pkp−k. (2.57)

The other term in the Hamiltonian may be treated similarly and we haveX

j

(xj+1 − xj)2 =

1

N

X

j

X

k

X

k′

xkxk′ei(k+k′)ja(eika − 1)(eik

′a − 1)

=X

k

xkx−k

„

4 sin2 ka

2

«

, (2.58)

where use has been made of the identity 1−cos ka = 2 sin2 ka2

. We can thus express33Notational point: For clarity, we willnow omit the tilde from xk and pk, butto remind ourselves that they are op-erators we will reinstate the ‘hats’, sothey become xk and pk.

the Hamiltonian as

H =X

k

»1

2mpkp−k +

1

2mω2

kxkx−k

–

, (2.59)

where ω2k = (4K/m) sin2(ka/2).

With our definitions we automatically have that p†k = p−k and x†k = x−k (whichfollow from the requirement that pj and xj are Hermitian), so we can write down

the commutation relation as [xk, p†k′

] = i~δkk′ and we can also write down creationand annihilation operators as follows:

ak =

rmωk

2~

„

xk +i

mωkpk

«

, (2.60)

a†k =

rmωk

2~

„

xk − i

mωkpk

«

, (2.61)

which have commutation relations [a†k, a†k′

] = [ak, ak′ ] = 0 and [ak, a†k′

] = δk,k′ . Wecan invert eqns 2.60 and 2.61 to yield

xk =

s

~

2mωk

“

ak + a†−k

”

, (2.62)

pk = −i

rm~ωk

2

“

ak − a†−k

”

. (2.63)

The Hamiltonian becomes

H =X

k

~ωk

2

“

aka†k + a†−ka−k

”

, (2.64)

and reindexing this gives

H =X

k

~ωk

2

“

aka†k + a†kak

”

. (2.65)

Using the commutator4 then produces4i.e. aka†k − a†kak = 1.

H =NX

k=1

~ωk

„

a†kak +1

2

«

. (2.66)

This example has demonstrated that the Hamiltonian for a set of coupledmasses on a chain can be expressed in terms of a sum over modes,labelled by wave vector k. Comparing this result (eqn 2.66) with thatobtained at the end of the previous section (eqn 2.44), one sees thatthey are identical. Thus these modes behave as if they are entirelyindependent and uncoupled simple harmonic oscillators. We call thesemodes phonons, and each phonon mode, labelled by k, can be giveninteger multiples of the quantum of energy ~ωk.

Exercises 27

This is the key point. Each phonon mode behaves like a simple har-monic oscillator and so can accept energy in an integer number of lumps,each lump being of size ~ωk. This is because the energy eigenvalues of asimple harmonic oscillator form a ladder of energy levels where the rung-size is fixed. These lumps of energy look like particles, and so we thinkof phonons themselves as particles. Later on, we play exactly the sametrick with the electromagnetic field and show that it behaves like a setof uncoupled simple harmonic oscillators and call the quanta photons.This is the heart of second quantization: wave problems have oscillatorsolutions and hence produce particles! The key insight in this chapteris essentially that the oscillator picture of physical systems leads to aparticle picture of those systems. This notion is summarized in Fig. 2.6.

Simple harmonic oscillator

in the nth level

En =n + 1

2

hωk

n particles of energy hωk

PARTICLE PICTURE

OSCILLATOR PICTURE

⇐⇒

Fig. 2.6 The equivalence of the oscil-lator and particle pictures.

Chapter summary

• First quantization shows that particles behave like waves; secondquantization shows that waves behave like particles.

• The simple harmonic oscillator has energy eigenvalues given byEn = (n + 1

2 )~ω. Eigenstates can be written in the occupationnumber representation.

• Using creation and annihilation operators the Hamiltonian for thesimple harmonic oscillator can be written H = ~ω(a†a+ 1

2 ).

• The phonon problem can be re-expressed as a sum over non-interacting modes, each one of which behaves like a simple har-monic oscillator.

Exercises

(2.1) For the one-dimensional harmonic oscillator, showthat with creation and annihilation operators de-fined as in eqns 2.9 and 2.10, [a, a] = 0, [a†, a†] = 0,[a, a†] = 1 and H = ~ω(a†a+ 1

2).

(2.2) For the Hamiltonian H = p2

2m+ 1

2mω2x2 + λx4,

where λ is small, show by writing the Hamiltonianin terms of creation and annihilation operators andusing perturbation theory, that the energy eigenval-ues of all the levels are given by

En =

„

n+1

2

«

~ω +3λ

4

„

~

mω

«2

(2n2 + 2n+ 1).

(2.67)

(2.3) Use eqns 2.46 and 2.62 to show that

xj =1√N

„

~

m

«

X

k

1

(2ωk)1/2[ake

ikja + a†ke−ikja].

(2.68)

(2.4) Using a|0〉 = 0 and eqns 2.9 and 2.10 together with〈x|p|ψ〉 = −i~ d

dx〈x|ψ〉, show that

0 =

„

x+~

mω

d

dx

«

〈x|0〉, (2.69)

and hence

〈x|0〉 =“mω

π~

”1/4

e−mωx2/2~ . (2.70)

3Occupation number

representation

3.1 A particle in a box 28

3.2 Changing the notation 29

3.3 Replace state labels with op-erators 31

3.4 Indistinguishability and sym-metry 31

3.5 The continuum limit 35

Chapter summary 36

Exercises 36

I am not a number. I am a free man.Patrick McGoohan in The Prisoner

In the previous chapter we considered simple harmonic oscillators,showed that such oscillators have particle-like eigenfunctions, and in-troduced en passant the occupation number representation to describesets of uncoupled oscillators. Now we are going to make further use ofthis idea to describe systems of many particles more generally. The useof the occupation number representation to describe a system of identi-cal particles is partly a matter of notational administration and partlya radical change in the way we look at the world. We (i) change theway we label our states (this is administration) and (ii) get rid of wavefunctions altogether (this is the radical part). We replace wave func-tions with a version of the creation and annihilation operators we usedto solve the quantum oscillator problems. This turns out to simplifythings enormously. In setting up this new method we will need to en-sure that we never violate the sacred laws of indistinguishable particles.These are as follows:

• There are two types of particle in the Universe: Bose1 particles1Satyendra Nath Bose (1894–1974)wrote a paper in 1924 which laid thefoundations for quantum statistics.

and Fermi2 particles.

2In addition to his many contribu-tions to other branches of physics, En-rico Fermi (1901–1954) made numerouscontributions to quantum field theory.His 1932 review of Quantum Electro-dynamics (Rev. Mod. Phys. 4, 87) hadenormous influence on the field.

• If you exchange two identical bosons you get the same state again.

• If you exchange two fermions, you get minus the state you startedwith.3

3A consequence of this principle is thatfermions obey the Pauli exclusion prin-ciple: It is impossible to put two in thesame quantum state. This affects thestatistical distribution among a set ofdiscrete energy levels of these indistin-guishable quantum particles, and so isreferred to as quantum statistics. InSections 3.1–3.3 we won’t worry aboutquantum statistics, but return to theissue in Section 3.4.

To make things simple we’ll confine our particles to a box for most ofthis chapter. Although this makes the notation ugly (and non-covariant)it has several advantages (not least that we can normalize the states ina simple way). We recap the physics of particle in a box in the nextsection.

3.1 A particle in a box

We will now choose units equivalent to setting ~ = 1. Therefore themomentum operator for motion along the x-direction p can be written

p = −i∂

∂x. (3.1)

3.2 Changing the notation 29

Solutions to the Schrodinger equation for a particle in a box of sizeL are running waves ψ(x) = 1√

Leipx, and eigenstates of the momentum

operator [pψ(x) = −idψdx = pψ(x)]. We use periodic boundary conditionswhich force us to have the same amplitude for our wave at equivalentpositions of each box, that is ψ(x) = ψ(x+L). Therefore eipx = eip(x+L),and hence eipL = 1. In order to satisfy this condition, we need pL =2πm, with m an integer. This imposes a quantization condition on thepossible momentum states that particles in the box can take. They mustsatisfy

pm =2πm

L. (3.2)

We’ll call the possible momentum states things like p1 and p2, etc. Theenergy of a single particle in a state pm is Epm

, which depends on themomentum state, but it doesn’t matter how yet.

p

np

p1 p2

1

2

3

Fig. 3.1 A two-particle state given by|p1p2〉.

To make a simple many-particle system we’ll put more and more non-interacting Bose particles in the box. How does this affect the totalmomentum and energy of the system? Applying the momentum opera-tor to the two-particle state (see Fig. 3.1) yields

p|p1p2〉 = (p1 + p2)|p1p2〉, (3.3)

and applying the Hamiltonian operator gives us

H|p1p2〉 = (Ep1 + Ep2)|p1p2〉. (3.4)

Since the particles don’t interact, having two particles in a particularenergy state (p3 say) costs an energy 2Ep3 , i.e. double the single-particleenergy. The total energy is given by

∑m npm

Epm, where npm

is thetotal number of particles in the state |pm〉.

3.2 Changing the notation

As advertised, the first thing we’ll do is change the notation. In conven-tional quantum mechanics we label each identical particle (e.g. particleA, particle B, particle C) and list its momentum. This gives rise to stateslabelled like |ABC〉 = |p1p2p1〉 for a three-particle state. We read thisas ‘particle A in momentum state p1, particle B in state p2 and particleC in state p1’. This is shown in Fig. 3.2 How could we streamline this

p

np

p1 p2

1

2

3

Fig. 3.2 A three-particle state givenby |p1p2p1〉 can be written as |2100 · · · 〉in the occupation number representa-tion.

notation? We haven’t yet used the facts that the particles are indistin-guishable (making the assignment A, B, C, etc. rather meaningless) andthat there are only certain allowable values of momentum (i.e. 2πm/L).We can therefore decide that instead of listing which particle (A, B, C,etc.) is in which state pm, we could just list the values of the momentumpm, saying how many particles occupy each momentum state. For ourexample, we’d say ‘two particles are in momentum state p1, one is inmomentum state p2’. We therefore define a state of N particles by list-ing how many are in each of the momentum states. We will write theseas |n1n2n3 . . .〉. Our example is written |2100 . . .〉. Specifying a state by

30 Occupation number representation

listing the number of identical particles in each quantum state is anotherinstance of the occupation number representation, which we metin the last chapter.

Example 3.1

Let’s compare the old and new notation for a system with only two possible states:p1 = 2π/L and p2 = 4π/L. Into these states we can put any number of Bose particlesthat we choose. The table in the margin compares the old and new notations.

Particles old new

0 |0〉 |00〉

1 |q1〉 |10〉1 |q2〉 |01〉

2 |q1q1〉 |20〉2 |q1q2〉 |11〉2 |q2q2〉 |02〉

3 |q1q1q1〉 |30〉...

.

.

....

and so on.

Table of old and new notations.

What happens when we act on a state in occupation number represen-tation with the Hamiltonian? As we saw above, the answer needs tobe

H|n1n2n3 . . .〉 =

[∑

m

npmEpm

]|n1n2n3 . . .〉, (3.5)

that is, we multiply the number of particles in each state by the energyof that state and sum over all of the states. This is obviously the rightthing to do to find the total energy.

Now for the reason why occupation numbers are useful. Equation 3.5can be made to look quite similar to something we’ve seen before. Recallthat a quantum harmonic oscillator had a total energy given by E =(n+ 1

2

)~ω with n being some integer. Dropping the zero-point energy

(which just amounts to a new choice in our zero of energy) we canwrite this as E = n~ω. In words, the oscillator has n quanta in it.Crucially, the energy levels are equally spaced, so having two quantain the system gives an energy which is twice that for a single quantumin the system. Now imagine N independent oscillators labelled by theindex k. These might not be identical, so we say that the energy levelspacing of the kth oscillator is ~ωk. The total energy of this systemis now E =

∑Nk=1 nk~ωk. So the kth oscillator has nk quanta in it,

contributing an energy nk~ωk to the sum. Compare this to our systemof identical particles for which the total energy is E =

∑pmnpm

Epm.

In words, this says that the momentum state pm has npmparticles in

it. Each particle contributes an energy Epmgiving a contribution of

npmEpm

for the momentum mode pm. These two systems thereforehave the same energy level structure.

We have built an analogy between two completely different systems:harmonic oscillators and identical particles.

SHO Identical particles

Quanta in oscillators → Particles in momentum stateskth oscillator → mth momentum mode pm

E =∑Nk=1 nk~ωk → E =

∑Nm=1 npm

Epm.

(3.6)

As we’ll see in the next section, the real value to this analogy is that itcan be pushed further.4

4We should be a little careful here;we haven’t physically replaced par-ticles with pendulums. We’ve sim-ply exploited the fact that the sim-ple harmonic oscillator energy levels areequally spaced, so, for example, twoquanta in an oscillator have an energythat is twice that of a single quantumin that oscillator.

3.3 Replace state labels with operators 31

3.3 Replace state labels with operators

The notation change to occupation number representation is just win-dow dressing. We’re now going to make a more dramatic change in anattempt to almost remove state vectors altogether! We’ll do this by hav-ing operators describing the physics rather than state vectors. We’ll haveto retain only one special state, the vacuum |0〉. Following eqn 2.43,5 5For the oscillator problem in Chap-

ter 2 we had the very useful technol-ogy of creation and annihilation oper-ators to add and remove quanta fromthe SHO system, and we use the sameidea here.

we will build up a general state of several harmonic oscillators by actingon the vacuum state |0〉 and write

|n1n2 · · · 〉 =∏

k

1

(nk!)12

(a†k)nk |0〉. (3.7)

Example 3.2

We can build up a state with quanta arranged among various oscillators. A state ofthe system in which there are two quanta in oscillator number 1 and one quantumin oscillator number 2 (written in occupation number representation as |21000 . . .〉)can be written as

|21000 . . .〉 =

»1√2!

(a†1)2– »

1√1!a†2

–

|0〉. (3.8)

However, what we are doing here is not just talking about creatingquanta in oscillators, but also particles in particular momentum states.We want a creation operator a†pm

to create one particle in a momentumstate |pm〉. It’s hopefully now just a matter of changing k → pm to gofrom operators describing oscillators to operators describing particles inmomentum states and we say that a†pm

creates a particle in momentumstate |pm〉 rather than a quantum in oscillator k. Before we can do this,we need to check that doing this doesn’t violate any of the symmetryprinciples we stated at the start of this chapter. That is, we need toknow how this new formulation deals with Bose and Fermi particles.

3.4 Indistinguishability and symmetry

We now have to make sure that introducing creation and annihilation ofstates will respect the exchange symmetry principles for identical par-ticles. It will turn out that the basic technology is supplied by thecommutation relations of the creation a†pm

and annihilation apmopera-

tors for a particular state. The key point is that it matters which orderyou use these operators to put particles into states.

Example 3.3

Consider a system with two momentum states p1 and p2 described in the occupationnumber representation as |n1n2〉 formed from creation operators acting on a vacuumstate |0〉. We’ll define

a†p1 |0〉 = |10〉, a†p2 |0〉 = |01〉. (3.9)


Now let’s add another particle in the unoccupied state. This gives us

a†p2 a†p1 |0〉 ∝ |11〉, a†p1 a

†p2 |0〉 ∝ |11〉. (3.10)

The proportionality sign is because the constant remains to be determined.

Whether you first put one particle in state p1 and then another in statep2, or do the same thing in the reverse order, you must end up with thesame final state: |11〉. This means that

a†p1 a†p2 = λa†p2 a

†p1 , (3.11)

where λ is a constant. We assume two possibilities,6 λ = ±1, and they6In Chapter 29 we examine the possi-bility of other choices. correspond to wave functions which are either symmetric or antisym-

metric under particle exchange. The particles are then called bosons

or fermions and we will find that the two cases correspond to two dif-ferent types of commutation relation for the creation and annihilationoperators. We will examine each case in turn.

Case I: λ = 1, bosons: In this case

a†p2 a†p1 = a†p1 a

†p2 . (3.12)

Rearranging (and generalizing the labels of the momentum states to iand j) we find [

a†i , a†j

]= a†i a

†j − a†j a

†i = 0, (3.13)

which means that the creation operators for different particle states com-mute. We also have [ai, aj ] = 0 and, importantly, we define

[ai, a

†j

]= δij , (3.14)

so that we can use our harmonic oscillator analogy. We conclude thatthese commutation relations are the same as those for simple harmonicoscillators above. We’ve succeeding in developing a system to describemany particles! The formalism is identical to that we developed formany independent oscillators in the previous section. We can thereforebuild up a general state of many particles in momentum states using

|n1n2 · · · 〉 =∏

m

1

(npm!)

12

(a†pm)npm |0〉. (3.15)

The particles this formalism describes are Bose particles: we can put anynumber into quantum states and, as we’ll see later, they are symmetricupon exchange of particles.

Finally we note that the commutation of different operators impliesthat

a†p1 a†p2 |0〉 = a†p2 a

†p1 |0〉 = |1p11p2〉, (3.16)

that is, it doesn’t matter which order you put particles in the states,you get exactly the same in either case.

3.4 Indistinguishability and symmetry 33

In general we can write

a†i |n1 · · ·ni · · · 〉 =√ni + 1|n1 · · ·ni + 1 · · · 〉, (3.17)

ai|n1 · · ·ni · · · 〉 =√ni|n1 · · ·ni − 1 · · · 〉. (3.18)

Case II: λ = −1, fermions: In this case we’ll write the fermion cre-ation operators c†i to distinguish them from the boson operators (writtena†i ). Plugging in λ = −1, we find that

c†i , c

†j

≡ c†i c

†j + c†j c

†i = 0, (3.19)

where the curly brackets indicate an anticommutator. Fermion oper-ators anticommute: when you change their order you pick up a minussign since c†i c

†j = −c†j c

†i . In particular, setting i = j we find

c†i c†i + c†i c

†i = 0 and hence c†i c

†i = 0. (3.20)

In other words, trying to shoehorn two fermions into the same state an-nihilates it completely. This is the Pauli exclusion principle7 which 7The Pauli exclusion principle is named

after Wolfgang Pauli (1900–1958), theAustrian theoretical physicist who wasone of the pioneers of quantum mechan-ics, and also of quantum field theoryand quantum electrodynamics, whichwere his main concerns after 1927. Hereferred to fellow-physicist WolfgangPaul (1913–1993) as his ‘real part’.Pauli was merciless with his criticism ofother physicists and their work, famousfor his put-downs, one of the most well-known of which was that a particularscientific paper was ‘not even wrong’.

means that each state can either be unoccupied or contain a singlefermion. We also have

ci, cj = 0. (3.21)

Finally we define an anticommutator

ci, c

†j

= δij , (3.22)

which enables us to use the simple harmonic oscillator analogy here too.Again we can describe putting particles in momentum states in the sameway as putting quanta in independent oscillators. The only difference isthat we use anticommutators rather than commutators.

Finally we note that since c†i c†j |0〉 = −c†j c

†i |0〉, it really does matter

which order you put the particles into the states.

Example 3.4

Let’s check that the number operator still works for fermions. The number operatoris ni = c†i ci. We’ll see the action of n1 on a state |11〉 ≡ c†1c

†2|0〉. We expect that this

is an eigenstate with eigenvalue 1 if all goes to plan because there is one particle inthe i = 1 state. We’ll now proceed to check this:

n1|11〉 = c†1c1|11〉= c†1c1(c†1c

†2|0〉)

= c†1(1 − c†1c1)c†2|0〉 (using the anticommutation relation)

= c†1c†2|0〉 ≡ |11〉, (3.23)

i.e. |11〉 is an eigenstate of n1 with eigenvalue 1 as we expected.


For fermions, in order to get the signs right, the normalization of creationand annihilation operators are defined as follows

c†i |n1 · · ·ni · · · 〉 = (−1)Σi√

1 − ni|n1 · · ·ni + 1 · · · 〉, (3.24)

ci|n1 · · ·ni · · · 〉 = (−1)Σi√ni|n1 · · ·ni − 1 · · · 〉, (3.25)

where

(−1)Σi = (−1)(n1+n2+···+ni−1). (3.26)

This gives us a factor of −1 for every particle standing to the left of thestate labelled ni in the state vector.

The square roots are redundant ineqns 3.24 and 3.25 since ni = 0 or 1,so they can be omitted if desired.

Example 3.5

Let’s check to see we’ve got it right when we come to exchange the particles. Remem-ber that for bosons we get the same state upon exchange and for fermions we pickup a minus sign. To exchange particles we need to imagine a sequence of operationswhich swap particles around. An example would be

|110〉 a state with particles in states 1 and 2 (3.27)

→ |101〉 move particle from state 2 to 3

→ |011〉 move particle from state 1 to 2

→ |110〉 move particle from state 3 to 1.

To move a particle we annihilate it from its original state and then create it in itsnew state. Moving a particle from state 2 to 3 is achieved using a†3a2|110〉 The fullsequence of events can be described with the string of operators

a†1a3a†2a1a

†3a2|110〉 = ±|110〉, (3.28)

where we want the + sign for bosons and the − sign for fermions.Let’s check that the commutation relations give the correct results for swapping

Bose particles. We can swap operators that refer to different states (since [ai, aj ] = 0)to get

a†1a3a†2a1a

†3a2|110〉 = a3a

†3a

†1a1a

†2a2|110〉. (3.29)

But a†i ai is just the number operator ni, counting the number of states in a mode i.

a3a†3n1n2|110〉 = (1) × (1) × a3a

†3|110〉, (3.30)

we need to turn a3a†3 into a number operator. Swap the order with the commutators

to get

(1 + a†3a3)|110〉 = (1 + n3)|110〉, (3.31)

and notice that n3|110〉 = 0 to obtain (1 + a†3a3)|110〉 = |110〉, as expected.Let’s now check that the commutation relations give the correct results for swap-

ping Fermi particles. Now when we swap any two fermion operators we pick up aminus sign. Carrying out the same steps as before

c†1c3c†2c1c

†3c2|110〉 = −c3c†3c

†1c1c

†2c2|110〉 (3.32)

= −(1 − c†3c3)|110〉= −|110〉.

Exchanging the particles gives us the required minus sign.

3.5 The continuum limit 35

3.5 The continuum limit

So far our particles have been confined in a box, but we will frequentlywish to work in the continuum limit of the formalism. As the box sizeincreases, the momentum states become more closely spaced until, in thelimit, the momentum becomes a continuous variable. The Kronecker δijfunctions we’ve been using become Dirac δ(3)(p) functions and the sumsbecome integrals. We have, for the continuous case, that

[ap, a†q] = δ(3)(p − q), (3.33)

and for energies

H =

∫d3pEpa

†pap. (3.34)

Now that we have a working theory where the commutation propertiesof the operators contain all of the information about the states, let’scheck that it works and gets the correct rules for inner products and forposition space wave functions.

Example 3.6

For single-particle states, we have

〈p|p′〉 = 〈0|ap a†p′ |0〉. (3.35)

We use the commutation relations to get

〈p|p′〉 = 〈0|h

δ(3)(p− p′) ± a†p′ ap

i

|0〉 (3.36)

= δ(3)(p− p′),

which is the right answer. We can check this by working out a position space wavefunction. We’ll start by making a change of basis8 |x〉 =

Rd3q φ∗q (x)|q〉, which in- 8The state |x〉 can be written in a new

basis by inserting the resolution of theidentity in the form

1 =

Z

d3q |q〉〈q|,

so that

|x〉 =

Z

d3q |q〉〈q|x〉,

and writing 〈x|q〉 = φq (x) [and hence〈q|x〉 = φ∗q (x)] yields

|x〉 =

Z

d3q φ∗q (x)|q〉.

volves expanding a position state in terms of momentum states. Using this expansion,we obtain

〈x|p〉 =

Z

d3q φq (x)〈q|p〉 = φp(x). (3.37)

This is clearly okay, but rather trivial. More interesting is the case of a two-particlestate

〈p′q′|qp〉 = 〈0|ap′ aq′ a†q a†p |0〉. (3.38)

Commuting, we obtain

〈p′q′|qp〉 = δ(3)(p′ − p)δ(3)(q′ − q) ± δ(3)(p′ − q)δ(3)(q′ − p), (3.39)

where we take the plus sign for bosons and minus sign for fermions. Again we canwork out the position space wave function by making a change of basis |xy〉 =

1√2!

Rd3p′d3q′ φ∗

p′ (x)φ∗q′ (y)|p′q′〉 (where the factor of 1√

2!is needed to prevent the

double counting that results from the unrestricted sums). This gives us

1√2!

Z

d3p′d3q′φp′ (x)φq′ (y)〈p′q′|pq〉 =1√2

[φp(x)φq (y) ± φq (x)φp(y)] , (3.40)

which is the well-known expression for two-particle states.

It works. We have successfully reformulated quantum mechanics. Theeffort we’ve spent here writing everything in terms of creation and anni-hilation operators will soon pay off. The machinery they represent is souseful that we’ll try to build the quantum fields we’re searching for outof these objects. As you might suspect, this search will be successful.


Chapter summary

• The occupation number representation describes states by listingthe number of identical particles in each quantum state.

• We focus on the vacuum state |0〉 and then construct many-particlestates by acting on |0〉 with creation operators.

• To obey the symmetries of many-particle mechanics, bosons aredescribed by commuting operators and fermions are described byanticommuting operators.

Exercises

(3.1) For boson operators satisfying

h

ap , a†q

i

= δpq , (3.41)

show that

1

VX

pq

ei(p·x−q·y)h

ap , a†q

i

= δ(3)(x − y), (3.42)

where V is the volume of space over which the sys-tem is defined. Repeat this for fermion commuta-tion operators.

(3.2) Show that for the simple harmonic oscillator:

(a) [a, (a†)n] = n(a†)n−1,

(b) 〈0|an(a†)m|0〉 = n!δnm,

(c) 〈m|a†|n〉 =√n+ 1δm,n+1,

(d) 〈m|a|n〉 =√nδm,n−1.

(3.3) The three-dimensional harmonic oscillator is de-scribed by the Hamiltonian

H =1

2m(p2

1+p22+p2

3)+1

2mω2(x2

1+x22+x2

3). (3.43)

Define creation operators a†1, a†2 and a†3 so that

[ai, a†j ] = δij and show that H = ~ω

P3i=1(

12+a†i ai).

Angular momentum can be defined as

Li = −i~ǫijka†j ak, (3.44)

so that for example L3 = −i~(a†1a2 − a†2a1). Definenew creation operators

b†1 = − 1√2(a†1 + ia†2),

b†0 = a†3,

b†−1 =1√2(a†1 − ia†2), (3.45)

and show that [bi, b†j ] = δij , the Hamiltonian is H =

~ωP1m=−1(

12

+ b†mbm) and L3 = ~P1m=−1mb

†mbm.

(3.4) Since for fermions we have that c†1c†2|0〉 = −c†2c†1|0〉,

then the state |11〉 depends on which fermion isadded first. If we put the first fermion into thestate ψi(r1) and the second fermion into the stateψj(r2), a representation for |11〉 is

Ψ(r1, r2) =1√2

[ψ1(r1)ψ2(r2) − ψ2(r1)ψ1(r2)] .

(3.46)Show that a generalization of this result for Nfermions is Ψ(r1 · · · rN ), which can be written outas

1√N !

˛

˛

˛

˛

˛

˛

˛

˛

˛

ψ1(r1) ψ2(r1) · · · ψN (r1)ψ1(r2) ψ2(r2) · · · ψN (r2)

......

...ψ1(rN ) ψ2(rN ) · · · ψN (rN )

˛

˛

˛

˛

˛

˛

˛

˛

˛

.

(3.47)This expression is known as a Slater determi-

nant.

4Making second

quantization work

4.1 Field operators 37

4.2 How to second quantize an op-erator 39

4.3 The kinetic energy and thetight-binding Hamiltonian 43

4.4 Two particles 44

4.5 The Hubbard model 46

Chapter summary 48

Exercises 48

In this chapter we will bridge the gap between first and second quan-tization. This will result in new operators that act on states given inthe occupation number representation. For this discussion we’ll limitourselves to non-relativistic particles in a box.1 By working in the non-

1We retain the ‘particle-in-a-box’normalization of Chapter 3. States aredefined in a box of volume V (usingperiodic boundary conditions). Forlater convenience, we list some usefulresults using this normalization:

A state |α〉 can be described in posi-tion coordinates ψα(x) = 〈x|α〉 or mo-mentum coordinates ψα(p) = 〈p|α〉,and since

〈p|α〉 =

Z

d3x 〈p|x〉〈x|α〉, (4.1)

then the Fourier transform

ψα(p) =1√V

Z

d3x e−ip·xψα(x),

(4.2)implies that

〈p|x〉 =1√V

e−ip·x . (4.3)

The inverse transform is

ψα(x) =1√VX

p

eip·xψα(p), (4.4)

since p takes discrete values. We alsohave

Z

d3x e−ip·x = Vδp,0 (4.5)

and

1

VX

p

eip·x = δ(3)(x). (4.6)

relativistic limit, we are taking a shortcut and thereby avoiding someimportant complications. A fuller, relativistic treatment will be pre-sented in Chapter 11, but the approach taken in this chapter gives aroute to some immediately useful results and in particular to the be-haviour of electrons in a solid.

4.1 Field operators

In the discussion so far we have introduced creation and annihilationoperators a†p and ap which create and annihilate particles into particu-lar momentum states. So if we have the vacuum state |0〉 and we applya†p to it, then there is a big puff of smoke and a flash of light and theresulting state a†p|0〉 contains a single particle sitting in the momentumstate p. Being a momentum eigenstate it is of course extended in space(but completely localized in momentum space). Now, by making appro-priate linear combinations of operators, specifically using Fourier sums,we can construct operators, called field operators, that create and an-nihilate particles, but this time they don’t create/annihilate particles inparticular momentum states but instead they create/annihilate parti-cles localized at particular spatial locations. Thus the operator ψ†(x)defined by

ψ†(x) =1√V∑

p

a†pe−ip·x, (4.7)

creates a particle at position x, while a†p creates a particle in a state

with three-momentum p. Similarly, the operator ψ(x) defined by

ψ(x) =1√V∑

p

apeip·x, (4.8)

annihilates a particle at position x, while ap annihilates a particle in astate with three-momentum p. In summary:

38 Making second quantization work

The field operators ψ†(x) and ψ(x) respectively create or destroy asingle particle at a point x.

The following examples explore some properties of field operators.

Example 4.1

We examine the state |Ψ〉 = ψ†(x)|0〉 and check that it does indeed correspond to asingle particle at a particular location. Note that

|Ψ〉 = ψ†(x)|0〉 =1√VX

p

e−ip·x a†p |0〉. (4.9)

To calculate the total number of particles in this state we can use the number operatornp = a†p ap (which measures the number of particles in state p) and then sum overall momentum states. Consider

X

q

nq |Ψ〉 =1√VX

qp

a†q aq a†p |0〉e−ip·x , (4.10)

and again using 〈0|aq a†p |0〉 = δpq we deduce that

X

q

nq |Ψ〉 = |Ψ〉, (4.11)

and the state Ψ is then an eigenstate of the number operator with eigenvalue 1.To find out exactly where the particle is created, first define |p〉 = a†p |0〉 and then

we can look at the projection of |Ψ〉 on another position eigenstate |y〉:We use the result

〈y|p〉 =1√V

eip·y ,

which follows from eqn 4.3 that statesthat the projection of a momentumeigenstate in position space is a planewave. The last equality in eqn 4.13 fol-lows directly from eqn 4.6.

〈y|Ψ〉 = 〈y|ψ†(x)|0〉 =1√VX

p

e−ip·x〈y|p〉

=1

VX

p

e−ip·(x−y) (4.12)

= δ(3)(x− y),

showing that the single particle created by ψ†(x) may only be found at y = x.

Example 4.2

The field operators satisfy commutation relations. For exampleh

ψ(x), ψ†(y)i

=1

VX

pq

ei(p·x−q·y)[ap , a†q ], (4.13)

and using [ap , a†q ] = δpq we have that

h

ψ(x), ψ†(y)i

=1

VX

pq

ei(p·x−q·y)δpq =1

VX

p

eip·(x−y) = δ(3)(x− y), (4.14)

where the last equality follows from eqn 4.6. By the same method you can showh

ψ†(x), ψ†(y)i

=h

ψ(x), ψ(y)i

= 0. (4.15)

These results are for boson operators, but we can derive analogous results for fermionoperators:

n

ψ(x), ψ†(y)o

= δ(3)(x− y),n

ψ†(x), ψ†(y)o

=n

ψ(x), ψ(y)o

= 0. (4.16)

4.2 How to second quantize an operator 39

4.2 How to second quantize an operator

How can we upgrade the operators that we had in first-quantized quan-tum mechanics into second-quantized operators? Let us first remindourselves what an operator A does. It maps a quantum state |ψ〉 (whichlives in a Hilbert space2 and describes the quantum mechanical state 2A Hilbert space is a complex vector

space endowed with an inner product,whose elements look like

P

n an|φn〉where an are complex numbers. It canthus describe particular superpositionsof states such as, just for an example,(|φ0〉 + i|φ1〉)/

√2 in which the system

might equally be found in state |φ0〉 or|φ1〉, but the quantum amplitudes forbeing in each state are π/2 out of phase.This is a superposition of possibilities,but it only describes a single particle.

of a single particle) into another state A|ψ〉. We will focus in this sectionon A being a single-particle operator, an example of which would bep1 = −i~∇ = −i~ ∂

∂x1, which operates on a single coordinate x1 that

can only describe the position of one particle. Thus we will ignore fornow objects like the Coulomb potential operator V (x1,x2) = q

4π|x1−x2|which depends on the coordinates of two particles. An operator thatacts on n particles is called an n-particle operator.

The operator A can be described quite generally using a variety ofcoordinate systems, as can be seen if you insert two resolutions of theidentity, 1 =

∑α |α〉〈α| and 1 =

∑β |β〉〈β| into the right-hand side of

the trivial identity A = A:

A =∑

αβ

|α〉〈α|A|β〉〈β| =∑

αβ

Aαβ |α〉〈β|, (4.17)

where the matrix element Aαβ = 〈α|A|β〉. This is all well and good, butquantum field theory is acted out on a larger stage than Hilbert space.We need a space that includes not only single-particle states, but thepossibility of many-particle states. This is known as Fock space.3

3Fock space F includes N -particlestates for all possible values of N .Mathematically, one can write this as

F =

∞M

N=0

FN = F0 ⊕F1 ⊕ · · · ,

where F0 = |0〉 is a set contain-ing only the vacuum state and F1 isthe single-particle Hilbert space. Thesubspaces FN≥2, which describe N -particle states with N ≥ 2, have tobe symmetrized for bosons and anti-symmetrized for fermions (so, for ex-ample, F2 is not simply F1 ⊗ F1, butonly the symmetric or antisymmetricpart of F1 ⊗ F1). Creation operatorsmove an element of FN into an ele-ment of FN+1, while annihilation op-erators perform the reverse manoeuvre.An element of F can describe a super-position of states with different particlenumbers.

Remarkably, the second-quantized many-body upgrade A of thesingle-particle operator A to Fock space is a very compact expression:

A =∑

αβ

Aαβ a†αaβ . (4.18)

This equation has the same single-particle matrix element Aαβ =〈α|A|β〉 that we had in eqn 4.17, even though it is now valid for themany-particle case. The interpretation of eqn 4.18 is beautifully simple.The operator A is a sum over all processes in which you use aβ to removea single particle in state |β〉, multiply it by the matrix element Aαβ , andthen use a†α to place that particle into a final state |α〉. This operator Athus describes all possible single-particle processes that can occur andoperates on a many-particle state (see Fig. 4.1). The proof of eqn 4.18is rather technical and is covered in the following example, which canbe skipped if you are happy to take the result on trust.

aβ

a†α

Aαβ

Fig. 4.1 A process represented by theoperator in eqn 4.18.

Example 4.3

To prove eqn 4.18, we need to introduce some formalism. Let us first write down anexpression for an N -particle state

|ψ1, . . . , ψN 〉 =1√N !

X

P

ξPNY

i=1

|ψP (i)〉. (4.19)


This is a sum over all N ! permutations of single-particle states |ψi〉. There are N !permutations of the sequence of integers i = 1, . . . , N and P (i) labels a permutationof this sequence. The factor ξ = 1 for bosons and ξ = −1 for fermions. For fermions,the notation ξP gives a plus or minus sign for an even or odd permutation respectively(and for bosons, of course, the factor can be ignored as it is always +1). The innerproduct of two such N -particles states is

〈χ1, . . . , χN |ψ1, . . . , ψN 〉 =1

N !

X

P

X

Q

ξP+QNY

i=1

〈χQ(i)|ψP (i)〉. (4.20)

We can rewrite this using P ′ = P + Q which spans all the permutations N ! times,see Fig. 4.2, and hence

〈χ1, . . . , χN |ψ1, . . . , ψN 〉 =X

P ′

ξP′NY

i=1

〈χi|ψP ′(i)〉. (4.21)

If desired, this rather complicated expression can be written rather compactly as a

P

Q

Fig. 4.2 For N = 3, a sum over per-mutations P and Q is equivalent toN ! copies of a sum over permutationsP ′ = P +Q.

determinant (for fermions, or a ‘permanent’ for bosons).44The right-hand side of eqn 4.21 canbe written det(〈χi|ψj〉) for fermions orper(〈χi|ψj〉) for bosons. The determi-nant of a matrix Aij is

det(A) =X

P

(−1)PNY

i=1

AiP (i).

The permanent of a matrix Aij is

per(A) =X

P

NY

i=1

AiP (i).

Let’s now try using a creation operator a†φ to act on one of these N -particle states:

a†φ|ψ1, . . . , ψN 〉 = |φ, ψ1, . . . , ψN 〉. (4.22)

Thus the cuckoo lays its |φ〉 egg in the |ψ1, . . . , ψN 〉 nest, see Fig. 4.3. Annihilationis somewhat harder to formulate, mainly because you have to cater for the possibilitythat the state you are annihilating is some admixture of states already present. Toproceed, it is helpful to realize that aφ is the Hermitian conjugate of a†φ. Thus

a†φ

φψ1

ψ2

ψN

Fig. 4.3 The action of a creation oper-

ator a†φ on a state |ψ1, . . . , ψN 〉.

〈χ1, . . . , χN−1|aφ|ψ1, . . . , ψN 〉 = 〈ψ1, . . . , ψN |a†φ|χ1, . . . , χN−1〉∗ (4.23)

= 〈ψ1, . . . , ψN |φ, χ1, . . . , χN−1〉∗

=

˛˛˛˛˛˛˛

〈ψ1|φ〉〈ψ1|χ1〉 · · · 〈ψ1|χN−1〉...

.

.

....

〈ψN |φ〉〈ψN |χ1〉 · · · 〈ψN |χN−1〉

˛˛˛˛˛˛˛

∗

ξ

.

This last line has been obtained by recognizing that the expression to evaluate is adeterminant (fermions) or permanent (bosons). It can be expanded as a series ofminors:

〈χ1, . . . , χN−1|aφ|ψ1, . . . , ψN 〉

=NX

k=1

ξk−1〈ψk|φ〉∗〈ψ1, . . . (no ψk) . . . ψN |χ1, . . . , χN−1〉∗

=NX

k=1

ξk−1〈φ|ψk〉〈χ1, . . . , χN−1|ψ1, . . . (no ψk) . . . ψN 〉.

(4.24)

This is true for any premultiplying 〈χ1, . . . , χN−1|, and so

aφ|ψ1, . . . , ψN 〉 =

NX

k=1

ξk−1〈φ|ψk〉|ψ1, . . . (no ψk) . . . ψN 〉. (4.25)

This means that a combination of annihilation of |β〉 and creation of |α〉 gives

a†αaβ |ψ1, . . . , ψN 〉 =NX

k=1

ξk−1〈β|ψk〉|α, ψ1, . . . (no ψk) . . . ψN 〉, (4.26)

but now we can simply translate the inserted state |α〉 into the position formerly heldby the annihilated state |ψk〉 and thereby remove the annoying ξk−1 factor, since

ξk−1|α, ψ1, . . . (no ψk) . . . ψN 〉 = |ψ1, . . . ψk−1, α, ψk+1, . . . ψN 〉 (4.27)

and hence

a†αaβ |ψ1, . . . , ψN 〉 =NX

k=1

〈β|ψk〉|ψ1, . . . ψk−1, α, ψk+1, . . . ψN 〉. (4.28)

4.2 How to second quantize an operator 41

After all this effort, we are now ready to tackle the real problem. Given a state|ψ1, . . . , ψN 〉 and a single-particle operator A =

P

αβ Aαβ |α〉〈β|, the obvious second-

quantized upgrade is to let A act separately on each individual component stateinside |ψ1, . . . , ψN 〉 and then sum the result. For example, if A is the single-particlemomentum operator, we would want A to be the operator for the total momentum,the sum of all the momenta of individual particles within the many-body state.

Let’s make it very simple to start with and set A = |α〉〈β|. In this case, for asingle-particle state |ψj〉 we would have A|ψj〉 = 〈β|ψj〉|α〉 and hence

A|ψ1 . . . ψN 〉 =NX

j=1

〈β|ψj〉|ψ1, . . . ψk−1, α, ψk+1, . . . ψN 〉. (4.29)

But this is simply a†αaβ |ψ1, . . . , ψN 〉 as we found in eqn 4.28. Hence A = a†αaβ . In

the general case of A =P

αβ Aαβ |α〉〈β|, we easily then arrive at

A =X

αβ

Aαβ a†αaβ . (4.30)

After the hard work of the previous example, which justified eqn 4.18,we now have the power at our fingertips to do some splendidly elegantthings.

Example 4.4

(i) The resolution of the identity 1 =P

α |α〉〈α| is a special case of A =P

αβ Aαβ |α〉〈β| with Aαβ = δαβ . Its second-quantized upgrade is immediately

n =X

α

a†αaα, (4.31)

the number operator, that counts one for every particle in the state on which itoperates.(ii) The usual momentum operator can be written A = p =

P

p p|p〉〈p| and thisupgrades immediately to

p =X

p

p a†p ap =X

p

p np . (4.32)

In much the same way, a function of the momentum operator A = f(p) becomes

A =X

p

f(p) a†p ap =X

p

f(p) np . (4.33)

One particular example of this is the free-particle Hamiltonian which one can writedown immediately in second-quantized form as

ap

a†p

p2

2m

Fig. 4.4 A process represented by the

Hamiltonian H =P

pp2

2mnp . Sum-

ming over all momentum states involvesrepeating this counting process whichcan be imagined as counting sheepcrossing through a gate (annihilate asheep on the left-hand side of the gate,count its energy in the sum, create thesheep again on the right-hand side ofthe gate, now move on to the next sheep. . . ).

H =X

p

p2

2mnp . (4.34)

Equation 4.34 can be thought about as the simple diagram shown in Fig. 4.4. Sincethe occupation number states (with which we want to work) are eigenstates of np thenso is H. We have found that the operator H is diagonal in the basis in which we’reworking. In fact, to diagonalize any Hamiltonian (which means to find the energiesof the eigenstates) we simply express it in terms of number operators. Having theHamiltonian in this form makes perfect sense. We’re saying that the total energy inthe system is given by the sum of the energies p2/2m of all of the particles.(iii) What happens if our operator is a function of x, rather than p? Well, eqn 4.18 isvalid even if our states |α〉 and |β〉 are position states. Our creation and annihilationoperators are simply field operators, and the sum over momentum values becomesan integral over space. We then can write down our second-quantized operator V as

V =

Z

d3x ψ†(x)V (x) ψ(x). (4.35)


If we want, we can express this back in momentum space using eqns 4.7 and 4.8, andhence

V =1

V

Z

d3xX

p1p2

a†p1e−ip1·xV (x) ap2

eip2·x =X

p1p2

Vp1−p2a†p1

ap2, (4.36)

where Vp = 1VR

d3xV (x)e−ip·x is the Fourier transform of V (x). The operator

V isn’t diagonal, since the a operators create and annihilate particles with differentmomenta. We can think of this term as representing a process of an incoming particlewith momentum p2 which interacts with a potential field (represented by Vp1−p2

)and then leaves again with momentum p1 (see Fig. 4.5).ap2

a†p1

Vp1−p2

Fig. 4.5 A process represented byeqn 4.36. The sequence of events is: an-nihilate a particle with momentum p2,count the potential energy Vp1−p2

, cre-ate a particle with momentum p1.

Example 4.5

Let’s have a look at the influence of the potential on a simple system described by aHamiltonian

H = E0

X

p

d†p dp − V

2

X

p1p2

d†p1dp2

. (4.37)

We’ll further constrain the system to have three energy levels, so states are expressedusing a basis |np1

np2np3

〉. For simplicity, let’s first set V = 0. In this case, if weput a particle into the system then the three possible eigenstates of the system are|100〉, |010〉, |001〉. Now we put V 6= 0, and so

V

2

X

p1p2

d†p1dp2

|100〉 =V

2(|100〉 + |010〉 + |001〉) , (4.38)

and similarly for the other states (try them).At this point it’s easiest to slip into matrix notation, for which the entire Hamil-

tonian takes the form

H =

2

4E0

0

@

1 0 00 1 00 0 1

1

A− V

2

0

@

1 1 11 1 11 1 1

1

A

3

5 . (4.39)

The eigenvalues for this equation are ε = E0, E0, E0 − 3V2

. The ground state of the

system (called |Ω〉) therefore has an energy E0 − 3V2

. The corresponding eigenstateis

|Ω〉 =1√3

(|100〉 + |010〉 + |001〉) . (4.40)

We conclude that the ground state is a special 1:1:1 superposition of the three stateswe started with.

Example 4.6

What is the equivalent of the probability density of the wave function in second-quantized language? The answer is that the number density of particles at a pointx is described by the number density operator ρ(x) given by

ρ(x) = ψ†(x)ψ(x)

=1

VX

p1p2

h

e−i(p1−p2)·xi

a†p1ap2

. (4.41)

This operator enables us to write the potential energy operator for a single particlein an external potential as

V =

Z

d3xV (x)ρ(x). (4.42)

4.3 The kinetic energy and the tight-binding Hamiltonian 43

4.3 The kinetic energy and the

tight-binding Hamiltonian

One application of this formalism is to the tight-binding Hamiltonianwhich is used in condensed matter physics. The basic idea behind thismodel is very straightforward: electrons don’t like being confined. Weknow this from the well-known problem of a one-dimensional infinitesquare well with width L. There the kinetic part of the energy is givenby En = 1

2m

(nπL

)2. The smaller we make L, the larger the energy.

Conversely, kinetic energy can be saved by allowing the electron to movein a larger volume. In the tight-binding model, one considers a lattice offixed atoms with electrons moving between them (as shown in Fig. 4.6).These electrons can lower their kinetic energies by hopping from latticesite to lattice site. To deal with the discrete lattice in the model, weneed to work in a basis where the fermion creation operator c†i createsa particle at a particular lattice site labelled by i. The kinetic energysaving for a particle to hop between points j and i is called tij . Clearlytij will have some fundamental dependence on the overlap of atomicwave functions. The Hamiltonian H contains a sum over all processesin which an electron hops between sites, and so is a sum over pairs ofsites:

H =∑

ij

(−tij)c†i cj . (4.43)

Each term in this sum consists of processes in which we annihilate aparticle at site j and create it again at site i (thus modelling a hoppingprocess), thus saving energy tij . For now, we’ll consider the simplestpossible case of tij = t for nearest neighbours and t = 0 otherwise. Thuswe rewrite H as

H = −t∑

iτ

c†i ci+τ , (4.44)

where the sum over τ counts over nearest neighbours.

j

i

Fig. 4.6 The operator c†i cj causing anelectron to hop from site j to i. Such aprocess saves kinetic energy tij .Example 4.7

The tight-binding Hamiltonian is not diagonal since we have a bilinear combinationof operators that look like c†i cj . We need to diagonalize it and that can be achievedby making a change of basis (again, essentially using a Fourier transform). Thus

ci =1√VX

k

eik·ri ck and c†i =1√VX

q

e−iq·ri c†q . (4.45)

Substituting this into the Hamiltonian yields

H = − t

VX

iτ

X

kq

e−i(q−k)·ri−iq·rτ c†q ck . (4.46)

We play our usual trick and exploit the fact that 1VP

i e−i(q−k)·ri = δqk , which

removes the sum over i. Lastly we do the sum over q. After jumping through thesefamiliar mathematical hoops we end up with

H = −tX

k

X

τ

eik·rτ c†kck , (4.47)

which is diagonal in momentum.


We can tidy up this result by writing the Hamiltonian as

H =∑

k

Ekc†kck, (4.48)

where the dispersion Ek is given by

Ek = −∑

τ

t eik·rτ . (4.49)

Example 4.8

For a two-dimensional square lattice, the sum over τ counts over vectors (a, 0),(−a, 0), (0, a) and (0,−a). This leads to a dispersion relation (see also Fig. 4.7)

Ek = −2t(cos kxa+ cos kya). (4.50)

-π/a

0

π/a

ky

-π/a 0 π/a

kx

Fig. 4.7 The tight-binding model fora two-dimensional square lattice. Con-tours showing lines of constant energyEk based on eqn 4.50 are plotted as afunction of kx and ky . The lowest Ek

occurs at (kx, ky) = (0, 0), the centre ofthis plot, for t > 0 and at the corners,(kx, ky) = (±π

a,±π

a) for t < 0.

4.4 Two particles

We have been focussing on single-particle operators. What happens forthe more interesting case of two particles interacting with each other?From the discussion above we can guess (quite correctly) that a second-quantized two-particle operator A is given by

A =∑

αβγδ

Aαβγδ a†αa

†β aγ aδ, (4.51)

where Aαβγδ = 〈α, β|γ, δ〉 is the matrix element for the Hilbert-spaceversion. This expression embodies the idea that two-particle operatorsinvolve a sum over all processes involving the annihilation of two parti-cles in particular states, multiplication by the matrix element, and cre-ation of two particles in two new states. We will frequently encountertwo-particle operators V written as a function of spatial coordinates,and the corresponding expression involving field operators is

V =1

2

∫d3xd3y ψ†(x)ψ†(y)V (x,y)ψ(y)ψ(x). (4.52)

The factor of 12 ensures that we don’t double count the interactions.

Notice the way we have arranged the operators: we put the creationoperators to the left and the annihilation operators to the right, a con-vention which is known as normal ordering. Why is this important?Normal ordering makes the evaluation of matrix elements much easierand in particular it makes sure that our operator, V , has zero vacuumexpectation value:

〈0|V |0〉 = 0. (4.53)

4.4 Two particles 45

This avoids us to having to blush over embarrassing infinite self-energyterms that can otherwise creep into our answers. With creation oper-ators to the left and annihilation operators to the right we make surethat 〈0|V |0〉 is fixed at zero.

An additional convention we have adopted in eqn 4.52 is that read-ing the coordinates from right-to-left gives us x, y, y, x. (This can beremembered as the dance step: left-in, right-in, right-out, left-out.) Forbosons the order would not matter, but fermions anticommute and so forexample interchanging the first two creation operators would introducea minus sign.

It is useful to see what would happen if we hadn’t introduced normalordering. For example, an alternative guess for the form of the two-particle interaction might have been

Vwrong =1

2

∫d3xd3y V (x,y)ρ(y)ρ(x). (4.54)

This form is wrong and, because we’ve left out the all-important normalordering, will lead to an extra self-energy term, as shown in the followingexample.

Example 4.9

We will consider the case of fermions and will need the anticommutation relationsn

ψ(x), ψ†(y)o

= δ(3)(x − y) andn

ψ(x), ψ(y)o

= 0. Now consider the operator

ρ(x)ρ(y), which is given by

ρ(x)ρ(y) = ψ†(x)ψ(x)ψ†(y)ψ(y). (4.55)

Notice that this is certainly not normal ordered, but we can make it so as follows.Swap the second and third operators. This involves using an anticommutator. Weobtain

ρ(x)ρ(y) = −ψ†(x)ψ†(y)ψ(x)ψ(y) + δ(3)(x− y)ψ†(x)ψ(y)

= ψ†(x)ψ†(y)ψ(y)ψ(x) + δ(3)(x− y)ψ†(x)ψ(y), (4.56)

where, in the last line, we’ve made another swap (this time of the two left-mostoperators, involving a second sign change). We are left with a normal ordered termplus a term involving a delta function. Comparing this to the correct expression forthe two-particle interaction, we have

Vwrong = V +1

2

Z

d3xV (x,x)ψ†(x)ψ(x). (4.57)

The ‘wrong’ form of the potential energy includes an unwanted self-energy term.Normal ordering ensures we dispose of this abomination.

As before, we continue the procedure of second quantization by substi-tuting the mode expansion into our equation

V =1

2

∫d3xd3y ψ†(x)ψ†(y)V (x,y)ψ(y)ψ(x)

=1

2V2

∫d3xd3y ei(−p1·x−p2·y+p3·y+p4·x)

∑

p1p2p3p4

a†p1a†p2

V (x − y)ap3ap4

.

(4.58)


Example 4.10

To deal with this, we define z = x− y and eliminate x to obtain

V =1

2V2

X

p1p2p3p4

a†p1a†p2

ap3ap4

Z

d3z V (z)ei(p4−p1)·zZ

d3y ei(−p1−p2+p3+p4)·y .

(4.59)The last integral gives us a Kronecker delta Vδp1+p2−p3,p4

, which can be used toeat up one of the momentum sums (the p4 one here) and sets p4 = p1 + p2 − p3.Now we have

1

2VX

p1p2p3

a†p1a†p2

ap3ap1+p2−p3

Z

d3z V (z)e−i(p3−p2)·z . (4.60)

The last integral is the Fourier transform of the potential

1

V

Z

d3z V (z)e−i(p3−p2)·z = Vp3−p2. (4.61)

Now for some interpretation: set q = p3 − p2 and eliminate p3, yielding

V =1

2

X

p1p2q

Vq a†p1a†p2

ap2+q ap1−q . (4.62)

Lastly we reindex the sum subtracting q from p2 and adding it to p1.

The result of these index acrobatics is a two-particle interaction potentialthat looks like

V =1

2

∑

p1p2q

Vqa†p1+qa

†p2−qap2

ap1. (4.63)

This can be interpreted as a scattering problem in momentum space.This scattering is illustrated in a cartoon known as a Feynman diagram.The diagram in Fig. 4.8 is translated into a story as follows: A particlecomes in with momentum p2. It sends out a force-carrying particlewith momentum q, reducing its final momentum to p2 − q. The force-carrying particle is absorbed by a second particle (which originally hasmomentum p1) and ends up with a final momentum of p1 + q. Noticewe conserve momentum at the vertices of the diagrams.

p1 p2q

p1 + q p2 − q

Fig. 4.8 A Feynman diagram for theprocess described in the text. Inter-preting this diagram, one can think oftime increasing in the vertical direction.

Our aim is going to be to turn all of our combinations of operators intonumber operators. This makes evaluating the energy a simple matter.The first thing to notice about eqn 4.63 is that it is not diagonal. Ingeneral, problems involving the potential energy (as in eqn 4.63) cannotbe solved exactly. Despite this, the two-particle potential leads to aremarkably rich seam of physics.5

5As we will see, not only does thistell us about seemingly pedestrian mat-ters concerning the energies of elec-trons in metals, but it also describesinteractions that lead to superfluidity,superconductivity, magnetism, chargedensity waves and pretty much everyother phenomenon you’ve ever encoun-tered in condensed matter physics. Ofcourse, the fact that eqn 4.63 cannotgenerally be diagonalized should giveus pause. We need a programme toturn the two-body potential diagonal.As we will see in Part X, we do thisin two stages: firstly we massage ourHamiltonian until it’s made up of bi-linear combinations of operators. Wemay then make transformations untilwe have number operators.

4.5 The Hubbard model

As this chapter’s final example of a second-quantized Hamiltonian, weturn to the Hubbard model.6 It is difficult to overemphasize the im-

6The Hubbard model is named afterJohn Hubbard (1931–1980).

portance of this model in solid state physics, due to the fact that itcaptures the essential physics of the competition between kinetic energy(favouring electrons to become delocalized) and potential energy (favour-ing electrons to become localized). Though relatively easy to state, the

4.5 The Hubbard model 47

Hubbard model is surprisingly complicated to solve and exact results areonly available in certain scenarios. The Hubbard model is based on atight-binding Hamiltonian (to model the kinetic energy) plus a two-bodypotential energy term (to model electron–electron interactions):

H =∑

ij

(−tij)c†i cj +1

2

∑

ijkl

c†i c†jVijklck cl. (4.64)

Recognizing that electrons carry a spin σ which can point up (| ↑〉) ordown (| ↓〉) this becomes

H =∑

ijσ

(−tij)c†iσ cjσ +1

2

∑

ijklσσ′

c†iσ c†jσ′Vijklckσ′ clσ. (4.65)

We assume in this model that no spin-flip processes can occur; thusspins never flip from up-to-down or vice versa on hopping. However, theinteraction between electrons is based upon the Coulomb interactionbetween their charges and so electrons with opposite spin will interactwith each other just as much as electrons with the same spins. Now forHubbard’s model: the Coulomb interaction is assumed to be significantonly between two electrons that are on the same site. These electronsinteract via a constant potential energy U = Viiii. However, the Pauliexclusion principle ensures that should two spins hop into the samesite they must have different spins, and this is where spin does make adifference. The Hamiltonian for the Hubbard model is then

H =∑

ijσ

(−tij)c†iσ cjσ + U∑

i

ni↑ni↓, (4.66)

which looks simple, but the eigenstates tend to be complex and highlycorrelated. It is often only necessary to consider hopping elements fornearest neighbours.

Example 4.11

Consider the Hubbard model with nearest-neighbour hopping only, and to makethings as simple as possible, consider a lattice of only two sites. If, in addition, weonly have one, spin-up electron, then that single electron can either be on the firstsite (a configuration we will denote as | ↑, 0〉, meaning ‘spin-up electron on site 1,nothing on site 2’) or on the second site (a configuration we will denote as |0, ↑〉).Hence a general state can be written as the superposition |ψ〉 = a| ↑, 0〉 + b|0, ↑〉,where a and b are complex numbers. In this case, there’s no possibility of potentialterms (because there is only one electron) and in this basis the Hubbard Hamiltoniancan be written

H =

„0 −t−t 0

«

. (4.67)

Solving gives us a ground state |ψ〉 = 1√2

(| ↑, 0〉 + |0, ↑〉) with energy E = −t and an

excited state |ψ〉 = 1√2

(| ↑, 0〉 − |0, ↑〉) with energy E = t.

Now we’ll have two electrons in the system. If they have the same spin the answer’ssimple and E = 0 (try it). If the electrons have different spins we write a generalstate as |ψ〉 = a| ↑↓, 0〉 + b| ↑, ↓〉 + c| ↓, ↑〉 + d|0, ↑↓〉 and in this basis the HubbardHamiltonian is

H =

0

BB@

U −t −t 0−t 0 0 −t−t 0 0 −t0 −t −t U

1

CCA. (4.68)


This can be diagonalized to give a ground state energy of E = U2− 1

2

Ù2 + 16t2

´ 12

which corresponds to a wave function |ψ〉 = N (| ↑↓, 0〉 +W | ↑, ↓〉 +W | ↓, ↑〉 + |0, ↑↓〉)with N some normalization constant and W = U

4t+ 1

4t

Ù2 + 16t2

´ 12 .

In this chapter we have considered models on discrete lattices which arerelevant to condensed matter physics. In the following chapter, we willconsider how to generalize this approach to the continuum. However,readers who are eager to see some uses of the machinery developed inthis chapter should now be able to tackle the first parts of the chaptersin Part X, which deal with some applications from condensed matterphysics.

Chapter summary

• Field operators ψ† and ψ are the analogues of creation and anni-hilation operators in position space.

• Second quantization of models can be carried out by writing downoperators as products of creation and annihilation operators thatencode the physical particle processes. We write these productsin normal-ordered form (meaning the field creation operators arewritten to the left and the field annihilation operators to the right).

• The vacuum expectation of normal-ordered products is always zero.

• Examples of this technique have been given for the tight-bindingmodel and the Hubbard model.

Exercises

(4.1) We can define a generalized commutator [A, B]ζ as

[A, B]ζ = AB − ζBA, (4.69)

so that ζ = 1 yields [A, B]ζ = [A, B] for bosonsand ζ = −1 yields [A, B]ζ = A, B for fermions.The generalized commutation relations can then bewritten

[ψ(x), ψ†(y)]ζ = δ(3)(x − y), (4.70)

and[ψ(x), ψ(y)]ζ = 0. (4.71)

Repeat the argument given in Example 4.9 usingthe generalized commutator and show that Vwrong

yields the same result for both bosons and fermions.

(4.2) One can define a single-particle density matrix

asρ1(x − y) = 〈ψ†(x)ψ(y)〉. (4.72)

By substituting the expansion of ψ†(x) and ψ(y) interms of a†q and aq into this expression, show that

ρ1(x − y) =1

VX

pq

e−i(q·x−p·y)〈a†q ap〉. (4.73)

(4.3) Evaluate the eigenvalues and eigenvectors of theHubbard Hamiltonian given in eqn 4.68 in Exam-ple 4.11.(a) What are the energy eigenvalues in the limitt = 0?(b) How do these energy levels change as t 6= 0 inthe limit t/U ≪ 1?

Part II

Writing down Lagrangians

An important step in writing down a physical model is to construct theappropriate Lagrangian. This part of the book is concerned with howto do that and is structured as follows:

• In Chapter 5 we describe how the arguments in the previous partwhich worked on discrete systems can be generalized to the con-tinuum limit. After reviewing Hamilton’s formulation of classicalmechanics and Poisson brackets, we shift our attention from theLagrangian L to the Lagrangian density L. We use the electromag-netic field as a first example of this approach.

• We have our first stab at constructing a relativistic quantum waveequation, the Klein–Gordon equation in Chapter 6. This turnsout to have some unsavoury characteristics that means that it isnot the right equation to describe electrons, but it is neverthelessilluminating and illustrates some of the issues we are going to comeacross later.

• In Chapter 7 we present a set of example Lagrangians and showhow they work. After reading this chapter the reader should havea good working knowledge of the simplest examples of quantumfield theories.

5 Continuous systems

5.1 Lagrangians and Hamiltoni-ans 50

5.2 A charged particle in an elec-tromagnetic field 52

5.3 Classical fields 54

5.4 Lagrangian and Hamiltoniandensity 55

Chapter summary 58

Exercises 58

In the last few chapters we have seen examples of how various systemscan be reduced to a set of simple harmonic oscillators, each one describ-ing a different normal mode of the system. Each normal mode may bethought of as a momentum state for non-interacting, identical particles,with the number of particles corresponding to the number of quantizedexcitations of that mode. We have defined operators that have createdor annihilated these particles. In the previous chapter, we particularlyfocussed on discrete models (tight-binding, Hubbard, etc.) in which par-ticles moved around on a lattice. In this chapter we want to extend thisnotion to the continuum limit, that is when the discretization of thelattice can be ignored. This will lead to the concept of a classical field.Before we scale those dizzy heights, there is some unfinished businessfrom classical mechanics to deal with, thereby introducing some usefulformalism.

5.1 Lagrangians and Hamiltonians

In this section, we return to the Lagrangian formulation of classicalmechanics and introduce the Hamiltonian and the Poisson bracket. TheHamiltonian will give us an expression for the energy which appears asa conserved quantity if the Lagrangian doesn’t change with time. ThePoisson bracket is also telling us about how quantities are conserved.Let’s begin by considering the Lagrangian L(qi, qi) for a system describedby coordinates qi. The rate of change of L is given by

dL

dt=∂L

∂qiqi +

∂L

∂qiqi, (5.1)

where the summation convention is assumed. Using the Euler–Lagrangeequation we can turn this into

dL

dt=

d

dt

(∂L

∂qi

)qi +

∂L

∂qiqi =

d

dt

(∂L

∂qiqi

). (5.2)

We define the canonical momentum by

pi =∂L

∂qi, (5.3)

and then eqn 5.2 can be written

d

dt(piqi − L) = 0. (5.4)

5.1 Lagrangians and Hamiltonians 51

This is clearly a conservation law, and we write it as

dH

dt= 0, (5.5)

thereby defining the Hamiltonian H by

H = piqi − L. (5.6)

We will see later that the Hamiltonian corresponds to the energy of thesystem. Since H is conserved it is instructive to look at variations in H.Thus

δH = piδqi + δpiqi −∂L

∂qiδqi −

∂L

∂qiδqi. (5.7)

The first and fourth terms in this expression cancel because of eqn 5.3,and so

δH = δpiqi −∂L

∂qiδqi. (5.8)

Since H is a function of qi and pi, we expect that

δH =∂H

∂qiδqi +

∂H

∂piδpi, (5.9)

and so matching terms in eqns 5.8 and 5.9 produces qi = ∂H/∂pi and∂H/∂qi = −∂L/∂qi = −pi, and these two equalities are known as Note that because pi = (∂L/∂qi) then

pi = d/dt(∂L/∂qi) = ∂L/∂qi, wherethe last equality follows from the Euler–Lagrange equation.

Hamilton’s equations:

∂H

∂pi= qi,

∂H

∂qi= −pi. (5.10)

Hamilton’s equations represent another way to find the equations ofmotion of a system.

We now turn to the other piece of nineteenth century classical mechan-ics which we want to review. We define the Poisson bracket A,BPB Simeon Denis Poisson (1781–1840)

by

A,BPB =∂A

∂qi

∂B

∂pi− ∂A

∂pi

∂B

∂qi. (5.11)

Now consider any function F which depends on the coordinates qi andpi. The rate of change of F is given by

dF

dt=∂F

∂t+∂F

∂qiqi +

∂F

∂pipi =

∂F

∂t+ F,HPB , (5.12)

where the last equality is achieved using Hamilton’s equations. Thus ifthe field is not itself a function of time, we have

dF

dt= F,HPB , (5.13)

so that if F,HPB = 0 then F is a constant of the motion. We havetherefore found a link between a conservation law (F is a conservedquantity) with the Poisson bracket of the Hamiltonian and F being zero.This is highly evocative of the commutator in quantum mechanics, butnote that this result is entirely classical!


Example 5.1

There is an interesting graphical interpretation of this result of a conserved F (seeFig. 5.1). If F = F (q,p), then the vector (q, p) = (∂H/∂p,−∂H/∂q) is tangent tothe surface F (q,p) = constant, because

∇F · (q, p) =

„∂F

∂q,∂F

∂p

«

· (q, p) =∂F

∂qi

∂H

∂pi− ∂F

∂pi

∂H

∂qi= F,HPB = 0, (5.14)

and (q, p) is also tangent to the surface H(q,p) = constant, because

∇H · (q, p) =

„∂H

∂q,∂H

∂p

«

· (q, p) = −p · q + q · p = 0, (5.15)

and so the trajectory of the system moves on the intersection of these two surfaces.

∇H∇F

Fis constant

H is constant

Fig. 5.1 The trajectory of the sys-tem moves on the intersection of sur-faces of constant H and F . Thisis because (q, p) is perpendicular to

both ∇F ≡“∂F∂q, ∂F∂p

”

(eqn 5.14) and

∇H ≡“∂H∂q, ∂H∂p

”

(eqn 5.15).

In quantum mechanics, the rate of change of the expectation value of anoperator F is given by

d〈F 〉dt

=1

i~〈[F , H]〉, (5.16)

and since this looks very much like the classical version

dF

dt= F,HPB , (5.17)

we are led to visualize the classical to quantum crossover as involvingthe replacement

F,HPB → 1

i~〈[F , H]〉, (5.18)

and in fact for any Poisson bracket

A,BPB → 1

i~〈[A, B]〉. (5.19)

Example 5.2

The Poisson bracket between position and momentum coordinates is

qj , pkPB =∂qj

∂qi

∂pk

∂pi− ∂pj

∂qi

∂qk

∂pi= δijδik − 0 × 0

= δjk. (5.20)

Using eqn 5.19 we have [qj , pk] = i~δjk, familiar from quantum mechanics.

5.2 A charged particle in an

electromagnetic field

To practise some of the ideas in the previous section, we will consider avery simple example: a free particle of mass m. Shortly we will give it a

5.2 A charged particle in an electromagnetic field 53

charge q and turn on an electromagnetic field, but for now we will keepall electromagnetic fields switched off. We want to include relativity,and so therefore the action S must be a Lorentz-invariant scalar. Wecan write S =

∫ τ2τ1Lγ dτ where τ is the proper time, and so γL must

be a Lorentz invariant. We can thus write L = constant/γ and we willchoose

L = −mc2

γ, (5.21)

where the constant mc2 is chosen so that L has the correct low-velocitylimit.1 The action S can now be written 1See Exercise 5.4.

S =

∫ τ2

τ1

−mc2dτ = −mc∫ b

a

ds, (5.22)

where the interval2 ds =√c2dt2 − dx2 − dy2 − dz2. By the principle of

2The interval ds is an invariant, andso takes the same form in all inertialframes. In the rest frame of the parti-cle ds = cdτ and so this is true in allinertial frames.

least action δS = 0 and so δ∫ ba

ds = 0. The integral∫ ba

ds takes its max-imum value (see Fig. 5.2) along a straight world-line (see Exercise 5.5)and so this implies reassuringly that free particles move along straightlines.

x

t

A

B

Fig. 5.2 The straight-line path from

A to B has a largerR ba ds than the wig-

gly one (because c dt <√c2 dt2 − dx2),

even though it looks longer.

Example 5.3

Find expressions for the momentum and energy of a free particle.Answer:

p =∂L

∂v=

∂

∂v

“

−mc2(1 − v2/c2)1/2”

= γmv, (5.23)

E = H = p · v − L = γmv2 +mc2

γ= γmc2

»v2

c2+

„

1 − v2

c2

«–

= γmc2. (5.24)

In special relativity, we can assemble the energy and momentum intoa momentum four-vector pµ =

(Ec ,p

)[or equivalently pµ =

(Ec ,−p

)].

To make our free particle more interesting, let’s give it a charge q andcouple it to an electromagnetic field. The electromagnetic field can be

described by a four-vector field Aµ(x) =(V (x)c ,A(x)

)where V (x) is the

scalar potential and A(x) is the magnetic vector potential. The actionbecomes3 3The potential energy of the electro-

magnetic field interaction is linear inthe field Aµ and takes the Lorentz-invariant form −qAµdxµ. Note

that Aµ =“Vc,−A

”

and dxµ =

(cdt,dx) = (c,v)dt and so −qAµdxµ =−q(V − A · v)dt.

S =

∫−mcds− qAµ dxµ =

∫ t2

t1

(−mc2γ

+ qA · v − qV

)dt. (5.25)

The Lagrangian is the integrand of this equation, and so the canonicalmomentum in this case becomes

p =∂L

∂v= γmv + qA. (5.26)

Now setting c = 1, the familiar E and B fields can be obtained fromthe four-vector potential Aµ = (V,A) using the equations

B = ∇ × A and E = −∂A

∂t− ∇V, (5.27)


but to put this in a relativistically covariant form we define the second-rank antisymmetric tensor Fµν byNote that:

∂µ =

„∂

∂t,−∇

«

,

∂µ =

„∂

∂t,∇

«

,

Aµ = (V,A),

Aµ = (V,−A),

Also, keep in mind that

E = (E1, E2, E3),

andB = (B1, B2, B3),

are ordinary three-vectors.

Fµν = ∂µAν − ∂νAµ. (5.28)

This is known as the electromagnetic field tensor and looks like afour-dimensional version of a curl. In components, this yields

Fµν =

0 E1 E2 E3

−E1 0 −B3 B2

−E2 B3 0 −B1

−E3 −B2 B1 0

. (5.29)

µ

ν

Similarly, the tensor

Fµν =

0 −E1 −E2 −E3

E1 0 −B3 B2

E2 B3 0 −B1

E3 −B2 B1 0

. (5.30)

Searching for a Lagrangian for the electromagnetic field, we again lookfor a Lorentz scalar. One very suitable choice isThe equation is FµνFµν =

2`B2 − (E/c)2

íf you include

the factors of c. FµνFµν = 2(B2 − E2). (5.31)

The Lagrangian may then be written

L = −1

4

∫d3xFµνF

µν , (5.32)

where the choice of the factor − 14 will be explained later in the chapter.44It’s useful to note that the components

of the electric field may be extractedfrom Fµν using

Ei = −F 0i = F i0, (5.33)

and the magnetic field is given by

Bi = −εijkF jk, (5.34)

where εijk is the antisymmetric symboland no sum over indices is implied inthe last equation.

Finally we note a particularly important feature of electromagnetismis that electric charges are locally conserved. This fact is enshrined inthe continuity equation relating the charge density ρ(x) and currentdensity J(x), which is written

∂ρ

∂t+ ∇ · J = ∂µJ

µ = 0, (5.35)

where the electromagnetic current four-vector Jµ = (ρ,J). The continu-ity equation prevents the destruction of charge followed by its creationat another arbitrarily distant point. Such a process would obey globalconservation of charge, but would allow information to travel at an ar-bitrary velocity, which we know from special relativity is forbidden.

5.3 Classical fields

We will now take the continuum limit of a simple model from classi-cal mechanics and see how this leads to the concept of a classical field.We consider the case of a classical system of balls connected by springs

5.4 Lagrangian and Hamiltonian density 55

(whose quantum mechanics was examined in Chapter 2). In the contin-uum limit we are going to make the distance ℓ between the balls shrinkto zero and simultaneously let the mass m of each ball shrink to zero,while keeping the mass per unit length ρ = m/ℓ along the chain constant.The displacement qj(t) of the jth mass at time t becomes a continuousvariable φ(x, t) which depends on both spatial position x and time t. Ateach point x we have an independent degree of freedom φ(x, t), so wesay that the field has a ‘continuous infinity’ of degrees of freedom. Thefunction φ(x, t) is one example of a classical field, but the concept can bemuch more general than this. For example, rather than describing justa simple displacement, our field can describe more complicated objectssuch as electromagnetic fields. In each case though, the field dependson position in spacetime. We can now make a working definition of theidea of a classical field:5 5We make the fussy assignment of a

classical field here to distinguish it froma quantum field. As we will see shortlythe latter inputs a position in spacetimeand outputs an operator rather than anamplitude.

A classical field is a machine that takes a position in spacetime andoutputs an object representing the amplitude of the field at that point.The output might be a scalar (in which case we refer to a scalar field),a complex number (a complex scalar field), a vector (in which casewe refer to a vector field), a tensor or something more complicated.

Fields are locally defined objects. The field ‘machine’ that we have de-scribed is like a computer subroutine which returns the value of φ at theposition x and time t that we’ve inputted. It is local in the sense thatit doesn’t tell you about the value of φ somewhere else.

Example 5.4

The temperature T (x, t) is an example of a scalar field. The E(x, t) and B(x, t)fields of electromagnetism are familiar examples of vector fields, which could betime-dependent. The four-vector potential Aµ(x) is an example of a four-vector field(and here we have written it as a function of spacetime coordinates xν). The objectdefined in eqn 5.28, Fµν(x), is an example of a second-rank tensor field.

5.4 Lagrangian and Hamiltonian density

The next task is to use this language of classical fields to formulate La-grangians and Hamiltonians. We will do this first for a simple example.

Example 5.5

In the discrete case considered in Section 2.4 we had a Hamiltonian given by

H =X

j

p2j

2m+

1

2K(qj+1 − qj)

2 (5.36)


and a Lagrangian given by

L =X

j

p2j

2m− 1

2K(qj+1 − qj)

2. (5.37)

In the continuum limit, these need to be replaced as the number of masses goes toinfinity while ℓ→ 0. The sums become integrals, so we make the substitution

X

j

→ 1

ℓ

Z

dx. (5.38)

Using the substitution in eqn 5.38 the kinetic energy term in H and L becomesX

j

1

2m

„∂qj

∂t

«2

→ 1

ℓ

Z

dx1

2m

„∂φ(x, t)

∂t

«2

=

Z

dx1

2ρ

„∂φ

∂t

«2

, (5.39)

where we’ve used the string density ρ = m/ℓ. For the potential energy term we alsomake a rule for the differences in coordinates: these become spatial derivatives andhence

(qj+1 − qj)

ℓ→ ∂φ(x, t)

∂x. (5.40)

This means that the potential energy term becomesX

j

1

2K(qj+1 − qj)

2 →X

j

1

2Kℓ2

„∂φ(x, t)

∂x

«2

=

Z

dx1

2T„∂φ

∂x

«2

(5.41)

where we’ve replaced the spring constant K and the lattice spacing ℓ with the ten-sion in the string T = Kℓ. Putting this all together (and generalizing to a three-dimensional system) gives a Hamiltonian

H =

Z

d3x

"

1

2ρ

„∂φ

∂t

«2

+1

2T (∇φ)2

#

, (5.42)

and a Lagrangian

L =

Z

d3x

"

1

2ρ

„∂φ

∂t

«2

− 1

2T (∇φ)2

#

. (5.43)

Our expressions for H and L are integrals over a position coordinateand so we can define a Hamiltonian density H by

H =

∫d3xH, (5.44)

and similarly a Lagrangian density L as

L =

∫d3xL. (5.45)

In general, both the Hamiltonian density H and the Lagrangian densityL are functions of φ, φ and φ′. We can define a conjugate momentum

π(x) in terms of the functional derivative66Just like pi =∂L

∂qi. For a careful jus-

tification of the last equality see Wein-berg, Chapter 7. π(x) =

δL

δφ=∂L∂φ

, (5.46)

and then this implies that H and L can be related by77In Example 5.5 the particular expres-sions are:

L =1

2ρ

„∂φ

∂t

«2

− 1

2T (∇φ)2 ,

H =1

2ρ

„∂φ

∂t

«2

+1

2T (∇φ)2 ,

and

π = ρ∂φ

∂t.

You can check that eqn 5.47 holds.

H = πφ− L. (5.47)

Now recall from eqn 1.34 that the action S is related to L by the four-dimensional integral S =

∫d4xL(φ, ∂µφ), and the action principle, given

by δS = 0, implies that (eqn 1.35)

∂L∂φ

− ∂µ

(∂L

∂(∂µφ)

)= 0, (5.48)

the four-vector version of the Euler–Lagrange equation.

5.4 Lagrangian and Hamiltonian density 57

Example 5.6

For the electromagnetic field, the Lagrangian density is written In SI units L = − 14µ0

FµνFµν and so

L = 12ǫ0E2 − 1

2µ0B2.

L = −1

4FµνF

µν =1

2(E2 − B2). (5.49)

In the absence8 of an electric potential (V = 0), we have L = 12(E2 − B2) = 8Recall eqn 5.27 for the relationship be-

tween fields and potentials.12(A

2 − B2). The conjugate momentum is πi = ∂L/∂(∂0Ai), and so π = −A = E,so that

H = πiAi − L =1

2(E2 + B2). (5.50)

The Euler–Lagrange equation (eqn 1.35) gives In SI units H = 12µ0

B2 + 12ǫ0E2.

∂L∂Aµ

− ∂λ

„∂L

∂(∂λAµ)

«

= 0. (5.51)

The first term in this expression is zero since L = − 14FµνFµν contains only deriva-

tives of Aµ. The second term can be rewritten as

∂λFλµ = 0, (5.52)

which is a compact way of writing two of Maxwell’s equations in free space, namely∇ · E = 0 and ∇ × B = E. The current density four-vector Jµ = (ρ,J) canalso be included, and couples linearly to the electromagnetic potential, and so theLagrangian becomes

L = −1

4FµνF

µν − JµAµ. (5.53)

In this case∂L∂Aµ

= −Jµ, (5.54)

and so

∂λFλµ = Jµ, (5.55)

which generates the other two Maxwell equations: ∇ · E = ρ and ∇× B = J + E. In SI units ∂λFλµ = µ0Jµ.

Solving the Euler–Lagrange equation will produce all the allowablemodes of oscillation and hence the allowed wave vectors, the particu-lar values kn, that may occur without dissipation. By the principle ofsuperposition the most general wave is one made up of a weighted sumof all of the possible modes

φ(x, t) =∑

kn

akne−i(ωt−kn·x). (5.56)

Thus we have seen how many classical continuous systems may be ex-pressed in a Lagrangian formalism so that the Euler–Lagrange equationallows us to derive the normal modes. Our experience with the simpleharmonic oscillator in Chapter 2 motivates the idea that these normalmodes are simply harmonic oscillators themselves and so can be quan-tized, leading to particle-like solutions. For the electromagnetic field,those quanta are of course photons.9 Now with this classical field the- 9We will put this on a firmer footing in

Chapter 14 and show in detail how pho-tons emerge from quantizing the classi-cal electromagnetic field.

ory under our belt, we are ready to return to the quantum world, andin the next chapter we will make our first stab at relativistic quantummechanics.


Chapter summary

• A field takes a position in spacetime and outputs an object repre-senting the amplitude (be it scalar, vector or complex) of the fieldat that point.

• The Lagrangian and Hamiltonian density are related by H = πφ−Lwhere π = ∂L/∂φ is the generalized momentum.

• The Lagrangian for the electromagnetic field is L = − 14FµνF

µν ,where Fµν = ∂µAν − ∂νAµ is the electromagnetic field tensor.

Exercises

(5.1) If the Lagrangian does depend explicitly on time,then

dL

dt=∂L

∂t+∂L

∂xixi +

∂L

∂xixi. (5.57)

In this case show that

∂L

∂t= −dH

dt. (5.58)

(5.2) Show that Poisson brackets anticommute

A,BPB = −B,APB , (5.59)

and also satisfy the Jacobi identity˘

A,BPB , CPB +˘

C,APB , B¯

PB

+˘

B,CPB , A¯

PB= 0,

(5.60)

and show that quantum mechanical commutatorsalso have the same properties.

(5.3) Show that the commutator of two Hermitian oper-ators A and B is anti-Hermitian, i.e. that

[A, B]† = −[A, B]. (5.61)

The factor of i in many commutator expressions(e.g. [x, p] = i~, [Lx, Ly] = i~Lz, and [A, B] =1i~A,BPB) makes sure that this property is

obeyed.

(5.4) The Lagrangian for a free particle is L = −mc2/γ.Find an expression for L, p and H when v ≪ c.

(5.5) Show thatR b

ads takes its maximum value when the

integration is done over a straight world-line be-tween a and b [NB ds2 = c2dt2 − dx2 − dy2 − dz2].

(5.6) Use the Lagrangian L = −mc2γ

+ qA · v − qV fora free particle of charge q and mass m in an elec-tromagnetic field to derive the Lorentz force, i.e.show

d

dt(γmv) = q(E + v × B). (5.62)

[Hint: ∇(a · b) = (a ·∇)b + (b ·∇)a + b× curl a +a × curl b.]

(5.7) Show that for the Lagrangian

L =−mc2γ

+ qA · v − qV, (5.63)

when v ≪ c the momentum becomes p = mv + qAand the energy becomes

E = mc2 +1

2m(p − qA)2 + qV. (5.64)

(5.8) Another Lorentz scalar that can be obtained fromFµν is

εαβγδFαβFγδ. (5.65)

Show that this leads to E · B being an invariant.

(5.9) Show that ∂µFµν = Jν reduces to two of Maxwell’s

equations, and that the other two can be generatedfrom the equation

∂λFµν + ∂µFνλ + ∂νFλµ = 0. (5.66)

(5.10) Show that ∂β∂αFαβ = 0 and hence that ∂µJ

µ = 0.Interpret this as a continuity equation.

6A first stab at relativistic

quantum mechanics

6.1 The Klein–––Gordon equation59

6.2 Probability currents and den-sities 61

6.3 Feynman’s interpretation ofthe negative energy states 61

6.4 No conclusions 63

Chapter summary 63

Exercises 63

Bohr: What are you working on?Dirac: I’m trying to get a relativistic theory of the electron.Bohr: But Klein has already solved that problem.(Conversation at the 1927 Solvay Conference.)

We now take a break from the formalism of classical mechanics and turnback to quantum mechanics. In this chapter, we will attempt to finda quantum mechanical equation of motion that is properly relativistic.1

1It’s a first stab because the right an-swer is due to Dirac, and we will dealwith that later. This chapter is aboutthe attempt introduced by Klein andalso by Gordon, which is why in theopening quote Bohr didn’t think Diracwas working on something worthwhile.

This will lead to an equation known as the Klein–Gordon equation,though in fact it was derived first by Schrodinger, who rejected it becauseit had what seemed like a fatal flaw (which we will explore later); hethen went on to a second attempt at the problem and produced whatis known as the Schrodinger equation, which works brilliantly but isnon-relativistic.

6.1 The Klein–––Gordon equation

We first review the logical chain of ideas which leads to the Schrodingerequation. In the non-relativistic quantum mechanics of a free particle,we start with the dispersion relationship, which links the energy andmomentum of the particle using the simple expression for kinetic energy

E = p2

2m . We turn the variables E and p into operators (E → E,p → p),

and then we make the substitutions2 E = i~ ∂∂t and p = −i~∇. This 2Just to make some of the equations a

bit more familiar, we will briefly rein-troduce all the factors of ~ and c.

then leads to the Schrodinger equation

i~∂φ(x, t)

∂t= − ~2

2m∇

2φ(x, t), (6.1)

where the object that the operators act on is φ(x, t), the wave function.The free-particle solutions to the Schrodinger equation are plane waves

of the form φ(x, t) = Ne−i(ωt−k·x), where N is a normalization constant.We can write this solution in four-vector form φ(x) = Ne−ip·x. We’llcall a wave with this sign combination in the exponential an incoming

wave. Operating on it with momentum and energy operators yields

pφ(x, t) = −i~∇(Ne−i(ωt−k·x)) = ~kφ(x, t), (6.2)

Eφ(x, t) = i~∂

∂t(Ne−i(ωt−k·x)) = ~ωφ(x, t).

60 A first stab at relativistic quantum mechanics

That is, an incoming wave has a positive momentum and energy.Now, to get a relativistic wave equation we’ll try the same trick. For

a relativistic particle we have the dispersion relationship

E = (p2c2 +m2c4)12 , (6.3)

and by making the same operator substitutions as we had before (E →E = i~ ∂

∂t ,p → p = −i~∇) we obtain

i~∂φ

∂t= (−~2c2∇2 +m2c4)

12φ. (6.4)

This has two big problems. Number one: it doesn’t look covariant (thespatial and temporal derivatives appear in different forms). Numbertwo: how do we cope with the square root sign which seems to implythat we take the square root of a differential operator?

Instead of tackling these problems we take a side-step. To avoid thesquare root we’ll simply square the dispersion relation and start again.The squared form of the dispersion is E2 = p2c2 + m2c4. Making thesame operator substitutions as before, we obtain

−~2 ∂2φ(x, t)

∂t2= (−~2c2∇2 +m2c4)φ(x, t). (6.5)

This equation of motion for a wave function is called the Klein–Gordon

equation. To make things as easy to read as possible we’ll revert toOskar Klein (1894–1977)Walter Gordon (1893–1939) using natural units where ~ = c = 1 from now on, and this gives us the

following form of the Klein–Gordon equation:

−∂2φ(x, t)

∂t2= (−∇

2 +m2)φ(x, t). (6.6)

The Klein–Gordon equation fits in nicely with our relativistically covari-ant language, since we can write

(∂2 +m2)φ(x) = 0, (6.7)

thereby amalgamating the space and time derivatives.33Note that ∂2 = ∂µ∂µ = ∂2

∂t2− ∇

2.

Example 6.1

Let’s solve the Klein–Gordon equation for a free particle. We’ll try a solution44In natural units k = p and ω = Eand since we’re mostly interested in theenergies and momenta of particles we’lluse p and E.

φ(t,x) = Ne−iEt+ip·x = Ne−ip·x. Upon substituting we find that φ(x) is indeed asolution, yielding the expected dispersion

E2 = p2 +m2. (6.8)

This looks fine, until we realize that we have to take the square root to get the energywhich gives E = ±(p2 +m2). We might be tempted to just throw away the negativesolutions as unphysical, but in fact the appearance of negative energy solutions isgoing to be telling us something.

6.2 Probability currents and densities 61

6.2 Probability currents and densities

One of the reasons that Schrodinger wasn’t happy with the Klein–Gordon equation after he’d derived it was that something rather nastyhappens when you think about the flow of probability density. Theprobability of a particle being located somewhere depends on φ∗(x)φ(x)and so if this quantity is time-dependent then particles must be sloshingaround. The probability density ρ and probability current density5 j

5Here j is the current density of par-ticles, whereas in Example 5.6 we usedJ as the current density of charge. Ifboth appear together, as they will doin Section 6.3 below, we will write theformer as j and the latter as Jem.

obey a continuity equation

dρ

dt+ ∇ · j = 0, (6.9)

which is more easily written in four-vector notation as

∂µjµ = 0. (6.10)

If, as is usual in non-relativistic quantum mechanics,6 we take the spatial

6In non-relativistic quantum mechan-ics, j is defined with some additionalconstants:

j = −(i~/2m)(ψ∗∇ψ − ψ∇ψ∗).

We dispense with the ~/2m constanthere.

part to bej(x) = −i [φ∗(x)∇φ(x) − φ(x)∇φ∗(x)] , (6.11)

then, for eqn 6.10 to work,7 we require the probability density to look

7It will work, and you can prove it asfollows. Take the Klein–Gordon equa-tion (eqn 6.5) and premultiply it byφ∗. Then take the complex conjugateof eqn 6.5 and premultiply by φ. Sub-tracting these two results will give anequation of the form of eqn 6.9 with j

and ρ as given.

like8

8Note that this is very differentfrom non-relativistic quantum mechan-ics where ρ = |ψ|2.

ρ(x) = i

[φ∗(x)

∂φ(x)

∂t− ∂φ∗(x)

∂tφ(x)

]. (6.12)

The resulting covariant probability current for the Klein–Gordon equa-tion is then given by

jµ(x) = i φ∗(x)∂µφ(x) − [∂µφ∗(x)]φ(x) , (6.13)

which, as the notation suggests, is a four-vector. Substituting in ourwave function φ(x) = Ne−ip·x gives the time-like component of the prob-ability current as

j0 = ρ = 2|N |2E. (6.14)

Since E can be positive or negative we can’t interpret ρ as a probabilitydensity, since there’s no such thing as negative probability! This seemsto ring the death knell for the Klein–Gordon equation as a half-decentsingle-particle quantum mechanical equation.

6.3 Feynman’s interpretation of the

negative energy states

Fortunately, all is not lost. Richard Feynman9 and, independently, Ernst

9Richard Feynman (1918–1988) in-vented much of the physics in this book.The bongo-playing theoretical physicistonce defined science as ‘the belief in theignorance of experts’. He was never sat-isfied by an explanation until he hadworked it out for himself, and it is per-haps fitting that his quirky shorthandfor doing quantum field theory calcu-lations, the little cartoons we now callFeynman diagrams, decorate the pagesof every book on quantum field the-ory, including this one. (He also notedthat bongo players didn’t feel the needto point out that he was a theoreticalphysicist.)

Stuckelberg10 came up with an interpretation of the negative energy

10Brilliant, eccentric and difficult, theSwiss mathematician and physicistErnst Stueckelberg (1905–1984) mayalso claim credit for the invention ofa significant amount of the physics inthis book, although the opaqueness ofmany of his arguments and his insis-tence on inventing novel notation haveprevented this from being widely recog-nized. His achievements include: thisinterpretation of antiparticles; the in-vention of spacetime (‘Feynman’) dia-grams; the explanation of (‘Yukawa’)force-carrying particles; the discoveryof the renormalization group and theconservation of baryon number. SeeSchweber, Chapter 10 for more details.

states as particles moving backward in time. We call these states an-

tiparticles. Consider the classical equation of motion for a chargedparticle in an electromagnetic field which looks like

md2xµ

dτ2= qFµν

dxν

dτ, (6.15)

62 A first stab at relativistic quantum mechanics

where τ is proper time, q is charge and Fµν is the electromagnetic fieldtensor. Changing the sign of the proper time τ has the same effect aschanging the sign of the charge. Therefore, a particle travelling back-ward in time looks the same as an oppositely charged antiparticle mov-ing forward in time. One strategy to eliminate negative energy statesis therefore to reverse the charge and momentum of all negative energysolutions, turning them into antiparticles. How does this eliminate thenegative sign of the energy? Since we write the phase of the wave func-tion as Et− p · x, then making the substitution t→ −t means that wehave to change −E → E for the first term to make sense. Of coursereversing the time (like playing a film backwards) reverses all momentaso we also need to swap p → −p for consistency.

We can go further by invoking some quantum mechanics. Let’s ex-amine the electromagnetic current density for a Klein–Gordon particlewhich we will define as Jµem = qjµ where q is the charge of the particle.The current four-vector Jµem for a positively charged, incoming particle1111Incoming implies inclusion of a factor

e−ip·x. with positive energy is given by

Jµem = (+q)2|N |2pµ (6.16)

= (+q)2|N |2 (E,p) . (6.17)

In contrast, a negative energy particle with positive charge has Jµem =(+q)2|N |2 (−E,p). Since we’ve decided that negative energies are dis-gusting because they mess up the probability, we want to turn the energypositive. We do this by taking the minus sign out to the front of theequation to yield

Jµem = (−q)2|N |2 (E,−p) . (6.18)

Notice that we have to swap the sign of charge and the three-momentump, changing an incoming particle into an outgoing antiparticle, in agree-ment with our argument above.

We now have a rule to turn unphysical negative energies into physicalpositive energies. Simply treat the particle as one with positive energy,but change the particle into an antiparticle by reversing the charge andswapping the sign of the three-momentum.

BEFORE AFTER

(i)

E, e,p

EQq

E + EQ + eq + p

(ii)E, e,p

EQq

E − EQ− eq − p

(iii)

−E, e,p

EQq

E − EQ + eq + p

(iv)E,−e,−p

EQq

E − EQ + eq + p

Fig. 6.1 (i) An incoming particle (en-ergy E, charge e (≡ |e|), momentump) is absorbed by a system (energy E,charge Q, momentum q). Also shownare (ii) emission of particle, (iii) absorp-tion of a particle in a negative energystate and (iv) emission of an antiparti-cle in a positive energy state.

Is this really permissible for thinking about interacting quantum par-ticles, which we know from quantum mechanics can be created and de-stroyed and emitted and absorbed? It is, as we’ll see from Fig. 6.1.In a typical process we might have an incoming particle absorbed bya system, donating its energy, momentum and charge to the system[Fig. 6.1(i)]. Alternatively a particle could be emitted, depleting thesystem of energy, momentum and charge [Fig. 6.1(ii)]. The key point isthat absorption of a particle in a negative energy state [Fig. 6.1(iii)] isequivalent to the emission of an antiparticle in a positive energy state[Fig. 6.1(iv)], at least as far as the system is concerned.

To summarize then, although the dispersion relation E = ±√

p2 +m2

admits positive and negative solutions, only positive energies are physi-cal. We therefore make all energies positive, but we can’t just ignore theformerly negative energy states. The formerly negative energy states are

Exercises 63

interpreted as positive energy antiparticles with momenta in the oppositedirection to the corresponding particle. A general solution to the Klein–Gordon equation with a particular positive energy now corresponds toa superposition of two states12 12The assignments ‘incoming’ and ‘out-

going’ are made to link up with theconstruction of Feynman diagrams inChapter 19.φ(x) =

Incoming positiveenergy particle∝ e−i(Et−p·x)

+

Outgoing positiveenergy antiparticle

∝ e+i(Et−p·x)

. (6.19)

This general solution is therefore not an inherently single-particle de-scription, because we are forced to consider particles and antiparticles.

6.4 No conclusions

Although the Klein–Gordon equation has now been made semi-respectable with Feynman’s interpretation it still has an uncomfortablefeeling about it. What is implied by the new interpretation? Does itcorrespond to anything sensible in real life? We’ll leave these questionsfor a moment. The Klein–Gordon equation will return later and we willfind that it is, in fact, a perfectly respectable equation but in a slightlydifferent context, namely that of describing the dynamics of fields whoseexcitations are spinless particles. Our next step is to systematize ourprocedure for writing down theories.

Chapter summary

• The Klein–Gordon equation is (∂2 +m2)φ = 0.

• It suffers from having negative energy states and negative proba-bilities.

• We can think of these negative energy states as positive energyantiparticles

Exercises

(6.1) This problem looks forward to ideas which we willconsider in more detail in the next chapter. Showthat the Lagrangian density L given by

L =1

2(∂µφ)2 − 1

2m2φ2, (6.20)

yields the Klein–Gordon equation when using theEuler–Lagrange equation. Also derive expressionsfor the momentum π = ∂L/∂φ and the Hamilto-nian density H = πφ − L for this Lagrangian andinterpret the result.

7Examples of Lagrangians,

or how to write down a

theory

7.1 A massless scalar field 64

7.2 A massive scalar field 65

7.3 An external source 66

7.4 The φ4 theory 67

7.5 Two scalar fields 67

7.6 The complex scalar field 68

Chapter summary 69

Exercises 69

I regard as quite useless the reading of large treatises of pureanalysis: too large a number of methods pass at once beforethe eyes. It is in the works of applications that one muststudy them; one judges their ability there and one apprisesthe manner of making use of them.Joseph-Louis Lagrange (1736–1813)

Theoretical physics is all about writing down models to describe thebehaviour of particular systems in the Universe. The insight that wehave gained from the previous chapters is that the main starting pointfor a field theory is writing down an appropriate Lagrangian.1 This is1Lagrangian density is a bit of a mouth-

ful, so we are using the universal short-hand of referring to Lagrangian densi-ties simply as ‘Lagrangians.’ Pedantsbeware.

such an important step in the process that we are devoting an entirechapter to it, partly to reiterate some of the ideas introduced earlier andpartly to demystify the act of plucking a Lagrangian out of thin air aspractised by many professional theorists.

Why do we choose a particular Lagrangian L? Usually because it’sthe simplest model that contains all the key physical ideas and has thecorrect symmetry for the problem.2 Its artful construction allows it to2Symmetry will turn out to be essen-

tial to our understanding of field the-ory. However, we postpone its discus-sion until Chapter 10.

spit out the right equation of motion once fed into the waiting jaws ofthe Euler–Lagrange equation, given by

∂L∂φ

− ∂µ

(∂L

∂(∂µφ)

)= 0. (7.1)

Then, the wavelike solutions of this equation of motion can be inter-preted as particles with the appropriate dispersion relation, i.e. the rela-tion between energy and momentum. This process of upgrading classicalmodes into quantum particles will be described in detail in the next partof the book. Here we will just assume that it can be done somehow.

In this chapter, we will examine a succession of simple examples ofchoices of Lagrangian, for massless and massive scalar fields, a scalar fieldcoupled to an external source, a couple of scalar fields and a complexfield (deferring vector and spinor fields to later in the book).

7.1 A massless scalar field

We start with the simplest possible Lagrangian, one for a massless33The assignment of masslessness will beexplained shortly.

7.2 A massive scalar field 65

scalar field4 φ(x). The scalar field assigns a scalar amplitude to each 4Throughout this book we assume thescalar field φ(x) outputs a real numberrather than a complex one. The com-plex scalar field ψ(x) will be discussedseparately.

position x in spacetime. It can change as a function of spacetime co-ordinate and its gradient is ∂µφ ≡ ∂φ(x)/∂xµ. The Lagrangian willdepend only on the rate of change of φ in time ∂0φ and in space ∇φ.We have to make a choice of L which is relativistically covariant and sowe choose

L =1

2∂µφ∂µφ =

1

2(∂µφ)

2. (7.2)

This can be expanded as L = 12 (∂0φ)2 − 1

2∇φ · ∇φ, which has the lookof5 L = (kinetic energy)−(potential energy). The factor of 1

2 is included 5Note that 12(∂µφ)2 is often called the

kinetic energy as a shorthand.so that later results come out nicely, as we shall see.

Example 7.1

Using the Lagrangian in eqn 7.2, we have

∂L∂φ

= 0,∂L

∂(∂µφ)= ∂µφ, (7.3)

and hence plugging into the Euler–Lagrange equation (eqn 7.1) we have

∂µ∂µφ = 0. (7.4)

This is the wave equation ∂2φ = 0, or

∂2φ

∂t2− ∇

2φ = 0, (7.5)

and has wave-like solutions

φ(x, t) =X

p

ape−i(Ept−p·x), (7.6)

with dispersion relation Ep = c|p| [though in our units c = 1], see Fig. 7.1.

p

Ep

Fig. 7.1 The dispersion relation Ep =|p|.

Note that Ep = 0 at |p| = 0 and so the dispersion relation is gapless.If we take this equation to describe the energies of the particles thatwill (later) represent quantum excitations of this system then, since thisis the m = 0 version of the relativistic dispersion Ep =

√p2 +m2,

we call the particles massless and the field a massless scalar field. Thewave solutions of this linear equation obey the principle of superposition.We say that the particles which correspond to these waves are free ornon-interacting. If we send these particles towards each other they willsimply pass on through each other without scattering.

7.2 A massive scalar field

The next step is to try and include mass in the problem. To do this, wemake L depend not only on ∂µφ (how the scalar field changes throughspacetime) but also on φ (the magnitude of the scalar field). This canbe done by introducing a potential energy term U(φ) (and since La-grangians are kinetic energy minus potential energy, this will come witha minus sign). The potential energy term expresses the cost of having a

66 Examples of Lagrangians, or how to write down a theory

field there at all, rather than just vacuum. By choosing U(φ) ∝ φ2 wewill have a quadratic potential energy with a minimum at φ = 0. Thuswe write down

L =1

2(∂µφ)

2 − 1

2m2φ2, (7.7)

where the 12m

2 factor is chosen to make the next bit come out right.The parameter m has yet to be shown to be anything physical, but willturn out (of course) to be the particle mass.

Example 7.2

Using the Lagrangian in eqn 7.7, we have

∂L∂φ

= −m2φ,∂L

∂(∂µφ)= ∂µφ, (7.8)

and hence plugging into the Euler–Lagrange equation (eqn 7.1) we have

(∂µ∂µ +m2)φ = 0. (7.9)

The equation of motion for this field theory is the Klein–Gordon equation! Thesolution of these equations is again φ(x, t) = ae−i(Ept−p·x), with dispersion E2

p =p2 +m2 (see Fig. 7.2).

p

Ep

Ep = |p|m

Ep

=√ p

2+m

2

Fig. 7.2 The dispersion relation Ep =p

p2 +m2 for a massive particle.

Obviously if m = 0 we revert to the case of Section 7.1. With m 6= 0 wehave a gap in the dispersion relation (at p = 0, Ep = ±m) correspondingto the particle’s mass. Again, the equations of motion are linear andtherefore the particles described by this theory don’t interact.

7.3 An external source

We now want to introduce interactions. The simplest way to do this is tohave the scalar field interact with an external potential. This potential isdescribed by a function known as a source current J(x) which interactswith the field, giving a potential energy term −J(x)φ(x). The resultingLagrangian is written

L =1

2[∂µφ(x)]2 − 1

2m2[φ(x)]2 + J(x)φ(x). (7.10)

The equations of motion become66A shorthand version of the same equa-tion is

(∂2 +m2)φ(x) = J(x). (∂µ∂µ +m2)φ(x) = J(x). (7.11)

This is now an inhomogeneous partial differential equation. The use ofsources (or sorcery) will later turn up as a method of probing quantumfields following a technique invented by Julian Schwinger.

7.4 The φ4 theory 67

7.4 The φ4 theory

How do we get particles to interact with other particles (or, equiva-lently, fields to interact with other fields)? Here’s the simplest recipe.The φ4 Lagrangian is like the scalar case, but with an extra potentialterm U(φ) proportional to φ4. This term makes the theory interacting.Unfortunately it also makes it unsolvable. As we’ll see, we have to usea sort of perturbation theory to make predictions from this theory. TheLagrangian is7 7The numerical coefficient of the φ4

term is often chosen with conveniencein mind. We will use several in thecourse of the book, but the coefficientλ4!

will be most common.

L =1

2∂µφ∂µφ− 1

2m2φ2 − 1

4!λφ4, (7.12)

which leads to the (not-very-memorable) equation of motion

(∂2 +m2)φ = − λ

3!φ3. (7.13)

The constant of proportionality λ is known as the interaction strength.Clearly, when λ = 0 we return to the massive scalar field Lagrangianand the fields don’t interact.

7.5 Two scalar fields

Here’s another way to make particles interact. Why not have two differ-ent types of particle in the Universe, described by two fields φ1(x) andφ2(x)? In our simple theory they have the same mass and will inter-act with themselves and each other via a potential energy U(φ1, φ2) =

g(φ2

1 + φ22

)2where g is the interaction strength here. Notice that mul-

tiplying out the bracket gives self-interacting φ4 terms and, crucially forus, a cross-term 2φ2

1φ22 which forces the two types of field to interact.

The resulting Lagrangian density is

L =1

2(∂µφ1)

2 − 1

2m2φ2

1 +1

2(∂µφ2)

2 − 1

2m2φ2

2 − g(φ2

1 + φ22

)2.

(7.14)

We notice that we get the same Lagrangian as in eqn 7.14 even if wechange our definition of the fields. (We say that the Lagrangian has asymmetry.) If we transform the two fields using the mapping φ1 → φ′1and φ2 → φ′2 where

(φ′1φ′2

)=

(cos θ − sin θsin θ cos θ

)(φ1

φ2

), (7.15)

then the Lagrangian is unchanged. This looks a lot like we’ve rotatedthe fields in φ1-φ2 space. We say that the particles have an internal

degree of freedom. It’s as if they carry round a dial labelled by θ andby turning this dial (i.e. changing θ) a φ1 particle may be turned intoa superposition of φ1 and φ2 particles. The invariance of the physics(via the Lagrangian) with respect to rotations by θ in the φ1-φ2 planeexpresses an SO(2) symmetry.8

8SO(2) is the two-dimensional specialorthogonal group, corresponding to thegroup of 2 × 2 orthogonal matriceswith unit determinant. This is a con-tinuous symmetry and we will latershow that continuous symmetries leadto conserved quantities. More of this inChapters 13 and 15.

68 Examples of Lagrangians, or how to write down a theory

7.6 The complex scalar field

We can also make a transformation that simplifies the two-scalar-fieldsLagrangian considered in the previous section. We define two new fieldsψ and ψ† given by

ψ =1√2[φ1 + iφ2],

ψ† =1√2[φ1 − iφ2]. (7.16)

This transformation can be pictured by an Argand diagram with φ1

along the real axis and φ2 along the imaginary axis [Fig. 7.3(a)]. Usingthe new fields we obtain a new Lagrangian

L = ∂µψ†∂µψ −m2ψ†ψ − g(ψ†ψ)2. (7.17)

This is known as a complex scalar field theory. Although we originallymade it up from two sorts of field, we can imagine that it describesone sort of complex-valued field ψ. This new field is constructed froma bit of φ1 and a bit of φ2 and retains two degrees of freedom. Thenew Lagrangian (eqn 7.16) is invariant with respect to rotations in thecomplex plane [Fig. 7.3(b)]

ψ → ψeiα, ψ† → e−iαψ†, (7.18)

which expresses a U(1) symmetry.99U(1) is the one-dimensional group ofunitary transformations. Mathemati-cally we note that SO(2) is isomorphicto U(1) (see Fig. 7.4), so nothing hasreally changed at a fundamental level.

Realφ1√

2

Imaginary

φ2√2

ψ

ψ†

eiα ψ

e−iα ψ†

α

α

Realφ1√

2

Imaginary

φ2√2

ψ

ψ†

(a)

(b)

Fig. 7.3 The complex scalar field onan Argand diagram.

SO(2) U(1)

rotate by θ Reψ

Imψ

θ

Fig. 7.4 The group SO(2), describ-ing rotations in two-dimensions, is iso-morphic to U(1), the one-dimensionalgroup of unitary transformations de-scribed by the phase of the complexnumber ψ = |ψ|eiθ.

Example 7.3

The ψ†ψφ theory: Finally, as a rather more complicated example, we consider atheory with three types of particle. We add together Lagrangians for the complexscalar field theory (with massm) and for the scalar field theory (with mass µ) and alsoarrange for the φ and ψ particles to interact via a potential gψ†ψφ. The Lagrangianis

L = ∂µψ†∂µψ −m2ψ†ψ

+1

2(∂µφ)2 − 1

2µ2φ2 − gψ†ψφ. (7.19)

As we will see a little later, this theory looks a lot like quantum electrodynamics!

The fields in this chapter have all been classical and although we’veclaimed that particles are the excitations of these fields we have yet tosee how they arise from quantum mechanical versions of the fields. Thisis the subject of the next part of the book.

Exercises 69

Chapter summary

• Lagrangians are deduced for a massless scalar field, massive scalarfield, the inclusion of a source, two scalar fields and a complex field.

• For example, the Lagrangian for a massive scalar field coupled toa source is

L =1

2[∂µφ(x)]2 − 1

2m2[φ(x)]2 + J(x)φ(x).

• The Euler–Lagrange equation can be used to deduce an equationof motion.

• For example, the equation of motion for a massive scalar field cou-pled to a source comes out to be

(∂2 +m2)φ(x) = J(x).

• The complex scalar field ψ = 1√2(φ1 + iφ2) possesses a U(1) sym-

metry.

Exercises

(7.1) For the Lagrangian L given by

L =1

2∂µφ∂µφ− 1

2m2φ2 −

∞X

n=1

λnφ2n+2, (7.20)

show that the equation of motion is given by

(∂2 +m2)φ+∞X

n=1

λn(2n+ 2)φ2n+1 = 0. (7.21)

(7.2) Consider a massive scalar field φ(x) coupled toa source J(x), described by the Lagrangian ofeqn 7.10. Show that the equations of motion arethose of eqn 7.11.

(7.3) Show that the equations of motion following fromthe Lagrangian in eqn 7.14 are the coupled equa-tions

∂µ∂µφ1 +m2φ1 + 4gφ1(φ

21 + φ2

2) = 0, (7.22)

∂µ∂µφ2 +m2φ2 + 4gφ2(φ

21 + φ2

2) = 0. (7.23)

(7.4) Show that eqn 7.7 is equivalent to

L =1

2φ2 − 1

2(∇φ)2 − 1

2m2φ2. (7.24)

Show that

π =∂L∂φ

= φ (7.25)

and

H =1

2φ2 +

1

2(∇φ)2 +

1

2m2φ2. (7.26)

Define the quantity

Πµ =∂L

∂(∂µφ), (7.27)

and show that Πµ = ∂µφ and Π0 = π = φ.

Part III

The need for quantum

fields

In this part we introduce quantum fields, how to construct quantum fieldtheories, and look at the transformations that can be applied to them.Understanding the underlying symmetries of a quantum field turns outto be crucial for exploring the properties of quantum field theory. Thispart is structured as follows:

• To study quantum fields, we must first describe how to includetime dependence, which is covered in Chapter 8. In doing this,we inadvertently stumble on a reason why single-particle quantummechanics is insufficient for fully describing the world in which welive.

• In Chapter 9 we discuss how to describe the response of quantummechanical states to transformations: translations, rotations andLorentz boosts.

• This sets up the treatment of Noether’s theorem in Chapter 10,demonstrating the fundamental relationship between symmetryand conservation laws.

• We are then ready in Chapter 11 to plunge into the deep waters ofconstructing quantum field theories out of classical field theories, amanufacturing process termed canonical quantization. We give anextended example of this process for complex scalar field theory inChapter 12

• These ideas are extended to multicomponent fields in Chapter 13and we work this through for the case of massive electromagnetism,thereby studying a spin-1 vector field.

• Transformations so far have been global. When examining localtransformations we find ourselves introducing gauge theory andthis is covered in Chapter 14.

• In Chapter 15 we cover discrete symmetries: C, P and T. We lookat the connection between group theory and the symmetries wehave been studying in Part III.

8 The passage of time

8.1 Schrodinger’s picture and thetime-evolution operator 72

8.2 The Heisenberg picture 74

8.3 The death of single-particlequantum mechanics 75

8.4 Old quantum theory is dead;long live fields! 76

Chapter summary 77

Exercises 78

The innocent and the beautifulHave no enemy but timeW. B. Yeats (1865–1939)

He that will not apply new remedies must expect new evils;for time is the great innovatorFrancis Bacon (1561–1626)

There is more than one way of putting time into quantum mechanics.In this chapter we’ll discuss Heisenberg’s version of quantum theory inwhich all the time dependence is located in the operators, while thewave functions stay fixed and unchanging. This contrasts with the morefamiliar version that Schrodinger formulated in which the wave func-tions carry the time dependence, the formulation that is taught in mostintroductory courses. Far from being a dry exercise in formalism, thiswill allow us to kill single-particle quantum mechanics1 stone-dead and1Single-particle quantum mechanics, as

the name suggests, describes the quan-tum behaviour of a single particle con-fined in some potential. As we will de-scribe, even in such a starting situationit is possible to have to worry aboutthe appearance of particle–antiparticlepairs, and thus the assumption that‘you are just dealing with a single par-ticle’ is not as straightforward as youmight think.

guide us in a direction towards quantum fields.

8.1 Schrodinger’s picture and the

time-evolution operator

The most familiar way of doing quantum mechanics is to think abouttime-dependent wave functions. The manner in which the wave functionchanges with time is given by the Schrodinger2 equation2This is of course named after Austrian

physicist Erwin Schrodinger (1887–1961).

i∂ψ(x, t)

∂t= Hψ(x, t), (8.1)

which is an equation of motion for the wave function. Dynamic variableslike momentum and position are accessed with operators p = −i∇ andx = x respectively, which act on the wave function. This way of doingquantum mechanics is known as the Schrodinger picture and is com-pletely different from classical mechanics, where the dynamic variablesdepend on time [e.g. p(t),x(t), etc.] and are themselves described byequations of motion. The root of this difference is that time itself isa funny thing in quantum mechanics. There is no operator that tellsus when some event occurred; instead time is a parameter that in theSchrodinger picture is located purely in our wave function ψ(x, t).

Although no operator tells us the time, there is an operator U(t2, t1)that can evolve a particle forward through time, from time t1 to t2. Thus

8.1 Schrodinger’s picture and the time-evolution operator 73

we can writeψ(t2) = U(t2, t1)ψ(t1), (8.2)

and we will call U(t2, t1) a time-evolution operator. Notice that thetime-evolution operator has two slots in which times can be inserted:the start-time t1 and the stop-time t2.

t1 t2 t3

t1 t2

time

timeU(t2, t1)

U(t1, t2)

U(t2, t1)

U(t3, t2)

U(t3, t1)

(b)

(a)

Fig. 8.1 (a) The composition law:

U(t3, t2)U(t2, t1) = U(t3, t1). (b) Theinverse of a time-evolution operator isequivalent to reversing the direction oftime: U(t1, t2) = U−1(t2, t1).

The time-evolution operator has five very useful properties:

(1) U(t1, t1) = 1.If the two times are the same, then nothing can happen and U isjust the identity operator.

(2) U(t3, t2)U(t2, t1) = U(t3, t1).This is the composition law and shows that we can build up a timetranslation by multiplying a set of smaller translations that do thesame evolution, but in steps [see Fig. 8.1(a)].

(3) id

dt2U(t2, t1) = HU(t2, t1).

The proof of this equation is simple. Differentiating eqn 8.2 withrespect to t2 gives

dψ(t2)

dt2=

dU(t2, t1)

dt2ψ(t1), (8.3)

but eqn 8.1 gives

idψ(t2)

dt2= Hψ(t2) = HU(t2, t1)ψ(t1), (8.4)

and the result follows. This shows that the time-evolution operatoritself obeys the Schrodinger equation.

(4) U(t1, t2) = U−1(t2, t1),so by taking the inverse of a time-evolution operator, one can turnback time [see Fig. 8.1(b)].

(5) U†(t2, t1)U(t2, t1) = 1,i.e. the time-evolution operator is unitary. The proof is short.When t2 = t1, the result is trivial. If we vary t2 away from t1,then U†(t2, t1)U(t2, t1) could change, but since3

3Note that in deriving eqn 8.5 we areusing property 3 of the time evolutionoperator in the form

dU

dt=HU

i,

dU†

dt= − U

†H

i.d

dt2

[U†(t2, t1)U(t2, t1)

]=

dU†

dt2U + U† dU

dt2

= − U†HU

i+U†HU

i= 0, (8.5)

it won’t. QED. One consequence of this result is that U†(t2, t1) =U−1(t2, t1), that is, the adjoint is the inverse.

Property number 3 allows us to write an explicit expression for U(t2, t1),since integrating this equation gives4

4This exponential expression for an op-erator is actually shorthand for an ex-pansion:

eA = 1+A+1

2!AA+

1

3!AAA+. . . (8.6)

We will meet many such exponentialexpressions.U(t2, t1) = e−iH(t2−t1). (8.7)


8.2 The Heisenberg picture

The Schrodinger picture puts all the time-dependence into the wavefunctions, but is that the best way to look at things? Consider theexpectation value of an operator

〈O(t)〉 = 〈ψ(t)|O|ψ(t)〉. (8.8)

This tells us the average value that we’ll get if we make a measurementof the quantity corresponding to the operator O at a time t. However weconstruct our quantum mechanics, we must end up with this equationfor expectation values to guarantee that we get the same prediction forthe result of any measurement.55Another example of a quantity we

want to agree on is the amplitude fora system being in state |φ(t2)〉 at atime t2, having been prepared in a state|ψ(t1)〉 at an earlier time t1, which isgiven by 〈φ(t2)|ψ(t1)〉. (Strictly weshould only insist on the probability be-ing the same, so we could allow a dif-ference in phase.)

We can also formulate things as follows by only worrying about thewave function at t = 0. After all, if we know ψ(0), we can alwayscalculate ψ(t) using

ψ(t) = U(t, 0)ψ(0) = e−iHtψ(0). (8.9)

Applying this to the expectation value in eqn 8.8, we can write

〈ψ(t)|O|ψ(t)〉 = 〈ψ(0)|U†(t, 0) O U(t, 0)|ψ(0)〉. (8.10)

We can interpret this equation in two ways. In the Schrodinger picturewe think of time-independent operators OS ≡ O and time-dependentwave function states |ψS(t)〉 ≡ U(t, 0)|ψ(0)〉, so that

〈ψ(0)|U†(t, 0)[O]U(t, 0)|ψ(0)〉 = 〈ψS(t)|OS|ψS(t)〉, (8.11)

where the ‘S’ subscripts stand for ‘Schrodinger picture’. To work out howthe states evolve with time, we use the Schrodinger equation, eqn 8.1.However, we can also move the square brackets in eqn 8.11, so that itreads

〈ψ(0)|[U†(t, 0) O U(t, 0)

]|ψ(0)〉 = 〈ψH|OH(t)|ψH〉, (8.12)

where the ‘H’ stands for the Heisenberg picture. We have made theWerner Heisenberg (1901–1976)

wave functions ψH ≡ ψ(0) time-independent (essentially by freezing thewave functions at their t = 0 values) and all the time dependence is nowformally located in the operators OH(t) which are given by

OH(t) ≡ U†(t, 0) OS U(t, 0), (8.13)

a situation which is analogous to the dynamical variables in classical me-chanics. The time dependence of OH(t) is simply given by differentiatingeqn 8.13, so that6

6Note that we are once again usingproperty 3 of the time evolution opera-tor in the form

dU

dt=HU

i,

dU†

dt= − U

†H

i. dOH(t)

dt=

dU†

dtOSU + U†OS

dU

dt=

1

i(−U†HOSU + U†OSHU), (8.14)

and hence we arrive7 at the result7We do this using eqn 8.13 and the fact

that U and H commute. dOH(t)

dt=

1

i~[OH(t), H], (8.15)

which is known as the Heisenberg equation of motion.

8.3 The death of single-particle quantum mechanics 75

8.3 The death of single-particle quantum

mechanics

The time-evolution operator gives us enough ammunition to kill offsingle-particle quantum mechanics forever. We ask the question: ‘whatis the amplitude for a particle to travel outside its forward light-cone?’(see Fig. 8.2), or in other words: ‘what is the amplitude for a particleat the origin in spacetime (x = 0, t = 0) to travel to position x at timet which would require it to travel faster than light?’ Faster than lightmeans |x| > t and the interval (x− 0, t− 0) is space-like.8 If the ampli- 8This means that |x| > ct in unnatural

units. Our convention is that if x =(t,x) then x2 = t2 − (x1)2 − (x2)2 −(x3)2 and so the interval is space-like ifx2 < 0.

tude is nonzero then there will be a nonzero probability for a particle tobe found outside its forward light-cone. This is unacceptable and wouldspell the death of quantum theory as we’ve known it so far.

t

x2

x1

timelike

spacelike

origin

(x 1) 2

+(x 2

) 2+

(x 3) 2

=t 2

Fig. 8.2 A particle shouldn’t be ob-served outside its forward light-cone. Ifit starts at the origin, it should only beable to make time-like journeys.

To answer the question we are going to evaluate the expression

A = 〈x|e−iHt|x = 0〉, (8.16)

for the space-like interval. We’ll work with a basis of momentum statesH|p〉 = Ep|p〉 where the particle has a relativistic dispersion Ep =√

p2 +m2. We’ll also need the transformation between position andmomentum bases:9 〈x|p〉 = 1

(2π)3/2 eip·x.

9This assumes a particular normaliza-tion. For our current purpose, where allwe will do is to show a particular quan-tity is nonzero, the value of the normal-ization constant is irrelevant.

We start by inserting a resolution of the identity10

10i.e. since we have a complete set of

states then 1 =R

d3p |p〉〈p|.

〈x|e−iHt|x = 0〉 =

∫d3p 〈x|e−iHt|p〉〈p|x = 0〉

=

∫d3p 〈x|p〉e−iEpt

1

(2π)3/2e−ip·0

=

∫d3p

1

(2π)3eip·xe−iEpt. (8.17)

This has been fairly straightforward so far. Now we need to do theintegral.

Example 8.1

The technique for doing an integral like this is very useful in quantum field theory,so we’ll go through it in detail. We start by converting to spherical polars

A =

Z 2π

0dφ

Z ∞

0

d|p|(2π)3

|p|2Z 1

−1d(cos θ) ei|p||x| cos θe−iEpt

=1

(2π)21

i|x|

Z ∞

0d|p| |p|

“

ei|p||x| − e−i|p||x|”

e−iEpt

=−i

(2π)2|x|

Z ∞

−∞d|p| |p|ei|p||x|e−it

√|p|2+m2

. (8.18)

To do this integral we need to resort to complex analysis.11 The integration path 11See Appendix B for a primer.is shown in Fig. 8.3(a). There are cuts on the imaginary axis starting at ±im andheading off to ±∞. We can deform the contour into a large semicircle in the upperhalf-plane that goes down the upper cut on the left side and up the cut on the right[Fig. 8.3(b)].


The term ei|p||x| decreases exponentially as we head off to large imaginary values

of |p| because we have |x| > 0. The term e−it√

|p|2+m2also decreases exponentially

as you head up the left side of the cut but increases as you go up on the right side.

(This is because Im“p

|p|2 +m2”

< 0 on the left side and Im“p

|p|2 +m2”

> 0

on the right side.) The increase on the right side isn’t a problem since it’s multiplied

by ei|p||x| and for our condition that |x| > t the product e−i|p||x|e−it√

|p|2+m2

decreases. Jordan’s lemma therefore tells us that as we make the half-circle largeit doesn’t contribute to the integral, therefore, and we only have to worry aboutthe part going up and down the cut. This allows us to deform the integral further[Fig. 8.3(c)]. We now substitute |p| = iz and the integral becomes

A =−i

(2π)2|x|

Z ∞

md(iz) ize−z|x|

„

et√z2−m2 − e−t

√z2−m2

«

=i

2π2|x| e−m|x|

Z ∞

mdz ze−(z−m)|x| sinh

“

tp

z2 −m2”

. (8.19)

The integrand is positive, definite and the integral is nonzero. This means that |A|2yields a nonzero number, heralding doom for the single-particle picture.

im

−im

∫∞−∞ · · ·

im

−im

im

−im

∫ im∞ left-side +

∫∞im right-side

Fig. 8.3 The contour integration.

We’ve discovered that the probability of finding a particle outside itsforward light-cone is proportional to e−m|x|. Although this is small forlarge |x| it’s certainly unacceptable that it exists at all. We are unableto reconcile single-particle quantum mechanics and special relativity.Single-particle quantum mechanics is dead!

There is another way of understanding why single-particle quantummechanics fails when faced with special relativity. Imagine squeezinga particle of mass m into a box with side smaller than the particle’sCompton wavelength.12 The uncertainty in position ∆x ≪ λ so ∆p ≫

12The Compton wavelength λ of a par-ticle is given by λ = h/mc, i.e. h/m inunits where c = 1.

h/λ = m and so the energy of the particle is now much greater than m.This is enough, according to special relativity, to lead to pair production(i.e. the creation of particle–antiparticle pairs). Thus, on average, thebox has more than one particle in it. Single-particle quantum mechanicscan never describe this situation, by definition!

8.4 Old quantum theory is dead; long live

fields!

A desperate disease requires a dangerous remedyGuy Fawkes (1570–1606)1313The same sentiment was also ex-

pressed by Hippocrates and by Shake-speare (in Hamlet). What are we to do instead? We resort to fields. In classical mechanics

we assume that an observer has access to all possible measurements ofa system. Special relativity teaches us that no one person can measureeverything. If information would have to travel faster than c for a mea-surement to be made it is impossible to make such a measurement. Infact, you make a measurement at a single point in spacetime where you,yourself, are localized. You can’t then make a measurement at a pointthat has a space-like separation from you. More simply, an observer can’tmake a measurements here on Earth and then make a measurement of anevent in Andromeda14 that occurred one second later. The Andromeda

14The Andromeda galaxy is some 2.5million light-years from Earth.

8.4 Old quantum theory is dead; long live fields! 77

information would have to travel to the Earthbound observer faster thanc, which is forbidden by special relativity.

Have we seen this limitation elsewhere in physics? Absolutely. Inelectromagnetism we calculate and measure the value of an electric ormagnetic vector at a spacetime point x. We call these E(x) and B(x)respectively and they are examples of fields. They are machines that takespacetime coordinates xµ = (t,x) and output the value of the electricor magnetic vector at that point. The field machines are defined locally,and so they only tell us about the electromagnetic vectors at the pointx. We know that we can’t design an experiment that measures E(x) forx = (now,here) and for x = (now,Andromeda).15 15Of course, if we solve some equa-

tions of motion for the fields we maybe able to work out what’s happeningelsewhere.

We need to fix up this idea of a locally defined field to make it quantummechanical. In quantum mechanics we have operators that correspondto observables. It seems, therefore, that we will need operator-valued

fields if we’re to reconcile quantum mechanics and special relativity.We know two things about these field operators already: (i) crucially,they must not allow information to travel faster than c; (ii) they mustbe Hermitian if they are to correspond to observables.16 We can state 16It is not essential that a general field

should necessarily correspond to an ob-servable. Indeed we will meet many ex-amples where this is not the case. Forelectric and magnetic fields we do seekHermitian fields.

the former property quantum mechanically. If we say that the operatorO1 corresponds to an observable measured at spacetime point x andO2 corresponds to an observable measured at y, then if x and y have aspace-like separation [(x− y)2 < 0] then the operators must commute:

[O(x), O(y)] = 0, (8.20)

that is, the result of one measurement won’t affect the result of the other.How could they if the separation between the experiments is space-like?

To make quantum fields we need to find operators that are defined atpositions in spacetime. We’ll call these φ(x). We hope that we can buildeverything from these locally defined fields.17 In the next few chapters 17This turns out to be the case!

we will continue our search for these objects.

Chapter summary

• The time-evolution operator is given by U(t2, t1) = e−iH(t2−t1),and in the Schrodinger picture acts to time-evolve states. Theoperators OS are time-independent.

• In the Heisenberg picture states don’t evolve in time, but the oper-ators do: OH(t) = U†(t, 0) OS U(t, 0). Operators obey the Heisen-

berg equation of motion dOH(t)dt = 1

i~ [OH(t), H].

• Squeezing a particle of mass m into a region smaller that itsCompton wavelength h/m can lead to pair production, invalidatingsingle-particle quantum mechanics.

• To fix this, we need operator-valued fields.


Exercises

(8.1) Show that the form of the time evolution operatorU(t2, t1) = exp[−iH(t2 − t1)] (as given in eqn 8.7)exhibits properties 1–5 in Section 8.1.

(8.2) For the Hamiltonian

H =X

k

Eka†kak, (8.21)

use the Heisenberg equation of motion to show thatthe time dependence of the operator a†k is given by

a†k(t) = a†k(0) eiEkt/~ , (8.22)

and find a similar expression for ak(t).

(8.3) For the Hamiltonian in the previous problem findan expression for time-dependence of the operatorX = Xℓma

†ℓ am.

(8.4) A spin- 12

particle is in a magnetic field aligned alongthe y-direction. The Hamiltonian can be written asH = ωSy where ω is a constant (proportional tothe magnetic field). Use the Heisenberg equation ofmotion to show that

dSzHdt

= −ωSxH, (8.23)

dSxHdt

= ωSzH, (8.24)

and give a physical interpretation of this result.

9Quantum mechanical

transformations

9.1 Translations in spacetime 79

9.2 Rotations 82

9.3 Representations of transfor-mations 83

9.4 Transformations of quantumfields 85

9.5 Lorentz transformations 86

Chapter summary 88

Exercises 88

When Gregor Samsar awoke one morning from uneasydreams he found himself transformed in his bed into a gi-ant insect.Frank Kafka (1883–1924), The Metamorphosis

Quantum mechanical states transform (though usually less dramaticallythan the transformation that affected Gregor Samsar) when you trans-late or rotate coordinates. In this chapter we are going to look at howquantum states, operators and quantum fields will transform when sub-jected to translations, rotations or Lorentz boosts. Where do we start?One thing we certainly know how to do is transform coordinates inspacetime. For example, we can translate a position three-vector x by athree-vector a, which is achieved simply by adding a to the vector. Toput things formally, we act on x with a translation operator T(a) andtransform it into x′ given by

x′ = T(a)x = x + a. (9.1)

We also know how to rotate a three-vector through an angle θ about thez-axis, using

x′ = R(θ)x =

cos θ − sin θ 0sin θ cos θ 0

0 0 1

x1

x2

x3

. (9.2)

We need to find operators that can transform quantum states, whichlive in Hilbert space (rather than the Euclidean space of three-vectorsor Minkowski space of four-vectors). Here we investigate what formthese operators take.

Remark about notation:We will be careful to distinguish be-tween, on one hand, operators T(a)and R(θ) which act on single vectors(like x) and, on the other hand, quan-tum mechanical transformation oper-ators like U(a) and U(θ) which acton quantum states (like |x〉) and also(as we shall see) on operators (likex) and on quantum fields (like φ(x)).These quantum mechanical transfor-mation operators will turn out to beunitary, so we shall use the symbol Ufor all of them.

9.1 Translations in spacetime

Let’s start with looking at the properties of single-particle quantumsystems when we translate them in space. Suppose we’re examining aparticle localized at a coordinate and we want to translate it to anotherpoint in space. We need a quantum mechanical operator U that takes astate localized at x and transforms it to a position x+a. The translationoperator is written as

U(a)|x〉 = |x + a〉. (9.3)

80 Quantum mechanical transformations

We will treat this as an active transformation where we move a particleto a new position (instead of changing the coordinates under the particle,which is known as a passive transformation)1 (see Fig. 9.1).

1The transformation in eqn 9.2 is aright-handed, active rotation or a left-handed, passive rotation.

x

y

x

y

x′y′

x ′

y ′

(b)

(a)

passive rotation

active rotation

Fig. 9.1 (a) A passive rotation acts onthe coordinates. (b) An active rotationacts on the object.

Next we examine the properties we want in a translation operator.Translating the particle shouldn’t change the probability density. Thus

〈ψ(x)|ψ(x)〉 = 〈ψ(x + a)|ψ(x + a)〉 = 〈ψ(x)|U†(a)U(a)|ψ(x)〉, (9.4)

which means that we may write

U†(a)U(a) = 1, (9.5)

that is, the translation operator is unitary.2

2A unitary matrix is defined as one for

which U† = U−1. Note that sometransformations in quantum mechanicsmust be performed by an antiunitaryoperator as discussed in Chapter 15.

We know that a translation through a distance a followed by onethrough a distance b can be represented by a single translation througha distance a + b. We want our operators to have this property too, sowe add to our wish list

U(a)U(b) = U(a + b). (9.6)

We should also have the trivial property that the transformation can donothing to the particle, that is U(0) = 1.

To recap then, we have identified three rules for our translation oper-ator:

• U(a)U†(a) = 1 (U is unitary),

• U(a)U(b) = U(a + b) (composition rule),

• U(0) = 1 (a zero translation does nothing).

Notice that these rules are telling us that our transformations shouldform a group.3 It is a rather special group because each element depends

3A group is a set G together withan operation • which connects two el-ements of the set (let’s say a, b ∈ G)and produces another element of the set(written a • b). This has to satisfy fourconditions:

(1) ∀a, b ∈ G, a • b ∈ G (closure).The symbol ∀ means ‘for all’.The first property written in fullis ‘For all members of the groupa and b, the combination a • b isalso a member of the group’.

(2) ∀a, b, c ∈ G, a• (b•c) = (a•b)•c(associativity).

(3) ∃e ∈ G such that ∀a ∈ G, a•e =e • a = a (identity).

(4) ∀a ∈ G, ∃a−1 ∈ G such that a •a−1 = a−1 • a = e (inverse).

Note that one of the conditions wasnot that a • b = b • a. If that condi-tion (commutativity) holds, then onehas a special type of group called anabelian group, but many groups (in-cluding ones which are important inquantum field theory) are non-abelian.

on a vector a, and thus the group is continuous, differentiable and hasan infinite number of elements. Such a group is called a Lie group.4

4Named after the Norwegian mathe-matician Sophus Lie (1842–1899).Note that Lie is pronounced ‘lee’.

Example 9.1

Let’s check that U(a) has the required conditions of a group. A group has fourproperties: the existence of an identity [U(0) = 1], closure (we can make any elementof the group by multiplying two other elements of the group), associativity [U(a +b)U(c) = U(a)U(b + c)] and the existence of an inverse for each element of thegroup. We also know that the inverse of a translation T−1(a), is just a translationthrough −a. Putting this together with the unitarity of the operator, we can writethat U−1(a) = U†(a) = U(−a).

Let’s now try out the new operator. It’s supposed to move a state though a andwe can use the position operator x to check where it is:

U(a)|x〉 = |x + a〉xU(a)|x〉 = x|x + a〉 = (x + a)|x + a〉

U†(a) xU(a)|x〉 = (x + a)U†(a)|x + a〉 = (x + a)|x〉, (9.7)

or comparing the effects of operators on the state |x〉, we see that

U†(a) x U(a) = (x + a). (9.8)

That is, the space translation operator can be thought of as a transformation of anoperator like x instead of a transformation of a wave function like |x〉.

9.1 Translations in spacetime 81

Example 9.2

Often translating will have no effect on a property that we determine using an oper-ator O. In that case the property probed by that operator is an invariant.5 In that 5See Chapter 10.case we would have

〈ψ(x)|O|ψ(x)〉 = 〈ψ(x)|U−1(a)OU(a)|ψ(x)〉. (9.9)

Pulling out the operators we see that the condition for an invariant is

U−1(a)OU(a) = O. (9.10)

Act on the left with U and thus invariance implies that OU = UO, or [O, U ] = 0.Recall from eqn 8.15 that operators that commute with the Hamiltonian operator Hrepresent constants of the motion, i.e. quantities that are conserved.

Can we come up with an explicit expression for a translation operatorthat acts on quantum states? We will do this for the special case of aposition wave function ψ(x) = 〈x|p〉 (the result will turn out to also allowus to transform general quantum mechanical states). We increment thepoint at which we evaluate ψ(x) by an infinitesimal amount δa in thex-direction, for which we may write

ψ(x+ δa) = ψ(x) +dψ(x)

dxδa+ . . . . (9.11)

Remembering that the momentum operator p = −i ddx , we have

ψ(x+ δa) = (1 + ipδa)ψ(x). (9.12)

We say that the momentum operator p is the generator for the spacetranslation. To make a translation through a distance a we can translatethrough δa a large number N times, such that

ψ(x+ a) = limN→∞

(1 + ipδa)Nψ(x)

= eipaψ(x). (9.13)

This gives us a space evolution operator. However, this is not quite theU(a) that we’re trying to find! We want to translate the entire quantumstate through a distance a not study how it evolves over a distance a!The translation operator is, therefore, U(a) = e−ip·a.

Example 9.3

Acting on a momentum state with our operator we obtain

U(a)|q〉 = e−ip·a |q〉= e−iq·a |q〉. (9.14)

Projecting along the coordinate direction gives us a translated wave function

〈x|U(a)|q〉 = 〈x|q〉e−iq·a =1√V

eiq·(x−a). (9.15)


Previously we had a time-evolution operator U(t2,t1) = e−iH(t2−t1). Wecan see how this relates to the argument in the previous section. Let’sevolve a system through a time δta. We obtain

ψ(t+ δta) = ψ(t) +dψ(t)

dtδta + . . . (9.16)

Now we need to remember that H = i ddt , which gives us

ψ(t+ δta) =(1 − iHδta

)ψ(t). (9.17)

Remembering the change of sign required to turn this from a time-evolution operator6 into a time translation operator we easily obtain6It might be helpful to view time evo-

lution as passive: one studies how ψchanges as coordinate t increases. Timetranslation is then active: one moves aquantum state through time.

U(ta) = eiHta . (9.18)

We can put the space and time translations together and define a space-time translation operator U(a) = eip·a = eiHta−ip·a, where we choosethe definition of the four-momentum operator to be p = (H, p). We willreturn to the consequences of this exponential form for the expressionfor U(a) later in the chapter.

Remark about notation:Spacetime translations, rotations,Lorentz boosts and combinations ofthe above are all elements of thePoincare group. They can all berepresented by unitary operators. Tokeep ourselves from an overload ofbrackets and indices, we will simplifythe notation by denoting all theseunitary operators by U and let theoperator’s argument indicate whetherwe are dealing with a translation U(a),rotation U(θ) or Lorentz boost U(φ).This should not cause confusion as thecontext will make it clear which typeof transformation we are dealing with.

9.2 Rotations

Next we ask how to rotate objects. A rotation matrix R(θ) acts on avector quantity, such as the momentum of a particle as follows: p′ =R(θ)p. We specify the rotation as R(θ), where the direction of thevector θ is the axis of rotation and its magnitude is the angle. Forrotations of quantum states, we propose an operator

|p′〉 = U(θ)|p〉 = |R(θ)p〉. (9.19)

As before, we require that U(θ) has: (i) unitarity, (ii) an identity elementand (iii) a composition rule. What we’re looking for then is that

• U†(θ)U(θ) = 1,

• U(0) = 1,

• U(θ1)U(θ2) = U(θ12) where R(θ12) = R(θ1)R(θ2),

and, just as before, our operators form a Lie group, called the rotationgroup.

Example 9.4

We can prove that the operator defined this way is unitary

U(θ)U†(θ) = U(θ)

„Z

d3p |p〉〈p|«

U†(θ)

=

Z

d3p |R(θ)p〉〈R(θ)p| (9.20)

9.3 Representations of transformations 83

and since p′ = R(θ)p and d3p′ = d3p we haveZ

d3p |R(θ)p〉〈R(θ)p| =

Z

d3p′ |p′〉〈p′| = 1, (9.21)

as required. Notice that our proof rests on the fact that d3p′ = d3p, which is truesince detR(θ) = 1 for a proper rotation and so the Jacobian is unity.

For the translation case we had

U†(a) x U(a) = x + a, (9.22)

and we find7 an analogous expression for the rotations, i.e.

7The proof for this goes through muchlike the version in Example 9.1 and isleft as an exercise.

U†(θ) p U(θ) = R(θ)p. (9.23)

Thus the momentum operator is transformed in just the same way asone would rotate a momentum vector.

Finally, we may find an explicit expression for the rotation operatorthat acts on wave functions. Let’s examine a rotation of a wave functionabout the z-axis and expand

ψ(θz + δθz) = ψ(θz) +dψ(θz)

dθzδθz + . . . (9.24)

Here we recall that Jz = −i ddθz , from which we see that rotations are

generated by the angular momentum operator. We obtain

ψ(θ + δθz) =(1 + iJzδθz

)ψ(θz), (9.25)

and finally, repeating N → ∞ times and changing the sign gives U(θz) =

e−iJzθz

, and for a rotation about an arbitrary axis we have

U(θ) = e−iJ·θ. (9.26)

5

5(a)

(b)

(c)

Fig. 9.2 (a) A spatial rotation sim-ply rotates the coordinates of the scalarfields but does not alter the value of thescalar. (b) Simply rotating the coordi-nates of a vector field is not a correcttransformation since (c) you have to ro-tate the vectors at each point in spaceas well.

9.3 Representations of transformations

We have seen how to transform spacetime coordinates by translationsand rotations. But spacetime is an empty coordinate system withoutfilling it with quantum fields. So we now have to ask how the quantumfields themselves transform. A quantum field φ(x) takes a position inspacetime and returns an operator whose eigenvalues can be a scalar,a vector (the W± and Z0 particles are described by vector fields), aspinor (the object that describes a spin-1

2 particle such as an electron),or a tensor. A spatial rotation simply rotates the coordinates for a scalarfield [Fig. 9.2(a)] but for a vector field simply rotating the coordinates isnot enough [Fig. 9.2(b)], you also have to rotate the vectors [Fig. 9.2(c)].

For this reason we need to think a bit more deeply about how trans-formations are represented, and for concreteness let us keep focussing


on rotations. Any rotation R(θ) can be represented by a square matrix

D(θ). This takes a very similar form to U(θ) = e−iJ·θ, namely

D(θ) = e−iJ·θ. (9.27)

In this equation, J is a square matrix, a representation8 of the operator8A representation of a group D isobtained by mapping each element giof the group G into a continuous lin-ear operator that acts on some vec-tor space. This mapping preserves thegroup product, so that if g1 • g2 = g3,then D(g1)D(g2) = D(g3).

J . Equation 9.27 can be rewritten as

J i = − 1

i

∂D(θi)

∂θi

∣∣∣∣θi=0

. (9.28)

Example 9.5

(a) Consider rotations about the z-axis. A trivial representation is D(θz) = 1. Thismeans that

Jz = −1

i

∂D(θz)

∂θz

˛˛˛˛θz=0

= 0. (9.29)

By extension Jx = Jy = 0. This representation is appropriate for a scalar field.9

9This makes perfect sense because ascalar can have no angular momentum.

(b) It turns out that the representation of a rotation about the z-axis for a spin- 12

particle is

D(θz) =

„e−iθz/2 0

0 eiθz/2

«

, (9.30)

and hence

Jz = −1

i

∂D(θz)

∂θz

˛˛˛˛θz=0

=1

2

„1 00 −1

«

, (9.31)

which we recognise as the operator for the z-component of angular momentum fora spin- 1

2particle. This representation of the rotation group will be useful for the

spinor fields introduced in Chapter 37.

The representation for a rotation ofa field with angular momentum quan-tum number j is frequently given thesymbolD(j)(θ). Thus in this example,part (a) describes D(0)(θ) = 1, part(b) describes D(1/2)(θ), and part (c)describes a representation related toD(1)(θ) which uses the Cartesian axesx, y and z as the basis of the spatialcomponents, rather than the three az-imuthal quantum numbers. The gen-eral formulae for the matrix elementsof D(j)(θ) are

[D(j)(θ)]m,m′ = 〈jm′|U(θ)|jm〉= 〈jm′|e−iJ ·θ |jm〉.

We can then use the standard relationsfor angular momentum operators:

Jz |jm〉 = m|jm〉,

J±|jm〉=p

(j∓m)(j+1±m)|j m±1〉,J± = Jx ± iJy .

(c) For rotations about the x-axis we have

R(θx) =

0

BB@

0 0 0 00 1 0 00 0 cos θx − sin θx

0 0 sin θx cos θx

1

CCA, (9.32)

so we obtain

Jx = −1

i

∂R(θx)

∂θx

˛˛˛˛θx=0

= i

0

BB@

0 0 0 00 0 0 00 0 0 −10 0 1 0

1

CCA. (9.33)

Repeating for the y- and z-axes yields

Jy = i

0

BB@

0 0 0 00 0 0 10 0 0 00 −1 0 0

1

CCA

Jz = i

0

BB@

0 0 0 00 0 −1 00 1 0 00 0 0 0

1

CCA. (9.34)

The important point about these representations is that they all sharethe same underlying algebraic structure as the rotation operator. Thisalgebra is called a Lie algebra, and can turn up whenever you have acontinuous group.10 The Lie algebra is encoded within the commutation

10This means that there are elements ofthe Lie group which are arbitrarily closeto the identity and so an infinitesimalgroup element can be written

g(α) = 1 + iαiT i +O(α2),

where T i are the generators of thegroup. The Lie algebra is expressed bythe commutator

[T i, T j ] = if ijkTk,

where f ijk are called structure con-stants. For rotations T i = Ji andf ijk = εijk.

relations of the generators. In other words, whether a generator of rota-tions rotates abstract quantum states in Hilbert space (like Jz = −i d

dθz ),real space vectors (as in the previous example) or complex spinors (as inChapter 37) each set of generators should have the same commutationrelations: [J i, Jj ] = iεijkJk.

9.4 Transformations of quantum fields 85

9.4 Transformations of quantum fields

We have seen that a transformation associated with a unitary operatorU will turn a locally-defined operator O (such as the position operator)into U†OU (see eqn 9.8). Now we need to determine how such transfor-mations affect quantum fields. We’ll start by examining a scalar field:an operator-valued field whose matrix elements are scalars.

Example 9.6

If we translate both a state and an operator by the same vector distance a thennothing should change. In equations

〈y|φ(x)|y〉 = 〈y + a|φ(x + a)|y + a〉. (9.35)

Since |y + a〉 = U(a)|y〉 we must have

U(a)φ(x)U†(a) = φ(x + a). (9.36)

Notice that this looks like U and U† have been bolted on the wrong way round whencompared to the expression

U†(a)xU(a) = (x + a), (9.37)

but these two expressions are talking about different things. The position operatorx tells us where a particle is localized, the field operator φ(x) acts on a localizedparticle at x. There’s no reason that they should transform in the same way.

〈 |φ(x)| 〉 = 1

〈 |φ(x)U(a)| 〉 = 0

〈 |U†(a)φ(x)U (a)| 〉 = 0at x

at x

at x

x− ax

x x + a

x

(c)

(b)

(a)

Fig. 9.3 The field operator φ(x) act-ing on a localized state is pictured as asniper’s bullet shooting a duck locatedat position x. (a) Dead on target, thestate of a dead duck has perfect overlapwith the state produced by acting φ(x)on the live duck at x. (b) Translatingthe state results in a miss, as does (c)translating the operator.

The previous example demonstrates that there are two ways of thinkingabout translations operators, and we shouldn’t confuse them. They canbe pictured:

• As acting on states: U(a)|x〉 = |x + a〉 moves a locally definedstate from being localized at x to being localized at x + a.

• As acting on locally defined operators:

U†(a)φ(x)U(a) = φ(x − a). (9.38)

These are not the same as we can see from Fig. 9.3. We can imaginea field operator φ(x) as a sniper’s rifle shooting a state at a position x

[Fig. 9.3(a)]. If we translate the state from |x〉 to |x+a〉, φ(x) misses thestate because we’ve moved the state [Fig. 9.3(b)]. If, on the other hand,we translate the operator from φ(x) to φ(x−a), we miss again, but thistime because we’ve changed the position we’re aiming at [Fig. 9.3(c)].Since we’re interested in the fields, changing the operator will be themost useful procedure and so the equation to focus on is eqn 9.38.

Example 9.7

The creation and annihilation operators are also actually quantum field operators,albeit in momentum space. That is, the operator a†m takes a momentum m andoutputs an operator that creates a particle at that position in momentum space.


We’ll examine what we get if we transform according to U†(a)a†m U(a), which isa translation in spacetime by a vector a. We can try out this translated operator ona momentum eigenstate |q〉 (assumed different to m):

U†(a)a†m U(a)|q〉 = eip·a a†me−ip·a |q〉= eip·a a†m |q〉e−iq·a

= eip·a |m,q〉e−iq·a

= |m,q〉ei(m+q)·ae−iq·a

= |m,q〉eim·a . (9.39)

So the transformed operator a†m still creates a state with momentum m, but theresult of the transformation is an additional phase eim·a . We conclude that

U†(a)a†m U(a) = eim·a a†m . (9.40)

This discussion has been for scalar fields, but we must remember thatfields come in many forms, including vector fields as discussed in thefollowing example.

dR(θ)d

d R(θ)d

V maxR(θ)V max

(a)

(c)

(b)

(d)

Fig. 9.4 (a) A classical scalar field hasa maximum at some point in space re-moved from the origin by a vector d.(b) Rotating the distribution with anoperator R(θ) moves the maximum toa position R(θ)d. (c) A classical vec-tor field also has a maximum value at d.(d) The transformed field has a maxi-mum at R(θ)d and this vector is ro-tated to R(θ)V max.

Example 9.8

A classical scalar field φ(x) (like the distribution of temperature across a metal block)has a large maximum φ(d) = φmax at some point in space removed from the originby a vector d. Rotating the distribution with an operator R(θ) moves the maximumto a position R(θ)d. This is embodied in the operator equation U†(θ)φ(x)U(θ) =φ(R−1(θ)x), which demonstrates that we need to enter the position x = R(θ)d inthe transformed field to find the maximum [see Fig. 9.4(a) and (b)].

For a classical vector field V , such as the velocity distribution in a liquid, let’simagine that we have a maximum velocity V (d) = V max. Being a vector, this hasa direction. A rotation R(θ) now moves the position of the maximum and alsoits direction. The analogous mathematical description is now U†(θ)V (x)U(θ) =R(θ)V (R−1(θ)x), demonstrating that the transformed field has a maximum atR(θ)d and that this vector is rotated to R(θ)V max [see Fig. 9.4(c) and (d)].

It is therefore important to note that if we transform a field in someway, we generally change the point at which the field is evaluated andalso its polarization. We can write a more general rotation of a generalfield operator Φ(x) as

U†(θ)Φ(x)U(θ) = D(θ)Φ(R−1(θ)x), (9.41)

where D(θ) is the appropriate representation of the rotation operator,as discussed in Section 9.3 (e.g. for scalar fields D(θ) = 1).

9.5 Lorentz transformations

A similar philosophy carries over to the case of Lorentz transformationsof states and quantum fields.11 Consider a boost of a four-vector in the

11As we are now explicitly consideringspacetime, we will now switch from la-belling our spatial axes x, y and z andgo back to using 1, 2 and 3, with 0 forthe time axis.

9.5 Lorentz transformations 87

x-direction whose Lorentz transformation is given by x′µ = Λ(β1)µνxν ,

where

Λ(β1) =

γ1 β1γ1 0 0β1γ1 γ1 0 0

0 0 1 00 0 0 1

. (9.42)

This transformation connects two inertial frames moving with relativespeed v = cβ1 along x. Using the substitutions γi = coshφi, γiβi =sinhφi and tanhφi = βi, where φi is called the rapidity, this matrixbecomes

Λ(φ1) =

coshφ1 sinhφ1 0 0sinhφ1 coshφ1 0 0

0 0 1 00 0 0 1

. (9.43)

We write an operator that acts in Hilbert space:12 U(φ)|p〉 = |Λ(φ)p〉. 12We will not pursue this operator fur-ther here. Unlike linear and angularmomentum the boost three-vector isnot conserved, so we do not use itseigenvalues to label states.

We also want a generalized matrix form like we had with the rotatione−iJ·θ. From our experience with rotations we suppose that we can writea generalized Lorentz transformation matrix as13

13Note the plus sign in the exponentcompared to the rotation case.

D(φ) = eiK·φ, (9.44)

and the generators of the Lorentz transformations are given by

Ki =1

i

∂D(φi)

∂φi

∣∣∣∣φi=0

. (9.45)

As shown in the exercises, we can find the explicit forms for these gen-erators for the case of vectors by taking D(φi) to be the Λ(φi) matricessuch as that in eqn 9.43 for a boost along x. We therefore require ourquantum fields to Lorentz transform according to

U†(φ)Φ(x)U(φ) = D(φ)Φ(Λ−1(φ)x). (9.46)

Finally, we recall that the fundamentally important feature of thegenerators of a transformation is their commutation relations. Uponworking out a set of generators for the Lorentz transformations, as you’reinvited to in the exercises, it comes as a shock to find

[K1,K2] = K1K2 −K2K1 = −iJ3. (9.47)

In other words, the difference between (i) boosting along x and thenboosting along y and (ii) boosting along y and then boosting along x isa rotation about z. Mathematically speaking, the fact that [K1,K2] =−iJ3 implies that the Lie algebra of the Lorentz transformations isn’tclosed and their generators do not form a group. Let’s see what happensif we examine some other commutators. Pressing on, we find

[J1,K1

]= 0, (9.48)[

J1,K2]

= iK3, (9.49)


along with cyclic permutations, that is to say[J i,Kj

]= iεijkKk. This

implies that the boosts and rotations taken together form a closed Liealgebra and it is this larger group that is called the Lorentz group.We may write a general Lorentz transformation combining both boostsand rotations as

D(θ,φ) = e−i(J·θ−K·φ), (9.50)

where, as always, the generators J and K are those appropriate for theobject being transformed. If you also include the spacetime translationsin the mix, you end up with an even larger group called the Poincare

group.

Chapter summary

• A transformation is associated with a unitary operator U whichacts on states via |x′〉 = U |x〉.

• A general spacetime translation has U(a) = eip·a.

• A rotation has U(θ) = e−iJ·θ.

• A Lorentz transformation has U(φ) = eiK·φ.

• Transformations can be represented by a matrix D(a,θ,φ) appro-priate to the type of field. For scalar fields D = 1.

• The quantum fields that we seek will have to have the followingtransformation properties:

U†(a)Φ(x)U(a) = D(a) Φ(x− a),

U†(θ)Φ(x)U(θ) = D(θ) Φ(R−1(θ)x),

U†(φ)Φ(x)U(φ) = D(φ) Φ(Λ−1(φ)x),

i.e. they obey the translation, rotation and Lorentz transformationproperties of a local operator.

Exercises

(9.1) Deduce that the generators of the translation oper-ator are given by

p = − 1

i

∂U(a)

∂a

˛

˛

˛

˛

˛

a=0

. (9.51)

(9.2) Show that explicit forms of the generators of theLorentz group for four-vectors are

K1 =1

i

∂U(φ1)

∂φ1

˛

˛

˛

˛

˛

φ1=0

= −i

0

B

B

@

0 1 0 01 0 0 00 0 0 00 0 0 0

1

C

C

A

,

(9.52)

Exercises 89

and similarly

K2 = −i

0

B

B

@

0 0 1 00 0 0 01 0 0 00 0 0 0

1

C

C

A

, (9.53)

and

K3 = −i

0

B

B

@

0 0 0 10 0 0 00 0 0 01 0 0 0

1

C

C

A

. (9.54)

(9.3) Show that an infinitesimal boost by vj along thexj-axis is given by the Lorentz transformation

Λµν =

0

B

B

@

1 v1 v2 v3

v1 1 0 0v2 0 1 0v3 0 0 1

1

C

C

A

. (9.55)

Show also that an infinitesimal rotation by θj aboutxj is given by

Λµν =

0

B

B

@

1 0 0 00 1 θ3 −θ20 −θ3 1 θ1

0 θ2 −θ1 1

1

C

C

A

. (9.56)

Hence show that a general infinitesimal Lorentztransformation can be written x′µ = Λµνx

ν whereΛ = 1 + ω where

ωµν =

0

B

B

@

0 v1 v2 v3

v1 0 θ3 −θ2v2 −θ3 0 θ1

v3 θ2 −θ1 0

1

C

C

A

. (9.57)

Show that ωµν = ωµλgλν and ωµν = gµλω

λν are

antisymmetric. An antisymmetric 4×4 matrix hassix independent parameters. This makes sense aswe are encoding three rotations and three boosts.

Show that the relationships between θi, vi and ωij

are

θi = −1

2εijkωjk and vi = ω0i. (9.58)

(9.4) A transformation of the Poincare group combinestranslations (by vector aµ), rotations and Lorentzboosts and can be written x′µ = aµ + Λµνx

ν . Aninfinitesimal transformation of the Poincare groupis thus x′µ = xµ + aµ + ωµνx

ν , where ωµν is givenin the previous problem. Therefore a function f(x)transforms to

f(x′) = f(x+ a+ ωx) (9.59)

= f(x) + aµ∂µf(x) + ωµνxν∂µf(x).

Use the antisymmetry of ωµν to write

f(x′) = [1 + aµ∂µ − 1

2ωµν(x

µ∂ν − xν∂ν)]f(x),

(9.60)and hence

f(x′) = [1 − iaµpµ +i

2ωµνM

µν ]f(x), (9.61)

where pµ = −i∂µ and Mµν = −i(xµ∂ν − xν∂µ) aregenerators of the Poincare group. Note that Mµν

is antisymmetric and is related to the generators J

and K by

J i =1

2εijkM jk, (9.62)

so that J = (M23,M31,M12), and

Ki = M0i, (9.63)

so that K = (M01,M02,M03). Also show that

Λ = exp(− i

2ωµνM

µν), (9.64)

with Λ = e−i(J·θ−K·φ).

10 Symmetry

10.1 Invariance and conservation90

10.2 Noether’s theorem 92

10.3 Spacetime translation 94

10.4 Other symmetries 96

Chapter summary 97

Exercises 97

The Universe is built on a plan the profound symmetry ofwhich is somehow present in the inner structure of our intel-lectPaul Valery (1871–1945)

In the previous chapter we saw that translations were achieved withan operator that contains the momentum operator, and rotations wereachieved using the angular momentum operator. A coincidence? Actu-ally not, and in fact we can go further. If a system possesses some kindof invariance (so that for example it is invariant to translations) thena particular quantity will be conserved. This idea is bound up with aresult called Noether’s theorem, named after the great mathematicianEmmy Noether.1 This chapter will explore this theorem in detail.1Emmy Noether (1832–1935). Her

name should be pronounced to rhymewith Goethe, Alberta and NorrisMcWhirter.

A point to note from the outset. In this chapter we will be using theword symmetry to mean global symmetry, i.e. a symmetry possessedby the entire system under consideration. Thus rotational symmetryrefers to a symmetry associated with rotation of the coordinates applyingto every point in our system.22Consideration of local symmetries,

where a transformation changes frompoint to point, will be postponed to alater chapter. 10.1 Invariance and conservation

Let’s get the terminology straight. A quantity is invariant when it takesthe same value when subjected to some particular transformation, and isthen said to exhibit a particular symmetry. A translationally invariantsystem looks the same however you translate it. A rotationally invariantsystem appears identical however it is rotated around a particular axis.Electric charge is said to be Lorentz invariant because it takes the samevalue when viewed in different inertial frames of reference. A system issaid to possess a particular symmetry if a property is invariant under aparticular transformation, so for example a cylinder possesses rotationalsymmetry about its axis. A quantity is said to be conserved whenit takes the same value before and after a particular event. Thus ina collision between particles, the four-momentum is conserved within aframe of reference but it is not invariant between frames of reference.

Having stressed that invariance and conservation are entirely differentconcepts, it is now time to point out that they are in fact connected!For example:

• Conservation of linear momentum is related to invariance underspatial translations.

10.1 Invariance and conservation 91

• Conservation of angular momentum is related to invariance underrotations.

• Conservation of energy is related to invariance under time trans-lations.

The general idea is clear: invariances lead to conservation laws.

Example 10.1

(i) The Euler–Lagrange equation (eqn 1.28) can be written as

∂L

∂xµ=

d

dt

„∂L

∂xµ

«

, (10.1)

and using the canonical momentum pµ = ∂L/∂xµ (eqn 5.3), we can writeeqn 10.1 as

∂L

∂xµ= pµ. (10.2)

This has the immediate consequence that if L does not depend on xµ, then∂L/∂xµ = 0 and hence pµ is a constant. In other words if the Lagrangianis invariant under a particular component of spacetime translation, the corre-sponding component of four-momentum is a conserved quantity.

(ii) Another example of this follows from eqn 5.58 of Exercise 5.1 which showedthat

∂L

∂t= −dH

dt. (10.3)

Thus if L does not depend explicitly on time, then the Hamiltonian H (i.e.the energy) is a conserved quantity.

Under a symmetry transformation, various quantities will change.3 One 3In this chapter, we are going to as-sume that the transformations are con-tinuous. Thus we are ignoring discretesymmetries which will be considered inChapter 15.

efficient way to describe this is to say that a field φ(xµ) will change undera symmetry transformation by an amount parameterized by a quantityλ. For example, for a translation along a spacetime vector aµ we couldwrite the transformation

φ(xµ) → φ(xµ + λaµ). (10.4)

The larger λ, the larger the degree the transformation is applied. It willbe useful to consider infinitesimal transformations, and for this purposewe adopt the notation

Dφ =∂φ

∂λ

∣∣∣∣λ=0

. (10.5)

An infinitesimal change in φ, induced by an infinitesimal δλ, can bewritten as

δφ = Dφδλ. (10.6)

Example 10.2

In the example of our spacetime translation φ(xµ) → φ(xµ + λaµ), we have

Dφ = aµ∂µφ. (10.7)

92 Symmetry

This result arises as follows: by writing yµ = xµ + λaµ, we have

∂φ

∂λ=

∂φ

∂yµ∂yµ

∂λ, (10.8)

and ∂yµ

∂λ= aµ. Taking the limit of λ→ 0 gives the required result.

10.2 Noether’s theorem

We now turn to Noether’s theorem, which shows that where we have acontinuous symmetry we also have a conservation law. To identify thisconservation law we will look for a divergenceless current (i.e. obeying∂µJ

µ = 0). Such a current is locally conserved (see eqn. 5.35 if in doubt)and its time-like component will give rise to a conserved charge.

Let the field φ(x) change to φ(x) + δφ(x), where the variationδφ(x) vanishes on the boundary of the region in spacetime that we’reconsidering.4 The change in the Lagrangian density L is then given by4This is a continuous transformation.

Noether’s theorem applies to these con-tinuous transformations but not to thediscrete transformations examined inChapter 15.

δL =∂L∂φ

δφ+∂L

∂(∂µφ)δ(∂µφ). (10.9)

We will slightly simplify this equation using the substitution

Πµ(x) =∂L

∂(∂µφ). (10.10)

This new object Πµ(x) is a generalization of what, in eqn 5.46, we calledthe conjugate momentum π(x) = δL/δφ. Our new object Πµ(x), themomentum density, is the four-vector generalization of π(x). In factπ(x) is the time-like component of Πµ(x), i.e.

Π0(x) = π(x). (10.11)

This substitution allows us to write eqn 10.9 as

δL =∂L∂φ

δφ+ Πµδ(∂µφ). (10.12)

Using δ(∂µφ) = ∂µ(δφ), and the simple differentiation of a product

∂µ(Πµδφ) = Πµ∂µ(δφ) + (∂µΠ

µ)δφ, (10.13)

we have

δL =

(∂L∂φ

− ∂µΠµ

)δφ+ ∂µ(Π

µδφ). (10.14)

The action S should not change under the transformation and so

δS =

∫d4x

(∂L∂φ

− ∂µΠµ

)δφ = 0, (10.15)

10.2 Noether’s theorem 93

where the second term in eqn 10.14 vanishes when integrated over alarge enough volume (using the divergence theorem), and so

∂L∂φ

= ∂µΠµ, (10.16)

which is the Euler–Lagrange equation. Now let’s play this argument theother way round and state that φ obeys the equation of motion. In thiscase eqn 10.14 becomes

δL = ∂µ(Πµδφ), (10.17)

and eqn 10.6 then allows us to write

δL = ∂µ(ΠµDφ)δλ. (10.18)

Because we are applying a symmetry transformation, the action S =∫d4xL is unchanged as we have said, and so we don’t expect L to

change, i.e. δL = 0. However, this condition is too restrictive and so insome circumstances we could be persuaded to allow L to change by thefour-divergence of some function Wµ(x), so that

A

nµ

Fig. 10.1 A region of spacetime withsurface A. The outward normal is nµ.

δL = (∂µWµ)δλ. (10.19)

This is because

δS =

∫d4x δL = δλ

∫d4x ∂µW

µ = δλ

∫dAnµW

µ, (10.20)

where A is the surface of the region of spacetime and nµ is the outwardnormal (see Fig. 10.1). Since the fields are held constant on the surface inorder to have a well-defined variational principle, the change in action isat most a constant, and so S′ =

∫d4x (δL+∂µW

µδλ) will be stationaryunder the same conditions that S =

∫d4x δL will be stationary. Putting

eqns 10.18 and 10.19 together leads to

∂µ(ΠµDφ−Wµ) = 0, (10.21)

or more simply ∂µJµN = 0 where

JµN(x) = Πµ(x)Dφ(x) −Wµ(x), (10.22)

is a locally conserved Noether current. We have deduced a form ofNoether’s theorem that states the following:

If a continuous symmetry transformation φ→ φ+Dφ only changes Lby the addition of a four-divergence (i.e. DL = ∂µW

µ) for arbitraryφ, then this implies the existence of a current JµN = ΠµDφ−Wµ(x).If φ obeys the equations of motion then the current is conserved, i.e.∂µJ

µN = 0.

94 Symmetry

Conserved currents are important because they give rise to conserved

charges QN =∫JµN dAµ (also called Noether charges). A helpful

way to apply this idea is to perform the integration over a surface atconstant time (see Fig. 10.2). In four dimensions, this ‘surface’ is athree-volume. Thus

QN =

∫d3xJ0

N, (10.23)

where J0N is the time-like component normal to the surface. Now since

time

J0N

J0N

Fig. 10.2 The integration is performedover a surface in spacetime at constanttime. The component of the Noethercurrent which is perpendicular to thissurface is J0

N, the time-like part.

∂µJµN = 0, then

0 =

∫d3x (∂µJ

µN) =

∫d3x (∂0J

0N + ∂kJ

kN). (10.24)

The second term in the integral is∫

d3x ∂kJkN =

∫dAk J

kN by the diver-

gence theorem, and this will vanish if the volume is big enough. Thus∫d3x ∂0J

0N = 0 and since ∂0 = d/dt, this implies dQN/dt = 0, i.e. a

conserved charge.

Using Noether’s theorem to findconserved charges

I: Find Dφ = ∂φ∂λ

˛˛˛λ→0

.

II: Find Πµ(x) = ∂L∂(∂µφ)

.

III: Find Wµ = DL.

IV: Write JµN = DφΠµ −Wµ.

V: Find QN =R

d3xJ0N.

Everything we have done so far is valid for classical field theories.We might wonder if this discussion can be carried over to the quantummechanical fields that are the main subject of this book. Fortunatelyfor us Noether’s theorem can be carried over to the quantum realm5

5This assumes that all symmetries ofthe classical theory will turn out to besymmetries of the quantum theory too.This is true for many important sym-metries. It turns out not to be the casefor a small number of cases, which areknown as anomalies and are discussedin more advanced books such as Zee orPeskin and Schroeder.

with one addition: the so-called ‘normal ordering interpretation’ whichwe introduce in the next chapter. Noether’s theorem then provides uswith a conserved charge operator QN, whose eigenvalues are (observable)conserved charges. In fact, there is also a quantum mechanical corollaryto Noether’s theorem which says that if we know the conserved chargeoperator QN for a quantum mechanical field φ then we can use it togenerate the symmetry transformation Dφ via6

6This result is easily seen by recogniz-ing that conserved charge may be writ-ten QN =

Rd3xΠ0Dφ and recognizing

that Π0 is the conjugate momentum toφ.

[QN, φ

]= −iDφ. (10.25)

This is illustrated in Fig. 10.3. For now we return to classical field theoryand continue to investigate the benefits we gain from Noether’s theorem.

10.3 Spacetime translation

The field theories that we will examine in this book will be sym-metric with respect to spacetime translations. That is to say thatthe physical consequences of the theories will be the same whetheran experiment takes place at a point xµ = (1066AD,Hastings) oryµ = (1863AD,Gettysburg). Here we examine the consequences ofthis using Noether’s theorem.

If we subject the coordinates to a variation, so that x′µ = xµ + δxµ,then we have that Dφ = aµ∂µφ and also that

DL = aµ∂µL = ∂µ(aµL). (10.26)

10.3 Spacetime translation 95

We recognize therefore that DL = ∂µWµ where Wµ = aµL and so we

can write down the conserved current

JµN = ΠµDφ−Wµ

= Πµaν∂νφ− aµL= aν [Πµ∂νφ− δµνL]

= aνTµν , (10.27)

where Tµν = Πµ∂νφ− gµνL is a quantity that is known as the energy-

momentum tensor. The conserved charges that arise from this currentcan be written

Pα =

∫d3xT 0α. (10.28)

Thus the time-like component of this conserved charge is

symmetry generator

conserved charge

commutatorNoethercurrent

Dφ

Q

Fig. 10.3 Noether’s theorem for thefield φ.P 0 =

∫d3xT 00 =

∫d3x [π(x)φ(x) − L(x)] =

∫d3xH, (10.29)

which we recognize as the energy. The space-like components give us

P k =

∫d3xT 0k =

∫d3xπ(x)∂kφ(x), (10.30)

which we recognize as the momentum. (Note that g00 = 1 and g0k = 0.)

Example 10.3

The energy-momentum tensor Tµν is not uniquely defined. If one adds a term∂λX

λµν to Tµν , where Xλµν = −Xµλν , then show that the new tensor is stilldivergenceless.Solution: ∂µ(Tµν+∂λX

λµν) = ∂µTµν+∂µ∂λXλµν = 0+0 = 0. This works because

(i) ∂µTµν = 0 because Tµν is a conserved current and (ii) ∂µ∂λXλµν = 0 follows

from the antisymmetry of Xλµν with respect to swapping the first two indices andthe fact that you are summing over both indices µ and λ.

The energy-momentum tensor Tµν , as we have constructed it, is notsymmetric, and that turns out to cause problems when it is used ingeneral relativity,7 but we can use the trick in the previous example 7Einstein’s field equations are

Rµν − 1

2gµνR = −8πG

c4Tµν ,

where Rµν is the Ricci tensor, express-ing the curvature of spacetime, and Ris the Ricci scalar. For further details,see any good book on general relativity,e.g. Penrose (2004).

to redefine Tµν → Tµν + ∂λXλµν and force it to be symmetric. This

symmetric nature of Tµν is also important when we come to look atLorentz transformations.

Example 10.4

We will extract some information from Tµν . For an infinitesimal Lorentztransformation8 we can write 8See Exercise 9.3.

δxµ = ωµνxν δλ, (10.31)

where ωµν is an antisymmetric tensor. Hence the change in field φ is given by

δφ = δxµ∂µφ = ωµνxν∂µφδλ, (10.32)

96 Symmetry

and soDφ = ωµνxν∂µφ = ωµνx

ν∂µφ (10.33)

so that the effect on the Lagrangian density L is99The second equality in eqn 10.34 fol-lows from the following argument. Dif-ferentiating the product xσL gives

∂ρ(xσL) = xσ∂ρL + L∂ρxσ ,

and because ∂ρxσ = δρσ , the antisym-metry of ωρσ implies that

ωρσ∂ρxσ = ωρσδ

ρσ = 0.

Hence

ωρσxσ∂ρL = ωρσ∂

ρ(xσL)

= ∂ρ(ωρσxσL).

DL = ωρσxσ∂ρL = ∂ρωρσx

σL = ∂µ[gµρωρσxσL] = ∂µW

µ, (10.34)

where Wµ = gµρωρσxσL. Thus we can go straight to our conserved current

JµN = ΠµDφ−Wµ

= ωρσ [Πµxσ∂ρφ− gµρxσL]

= ωρσxσT ρµ

= ωρσ(Jµ)ρσ , (10.35)

where (Jµ)ρσ = xσT ρµ. There are six parameters in ωρσ (remember that it isantisymmetric) and so there are six conserved currents. We will write

(Jµ)ρσ = xρTµσ − xσTµρ, (10.36)

where the currents have been antisymmetrized10 compared with eqn 10.35. These10We have turned (Jµ)ρσ into (Jµ)ρσ

using (Jµ)ρσ = (Jµ)ρσ − (Jµ)σρ. are conserved currents and so ∂µ(Jµ)ρσ = 0, although this only works because Tµν

is symmetric. The conserved charges11 are11These are Noether conserved charges,but we drop subscript N’s here to avoidclutter.

Qλρ =

Z

d3x (Jλ)ρ0, (10.37)

from which we can extract the angular momentum components

Qij =

Z

d3x (xiT j0 − xjT i0) (10.38)

and the boosts

Q0i =

Z

d3x (x0T i0 − xiT 00). (10.39)

Note that since dQ0i/dt = 0 and x0 = t, then

0 =

Z

d3xT i0+t

Z

d3x T i0− d

dt

Z

d3xxiT 00 = P i+tP i− d

dt

Z

d3xxiT 00, (10.40)

and hence ddt

Rd3xxiT 00 is a constant, showing that the centre of mass of the system

moves in uniform motion.

10.4 Other symmetries

We have seen how translations, rotations and boosts can transformspacetime coordinates and also Lagrangians. However, the only effecton the fields has been due to the fact that the fields are functions of thespacetime coordinates. A field, you will remember, is a machine thatwhen fed a coordinate xµ returns a number φ(xµ). What we have beenconsidering is a scalar field, but of course when the field is more compli-cated (for example, a vector or a spinor) then we have to worry abouthow the symmetry transformation affects the field itself, in addition tothe effect on the spacetime coordinates of which the field is a function.That will lead to additional layers of complexity which we will save forlater. One of the most exciting consequences of this is that if we havea complex scalar field we will be able to show that the U(1) symmetrywe noticed in Section 7.6 leads to the conservation of particle number.We will save that gem for Chapter 12, but before that, in the followingchapter, we will outline a procedure for quantizing our field theories.

Exercises 97

Chapter summary

• Noether’s theorem states that if a continuous symmetry transfor-mation φ → φ + Dφ only changes L by the addition of a four-divergence (i.e. DL = ∂µW

µ) for arbitrary φ, then this implies theexistence of a current JµN = ΠµDφ − Wµ(x) and if φ obeys theequations of motion then the current is conserved, i.e. ∂µJ

µN = 0.

• In short, Noether’s theorem states that continuous symmetries leadto conserved currents.

Exercises

(10.1) Show that [φ(x), Pα] = i∂αφ(x), where Pα isthe conserved charges from spacetime translation(eqn 10.28).

(10.2) Consider a system characterized by N fieldsφ1, . . . , φN . The Lagrangian density is thenL(φ1, . . . , φN ; ∂µφ1, . . . , ∂µφN ;xµ). Show that theNoether current is

Jµ =X

a

ΠµaDφ

a −Wµ(x), (10.41)

where DL = ∂µWµ.

(10.3) For the Lagrangian

L =1

2(∂µφ)2 − 1

2m2φ2, (10.42)

evaluate Tµν and show that T 00 agrees with whatyou would expect from the Hamiltonian for this La-grangian. Show that ∂µT

µν = 0. Derive expres-sions for P 0 =

R

d3xT 00 and P i =R

d3xT 0i.

(10.4) For the Lagrangian

L = −1

4FµνF

µν =1

2(E2 − B

2), (10.43)

show that

Πσρ ≡ ∂L∂(∂σAρ)

= −Fσρ. (10.44)

Hence show that the energy-momentum tensor

Tµν = Πµσ∂νAσ − δµνL, (10.45)

can be written as

Tµν = −Fµσ∂νAσ +1

4gµνFαβFαβ . (10.46)

This tensor is not symmetric but we can symmetrizeit by adding ∂λX

λµν where Xλµν = FµλAν . Showthat Xλµν = −Xµλν . Show further that the sym-metrized energy-momentum tensor Tµν = Tµν +∂λX

λµν can be written

Tµν = FµσFσν +

1

4gµνFαβFαβ . (10.47)

Hence show that T 00 = 12(E2 + B2), the energy

density in the electromagnetic field, and T i0 =(E × B)i, which is the Poynting vector and de-scribes the energy flow.

11Canonical quantization of

fields

11.1 The canonical quantizationmachine 98

11.2 Normalizing factors 101

11.3 What becomes of the Hamil-tonian? 102

11.4 Normal ordering 104

11.5 The meaning of the mode ex-pansion 106

Chapter summary 108

Exercises 108

You’ve got to have a systemHarry Hill (1964– )

Quantum field theory allows us to consider a Universe in which there ex-ist different, yet indistinguishable, copies of elementary particles. Theseparticles can be created and destroyed by interactions amongst them-selves or with external entities. Quantum field theory allows us to de-scribe such phenomena because particles themselves are simply excita-tions of quantum fields.

To see how particles emerge from fields we need to develop a wayto quantize the classical fields we have looked at previously. The goodnews is that there is a machine available that takes a classical field the-ory and, after we turn the handle, it spits out a quantum field theory:that is, a theory where quantities are described in terms of the numberof quantum particles in the system. The name of the machine is canon-

ical quantization. In developing canonical quantization we’ll see thatparticles are added or removed from a system using field operators11Field operators were introduced in

Chapter 4. and these are formed from the creation and annihilation operators wefound so useful in previous chapters.

11.1 The canonical quantization machine

Canonical quantization is the turn-the-handle method of obtaining aquantum field theory from a classical field theory. The method runs likethis:

• Step I: Write down a classical Lagrangian density in terms offields. This is the creative part because there are lots of possibleLagrangians. After this step, everything else is automatic.

• Step II: Calculate the momentum density and work out theHamiltonian density in terms of fields.

• Step III: Now treat the fields and momentum density as opera-tors. Impose commutation relations on them to make them quan-tum mechanical.

• Step IV: Expand the fields in terms of creation/annihilation op-erators. This will allow us to use occupation numbers and staysane.

11.1 The canonical quantization machine 99

• Step V: That’s it. Congratulations, you are now the proud ownerof a working quantum field theory, provided you remember thenormal ordering interpretation.

We’ll illustrate the method with one of the simplest field theories: thetheory of the massive scalar field.

Step I: We write down a Lagrangian density for our theory. Formassive scalar field theory this was given in eqn 7.7, which we rewriteas

L =1

2[∂µφ(x)]

2 − 1

2m2 [φ(x)]

2. (11.1)

[Recall that the equation of motion for this theory is the Klein–Gordonequation (∂2 +m2)φ = 0 leading to a dispersion E2

p = p2 +m2.]Step II: Find the momentum density (eqn 10.10) Πµ(x) given by

Πµ(x) =∂L

∂(∂µφ(x)). (11.2)

For our Lagrangian in eqn 11.1 this gives Πµ(x) = ∂µφ(x). The time-likecomponent2 of this tensor is Π0(x) = π(x) = ∂0φ(x). This allows us to 2The metric (+ − −−), which allows

us to say A0 = A0, lets us swap upand down indices for the zeroth com-ponent of any object without incurringthe penalty of a minus sign.

define the Hamiltonian density in terms of the momentum density

H = Π0(x)∂0φ(x) − L, (11.3)

and using our Lagrangian3 leads to a Hamiltonian density 3which may be helpfully rewritten as

L =1

2

(„∂φ

∂t

«2

− (∇φ)2 −m2φ2

)

.

(11.4)

H = ∂0φ(x)∂0φ(x) − L=

1

2[∂0φ(x)]

2+

1

2[∇φ(x)]

2+

1

2m2 [φ(x)]

2. (11.5)

The last line in eqn 11.5 tells us that the energy has contributions from(i) a kinetic energy term reflecting changes in the configuration in time,(ii) a ‘shear term’ giving an energy cost for spatial changes in the fieldand (iii) a ‘mass’ term reflecting the potential energy cost of there beinga field in space at all. Taken together this has a reassuring look ofE = (kinetic energy)+(potential energy), which is what we expect froma Hamiltonian.

Step III: We turn fields into field operators. That is to say, we makethem operator-valued fields: one may insert a point in spacetime intosuch an object and obtain an operator. We therefore take φ(x) → φ(x)and Π0(x) → Π0(x). To make these field operators quantum mechanicalwe need to impose commutation relations between them. In single-particle quantum mechanics we have [x, p] = i~. By analogy, we quantizethe field theory by defining the equal-time commutator for the fieldoperators

[φ(t,x), Π0(t,y)] = iδ(3)(x − y). (11.6)

As the name suggests, this applies at equal times only and otherwisethe fields commute. We also have that [φ(x), φ(y)] = [Π0(x), Π0(y)] = 0(and likewise for the daggered versions). Expressed in terms of thesefields, the Hamiltonian density H is now an operator H which acts on

100 Canonical quantization of fields

state vectors. This is all well and good, except that we don’t know howoperators like φ(x) act on occupation number states like |n1n2n3 . . .〉.What we do know is how creation and annihilation operators act onthese vectors. If only we could build fields out of these operators!

Step IV: How do we get any further? The machinery of creation andannihilation operators is so attractive that we’d like to define everythingin terms of them. In particular, we found a neat analogy between par-ticles in momentum eigenstates and quanta in oscillators. What we do,therefore, is to expand the field operators in terms of the creation andannihilation operators. We’ve already seen this in the coupled oscilla-tor problem (eqn 2.68) where we obtained a time-independent positionoperator of the form

xj =

(~

m

) 12 ∑

k

1

(2ωkN)12

(akeijka + a†ke

−ijka). (11.7)

By analogy we write down a time-independent field operator for thecontinuous case which will look likeWe are rewriting our momenta k → p

and our energies ωk → Ep , so thefactor(2ωk)1/2 → (2Ep)1/2. The nor-malization will be discussed below.

φ(x) =

∫d3p

(2π)32

1

(2Ep)12

(apeip·x + a†pe−ip·x) , (11.8)

with Ep = +(p2+m2)12 (that is, we consider positive roots only, a feature

discussed below) and where, as before, our creation and annihilationoperators have a commutation relation [ap, a

†q] = δ(3)(p − q).

The field operators that we’ve written down are intended to work inthe Heisenberg picture. To obtain their time dependence, we hit themwith time-evolution operators. We use the Heisenberg prescription formaking an operator time dependent:

φ(x) = φ(t,x) = U†(t, 0)φ(x)U(t, 0) = eiHtφ(x)e−iHt. (11.9)

The only part that the U(t, 0) = e−iHt operators affect are the creationand annihilation operators, and we have

U†(t, 0)apU(t, 0) = e−iEptap, (11.10)

which shows that the ap picks up a e−iEpt. Similarly, the a†p part willpick up a eiEpt.

Example 11.1

To see this unfolding we’ll consider a simple example of one component of the fieldacting on an example state. The component we’ll consider is aqeiq·x and the examplestate is |npnqnr 〉. Taking it one step at a time:

eiHtaqeiq·xe−iHt|npnqnr 〉= eiHtaq |npnqnr 〉eiq·xe−i(npEp+nqEq+nrEr)t

=√nqeiHt|np(nq − 1)nr 〉eiq·xe−i(npEp+nqEq+nrEr)t

=√nq |np(nq − 1)nr 〉ei(npEp+(nq−1)Eq+nrEr)teiq·xe−i(npEp+nqEq+nrEr)t

=√nq |np(nq − 1)nr 〉e−iEqteiq·x . (11.11)

11.2 Normalizing factors 101

Part of what we’re left with, namely√nq |np(nq − 1)nr 〉, is exactly the same as if

we’d just acted on the original state with aq . The effect of dynamicizing this operatorhas just been to multiply by a factor e−iEqt, so we conclude that the operator weseek is aqe−i(Eqt−q·x) = aqe−iq·x.

In summary, what we call the mode expansion of the scalar field isgiven by4 4This looks a little different from what

we had in Chapter 4 when we intro-duced field operators. We will explainthe reason for the difference in Sec-tion 11.5.φ(x) =

∫d3p

(2π)32

1

(2Ep)12

(ape−ip·x + a†peip·x) , (11.12)

with Ep = +(p2 +m2)12 .

Note that by expanding out the position field we’ll get the momentumexpansion for free, since for our scalar field example Πµ(x) = ∂µφ(x).Also note that because the field in our classical Lagrangian is a realquantity, the field operator φ(x) should be Hermitian. By inspectionφ†(x) = φ(x), so this is indeed the case.

11.2 Normalizing factorsThis section can be skipped if you arehappy to take eqn 11.12 on trust. Thepurpose here is simply to justify the fac-tors (2π)3/2(2Ep)1/2 which otherwiseseem to appear by magic.

Before proceeding, we will justify the normalization factors in eqn 11.12.In evaluating integrals over momentum states we have the problem thatd3p is not a Lorentz-invariant quantity. We can use d4p where p = (p0,p)is the four-momentum, but for a particle of mass m then only values ofthe four-momentum which satisfy5 p2 = m2 need to be considered. This 5The condition p2 = m2 means that

(p0)2 − p2 = E2p − p2 = m2.is known as the mass shell condition (see Fig. 11.1). Consequently

we can write our integration measure

d4p δ(p2 −m2) θ(p0). (11.13)

We have included a Heaviside step function θ(p0) to select only positive

p2 = 0p2 = m2

mass shell light

m

E = p0

p1

Fig. 11.1 The mass shell p2 = m2 is ahyperboloid in four-momentum space.Also shown is the equivalent surface forlight, p2 = 0.

mass particles.

Example 11.2

Show that δ(p2 −m2) θ(p0) = 12Ep

δ(p0 − Ep)θ(p0).

We use the identity

δ[f(x)] =X

x|f(x)=0

1

|f ′(x)| δ(x), (11.14)

where the notation tells us that the sum is evaluated for those values of x thatmake f(x) = 0. We take x = p0 and f(p0) = p2 − m2 = (p0)2 − p2 − m2. Thisgives us that |f ′(p0)| = 2|p0| and we use the fact that the zeros of f(p0) occur for

p0 = ±(p2 +m2)12 = ±Ep to write

δ(p2 −m2) θ(p0) =1

2Ep

[δ(p0 − Ep)θ(p0) + δ(p0 + Ep)θ(p0)], (11.15)

and so the result follows (since the second term in eqn 11.15 is zero).


Thus we will write our Lorentz-invariant measure as

d3p

(2π)32Ep

, (11.16)

where the additional factor δ(p0 −Ep)θ(p0) is there in every calculationand so we suppress it, and we have included the factor 1/(2π)3 becausethe mode expansion is essentially an inverse Fourier transform (we haveone factor of 1/(2π) for every component of three-momentum). We arenow in a position to write down integrals, for example:

1 =

∫d3p

(2π)32Ep

|p〉〈p|. (11.17)

This requires us to have Lorentz-covariant four-momentum states |p〉.We previously normalized momentum states according to

〈p|q〉 = δ(3)(p − q), (11.18)

so our new four-momentum states |p〉 will need to be related to thethree-momentum states |p〉 by

|p〉 = (2π)3/2(2Ep)1/2|p〉, (11.19)

and then their normalization can be written

〈p|q〉 = (2π)32Epδ(3)(p − q). (11.20)

Similarly, to make creation operators α† appropriately normalized sothat they create Lorentz-covariant states, we must define them by

α†p = (2π)3/2(2Ep)1/2a†p, (11.21)

so that α†p|0〉 = |p〉. In this case our mode expansion would take the

form of a simple inverse Fourier transform using our Lorentz-invariantmeasure

φ(x) =

∫d3p

(2π)31

(2Ep)

(αpe−ip·x + α†

peip·x) , (11.22)

or writing in terms of a†p and ap rather than α†p and αp:

φ(x) =

∫d3p

(2π)32

1

(2Ep)12

(ape−ip·x + a†peip·x) , (11.23)

which is identical to eqn 11.12.

11.3 What becomes of the Hamiltonian?

We can now substitute our expansion of the field operator φ(x) intothe Hamiltonian to complete our programme of canonical quantization.This will provide us with an expression for the energy operator in terms

11.3 What becomes of the Hamiltonian? 103

of the creation and annihilation operators. The Hamiltonian is given bythe volume integral of the Hamiltonian density,

H =

∫d3x

1

2

[∂0φ(x)

]2+ [∇φ(x)]2 +m2[φ(x)]2

. (11.24)

All we need do is substitute in the mode expansion and use the com-mutation relations to simplify. We’ll start by computing the momentumdensity Πµ(x) = ∂µφ(x) which is given by

Πµ(x) = ∂µφ(x) =

∫d3p

(2π)32 (2Ep)

12

(−ipµ)(ape−ip·x − a†peip·x) .

(11.25)To obtain an expression for ∂0φ(x), simply take the time-like componentof Πµ(x):

∂0φ(x) =

∫d3p

(2π)32 (2Ep)

12

(−iEp)(ape−ip·x − a†peip·x) , (11.26)

while the space-like components provide us with ∇φ(x):

∇φ(x) =

∫d3p

(2π)32 (2Ep)

12

(ip)(ape−ip·x − a†peip·x) . (11.27)

We now have all of the ingredients to calculate the Hamiltonian.

Example 11.3

Substitution of the mode expansion leads to a multiple integral of the form

H =1

2

Zd3x d3pd3q

(2π)3(2Ep)12 (2Eq )

12

h

(−EpEq − p · q)[ape−ip·x − a†peip·x]

× [aqe−iq·x − a†qeiq·x] +m2[ape−ip·x + a†peip·x][aqe−iq·x + a†qeiq·x]i

. (11.28)

We use our favourite trick again: first, we integrate over x and useR

d3x eip·x =(2π)3δ(3)(p) and obtain

H =1

2

Zd3pd3q

(2Ep)12 (2Eq )

12

(11.29)

×h

δ(3)(p− q)(EpEq + p · q +m2)[a†p aqe−i(Ep−Eq)t + ap a†qe−i(Ep−Eq)t]

+ δ(3)(p + q)(−EpEq − p · q +m2)[a†p a†qe−i(Ep+Eq)t + ap aqe−i(Ep+Eq)t]

i

.

Then we do the integral over q to mop up the delta functions:

H =1

2

Z

d3p1

2Ep

h

(E2p + p2 +m2)(a†p ap + ap a

†p)

+ (−E2p + p2 +m2)(a†p a

†−pe2iEpt + ap a−pe−2iEpt)

i

, (11.30)

and since E2p = p2 +m2 this quickly simplifies to

H =1

2

Z

d3pEp(ap a†p + a†p ap). (11.31)


Using the commutation relations on the result in eqn 11.31 we obtain

E =

∫d3pEp

(a†pap +

1

2δ(3)(0)

). (11.32)

The last term should fill us with dread. The term 12

∫d3p δ3(0) will give

us an infinite contribution to the energy! However we should keep inmind that by evaluating the total energy E we’re asking the theory totell us about something we can’t measure. This is a nonsensical questionand we have paid the price by getting a nonsensical answer. What wecan measure is differences in energy between two configurations. Thisis fine in our formalism, since upon taking the differences between twoenergies the 1

2δ(3)(0) factors will cancel obligingly. However, it’s still

rather unsatisfactory to have infinite terms hanging around in all of ourequations. We need to tame this infinity.

11.4 Normal ordering

To get around the infinity encountered at the end of the last section wedefine the act of normal ordering. Given a set of free fields, we definethe normal ordered product as

N[ABC† . . . X†Y Z

]=

(Operators rearranged with allcreation operators on the left

). (11.33)

The N is not an operator in the sense that is doesn’t act on a state toprovide some new information. Instead, normal ordering is an interpre-tation that we use to eliminate the meaningless infinities that occur infield theories. So the rule goes that ‘if you want to tell someone abouta string of operators in quantum field theory, you have to normal orderthem first’. If you don’t, you’re talking nonsense.66The origin of the infinity in eqn 11.32

is due to an ordering ambiguity in theclassical Lagrangian. Consider a gen-eral classical theory formed from com-plex values fields ψ(x). The Lagrangiandescribing these fields will contain bilin-ear terms like ψ†ψ, which could equallywell be written in a classical theory asψψ†. When we quantize and insert themode expansion the order that was cho-sen suddenly becomes crucial. Normalordering removes the ambiguity to givea meaningful quantum theory.

Rearranging operators is fine for Bose fields, but swapping the orderof Fermi fields results in the expression picking up a minus sign. Wetherefore need to multiply by a factor (−1)P , where P is the number ofpermutations needed to normally order a product of operators.

Example 11.4

Some examples of normal ordering a string of Bose operators follow. We first considera single mode operator

Nh

aa†i

= a†a, Nh

a†ai

= a†a, (11.34)

andNh

a†aaa†a†i

= a†a†a†aa. (11.35)

Next, consider the case of many modes

Nh

ap a†q ar

i

= a†q ap ar . (11.36)

The order of ap and ar doesn’t matter since they commute.

For Fermi fields, N [cp c†q cr ] = −c†q cp cr , the sign change occurring because we have

performed a single permutation (and hence an odd number of them).

11.4 Normal ordering 105

Step V: We can finally complete the programme of canonical quantiza-tion. With the normal ordering interpretation the old expression

H =1

2

∫d3pEp(apa

†p + a†pap), (11.37)

becomes

N [H] =1

2

∫d3pEpN [apa

†p + a†pap]

=1

2

∫d3pEp2a†pap

=

∫d3pEpnp, (11.38)

where np = a†pap is the number operator. Acting on a state it tells youhow many excitations there are in that state with momentum p.

We now have a Hamiltonian operator that makes sense. It turns outthat the Hamiltonian for the scalar field theory is exactly that which weobtained for independent particles in Chapter 3. This isn’t so surprisingsince we started with a Lagrangian describing waves that didn’t interact.What we’ve seen though, is that the excited states of the wave equationcan be thought of as particles possessing quantized momenta. Theseparticles could be called scalar phions.7 They are Bose particles with

7This is because they are excitations ofthe scalar field φ (Greek letter ‘phi’).

spin8 S = 0.

8We shall see that spin informationis encoded by multiplying the creationand annihilation operators by objectsthat tell us about the particle’s spinpolarization. These objects are vectorsfor S = 1 particles and spinors for aS = 1/2 particles. For S = 0 they donot feature.

Example 11.5

We have claimed that the vacuum energy is unobservable, so may be ignored. How-ever changes in vacuum energy terms are physically significant and lead to measurableeffects. Such a change results if you adjust the boundary conditions for your fieldand this is the basis of the Casimir effect.9 This is a small, attractive force be-

9The effect was predicted by the Dutchphysicist Hendrik Casimir (1909–2000).

tween two closely spaced metal plates, which results from the vacuum energy of theelectromagnetic field. The essential physics can be understood using a toy model10

10This approach follows Zee. A moredetailed treatment may be found inItzykson and Zuber, Chapter 3.

involving a massless scalar field in one dimension.

x L− x

I III II

Fig. 11.2 Two metal plates (I and II)separated by a distance L, with a thirdplate (III) inserted in between.

Consider two metal plates I and II separated by a distance L. We put a thirdplate (III) in between them, a distance x from plate I (see Fig. 11.2). We will derivethe force on plate III resulting from the field on either side of it. The presence of theplates forces the field to be quantized according to kn = nπ/x or nπ/(L − x). Thedispersion is En = kn and so the total zero-point energy is given by

E =∞X

n=1

»1

2

“nπ

x

”

+1

2

„nπ

L− x

«–

= f(x) + f(L− x), (11.39)

that is, 12

~ωn per mode.11 These sums both diverge, just as we expect since we are

11Remember, for photons ωn = ckn,

and the zero-point energy is 12

~ωn, so inunits in which ~ = c = 1 this becomes12kn, and so E =

P

n12kn. Equa-

tion 11.39 contains two such sums.

evaluating the infinite vacuum energy.However, real plates can’t reflect radiation of arbitrarily high frequency: the high-

est energy modes leak out. To take account of this we cut off these high-energy modesthus:

nπ

2x→ nπ

2xe−nπa/x, (11.40)

which removes those modes with wavelength significantly smaller than the cut-offa. The value of this cut-off is arbitrary, so we hope that it won’t feature in anymeasurable quantity.


The sums may now be evaluated:

f(x) =X

n

nπ

2xe−nπa/x

=1

2

∂

∂a

X

n

e−nπa/x

=1

2

∂

∂a

1

1 − eπa/x

=π

2x

eπa/x

(1 − eπa/x)2≈ x

2πa2− π

24x+O(a2). (11.41)

The total energy between plates I and II is

E = f(x) + f(L− x) =L

2πa2− π

24

„1

x− 1

L− x

«

+O(a2). (11.42)

If x is small compared to L, then we find a force

F = −∂E∂x

= − π

24x2, (11.43)

which is independent of a, as we hoped. Thus, there is an attractive force betweenthe closely spaced plates I and III. This is the Casimir force. We can understandthis force intuitively by realizing that as the two plates are pulled together we losethe high-energy modes. This reduces the energy between the plates and leads toan attractive force. A more quantum-field-theory friendly interpretation is that theeffect results from quantum fluctuations in the vacuum, in which particles are spon-taneously created and annihilated. These processes give rise to the vacuum Feynmandiagrams described in Chapter 19.

11.5 The meaning of the mode expansion

In the next chapter we’ll turn the crank on our canonical quantizationmachine for the second simplest field theory that we can imagine: thatof the complex scalar field. Before doing that, let’s have a closer look atthe mode expansion for this theory. Our first guess for the field operatormight have been the simple Fourier transform

φ−(x) =

∫d3p

(2π)32

1

(2Ep)12

ape−ip·x, (11.44)

which looks like the one we had in Chapter 4, where the e−ip·x tellsus that the particles are incoming. Unfortunately, this expansion won’twork for the relativistic theory we’re considering. The problem is theexistence of negative energy solutions in the relativistic equations ofmotion: each momentum state gave rise to two energies Ep = ±(p2+m2)and we can’t just leave out half of the solutions.12 Looking back at our12Why not? The result would be a field

for which [φ(x), φ(y)] 6= 0 for space-like|x − y|. See Weinberg, Chapter 5 forthe details.

discussion of the Klein–Gordon equation we saw the resolution of thisproblem, and we’ll employ the same solution here.

What we need is a mode expansion that includes these negative energymodes

φ(x) =∑

p

(positive Ep

mode

)+∑

p

(negative Ep

mode

). (11.45)

11.5 The meaning of the mode expansion 107

The expansion is carried out in terms of incoming plane waves e−ip·x.Now recall Feynman’s interpretation of the negative frequency modes inwhich negative energies are assumed to be meaningless, all energies areset to be positive and the signs of three-momenta for such modes areflipped. In this picture, the formerly negative energy states are inter-preted as outgoing antiparticles. The positive energy modes continue torepresent incoming particles. Incoming here means, not only a factore−ip·x, but also that the particle is annihilated: it comes into the systemand is absorbed by it. Conversely, outgoing means a factor eip·x andthat the particle is created. We therefore interpret the mode expansionas

φ(x) =∑

p

(incoming positive Ep

particle annihilated

)+∑

p

(outgoing positive Ep

antiparticle created

).

(11.46)The resulting expansion of a field annihilation operator is then

φ(x) =

∫d3p

(2π)32

1

(2Ep)12

(ape−ip·x + b†peip·x

), (11.47)

where the particle and antiparticle energies are given by Ep = +(p2 +m2)

12 . The rules are that ap annihilates particles, a†p creates particles

and the operators bp and b†p respectively destroy and create antiparticles.(For our scalar field theory the particles and antiparticles are the same:we say that ‘each particle is its own antiparticle’. This means that ap

can be thought of as either an operator that annihilates a particle or onethat annihilates an antiparticle.)13

13As we will see, in other theories wewill need separate operators to performthese two distinct roles, but in scalarfield theory there is no difference. Thusin this case ap = bp , because in a scalarfield theory a particle is its own antipar-ticle: particles and antiparticles are thesame. This will not necessarily be thecase for other field theories, an exampleof which will be described in the nextchapter.

Example 11.6

Let’s see what happens if we hit the vacuum state with our field annihilation operator.Since a†p |0〉 = |p〉 we have

φ(x)|0〉 =

Zd3p

(2π)32 (2Ep)

12

eip·x|p〉. (11.48)

We have therefore created an outgoing superposition of momentum states. Pick outone of these states and make an amplitude by folding in a relativistically normalized

state 〈q| = (2π)32 (2Ep)

12 〈q|:

(2π)32 (2Ep)

12 〈q|φ(x)|0〉 =

Z

d3p eip·x〈q|p〉 =

Z

d3p eip·xδ(3)(q − p) = eiq·x.

(11.49)In the language of second quantization, eiq·x tells us how much amplitude there is inthe qth momentum mode if we create a scalar particle at spacetime point x. Notethat eip·x ≡ ei(Ept−p·x) and so

Z

d3p eip·xδ(3)(q − p) = ei(Eqt−q·x) = eiq·x, (11.50)

demonstrating once again that our integral over three-momentum coordinates nev-ertheless results in a single particle in a four-momentum state.


It is important to note that canonical quantization will not succeedin diagonalizing all field theories. Roughly speaking it works only forthose Lagrangians which can be written as quadratic in a field and itsderivatives.14 (We will revisit the mathematics of this point in Chap-

14or, for the complex valued fields, asbilinear in the field and its adjoint.

ter 23.) The result of canonical quantization is a system described bysingle particles in momentum states which don’t interact with each other.For this reason Lagrangians that may be canonically quantized are callednon-interacting theories. In contrast, those theories which cannot bediagonalized with canonical quantization are called interacting theo-

ries; these will be described in terms of single particles in momentumstates which do interact with each other. Interacting theories are thesubject of much of the rest of this book. For now we will continueto look at non-interacting theories and in the next chapter we examinesome more uses of the canonical quantization routine that we have built.

Canonical quantization for ascalar field:

I Write down a Lagrangian den-sity L.

II Evaluate the momentum den-sity Πµ(x) = ∂L/∂(∂µφ) andthe Hamiltonian density

H = Π0∂0φ− L.

III Turn fields into operators andenforce the commutation rela-tion [φ(x), Π0(y)] = iδ(3)(x−y)at equal times.

IV Express the field in terms of amode expansion of the form

φ(x) =

Zd3p

(2π)32

1

(2Ep)12

×“

ape−ip·x + a†peip·x”

,

where Ep = +(p2 +m2)12 .

V Evaluate H and normal orderthe result, leading to an expres-sion like:

N [H] =

Z

d3pEp a†p ap .

Chapter summary

• Canonical quantization is an automated method for turning a clas-sical field theory into a quantum field theory.

• Field operators are formed from mode expansions with amplitudesmade of creation and annihilation operators.

• Normal ordering is needed to make sense of quantum field theories.

• The results of each step of the canonical quantization procedurefor the scalar field theory are shown in the box in the margin.

Exercises

(11.1) One of the criteria we had for a successful theoryof a scalar field was that the commutator for space-like separations would be zero. Let’s see if our scalarfield has this feature. Show that

[φ(x), φ(y)]=

Z

d3p

(2π)31

2Ep

“

e−ip·(x−y)−e−ip·(y−x)”

.

(11.51)For space-like separation we are able to swap (y−x)in the second term to (x − y). This gives us zero,as required.

(11.2) Show that, at equal times x0 = y0,

[φ(x), Π0(y)]=i

2

Z

d3p

(2π)3

“

eip·(x−y)+e−ip·(x−y)”

.

(11.52)In this expression there’s nothing stopping us swap-ping the sign of p in the second term, and show thatthis leads to

[φ(x), Π0(y)] = iδ(3)(x − y). (11.53)

12Examples of canonical

quantization

12.1 Complex scalar field theory109

12.2 Noether’s current for com-plex scalar field theory 111

12.3 Complex scalar field theoryin the non-relativistic limit

112

Chapter summary 115

Exercises 116

Examples are like cans of over-strength lager. One is bothtoo many and never enough.Anonymous

In this chapter we will apply the methods from the previous chapter tothe complex scalar field. This theory illustrates several ideas that arevery important in quantum field theory. We will also find the Noethercurrent and finally, we will take the non-relativistic limit of the theory.This last point is particularly important for thinking about applicationsto condensed matter, where much of the physics is firmly in the non-relativistic domain.

12.1 Complex scalar field theory

The Lagrangian for the complex scalar field1 is an example of a the- 1Recall where it came from. We con-sidered the addition of two scalar fields

L =1

2[∂µφ1(x)]2 − 1

2m2 [φ1(x)]2

+1

2[∂µφ2(x)]2 − 1

2m2 [φ2(x)]2 .

We noticed that we can make atransformation that simplified this La-grangian. We defined

ψ = 1√2[φ1(x) + iφ2(x)],

ψ† = 1√2[φ1(x) − iφ2(x)],

and we obtained the Lagrangian ineqn 12.1. This Lagrangian looks ratherlike that of the massive scalar field, butwithout the factors of 1/2.

ory with more than one component. Here we have two: ψ(x) and ψ†(x).We’ll now canonically quantize this Lagrangian using the procedure pre-sented in Chapter 11.

Step I: We start by writing down the Lagrangian we want to quantize.The Lagrangian for complex scalar field theory is

L = ∂µψ†(x)∂µψ(x) −m2ψ†(x)ψ(x). (12.1)

Step II: Identify the momentum density and write down the Hamilto-nian. Each component2 σ of the field (i.e. σ = ψ and ψ†) has a different

2Note carefully that σ isn’t a tensorcomponent, it simply labels the compo-nent of the field: ψ or ψ†. This impliesthat Πµσ = Πµσ .

momentum density. The momentum of the ψ-field and of the ψ†-field aregiven, as usual by the time-like (zeroth) component of the momentumdensity Πµ

σ:

Π0σ=ψ =

∂L∂(∂0ψ)

= ∂0ψ†, Π0σ=ψ† =

∂L∂(∂0ψ†)

= ∂0ψ. (12.2)

Now that we have the momenta we can write the Hamiltonian. For thecase of more than one field component we can write down

H =∑

σ

Π0σ(x)∂0ψ

σ(x) − L

= ∂0ψ†(x)∂0ψ(x) + ∇ψ†(x) · ∇ψ(x) +m2ψ†(x)ψ(x). (12.3)

110 Examples of canonical quantization

Step III: Make the fields into quantum mechanical operators. Wepromote the fields to operators and impose equal-time commutation re-lations thus

[ψ(t,x), Π0

ψ(t,y)]

=[ψ†(t,x), Π0

ψ†(t,y)]

= iδ(3)(x − y), (12.4)

and all other commutators vanish.Step IV: Expand the fields in terms of modes. For the case of the

real scalar field, we found that the field operator3 φ(x) =∫

p(ape−ip·x +3Here we abbreviate

R

p≡R d3p

(2π)32

1

(2Ep)12

. a†peip·x), so that it is Hermitian [i.e. self-adjoint: φ†(x) = φ(x)]. For ourcomplex scalar field there is no reason why the field operator should beHermitian. The appropriate mode expansion of the complex scalar fieldoperators take the more general form

ψ(x) =

∫d3p

(2π)32

1

(2Ep)12


),

ψ†(x) =

∫d3p

(2π)32

1

(2Ep)12

(a†peip·x + bpe−ip·x

), (12.5)

where Ep = +(p2 + m2)12 . Here the operators ap and bp annihilate

two different types of particle. They satisfy the commutation relations[ap, a

†q

]=[bp, b

†q

]= δ(3)(p − q) and all others combinations vanish.

Substituting these into the Hamiltonian we find that, after some algebraand then (Step V) normal ordering, that we obtain:

N[H]

=

∫d3pEp

(a†pap + b†pbp

)

=

∫d3pEp

(n(a)

p + n(b)p

), (12.6)

where n(a)p counts the number of a-particles with momentum p and, sim-

ilarly, n(b)p counts the number of b-particles. We notice that the a and b

particles have the same energy Ep, and so we interpret a and b as par-ticles and antiparticles respectively. This is justified in the next section.Thus ψ(x) involves a sum over all momenta of operators that annihilateparticles (ap) and create antiparticles (b†p). We therefore understandthese new operators as follows:

• a†p is the creation operator for a particle with momentum p,

• b†p is the creation operator for an antiparticle with momentum p.

We’ve quantized the theory and we’ve found that the excitations of thefield are scalar particles and antiparticles.

It is worthwhile justifying how we have been able to treat ψ and ψ†

as if they are independent fields. We began with

ψ(x) =1√2[φ1(x)+ iφ2(x)] and ψ†(x) =

1√2[φ1(x)− iφ2(x)], (12.7)

12.2 Noether’s current for complex scalar field theory 111

and this implies that

φ1 =1√2[ψ + ψ†] and φ2 =

−i√2[ψ − ψ†]. (12.8)

Thus all we have done in moving from a description based on φ1 and φ2

to one based on ψ and ψ† is to perform a change of basis. Allowing theaction S to vary with respect to ψ and ψ† gives as much freedom as ifwe were varying φ1 and φ2. For example, we can adjust φ1 alone usingδψ = δψ† and we can adjust φ2 alone using δψ = −δψ†.

12.2 Noether’s current for complex scalar

field theory

The complex scalar field has an internal U(1) symmetry.4 This means 4Internal symmetries are discussed inmore detail in the next chapter.that the global transformations of the fields

ψ → eiαψ, ψ† → e−iαψ† (12.9)

have no effect on the Lagrangian. Remember that Noether’s theoremsays that every symmetry begets a conserved current. This is our firstopportunity to see what is conserved for an internal symmetry. In fact,we will see that the conserved Noether charge is particle number.

To get to Noether’s current for the internal U(1) symmetry, it’s easiestto write out the transformation for an infinitesimal change in the phaseα:

ψ → ψ + iψ δα, Dψ = +iψ,ψ† → ψ† − iψ†δα, Dψ† = −iψ†.

(12.10)

Note that DL = 0 and since DL = ∂µWµ this implies that we can take5 5The property Wµ = 0 turns out to be

true of all internal symmetries.Wµ = 0. The Noether current is therefore given by JµN =∑σ Πµ

σDσ,(because Wµ = 0) where the index σ again labels the component of thefield: in this case whether it’s σ = ψ or ψ†. As a result we have

JµN =∑

σ

ΠµσDσ = Πµ

ψDψ + Πµψ†Dψ

†

= i[(∂µψ†)ψ − (∂µψ)ψ†] . (12.11)

We may make this a Noether current operator JµN by upgrading all of ourfields into field operators. This creates an ordering ambiguity in writingthe fields and their derivatives. We will remove this ambiguity by normalordering. Substituting the mode expansion for the fields allows us to findthe conserved charge operator QN.

Example 12.1

The charge operator is

QN =

Z

d3xJ0N = i

h“

∂0ψ†”

ψ − (∂0ψ)ψ†i

. (12.12)


Of course, there’s the ordering ambiguity here, but we’ll press on and insert the modeexpansion to yield

QN =1

2

Z

d3p“

−a†p ap + bp b†p − ap a

†p + b†p bp

”

. (12.13)

Finally, we normal order to give a conserved charge of

Nh

QN

i

=

Z

d3p“

b†p bp − a†p ap

”

. (12.14)

Noether’s theorem: U(1) internalsymmetry

Dψ = iψ Dψ† = −iψ†

Πµψ = ∂µψ† Πµψ† = ∂µψ

DL = 0 Wµ = 0JµN = i

ˆ(∂µψ†)ψ − (∂µψ)ψ†˜

QNc = −Nh

QN

i

=R

d3p (n(a)p − n

(b)p ).

The conserved charge is given by the difference in the number of an-tiparticles n

(b)p and the number of particles n

(a)p . This reason for this is

quite simple. Particles, which are excitations in the field, carry Noethercharge of one sign. To have a conserved charge we need other sorts ofparticle excitation carrying negative charge to cancel out the contribu-tion from the particles. We are left with the conclusion that the reasonfor the existence of antiparticles is that charge wouldn’t be conserved ifthey didn’t exist.

Obviously, if a current JµN is conserved, then so is −JµN and so ourassignment of which excitations carry positive Noether charge and whichare negative is quite arbitrary. It is only when we attach some observableproperty, such as electric charge q, to each that we decide which is which.By convention, the number current JµNc is defined to be positive forparticles (created by a†p operators) and negative for antiparticles (created

by b†p operators) and so we will define6 JµNc = −N[JµN

], QNc =

∫d3x J0

Nc6With a subscript c for conventional,just like conventional current directionin circuit theory! and therefore QNc = −N [QN]. Hence using eqn 12.14 and recalling that

n(a)p = a†pap and n

(b)p = b†pbp, we can conclude that

QNc =

∫d3p

(n(a)

p − n(b)p

). (12.15)

12.3 Complex scalar field theory in the

non-relativistic limit

In the final part of this chapter, we will road test our new complex scalarfield theory and see how it works out in the non-relativistic limit andwhether or not it will regenerate the familiar results of non-relativisticquantum mechanics.

In the non-relativistic domain the excitation energies of particles aresmall compared to the particle mass contribution mc2, where here we’lltemporarily reinstate factors of c and ~. This means that the mass termEp=0 = mc2 provides by far the largest fraction of the energy of anexcitation, that is E = mc2 + ε, where ε is small. To make a theorynon-relativistic, the strategy is to replace the relativistic field φ with

φ(x, t) → Ψ(x, t)e−imc2t/~, (12.16)

allowing us to factor out the enormous rest energy.

12.3 Complex scalar field theory in the non-relativistic limit 113

Example 12.2

To get an idea of how this works, we’ll take the non-relativistic limit of the Klein–Gordon equation. We start (in natural units) with

(∂2 +m2)Ψ(x, t)e−imt = 0, (12.17)

and moving to unnatural units we have„

~2 ∂

2

∂t2− ~

2c2∇2 +m2c4

«

Ψ(x, t)e−imc2t/~ = 0. (12.18)

The first term on the left yields

~2 ∂

2

∂t2Ψ(x, t)e−imc2t/~ = ~

2

„∂2Ψ

∂t2− 2imc2

~

∂Ψ

∂t− m2c4

~2Ψ

«

e−imc2t/~ , (12.19)

and substitution gives us

~2 ∂

2Ψ

∂t2− 2imc2~

∂Ψ

∂t− ~

2c2∇2Ψ = 0. (12.20)

Lastly we notice that the first term, with its lack of factors of c2, will be much smallerthan the other two. We can therefore drop the first term and we end up with

i~∂

∂tΨ(x, t) = − ~2

2m∇

2Ψ(x, t). (12.21)

The result is, of course, that we recover the Schrodinger equation for a free particle.

To take the non-relativistic limit of the complex scalar field Lagrangian7 7We’ll use the full interacting form

L = ∂µψ†(x)∂µψ(x) −m2ψ†(x)ψ(x) −g[ψ†(x)ψ(x)]2 and return to our con-vention that ~ = c = 1.

we substitute in ψ = 1√2m

e−imtΨ (with an extra normalization factor

1/√

2m added with malice aforethought).

Example 12.3

As in the last example, the guts of taking the limit may be found in the time deriva-tives. We find

∂0ψ†∂0ψ =

1

2m

h

∂0Ψ†∂0Ψ − im“

Ψ†∂0Ψ − (∂0Ψ†)Ψ”

+m2Ψ†Ψi

. (12.22)

The first term going as 1/m is negligible in comparison with the others. The thirdterm cancels against the mass term in the original Lagrangian. The dynamic partof the theory is therefore contained in the second term. We can see that a timederivative of Ψ (which we may expand in plane wave modes as something likePape−ip·x) will bring down a factor −iEp . Since Ψ† is obtained from Ψ through

taking an adjoint, the time derivative of Ψ† will therefore involve bringing down afactor iEp . The result is that (Ψ†∂0Ψ − Ψ∂0Ψ†) may be replaced8 by −2Ψ†∂0Ψ. 8We could have replaced this with

2Ψ(∂0Ψ†) if we had chosen modes vary-ing as e+ip·x, but we choose to favourmatter over antimatter with our choice.After these manipulations, we end up with

L = iΨ†(x)∂0Ψ(x) − 1

2m∇Ψ†(x) · ∇Ψ(x) − g

2[Ψ†(x)Ψ(x)]2, (12.23)

with g = λ/2m2. This looks a lot less neat and covariant than therelativistic form, because of the asymmetry between the matter and an-timatter fields that resulted from the choice we made in simplifying thetime derivative term. This is acceptable because life is a lot less sym-metrical in the non-relativistic world, but we will see the consequenceslater. Anyway, we will now continue and feed this Lagrangian into thecanonical quantization machine.


Example 12.4

Let’s attempt to quantize the non-relativistic Lagrangian, without the self-interaction(g = 0), but in the presence of an external potential V (x) which enters as VΨ†Ψ.There’s nothing for it but to press on with the five point programme.In this example we begin to see some

of the limitations of canonical quantiza-tion. For a general potential V (x) theprocedure does not return a diagonal-ized Hamiltonian (although it will workfor a constant potential V (x) = V0).We will see in Chapter 16 how moregeneral potentials may be dealt with us-ing perturbation theory.

Step I: The Lagrangian density with the external potential is given by

L = iΨ†(x)∂0Ψ(x) − 1

2m∇Ψ†(x) · ∇Ψ(x) − V (x)Ψ†(x)Ψ(x). (12.24)

Feeding this through the Euler–Lagrange equations yields the Schrodinger equation

and, for V (x) = 0, a dispersion Ep = p2

2m(Exercise 12.5). This only has positive

energy solutions so we might expect that the negative frequency part of the modeexpansion won’t be needed here. This will turn out to be the case.

Step II: We calculate the momentum densities thus:

Π0Ψ = ∂L

∂(∂0Ψ)= iΨ†, Π0

Ψ† = ∂L∂(∂0Ψ†)

= 0. (12.25)

Notice that there’s no momentum density conjugate to the field Ψ†, which is theconsequence of our choice in Example 12.3. The momentum allows us to calculatethe Hamiltonian density:

H = Π0Ψ∂0Ψ − L

=1

2m∇Ψ†(x) · ∇Ψ(x) + V (x)Ψ†(x)Ψ(x). (12.26)

Lo and behold, we have obtained a Schrodinger-like equation for the Hamiltoniandensity.

Step III: The equal-time commutation relation between the position and momen-tum operators is

h

Ψ(t,x), Π0Ψ(t,y)

i

= iδ(3)(x− y), (12.27)h

Ψ(t,x), Ψ†(t,y)i

= δ(3)(x− y),

where the second equality follows from inserting our expression Π0Ψ = iΨ†.

Step IV: A mode expansion with positive and negative frequency parts won’tobey the commutation relation above. The mode expansion which does the trick issimply

Ψ(x) =

Zd3p

(2π)32

ape−ip·x, (12.28)

with Ep = p2

2m.

Step V: Substituting the mode expansion gives us a Hamiltonian operator

H =

Z

d3p

„p2

2ma†p ap

«

+

Zd3xd3pd3q

(2π)3

“

V (t,x)e−i(Ep−Eq)tei(p−q)·x a†p aq

”

.

(12.29)The time-dependent part will guarantee conservation of energy. It forces the potentialto have a time dependence of the form V (t,x) = ei(Ep−Eq)tV (x) if the momentumof the particle is going to change. With this constraint we obtain a Hamiltonian

H =

Z

d3p

„p2

2ma†p ap

«

+

Z

d3p d3q“

V (p− q)a†p aq

”

, (12.30)

where V (p−q) =R

d3x 1(2π)3

V (x)ei(p−q)·x and the potential must impart an energy

Ep−Eq . Notice that this is equivalent to the Hamiltonian we argued for in Chapter 4.

We can look at the non-relativistic complex scalar field in another wayby considering a different form for the field. We first note that the non-relativistic form of the Lagrangian is still invariant with respect to globalU(1) transformations. Motivated by this, we write

Ψ(x) =√ρ(x)eiθ(x), (12.31)

12.3 Complex scalar field theory in the non-relativistic limit 115

so that Ψ(x) is written in terms of amplitude and phase. The U(1)transformation is now enacted by θ → θ + α (see the box for the con-served charges). Whereas before we had two fields, φ1(x) and φ2(x),


Dθ = 1Π0θ = −ρ Πiθ = ρ

m∂iθ

DL = 0 Wµ = 0J0N = −ρ(x) JN = − ρ

m∇θ

QNc =R

d3x ρ(x) JNc = ρm

∇θ

now we have two fields, ρ(x) and θ(x). Substituting in the Lagrangianin eqn 12.23, we obtain (Step I) the Lagrangian in terms of ρ and θfields

L =i

2∂0ρ− ρ∂0θ −

1

2m

[1

4ρ(∇ρ)2 + ρ(∇θ)2)

]− g

2ρ2. (12.32)

Now we turn off interactions (by setting g = 0) and feed the resultingLagrangian into the canonical quantization machine and make selectiveuse of its features. First we find the momenta (Step II):

Π0ρ(x) =

∂L∂(∂0ρ(x))

=i

2,

Π0θ(x) =

∂L∂(∂0θ(x))

= −ρ(x), (12.33)

from which we see that the Π0θ is most interesting since it’s not a con-

stant. Let’s make this quantum mechanical (Step III) by imposingcommutation relations

[θ(x, t), Π0

θ(y, t)]

= −[θ(x, t), ρ(y, t)

]= iδ(3)(x − y). (12.34)

The box shows that the time-like component of the conserved currentis ρ(x). We therefore define the total number of particles as N(t) =∫

d3x ρ(x, t) and by integrating9 the commutation relation equation we 9This calculation simply usesZ

d3y [ρ(y, t), θ(x, t)] = [N(t), θ(x, t)]

andZ

d3y δ(3)(x− y) = 1.

obtain [N(t), θ(x, t)

]= i, (12.35)

which is an important result.10 It tells us that in the sort of coherent

10Equation 12.35 is called the number-phase uncertainty relation.

condensed matter system that can be described by the non-relativisticcomplex scalar field theory the operator for the number of excitationsis conjugate to the operator for the phase angle of the field. We willreturn to this important idea when we come to discuss superfluids andsuperconductors.

Chapter summary

• Quantization of the complex scalar field leads to two types of ex-citation: particles and antiparticles. The conserved charge is thenumber of particles minus the number of antiparticles.

• The non-relativistic limit of the theory may be taken. From thiswe can derive a number-phase uncertainty relation.


Exercises

(12.1) Fill in the missing steps of the algebra that led toeqn 12.6.

(12.2) Evaluate (a) [ψ(x), ψ†(y)] and (b) [Ψ(x), Ψ†(y)] us-ing the appropriate mode expansions in the text.

(12.3) Consider the Lagrangian for two scalar fields fromSection 7.5.(a) Evaluate [QN, φ1] and (b) [QN, φ2], where QN isthe Noether charge.Hint: You could do this by brute force and evalu-ate QN and then find the commutator. A preferablemethod is just to use: [QN, φi] = −iDφi from Chap-ter 10.(c) Use these results to show that [QN, ψ] = ψ.

(12.4) Consider the theory described by the Lagrangianin eqn 12.32. Use Noether’s theorem in the form[QN, φ] = −iDφ to provide an alternative deriva-tion of eqn 12.35.

(12.5) Apply the Euler–Lagrange equations to eqn 12.24and show that it yields Ep = p2/2m when V = 0.

(12.6) Find the Noether current for the Lagrangian ineqn 12.24. Check that it is what you expect fornon-relativistic quantum mechanics.

(12.7) Consider the complex scalar field again. The inter-nal transformation operator may be written U(α) =

eiQNcα, where QNc is the conserved number chargeoperator. Show that U†(α)ψ(x)U(α) = eiαψ(x).

13Fields with many

components and massive

electromagnetism

13.1 Internal symmetries 117

13.2 Massive electromagnetism120

13.3 Polarizations and projec-tions 123

Chapter summary 125

Exercises 125

In the previous chapter we described a theory with two field components.Here we expand the discussion to theories with more components andwe will see that the interesting thing about these components is theirsymmetries. We’ll examine two cases, the first of which has an internalsymmetry and a second where the theory is described in terms of four-vectors.

13.1 Internal symmetries

When we find families of closely-related particles in Nature, there isan inevitable lurking suspicion that these particles are really a singlephysical entity but that there is some internal dial which can be rotatedto turn one member of the family into another. The invariance of theLagrangian with respect to turns on this internal dial is described as aninternal symmetry, quite unrelated to the symmetry of spacetime inwhich the particles live, move and have their being.1 1We saw the consequence of a U(1) in-

ternal symmetry in the complex scalarfield in the last chapter.

An early example of the idea of an internal symmetry was isospin.The neutron and proton have almost the same mass (mp ≈ mn) and thisled Heisenberg to wonder if they were the same particle in two differentstates with some kind of internal symmetry that he termed isospin. Theidea is that neutrons and protons both have isospin I = 1

2 but that, as for

conventional spin,2 there are two possible eigenvalues of the Iz operator: 2Note that isospin has nothing to dowith the actual spin angular momen-tum of the particles.

Iz = 1/2 (the proton) and Iz = −1/2 (the neutron). By analogy withspin angular momentum, we may arrange these into a two-component

object

(pn

), known as an isospin doublet. We can rotate these isospin

doublets with the same matrices3 that rotate spins-1/2. Before we get 3As explored in Chapter 37, these rota-tions form a representation of the groupSU(2).

too carried away, we should remember that, strictly, mp 6= mn so thatisospin is only an approximate description.4

4Of course protons and neutrons are ac-tually each composed of quarks.

The idea of an internal symmetry nevertheless has wide applicability.We will explore this idea using the example of three scalar particles. Weimagine arranging them into a vector (t, d, h) which transforms internallyaccording to the three-dimensional rotation group SO(3). Rotating theinternal dial by 90 about the h-axis turns a t particle, denoted (1, 0, 0),into a d particle (0, 1, 0) (see Fig. 13.1). This also implies that super-

118 Fields with many components and massive electromagnetism

positions of particles, obtained by some arbitrary rotation of the dial,have just as much validity as the original particles do. Since particlesare excitations in fields, we should be able to examine isospin in fieldtheory. In the field case we arrange the fields that generate these par-ticles into a vector such as (φ1, φ2, φ3). Subjecting this column vectorto internal rotations by turning the internal dial, one can carry out theequivalent act of turning a φ1 field into a φ2 field by rotating aboutthe φ3-axis. If the Lagrangian describing the theory of these particles isinvariant with respect to these rotations, then Noether’s theorem givesus a conservation law describing the charges of the fields.

To simplify our notation, let’s write the three scalar fields as a columnvector Φ(x) as follows

Φ(x) =

φ1(x)φ2(x)φ3(x)

. (13.1)

The free Lagrangian (step I of the Canonical quantization process) for

I’m an h particle!

I’m just a d.

My internal pointersays I’m a t.

Fig. 13.1 In our toy model, particlescan be imagined as having an internaldial. Whether the dial points along(1, 0, 0), (0, 1, 0) or (0, 0, 1) determineswhich kind of particle it is. The po-sition of the dial is not related to thedirections in spacetime through whichparticles move, but expresses a direc-tion in abstract space describing the in-ternal symmetry.

this theory is given by

L =1

2(∂µΦ) · (∂µΦ) − m2

2Φ · Φ, (13.2)

which is short for

L =1

2

[(∂µφ1)

2 −m2φ21 + (∂µφ2)

2 −m2φ22 + (∂µφ3)

2 −m2φ23

], (13.3)

that is, the sum of the Lagrangians for the individual fields. Note care-fully that the object Φ(x) is not a vector field that lives in Minkowskispace like the vectors xµ and pµ do. This point is examined in thefollowing example.

Example 13.1

Here, unlike elsewhere, the dot products aren’t short for a sum gµνAµAν . Thevector Φ doesn’t live in Minkowski space at all, so has no dealings with the tensorgµν . Instead, the dot products here are short for

Φ · Φ = φ1φ1 + φ2φ2 + φ3φ3, (13.4)

and∂µΦ · ∂µΦ = ∂µφ1∂µφ1 + ∂µφ2∂µφ2 + ∂µφ3∂µφ3. (13.5)

This is to say that the α in Φα isn’t a tensor index and there is consequently nodifference between Φα and Φα.

Our example has an SO(3) internal symmetry: we can transform thevector Φ using a three-dimensional rotation matrix and the Lagrangianwon’t change one iota. That is to say that there is no effect on theLagrangian on transforming to a field Φ′ via a transformation such as arotation about the φ3-axis:

φ′1φ′2φ′3

=

cos θ − sin θ 0sin θ cos θ 0

0 0 1

φ1

φ2

φ3

. (13.6)

13.1 Internal symmetries 119

The canonical quantization for this theory proceeds as before, but wenow implement our new notation. This leads to a Hamiltonian (stepII) Note that Πµα = ∂L

∂(∂µΦα).

H =∑

α

[1

2(∂0φα)2 +

1

2(∇φα)2 +

1

2m2φ2

α

]. (13.7)

The equal-time commutation relations (step III) become

[Φα(t,x), Π0

β(t,y)]

= iδ(3)(x − y)δαβ , (13.8)

including a Kronecker delta because each component of Φ has a nonzerocommutator with the same component of its own momentum field only.The mode expansion (step IV) is then given by

Φ(x) =

∫d3p

(2π)32

1

(2Ep)12

ap1e−ip·x + a†p1e

ip·x


ip·x


ip·x

, (13.9)

where apα are the annihilation operators for the α field and we have the

commutators [apα, a†qβ ] = δ(3)(p − q)δαβ and all others vanish.

Example 13.2

We could also abbreviate the notation slightly by writing this as

Φ(x) =

Zd3p

(2π)32

1

(2Ep)12

3X

α=1

hα

“

apαe−ip·x + a†pαeip·x”

, (13.10)

where

h1 =

0

@

100

1

A , h2 =

0

@

010

1

A , h3 =

0

@

001

1

A , (13.11)

are internal vectors that tell us about the polarization of the field in its internal space.

Substituting into the Hamiltonian for this theory then leads5 to the 5One needs to substitute the mode ex-pansion and normal ordering steps IVand V.

result that

H =

∫d3p Ep

3∑

α=1

a†pαapα, (13.12)

and we observe that the energy of the particles may be found by summingover all momenta and polarizations.

Example 13.3

Noether’s theorem reminds us that a symmetry gives us a conserved quantity, so we

Noether’s theorem: SO(3) inter-nal symmetry

DbΦa = −εabcΦc Πµa = ∂µΦa

DL = 0 Wµ = 0JaµNc = εabcΦb(∂µΦc)

QNc =R

d3x (Φ × ∂0Φ)

Nh

QaNc

i

= −iR

d3p εabca†bp acp .examine the internal SO(3) symmetry of our fields. The Lagrangian is symmetricwith respect to the transformation Φ → Φ − θ × Φ. For example (and assumingsums over repeated indices), rotations about the φ3-axis give Φa → Φa − εa3cθ3Φc

and we haveD3φ1 = φ2, D3φ2 = −φ1, D3φ3 = 0, (13.13)


where we call the symmetry from rotation around the b-axis DbΦa. Using the factthat, as usual, Wµ = 0 for this internal transformation (since DL = 0), this yields aNoether current J3µ

N for rotation around the internal 3-axis:

J3µN = ΠaµD3Φa = (∂µφ1)φ2 − (∂µφ2)φ1. (13.14)

Swapping signs over, the conserved charge is given in terms of (normally ordered6)6We dispense with the normal orderingsymbol from this point. It should beassumed in all calculations of energiesand conserved charges.

creation and annihilation operators by

Q3Nc = −i

Z

d3p“

a†1p a2p − a†2p a1p

”

. (13.15)

Rotations about the 1- and 2-axes will lead to similar conserved charges in terms ofthe other field operators. In general the conserved charge will be given by

QNc =R

d3x (Φ × ∂0Φ) and QaNc = −iR

d3p εabca†bp acp . (13.16)

This conserved charge is isospin. As shown in Exercise 13.1, isospin has the structureof angular momentum: in this case an isospin-1. Remember that, despite the similar-ity between the angular momentum in Exercise 3.3 (which arises since the symmetryis with respect to spatial rotations) and the conserved charge in this example, thesymmetry here is internal and so this angular momentum-like charge is internal tothe field.

13.2 Massive electromagnetism

We now turn to the interesting case of the massive vector field. Thistheory describes an intriguing world7 in which photons have mass! The7The theory described in this section

is rather like quantum electrodynam-ics but, it turns out, without the inter-esting complication of gauge invariance,which we leave until the next chapter.

theory is described in terms of Aµ(x) which is called a ‘vector field’because, when we input a spacetime point x, the field outputs a four-vector Aµ corresponding to that point. The key observation about Aµ

is that it lives in Minkowski space and so transforms in the same way asa four-vector does under the Lorentz transformation.8 We’ll see that the8This is in contrast to our example of

the internal SO(3) field Φ(x). particles created by the quantum version of this field have three possiblepolarizations, which will lead to the excitations carrying a spin of S = 1.The Lagrangian for this theory (Step I) is given by

L = −1

4FµνF

µν +1

2m2AµA

µ, (13.17)

where Fµν = ∂µAν − ∂νAµ. This is just the Lagrangian for the electro-magnetic field but with a Klein–Gordon-esque mass term on the right-hand side. Note that the sign of the second term that contains the factorm2 is positive. In the Klein–Gordon equation the m2 term was nega-tive, but the reason for our sign choice here will become apparent in thefollowing example.

Example 13.4

We can find the equations of motion for the field in this theory by plugging theLagrangian (eqn 13.17) into the Euler–Lagrange equation. The result is that9

9Equation 13.18 can of course be writ-ten out purely in terms of Aµ as

∂µ (∂µAν − ∂νAµ) +m2Aν = 0.∂µF

µν +m2Aν = 0. (13.18)

13.2 Massive electromagnetism 121

This is known as the Proca equation. Taking the divergence gives us zero from the Alexandru Proca (1897–1955) devel-oped the theory of massive spin-1 par-ticles using field theory.

first term (since ∂µ∂νFµν = 0) and we obtain from the second term m2∂νAν = 0.Since m2 6= 0, we have the additional condition that the field obeys ∂µAµ = 0. Thisallows us to write the Proca equations in another form

`∂2 +m2

Áν = 0, (13.19)

which looks a lot like the Klein–Gordon equation10 for each of the four components 10This shows that we did indeed makea sensible sign choice in eqn 13.17.of the field Aν(x). The dispersion relation is therefore E2

p = p2 + m2 for eachcomponent.

We now continue the procedure of quantizing this theory.11 Finding the 11The rest of this chapter works care-fully through the nuts and bolts of get-ting from the Lagrangian in eqn 13.17to the final result of the Hamil-tonian in eqn 13.31, which is ourfamiliar second-quantized harmonic-oscillator type Hamiltonian.

momentum density Π0σ(x) for each field component σ is the first part of

Step II, which is easily done. Extending the definition of Πµσ(x) from

the last chapter to include the vector index on Aν , we have:

Πµν(x) =∂L

∂(∂µAν)= ∂νAµ − ∂µAν . (13.20)

Note that Πµν is now a second-rank tensor whose two indices can beraised or lowered using the metric tensor gµν . We then have momentumdensities

Π00 = 0, Π0i = ∂iA0 − ∂0Ai = Ei, (13.21)

where E is the ‘electric field’ of the theory (a useful name for notationalsimplicity, but that’s all since this is not electromagnetism!). Notice thatΠ00 = 0 means that A0 has no dynamics of its own.12 We will eliminate 12This lack of dynamics might cause us

to worry that the quantization routinewon’t work on this field component. Infact, we will eliminate A0 and so thislack of a momentum won’t turn out tobe a problem. This is quite unlike thecase of QED where the problem of hav-ing Π00 = 0 is a knotty one. See Aitchi-son and Hey, Section 7.3 for a discus-sion.

it in the next step (and justify this in step IV). We find the Hamiltoniandensity to be13

13 In four steps:I. Write

H = −E · A− 1

2(E2 − B2)

−m2

2

ˆ(A0)2 − A2

˜.

II. Use eqn 13.21 along withE · ∇A0 = ∇ · (EA0) −A0(∇ · E).III. Then use the equations of motion(eqn 13.18) to eliminate A0 via

A0 = − 1

m2∇ · E.

IV. The Hamiltonian includes a term∇ · (EA0), but this is a total deriva-tive. We constrain all of our fields tovanish at infinity, so we may drop thetotal derivative from the Hamiltoniandensity.

H =1

2

(E2 + B2 +m2A2 +

1

m2(∇ · E)2

). (13.22)

Notice the positive contribution from the mass term, which justifies thechoice of sign in the Lagrangian in eqn 13.17.

Next we need some equal-time commutation relations (step III) tomake all of this quantum mechanical. Clearly we want a way to deal withthe different components of the vector field which, when we substitute ina sensible mode expansion, will yield up a total energy which is the sumof the positive energies of all of the particles. Since we have eliminatedA0 from the Hamiltonian in eqn 13.22 we only need relations betweenthe space-like components (i and j). We choose

[Ai(t,x), Ej(t,y)

]= −iδ(3)(x − y)gij , (13.23)

which has the form we had for previous cases (namely [φ, ∂0φ] = iδ sincegij = −δij) for each spatial component.

We may now expand the field Aµ (step IV) in terms of planewaves. Since these will look something like A ∼ ape−ip·x, the condi-tion ∂µA

µ = 0 implies that pµAµ = 0. This tells us that the four-vector

Aµ is orthogonal to pµ, just as we have in three dimensions if c · d = 0for vectors c and d. Moreover, the condition pµA

µ = 0 introduces a con-straint on the components of Aµ: they are no longer independent and


we may express one of the components of Aµ in terms of the others. Thetheory therefore describes a field (and particles) with three independentpolarization degrees of freedom. The polarization degrees of freedomare taken to be Ai with i = 1, 2, 3. The component A0 is completelydetermined via p · A = p0A0, which shows that we were justified ineliminating it from eqn 13.22.

Now for the mode expansion. As in the case of the SO(3) fields whichwe wrote as Φ, we’ll need separate creation and annihilation operatorsfor each polarization. Unlike that case, however, we need to multiplyeach by a polarization vector14 ǫµλ(p) which lives in Minkowski space

14The polarization of the vector parti-cle tells us about its spin angular mo-mentum. Vector particles have S = 1and the three degrees of freedom cor-respond to Sz = 1, 0 and −1. This isexamined further in the next chapter.

and whose components depend on the value of the momentum of theparticle that we’re considering.15

15Unlike µ, the label λ on the polar-ization vector is not a tensor index, it’spart of the name that tells us the polar-ization vector to which we’re referringWe could equally have called the vec-tors Tom, Dick and Harry. However,as we have done here, it’s often usefulto use λ as a shorthand to also tell usalong which direction the polarizationvector points in the rest frame of theparticle.

Since there are only three degrees of freedom there will be three po-larizations λ = 1, 2, 3. We choose that there is no λ = 0 polarization[that is, we will force the field to be orthogonal to the 0-direction inthe rest frame of a particle, so we won’t require a time-like polarizationbasis vector ǫµλ=0(m, 0) = (1, 0, 0, 0) (see below)]. The resulting modeexpansion is written

Aµ(x) =

∫d3p

(2π)32

1

(2Ep)12

ǫ01(p)ǫ11(p)ǫ21(p)ǫ31(p)

ap1e

−ip·x +

ǫ0∗1 (p)ǫ1∗1 (p)ǫ2∗1 (p)ǫ3∗1 (p)

a†p1e

ip·x

+

ǫ02(p)ǫ12(p)ǫ22(p)ǫ32(p)

ap2e

−ip·x +

ǫ0∗2 (p)ǫ1∗2 (p)ǫ2∗2 (p)ǫ3∗2 (p)

a†p2e

ip·x

+

ǫ03(p)ǫ13(p)ǫ23(p)ǫ33(p)

ap3e

−ip·x +

ǫ0∗3 (p)ǫ1∗3 (p)ǫ2∗3 (p)ǫ3∗3 (p)

a†p3e

ip·x

, (13.24)

or, more compactly1616The commutation relations betweenthe creation and annihilation operatorsareh

apλ, a†qλ′

i

= δ(3)(p−q)δλλ′ . (13.25)Aµ(x) =

∫d3p

(2π)32

1

(2Ep)12

3∑

λ=1

(ǫµλ(p)apλe

−ip·x + ǫµ∗λ (p)a†pλeip·x).

(13.26)Since pµA

µ = 0, we’ll require pµǫµλ(p) = 0, which shows how the polar-

ization vectors depend on the momentum.

Example 13.5

The next job is to work out what we want the polarization vectors to be. Being vectorsand living in Minkowski space, these will have to transform like vectors under theLorentz transformations. It will be a good idea to have these vectors orthonormal sowe want

ǫ∗λ(p) · ǫλ′ (p) = gµνǫµ∗λ ǫνλ′ = −δλλ′ , (13.27)

13.3 Polarizations and projections 123

where the minus sign comes from our metric.17 Consider a particle in its rest frame, 17Note that despite λ not being a ten-sor index, the orthogonality conditionis often written gλλ′ , where gµν is theMinkowski tensor.

where it has momentum pµ = (m, 0, 0, 0). As stated above, we need pµǫλµ(p) = 0,for all λ. That is, we want the polarization vectors in this frame to be normal to thispµ. One possible choice is to use linear polarization vectors, given by

ǫ1(m, 0) =

0

BB@

0100

1

CCA, ǫ2(m, 0) =

0

BB@

0010

1

CCA, ǫ3(m, 0) =

0

BB@

0001

1

CCA. (13.28)

We can now work out the value of ǫλ(p) in an arbitrary frame of reference by boostingwith Lorentz transformation Λ(p). For example, a particle moving with momentumpz = |p| along the z-direction has pµ = (Ep , 0, 0, |p|) which is achieved with a boostmatrix

Λµν(p) =1

m

0

BB@

Ep 0 0 |p|0 m 0 00 0 m 0|p| 0 0 Ep

1

CCA, (13.29)

yielding the polarization vectors

ǫ1(Ep , 0, 0, |p|) =

0

BB@

0100

1

CCA, ǫ2(Ep , 0, 0, |p|) =

0

BB@

0010

1

CCA, ǫ3(Ep , 0, 0, |p|) =

0

BB@

|p|/m00

Ep/m

1

CCA.

(13.30)

Finally, using the mode expansion,18 the Hamiltonian after normal or- 18The expansion is rather tedious. Forhelp see the careful explanation inGreiner and Reinhardt, Chapter 6.

dering (step V) is diagonalized in the form

H =

∫d3pEp

3∑

λ=1

a†pλapλ, (13.31)

meaning that the total energy is the energy of all particles in all polariza-tions as we might have expected. Although the complication of havingthree field components is irritating, the algebra involves little more thanthe tedious job of keeping track of some new indices and the polarizationvectors.

13.3 Polarizations and projections

Clearly, the added complexity of the vector field arises because of thedifferent polarizations which are possible. In this section we’ll developa tensor toolkit to deal with this feature.19

19This section can be skipped on a firstreading.

Let’s start with something very simple. To project a three-vector X

along the direction of a vector p (see Fig. 13.2), then all we have to do isto take the scalar product between X and p = p/|p| (to find the lengthof the component), and then multiply the result by p to point it in thecorrect direction. The projected vector is then XL where

XL =(p · X)p

|p|2 , (13.32)

where the ‘L’ superscript stands for ‘longitudinal’. The transverse part

XT

XL

X

p

Fig. 13.2 Projecting a three-vector X

along the direction of a vector p.

of X is then XT = X−XL. We can write these equations in componentform as20

20Assume summation over repeated in-dices.


XiL = P ijL X

j and XiT = P ijT X

j , (13.33)

where the longitudinal and transverse projection matrices P ijL andP ijT are given by

P ijL =pipj

|p|2 , (13.34)

and

P ijT = δij −pipj

|p|2 . (13.35)

The four-dimensional upgrades of these are immediate: we define pro-jection tensors (returning to tensor notation)

PµνL =pµpν

p2and PµνT = gµν − pµpν

p2. (13.36)

For our massive spin-1 modes pµ represents the energy-momentum four-vector, so p2 = m2. If we work in a basis of linear polarizations, thetransverse projection tensor can be related to the polarization vectorsvia the following equality2121If we work in a basis of circular po-

larizations a slightly more complicatedidentity is needed.

1

2

X

λ

ˆǫµλ(p)ǫ∗νλ(p) +ǫ∗µλ(−p)ǫνλ(−p)

i

= −PTµν .

(13.37)

See Maggiore, Chapter 4.

3∑

λ=1

ǫλµ(p)ǫλν(p) = −PTµν . (13.38)

This relationship is proved in the following example:

Example 13.6

Method 1: In the rest frame pµ = (m, 0, 0, 0) and the polarization vectors are givenby eqn 13.28 and hence we have

3X

λ=1

ǫλµ(m, 0)ǫλν(m, 0) =

0

BB@

0 0 0 00 1 0 00 0 1 00 0 0 1

1

CCA. (13.39)

One can then boost this second-rank tensor using the matrix of eqn 13.29 to give

3X

λ=1

ǫλµ(p)ǫλν(p) = ΛρµΛσν

3X

λ=1

ǫλρ(m, 0)ǫλσ(m, 0) (13.40)

=1

m2

0

BB@

|p|2 0 0 −Ep |p|0 m2 0 00 0 m2 0

−Ep |p| 0 0 E2p

1

CCA

= −gµν +pµpν

m2.

Method 2: The quantityP3λ=1 ǫλµǫλν is a tensor and so it must be built up from

other tensors. Since the only other tensors we can build from the variables inthe problem are gµν and pµpν we’ll assume it has to be a linear combination ofthese: Agµν + Bpµpν . For a particle at rest we need A + Bm2 = 0 in order thatP3λ=1 ǫλ0ǫλ0 = 0, which fixes B = −A/m2. Next, for a particle at rest we need the

diagonal space-like components to equal 1, giving us A = −1.

Exercises 125

Chapter summary

• Putting more components into our theory introduces the notion ofthe polarization of our field. The Hamiltonian for a simple non-interacting Lagrangian again yields harmonic oscillator solutions,but this time the energy requires a sum over all momenta and overpolarizations.

• We have applied this to a Lagrangian for massive electromag-netism,

L = −1

4FµνF

µν +1

2m2AµA

µ,

and have shown that the polarization vectors ǫµ(p) are related tothe transverse projection tensor.

Exercises

(13.1) (a) Show that the conserved charge in eqn 13.16may be written

QNc =

Z

d3p A†pJAp , (13.41)

where Ap = (a1p , a2p , a3p) and J = (Jx, Jy, Jz) arethe spin-1 angular momentum matrices from Chap-ter 9.(b) Use the transformations from Exercise 3.3 tofind the form of the angular momentum matri-ces appropriate to express the charge as QNc =R

d3p B†pJBp where Bp = (b1p , b0p , b−1p).

(13.2) (a) Confirm that eqn 13.29 is the appropriate ma-trix to boost a particle along the z-direction.(b) Show that the boosted vectors in eqn 13.30 arestill correctly normalized.(c) Consider the circular polarization vectorsǫµ∗λ=R = − 1√

2(0, 1, i, 0), ǫµ∗λ=L = 1√

2(0, 1,−i, 0),

ǫµ∗λ=3 = (0, 0, 0, 1). Show that these are correctlynormalized according to gµνǫ

µ∗λ ǫνλ′ = −δλλ′ .

(13.3) Show that PL and PT are indeed projection opera-tors by showing that P 2 = P .

(13.4) The Lagrangian for electromagnetism in vacuo isL = − 1

4FµνFµν . Show that this can be rewritten

as

L = −1

2(∂µAν∂

µAν − ∂µAν∂νAµ), (13.42)

and hence show that using the transverse projectionoperator, it may be expressed as

L =1

2AµPT

µν∂2Aν . (13.43)

This shows that L only includes the transversecomponents of the field, squaring with the idea ofelectromagnetic waves only representing vibrationstransverse to the direction of propagation.

14Gauge fields and gauge

theory

14.1 What is a gauge field? 126

14.2 Electromagnetism is the sim-plest gauge theory 129

14.3 Canonical quantization ofthe electromagnetic field 131

Chapter summary 134

Exercises 134

Nobody ever reads a paper in which someone has done anexperiment involving photons with the footnote that says‘this experiment was done in Coulomb gauge’.Sidney Coleman (1937–2007)

14.1 What is a gauge field?

We continue our discussion of invariances possessed by physical systemsand arrive at gauge invariance. The idea here is that our field theory mayadmit different configurations of the fields which yield identical observ-ables. Our physical description thus contains some inherent vaguenessand so we can make a choice about which particular formulation, outof the many possible ones, to adopt in a given situation (this is called‘a choice of gauge’). A transformation from one description to anotheris called a gauge transformation, and the underlying invariance iscalled a gauge invariance.1 Note that gauge invariance is not a sym-1‘Gauge’ is an awful bit of terminol-

ogy with which we are unfortunatelystuck. Einstein’s general relativityshowed that spacetime geometry hasa dynamical role and Hermann Weylwondered if the scale of length coulditself be dynamical, varying throughspacetime. In this picture, one couldmake a choice of gauge which wouldbe a choice of scale-length: metal wirecomes in different thickness or gauges,so the term seemed entirely appropri-ate. He later adapted his scale argu-ment to one involving phase, as out-lined here, but the name ‘gauge’ stuck.

metry. Particles do not carry around a knob called ‘gauge’ that allowus to change them into other particles. Gauge invariance is merely astatement of our inability to find a unique description of a system.

Example 14.1

(i) In electrical theory we have to choose a zero for our potential V (i.e. choosewhich potential we call ‘ground’). This is because we only measure potentialdifferences (e.g. with a voltmeter) and so the choice of zero potential is anarbitrary one.

(ii) In quantum mechanics we have to make a choice of what we mean by the zeroof phase of a wave function. The choice is arbitrary because we only measurephase differences (e.g. in an interference experiment). There is nothing stop-ping one transforming all wave functions according to ψ(x) → ψ(x)eiα andthe physics is unchanged.

(iii) In electromagnetism, the magnetic vector potential A can be transformedaccording to A → A + ∇χ, where χ(x) is a function of position, and B =∇×A will be unaffected (because ∇×∇χ = 0). Thus one has to choose theform of A.

14.1 What is a gauge field? 127

The preceding example raises some important issues.

• The potential V and A, and the quantum mechanical phase, allpossess an intrinsic vagueness in description, and we have to makespecific choices in each case. We might wonder if the choices areconnected. It will turn out that they are.

• We thought about our choice of zero for V and phase as beingsomething we make once for our particular description of a particu-lar problem, but which is valid for the whole Universe. If zero voltsis set differently in New York and in San Francisco, then problemswill occur if you run a conducting wire between their respectiveelectrical grounds. However, A(x) is chosen with a function χ(x)which varies from place to place. Thus we will need to distinguishbetween global (affecting everywhere in the Universe) and local(affecting only a particular point) gauge transformations.

With these points in mind, it is time to start exploring this subtle in-variance, and by doing so we will shed new light on the origin of elec-tromagnetism.

We start with another look at the complex scalar field theory

L = (∂µψ)†(∂µψ) −m2ψ†ψ. (14.1)

Recall from Section 7.6 that this has a U(1) symmetry. We can makethe replacement

ψ(x) → ψ(x)eiα, (14.2)

and the Lagrangian (and by extension the equations of motion) aren’tchanged. The most important point to note about eqn 14.2 is that it’s aglobal transformation: it changes the field ψ(x) by the same amountat every spacetime point. We say that the theory, given in eqn 14.1, isinvariant with respect to global U(1) transformations.

It turns out that an incredibly rich seam of physics is revealed ifone asks what happens if the value of α given in eqn 14.2 depends onspacetime. That is, we investigate the transformation

ψ(x) → ψ(x)eiα(x), (14.3)

where the function α(x) allows us to transform the field by a differentamount at every point in spacetime. This is a very significant change.Your first reaction to this might be to say that surely no theory couldbe invariant with respect to a different change in phase at every pointin spacetime! Let’s see what actually happens.

Example 14.2

The good news is that the mass term m2ψ†ψ doesn’t change when we transformlocally. The bad news is that the term involving the derivatives does. This is becausethe derivative now acts on α(x). In fact, we have

∂µψ(x) → ∂µψ(x)eiα(x)

= eiα(x)∂µψ(x) + ψ(x)eiα(x)i∂µα(x)

= eiα(x) [∂µ + i∂µα(x)]ψ(x). (14.4)

128 Gauge fields and gauge theory

Similarly

∂µψ†(x) → e−iα(x) [∂µ − i∂µα(x)]ψ†(x). (14.5)

The first term in the Lagrangian becomes

(∂µψ†)(∂µψ) − i(∂µα)ψ†(∂µψ) + i(∂µψ†)(∂µα)ψ + (∂µα)(∂µα)ψ†ψ, (14.6)

which is certainly not what we started with.

As expected, having α depend on x has robbed the theory of its U(1)symmetry. We say that it is not invariant with respect to local U(1)transformations. Undaunted, we’re going to see if we can restore thislocal symmetry. We’ll do this by adding a new field Aµ(x), designedto roll around spacetime cancelling out terms that stop the theory frombeing invariant. The way this happens will depend on the manner inwhich the new field transforms. The way to proceed is to introduce thefield Aµ(x) via a new object Dµ given by

Dµ = ∂µ + iqAµ(x). (14.7)

This new object Dµ is called the covariant derivative. This will fixup the local U(1) symmetry if we insist that the new field Aµ transformsaccording to Aµ → Aµ − 1

q∂µα(x). The parameter q is known as thecoupling strength and will later tell us how strongly the Aµ field interactswith other fields.22For the case of QED examined in

Chapter 39 we will be able to identify qas the electromagnetic charge. For nowwe treat it as a parameter.

Example 14.3

If ψ(x) → ψ(x)eiα(x), then ∂µψ → (∂µψ)eiα + i(∂µα)ψ and so

Dµψ = (∂µ + iqAµ)ψ → (∂µψ)eiα + i(∂µα)ψ + iqAµψeiα − i(∂µα)ψ

= Dµ(ψeiα). (14.8)

This property makes the whole Lagrangian invariant if we replace ordinary derivativesby covariant ones:

L = (Dµψ)†(Dµψ) −m2ψ†ψ, (14.9)

since now with Dµψ → Dµψeiα, the first term is invariant.

Thus we conclude that we need our theory to be invariant with respectto two sets of transformations, which must be implemented together:

ψ(x) → ψ(x)eiα(x), (14.10)

Aµ(x) → Aµ(x) −1

q∂µα(x). (14.11)

A theory which has a field Aµ(x) introduced to produce an invariancewith respect to local transformations is known as a gauge theory.

The field Aµ(x) is known as a gauge field.

We might expect the field Aµ(x) could make a contribution to the La-grangian itself. It only exists in our description because we’ve invented itto satisfy our demand for a locally invariant theory, but if such ambitionshave any groundings in reality then the field Aµ should have dynamics ofits own! As we will see in the next section, it gives us electromagnetism.

14.2 Electromagnetism is the simplest gauge theory 129

14.2 Electromagnetism is the simplest

gauge theory

What are the possible forms of the part of the Lagrangian describing thefield Aµ(x)? It must be a theory for which the Lagrangian is invariantunder transformations of the form Aµ(x) → Aµ(x) − 1

q∂µα(x). Electro-magnetism provides an example of such a theory. It is, after all, a theorydescribed in terms of a vector field Aµ(x) = (V (x),A(x)) which is usedto form a Lagrangian

L = −1

4(∂µAν − ∂νAµ)(∂

µAν − ∂νAµ) − JµemAµ, (14.12)

from which the equations of motion follow as

∂2Aν − ∂ν(∂µAµ) = Jνem, (14.13)

which are known as the first two Maxwell equations.3 3Recall that we wrote these in Chap-

ter 5 as ∂λFλν = Jνem (where Fλν =

∂λAν − ∂νAλ), which is clearly thesame. More interestingly, it is worthconsidering the role of the Maxwellequations in this context. In the nextsection we will quantize the theoryin terms of electromagnetic particlescalled photons, which are excitations ofthe quantum field A(x). The Maxwellequations are the equations of motionfor A(x) and the closest we have to aSchrodinger equation for the photon.As it is not possible to construct a prob-ability density from the field A(x), thephotons that the quantum field theoryof electromagnetism describes do nothave an interpretation in terms of aprobability amplitude of their spatiallocalization. See Berestetskii, Lifshitzand Pitaevskii (Introduction and Chap-ter 1) for further details.

One of the most important observations about this formulation ofelectromagnetism is that neither the Lagrangian nor the equations ofmotion are changed if one makes the swap

Aµ(x) → Aµ(x) − ∂µχ(x), (14.14)

which is short for the changes

V → V − ∂0χ,

A → A + ∇χ. (14.15)

This observation, which is known as gauge invariance, means that if Aµdoes the job of describing the electromagnetic fields in some situationthen so does Aµ−∂µχ. This also means that electromagnetism is a gaugetheory since we can call χ(x) by the name α(x)/q and the condition ineqn 14.11 will clearly be satisfied.

Example 14.4

The theory can also be written in terms of the tensor Fµν = ∂µAν − ∂νAµ with aLagrangian written as

L = −1

4FµνF

µν . (14.16)

Gauge invariance amounts to the statement that we can change the field accordingto Aµ → Aµ − ∂µχ(x) and Fµν is unchanged.

It may be helpful to think of the freedom of the choice of gauge as achoice of language. We can speak French, German or Venusian, butwe are able to communicate the same underlying message no matterwhat language we speak. If we’re sensible, we can choose χ(x) (that is,the language we’re speaking) in such a way as to make whatever we’re


describing as simple as possible. There are several common choices, themost famous of which are listed in the box in the margin.

Let’s try making a choice of χ(x). We’ll try choosing χ(x) in such away that

∂µAµ(x) = 0, (14.17)

which is known as the Lorenz condition or as Lorenz gauge.4

4 Actually it’s more usually (and incor-rectly) known as the Lorentz conditiondue to its misattribution to HendrikLorentz (1853–1928) rather than to theless famous Ludvig Lorenz (1829–1891)who used it first. See J. D. Jackson andL. B. Okun, Rev. Mod. Phys. 73, 663(2001) for details of the history.

Example 14.5

To do this we write Aµ → A′µ = Aµ − ∂µχ. We want

∂µA′µ = ∂µAµ − ∂µ∂µχ = 0, (14.18)

which is achieved by setting ∂2χ = ∂µAµ. This results in the desired condition∂µA′µ = 0.

Common gauges for electromag-netism

Lorenz gauge ∂µAµ = 0Coulomb gauge ∇ · A = 0

Axial gauge A3 = 0Weyl gauge A0 = 0

Lorenz gauge is useful because, in the absence of a current Jµem, it re-sults in a massless Klein–Gordon equation for each component of theelectromagnetic field. That is, with ∂µA

′µ = 0, eqn 14.13 becomes

∂2A′µ = 0, (14.19)

whose solutions are plane waves of the form Aµ = ǫµ(p)e−ip·x with Ep =|p|. The Lorenz condition therefore makes electromagnetism resemblethe case of vector field theory. Recall that there the Lorenz condition∂µA

µ wasn’t a choice, it was mandated by the mass term m2AµAµ in theLagrangian of that theory. In any case, the result is the same: it reducesthe number of independent components of A′µ from four to three.

However, this still doesn’t make A′µ unique! This is because we canmake a further shift A′

µ → A′′µ = A′

µ − ∂µξ as long as ∂2ξ = 0 (so thatboth A′

µ and A′′µ satisfy the Lorenz condition). To make A′′µ unique, we

will choose

∂0ξ = A′0, (14.20)

which implies A′′0 = 0. With this further choice, the Lorentz condition

then reduces to ∇ · A′′ = 0. This is known as Coulomb gauge andfurther reduces the number of independent field components by one.This makes it clear that although the field Aµ has four components,the physics allows only two independent components. This can lead totrouble when the field is quantized.

Example 14.6

The equations of motion in Lorenz gauge read ∂2Aµ = 0, which, with A0 = 0,have plane wave solutions5 A = ǫe−ip·x. The equation encoding the Coulomb gauge

5Remember that the polarization vec-tors introduced here will carry the in-formation about the spin state of thephoton.

condition, ∇ · A = 0, leads to

p · A = p · ǫ = 0, (14.21)

14.3 Canonical quantization of the electromagnetic field 131

which tells us that the direction of propagation of the wave is perpendicular to thepolarization. For a wave propagating along z with momentum qµ = (|q|, 0, 0, |q|) wecould have

ǫ1(q) =

0

@

100

1

A , ǫ2(q) =

0

@

010

1

A , (14.22)

corresponding to linear polarization. We could choose the waves to have circularpolarization instead, in which case we could have

ǫ∗R(q) = − 1√2

0

@

1i0

1

A , ǫ∗L(q) = 1√2

0

@

1−i0

1

A . (14.23)

In order to observe the effects of electromagnetism, the electromagneticfield must couple to a matter field. We might wonder how to write downthis coupling. It turns out that the most simple form may be achievedwith a simple recipe known as the minimal coupling prescription.This simply involves swapping derivatives ∂µ of the matter field for co-variant derivatives Dµ and is illustrated in the example below.

Example 14.7

Consider complex scalar field theory in the presence of an electromagnetic field. TheLagrangian is written as the sum of the Lagrangians for the two separate theories:

L = (∂µψ)†(∂µψ) −m2ψ†ψ − 1

4FµνF

µν . (14.24)

To enact the minimal coupling prescription we upgrade derivatives of the matter fieldψ to covariant derivatives:

L = (Dµψ)†(Dµψ) −m2ψ†ψ − 1

4FµνF

µν

= (∂µψ† − iqAµψ†)(∂µψ + iqAµψ) −m2ψ†ψ − 1

4FµνF

µνv

= ∂µψ†∂µψ −m2ψ†ψ − 1

4FµνF

µν

+“

−iqAµψ†(∂µψ) + iq(∂µψ†)Aµψ + q2ψ†ψAµAµ”

. (14.25)

The final line shows the coupling between the Aµ field and the ψ and ψ† fields. Thestrength of the coupling is set by the coupling strength q, which is also known aselectromagnetic charge.

The notion that a gauge field, introduced to guarantee a local symmetry,will dictate the form of the coupling, or interactions, in a theory is knownas the gauge principle and is a philosophy to which we will returnlater.6 6See Chapter 46.

14.3 Canonical quantization of the

electromagnetic field

The previous section reminded us that even though Aµ has four compo-nents, only two are needed for a physical description (which makes good


physical sense because light is a transverse wave and therefore has twocomponents).7

7The gauge field Aµ should not, there-fore, be confused with the vector fielddiscussed in the last chapter. The vec-tor field was massive and this propertyrobs it of the possibility of gauge invari-ance. It also led to its having three de-grees of freedom. The massless natureof the gauge field leads to its gauge in-variance and its only having two freecomponents.

Having two redundant components of Aµ flapping around, reflectingthe freedom to choose which language that we speak, is something wemight worry about when we quantize the theory. There are now twoways to go. We can fix the gauge from the outset in order to reducethe number of degrees of freedom from four to two, making a choiceof which language we’re going to speak throughout, or we can continueand see how far we get before the redundancy comes back to bite us.8

8This second method may be found inthe book by Aitchison and Hey and theone by Maggiore.

We choose the first option in the following example, and working in theCoulomb gauge we will canonically quantize electromagnetism simplyby following the usual five steps.

Example 14.8

Step I: The Lagrangian is still

L = −1

4FµνF

µν = −1

4(∂µAν − ∂νAµ)(∂µAν − ∂νAµ), (14.26)

although we’ll need to remember that our choice of gauge dictates A0 = 0 and∇ · A = 0. Actually, this will be implemented at step III.Step II: We find that the Πµν tensor has components

Πµν =∂L

∂(∂µAν)= −(∂µAν − ∂νAµ), (14.27)

just as in the case of massive vector field theory. Again, the momentum density ofthe zeroth component Π00 = 0, but we don’t care since we have decided to eliminateA0 = 0. The momentum component conjugate to the ith component of the field isΠ0i = Ei, that is, the electric vector field E(x). The Hamiltonian9 is then9You should find

H =1

2(E2 + B2) + E · ∇A0.

The last term is removed using thesame method as employed in the pre-vious chapter, noting that ∇ · E = 0and that all fields should vanish at in-finity.

H =1

2(E2 + B2). (14.28)

Step III: Next we need commutation relations. Our first guess might be thath

Aµ(x), Π0ν(y)i

= iδ(3)(x − y)gµν , similar to the massive vector field. This won’t

work here though. Rewriting yieldsh

Ai(x), Ej(y)i

= −iδ(3)(x− y)gij , which looks

fine until you take the divergence (with respect to x) of this equation and geth

∂iAi(x), Ej(y)

i

= iδ(3)(x− y)∂j . (14.29)

But this should be zero! (This is because ∂iAi = ∇ · A = 0.) The problem is that

we haven’t yet ensured that only transverse components of A are permitted to enterthe solution. We need a way of projecting out the unnecessary ones to implementthe gauge condition. The answer to this problem turns out to necessitate the use ofour projection tensor PT (see Chapter 13). The commutation relation we need isgiven by

h

Ai(x), Ej(y)i

= i

Zd3p

(2π)3eip·(x−y)

„

δij − pipj

p2

«

(14.30)

= iδ(3)tr (x− y),

which gives zero on the right when you take the divergence (exercise). The symbol

δ(3)tr is known as a transverse delta function due to the inclusion of the transverse

projection tensor PT. This guarantees that our creation and annihilation operatorssatisfy the required relation [apλ, a

†qλ′ ] = δ(3)(p− q)δλλ′ .

14.3 Canonical quantization of the electromagnetic field 133

Step IV: Next the mode expansion, which only contains those two polarizations thatare transverse to the direction of propagation:10 10Note that the two polarization vec-

tors obey pµǫµλ(p) = 0 and are nor-

malized according to gµνǫµ∗λ (p)ǫνλ(p) =

−δλλ′ = gλλ′ .Aµ(x) =

Zd3p

(2π)32

1

(2Ep)12

2X

λ=1

“

ǫµλ(p)apλe−ip·x + ǫµ∗λ (p)a†pλeip·x

”

. (14.31)

Inserting this and (Step V) normally ordering, we obtain the expected final result:

H =

Z

d3p2X

λ=1

Ep a†pλapλ, (14.32)

with Ep = |p|. We conclude that the excitations of the electromagnetic field arephotons, which may be observed with two transverse polarizations. Apart from theslight complication of making the commutation relations compatible with the choiceof gauge and the hassle of the indices, there’s nothing hard about the quantizationof electromagnetism in Coulomb gauge.

The last example showed us that quantized particles of the electromag-netic field are photons. These particles have spin S = 1 and come in twotypes: a†p1|0〉 and a†p2|0〉, corresponding to the two transverse polariza-tions of the electromagnetic field. Consider a photon propagating alongthe z-direction with momentum qµ = (|q|, 0, 0, |q|). If we work in a basisof circularly polarized vectors, we may write ǫ∗λ=R(q) = − 1√

2(0, 1, i, 0)

(corresponding to Sz = 1) and ǫλ=L(q) = 1√2(0, 1,−i, 0) (corresponding

to Sz = −1). There are no photons with Sz = 0 as this would corre-spond to an unphysical longitudinal polarization ǫ∗λ=3(p) = (0, 0, 0, 1).We examine this in Exercise 14.2 and in the following example.

Example 14.9

We calculated the rotation matrices J for the vector field in Chapter 9. These alsofunction as angular momentum operators for photon states. We will create a photontravelling along the z-direction, so that qµ = (|q|, 0, 0, |q|). We write

|γλ〉 = ǫ∗λ(q)a†qλ|0〉. (14.33)

Let’s calculate the helicity11 of photons with λ = R. The helicity operator is h = 11Helicity, the projection of the spinalong the momentum direction, is ex-plained in detail in Chapter 36. Sincethe photon does not have a rest framethe helicity is the most useful quantityto use in talking about its spin.

J · q/|q|, leading to an operator h = Jz . Operating with the angular momentummatrix Jz suitable for S = 1 spins, we have

Jz |γλ=R〉 = − 1√2

0

BB@

0 0 0 00 0 −i 00 i 0 00 0 0 0

1

CCA

0

BB@

01i0

1

CCAa†

qR|0〉 = +|γλ=R〉. (14.34)

We see that the right-circularly polarized photon is in an Sz = 1 eigenstate andalso the h = +1 (positive) eigenstate of helicity. It’s easy to show that the left-circularly polarized photon is in the Sz = −1 state and has negative helicity. Youmay confirm that the disallowed polarization ǫ∗λ=3(q) = (0, 0, 0, 1) would correspondto a longitudinally polarized photon. Note that this polarization is allowed for thevector particle from Chapter 13 as vector fields have three degrees of freedom


Chapter summary

• A theory which has a field Aµ(x) introduced to produce an in-variance with respect to local transformations is known as a gaugetheory. The field Aµ(x) is known as a gauge field. Theories mustbe invariant with respect to the transformations:

ψ(x) → ψ(x)eiα(x),

Aµ(x) → Aµ(x) −1

q∂µα(x).

(14.35)

• Applied to the Lagrangian for electromagnetism, L = − 14FµνF

µν ,we have canonically quantized electromagnetism.

Exercises

(14.1) Fill in the algebra leading to eqn 14.32.

(14.2) A demonstration that the photon has spin-1, withonly two spin polarizations.

A photon γ propagates with momentum qµ =(|q|, 0, 0, |q|). Working with a basis where thetwo transverse photon polarizations are ǫµλ=1(q) =(0, 1, 0, 0) and ǫµλ=2(q) = (0, 0, 1, 0), it may beshown, using Noether’s theorem, that the opera-tor Sz, whose eigenvalue is the z-component spinangular momentum of the photon, obeys the com-mutation relationh

Sz, a†qλ

i

= iǫµ=1∗λ (q)a†qλ=2 − iǫµ=2∗

λ (q)a†qλ=1.

(14.36)(i) Define creation operators for the circular polar-izations via

b†qR = − 1√2

“

a†q1 + ia†q2

”

,

b†qL =1√2

“

a†q1 − ia†q2

”

. (14.37)

Show that

h

Sz, b†qR

i

= b†qR,h

Sz, b†qL

i

= −b†qL. (14.38)

(ii) Consider the operation of Sz on a state |γqλ〉 =b†qλ|0〉 containing a single photon propagating alongz:

Sz|γλ〉 = Sz b†qλ|0〉, λ = R,L. (14.39)

Use the results of (i) to argue that the projection ofthe photon spin along its direction of propagationmust be Sz = ±1.See Bjorken and Drell Chapter 14 for the full ver-sion of this argument.

15Discrete transformations

15.1 Charge conjugation 135

15.2 Parity 136

15.3 Time reversal 137

15.4 Combinations of discrete andcontinuous transformations

139

Chapter summary 142

Exercises 142

In Chapter 9 we have explored how quantum fields behave under variousspacetime translations and rotations. Such transformations are contin-uous and are thus represented by continuous groups (Lie groups). How-ever, discrete transformations are also possible, represented this time byfinite groups, and this chapter examines some of the most commonlyencountered and important discrete transformations.

C

Fig. 15.1 The charge conjugation oper-ator C turns a particle into its antipar-ticle, reversing its charge and all other‘charge-like’ numbers.

15.1 Charge conjugation

A particularly interesting discrete transformation is one which changesall particles into their antiparticles (see Fig. 15.1). This is accomplishedby an operator C: this operator flips not only the sign of a particle’scharge, but also its lepton number, its hypercharge, and all the ‘charge-like’ numbers which characterize a particular particle. We can write

C|p〉 = |p〉, (15.1)

where p is the antiparticle of a particle p. Let’s say that p has charge1

1By ‘charge’ we could mean any prop-erty like charge, or lepton number, orbaryon number, etc. These are generi-cally called quantum charges.

q. If this charge can be measured by an operator Q then Q|p〉 = q|p〉,whereas Q|p〉 = −q|p〉. Thus CQ|p〉 = qC|p〉 = q|p〉, but QC|p〉 = Q|p〉 =−q|p〉. This implies that QC = −CQ, or equivalently

C−1QC = −Q. (15.2)

For the operator C to exchange particles and antiparticles we need

C−1apC = bp and C

−1b†pC = a†p. (15.3)

Since a complex scalar field ψ(x) can be written ψ(x) =∫

p(ape−ip·x +

b†peip·x), we must have2 C−1ψC = ψ†. 2Note that

ψ† =

Z

p

(a†peip·x + bpe−ip·x).

Example 15.1

Since C2 = I, the eigenvalues of C can only be ±1. Most particles are not eigenstatesof C, since if they were then C|p〉 = |p〉 = ±|p〉, but this would mean that |p〉 is thesame state as |p〉 and so the particle is its own antiparticle. This is true in some cases,specifically for particles which have no quantum charges.3 In the case of the photon 3In other words for particles which have

zero electrical charge, zero lepton num-ber, zero baryon number, etc.

(γ), it is an eigenstate of C with eigenvalue −1, since if you change all particles totheir antiparticles, the electromagnetic field reverses (Aµ → −Aµ). It is also thecase with the neutral pion π0 but here the eigenvalue is +1. This explains why thereaction π0 → γ+ γ is allowed [the eigenvalues of C go from +1 → (−1)× (−1)], butyou cannot have π0 → γ + γ + γ.

136 Discrete transformations

15.2 Parity

scalar

vector

pseudovectorc = a × b

−a

a

ba

c

π

π

πb′

a′

a′×b′−b

−a

c

Fig. 15.2 A parity operation consistsof reflection in a mirror followed by aπ-rotation about an axis perpendicularto the mirror. The mirror reverses asingle spatial coordinate (the one per-pendicular to the plane of the mirror),the rotation reverses the other two. Ascalar is unaffected by this transforma-tion, whereas an ordinary (polar) vec-tor changes sign. A pseudovector (axialvector), such as a vector product of twoordinary vectors, is unchanged.

L L

E

−E

* *

*B

B

* *

*

πmirror

Fig. 15.3 Electric field E behaves likean ordinary (polar) vector, while mag-netic field B and angular momentum L

(derived from vector products of polarvectors) are pseudovectors (axial vec-tors). The asterisk indicates a partic-ular vector in the electric field from apoint charge, and in the magnetic fieldfrom a current loop, so that you canfocus on how a specified vector trans-forms.

When you look in a mirror, things look the same but there are subtledifferences. Five screws in your hand are reflected into five screws, sonumbers of things (scalars) don’t change. But look carefully and you’llsee that the threads in the mirror screws turn in the opposite senseto those in the real world. A mirror image has the spatial coordinateperpendicular to the surface of the mirror reversed, with the other twospatial coordinates left the same. This effect cannot be achieved by a ro-tation. If you follow a mirror operation by a 180 rotation about an axisperpendicular to the mirror, you will have reversed all three coordinatesand this transformation is known as inversion. This transformationcan be performed using the parity operator P which acts to invert spa-tial coordinate, i.e. mapping x → −x. This means that the positionoperator x will anticommute with the parity operator:

xP = −Px, (15.4)

or equivalentlyP−1xP = −x. (15.5)

The effect on the momentum coordinates is also p → −p, and hence

P−1pP = −p. (15.6)

However, the commutation relation [x, px] = i will be preserved. SinceP is a Hermitian operator, and since also it is its own inverse (P2 = I,where I is the identity) then P is also unitary. The parity operation hasno effect on scalars but reverses vectors (see Fig. 15.2, and for physicalexamples see Fig. 15.3). In fact, there are a class of scalars and vec-tors for which this is not true: a pseudovector (sometimes called anaxial vector) is formed by a cross product between two ordinary vec-tors (which are sometimes called polar vectors) and a pseudoscalar isformed via a scalar triple product; these behave oppositely. In summary:

P(scalar) = scalar P(vector) = −vector

P(pseudoscalar) = −pseudoscalar P(pseudovector) = pseudovector.

Since P2 = I then the group formed by the elements I,P under mul-tiplication is isomorphic to Z2, the cyclic group of order 2, pretty muchthe simplest non-trivial group you can imagine. As for C, this means

The group multiplication table for theparity operator P and the identity op-erator I:

I P

I I P

P P I

the eigenvalues are ±1. Scalars and pseudovectors have parity eigen-value +1, while vectors and pseudoscalars have parity eigenvalue −1.For example, the photon is an excitation in a massless vector field andtherefore has intrinsic parity −1. The pion is described by a pseudoscalarfield and also has parity −1. We will later find (from using the Diracequation and considering what are called spinor fields) that the parityof a fermion is opposite to that of its antiparticle.

In general, the problem we face is that our quantum fields can bescalar-valued fields or vector-valued fields or even more complicated ob-

15.3 Time reversal 137

jects. Thus a parity transformation will affect the nature of the fielditself, as well as acting on the coordinates in which the field is defined.

Example 15.2

For now, let’s just keep things simple and focus on a scalar field. If we operate on ascalar field then φ(t,x) → φ(t,−x). Let’s look at what it does to the creation andannihilation operators. We have

P−1φ(t,x)P = φ(t,−x) =

Z

p

ape−i(Et+p·x) + a†pei(Et+p·x), (15.7)

and this will work if

P−1apP = a−p , P−1a†pP = a†−p , (15.8)

which means that in the case of scalar fields the parity operator simply reverses themomentum on the creation and annihilation operators.

15.3 Time reversal

‘If I could turn back time. . .’ With the time-reversal operator T youcan! It maps a scalar field φ(t,x) → φ(−t,x) and so while it leaves theposition operator well alone:

T−1xT = x, (15.9)

it reverses momentum coordinates:

T−1pT = −p. (15.10)

This means that the commutation relation [x, px] = i will only be pre-served if T−1iT = −i, and hence T must be antiunitary.4 The archetypal

4An antiunitary operator has T2 = −I(compare this with a unitary operatorU which satisfies U2 = I). Note alsothat the condition

T−1iT = −i

can be written iT = −Ti, meaning thati anticommutes with T. This looks non-sensical at first sight: how can a simplenumber like i anticommute with any-thing? However, T operates on every-thing to the right of it, and if there isa factor of i there, its sign will be re-versed. In the simple case that T = K,where K is the complex conjugation op-erator, this condition is easily demon-strated since

T(iψ) = −iTψ,

where ψ is any function.

antiunitary operator is K, the complex conjugation operator and in factyou can make a general antiunitary operator by forming the product ofK and U, where U is a unitary operator. Accordingly we write

T = UK, (15.11)

which means that5 U = TK.

5By postmultiplying both sides ofeqn 15.11 by K.

Example 15.3

(i) For spinless particles you can take U to be the identity and T = K. This is not sosurprising since the complex conjugate of the Schrodinger equation

Hψ = i∂ψ

∂t(15.12)

is

Hψ∗ = −i∂ψ∗

∂t= i

∂ψ∗

∂(−t) , (15.13)


and so the combined operation of time reversal and complex conjugation leaves theform of the Schrodinger equation invariant. Thus indeed it looks like K and T arethe same thing.(ii) Adding spin complicates things because angular momentum is reversed when timereverses (it corresponds in some general sense to things going round in a particularsense as a function of time). Thus

T−1ST = −S, (15.14)

where S is the spin operator. The operation of complex conjugation is a bit morecomplicated since66Remember the form of the Pauli spin

matrices in which only σy has imagi-nary components, while σx and σz arereal.

K−1SxK = Sx, K

−1SyK = −Sy , K−1SzK = Sz . (15.15)

An appropriate form for U is U = exp(−iπSy) so that

T = exp(−iπSy)K. (15.16)

This operator has the property7 T2 = ±1. For a system containing many particles7The proof of this goes as follows: T2 =UKUK = UU

∗ but since U is unitaryU† = (U∗)T = U−1 and so T2 =U(UT)−1 = X where X is a diagonalmatrix containing only phase factors.Thus U = XUT and thus UT = UXT =UX and thus U = XUX and hence thediagonal elements of X are only ±1.Thus T2 = ±1.

with spin, the time-reversal operator can be written

T =Y

i

exp(−iπSiy)K, (15.17)

where the product is evaluated over all the particles. Let’s put some flesh on thesebones. For a single electron, the spin is one-half and so exp(−iπSy) will have eigen-values ±i and so T2 = −1. This will also be the case for an odd number of electronsbut if the number of electrons is even then T2 = 1.

These expressions have some interesting consequences for the case in which T2 =−1 (for an odd number of electrons). For a Hamiltonian H that is invariant undertime-reversal (so that H commutes with T), both a state |ψ〉 and its time-reversedversion T|ψ〉 will have the same energy. Are they the same state? If they were, thenT|ψ〉 = α|ψ〉 where α is a complex number. But in that case8 T2|ψ〉 = Tα|ψ〉 =8We are using the fact that T is antiuni-

tary and this implies Tα|ψ〉 = α∗T|ψ〉. α∗T|ψ〉 = |α|2|ψ〉 but since we have assumed T2 = −1 then we would deduce |α|2 =−1 and that’s not possible since |α|2 > 0. We have arrived at a contradiction of ourinitial assumption, so we can deduce that |ψ〉 and T|ψ〉 are linearly-independent statesand are known as a Kramers doublet. We have deduced Kramers’ theoremwhich states that the energy levels of a time-reversal invariant system with an oddnumber of electrons are n-fold degenerate where n is even. Essentially the energylevels come in pairs of Kramers doublets, and you can only split these pairs byintroducing a perturbation that breaks time-reversal, such as a magnetic field.Hendrik Kramers (1894–1952).

Example 15.4

Reversing time t using T and reversing spatial coordinate x using P produces areversal of spacetime x. On a scalar field this can be written

(PT)−1φ(x)(PT) = φ(−x). (15.18)

This operation leaves the creation and annihilation operators unchanged:

(PT)−1ap(PT) = ap , (PT)−1a†p(PT) = a†p , (15.19)

and this occurs because a parity transformation flips all the momenta and the timereversal flips them all back. In the mode expansion of the scalar field the only effectis to flip the sign of i in the exponentials. It thus acts as an operator to perform acomplex conjugation of the scalar field.

15.4 Combinations of discrete and continuous transformations 139

Each of C, P and T are conserved in many-particle physics processes,but not all. Famously, P is ‘violated’ in the weak interaction. However,a quantum field theory that satisfies a fairly minimal set of assump-tions (ones in which L is Lorentz invariant, local, Hermitian and normalordered) possesses the symmetry CPT. In other words, if you reversespacetime and replace particles by antiparticles, the theory should beinvariant. The proof of this CPT theorem is (in brief outline only)based on showing that (CPT)−1L(x)(CPT) = L(−x), and hence deduc-ing that CPT commutes with the Hamiltonian and is hence a symmetry.9

9The first part can be deduced usingthe fact that L is a Lorentz scalar andhence any term in it contains terms inwhich all tensor indices are contracted,so the terms contain an even numberof tensor indices and under CPT theseproduce an even number of minus signs.More details can be found in the booksby Itzykson and Zuber, Srednicki andWeinberg.

A consequence of the CPT theorem is that the mass and lifetime of anyparticle is identical to that of its antiparticle. If experiment ever providesconclusive evidence of a deviation from this then CPT symmetry willhave been shown not to hold in some particular case, casting doubt onLorentz invariance. So far, CPT symmetry has survived the test.

15.4 Combinations of discrete and

continuous transformations

We have previously examined the group of three-dimensional rotations.These can be represented by orthogonal 3×3 matrices (let’s call them R)but with the condition that the determinant of the matrix is detR =+1. The group which describes this is known as SO(3), the special

orthogonal group, the ‘special’ being the determinant being +1. Theserotations are called proper rotations.

The parity transformation can be represented by a matrix−1 0 00 −1 00 0 −1

, (15.20)

[or, for short, diag(−1,−1,−1)] because it maps (x, y, z) → (−x,−y,−z)and this matrix clearly has determinant −1. By combining this singletransformation with our group SO(3), we get a much larger group O(3),the group of all orthogonal 3 × 3 matrices.10 This includes the proper

10Note that orthogonality implies

RTR = I. Hence taking the de-terminant, detR × detRT = 1.Now detR = detRT and hence(detR)2 = 1 or detR = ±1, showingthat O(3) contains matrices with bothdetR = +1 and detR = −1.

rotations of SO(3) (with detR = +1) as well as improper rotations

in which detR = −1. Improper rotations are ordinary, proper rotationsfollowed by a parity operation (or equivalently by a reflection).

det= +1 det= −1

Fig. 15.4 The group O(3) consists oftwo disjoint sets, which can be picturedas two disconnected islands. These setsare distinguished by their elements ei-ther having determinant +1 [this islandis SO(3)] and −1. Only the first is-land is independently a group becausethe identify transformation has deter-minant +1 and lives exclusively on theleft island.

Example 15.5

(i) Reflection by a mirror in the x-y plane is accomplished by diag(1, 1,−1). Thishas determinant −1 (an improper rotation).

(ii) Rotation about the z-axis by π is accomplished by diag(−1,−1, 1). This hasdeterminant +1 (proper rotation).

(iii) The parity operation, diag(−1,−1,−1), is accomplished by the product of (i)and (ii). It is therefore an improper rotation.


The group SO(3) is a connected group, meaning that you can imaginecontinuously wandering through all the elements of the group; the groupis not the union of two or more disjoint sets. The same cannot be said ofO(3). It consists of two disjoint sets, one set with determinant +1 [whichis SO(3)] and the other with determinant −1 and is a disconnected

group (see Fig. 15.4).We find something similar when we consider the Lorentz group, the

group containing all rotations, reflections and Lorentz boosts. This isoften given the symbol O(1, 3), emphasizing that it has time-like (one-dimensional) and space-like (three-dimensional) parts. This consists offour topologically separated components because as well as P we haveto consider T. In a four-dimensional representation we have

P = diag(1,−1,−1,−1), T = diag(−1, 1, 1, 1). (15.21)

The subgroup of the Lorentz group that doesn’t reverse spatial or tem-poral coordinates is the proper, orthochronous Lorentz subgroup

SO+(1,3). This connected group is one of the four components of theLorentz group. The other three components can be accessed by takingelements of SO+(1,3) and (i) including P, or (ii) including T, or (iii)including both P and T (see Fig. 15.5).

SO+(1, 3)

PP

T

T

Fig. 15.5 The Lorentz group O(1, 3)consists of four separated components.The identity element resides in thecomponent SO+(1, 3), the proper, or-thochronous Lorentz subgroup. Theother ‘islands’ can be reached using theoperators P, T or the combination PT.

While wading in these deep topological waters, it is worth makingsome important remarks about simple rotations which will have impor-tant consequences later. Let’s consider proper rotations which we havebeen representing using the group SO(3). Let’s think about the topol-ogy of this. A rotation is characterized by a rotation axis and an angle.Thus we could imagine a sphere with radius π and then any point in-side the sphere could represent a rotation: the direction from the originwould determine the rotation axis and the distance from the origin woulddetermine the rotation angle. However, a rotation of π about a partic-ular axis is equivalent to a rotation of −π about the same axis. Thusthe topology of SO(3) is equivalent to that of a ball in which antipodalpoints on the surface are identified.

(a)

(b)

Fig. 15.6 Closed paths in the spaceSO(3), represented as a ball in whichantipodal points on the surface areidentified. (a) A path from the northto south pole cannot be continuouslydeformed to a point, while (b) a doublepath can.

This topological space is clearly connected but is not simply con-

nected, which means that a closed path through the space cannot becontinuously shrunk to a point. To show this consider a path runningfrom the north pole to the south pole of the sphere [Fig. 15.6(a)]. Youcan’t deform this continuously to a point. If you move the path nearthe north pole to the right, the point near the south pole moves to theleft. You can’t get the two to join up. However, if you run the pathtwice from the north pole to the south pole, you can do it [Fig. 15.6(b)].This shows that a 4π rotation is continuously deformable to a point,while a 2π rotation is not. This is the basis of the famous Dirac scis-sors parlour-game trick, an amusing demonstration in which you threadstring around the eyes of a pair of scissors and a nearby chair. Two fullrotations of the scissors leave the string apparently highly tangled butwith a few deft movements which do not involve further rotations youcan untangle the string. The string serves to make the trick more com-plicated than it need be, and a more transparent version can be carried

15.4 Combinations of discrete and continuous transformations 141

out using a scarf or a belt, with one end fixed by, e.g. placing it under abook, as shown in Fig. 15.7(a). The free end is rotated by two full turns(i.e. by 720 degrees) in the same sense and looks very twisted, as shownin Fig. 15.7(b). It can be untwisted without rotating the free end in theopposite sense, simply by passing the free end around the middle of thebelt and pulling taut, see Fig. 15.7(c)-(f). Try it!

To show this point up more clearly in group theory, we could alsouse another group to represent rotations: SU(2). This is the special

unitary group of 2 × 2 matrices, where again the ‘special’ means thedeterminant is one. This can be used to rotate spinors.

(f)

(d) (e)

(c)

(b)

(a)

Fig. 15.7 Dirac’s scissor trick is moreeasily demonstrated with a belt withone end free and the other end heldfixed by placing it under a book. [Fig-ure from S. J. Blundell, Magnetism: AVery Short Introduction, OUP (2012).]

Example 15.6

To rotate by an angle θ about an axis defined by n we can use the rotation matrixwhich we’ve previously written as R(θ). Here we will write θ = θn, and we willemploy a useful identity which is

σ · n =

„nz nx − iny

nx + iny −nz

«

, (15.22)

where σ = (σx, σy , σz) are the Pauli spin matrices. Our rotation matrix, which wewill now write as R(n, θ), is given by

R(n, θ) = exp

„

−iθ

2σ · n

«

= I cosθ

2− i sin

θ

2σ · n, (15.23)

where I is the identity matrix. These matrices have an interesting feature, which isthat while as you would expect

R(n, 0) = I, (15.24)

you also haveR(n, 2π) = −I, (15.25)

and you need a full 4π rotation to recover

R(n, 4π) = I. (15.26)

The previous example shows that SU(2) is actually a double cover ofSO(3), meaning that for a particular rotation there are two represen-tations in SU(2) for every one representation in SO(3). For example,the identity (no rotation at all) is just diag(1, 1, 1) in SO(3), but isrepresented by both diag(1, 1) and diag(−1,−1) in SU(2).

A spinor can be written as a two-component entity

(ab

)where a and

b are complex numbers and |a|2 + |b|2 = 1. Writing a = x0 + ix1 andb = x2 + ix3 where xi are real we see that this condition is equivalentto x2

0 + x21 + x2

2 + x23 = 1 and so SU(2) is isomorphic to S3, that is

the 3-sphere.11 Thus SU(2) is simply connected, in contrast to SO(3).

11It may be helpful to remember:

• S1 is a circle, a one-dimensionalspace that we normally draw ona two-dimensional piece of pa-per. It can be described with theequation x2 + y2 = 1.

• S2 is a 2-sphere, or more com-monly just a sphere (meaningthe surface of a ball), a two-dimensional space that can beembedded in three-dimensions.It can be described with theequation x2 + y2 + z2 = 1.

• S3 is a 3-sphere, a three-dimensional space that can beembedded in four-dimensions. Itcan be described with the equa-tion x2 + y2 + z2 + w2 = 1.

For more details on topological nota-tion, see Chapter 29.

The formal relationship between the two groups is SO(3) ∼= SU(2)/Z2,meaning that SO(3) is a quotient group.

These arguments can be generalized to the connected component ofthe Lorentz group: SO(1, 3) ∼= SL(2,C)/Z2, where SL(2,C) is the groupof 2 × 2 complex matrices with unit determinant. This gives you someflavour of the way in which the mathematical structure of the spacesdescribing the spacetime transformations can be described using grouptheory.


circle S1 (x0)2 + (x1)2 = 1 [S1 ∼= U(1)]sphere S2 (x0)2 + (x1)2 + (x2)2 = 1n-sphere Sn (x0)2 + (x1)2 + · · · + (xn)2 = 1n-torus Tn S1 × S1 × · · · × S1 (n ≥ 2)

real n× n matrices

general linear group GL(n,R) n× n real matricesspecial linear group SL(n,R) GL(n,R) with detM = 1orthogonal group O(n) GL(n,R) with MMT = MTM = Ispecial orthogonal group SO(n) O(n) with detM = 1Lorentz group O(1, 3) GL(n,R) with MgMT = gspecial Lorentz group SO(1, 3) O(1, 3) with detM = 1

complex n× n matrices

general linear group GL(n,C) n× n complex matricesspecial linear group SL(n,C) GL(n,C) with detM = 1unitary group U(n) GL(n,C) with MM† = M†M = Ispecial unitary group SU(n) U(n) with detM = 1

Table 15.1 Some useful mathematical objects.

Chapter summary

• Discrete symmetries include charge conjugation C (which changesall particles into their antiparticles and vice versa), parity P whichinverts spatial coordinates and time-reversal T which reverses time.

• The combined symmetry CPT holds for a Lorentz invariant, local,Hermitian quantum field theory.

• For reference, some useful mathematical objects are tabulated inTable 15.1.

Exercises

(15.1) Why is the reaction π0 → γ + γ + γ not allowed?

(15.2) Classify the following as scalars, pseudoscalars, vec-tors (polar vectors) or pseudovectors (axial vec-tors): (a) magnetic flux; (b) angular momentum;(c) charge; (d) the scalar product of a vector and a

pseudovector; (e) the scalar product of two vectors;(f) the scalar product of two pseudovectors.

(15.3) Find representations for the spinor rotation matri-ces (a) R(x, θ), (b) R(y, θ) and (c) R(z, θ).

Part IV

Propagators and

perturbations

Quantum field theory is particularly challenging when one has to dealwith interacting systems. One method to calculate results uses perturba-tion theory and this can be evaluated using various techniques involvingGreen’s functions and Feynman diagrams. This part introduces some ofthese techniques.

• In Chapter 16 we introduce Green’s functions and show how theyare related to a propagator, the amplitude that a particle in somespacetime position will be later found at another spacetime posi-tion.

• We connect Green’s functions to propagators in a quantum fieldpicture in Chapter 17 and introduce the Feynman propagator whichinvolves time-ordering of operators. This is applied to Yukawa’smodel of virtual particle exchange.

• Interactions are added to the picture in Chapter 18 and in partic-ular the idea of the S-matrix. We work this out in the interactionrepresentation, another way of writing down states and operatorswhich turns out to be very useful for these problems. These ideasare applied to a perturbation expansion and use is made of Wick’stheorem which provides a method for simplifying products of op-erators.

• In Chapter 19 we introduce Feynman diagrams which provide awonderful pictorial way of visualizing terms in the perturbationexpansion. Rather than evaluating umpteen integrals we can sum-marize many complex calculations in diagrammatic form and eval-uate the results using Feynman rules for a particular situation.

• These ideas are applied to scattering theory in Chapter 20 for asimple ψ†ψφ model which serves as a toy model of QED and thescattering cross-section is defined and then evaluated.

16Propagators and Green’s

functions

16.1 What is a Green’s function?144

16.2 Propagators in quantum me-chanics 146

16.3 Turning it around: quantummechanics from the propaga-tor and a first look at pertur-bation theory 149

16.4 The many faces of the prop-agator 151

Chapter summary 152

Exercises 152

When will the world know that peace and propagation arethe two most delightful things in it?Horace Walpole, 4th Earl of Oxford (1717–1797)

One way to do quantum mechanics is to calculate a wave function andoperate on it with quantum operators. Another way is to directly con-sider amplitudes for a given process, such as ‘the amplitude that myparticle starts at point y at a time ty and ends up at point x at timetx’, which we write down1 as 〈x(tx)|y(ty)〉. This amplitude is known

1As Feynman notes, quantum ampli-tudes are like Hebrew in that you readthem right-to-left.

as a propagator. We’ve spent time in previous chapters removing thewave function from its role as the most useful entity in quantum physicsand we’ll see that propagators represent an alternative to wave functionsand enable us to extract all of the same information. In fact, we’ll seethat wave functions are actually special cases of propagators and con-tain less information! For our purposes, propagators represent the mosteconomical way to calculate all of the properties of quantum fields in aninteracting system of many particles. In addition propagators for sin-gle particles have a neat mathematical property: they are the Green’s

functions2 of the equation of motion for a particle.2These are named after George Greenof Sneinton, Nottinghamshire (1793–1841) who was perhaps the greatestself-taught mathematical physicist ofmodern times. His achievements are allthe more remarkable since, until 1829,his day job consisted of running a wind-mill. While still a miller he publishedhis most celebrated work ‘An Essay onthe Application of Mathematical Anal-ysis to the Theories of Electricity andMagnetism’ (1828). The essay includedthe first use of what are now calledGreen’s functions.

The plan for this chapter is as follows: (i) we’ll explain what’s meantby a Green’s function; (ii) we’ll demonstrate that propagator amplitudesare the Green’s’ functions of quantum mechanics: they tell us the am-plitudes for a particle to start at a spacetime point (y, ty) and then bedetected at a point (x, tx); (iii) finally we’ll show that quantum mechani-cal perturbation theory is conveniently expressed in terms of propagatorsand that these propagators are most conveniently expressed in terms ofcartoons.

16.1 What is a Green’s function?

Here’s a very general looking differential equation formed from a lineardifferential operator L:

L x(t) = f(t). (16.1)

The Green’s function G(t, u) of the linear operator L is defined by theequation

LG(t, u) = δ(t− u). (16.2)

16.1 What is a Green’s function? 145

Notice that the Green’s function G(t, u) is a function of two variables, tand u.

Example 16.1

Let’s consider a familiar example from mechanics. We’ll take an oscillator of mass mand spring constant K evolving under the influence of a time-dependent force f(t).The equation of motion for this is

md2

dt2x(t) +Kx(t) = f(t), (16.3)

so here the linear operator is L = m d2

dt2+K. We can build up a force function f(t)

by adding together lots of delta functions.

f(t) =

Z ∞

0du f(u)δ(t− u), (16.4)

where u us a dummy variable which allows us to break f(t) into deltas. This is, ofcourse, merely an example of superposition and is allowed because our differentialequation is linear.

Instead of hitting the problem head-on, we sidestep and solve eqn 16.3 for just oneof the delta functions. We’ll call this solution the Green’s function G(t, u):

»

md2

dt2+K

–

G(t, u) = δ(t− u). (16.5)

The complete solution to eqn 16.3 is then given by the integral of the Green’s function,weighted by the force function f(u):

x(t) =

Z ∞

0duG(t, u)f(u). (16.6)

The solution is therefore simply a sum of the responses of the system in x to thestimuli f(u).

Why is the solution to eqn 16.3 given by eqn 16.6? Start by acting on eqn 16.6with the operator L (which passes through the integral sign on the right-hand side),

L x(t) =

Z

du LG(t, u)f(u)

=

Z

du δ(t− u)f(u) by eqn 16.2

= f(t) by eqn 16.4, (16.7)

which is where we started! This implies we can solve an inhomogeneous differentialequation by finding the Green’s function G(t, u) and then integrating over f(u) toget the solution x(t).

It’s now clear why the Green’s function G(t, u) needs two arguments:one is the variable we’re interested in (here it’s the time t), the other uis the variable that describes the position of the delta function δ(t− u)that we use to define the Green’s function via LG(t, u) = δ(t− u). Theinhomogeneous part of the equation [here f(t)] is built up from a setof delta functions weighted by an amplitude function f(u), that is afunction of the dummy variable u.

Now we know what use they are, we can find the Green’s functionsfor some commonly encountered L operators.3

3Remember that δ(x) = δ(−x) so wecan swap the order of the arguments inthe delta function for convenience.

146 Propagators and Green’s functions

Example 16.2

Examples of finding some Green’s functions.

• The Green’s function for L = ∇2 can be read off using an argument

from electromagnetism. We know Poisson’s equation (reverting to SI units):

∇2V (x) = − ρ(x)

ǫ0, which for a point charge of magnitude unity, located at u,

readsǫ0∇

2V (x) = −δ(3)(x− u). (16.8)

The potential that solves this equation is known from electromagnetism to beV (x) = 1

4πǫ0|x−u| , and so

∇2

»1

4π|x− u|

–

= −δ(x− u), (16.9)

or

G(x,u) = − 1

4π|x− u| . (16.10)

• The Green’s function for L = (∇2 + k2)is defined by`∇

2 + k2´Gk(x,u) = δ(3)(x− u). (16.11)

The solution which will interest us is

Gk(x− u) = − ei|k||x−u|

4π|x− u| , (16.12)

which describes the amplitude of an outgoing, spherical wave.

16.2 Propagators in quantum mechanics

We know that the equation that governs the change of a wave func-tion φ(x, t) is the Schrodinger equation Hφ(x, t) = i∂φ(x,t)

∂t , where H isan operator function of x and we’ll only consider (1 + 1)-dimensionalspacetime for now.4 Why might the Green’s function of the Schrodinger

4(1+1)-dimensional spacetime meansone spatial dimension and one timedimension. We normally think about(3+1)-dimensional spacetime withthree spatial dimensions and one timedimension.

equation be useful, and what interpretation might such a function have?The real beauty of a Green’s function is the property that we had in

eqn 16.6, namely that

φ(x, tx) =

∫dy G+(x, tx, y, ty)φ(y, ty). (16.13)

(The + sign notation will be explained shortly.) Here the Green’s func-tion takes a wave function at some time and place and evolves it toanother time and place. To find out how it does this we need to inte-grate over space (but not time).5

5We don’t need to integrate overall time because the multiplicationG(x, tx, y, ty)φ(y) does that part au-tomatically, as in the case of thetime-evolution operator which, as we’llshortly see, is closely related to theGreen’s function.

In fact, we say that the Green’s function propagates the particle fromthe spacetime point (y, ty) to (x, tx), which explains why we call G+ apropagator. Here the plus-sign superscript in G+ means that we con-strain tx > ty and we define G+ = 0 for tx < ty, which prevents particlesgoing back in time.6

6With this in mind we call G+ thetime-retarded Green’s function,defined as

G+ =

(

G tx > ty

0 tx < ty ,(16.14)

or more simply

G+ = θ(tx − ty)G. (16.15)

We can also define the time-advancedGreen’s function by

G− =

(

0 tx > ty

G tx < ty ,(16.16)

or more simply

G− = θ(ty − tx)G. (16.17)Looking at eqn 16.13 we can interpret φ(y, ty) as the amplitude to find

a particle at (y, ty) and φ(x, tx) as the amplitude to find a particle at

16.2 Propagators in quantum mechanics 147

(x, tx). It follows that the propagator G+(x, tx, y, ty) is the probabilityamplitude that a particle in state |y〉 at time ty, ends up in a state |x〉at time tx. The interpretation means that the Green’s function may bewritten

G+(x, tx, y, ty) = θ(tx − ty)〈x(tx)|y(ty)〉. (16.18)

Note that, using this language, our old friend the wave functionφ(x, tx) = 〈x|φ(t)〉 is simply the amplitude that a particle is found at(x, tx) irrespective of where it started. The propagator therefore con-tains more information, since it cares where the particle started.

Example 16.3

From the basic definition of the propagator G+(x, tx, y, ty) = θ(tx − ty)〈x(tx)|y(ty)〉we can invoke the time-evolution operator so that the time dependence is taken awayfrom the states

G+(x, tx, y, ty) = θ(tx − ty)〈x|U(tx − ty)|y〉= θ(tx − ty)〈x|e−iH(tx−ty)|y〉. (16.19)

We can also expand the amplitudes in terms of the eigenstates of the operator H,which we call |n〉, which have eigenvalues En,

G+(x, tx, y, ty) = θ(tx − ty)〈x|e−iH(tx−ty)|y〉= θ(tx − ty)

X

n

〈x|e−iH(tx−ty)|n〉〈n|y〉. (16.20)

Remembering that 〈x|n〉 is just a fancy way of writing a wave function φn(x), we seethat, in general, the propagator may be written in terms of the eigenfunctions of Has

G+(x, tx, y, ty) = θ(tx − ty)X

n

φn(x)φ∗n(y)e−iEn(tx−ty), (16.21)

allowing us to relate wave functions and propagators.

So far we’ve assumed the propagator G+(x, tx, y, ty) is some sort ofGreen’s function of the Schrodinger equation: now we’ll confirm this.We define the Green’s function for the Schrodinger equation by consid-ering G+(x, tx, y, ty) as a function of x and tx only and say[Hx − i

∂

∂tx

]G+(x, tx, y, ty) = −iδ(x− y)δ(tx − ty) = −iδ(2)(x− y),

(16.22)where Hx only touches x-coordinates and not y-coordinates. Here wehold y and ty fixed as dummy variables. Finally, we want to confirmthat the amplitude 〈x(tx)|y(ty)〉 is truly the Green’s function of theSchrodinger equation.

Example 16.4

Here’s how is goes. The Green’s function is given by

G+(x, tx, y, ty) = θ(tx − ty)〈x(tx)|y(ty)〉 = θ(tx − ty)X

n

φn(x)φ∗n(y)e−iEn(tx−ty),

(16.23)


where we’ve explicitly included the θ function to ensure that the particle can’t prop-agate backward in time. Substituting the Green’s function into eqn 16.22, we findthat we want to work out

»

Hx − i∂

∂tx

–

θ(tx − ty)X

n

φn(x)φ∗n(y)e−iEn(tx−ty). (16.24)

Take this in two stages. Stage I: We use the fact that ∂∂tx

θ(tx − ty) = δ(tx − ty),to find that the time derivative acting on the retarded Green’s function gives us

i∂

∂txG+ = iδ(tx − ty)

X

n

φn(x)φ∗n(y)e−iEn(tx−ty)

+θ(tx − ty)X

n

Enφn(x)φ∗n(y)e−iEn(tx−ty). (16.25)

Stage II: Now considering Hx acting on the Green’s function, we note thatHxφn(x) = Enφ(x) is the only thing that is affected by the Hx operator. We have

HxG+ = Hxθ(tx − ty)〈x(tx)|y(ty)〉 = θ(tx − ty)

X

n

Enφn(x)φ∗n(y)e−iEn(tx−ty).

(16.26)Putting all of this together, we obtain

»

Hx − i∂

∂tx

–

G+(x, tx, y, ty) = −iδ(tx − ty)X

n

φn(x)φ∗n(y)e−iEn(tx−ty)

= −iδ(tx − ty)δ(x− y), (16.27)

which is what we set out to prove.

Now let’s try to find a propagator for free particles. Our first stop willbe the free-particle propagator in position space and the time domain,which we call G+(x, tx, y, ty).

Example 16.5

A non-relativistic, free particle has a Hamiltonian H = p2/2m, with eigenfunctions

φ(x) = 1√L

eipx and eigenvalues Ep = p2

2m. Treating p as a continuous variable, we

start with eqn 16.21, and transforming the sum into an integral7 we have7We use the replacement

X

n

→ L

Zdp

2π.

Note for later that we will often makethe replacement in three dimensions

X

p

→ VZ

d3p

(2π)3,

where V is the volume.

G+(x, tx, y, ty) = θ(tx − ty)L

Zdp

2πφp(x)φ

∗p(y)e

−iEp(tx−ty)

= θ(tx − ty)

Zdp

2πeip(x−y)e−i p2

2m(tx−ty). (16.28)

This integral can be done: it’s a Gaussian integral, of which, much more later (see

Chapter 23). For now, we use the resultR∞−∞ dx e−

ax2

2+bx =

q2πa

eb2

2a with a =i(tx−ty)

mand b = i(x− y). This gives

G+(x, tx, y, ty) = θ(tx − ty)

rm

2πi(tx − ty)e

im(x−y)2

2(tx−ty) . (16.29)

There are more examples of finding propagators in the exercises at theend of this chapter.

16.3 Turning it around 149

16.3 Turning it around: quantum

mechanics from the propagator and a

first look at perturbation theory

So far we’ve used things we know about quantum mechanics to derivesome properties of a single-particle propagator. In fact, we want to turnthis around. If we start from the propagator, what can we learn aboutthe particle? This is most easily seen by considering yet another formof the Green’s function, again as a function of space but this time in thefrequency/energy domain. This one looks like

G+(x, y,E) =∑

n

iφn(x)φ∗n(y)

E − En. (16.30)

(You’re invited to prove this relation in Exercises 16.2.)We can notice two things about this equation when we consider it as

a function of the complex8 variable E: (i) the singularities (or poles) on 8Readers unfamiliar with complexanalysis can refer to Appendix B.the real axis occur when E = En, that is, when the parameter E equals

the energies of the eigenstates φn(x); and (ii) the residues at the polesare (i times) the wave functions. We see that by being able to writedown the Green’s function of a system we have access to the energiesof the system and its wave functions.9 Propagators, with their ‘from 9Actually, to ensure causality it’s nec-

essary to replace En with En− iǫ so wehave

G+(x, y, E) = limǫ→0+

X

n

iφn(x)φ∗n(y)

E − En + iǫ.

(16.31)

here to there’ definition, also have the appealing property that theycan be drawn in a cartoon form showing a particle travelling from y tox. This isn’t quite as trivial as it sounds: Bohr was doubtful whetherthe trajectory of a quantum particle could be thought about due toposition–momentum uncertainty, but it can. The quantum propagator,which gives this trajectory some meaning, is illustrated in Fig. 16.1.

y, ty

x, tx

G+(x, y)

Fig. 16.1 A cartoon representation ofthe propagator. It will turn out to bethe basic unit of the Feynman diagram.

This doesn’t explain what all the fuss is about. Why are Green’sfunctions so useful? One place to look is perturbation theory. In mostinteresting cases we can’t solve a quantum mechanical problem exactly.What we do is write the problem by splitting up the Hamiltonian intotwo parts: H = H0 + V , where H0 is the solvable part and V is theperturbing potential. Perturbation theory involves the tedious task ofevaluating changes to eigenfunctions and eigenvalues in a series of in-creasingly complicated corrections.

Let’s look again at our Green’s functions and see if they lend anyinsight to the perturbation problem. For simplicity, we’ll write theoriginal Green’s function equation in a symbolic, matrix-like form as(H − E)G = −1.10 (Remember throughout that G just describes a par-

10Objects like G should be treated withcaution: they’re not well defined math-ematical objects, just symbols com-bined in the shape of the original equa-tions that we’re using to suggest rela-tionships. Needless to say these argu-ments are repeated rigorously in the lit-erature.

ticle propagating from y to x.) The symbolic Green’s function equationis solved by writing G = 1

E−H, which bears a resemblance to the form in

eqn 16.31.Green’s functions allow us to interpret a perturbation problem in

terms of a propagating particle. We think of the solvable part of theproblem in terms of a particle propagating from point to point. Wethink of the perturbation V as a scattering process that interruptsthe propagation. To visualize this we’ll write a symbolic expression for


the Green’s function in the presence of a perturbation as G = 1E−H0−V

,where we’ve used H = H0+V. In perturbation problems G (with no sub-script) is called the full propagator. Just considering the solvable partof the Hamiltonian H0 we can write G0 = 1

E−H0, that is, we can find the

Green’s function for a solvable problem. The ‘0’ subscript means thatwe’re talking about a particle freely propagating, with no scattering andso, in general, G0 is known as the free propagator. If we now considerthe perturbed problem and try to expand the full propagator G as afunction of the free propagator G0 we can sayThere’s a matrix identity that says

1

A+B=

1

A− 1

AB

1

A+

1

AB

1

AB

1

A− . . .

G =1

E − H0 − V=

1

E − H0+

1

E − H0V

1

E − H0

+1

E − H0V

1

E − H0V

1

E − H0+ . . . (16.32)

The point here is that, in general, we can’t calculate 1/(E − H0 − V).However, we can calculate each of the terms on the right-hand side ofthis expression and we might hope that G may be well approximated byjust the first few of these.

Switching back from symbolic relations to real functions11 shows that

11In the rest of this section we willwrite the functions as G, rather thanG+, since the expressions in eqns 16.33and 16.34 apply generally to G, G+ andG−.

we’ve succeeded in writing the full Green’s function as an expansion infree propagators and potentials. The form of the expansion is

G = G0 +G0V G0 +G0V G0V G0 + . . . (16.33)

This is a memorable equation and easily turned into pictures as shownin Fig. 16.2. The perturbation problem, previously involving an infiniteseries of unmemorable terms, has been turned into a vivid picture ofparticle scattering. The amplitude for a particle to go from y to x isG. This is a superposition of the amplitude for moving freely from yto x (which is G0); added to the amplitude for making the trip with asingle scattering event at some point along the way G0V G0; added to theamplitude for making the trip with two scatters G0V G0V G0 and so on.Thus the algebra of Green’s functions is highly amenable to visualizationof the physical processes.

G G0 G0V G0 G0V G0V G0

G0

G0

G0

V

G0

G0

G0

V

V

Fig. 16.2 The scattering processesthat lead to the full propagator in per-turbation theory.

Example 16.6

The perturbation expansion in eqn 16.33 is just a geometric series which we canrewrite as

G = G0 (1 + V G0 + V G0V G0 + . . .)

=G0

1 − V G0=

1

G−10 − V

. (16.34)

This is known as Dyson’s equation. This equation is also derived from the algebraicmanipulation of the cartoons, as shown in Fig. 16.3. This concept will turn out to bevery useful. Notice that it is non-perturbative: we don’t consider just the first fewof the terms in a perturbation series, we sum all of them to infinity.

V

V

V

V

VV

V

V

−1

Fig. 16.3 The derivation of Dyson’sequation in graphical form.

16.4 The many faces of the propagator 151

16.4 The many faces of the propagator

So far we have encountered the free-particle propagator in the spaceand time domain G+

0 (x, tx, y, ty) and in the space and energy domainG+

0 (x, y,E). There are other useful forms that we can derive.

Example 16.7

The physical interpretation of G+ allows us to work out another expression for theGreen’s function, this time in the momentum space and in the time domain, so wewrite G+

0 (p, tx, q, ty). Remember the interpretation: we want the amplitude that aparticle that starts off in a state |q〉 will, after a time tx − ty , end up in a state |p〉.Thus12 12Note that, for simplicity in these ex-

amples, we’re using the non-relativisticnormalization so that 〈p|q〉 = δ(p− q).G+

0 (p, tx, q, ty) = θ(tx − ty)〈p|U(tx − ty)|q〉= θ(tx − ty)〈p|q〉e−iEq(tx−ty)

= θ(tx − ty)δ(p− q)e−iEq(tx−ty). (16.35)

We see that the free particle cannot change its momentum state, so having both pand q is redundant, so we can write the same equation in the following shorthand:

G+0 (p, tx, ty) = θ(tx − ty)e

−iEp(tx−ty). (16.36)

Notice that we often miss out the delta functions if we decrease the number of argu-ments we list in G+. Strictly we should write all arguments and delta functions.

For dealing with perturbation theory problems, propagators are oftenmost useful to know in the momentum and energy domain.

Example 16.8

To get to G+0 (p,E) we’ll take a Fourier transform of G+

0 (p, t, 0)

G+0 (p,E) =

Z

dt eiEtG+0 (p, t, 0)

=

Z

dt eiEtθ(t)e−i(Ep−iǫ)t

=−iei(E−Ep+iǫ)t

E − Ep + iǫ

˛˛˛˛˛

∞

0

=i

E − Ep + iǫ. (16.37)

Remember again, that if we are being strict we should write this as

We make the replacement Ep → Ep−iǫto ensure the convergence of the inte-gral at t = ∞. It also effectively ensurescausality is obeyed. See Appendix Band Exercises 16.2 and 16.4 for the de-tails.

G+0 (p,E, q, E′) =

i

E − Ep + iǫδ(p− q), (16.38)

telling us the amplitude for a particle with energy E′ and momentum q to enter anda particle with energy E and momentum p to leave. The delta function conservesenergy and momentum.


As a helpful summary, we assemble a table of all of the flavours of Green’sfunction that we’ve introduced so far.

G+0 (x, tx, y, ty) = θ(tx − ty)

∑

n

φn(x)φ∗n(y)e

−iEn(tx−ty)

G+0 (x, y,E) =

∑

n

iφn(x)φ∗n(y)

E − En + iǫ

G+0 (p, tx, ty) = θ(tx − ty)e

−iEp(tx−ty)

G+0 (p,E) =

i

E − Ep + iǫ

Chapter summary

• Green’s functions are used as propagators which express the am-plitude that a particle goes from spacetime point y to spacetimepoint x. Various forms of the Green’s function are shown in thetable above.

• Dyson’s equation for the full propagator G is the perturbationexpansion: G = G0 +G0V G0 + · · · = (G−1

0 −V )−1 where G0 is thefree propagator and V describes a scattering process.

Exercises

(16.1) (a) Solve the Schrodinger equation to find the wavefunctions φn(x) for a particle in a one-dimensionalsquare well defined by V (x) = 0 for 0 ≤ x ≤ a andV (x) = ∞ for x < 0 and x > a.(b) Show that the retarded Green’s function for thisparticle is given by

G+(n, t2, t1) = θ(t2− t1)e−i

„

n2π2

2ma2

«

(t2−t1). (16.39)

(c) Find G+(n, ω) for the particle.

(16.2) (a) Using a damping factor e−ǫt, show that

G+0 (x, y, E) =

P

n

iφn(x)φ∗n(y)

E−En+iǫ.

(b) The θ-function may be written as an integral

θ(t) = i

Z ∞

−∞

dz

2π

e−izt

z + iǫ. (16.40)

Use this to derive G+0 (p,E) = i

E−Ep+iǫ, without

recourse to a damping factor.

(16.3) Consider the one-dimensional simple harmonicoscillator with a forcing function f(t) =F (ω)e−iω(t−u) described by an equation of motion:

m∂2

∂t2A(t− u) +mω2

0A(t− u) = F (ω)e−iω(t−u).

(16.41)(a) Show that the solution to this equation is

A(t− u) = − F (ω)

m

e−iω(t−u)

ω2 − ω20

+B(t), (16.42)

whereB(t) is the solution to the homogeneous equa-tion of motion.(b) Use this result to show that the general form of

Exercises 153

the Green’s function G(t, u) for a simple harmonicoscillator is given by

G(t, u) = − 1

m

Z

dω

2π

e−iω(t−u)

ω2 − ω20

+B(t). (16.43)

(c) If G(t, u) is subject to the boundary conditionsthat at t = 0 we have G = 0 and G = 0, show that

G+(t, u) =1

mω0sinω0(t− u), (16.44)

for 0 < u < t. Hint: This can be done with aLaplace transform of the differential equation forthe Green’s function, or by using complex analysisto do the integral in the previous problem subject tothe boundary conditions.(d) Show that the trajectory of a particle in a sim-ple harmonic potential subject to a force f(t) =F0 sinω0t is given by F0

2mω20(sinω0t− ω0t cosω0t).

(16.4) (a) By taking the Fourier transform of the equation

(∇2 + k2)Gk(x) = δ(3)(x), (16.45)

show that the momentum Green’s function impliedis

Gk(q) =1

k2 − q2. (16.46)

Note that Gk(q) is undefined when q2 = k2. Thisreflects the fact that there is an ambiguity in Gk(q)in that we don’t know if the wave is incoming, out-going or a standing wave. This is sorted out with

an iǫ factor as shown in the next part.

(b) Take the Fourier transform of G+k (x) = − ei|k||x|

4π|x|to show that, for outgoing waves,

G+k (q) =

1

k2 − q2 + iǫ. (16.47)

Hint: To ensure outgoing waves, you could includea damping factor e−ǫ|x| to damp out waves for|x| → ∞.*(c) (For aficionados of complex analysis.) Take aninverse Fourier transform of G+

k (q) = 1k2−q2+iǫ

toshow that it corresponds to the outgoing wave so-

lution G+k (x) = − ei|k||x|

4π|x| .Hint: On doing the angular part of the integral youshould obtain:

G+k (x)=

−1

4π2

Z ∞

0

|q| d|q|q2−k2−iǫ

„

ei|q||x|−e−i|q||x|

i|x|

«

.

(16.48)Extend the lower limit of the integration to |q| =−∞ and identify the positions of the poles in thecomplex |q| plane. For an outgoing wave, the con-tour over which the ei|q||x| part is integrated mustbe completed in the upper half-plane; the contourover which the e−i|q||x| part is integrated must becompleted in the lower half-plane.*(d) What do you expect for an incoming wave so-lution?

17 Propagators and fields

17.1 The field propagator in out-line 155

17.2 The Feynman propagator156

17.3 Finding the free propagatorfor scalar field theory 158

17.4 Yukawa’s force-carrying par-ticles 159

17.5 Anatomy of the propagator162

Chapter summary 163

Exercises 163

An error does not become truth by reason or multiplied prop-agation, nor does truth become error because nobody sees it.Mahatma Gandhi (1869–1948)

We describe the Universe by combining fields to form a Lagrangian den-sity L[φ(x)]. Our canonical quantization process often allows us to quan-tize these fields leading to a Universe pictured as a vacuum acted on byfield operators like a†p. The excitations of the vacuum that the fieldoperators produce are particles and antiparticles.

Our next task is to come up with a scheme to keep track of theseparticles. Having done away with the wave functions we need someobjects that contain the information about excitations of the system.These objects are propagator amplitudes denoted by the letter G. Thesepropagators are analogous to the Green’s function amplitudes discussedin the previous chapter. The propagators describe the fate of particlesin a field theory.

Interacting and non-interactingtheories

Non-interacting Interacting

H0 H = H0 + H′

|0〉 |Ω〉G0(x, y) G(x, y)

No scattering Scattering An important distinction that we will make in this chapter is betweennon-interacting and interacting theories. Non-interacting theories arethose which can be diagonalized using canonical quantization.1 Canoni-1See Chapter 11.

cal quantization allows us to describe the system in terms of a vacuum |0〉and non-interacting particles in momentum states |p〉 created with oper-ators like a†p. The Hamiltonian for a non-interacting theory is called H0

which may be written as H0 =∑

pEpa†pap, and we have H0|p〉 = Ep|p〉,

where Ep gives the energies of the non-interacting particles. In contrast,interacting theories cannot be diagonalized with canonical quantization.Put another way, these theories contain extra terms describing interac-tions between particles that we cannot transform away with canonicalquantization. We call the ground states of interacting theories |Ω〉 andthe Hamiltonian H. If we act on |Ω〉 with the operator a†p we won’tnecessarily get a state |p〉. Instead we might produce a superposition ofmany particles (whose momenta sum to p). Since most interacting fieldtheories cannot be exactly solved to give, for example, the energies of theexcitations or particles of the theory, we need to develop a perturbationprocess to make approximate calculations. This can be done by analogywith the perturbation process we suggested in the last chapter, whichwas based around a series written out in terms of propagators.

17.1 The field propagator in outline 155

17.1 The field propagator in outline

The definition of the field propagator involves a simple thought experi-ment. We start with our interacting system in its ground state, which wedenote by |Ω〉. The thought experiment works as follows: we introducean extra particle of our choice. We will use this extra particle to probethe system. The new particle is introduced (created) at a spacetimepoint (y0,y). It interacts with the system, possibly causing excitationsin the fields and all manner of complicated things. Then we remove(annihilate) the particle at spacetime point (x0,x) and ask if the systemhas remained in the interacting ground state |Ω〉. We’re interested in theamplitude G+(x, y) for the experiment, given by

G+(x, y) =

⟨Ω

∣∣∣∣(

Particle annihilatedat (x0,x)

)(Particle created

at (y0,y)

)∣∣∣∣Ω⟩.

(17.1)That is, the probability amplitude that the system is still in its groundstate after we create a particle at y and later annihilate it at x. Theamplitude for this process will depend on the complicated interaction ofthe probe particle with the system and, it will turn out, will tell us agreat deal about the system itself. By analogy with the results of theprevious chapter, the amplitude G+(x, y) is called the Green’s functionor propagator.2 So much for definitions. The mathematical object that 2We call the propagator for an interact-

ing theory G(x, y) and its ground state|Ω〉. For a free theory (with no interac-tions) we call the propagator G0(x, y)[or sometimes we will use the equiva-lent notation ∆(x, y)] and the groundstate |0〉.

tells this story of the propagator outlined above is

G+(x, y) = θ(x0 − y0)〈Ω|φ(x)φ†(y)|Ω〉, (17.2)

where the + sign on the left tells us that the particle is created at ybefore being annihilated at x and the θ-function on the right guaranteesthis.

Example 17.1

To see that this function does the job we can break down the action of the opera-tors on the states into stages. Remembering that a Heisenberg operator has a time

dependence defined by φ(t,x) = e+iHtφ(x)e−iHt, and substitution of this gives anexpression for the propagator

G(x, y) = 〈Ω|eiHx0φ(x)e−iH(x0−y0)φ†(y)e−iHy0 |Ω〉. (17.3)

Taking this one stage at a time (and temporarily ignoring the possibility of creatingand annihilating antiparticles), we see that:

• e−iHy0 |Ω〉 is the state |Ω〉 evolved to a time y0.

• φ†(y)e−iHy0 |Ω〉 is that state with a particle added at time y0 at a position y.

• e−iH(x0−y0)φ†(y)e−iHy0 |Ω〉 is that state time evolved to time x0.

• Now hitting this state with eiHx0φ(x) removes the particle at time x0.

• We terminate this string on the left with a 〈Ω| to find out how much of theoriginal state |Ω〉 is left in the final state.

The propagator is, therefore, the amplitude that we put a particle in the system atposition y at time y0 and get it out at position x time x0.


17.2 The Feynman propagator

We don’t always want to deal with the particle propagator G+ (which isconstrained with the θ-function so that x0 > y0) when thinking aboutfields. The reason is that although G+ accounts for the movement ofparticles it misses out the essential information about antiparticles.

Example 17.2

Consider, for example, the field creation operator for complex scalar field theory

ψ†(x) =R d3p

(2π)32

1

(2Ep)12

“

a†peip·x + bpe−ip·x”

, which creates a particle and annihi-

lates an antiparticle. If this acts on the free vacuum ψ†(y)|0〉 (no interactions, sonothing unpredictable can occur) then we create a particle at y, but the antiparticlepart annihilates the vacuum bp |0〉 = 0. Similarly, only the particle part contributesto 〈0|ψ(x), so the propagator 〈0|ψ(x)ψ†(y)|0〉 describes a particle propagating fromy to x, but tells us nothing about antiparticles.

Notation for propagators

In general G(x, y)Free propagator G0(x, y)Free scalar fields ∆(x, y)

Photon fields D(x, y)Fermion fields G(x, y) or iS(x, y)

Richard Feynman struggled over how to include the antiparticle contri-bution for some time before deciding that the most useful form of thepropagator was one that summed both particle and antiparticle parts.This form of the propagator makes up the guts of most quantum fieldcalculations and is called the Feynman propagator. How do we getthe propagator Feynman used which contains both particle and antipar-ticle parts? We need to introduce a new piece of machinery: the Wick

time-ordering symbol T . This isn’t an operator, but is much moreGian-Carlo Wick (1909–1992)

like the normal ordering symbol N in that it is just an instruction onwhat to do to a string in order that it makes sense. The time-orderingsymbol is defined for scalar fields3 as3This form applies only to the bosonic

(i.e. commuting) scalar fields we’re con-sidering in this chapter. For an anti-commuting Fermi field ψ, we pick upa minus sign for every permutation, sowe have

T ψ(x0)ψ(y0)

=

(

ψ(x0)ψ(y0) x0 > y0

−ψ(y0)ψ(x0) x0 < y0.

T φ(x0)φ(y0) =

φ(x0)φ(y0) x0 > y0

φ(y0)φ(x0) x0 < y0,(17.4)

so that the scalar fields are always arranged earliest on the right, lateston the left. The Feynman propagator is then defined as

G(x, y) = 〈Ω|T φ(x)φ†(y)|Ω〉= θ(x0 − y0)〈Ω|φ(x)φ†(y)|Ω〉 + θ(y0 − x0)〈Ω|φ(y)†φ(x)|Ω〉,

(17.5)

where |Ω〉 is the interacting ground state of the system. The propagatoris therefore made up of two parts. The first part applies for x0 laterthan y0: it creates a particle at y and propagates it to x where it isdestroyed. The second part applies when y0 is later than x0: it createsan antiparticle at x and propagates the system to point y. Both processesare included in the total propagator.

If, as in the example above, the system doesn’t contain any interac-tions then particles just move around passing through each other. In

17.2 The Feynman propagator 157

this case we call the ground state |0〉 and we have the free propagator

sometimes called G0(x, y) or, for scalar fields, more usually just ∆(x, y)where

∆(x, y) = 〈0|T φ(x)φ†(y)|0〉. (17.6)

The free-particle propagator is depicted in Fig. 17.1. The free propagator

∆

Fig. 17.1 The free-particle propagator∆(x, y) is a line without an interactionblob since a free particle doesn’t inter-act with any other particles on its wayfrom y to x.

forms an essential part of the structure of all perturbation calculations.This is because we think of interactions as events that take place atparticular spacetime points (the V -blobs in Fig. 17.2) and we imaginethat the particles propagate freely between interactions with the freepropagator.

∆

∆

∆

∆

V

V

V

Fig. 17.2 A third-order term inthe perturbation expansion of G interms of ∆(x, y). The amplitudefor the process is proportional to∆(x− z3)V (z3)∆(z3 − z2)V (z2)∆(z2 −z1)V (z1)∆(z1 − y).

We’ll now derive an expression for the free Feynman propagator∆(x, y).

Example 17.3

Let’s get an expression for the free propagator. If we use the general expression forthe annihilation field acting on the free vacuum φ†(y)|0〉, we have

φ†(y)|0〉 =

Zd3p

(2π)32 (2Ep)

12

“

a†p |0〉eip·y + bp |0〉e−ip·y”

. (17.7)

Only the particle creation part contributes (since bp |0〉 = 0) and gives

φ†(y)|0〉 =

Zd3p

(2π)32 (2Ep)

12

|p〉eip·y . (17.8)

How do we get the other half? Take a complex conjugate of eqn 17.8 (swappingy → x and p→ q) to get

〈0|φ(x) =

Zd3q

(2π)32 (2Eq )

12

〈q|e−iq·x. (17.9)

Sandwiching together eqn 17.8 and eqn 17.9 we obtain

〈0|φ(x)φ†(y)|0〉 =

Zd3p d3q

(2π)3(2Ep2Eq )12

e−iq·x+ip·yδ(3)(q − p)

=

Zd3p

(2π)3(2Ep)e−ip·(x−y), (17.10)

which corresponds to a particle being created at (y0,y) and propagating to (x0,x)where it is annihilated at the later time.

We also need to consider the reverse order for the advanced part of the propagator:

φ(x)|0〉 =

Zd3p

(2π)32 (2Ep)

12

“

ap |0〉e−ip·x + b†p |0〉eip·x”

. (17.11)

Here, only the antiparticle creation part contributes (since ap |0〉 = 0), to give

φ(x)|0〉 =

Zd3p

(2π)32 (2Ep)

12

|p〉eip·x. (17.12)

Again we take the complex conjugate (and change x→ y and p→ q):

〈0|φ†(y) =

Zd3q

(2π)32 (2Eq )

12

〈q|e−iq·y , (17.13)

yielding, finally

〈0|φ†(y)φ(x)|0〉 =

Zd3p

(2π)3(2Ep)eip·(x−y). (17.14)

This corresponds to an antiparticle being created at point (x0,x) and propagatingto (y0,y), where it is destroyed at the later time. Putting the two halves (eqns 17.10and 17.14) together we have our final answer for the free propagator ∆(x, y).


We have the final result that the Feynman propagator is given by

∆(x, y) = 〈0|T φ(x)φ†(y)|0〉 (17.15)

=

∫d3p

(2π)3(2Ep)[θ(x0 − y0)e−ip·(x−y) + θ(y0 − x0)eip·(x−y)].

Notice the way that this came together. We added together a particlepart from the retarded half of the propagator and an antiparticle fromthe advanced part. This is illustrated in Fig. 17.3.

y x y x

ty

tx ty

tx

Fig. 17.3 The Feynman propagator ismade up of a sum of particle and an-tiparticle parts.

We’ve been calling these propagators Green’s functions, but are theyreally Green’s functions in the sense of Chapter 16? The answer is nofor the interacting case. However, the free-propagator functions are theGreen’s functions of the equations of motion. So for the scalar field wehave

(∂2 +m2)∆(x− y) = −iδ(4)(x− y). (17.16)

This is examined in Exercise 17.2. In general, propagators are definedin such a way that they are true Green’s functions in the absence of in-teractions. That is, free propagators ∆(x, y) are true Green’s functions.

17.3 Finding the free propagator for scalar

field theory

Our expression for the free Feynman propagator for scalar field the-ory in eqn 17.15 is not very useful in its current form. We would likean expression which doesn’t include the rather non-covariant lookingHeaviside step functions. A simplification is achieved by using complexanalysis, as shown in the next example.

Example 17.4

How does the computation unfold? One of the tools we’ll need is the expansionof θ(x0 − y0) in the complex plane.4 Consider the first term in eqn 17.15, call it4This is given by

θ(x0 − y0) = i

Z ∞

−∞

dz

2π

e−iz(x0−y0)

z + iǫ.

[∆(x, y)](1), and substitute this expression for the θ function:

[∆(x, y)](1) ≡ θ(x0 − y0)〈0|φ(x)φ†(y)|0〉

= θ(x0 − y0)

Zd3p

(2π)3(2Ep)e−iEp(x0−y0)+ip·(x−y)

= i

Z ∞

−∞

dzd3p

(2π)4(2Ep)

e−i(Ep+z)(x0−y0)+ip·(x−y)

z + iǫ. (17.17)

Next we make the substitution z′ = z + Ep , leading to the integral

[∆(x, y)](1) = i

Z ∞

−∞

dz′d3p

(2π)4(2Ep)

e−iz′(x0−y0)+ip·(x−y)

z′ − Ep + iǫ. (17.18)

Then we change the definition of our four-momentum in the integral. We treat z′ asthe new p0, redefining p = (z′,p) = (p0,p) leading to

[∆(x, y)](1) = i

Z ∞

−∞

d4p

(2π)4(2Ep)

e−ip·(x−y)

p0 − Ep + iǫ. (17.19)

17.4 Yukawa’s force-carrying particles 159

With the redefinition, p0 = E doesn’t equal (p2 +m2)12 any more. It’s now just an

integration variable that we happen to have grouped with the three-momentum to

clean up the integral. Note that we still have that the energy Ep = (p2 +m2)12 .

We can carry the same process out on the second term:

[∆(x, y)](2) ≡ θ(y0 − x0)〈0|φ†)(y)φ(x|0〉

= θ(y0 − x0)

Zd3p

(2π)3(2Ep)e−iEp(y0−x0)+ip·(y−x)

= i

Z ∞

−∞

dzd3p

(2π)4(2Ep)

e−i(Ep+z)(y0−x0)+ip·(y−x)

z + iǫ. (17.20)

The substitution is now z′ = z + Ep , giving

[∆(x, y)](2) = i

Z ∞

−∞

dz′d3p

(2π)4(2Ep)

e−iz′(y0−x0)+ip·(y−x)

p0 − Ep + iǫ. (17.21)

Lastly, make the substitution p→ −p, which is to say (p0,p) → (−p0,−p), we get

[∆(x, y)](2) = −i

Z ∞

−∞

dz′d3p

(2π)4(2Ep)

e−ip0(x0−y0)+ip·(x−y)

p0 + Ep − iǫ

= −i

Z ∞

−∞

d4p

(2π)4(2Ep)

e−ip·(x−y)

p0 + Ep − iǫ. (17.22)

The full propagator ∆(x, y) = [∆(x, y)](1) + [∆(x, y)](2) is, therefore, the sum ofeqns 17.19 and 17.22:

∆(x, y) = i

Zd4p

(2π)4(2Ep)e−ip·(x−y)

„1

p0 − Ep + iǫ− 1

p0 + Ep − iǫ

«

=

Zd4p

(2π)4e−ip·(x−y) i

(p0)2 − E2p + iǫ

. (17.23)

Note further that using E2p = p2 +m2 the term in the denominator of the integrand

is (p0)2 − E2p + iǫ = p2 −m2 + iǫ.

The final result of this example yields the key result that the free Feyn-man propagator5 ∆(x, y) is given by 5Because ∆(x, y) is a function of x− y,

it is sometimes written ∆(x− y).

∆(x, y) = 〈0|T φ(x)φ(y)|0〉 =

∫d4p


p2 −m2 + iǫ.

(17.24)

We can immediately extract ∆(p), the Fourier component6 of the Feyn- 6The connection between the quan-

tities ∆(x, y) and ∆(p) is a simpleFourier transform:

∆(x, y) =

Zd4p

(2π)4e−ip·(x−y)∆(p).

man propagator corresponding to a particle with momentum p:

∆(p) =i

p2 −m2 + iǫ. (17.25)

This turns out to be a very useful equation.

17.4 Yukawa’s force-carrying particles

One of the most interesting thing about particles is that they interactwith each other. Hideki Yukawa’s great insight was that this interac- Hideki Yukawa (1907–1981)


tion process itself involved particles. These force-carrying particles arequite different to their more familiar cousins with whom we’re alreadyacquainted. They are, however, still described by propagators and pro-vide an immediate application of our field theory propagators. Yukawa’sidea centres around one key notion:

Particles interact by exchanging virtual, force-carrying particles.

Virtual particles are defined as particles existing ‘off mass-shell’. Themass-shell is the four-dimensional surface obeying the equation p2 =E2

p −p2 = m2 [see Fig. 11.1]. This, of course, is the usual dispersion fora relativistic particle. For an off mass-shell particle we have p2 6= m2.How can this be possible? The argument is that quantum mechanicsallows us to violate this classical dispersion, as long as we don’t do it fortoo long! By invoking energy-time uncertainty ∆E∆t ∼ ~, we can saythat particles of energy E are allowed to exist off the mass-shell as longas they last for a short time ∆t . ~/E. Virtual particles, therefore, musthave a finite range since (i) they can’t live forever and (ii) they travelat finite velocity. Yukawa guessed that the potential U(r) mediated bythe virtual particle would have the form

U(r) ∝ −e−|r|/a

4π|r| , (17.26)

where a has the dimensions of length.

Example 17.5

Looking back, we see that eqn 17.26 bears a resemblance to G+k

(r) = − ei|k||r|

4π|r| , which

is the Green’s function for the operator L = (∇2 + k2). From this, we see that if

we take i|k| = −m, then we have that the Yukawa potential U(r) ∝ − e−m|r|

4π|r| is the

Green’s function for the equation`∇

2 −m2Ú(r) = δ(3)(r), (17.27)

which is a time-independent version of the Klein–Gordon equation. What we’veshown here is that Yukawa’s potential is actually the field propagator describing thepropagation of virtual scalar field particles. The price that’s paid for being off themass shell is that the virtual particles can travel a distance fixed7 by their mass m.

7In our units, the length a = 1/m. (In

SI units, a = ~/mc.) For a pion (π0)with mass 135 MeV/c2, a ≈ 1.5 fm,roughly the size of a nucleus.

t

t

y

y x

x

Fig. 17.4 (a) A virtual particle prop-agates from y to x mediating a force.(b) The same process occurs with a vir-tual antiparticle propagating from x toy. (c) The processes are summed tomake the Feynman propagator.

The Yukawa particle exchange process is illustrated in Fig. 17.4. We canimagine spacetime processes such as that shown in Fig. 17.4(a) whereat time y0 particle A emits a virtual particle Q, with mass mQ fromposition y that at a later time x0 collides with particle B at position x.The energetics of this process may be written

EA = E′A + EQ, (17.28)

that is, particle A loses energy EQ = (p2Q + m2

Q)1/2. Of course in thisprocess particle B gains energy EQ.

17.4 Yukawa’s force-carrying particles 161

We may be in serious trouble if Yukawa exchange involves a net trans-fer of energy from particle A. It makes the interaction rather one-sidedfor one thing. To maintain the symmetry of the interaction, we mustalso imagine the additional process shown in Fig. 17.4(b) where particleB (at position x) emits an identical particle Q at time x0, which at latertime y0 collides with particle B at y. Both processes are necessary inorder to conserve energy so we must have an identical virtual particleheading back to A, which gives back the same amount of energy.

What does the Feynman propagator have to say about Yukawa’s no-tion of force-carrying virtual particles? Actually, the exchange of vir-tual particles idea is exactly what the Feynman propagator is describing.Recall that the Feynman propagator, describing a particle propagatingfrom spacetime point x to spacetime point y, looks like

∆(x, y) = θ(x0−y0)〈0|φ(x)φ†(y)|0〉+θ(y0−x0)〈0|φ(y)†φ(x)|0〉. (17.29)

The first term describes a particle travelling from y to x for x0 > y0,while the second describes an antiparticle travelling from x to y fory0 > x0. These situations are what we showed in Fig. 17.4. and we seethat both processes are included in the Feynman propagator. If we nowturn to the Feynman propagator in momentum space we have

∆(p) =i

p2 −m2 + iǫ=

i

(p0)2 − p2 −m2 + iǫ. (17.30)

Notice that the denominator is always off mass shell, that is to say thatthe expression is only well behaved if p0 6=

√p2 +m2. The message is

that the Green’s function describes the propagation of virtual particles.i

p2−m2+iǫ

Fig. 17.5 A Feynman diagram fortwo particles scattering. The ampli-tude for the process includes a factori/(p2 − m2 + iǫ), which describes thepropagation of a virtual, force-carryingparticle which can be thought of as me-diating a Yukawa force.

When we come to calculating Feynman diagrams in a few chapters’time, we’ll see that the amplitude A for two particles to scatter off eachother as shown in Fig. 17.5 contains a term ∝ i/(p2 −m2 + iǫ). We cancheck this immediately. If we want to calculate the scattering amplitudefor two particles in non-relativistic quantum mechanics we can use theBorn approximation, which tells us that the amplitude is proportionalto the Fourier transform of the scattering potential:

A ∝ U(p) =

∫d3r U(r)e−ip·r, (17.31)

where p is a scattering vector. If we use Yukawa’s suggestion for the

interaction potential U(r) ∝ − e−m|r|

4π|r| corresponding to an interaction

with a range of a, then doing the integral8 yields 8The integral can be evaluated using

Z

d3re−ip·re−m|r|

|r| =

Z ∞

0dr |r|2

×Z 2π

0dφ

Z π

0dθ sin θ

e−i|p||r| cos θ−m|r|

|r|

=4π

p2 +m2.

A ∝ 1

−|p|2 −m2, (17.32)

which looks rather like the propagator ∆(p) with p0 = 0. Our result forthe propagator ∆(p) is the four-dimensional, relativistic generalizationof this result. Indeed, as we saw previously, the propagator 1

−|p|2−m2 isthe Green’s function for the time-independent Klein–Gordon equation(−∇

2 + m2)φ(x) = 0, while ∆(p) is the Green’s function for the four-dimensional Klein–Gordon equation (∂2 +m2)φ(x) = 0.


17.5 Anatomy of the propagator

Let’s take a closer look at the anatomy of the free propagator9 ∆(x, y) =9Remember that for a real scalar field

φ(x) = φ†(x) and so we may write thescalar field propagator without the dag-gers: ∆(x, y) = 〈Ω|T φ(x)φ(y)|Ω〉. Forsimplicity’s sake, we will do this laterin the book.

〈0|T φ(x)φ†(y)|0〉 for scalar field theory. It arose from the amplitude thata single field excitation is created at spacetime point y and is found atspacetime point x. It contains the sum of amplitudes corresponding tothe possibility that this excitation might be a particle or an antiparticle.In position space we have that

∆(x, y) = 〈0|T φ(x)φ†(y)|0〉 =

∫d4p


p2 −m2 + iǫ. (17.33)

It is important to notice that we have a singularity (or pole10) in the10Another way of making sense of theiǫ is that it is meant to ensure thatthe integral never hits this pole. It isan instruction to avoid it by adding aninfinitesimal imaginary part to the de-nominator. This is discussed in Ap-pendix B.

function when the particle is on mass shell, that is, when p2 = E2p−p2 =

m2.Now let’s return to the full propagator for a general interacting the-

ory G(x, y) = 〈Ω|φ(x)φ†(y)|Ω〉. (Remember that we call the interactingground state |Ω〉.) The full name for G(x, y) is the ‘single-particle prop-agator’ or the ‘two-point Green’s function’. We can draw a diagramillustrating G(x, y) as shown in Fig. 17.6(a). The two stumps corre-spond to the incoming or outgoing particles. The hatched blob in themiddle depicts interactions: the adventures (involving other particles)that our test particle might have between its creation and annihilation.The name ‘two-point’ comes from the fact that a single particle is cre-ated and a single particle is annihilated, that is, the sum of particlesentering or leaving the system, i.e. (number in)+(number out) equals 2.We often emphasise this by writing the propagator as G(2)(x, y) witha superscript ‘(2)’. We can go further and define the n-point propaga-tor, where n equals the sum of incoming and outgoing lines. These willbe denoted G(n). For example we might want to study the four-pointpropagator [Fig. 17.6(b)]

G(4)(x1, x2, x3, x4) = 〈Ω|T φ(x1)φ(x2)φ†(x3)φ

†(x4)|Ω〉, (17.34)

which is useful when we’re considering two particles coming in and twoparticles leaving as might occur in a two-particle scattering experiment.

Fig. 17.6 (a) The two-point propaga-

tor G(2). The blob represents the in-teractions that the particle undergoeswith other particles in the system. (b)A four-point propagator G(4). (c) Ann-point propagator G(n).

The aim of much quantum field theory is to find the full interactingGreen’s functions G(n). One reason is that G(n) can be used to calculatescattering amplitudes using an equation known as the LSZ reductionformula.11 Another reason is that the form of the interacting Green’s

11See Peskin and Schroeder for a dis-cussion of the LSZ reduction formula.

function tells us a lot about the system. For example, the two-pointpropagator, G(2)(x, y) = 〈Ω|φ(x)φ†(y)|Ω〉, that tells us about the fate ofsingle particles, is especially useful.

To see this we can consider what happens if we start from a freescalar theory and slowly turn on the interactions to form an interactingtheory. Before we turn on the interactions we have the free propagator∆(p) = i

p2−m2+iǫ . The mass of the particle is given by the position ofthe pole. The residue at the pole (which is i) tells us that the operator

Exercises 163

φ†(y) creates precisely one quantum from the vacuum |0〉 (and later wedestroy it). When we turn on the interactions it turns out12 that the full 12The details of the full propagator are

discussed in Chapter 33.propagator resembles the free version. The general form of the two-pointpropagator in momentum space is

G(2)(p) =iZp

p2 −m2 − Σ(p) + iΓp

+

(Multiparticle

parts

), (17.35)

where

• Zp tells us how likely it is for a particle with momentum p to existstably without being destroyed by the interactions,

• Σ(p) tells us about how the particle of momentum p interacts withthe vacuum,

• (2Γp)−1 is the particle’s lifetime,

• The full propagator also contains a contribution from short-livedmultiparticle states which can be scattered out of the vacuum.

Chapter summary

• The Feynman propagator for single-particle excitations in the fieldis given by G(x, y) = 〈Ω|T φ(x)φ†(y)|Ω〉.

• For the scalar field theory the momentum space propagator forsingle particles is ∆(p) = i

p2−m2+iǫ .

Exercises

(17.1) Show that the retarded field propagator for a freeparticle in momentum space and the time domain:G+

0 (p, tx, q, ty) = θ(tx − ty)〈0|ap(tx)a†q(ty)|0〉 is

given by θ(tx − ty)e−i(Eptx−Eqty)δ(3)(p − q).

(17.2) Demonstrate that the free scalar propagator is theGreen’s function of the Klein–Gordon equation.That is, in (1 + 1) dimensions, show

„

∂2

∂(x0)2− ∂2

∂x2+m2

«

〈0|T φ(x0, x)φ†(y0, y)|0〉

= −iδ(2)(x− y). (17.36)

(17.3) Consider a scalar field theory defined by the La-grangian

L =1

2[∂µφ(x)]2 − m2

2[φ(x)]2. (17.37)

By considering the Fourier transform of the field,show that the action may be written

S =1

2

Z

d4p

(2π)4φ(−p)

`

p2 −m2´ φ(p). (17.38)

This provides us with an alternative method foridentifying the free propagator G0(p) as (i/2 times)the inverse of the quadratic term in the momentum-space action.

(17.4) Show that the Feynman propagator for the quan-tum simple harmonic oscillator with spring constantmω2

0 is given by

G(ω) =

„

1

m

«

i

ω2 − ω20 + iǫ

. (17.39)


(17.5) (a) Consider a one-dimensional system with La-grangian

L =1

2

„

∂φ(x)

∂x

«2

+m2

2[φ(x)]2. (17.40)

The choice of sign makes this a Euclidean the-ory (described in Chapter 25). Descretize thistheory (that is, put it on a lattice) by definingφj = 1√

Na

P

p φpeipja, where j labels the lattice

site, a is the lattice spacing and N is the number oflattice points. Using the method in Exercise 17.3show that the action may be written

S =1

2

X

p

φ−p

„

2

a2− 2

a2cos pa+m2

«

φp,

(17.41)

and read off the propagator for this theory.(b) The Lagrangian for a one-dimensional elas-tic string in (1+1)-dimensional Minkowski space iswritten

L =1

2

ˆ

(∂0φ)2 − (∂1φ)2˜

. (17.42)

Descretize the theory by defining φj(t) =1√Na

P

p

R

dω2πφp(ω)e−iωteipja. Show that the prop-

agator for the (phonon) excitations that this theorydescribes is given by

G(ω, p) =i

ω2 − ω20(1 − cos pa)

, (17.43)

where ω20 = 2/a2.

18The S-matrix

18.1 The S-matrix: a hero for ourtimes 166

18.2 Some new machinery: the in-teraction representation 167

18.3 The interaction picture ap-plied to scattering 168

18.4 Perturbation expansion ofthe S-matrix 169

18.5 Wick’s theorem 171

Chapter summary 174

Exercises 174

The Matrix is a system, Neo. That system is our enemy.But when you’re inside, you look around, what do you see?Businessmen, teachers, lawyers, carpenters. The very mindsof the people we are trying to save. But until we do, thesepeople are still a part of that system, and that makes themour enemy.Morpheus (Laurence Fishburne) in The Matrix

Canonical quantization works well for a small number of theories, whereit results in a diagonalized Hamiltonian describing the excitations of thequantum field in terms of non-interacting, or ‘free’ particles. In fact, thetheories that can be solved exactly using the methods of quantum fieldtheory generally describe free particles with no prospect for interactions.A world of free particles would be a very boring place since particleinteractions are at the heart of the workings of Nature. Most interestingtheories involve particle interactions and cannot be solved exactly. Wetherefore need to resort to some sort of approximate method, like theperturbation theory of quantum particle mechanics.

We will explore a perturbation theory for quantum fields in the follow-ing chapters. The Lagrangian describing the field is often the additionof two parts: a solvable part (frequently describing non-interacting par-ticles only and therefore called the free part) and a part that makesthe problem unsolvable (frequently describing interactions and calledthe interacting part). As before, the free part is able to be solved viacanonical quantization and results in non-interacting particles. The in-teraction part is more interesting because of one key fact:

Interactions involve the creation or destruction of particles.

One example of an interaction process is particle scattering. Particles arefired at each other. At the start of the experiment they are far from eachother so don’t interact: they are free. When they’re smashed togetherthey interact but only for a short time. In this chapter we’ll describea method of dealing with the interactions in a scattering process. Thiswill be time well spent. It turns out that the machinery developed herewill be central to creating a more general perturbation theory that candeal with other interactions.

166 The S-matrix

18.1 The S-matrix: a hero for our times

One of the basic building blocks of quantum field theory is the scatteringor S-matrix. The idea was one of John Wheeler’s and it’s a greatJohn Wheeler (1911–2008) made nu-

merous contributions to theoreticalphysics and is also famous for coiningthe terms ‘black hole’ and ‘wormhole’and ‘it from bit’.

one. The S-matrix is rather like the time-evolution operator, but for animportant special case: it describes a Scattering experiment.

In a prototypical scattering experiment we start off with well separatedparticles [Fig. 18.1(a)]. We fire the particles at each other and theyinteract in some complicated way governed by the Hamiltonian of thereal world H [Fig. 18.1(b)]. The particles then recede from each otherand end up well separated [Fig. 18.1(c)]. The first thing to say about thisis that the complicated interactions of the real world make this processimpossible to analyse. Working in the Heisenberg picture, we can’t evenwrite down convincing expressions for the time-dependent operators. Allappears to be lost! What do we do?

(a)

(b)

(c)

Fig. 18.1 A prototypical scattering ex-periment. (a) The particles begin wellseparated. (b) They interact in a com-plicated way. (c) They recede from eachother and end up well separated.

Wheeler says that we start by imagining a world without compli-cated interactions. That is, we split H into two parts H = H0 + H ′,where H0 describes a simple world of non-interacting particles describedby some set of state vectors. Remember that we’re working in theHeisenberg picture here: these state vectors don’t change at all. Fortwo particles one simple-world state might be a momentum state like|ψ〉 = |p2p1〉simpleworld. Now we look at the real world and ask ‘Is therea real-world state that looks like the simple-world state |p2p1〉simpleworld?’There is such a state at the start of the scattering experiment (at a timet→ −∞) when the particles are very far apart. We’ll call this real-worldstate |p2p1〉inrealworld or an ‘in’ state, for short. Similarly, we pick out an-other simple-world state, e.g. |φ〉 = |q2q1〉simpleworld and ask if there’s areal-world state that looks like this one. There is, but this time at theend of the experiment (t → ∞) when the particles are well separatedafter their interaction. We’ll call this one |q2q1〉out

realworld. Notice that thesimple-world states only describe the real world in the limits t→ ±∞.

For the real-world scattering process in which we’re interested theamplitude A for starting with |p2p1〉inrealworld and ending up with|q2q1〉out

realworld is given by

A =outrealworld 〈q1q2|p2p1〉inrealworld. (18.1)

We must recreate this amplitude using the simple-world states which,after all, are the only ones we can work with. How can this be achieved?Wheeler’s answer is the S-matrix. We define

A =outrealworld 〈q1q2|p2p1〉inrealworld =simpleworld 〈q1q2|S|p2p1〉simpleworld.

(18.2)So the S-matrix (that is, the matrix elements of the S-operator) containsthe amplitudes for starting with a particular ‘in’ state and ending upwith a particular ‘out’ state.

Let’s take this further and calculate a real amplitude. We now needtwo things:

18.2 Some new machinery: the interaction representation 167

• A way of getting a suitable H0 to describe some useful simple-worldstates which resemble the ‘in’ and ‘out’ states.

• A way of calculating an expression for the S-operator. We canthen use the eigenstates of the simple Hamiltonian to work out anamplitude.

In the next section we’ll examine the machinery needed to get H0, whichinvolves yet another new formulation of quantum mechanics. This oneis called the interaction representation.

18.2 Some new machinery: the interaction

representation

So far in our quantum mechanics we’ve either had a formulation withtime-dependent states and time-independent operators (the Schrodingerpicture) or one with time-independent states and time-dependent oper-ators (the Heisenberg picture). A third way is available, called the in-teraction representation, where both the states and the operators havesome time dependence. It turns out that this combination is the one weneed to obtain H0, the simple-world Hamiltonian.

As stated above, to get H0 we split up the Hamiltonian into two partsH = H0 + H ′. We call these a free part H0 and an interaction part H ′.The free part will generally be one that is time-independent and can beeasily solved. We then say that operators in the interaction picture OI

evolve in time via the free part H0 of the Hamiltonian

OI(t) = eiH0tOe−iH0t. (18.3)

This is just like the Heisenberg version of quantum mechanics, exceptthat we’re only using the free part of the Hamiltonian. We thereforehave a Heisenberg-like equation of motion for the operators

idOI

dt= [OI(t), H0], (18.4)

but again we emphasize that it just involves H0.So far we’ve not included H ′. Its inclusion results in the wave function

having some time dependence, which it wouldn’t in the non-interactingcase. To see what happens, we compare a matrix element from theSchrodinger picture to one in the interaction picture

〈φ(t)|O|ψ(t)〉 = 〈φI(t)|eiH0tOe−iH0t|ψI(t)〉, (18.5)

where the interaction picture states are labelled with a subscript I. Wecan see that, for the matrix elements to be the same as in the Schrodingerpicture, we’ll need to define

|ψI(t)〉 = eiH0t|ψ(t)〉. (18.6)

168 The S-matrix

We can now get to the equation of motion for the interaction picturewave function by differentiating eqn (18.6) with respect to time. We find

id

dt|ψI(t)〉 = eiH0t

(−H0 + i

d

dt

)|ψ(t)〉

= eiH0t(−H0 + H

)|ψ(t)〉, (18.7)

and recalling that H = H0 + H ′ and eqn 18.6 we have

id

dt|ψI(t)〉 = eiH0tH ′e−iH0t|ψI(t)〉, (18.8)

which we can rewrite in a recognizable form as

id

dt|ψI(t)〉 = HI(t)|ψI(t)〉, (18.9)

where HI(t) = eiH0tH ′e−iH0t. This then completely defines the interac-tion picture.

To recap the description of the interaction picture:

• Both the operators and states can evolve in time.

• The operators evolve by the free part of the Hamiltonian: φI(t) =

eiH0tφ e−iH0t.

• The states evolve according to the interaction part of the Hamil-tonian: i ∂∂t |ψ(t)〉I = HI(t)|ψ(t)〉I where HI(t) = eiH0tH ′e−iH0t.

Note that all of our different representations coincide at t = 0.

18.3 The interaction picture applied to

scattering

t

λ

Fig. 18.2 Turning on an interaction

Hamiltonian H′. We multiply it by λ(t)which has this profile.

Why is the interaction picture useful for the scattering problem? Thepoint is that the interaction part of the Hamiltonian is zero at the startand end of the problem. In fact, we imagine that the interaction part ofthe Hamiltonian H ′ is turned on and off slowly and smoothly as shownin Fig. 18.2. When HI = 0 we just have the Heisenberg picture for thefree part of the Hamiltonian H0. This is good because we choose H0 sothat we can solve it using our canonical quantization machine and wecertainly then know its eigenstates.

What becomes of the states? We can identify the simple-world statesas those of the interaction picture at the start and end:

|φ〉simpleworld = |φI(±∞)〉. (18.10)

They are eigenstates of H0, which is vital since it allows us to buildthem up from the state |0〉 which is the vacuum of H0 (called variouslythe non-interacting, free or bare vacuum) using our beloved creationand annihilation operators. During the interaction process we have that

18.4 Perturbation expansion of the S-matrix 169

HI is nonzero so the states evolve in some mysterious and complicatedway. However, at the end of the experiment HI is zero again and thestates freeze. It’s a bit like musical chairs in this respect. The states areinitially sitting down, they rearrange themselves during the music andthen freeze on the chairs when the music stops.

We’ve thought about the states, but what about the operators? Sim-ple: the operators time-evolve according to H0 at all times. We cantherefore use the freely evolving field operators like φ(x) that we’ve usedpreviously. These also enjoy an expansion in terms of creation and an-nihilation operators.

Finally we need to figure out how to work out the S-operator. We areassisted by the useful fact that all of our quantum mechanical picturesare defined so that they coincide at t = 0, so that we’re free to write

simpleworld〈φ|S|ψ〉simpleworld =outrealworld 〈φ|ψ〉inrealworld = 〈φI(0)|ψI(0)〉.

(18.11)Then, if we knew the time-evolution operator in the interaction pictureUI(t2, t1), we could say that

simpleworld〈φ|S|ψ〉simpleworld = 〈φI(∞)|UI(∞, 0)UI(0,−∞)|ψI(−∞)〉= 〈φI(∞)|UI(∞,−∞)|ψI(−∞)〉 (18.12)

= simpleworld〈φ|UI(∞,−∞)|ψ〉simpleworld.

We get the important result that the S-operator is the time-evolutionoperator for the interaction-picture UI(t,−t) as t→ ∞.

18.4 Perturbation expansion of the

S-matrix

As in the Schrodinger picture, we have an equation of motion for theinteraction picture time-evolution operator

id

dt2UI(t2, t1) = HI(t2)UI(t2, t1), (18.13)

where U(t, t) = 1. We might be tempted to treat HI(t) as a number,and if that were valid we could then integrate and write U(t2, t1) =

e−iR t2

t1dt HI(t), but this is wrong. The interaction Hamiltonian HI(t)

doesn’t necessarily commute with itself when evaluated at differenttimes. That is, in general, [HI(t2), HI(t1)] 6= 0. To circumvent thisproblem, we’ll define a new version of the exponential which will makethis equation right. The solution turns out to use the time-ordered

product

T [A1(t1)A2(t2) . . . An(tn)], (18.14)

defined as the string arranged so that the later operators are on the left.1

1Just remember ‘later on the left’.Note that the anticommuting nature ofFermi fields causes us to pick up an ad-ditional factor (−1)P on time ordering,where P is the number of permutationsrequired to rearrange the operators.From here until Chapter 35 we will fo-cus our discussion on (bosonic) scalarfields. Fermi fields, whose behaviour isbased on the same general principles wewill examine, are discussed specificallyin Part IX.

As discussed in the previous chapter, the Wick time ordering symbolsymbol T [ ] is a little like the normal ordering procedure N [ ] in that it is

170 The S-matrix

not an operator, it’s an instruction on what to do to a string of operators.The reason that using T [ ] helps us here is that everything within a time-ordered product commutes, so we may now take a derivative. We cantherefore write an expression for UI(t2, t1) that incorporates correctlytime-ordered operators as follows:

UI(t2, t1) = T[e−i

R t2t1

dt HI(t)]. (18.15)

This expression is known as Dyson’s expansion.Freeman Dyson (1923– )

Example 18.1

We’ll pause to justify eqn 18.15. Using the fact that everything within a time-orderedproduct commutes we can treat UI(t2, t1) as a function of t2 and take a derivativewith respect to this time to obtain

id

dt2T

»

e−iR t2

t1dt HI(t)

–

= T

»

HI(t2)e−iR t2t1

dt HI(t)–

. (18.16)

Next we notice that t2 is the latest time in the problem (since U evolves the systemfrom t1 to t2) and the time ordering therefore puts the operator HI(t2) left-most.We can then pull HI(t2) out of the time-ordered product to obtain

id

dt2T

»

e−iR t2

t1dt HI(t)

–

= HI(t2)T

»

e−iR t2t1

dt HI(t)–

. (18.17)

Comparing this with eqn 18.13 we see that the time-evolution operator in the inter-action picture is given by

UI(t2, t1) = T

»

e−iR t2t1

dt HI(t)–

. (18.18)

We also know that the S-operator is the limit S = UI(t2 → ∞, t1 →−∞). Using this, we are left with Dyson’s expansion of the S-operator

S = T[e−i

R∞−∞

d4x HI(x)], (18.19)

where we’ve replaced∫

dt HI(t) with∫

d4x HI(x), an integral of theHamiltonian density.

The exponential form of Dyson’s expansion in eqn 18.19 is beautifullymemorable, but not very useful. Usually the integral in the exponentcannot be done exactly so we have to expand out the exponential inDyson’s expansion thus:

S = T

[1 − i

∫d4z HI(z) + (−i)2

2!

∫d4y d4w HI(y)HI(w) + . . .

].

(18.20)Provided that the interaction Hamiltonian HI(x) is small compared tothe full Hamiltonian density, then this provides the basis for a perturba-tion expansion of the S-matrix. We’ll deal with this expansion in detailin the next chapter. Notice that in eqn 18.20 we integrate over differentspacetime coordinates each time the interaction Hamiltonian appears.


18.5 Wick’s theorem

To make progress in digesting long strings of operators that appear inDyson expansions, we are faced with solving the following conundrum:

• We will frequently have to evaluate a term like

〈0|T [ABC . . . Z]|0〉, (18.21)

a vacuum expectation value (or VEV for short) of a time-ordered string of operators. This is hard.

• On the other hand, what is very easy to evaluate is

〈0|N [ABC . . . Z]|0〉, (18.22)

a VEV of a normal-ordered string of operators. This is trivialbecause normal ordering places annihilation operators on the rightand creation operators on the left, and so this VEV is identicallyzero. So if we could find a way of relating N [stuff] to T [stuff] thenthe problem would be solved.

The next thing to notice is that the mode expansion of a field opera-tor φ contains two parts, an annihilation part and a creation part (seeeqn 11.8) and so can be written φ = φ− + φ+. Thus we have2 φ−|0〉 = 0 2Our definition assigns the superscript

+ to the creation part and − to theannihilation part. Some books, e.g. Pe-skin and Schroeder, choose an oppositeconvention.

and 〈0|φ+ = 0. Let’s now take the simplest non-trivial case of a stringof operators and consider two field operators A and B. Their productis given by

AB = (A+ + A−)(B+ + B−) = A+B+ + A−B− + A+B− + A−B+.(18.23)

If we normal-order this product, we get

N [AB] = A+B+ + A−B− + A+B− + B+A−, (18.24)

where we have only needed to swap operators in the final term. ThusN [AB] only differs from AB by

AB −N [AB] = A−B+ − B+A− = [A−, B+], (18.25)

i.e. by a simple commutator of operators (which we know from previousexperience3 will just be a complex number, or c-number for short). Now 3Commutators turn out to be quanti-

ties such as i~.the fields A and B are functions of spacetime coordinates, so we couldalso write down the time-ordered product

T [A(x)B(y)] =

A(x)B(y) x0 > y0

B(y)A(x) x0 < y0,(18.26)

and therefore

T [A(x)B(y)] −N [A(x)B(y)] =

[A−(x), B+(y)] x0 > y0

[B−(y), A+(x)] x0 < y0.(18.27)

172 The S-matrix

Since the VEV of a normal ordered product is zero, we can immediatelywrite down

〈0|T [A(x)B(y)]|0〉 =

〈0|[A−(x), B+(y)]|0〉 x0 > y0

〈0|[B−(y), A+(x)]|0〉 x0 < y0.(18.28)

If we choose A = B = φ, then the quantity on the left is simply aFeynman propagator.

Subtracting out a normal-ordered string of operators from the time-ordered version is clearly a useful thing to do and so we formalize thisand give it a special name: a contraction, defined as

AB = T [AB] −N [AB]. (18.29)

The contraction AB is simply a commutator (see eqn 18.27) becausethe only difference between normal ordering and time ordering is thatwe have shuffled various creation and annihilation operators, therebyaccumulating various c-numbers. Thus the contraction is a c-number,and it takes the value

AB = AB〈0|0〉 = 〈0|AB|0〉 = 〈0|T [AB]|0〉. (18.30)

Moreover, since it is only a c-number, normal ordering has no effect onit and we can write

T [AB] = N [AB] + AB = N [AB + AB]. (18.31)

This result can be generalized4 to longer strings of operators to yield4The proof is by induction, and canbe found on page 90 of Peskin andSchroeder.

Wick’s theorem, which can be stated as follows:

T [ABC . . . Z] = N

[ABC . . . Z +

all possible contractions of

ABC . . . Z

].

(18.32)

It’s worth noting that Wick’s theoremapplies for free fields only, such as thosein the interaction picture. This will beimportant in Chapter 22.

Example 18.2

Wick’s theorem can be illustrated for the case of four operators

T [ABCD] = Nh

ABCDi

+N

"

ABCD

#

+N

"

ABCD

#

+N

"

ABCD

#

+N

"

ABCD

#

+N

"

ABCD

#

+N

"

ABCD

#

+N

"

ABCD

#

+N

2

4ABCD

3

5+N

2

4ABCD

3

5 . (18.33)

This is a sum of normal ordered terms including zero, one or two contractions.


Let us look at a particular term involving a single contraction in a Wick

expansion of T [ABC . . . Z], say for example N [ABCDEF G . . .]. Sincethe contracted part is a c-number, we can factor it out of the normalordered product5 and write 5There could be a sign change in

this expression if the operators arefermionic, so in that case you have topay attention to any sign changes ac-cumulated with swapping the order ofoperators.

N [ABCDEF G . . .] = CF ×N [ABDEG . . .]. (18.34)

However, when we evaluate the VEV of this term it will vanish because〈0|N [anything]|0〉 = 0. Therefore, the only terms which survive when wework out the VEV of T [ABC . . . Z] are the ones involving contractionsof all operators.6 Thus 6By extension, the VEV of an odd-

numbered string of operators will bezero, since you can only contract pairsof operators and so there will always beone operator left over.

〈0|T [ABC . . . Y Z]|0〉 = 〈0|T [ABCDEF . . . Y Z]|0〉 (18.35)

+〈0|T [ABCDEF . . . Y Z]|0〉

+

(all other combinations involving

contractions of every pair of operators

),

and using eqn 18.30 this reduces to

〈0|T [ABC . . . Y Z]|0〉= 〈0|T [AB]|0〉〈0|T [CD]|0〉〈0|T [EF ]|0〉 . . . 〈0|T [Y Z]|0〉

+ 〈0|T [AC]|0〉〈0|T [BD]|0〉〈0|T [EF ]|0〉 . . . 〈0|T [Y Z]|0〉+ . . .

(18.36)

that is, the VEV of a time-ordered string of operators is given by thesum of products of all possible combinations of VEVs of time orderedpairs.

Example 18.3

For the case of four bosonic operators.

〈0|Th

ABCDi

|0〉 = 〈0|T"

ABCD

#

|0〉 + 〈0|T

2

4ABCD

3

5 |0〉 + 〈0|T

2

4ABCD

3

5 |0〉

= 〈0|T [AB]|0〉〈0|T [CD]|0〉 + 〈0|T [AC]|0〉〈0|T [BD]|0〉+〈0|T [AD]|0〉〈0|T [BC]|0〉. (18.37)

In particular

〈0|φ(x1)φ(x2)φ(x3)φ(x4)|0〉 = ∆(x1 − x2)∆(x3 − x4) + ∆(x1 − x3)∆(x2 − x4)

+∆(x1 − x4)∆(x2 − x3), (18.38)

where we have used the Feynman propagator ∆(x1 − x2) = 〈0|T φ(x1)φ(x2)|0〉.

Wick’s theorem is an essential tool for expanding S in perturbationtheory which we discuss in the next chapter.

174 The S-matrix

Chapter summary

• Real-world scattering processes can be described by an S-matrixelement via out

realworld〈φ|ψ〉inrealworld =simpleworld 〈φ|S|ψ〉simpleworld.

• The S-matrix works in the interaction picture where operatorstime-evolve according to the free part of the Hamiltonian H0 andstates time-evolve according to the interaction part HI(t).

• The S-operator is given by the Dyson equation S =T [e−i

R

d4xHI(x)].

• Wick’s theorem allows us to grind down long time-ordered stringsof operators. A vacuum expectation value (VEV) of a time-orderedstring is a sum of products of VEVs of time-ordered pairs.

Exercises

(18.1) Although the interaction picture is most useful tous as a step towards a powerful version of pertur-bation theory, it is also useful in itself for someproblems, such as this famous example from mag-netic resonance. Consider a Hamiltonian describinga spin-1/2 particle in a large, static magnetic fieldB0 which is subject to a perturbing, perpendicular,oscillating field B1:

H = γB0Sz + γB1(Sx cos γB0t+ Sy sin γB0t).(18.39)

Here γ is the particle’s gyromagnetic ratio. Noticethat the frequency γB0 of the oscillating field withamplitude B1 is chosen so that it matches the en-ergy level separation caused by the static field B0.(a) Write the problem in the interaction represen-tation by splitting the Hamiltonian into free andinteracting parts.(b) Use the identities

S± = Sx ± iSy,

S+eiωt = eiωSztS+e−iωSzt,

S−e−iωt = eiωSztS−e−iωSzt, (18.40)

to simplify the interaction Hamiltonian HI(t). Youshould find that you’re able to remove the time de-

pendence completely.(c) Find the interaction picture evolution operatorUI(t2, t1).(d) What is the probability that a particle initiallyprepared in a state | ↑〉 at t = 0 will still be in thatstate at time t. What is the probability that it willbe found in the | ↓〉 state?(e) Find the expectation value of the Sz operator.

(18.2) Show that |ψ(t = ∞)〉I =P

φ〈φ|S|ψ〉|φ〉, where thestates on the right-hand side are the simple-worldstates defined in the chapter.

(18.3) Use Wick’s theorem to express the string of Boseoperators ap a

†q ak in terms of normal ordered fields

and contractions.

(18.4) (a) Normal order the string of Bose operatorsbgbb†b† using the usual Bose commutation relations.(b) Show that Wick’s theorem gives the same an-swer.

(18.5) Use Wick’s theorem to simplify

〈0|c†p1−q c†p2+q cp2

cp1|0〉,

where the operators c†p and cp create and annihilatefermions respectively.

19Expanding the S-matrix:

Feynman diagrams

19.1 Meet some interactions 176

19.2 The example of φ4 theory 177

19.3 Anatomy of a diagram 181

19.4 Symmetry factors 182

19.5 Calculations in p-space 183

19.6 A first look at scattering 186

Chapter summary 186

Exercises 187

They were funny-looking pictures. And I did think con-sciously: Wouldn’t it be funny if this turns out to be usefuland the Physical Review would be all full of these funnylooking pictures. It would be very amusing.Richard Feynman (1919–1988)

Like the silicon chips of more recent years, the Feynmandiagram was bringing computation to the masses.Julian Schwinger (1918–1994)

(a)

(b) (c)

(d) (e)

tim

e = particle = antiparticle

t = t0

t = t1

t = t3

t = t2

Fig. 19.1 The Feynman diagram. (a)A particle is represented by a line withan arrow going in the direction of time(conventionally up the page). An an-tiparticle is represented by a line withan arrow going the opposite way. (b)A particle and antiparticle created attime t = t0. (c) A particle and antipar-ticle are annihilated at time t = t1. (d)Trajectory of a single particle can beinterpreted as follows: (e) A particle–antiparticle pair are created at t2, theantiparticle of which annihilates with athird particle at t = t3.

One of Richard Feynman’s greatest achievements was his invention of(what are now called) Feynman diagrams. These are cartoons thatrepresent terms in the perturbation expansion of the S-matrix. We willsee that Feynman diagrams exist in several related forms, but the moststraightforward of these are simply spacetime diagrams describing thetrajectory of particles. We imagine time running up the page and rep-resent particles by lines with an arrow going in the direction of time[Fig. 19.1(a)]; antiparticles are represented by lines with arrows going inthe opposite direction. (This is consistent with Feynman’s interpreta-tion of negative energy states as positive energy antiparticles travellingbackwards in time.) We can draw particle and antiparticle pairs beingcreated at some time [Fig. 19.1(b)] or destroyed [Fig. 19.1(c)] at someother time. We can also tell stories: the line in Fig. 19.1(d) may looklike it describes a single particle looping backward in time, but Feynmanhas another interpretation. Just as it’s unclear to a bomber pilot flyingover a curving road whether (s)he is above a single road or a numberof them, we’re asked to fly over the diagram experiencing everything intime order. From the point of view of the bombardier, we start witha particle for t < t2, then at t = t2 a new particle–antiparticle pair iscreated. The new particle and antiparticle coexist with the original par-ticle in the interval t2 < t < t3. At t = t3 the newly created antiparticlecollides with our original particle and annihilates. The newly createdparticle continues out of the diagram for t > t3 [see Fig. 19.1(e)]. Sincethe particles are identical, no-one can tell the difference.

In this chapter we introduce the perturbation expansion of the S-matrix and its representation in terms of Feynman diagrams. Recallfrom the previous chapter that an expression for the S-operator is given

176 Expanding the S-matrix: Feynman diagrams

by Dyson’s expansion

S = T[e−i

R

d4z HI(z)]. (19.1)

To use this we must expand it as a power series in the operator HI(z).We’ll now consider some of the forms of HI(z) that are encountered inquantum field theory and see how the expansion of Dyson’s equationproceeds. The interaction Hamiltonians will generally involve the prod-ucts of free-field operators localized at some point z. We’ll see laterthat if we interpret the field operators as describing particles then theinteraction1 can be thought of as a collision of the particles at spacetime

1Notice also that we integrate over allz in eqn 19.1 thus taking into accountthe possibility of the interaction takingplace at every point in spacetime.

point z.

19.1 Meet some interactions

There are a number of models of interactions to try. For each inter-action we will draw a diagram in spacetime illustrating the interactionprocesses. This will eventually lead to the famous diagrams invented byRichard Feynman. At present we won’t make any interpretations of thediagrams in terms of particles: this will only be possible a little laterwhen the field operators are made to act on states.

Example 19.1

(i) The simplest interaction Hamiltonian HI(z) involves a scalar field φ(z) interactingwith an external source field J(z) at a spacetime point z:

HI(z) = J(z)φ(z). (19.2)

We draw the source J(z) as a blob at a point z. The field is drawn as a stumpy lineattached to the source blob [see Fig. 19.2(a)].

(ii) The simplest self-interaction of fields is φ4 (say ‘phi-fourth’ or ‘phi-four’) theorydescribed by a Hamiltonian

HI(z) =λ

4!φ(z)4. (19.3)

This interaction causes four scalar φ-fields to meet at a spacetime point z [seeFig. 19.2(b)]. The diagram shows four stumpy lines meeting at z.

(iii) Another simple interaction is

HI(z) = gψ†(z)ψ(z)φ(z), (19.4)

where φ(z) is, again, a scalar field and ψ(z) is a complex scalar field. This interaction2

2This has a very similar structure tothe interaction used in QED. There theelectron corresponds to the excitationsin the psi-field and the photon to exci-tations in the phi-field.

causes an adjoint psi-field [represented by ψ†(z)], a psi-field [ψ(z)] and a scalar phi-field [φ(z)] to meet at z. To keep track of fields and their adjoints we use theconvention that a field ψ is shown by a stumpy line with an arrow pointing towardsthe interaction vertex [see Fig. 19.2(c)]. The adjoint field ψ† is shown with an arrowpointing away from the interaction.

(iv) Another very useful interaction is non-relativistic and describes the Coulombinteraction

HI(x− y) =1

2ψ†(x)ψ†(y)V (x− y)δ(x0 − y0)ψ(y)ψ(x). (19.5)

This one is delocalized. It says that a psi-field and an adjoint psi-field meet at x.These interact, via a potential V (x − y), with a psi-field and adjoint psi-field at y.The δ-function ensures that the interaction is instantaneous (that is, occurs at thesame time). [This is shown in Fig. 19.2(d).]

(a)

(b)

(c)

(d)

J φ(z)

φ(z) φ(z)

φ(z)φ(z)

φ(z)ψ†(z)

ψ(z)

ψ(x)

ψ†(x)

ψ(y)

ψ†(y)

V (x − y)

Fig. 19.2 Spacetime diagrams rep-resenting some commonly encounteredinteractions.


19.2 The example of φ4 theory

We’ll now carry out the full procedure of calculating an S-matrix elementfor the simple case of φ4 theory. The Lagrangian density describing theinteracting theory is given by

L =1

2[∂µφ(x)]2 − m2

2φ(x)2 − λ

4!φ(x)4. (19.6)

The free part is given by the first two terms L0 = 12 (∂µφ)2−m2

2 φ2, which

give rise to a free Hamiltonian operator upon canonical quantization

H0 =1

2

(∂φ

∂t

)2

+(∇φ)2

+m2φ2

. (19.7)

When we use the S-operator we work in the interaction picture, so thefree Hamiltonian time evolves the field operators φ(x). The states evolvevia the interaction Hamiltonian which we read off from the interactingpart of the Lagrangian LI = − λ

4!φ(x)4 as HI = λ4! φ(x)4. Next we need a

programme to expand the S-matrix using this interaction Hamiltonian.It’s a glorious five-step plan.

Step I

Decide what S-matrix element to calculate and write it as a vacuumexpectation value (VEV) of the free vacuum |0〉. We do this so we canuse the simple form of Wick’s theorem later.

As an example we’ll take our ‘in’ state to be a single particle in amomentum state p and the ‘out’ state will be a single particle in amomentum state q. We’ll be calculating the amplitude

A = out〈q|p〉in = 〈q|S|p〉= (2π)3(2Eq)

12 (2Ep)

12 〈0|aqSa

†p|0〉, (19.8)

where we recall that the relativistic normalization of our states meansthat |p〉 = (2π)

32 (2Ep)

12 a†p|0〉.

Step II

We expand the S-operator using Dyson’s expansion

S = T

[exp

(−i

∫d4z HI(z)

)]

= T

[1 − i

∫d4z HI(z) +

(−i)2

2

∫d4yd4w HI(y)HI(w) + . . .

](19.9)

= T

[1 − iλ

4!

∫d4z φ(z)4 +

(−i)2

2!

(λ

4!

)2 ∫d4yd4w φ(y)4φ(w)4 + . . .

].


Step III

Plug the resulting expression for the S-operator into the expression forthe S-matrix element that we’re trying to calculate

A = 〈q|S|p〉

= (2π)3(2Eq)12 (2Ep)

12 T

[〈0|aqa

†p|0〉 +

(−iλ

4

)∫d4z 〈0|aqφ(z)4a†p|0〉

+(−i)2

2!

(λ

4!

)2 ∫d4yd4w 〈0|aqφ(y)4φ(w)4a†p|0〉 + . . .

].

(19.10)

Since the amplitude is a sum of terms ordered by powers of λ, it makessense to label the terms A = A(0) + A(1) + A(2) . . ., where A(n) is theterm proportional to λn.

Step IV

Use Wick’s theorem to grind down the terms. That is, for each term wewrite the string of operators bookended by the vacuum states as a sumof all of the contractions of pairs.

Let’s first consider A(1), the first-order part of the expansion(the part proportional to λ). Using Wick’s theorem on the string〈0|aqφ(x)4a†p|0〉 ≡ 〈0|aqφ(z)φ(z)φ(z)φ(z)a†p|0〉, will yield up two sorts

of term. The first contains contractions of the a-operators and the φ-operators separately, like3

3Note that contractions involving a-operators don’t need T symbols sincehere the ap -operators are understoodto create particles at t = −∞ only,while the ap-operators destroy particlesat t = ∞ only, which fixes the time or-dering unambiguously.

〈0|aqφ(z)φ(z)φ(z)φ(z)a†p|0〉= 〈0|aqa

†p|0〉〈0|T φ(z)φ(z)|0〉〈0|T φ(z)φ(z)|0〉.

(19.11)

There are two other combinations of contractions of the φ-fields thatgive this term, making three in total.

The second type of term contracts a-operators with φ operators.There are twelve ways of doing this. One example is

〈0|aqφ(z)φ(z)φ(z)φ(z)a†p|0〉= 〈0|aqφ(z)|0〉〈0|T φ(z)φ(z)|0〉〈0|φ(z)a†p|0〉.

(19.12)

The result of this set of manipulations is that the first-order term in theexpansion of the S-matrix element for 〈q|S|p〉 is given by

A(1) = − iλ

4!

∫d4z

[3〈0|aqa

†p|0〉〈0|φ(z)φ(z)|0〉〈0|φ(z)φ(z)|0〉 (19.13)

+ 12〈0|aqφ(z)|0〉〈0|T φ(z)φ(z)|0〉〈0|φ(z)a†p|0〉].

Here are some rules for making sense of the contractions:

• Contractions between two fields make a free propagator4 for that

4Remember that for a scalar field

φ(x) = φ†(x) and

∆(y − z) =

Zd4q

(2π)4ie−iq·(y−z)

q2 −m2 + iǫ,

where q is a dummy momentum, overwhich we integrate. field, that is, φ(y)φ(z) = 〈0|T φ(y)φ(z)|0〉 = ∆(y − z).


• The contraction between a field and the creation operator

from the initial particle state φ(z)a†p corresponds to a factor1

(2π)32

1

(2Ep)12e−ip·z.

Example 19.2

To see this, we use the expansion on the field φ(z):

〈0|φ(z)a†p |0〉 =

Zd3q

(2π)32

1

(2Eq )12

〈0|“

aqe−iq·z + a†qeiq·z”

a†p |0〉

=

Zd3q

(2π)32

1

(2Eq )12

〈0|“

aqe−iq·z + a†qeiq·z”

|p〉

=

Zd3q

(2π)32

1

(2Eq )12

e−iq·zδ(3)(q − p)

=1

(2π)32

1

(2Ep)12

e−ip·z . (19.14)

Notice that, following our conventions, the factor e−ip·z cor-responds to an incoming particle. Notice also that the fac-tors 1

(2π)32

1

(2Ep)12

from the contraction exactly cancel against the

factors from the relativistic normalization of the states |p〉 =

(2π)32 (2Ep)

12 , with the result that the contraction φ(x)|p〉 = e−ip·x.

• Similarly, the contraction between a final state annihilation oper-

ator and a field aqφ(z) corresponds to a factor 1

(2π)32

1

(2Eq)12eiq·z.

This is the outgoing particle.

• Contractions between initial and final particles aqa†p = 〈0|aqa

†p|0〉

yield a delta function δ(3)(q − p).

Example 19.3

As an example of the use of the rules, consider the term:

12〈0|aq φ(z)φ(z)φ(z)φ(z)a†p |0〉.Using the rules, this is represented by

12

"

1

(2π)32

1

(2Eq )12

eiq·z#"Z

d4k

(2π)4e−ik·(z−z)

k2 −m2 + iǫ

#"

1

(2π)32

1

(2Ep)12

e−ip·z#

. (19.15)

Remember that to compute the contribution of this term to the S-matrix elementwe will need to integrate the variable z over all space. That is, on including thenormalization factors, we have that A, which is what we’ll call the contribution tothe amplitude 〈q|S(1)|p〉 from the above contraction, is given by

A = 12 × (2π)3(2Eq )12 (2Ep)

12

(−iλ)

4!

Z

d4z

"

1

(2π)32

1

(2Eq )12

eiq·z

× d4k

(2π)4e−ik·(z−z)

k2 −m2 + iǫ

1

(2π)32

1

(2Ep)12

e−ip·z#

=(−iλ)

2

Z

d4zd4k

(2π)4eiq·z

1

k2 −m2 + iǫe−ip·z . (19.16)


Notice that the normalization factors of (2π) and (2Ep) have cancelled. This was oneintention of the conventions for decorating the earlier equation with these factors.

We could, at this point, collect the terms together and prepare to in-tegrate the position of the interactions over all spacetime. However,there’s still one more part of the plan to carry out.

Step V

Make sense of a term by drawing a Feynman diagram. A Feynman di-agram represents an amplitude in the expansion of the S-matrix. Aparticular term in the expansion will comprise a certain number of in-teractions. This number is the same as the order of expansion of theS-operator. A second-order term has two interactions, a third-orderthree, etc. To represent the interaction in a diagram we draw the inter-action vertices exactly as shown in Fig. 19.2. Each line segment (we willcall these legs) emerging from an interaction vertex represents an uncon-tracted field operator φ(z). The Wick contractions (that we carried outin step IV) join the legs of the interaction to each other [for contractions

like φ(x)φ(y)] or to external particles (for contractions like φ(x)a†p).Here is a list of rules for drawing a diagram.

• Draw the interaction vertices and label them with their spacetimecoordinates [Fig. 19.3(a)].

• Contractions between the initial state and a field, i.e. φ(x)a†p, aredrawn as incoming lines connecting to one of the legs of the vertex[Fig. 19.3(b)]. This corresponds to a real, on-mass-shell particlecoming into the story.

• Propagators resulting from the field–field contractions φ(x)φ(y)are drawn as lines linking the points [Fig. 19.3(c)]. We can thinkof these as virtual particles which are internal to the story thediagram is telling.

• Contractions between the final state and a field aqφ(x) are drawnas an outgoing lines [Fig. 19.3(d)]. These correspond to on-mass-shell outgoing particles.

(a)

(b)

(c)

(d)

φ φ

φ φ

z (−iλ)

φ(z)a†pa

†p

φ

φφ

φ

φ

aqφ(z)aq

φ

Fig. 19.3 Steps in drawing a diagram.

Example 19.4

Let’s again consider the term

−12iλ

4!

Z

d4z〈0|ap φ(z)φ(z)φ(z)φ(z)a†p |0〉. (19.17)

The process of drawing a diagram is shown in Fig. 19.3.

• It’s a first-order diagram (it contains one copy of HI) so we draw one interac-tion vertex at position z [Fig. 19.3(a)].

19.3 Anatomy of a diagram 181

• The contraction φ(z)a†p is represented by an incoming line that grabs one ofthe vertex legs [Fig. 19.3(b)].

• The contraction φ(z)φ(z) ties together two vertex legs [Fig. 19.3(c)].

• The contraction aq φ(z) grabs the remaining vertex stub and forms a lineleaving the diagram [Fig. 19.3(d)].

The diagrams that result from our Wick expansion of the S-matrix tofirst order are shown in Fig. 19.4. The expression for eqn 19.11 is shownin Fig. 19.4(a), that for eqn 19.12 is shown in Fig. 19.4(b).

(a) (b)

φφ

φφ

φφ

aqφ

φa†paqa

†p

Fig. 19.4 First-order contributions in

φ4 theory. (a) A disconnected diagramwith two connected parts. (b) Con-nected diagram with one loop.

19.3 Anatomy of a diagram

The diagrams we’ve drawn so far could correspond to stories describingthe interaction of particles if you think of time running upwards. Acontraction gives you a full diagram. This may be made of many pieces.Figure 19.4(a) is a diagram made of two pieces, while Fig. 19.4(b) is adiagram made of a single piece. We call a piece of a diagram a con-nected diagram. Figure 19.4(a) is a disconnected diagram made up oftwo connected diagrams. Disconnected processes cannot influence eachother (because they are not connected to each other!) and so physicalintuition tells us that we will probably only have to consider connecteddiagrams (that expectation will turn out to be right on the money).5

5The cluster decomposition prin-ciple guarantees that distant experi-ments yield uncorrelated results. AnS-matrix constructed out of creationand annihilation operators turns outautomatically to satisfy this principle,which therefore provides the deep rea-son why these operators are required inquantum field theory, rather than justbeing a convenience. The cluster de-composition principle may be justifiedfor connected and disconnected Feyn-man diagrams using the linked clustertheorem, discussed in Chapter 22. SeeWeinberg, Chapter 4 for a detailed dis-cussion.

There’s more vocabulary to be learnt in order to talk diagrams with theprofessionals:

• External lines have one end which appears not to be connectedto anything. (Actually, they indicate a connection with the worldexterior to the physical process.) Figure 19.4(b) or the left-handpiece of Fig. 19.4(a) are connected diagrams with external lines.

• A vacuum diagram has no external lines. The right-hand piece ofFig. 19.4(a) is a (connected) vacuum diagram. Vacuum diagramsdon’t affect transition probabilities because they are not connectedto the incoming or outgoing particles; they only affect 〈0|S|0〉 (andtherefore only contribute a phase eiφ to transition amplitudes).

Vacuum

dia

gram

Incomingexternalline

Outgoingexternalline

inte

rnal

lines vertex

Fig. 19.5 The anatomy of a discon-nected third-order diagram.

A particular connected diagram might contain:

• Vertices where lines join together. These represent interactions.

• External incoming lines (we draw these below the vertices). Theyrepresent on-mass-shell particles entering the process.

• External outgoing lines (we draw these above the vertices). Theyrepresent on-mass-shell particles leaving the process.

• Internal lines (joining two vertices). These represent virtual par-ticles which are off-mass-shell and therefore exist internally withinthe diagram.


These various properties are illustrated in Fig. 19.5.Why do we draw Feynman diagrams? Simple: knowing how the di-

agrams relate to the contractions means that we can simply draw dia-grams for a particular interaction and, instead of going through all of theleg work of doing an expansion, we just write down the equation to whicha diagram corresponds. If someone were now just to give us a Feynmandiagram, how could we translate it into a term in the expansion? Hereare the rules:Remember that these are the rules for

φ4 theory. Other theories will haveslightly different rules.

Feynman rules for φ4 theory in position space

To calculate an amplitude in the S-matrix expansion, translate aFeynman diagram into equations as follows:

• Each vertex contributes a factor −iλ.

• Each line gives a propagation factor ∆(x−y), where x and y arethe start and end points of the line.

• External lines contribute incoming (−ip ·x) or outgoing (+ip ·x)waves e±ip·x.

• Integrate the positions of the vertices over all spacetime.

• In order to get the right coefficient in front of the term divide bythe symmetry factor D.

This last point is treated in the following section.

19.4 Symmetry factors

A potentially tricky thing is how to work out what number D we divideby in working out our Feynman diagram amplitude. The number arisesfrom the number of ways there are to produce a certain diagram throughcontractions, divided by 4! (In fact the reason for the 4! is to simplifythis number as much as possible.) Rather than deriving them we’llquote the result which the interested (and/or masochistic) reader caneasily justify by working through the combinatorics.6 The general rule6Often the symmetry factors aren’t re-

quired to understand the physics ofwhat a diagram is telling you, but whencombining several diagrams it’s usefulto know them.

is as follows: if there are m ways of arranging vertices and propagatorsto give identical parts of a diagram (keeping the outer ends of externallines fixed and without cutting propagator lines) we get a factor Di = m.The symmetry factor is given by the product of all symmetry factorsD =

∏iDi.

Two very useful special cases, which aren’t immediately apparent fromthe general rule, are as follows:

• Every propagator with two ends joined to one vertex (a loop) givesa factor Di = 2.

• Every pair of vertices directly joined by n propagators gives afactor Di = n!


Example 19.5

Figure 19.6 shows several diagrams to which we can apply the rules.

• (a) The bubble propagator has D = 2 since it contains a propagator with twoends joined to a vertex. Remember you are not allowed to move the ends ofexternal lines, so there are no more factors to worry about.

• (b) The two-bubble propagator has D = 4. Each bubble gives a factor of2, so D = 2 × 2. The bubble propagators in (a) and (b) are examples ofself-energy diagrams. They are basically free propagators with extra loopsin, but don’t interact with anything else. We will see later that these extraloops just change the constants in the free propagator, or to use the lingo ofquantum field theory, they ‘renormalize the free propagator’.

• (c) The double-bubble vacuum graph has two bubbles (or loops), each con-tributing a factor 2, but the bubbles can be swapped over (rotating the diagram180 about a vertical axis passing through the middle of it), and we get thesame diagram, giving us an extra factor m = 2. Therefore, we have D = 8.

• The Saturn diagram (d) has a pair of vertices joined by three lines, so D =3! = 6.

• Diagram (e) has a factor of 2 from the bubble. It also has two vertices joinedby two lines, contributing a factor 2!, so D = 2 × 2! = 4.

• Diagram (f) has three vertices. We think of these as being grouped in twopairs (with the middle vertex counted twice). Each pair is joined togetherby two lines (and so each contributes a factor 2!). In total we therefore haveD = 2 × 2! = 4.

• Diagram (g) has a bubble, so D = 2.

• Diagram (h) has two bubbles but the top and bottom parts between the ver-tices can be swapped giving an extra factor of 2, so D = 2 × 2 × 2 = 8.

• Diagram (i) has a pair of vertices joined by two lines so D = 2.

Other theories will have slightly different rules for their symmetry factors.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

Fig. 19.6 Examples of diagrams in φ4

theory.

The derivation of the Feynman rules represents the point where we canstart calculating scattering amplitudes for physical processes. A largepart of formulating a useful quantum field theory may be reduced tothe process of deriving the Feynman rules following the path shown inFig. 19.7.

Lagrangian

free part interacting part

Canonical

quantization

Dyson’s

expansion

for S

Wick expansion

Feynman’s rules

calculate

Amplitudes

Fig. 19.7 The process of deriving theFeynman rules.

19.5 Calculations in p-space

Calculations turn out to be far easier in momentum space, as illustratedby the following example: the second-order ‘Saturn diagram’.

Example 19.6

Let’s see how the Saturn diagram arises. Consider the second-order O(λ2) term inthe expansion of 〈q|S|p〉 (i.e. two copies of HI(x) in the expansion),

S(2) =(−i)2

2!

Z

d4yd4w HI(y)HI(z)

=(−iλ)2

2!(4!)2

Z

d4yd4w φ(y)φ(y)φ(y)φ(y)φ(w)φ(w)φ(w)φ(w). (19.18)


We now apply Wick’s theorem and obtain the diagrams shown in Fig. 19.8. The onein which we’re interested is Fig. 19.8(f), which comes from the term

〈q|S(2)|p〉 = (2π)3(2Eq )12 (2Ep)

12

× (−iλ)2

D〈0|aq φ(y)φ(y)φ(y)φ(y)φ(w)φ(w)φ(w)φ(w)a†p |0〉,

(19.19)

where D = 6 is the symmetry factor here. The amplitude for the Saturn Feynman

(b) (c)(a)

(d) (e) (f)

Fig. 19.8 All of the second-order φ4

diagrams.

w

y

q

k1 k2 k3

p

Fig. 19.9 (a) Position and (b) momen-tum space Feynman diagrams.

diagram in Fig. 19.9(a) is given by

−λ2

6

Z

d4yd4w eiq·y∆(y − w)∆(y − w)∆(y − w)e−ip·w. (19.20)

Each spacetime propagator ∆ gives us a factor of

∆(y − w) =

Zd4k

(2π)4e−ik·(y−w) i

k2 −m2 + iǫ. (19.21)

So, upon collecting all of the exponentials together, we have

Saturn = −λ2

6

Zd4k1

(2π)4d4k2

(2π)4d4k3

(2π)4i

(k21 −m2 + iǫ)

i

(k22 −m2 + iǫ)

i

(k23 −m2 + iǫ)

×Z

d4yd4w“

eiq·ye−i(k1+k2+k3)·y”“

ei(k1+k2+k3)·we−ip·w”

.

Next we notice that some of the integrals will give us delta functions, which simplifythings considerably. Integrating y over all space gives us an additional factor ofR

d4y e−i(k1+k2+k3−q)·y = (2π)4δ(4)(k1 + k2 + k3 − q) and we have

Saturn = −λ2

6

Zd4k1

(2π)4d4k2

(2π)4d4k3

(2π)4i

(k21 −m2 + iǫ)

i

(k22 −m2 + iǫ)

i

(k23 −m2 + iǫ)

×Z

d4w“

ei(k1+k2+k3)·we−ip·w”

(2π)4δ(4)(k1 + k2 + k3 − q),

which means that we may set k1 = q − k2 − k3 upon doing the k1 momentumintegral. We see that the integral has given us a constraint on the allowed values ofthe momenta. Translated into real life, this means that the sum of four-momentaat each vertex is zero. Energy-momentum is therefore conserved at each interactionvertex. So now we have

Saturn = −λ2

6

Zd4k2

(2π)4d4k3

(2π)4i

[(q − k2 − k3)2 −m2 + iǫ]

i

(k22 −m2 + iǫ)

i

(k23 −m2 + iǫ)

×Z

d4w“

eiq·we−ip·w”

.

The w integration givesR

d4w ei(q−p)·w = (2π)4δ(4)(q − p). The overall energy-momentum is conserved and the entire diagram carries an overall energy-momentumconserving delta function around with it. We’re left with two momentum integralsto do. Finally we arrive at the momentum space result:

Saturn =−λ2

6(2π)4δ(4)(q − p)

Zd4k2

(2π)4d4k3

(2π)4i

[(q − k2 − k3)2 −m2 + iǫ]

× i

(k22 −m2 + iǫ)

i

(k23 −m2 + iǫ)

. (19.22)

Although we’ve ended up with the same number of integrals as with the positionspace version, we’ve expunged all of the exponential factors. In addition, since theenergy-momentum conserving delta function is common to all diagrams, we oftenchoose not to carry it around, but to understand that energy-momentum is im-plicitly conserved (and reinstate the factor when it’s needed in the mathematics!).


We may now write down a set of Feynman rules for φ4 theory in mo-mentum space:

Feynman rules for φ4 theory in momentum space

• Each vertex contributes a factor −iλ [Fig. 19.10(a)].

• Label each internal line with a momentum q flowing along it anddescribe it by a propagator i

q2−m2+iǫ [Fig. 19.10(b)].

• Force the sum of each momentum coming into a vertex to beequal to the momentum leaving it.

• Integrate over unconstrained internal momenta with a measured4q

(2π)4 .

• External lines contribute a factor 1 [Fig. 19.10(c)].

• Divide by the symmetry factor.

• Include an overall energy-momentum conserving delta functionfor each diagram.

Remember that the external lines don’t contribute a propagator. In

momentum space the external line parts are just apφ(x) = 1. Rememberalso that you only need to integrate over unconstrained momenta. Thatis, only integrate over momenta which are not already determined by theconditions that momentum is conserved both at the vertices and overallfor a diagram. (If in doubt you can include a delta function for eachvertex and integrate over all momenta.)

(a)

(b)

(c)

= −iλ

=i

q2 −m2 + iǫq

= 1

= 1

Fig. 19.10 Momentum space Feyn-man rules for φ4 theory.

Example 19.7

Let’s now apply our rules to some example Feynman diagrams. Some terms in theexpansion of 〈q|S|p〉 up to second order are shown in Fig. 19.11. Applying the rulesto some of these yield the following for their contribution to 〈q|S|p〉:

(b) = (2π)4δ(4)(q − p)(−iλ)

2

Zd4k

(2π)4i

k2 −m2 + iǫ,

(f) = (2π)4δ(4)(q − p)(−iλ)

2

Zd4k3

(2π)4i

k23 −m2 + iǫ

×

(−iλ)

8

Zd4k1

(2π)4d4k2

(2π)4i

(k21 −m2 + iǫ)

i

(k22 −m2 + iǫ)

,

(g) = (2π)4δ(4)(q − p) ×(−iλ)2

4

Zd4k1

(2π)4d4k2

(2π)4i

(k21 −m2 + iǫ)

i

(p2 −m2 + iǫ)

i

(k22 −m2 + iǫ)

,

(h) = (2π)4δ(4)(q − p) ×(−iλ)2

6

Zd4k1

(2π)4d4k2

(2π)4i

(k21 −m2 + iǫ)

i

(k22 −m2 + iǫ)

i

[(p− k1 − k2)2 −m2 + iǫ].

(a) (b) (c)

(d) (e) (f)

(g) (h)

q q q

p p p

kk1

k2

p

p

q

k1

k2

q

p

p−k 1−k 2

k1 k2

Fig. 19.11 Examples of Feynman di-agrams for φ4 theory, up to second or-der in the interaction strength, for theS-matrix element 〈q|S|p〉.

In summary: the amplitude A =out 〈q|p〉in = 〈q|S|p〉 may be writtenas a sum of diagrams. Each diagram stands for an integral. We call


diagrams which describe how interactions affect the amplitudes of singleparticles self-energy diagrams.

We haven’t yet discussed how to finally do the integral to finally getthe numbers out. This is because often the integrals give us divergent(i.e. infinite) results! Taming these fearsome infinities reveals a hugeamount about the physics lying behind quantum field theory and wediscuss that in Chapter 32. Some diagrams don’t diverge, of course, andwe examine some of these in Chapter 20. For now, we need some moreexperience getting used to how the expansions go.

19.6 A first look at scattering

We have looked at diagrams describing single particles. What about aprocess of two particles entering and then leaving? The amplitude we’reafter is

〈q1q2|S|p2p1〉 = (2π)6(16Ep1Ep2

Eq1Eq2

)12 〈0|aq1

aq2Sa†p2

a†p1|0〉,(19.23)

describing particles entering in momentum states p1 and p2 and leavingin states q1 and q2. We could of course now start the process again andwork through all of the contractions. However, this is now unnecessarysince we have Feynman diagrams for the theory.

Example 19.8

Some diagrams contributing to the matrix element 〈q1q2|S|p2p1〉 are shown inFig. 19.12. The first two yield up

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

(j) (k) (l)

p1 p2

q1 q2

p1 p2

q1 q2

p1 p2

q1 q2

q1 q2

p1p2

k

p 1+p 2

−k

Fig. 19.12 Examples of Feynmandiagrams for the S-matrix element〈q1q2|S|p2p1〉 up to second order in theinteraction strength.

Diagram (a) = (2π)6 (16Ep1Ep2

Eq1Eq2

)12 δ(3)(q1 − p1) δ(3)(q2 − p2),

Diagram (b) = (2π)6 (16Ep1Ep2

Eq1Eq2

)12 δ(3)(q1 − p2) δ(3)(q2 − p1).

These clearly aren’t very interesting since no interactions occur. They just arisefrom the zeroth-order part of the S-operator, S(0) = 1. They don’t contribute toscattering, and therefore aren’t measured in an experiment. We’ll ignore them.

Some more interesting examples of amplitudes are the following:

Diagram (c) = (2π)4δ(4)(q1 + q2 − p1 − p2)(−iλ),

Diagram (f) = (2π)4δ(4)(q1 + q2 − p1 − p2)(−iλ)2

2

×Z

d4k

(2π)4i

[k2 −m2 + iǫ]

i

[(p1 + p2 − k)2 −m2 + iǫ].

Chapter summary

• The S-matrix may be expanded in a five-step process.

• The terms in the expansion can be encoded in Feynman diagrams.

• It’s easiest to work in momentum space.

Exercises 187

Exercises

(19.1) Write down the momentum space amplitudes forthe processes shown in the Feynman diagrams inFig. 19.6.

(19.2) (a) Draw the interaction vertex for φ3 theory,which is a scalar field theory with Lagrangian L =12(∂µφ)2 − m2

2φ2 − η

3!φ3.

(b) By expanding the S-matrix find the contribu-tions to the amplitude 〈q|S|p〉 up to second orderin the interaction strength. Draw the correspond-ing Feynman diagrams. What are the symmetryfactors?(c) Draw the new, connected diagrams that con-tribute to 〈q|S|p〉 when you consider the fourth-order contribution.(d) Which connected diagrams contribute to thetwo-particle S-matrix element 〈q1q2|S|p2p1〉 atfourth order in the interaction.For simplicity, just draw topologically distinct dia-grams, ignoring new ones produced by permutationsof external lines.

(19.3) Consider the ABA theory, defined by the La-grangian

L =1

2(∂µφA)2 − m2

A

2φ2

A (19.24)

+1

2(∂µφB)2 − m2

B

2φ2

B − g

2φAφBφA.

This theory describes the interaction of two scalarfields.

(a) Draw the interaction vertex using differentstyles of line for the different fields.(b) Draw the Feynman diagrams that contribute tothe scattering amplitude 〈qA|S|pA〉 up to fourth or-der (here pA refers to an A-particle in momentumstate p). Write expressions for the amplitudes forthese diagrams. You may ignore symmetry factors.(c) Draw the Feynman diagrams that contribute tothe scattering 〈qB|S|pB〉 up to fourth order in theinteraction. Write expressions for the amplitudesfor these diagrams. Again, you may ignore symme-try factors.(d) Draw Feynman diagrams that contribute tothe A-particle scattering 〈qA1qA2|S|pA2pA1〉 up tofourth order.(e) What are the rules for the symmetry factors ofthis theory?

(19.4) (a) Show that the amplitude for the double bubblediagram [Fig. 19.6(c)] is given by

A =−iλ

8

»Z

d4p

(2π)4i

p2 −m2 + iǫ

–2

δ(4)(x = 0).

(19.25)Here the δ(4)(x = 0) factor gives us a quantity pro-portional to the volume of the system V multipliedby the total time T . This factor is infinite, but wewill show later that this will not worry us.(b) Argue that this factor arises for all vacuum di-agrams.

20 Scattering theory

20.1 Another theory: Yukawa’sψ†ψφ interactions 188

20.2 Scattering in the ψ†ψφ the-ory 190

20.3 The transition matrix andthe invariant amplitude 192

20.4 The scattering cross-section193

Chapter summary 194

Exercises 194

And the LORD shall cause his glorious voice to be heard, andshall shew the lighting down of his arm, . . . with scattering,and tempest, and hailstones.Isaiah 30:30

One important application of quantum field theory is the calculation ofscattering cross-sections. It is important since scattering cross-sectionsare measured in many experiments. In this chapter we examine scat-tering using Hideki Yukawa’s ψ†ψφ theory, which is an illuminating toymodel describing the interactions of scalar fields. The joy of this model isthat it bears a strong resemblance to quantum electrodynamics (QED),which describes electrodynamics to an astounding degree of accuracy.Specifically, in ψ†ψφ theory the phion excitations in the φ-field take therole of (massive, scalar) photons and the psion excitations in the ψ-fielddescribe (complex scalar, bosonic) electrons.

20.1 Another theory: Yukawa’s ψ†ψφinteractions

ψ(x)

x

ψ†(x)

φ(x)

Fig. 20.1 The ψ†ψφ interaction ver-tex.

The theory describes a complex scalar field ψ and a real scalar field φinteracting. The full Lagrangian for this theory is

L = ∂µψ†∂µψ −m2ψ†ψ +1

2(∂µφ)2 − 1

2µ2φ2 − gψ†ψφ. (20.1)

Here psions have mass m and phions have mass µ. This Lagrangian isthe sum of the free scalar field and free complex scalar fields Lagrangianwith an interaction part, shown in Fig. 20.1, given by LI = −gψ†ψφ.We start by writing down the mode expansions of each of the free fields:

ψ(x) =

∫d3p

(2π)32

1

(2Ep)12


),

ψ†(x) =

∫d3p

(2π)32

1

(2Ep)12

(a†peip·x + bpe−ip·x

),

φ(x) =

∫d3q

(2π)32

1

(2εq)12

(cqe−iq·x + c†qeiq·x) , (20.2)

where Ep = (p2 + m2)12 and εq = (q2 + µ2)

12 . Here the a-operators

describe the creation and annihilation of psions, the b-operators de-scribe the creation and annihilation of antipsions and the c-operators

20.1 Another theory: Yukawa’s ψ†ψφ interactions 189

create and destroy scalar phions. The interaction Hamiltonian to usein Dyson’s equation is1 HI(z) = gψ†(z)ψ(z)φ(z), whose interaction dia-

1Remember that HI = −LI so LI =

−gψ†ψφ implies HI = gψ†ψφ.

gram is shown in Fig. 20.1. We decorate the interaction diagram with anarrow pointing toward the interaction blob for the ψ-field and an arrowpointing away for the ψ†-field. The meaning of these arrows will be toshow particle number flow, as discussed a little later.

To calculate S-matrix elements we’ll work through our five-point plan.The most important new feature of this theory is that now we have thepossibility of antiparticles in our interactions.Step I: We decide what to calculate. The one psion in, one psion outamplitude is

A = 〈q|S|p〉 = (2π)3(2Eq)12 (2Ep)

12 〈0|aqSa

†p|0〉. (20.3)

Step II: We expand the S-operator using Dyson’s expansion

S = 1 + (−ig)

∫d4z ψ†(z)ψ(z)φ(z) (20.4)

+(−ig)2

2!

∫d4yd4w

[ψ†(y)ψ(y)φ(y)

] [ψ†(w)ψ(w)φ(w)

]+ . . .

Step III: The next stage is to plug in to get the amplitudes for thevarious processes we’re going to calculate. All first-order terms will turnout to give zero (try it!) so we’ll consider the second-order term

A(2) = (−ig)2

2! (2π)3(2Ep)12 (2Eq)

12 〈0|aqψ

†(y)ψ(y)φ(y)ψ†(w)ψ(w)φ(w)ap|0〉.(20.5)

Step IV: We use Wick to digest the matrix elements. Here are the rulesfor the nonzero contractions for this theory:

ψ(x)ψ†(y) =∫

d4p(2π)4

ie−ip·(x−y)

p2−m2+iǫ , φ(x)φ(y) =∫

d4q(2π)4

ie−iq·(x−y)

q2−µ2+iǫ ,

apψ†(x) = 1

(2π)32

1

(2Ep)12eip·x, ψ(x)a†p = 1

(2π)32

1

(2Ep)12e−ip·x,

bpψ(x) = 1

(2π)32

1

(2Ep)12eip·x, ψ†(x)b†p = 1

(2π)32

1

(2Ep)12e−ip·x,

cqφ(x) = 1

(2π)32

1

(2εq)12eiq·x, φ(x)c†q = 1

(2π)32

1

(2εq)12e−iq·x.

(20.6)All other contractions give zero. Finally, we’ll agree to work in momen-tum space and note that the symmetry factor for all diagrams in ψ†ψφtheory is D = 1.

Example 20.1

Using these contractions let’s see what sort of diagrams are thrown up. Consider thesecond-order term with the following contraction

〈0|aq ψ†(y)ψ(y)φ(y)ψ†(w)ψ(w)φ(w)a†p |0〉. (20.7)


This leads to the Feynman diagram shown in Fig. 20.2(a), which is known as atadpole.2 In a theory with no particles in the ground state this diagram gives zero.

2This name was coined by Sidney Cole-man. When the journal Physical Re-view objected to the name, Colemansuggested the alternative ‘Spermion’.Physical Review relented.

Consider another set of contractions:

〈0|aq ψ†(y)ψ(y)φ(y)ψ†(w)ψ(w)φ(w)a†p |0〉. (20.8)

This leads to the Feynman diagram shown in Fig. 20.2(b), known as an oyster, whichwhen translated into equations gives an amplitude Aoyster:

(−ig)2Z

d4k

(2π)4i

(k2 − µ2 + iǫ)

i

[(p− k)2 −m2 + iǫ](2π)4δ(4)(q − p). (20.9)

There are also two disconnected contributions involving vacuum diagrams, shown inFig. 20.3.

(a)

(b)

aψ†

ψψ†

ψa†

φφ

aψ†

ψψ†

ψa†

φφ

Fig. 20.2 (a) A tadpole diagram. (b)An oyster diagram.

(a)

(b)

ψψ†

ψψ†

ψψ† ψψ†

φφ

φφ

aa†

aa†

Fig. 20.3 Disconnected diagramsmaking second-order contributions.

There are a few new things to note about the Feynman diagrams in thistheory. The first is that we include arrows on the psion and antipsionlines of our Feynman diagrams, but not the phi lines. Psions have arrowsgoing in the direction of ‘time’ (that is, up the page), antipsions havearrows going against the ‘time’ direction. These arrows don’t representthe direction of momentum. Actually arrows on the lines represent the(conventional) direction of Noether current JNc (i.e. conserved particle

number flow). Motivated by the expression QNc =∫

d3p (n(a)p − n

(b)p )

we say that incoming particles increase particle number and correspondto lines inwards on the diagram whilst incoming antiparticles reduceparticle number resulting in outward going lines in the diagram. Toavoid confusion, it sometimes helps to draw extra lines showing thedirections of momenta (see, e.g. Fig. 20.6). Notice that this implies thatan antiparticle has momentum in the opposite direction to its numberflow.

20.2 Scattering in the ψ†ψφ theory

So far we’ve only considered one particle coming in and one particle leav-ing. Now let’s consider the process of two psions coming in, scattering,then two psions leaving. The S-matrix element we’re after is

〈q1q2|S|p2p1〉 = 〈0|aq1aq2

Sa†p2a†p1

|0〉. (20.10)

It’s fairly obvious that all first-order terms are zero (since it’s impos-sible to draw two-in two-out Feynman diagrams with only one of ourinteraction vertices). We’ll consider the second-order terms.

Example 20.2

From the contraction

〈0|aq1aq2

ψ†(y)ψ(y)φ(y)ψ†(w)ψ(w)φ(w)a†p2a†p1

|0〉, (20.11)

20.2 Scattering in the ψ†ψφ theory 191

we have the diagram shown in Fig. 20.4(a). Two psions enter, one emits a force-carrying, virtual phion which collides with the second psion. The two psions thenleave. This is known in particles physics as a t-channel process (following notationintroduced by Stanley Mandelstam in 1958 in which the three possible processes wereassigned the letters ‘s’, ‘t’ and ‘u’).

Stanley Mandelstam (1928– )

A closely related process arises from the following contraction

〈0|aq1aq2

ψ†(y)ψ(y)φ(y)ψ†(w)ψ(w)φ(w)a†p2a†p1

|0〉, (20.12)

represented by the diagram in Fig. 20.4(b) and is known as a u-channel process. It’sthe same as the t-channel process, except that the initial and final particles havechanged places after the interaction. To obtain the amplitude 〈q1q2|S|p2p1〉 for thescattering of indistinguishable particles at second order add the amplitudes from thet- and u-channel processes.

What about the following process describing psions interacting with antipsions(denoted |p〉)? The expression

〈q1q2|S|p2p1〉 = 〈0|bq1aq2

Sa†p2b†p1

|0〉 (20.13)

describes a psion and antipsion entering, interacting, then leaving. We can still have

(a)

(b)

(c)

aq1ψ†(y) aq2ψ

†(w)

ψ(y)a†p1

ψ(w)a†p2

φ(y)φ(w)

φ(y)φ(w)

φ(y)φ(w)

aq1ψ†(w) aq2ψ

†(y)

ψ(y)a†p1

ψ(w)a†p2

ψ(w)a†p2

ψ†(w)b†p1

bq1ψ(y)aq2

ψ†(y)

Fig. 20.4 (a) A t-channel scatteringprocess. (b) A u-channel process. (c)An s-channel process involving parti-cles scattering from antiparticles.

the contraction which leads to a process analogous to Fig. 20.4(a), although this willinvolve antipsion lines with arrows pointing against the direction of time (i.e. downthe page) attached to one vertex. In addition, another interesting new contraction

(a)

(b)

(c)

(d)

= −ig

=i

q2 − µ2 + iǫq

= 1

= 1

p

p′q = (p − p′)

Fig. 20.5 Feynman rules for ψ†ψφ the-ory.

possibility occurs here. Look at

〈0|bq1aq2

ψ†(y)ψ(y)φ(y)ψ†(w)ψ(w)φ(w)a†p2b†p1

|0〉, (20.14)

which corresponds to the diagram in Fig. 20.4(c). This depicts a psion and anantipsion coming in and annihilating. They become a virtual phion which decaysinto a psion and antipsion. This is known in particle physics as an s-channel process.

Note that the u-channel process now involves the exchange of the particle and an-tiparticle, so becomes distinguishable from the t- and s-channel processes. It there-fore contributes to 〈q1q2|S|p2p1〉, rather than 〈q1q2|S|p2p1〉. To obtain the amplitude〈q1q2|S|p2p1〉 at second order we therefore need to add the amplitudes from the t-and s-channel processes only.

The idea behind all of this exploration is to write down a set of Feynmanrules for this theory, which we now do.

Feynman rules for ψ†ψφ theory

• Each vertex contributes a factor −ig [Fig. 20.5(a)].

• For each phion internal line carrying momentum q include a prop-agator i

q2−µ2+iǫ [Fig. 20.5(b)]. For a psion internal line include

a propagator iq2−m2+iǫ .

• Integrate over all undetermined momenta.

• Incoming and outgoing lines contribute a factor 1 [Fig. 20.5(c-d)].

• All symmetry factors are 1.

• Include an overall energy-momentum conserving delta functionfor each diagram.


Example 20.3

The amplitude for the second-order scattering diagrams (including the overall energy-momentum conserving delta function) is

A(2) = (−ig)2i

q2 − µ2 + iǫ(2π)4 δ(4)

“X

pf −X

pi

”

. (20.15)

Here we don’t integrate over q, as it is completely determined by momentum conser-vation for these diagrams. Referring to Fig. 20.6 we have the following amplitudes:

• For the t-channel process q2 = (p′ − p)2 = t and we have

At = (−ig)2i

t− µ2 + iǫ(2π)4 δ(4)(p′ + k′ − p− k). (20.16)

• For the u-channel process we have q2 = (p′ − k)2 = u and

Au = (−ig)2i

u− µ2 + iǫ(2π)4 δ(4)(p′ + k′ − p− k). (20.17)

• For the s-channel process we have q2 = (p+ p′)2 = s and

As = (−ig)2i

s− µ2 + iǫ(2π)4 δ(4)(k + k′ − p− p′). (20.18)

(a)

(b)

(c)

p

p′

k

k′

p p′

k k′

p

p′

k

k′

q = p′ − p

q = p′ − k

q = p+ p′

Fig. 20.6 The t-, u- and s-channel scattering processes in momen-tum space. 20.3 The transition matrix and the

invariant amplitude

One thing that one immediately notices about the expansion of the S-operator is that the first term is unity. This means that the only nonzeroamplitude from this term in the expansion has identical initial and finalstates. This is a rather dull result as far as scattering is involved (sincenothing happens) so it’s commonly removed by writing S = 1+iT , wherethe matrix elements of T are often called the transition or T -matrix.Another irritation is that diagrams carry the overall energy-momentumconserving delta function. This is often factored out to make the so-called invariant amplitude M, defined for a two-in, two-out process as

〈p1fp2f |iT |p2ip1i〉 = (2π)4δ(4)(p1f + p2f − p2i − p1i)iM, (20.19)

which at least prevents us from having to continuously write down thedelta function.33The factor of i is included so that the

amplitudes match up with the resultsfrom the conventions of non-relativisticscattering theory.

We’re now ready to reveal a wonderful simplification. Remember thatdiagrams fall into two classes: connected diagrams and disconnected di-agrams. It turns out that only fully connected diagrams contribute to theT -matrix, a point which makes good physical sense since disconnecteddiagrams correspond to physical processes which do not influence eachother. We also note that there are a class of diagrams which have loopsattached to the external legs. These are bad news and often lead tounpleasant and non-trivial infinities. The good news is that they don’tcontribute to the T -matrix either, so we can remove them.4 We are left

4To remove them, we define the act ofamputation, which involves choppingoff all such carbuncles on the externallegs.

20.4 The scattering cross-section 193

with the appealing result that

iM(2π)4 δ(4)(∑

pf −∑

pi

)=∑

All connected, amputated Feynman

diagrams with incoming momentum pi

and outgoing momentum pf

.

(20.20)We now need to design an experiment which measures M and that

turns out to be a scattering experiment.

20.4 The scattering cross-section (a)

(b)

Fig. 20.7 Scattering cross-sections.(a) σsheep > σfield mouse. (b)σcow, side > σcow, front.

When firing one particle at another, we want to think about the ampli-tude of scattering of our incoming particle in terms of how big the otherparticle appears to be in cross-sectional area terms. Imagine driving atnight, where your car headlights give you a beam of particles with whichyou probe the inky blackness. A sheep shows up more in your headlightsthan a field mouse, simply because it’s bigger and scatters more photons[Fig. 20.7(a)]. Similarly a cow will scatter more light if seen from theside than from the front, because it offers a larger cross-sectional areain the former case [Fig. 20.7(b)]. However it’s not just about area –white sheep are more visible than black sheep – but we fold this all into our definition of the scattering cross-section σ in which we measurethe rate R of a scattering process occurring as R = σL, where L is theluminosity5 of our incoming beam of particles.

5Luminosity is a quantity with dimen-

sions Time−1× Area−1 to be evaluatedfor the case under study.

Scattering can occur in all directions, and so in each bit of solid angledΩ there will be a bit of scattering dσ. A detector usually only covers arange of solid angle, so it is usual to think in terms of the differential

cross-section6 defined by dσ/dΩ. Fermi’s golden rule allows us to6The relationship between σ anddσ/dΩ is given by

Z 2π

0dφ

Z 1

−1d(cos θ)

dσ

dΩ= σ.

relate the scattering cross-section to the modulus squared of a matrixelement. After taking proper account of all the normalization factorsone can show,7 for example, that for a scattering process involving two

7This result is discussed in Peskin andSchroeder, Chapter 4.

equal-mass particles scattering off each other then

dσ

dΩ=

|M|264π2E2

CM

, (20.21)

where E2CM is the total energy in the centre of mass frame.

Example 20.4

We’re finally going to work out some scattering amplitudes and cross-sections forψ†ψφ theory. We start with the scattering of two distinguishable psion particles. Inthis case only the t-channel diagram [Fig. 20.6(a)] contributes. Let’s further simplifyby considering only a non-relativistic case. This means that we can take the four-momentum p = (Ep ,p) to be p ≈ (m,p). We have

p1i ≈ (m,p), p2i ≈ (m,k),p1f ≈ (m,p′), p2f ≈ (m,k′),

(20.22)


so t may be written t = q2 = (p1f − p1i)2 ≈ −|p′ − p|2 = −|q|2, where q = p′ − p is

the three-momentum transfer. The scattering amplitude is given by evaluating thet-channel Feynman diagram yielding

iM =ig2

|q|2 + µ2, (20.23)

where we ignore the iǫ, since it’s unnecessary here. Plugging into our expression forthe cross-section and noting that ECM = 2m, we obtain

dσ

dΩ=

1

256π2m2

„g2

|q|2 + µ2

«2

. (20.24)

We have a measurable prediction from the theory. Interestingly this calculationcan also be done in non-relativistic quantum mechanics using Born’s approximation,which says that the amplitude for scattering from a potential V (r) is 〈p′|iT |p〉 =

Max Born (1882–1970) −iV (q)(2π)δ(Ep′ − Ep), where V (q) = 〈p′|V (r)|p〉 =R

d3r ei(p−p′)·rV (r). Com-paring with our quantum field theory result, we see that Born says that our scatteringpotential is

V (q) =−g2

|q|2 + µ2. (20.25)

What function has a Fourier transform that looks like this? The answer is

V (r) = − g2

4π|r| e−µ|r|; (20.26)

the potential dies away over a distance 1/µ. But this is just Yukawa’s potential!88See Chapter 17.This is the reason why the ψ†ψφ theory is often called Yukawa theory.

We now have enough formalism to start calculating amplitudes relevantin real systems.9 However, in the next chapter we will take a break9Readers interested in fermions, pho-

tons and quantum electrodynamicsmay now proceed to Chapter 36. How-ever, the main route of the book will bedifferent.

from quantum mechanics and discuss the seemingly unrelated subjectof statistical physics. This subject has an unexpected connection toquantum field theory that will provide some powerful new insights.

Chapter summary

• Yukawa’s theory involves an interaction term HI = gψ†ψφ.

• We have introduced two new diagrams: tadpole and oyster.

• The scattering matrix and cross-section can be related to the in-variant amplitude M. The only contribution to M comes fromconnected, amputated Feynman diagrams.

Exercises

(20.1) Verify the rules for the contractions in eqn 20.6.

(20.2) (a) Verify that the Fourier transform of V (r) =

− g2

4π|r|e−µ|r| is V (q) = −g2

|q|2+µ2 .

(b) By taking an appropriate limit, find the form ofthe Fourier transform of the Coulomb potential.

Part V

Interlude: wisdom from

statistical physics

Statistical physics is the study of large assemblies of atoms or particlesand allows one to extract thermodynamic quantities through a processof averaging. Typically you write down the partition function Z forthe system and manipulate it to generate the quantities you want. Inquantum field theory there is an analogous process which involves agenerating function Z[J ] which can be processed to bring forth Green’sfunctions.

• A rapid crash course in basic statistical physics is given in Chap-ter 21, showing how the partition function can be used to manufac-ture any desired thermodynamic quantity or correlation function.

• Chapter 22 develops these ideas for quantum field theory, intro-ducing the generating functional Z[J ] and linking this with theS-matrix via the Gell-Mann–Low theorem and showing how theseconcepts are connected with diagrams.

21Statistical physics: a crash

course

21.1 Statistical mechanics in anutshell 196

21.2 Sources in statistical physics197

21.3 A look ahead 198

Chapter summary 199

Exercises 199

Relax, you know more than you think you do.Benjamin Spock (1903–1998), Baby and Child Care

21.1 Statistical mechanics in a nutshell

Consider a system with a Hamiltonian H0 that has energy eigenvaluesEλ. The probability of the system being in some particular state |λ〉with energy Eλ at a temperature T is given by the Gibbs distribution

pλ =e−βEλ

Z, (21.1)

where β = 1/kBT and Z is known as the partition function which isdefined as a sum over states Z =

∑λ e−βEλ , or equivalently

Z =∑

λ

〈λ|e−βH0 |λ〉 = Tr[e−βH0

]. (21.2)

The term Z is needed to guarantee that∑λ pλ = 1, which is a require-

ment of any probability distribution. However, the partition functionZ is more important than a mere normalization constant. It turns outthat all of the information about the system is contained in Z and forthis reason Z is also known as a generating function.

i = 1 2 3 4 5 6 7 8 9 10

...

Fig. 21.1 A simple model of a one-dimensional magnet.

Consider a one-dimensional magnet which is made up of a lattice ofN sites, each one labelled by an index i, decorated with spins (shown inFig. 21.1). The spins have magnitude S = 1

2 and can point up (Sz = + 12 )

or down (Sz = − 12 ).1 We define the operator-valued field φi. This is the

1This, of course, is shorthand. Wemean that the spin states are eigen-states of the Sz operator, which haseigenvalues Sz = ± 1

2.

same sort of field as we had before: we input a lattice point position iand get out an operator which we use to act on a state of the system.In this case the operator is Szi which we understand to act only on thespin in the ith position.

Example 21.1

A state-vector |A〉 lists the values of each of the spins in our one-dimensional magnet,e.g. |A〉 = | ↑↑↓↓↓↓ . . .〉. We input a position (e.g. i = 4) and obtain an operatorcorresponding to that position φ4. Acting on the state |A〉 with φ4 yields

φ4| ↑↑↓↓↓↓ . . .〉 = −1

2| ↑↑↓↓↓↓ . . .〉, (21.3)

21.2 Sources in statistical physics 197

since the spin at position i = 4 has eigenvalue Sz4 = − 12.

Acting on |λ〉 with φi yields S(λ)zi |λ〉, where S

(λ)zi is the eigenvalue of

the operator Sz applied to the ith spin for the state |λ〉. As usual,the expectation value of the φi operator for the state |λ〉 is 〈λ|φi|λ〉.At a temperature T we might want to know the thermal expectationvalue of the ith spin, denoted 〈φi〉t which is found by summing overthe expectation values weighted by the probabilities of finding a state ofenergy Eλ thus:

〈φi〉t =∑

λ

S(λ)zi pλ =

1

Z

∑

λ

S(λ)zi e−βEλ

=1

Z

∑

λ

〈λ|φi|λ〉e−βEλ

=1

Z

∑

λ

〈λ|φie−βH0 |λ〉.

=Tr[φie

−βH0

]

Z. (21.4)

The quantity ρ = e−βH0/Z is known as the probability density operator.Its matrix elements are known collectively as the density matrix. Weconclude that to find the thermal expectation value of an operator instatistical physics we need to multiply it by the density matrix and thentake a trace over all of the states. That’s pretty much all there is tostatistical physics.

In the absence of a magnetic field, asystem of non-interacting spins at someparticular temperature is expected tohave 〈φi〉t = 0 for all i. This reflectsthe fact that spin i has equal proba-bility to be found with Sz = 1/2 or−1/2. If however a system shows spinorder (more usually called magnetic or-der) then we would expect 〈φi〉t 6= 0 forall i.

21.2 Sources in statistical physics

There’s a neater way of getting the thermal average of the field 〈φi〉t.It involves adding to the Hamiltonian H0 a fictional term Hs =− 1β

∑k Jkφk. Such a term, which involves coupling a field Jk to the

spin at position k, is known as a source term for reasons which willcome apparent later.2 With the inclusion of the source term, the parti- 2The source term for a magnet is actu-

ally proportional to the magnetic fieldBi, i.e. the value of the B-field evalu-ated at position i.

tion function is now

Z(J) = Tr[e−βH

]= Tr

[e−βH0+

P

k Jkφk

]. (21.5)

The point of this procedure is that we can now use the partition functionZ(J) to work out the thermal average 〈φi〉t by differentiating Z(J) withrespect to Ji, and evaluating this at Ji = 0 and dividing by Z(J = 0).That is to say

〈φi〉t =1

Z(J = 0)

∂Z(J)

∂Ji

∣∣∣∣Ji=0

=Tr[φie

−βH0

]

Z(J = 0), (21.6)

as we had before.

198 Statistical physics: a crash course

But that’s not all. We can also use this philosophy to calculate cor-

relation functions. A useful question in statistical physics is to askabout paired averages. For example, if we know the value of the spin atposition i, we’d like to know the probability that the spin at position j isin the same state. The quantity we’re after is Gij = 〈φiφj〉t. This tellsus about the degree of correlation between spins in different parts of thesystem. If the spins are completely random, then Gij = 0, whereas ifthey are completely aligned then Gij = 1

4 (because the spin eigenvaluesare ± 1

2 ). More interesting is when the correlation is partial and thenwe might expect to find Gij = 1

4 when |i − j| is small but Gij → 0 as|i − j| → ∞. Sometimes it is interesting to look for deviations fromorder, which is the situation when 〈φi〉t 6= 0. In that case, one canexamine the connected correlation function

Gcij = 〈φiφj〉t − 〈φi〉t〈φj〉t. (21.7)

Our notation makes these correlation functions look rather like cousinsof the Green’s functions that we’ve been considering previously. Someexample correlations we can find include

Gij = 〈φiφj〉t =1

Z(0)

∂2Z(J)

∂Ji∂Jj

∣∣∣∣J=0

,

Gijk = 〈φiφj φk〉t =1

Z(0)

∂3Z(J)

∂Ji∂Jj∂Jk

∣∣∣∣J=0

, (21.8)

and in general

〈φi1 ...φin〉t =1

Z(0)

∂nZ(J)

∂Ji1 ...∂Jin

∣∣∣∣J=0

. (21.9)

The moral of this story is that by linearly coupling an operator-valuedfield φi to a source Ji our partition function becomes a generating func-tion. Once we’ve found the generating function we gain access to theaverage value of φi and a vast set of related quantities.

21.3 A look ahead

There is a close analogy between statistical physics, as exemplified bythe magnet, and quantum field theory. In fact everything we’re lookedat in this chapter has an analogue in field theory. In going to field theorywe’ll take a continuum limit, turning the lattice label i into the positionx so that φi → φ(x). The correlation functions of statistical physicsbecome propagators in field theory and our goal is to find these.

This is all very nice and certainly deserves further examination,3 but3See Chapter 25.

the point of this chapter is to tell you that to reach the goal of findingpropagators for a field theory we can use a generating functional Z[J ]similar to that used in statistical physics. The generating functional forfields will allow us, through (functional) differentiation, to find all of the

Exercises 199

Statistical physics Field theory(on a lattice) (continuum)

Source Ji J(x)Generating function Z(J) Z[J(x)]

Green’s functions G(n)i1...in

= 〈φi1 ...φin〉t G(n)(x1, ..., xn) = 〈Ω|T φ(x1)...φ(xn)|Ω〉Differentiation recipe G

(n)i1...in

= 1Z(J=0)

∂nZ(J)∂Ji1

...∂Jin

∣∣∣J=0

G(n)(x1, ..., xn) = 1in

1Z[J=0]

δnZ[J]δJ(x1)...δJ(xn)

∣∣∣J=0

Order 〈φi〉t 6= 0 〈Ω|φ(x)|Ω〉 6= 0

Table 21.1 Analogies between quantities in statistical physics on a lattice and thosein quantum field theory.

propagators. In equations, this can be written as

G(x1, ..., xn) = 〈Ω|T φ(x1)...φ(xn)|Ω〉

=1

in1

Z[J = 0]

δnZ[J ]

δJ(x1)...δJ(xn)

∣∣∣∣J=0

. (21.10)

One of the marvellous things about field theory is that even whenwe can’t solve a theory exactly, the generating functional Z[J ] and theGreen’s functions G(n)(x1, ..., xn) can be expressed in terms of Feynmandiagrams. We will examine this in the next chapter.

Chapter summary

• Correlation functions in statistical mechanics can be extracted bydifferentiating the partition function Z with respect to the sourceterm J . This approach carries over into quantum field theory be-cause Green’s functions can be extracted from a generating func-tional Z[J ]. Some analogies between statistical mechanics andquantum field theory are tabulated in Table 21.1.

Exercises

(21.1) Define the density operator as ρ = e−βH

Z. Us-

ing the rule that thermal averages are given by

〈A〉t = Trh

ρAi

, show that for a quantum oscillator,

with Hamiltonian H = ωa†a, the thermal average

of the number of excitations is given by

〈n〉t = 〈a†a〉 =1

eβω − 1. (21.11)

200 Statistical physics: a crash course

(21.2) Response functions or susceptibilities are ubiquitousin physics. Consider the forced quantum oscillator

L =1

2mx(t)2 − 1

2mω2x(t)2 + f(t)x(t), (21.12)

and take the definition of the response function tobe

〈ψ(t)|x(t)|ψ(t)〉 =

Z ∞

−∞dt′χ(t− t′)f(t′). (21.13)

(a) Using the interaction representation and treat-ing H ′ = −f(t)x(t) as the interaction part, showthat, to first order in the force function fI(t), wehave

|ψI(t)〉 = |0〉 + i

Z t

−∞dt′fI(t

′)xI(t′)|0〉. (21.14)

(b) Using the previous result, show that

χ(t− t′) = iθ(t− t′)〈0|ˆ

xI(t), xI(t′)˜

|0〉. (21.15)

(c) Using the fact that xI(t) =`

12mω

´ 12 (ae−iωt +

a†eiωt), calculate the response function for thequantum oscillator at temperature T = 0.(d) Using the fact that at nonzero T we have〈n〉t = (eβω − 1)−1, calculate the response func-tion for the quantum oscillator at nonzero T .

(e) Compare the behaviour of the response functionwith that of the correlation function S defined as

S =1

2〈xI(t

′), xI(t)〉 =1

2〈xI(t

′)xI(t)+xI(t)xI(t′)〉.

(21.16)

(21.3) Diffusion problems are common in thermodynam-ics. Here we meet the diffusion equation. The dif-fusion equation for the number density of particlesn(x, t) is

∂n(x, t)

∂t−D∇

2n(x, t) = 0. (21.17)

Consider a point source at time t = 0 and po-sition x = y, giving us the boundary conditionsG(x−y, t = 0) = δ(3)(x−y). Show that, for timest > 0 the Green’s function for this equation is givenby

G(ω, q) =1

−iω +D|q|2 , (21.18)

and identify the position of the pole in the complexω plane.Hint: one way to do this problem which makes theboundary conditions clear is to take a Laplace trans-formation in time and a Fourier transform in space.See Chaikin and Lubensky, Chapter 7, for a discus-sion.

22The generating functional

for fields

22.1 How to find Green’s func-tions 201

22.2 Linking things up with theGell-Mann–––Low theorem 203

22.3 How to calculate Green’sfunctions with diagrams 204

22.4 More facts about diagrams206

Chapter summary 207

Exercises 208

Since the finished product cannot be served to the table as itis, this isn’t so much a recipe, a method or a list of ingredi-ents: it is a way of life, for if there is fresh stock made everyweek, a reputation as a good cook will follow.Len Deighton (1929– ), Action Cook Book

In this chapter we will present a general method of finding Green’s func-tions based on the philosophy of generating functionals. We will explainthe meaning of the generating functional of quantum field theory, Z[J ],and show that all Green’s functions may be extracted from it thoughdifferentiation. This will allow us to relate together the S-matrix andGreen’s functions, linking two of the most important topics in quantumfield theory.

Although this chapter contains a lot of formal material the ideas arequite simple (and have much in common with the world of statisticalphysics). In fact the rules that emerge for perturbation theory allow us towrite generating functionals and Green’s functions as sums of Feynmandiagrams, relating formal objects such as G(x, y) and Z[J ] to a series ofcartoons.

22.1 How to find Green’s functions

All of the information about a field theory is available in the form ofGreen’s functions. Happily there’s an easy way to work out Green’sfunctions which is based on a trick which enables us to lump all of theinformation about a quantum field theory in a single entity called thegenerating functional or Z[J ]. The generating functional is repletewith all that juicy data in a zipped up and compressed form and there’sa simple procedure for extracting it. This is identical to the case ofstatistical physics, where all correlation functions were obtainable fromthe partition function Z(J).

Just as in statistical physics we make use of the philosophy of sources.In order to make a generating functional we’ll add a source function J(x)to our Lagrangian density, which couples linearly1 with the field φ(x): 1Note that by adding Jφ to L, we will

end up subtracting Jφ from H.L[φ(x)] → L[φ(x)] + J(x)φ(x). (22.1)

The reason why J(x) is referred to as a source is that in quantum fieldtheory this term will generate particle excitations in the field φ(x). By

202 The generating functional for fields

turning on a source we are, essentially, grabbing the field and shakingexcitations out.

We now define the generating functional for quantum fields as

Z[J ] = 〈Ω|U(∞,−∞)|Ω〉J , (22.2)

where U is the time-evolution operator for the full Hamiltonian H of thetheory and the J subscript on the right-hand side reads ‘in the presenceof a source J ’ when said out loud. We also have that H|Ω〉 = 0. Thatis |Ω〉 is the physical ground state of the system and the ground stateenergy is defined to be zero. Turning eqn 22.2 into words we see thatthe generating functional Z[J ] tells us the amplitude

Z[J ] =

⟨no particlesat x0 = ∞

∣∣∣∣no particlesat y0 = −∞

⟩

J

, (22.3)

that is, in the presence of a source, we start with no particles and endwith no particles. We could call Z[J ] the ‘no particle propagator’.

As we have defined it, Z[J ] allows two classes of process apart from therather trivial option of nothing happening at all: (i) those where a par-ticle is created by the source J and later absorbed by J ; (ii) those wherea particle spontaneously appears and then disappears. The latter class,corresponding to Z[J = 0], is described by vacuum diagrams and needsto be removed by normalization in order for the generating functionalto work. We therefore define a normalized generating functional:

Z[J ] =Z[J ]

Z[J = 0]. (22.4)

This guarantees that Z[J = 0] = 1 or, in other words, that the amplitudeto start and end with no particles in the absence of sources is unity.

We now need to find a way of calculating Z[J ]. Writing this in termsof the time-evolution operator as Z[J ] = 〈Ω|U(∞,−∞)|Ω〉J turns thisinto a problem of finding U(∞,−∞). A clue is provided by the wayin which we worked out the S-matrix in the last few chapters. In theinteraction picture, the time-evolution operator UI(∞,−∞) is known asthe S-operator and the Dyson expansion gives us a way to calculateS via S = T e−i

R

d4x HI . Things aren’t so simple here though, sinceU(∞,−∞) in eqn 22.2 is defined in the Heisenberg picture. The prob-lem is that we don’t have free fields in the Heisenberg picture so wecan’t use Wick’s theorem to grind down long strings of operators (sinceWick’s theorem only works on free fields). However we can still writea Dyson-like equation for the time evolution in terms of the Heisenbergfields φH. If we treat the source term like we would an interaction bysaying HI = −JφH, then the time-evolution operator is given by Dyson’sexpression U(∞,−∞) = T e−i

R

d4x HI . Splitting this up in this way willonly generate processes involving the sources and not the vacuum dia-grams, and so will result in the normalized generating functional. Wecan therefore write

Z[J ] = 〈Ω|T eiR

d4x J(x)φH(x)|Ω〉, (22.5)

22.2 Linking things up with the Gell-Mann–––Low theorem 203

where the fields φH(x) evolve according to the full Hamiltonian (ratherthan a free Hamiltonian H0 used in the interaction picture) and we usethe interacting ground state |Ω〉 rather than the free ground state |0〉.Again we stress this is all still permissible, as long as we remember thatwe wouldn’t be able to use Wick’s theorem. We expand Z[J ] to obtain

Z[J ] = 1 +

∞∑

n=1

in

n!

∫d4x1...d

4xn J(x1)...J(xn)〈Ω|T φH(x1)...φH(xn)|Ω〉.

(22.6)Recalling that Green’s functions are defined by

G(n)(x1, ..., xn) = 〈Ω|T φH(x1)...φH(xn)|Ω〉, (22.7)

we notice that we can get access to them through a manipulation of thegenerating function Z[J ] by differentiating eqn 22.6 as follows:

G(n)(x1, ..., xn) =1

inδnZ[J ]

δJ(x1)...δJ(xn)

∣∣∣∣J=0

. (22.8)

That is, we get the Green’s functions (which contain all of the informa-tion about a system) by differentiating a single functional Z[J ]. Thismethod of doing quantum field theory is illustrated in Fig. 22.1. In termsof the unnormalized generating functional Z[J ] we have the equivalentrule

G(n)(x1, ..., xn) =1

in1

Z[J = 0]

δnZ[J ]

δJ(x1)...δJ(xn)

∣∣∣∣J=0

, (22.9)

which closely resembles the version from statistical physics.

Lagrangian

Generating

functional

Z[J ]

Green’s

functions

Physical

predictions

add a source current J

differentiate

Fig. 22.1 Doing quantum field theoryusing generating functionals

We’ve done what we set out to do: we have found a single functionalthat contains all of the information about the system, which can be usedto find Green’s functions. All we need to do now is find an automatedway of calculating Z[J ] for any theory. It turns out that there are twoways: the first is by relating Z[J ] to the S-matrix; the second usesFeynman’s path integral. In this chapter we pursue the first approach,leaving the second for later (Chapter 24).

22.2 Linking things up with the

Gell-Mann–––Low theorem

We’ll examine the first method: calculating Z[J ] from the S-matrix. Westress that the S-matrix is defined for the interaction picture, so we’llneed an equation for Z[J ] given explicitly in terms of interaction picturefields. We’ll pluck the answer from the air and then prove we’ve pluckedcorrectly. We propose a form for the normalized generating functionZ[J ] which looks like:

Z[J ] =Z[J ]

Z[0]=

〈0|T e−iR

d4x[HI−J(x)φI(x)]|0〉〈0|T e−i

R

d4xHI |0〉, (22.10)


which looks like a good guess because it has the form of Z[J ] in eqn 22.5,but this time writing the interaction Hamiltonian in the form HI − JφI.This isn’t quite in terms of an expression for the S-matrix yet as we havethe source term in the numerator. Using the rule that you differentiaten times, divide by in and set J = 0 to yield a Green’s functions, we havethe Green’s function

G(n)(x1, ..., xn) =〈0|T φI(x1)...φI(xn)e

−iR

d4xHI |0〉〈0|T e−i

R

d4xHI |0〉, (22.11)

and this can be written in terms of the S-operator as

G(n)(x1, ..., xn) =〈0|T φI(x1)...φI(xn)S|0〉

〈0|S|0〉. (22.12)

All that’s left is now is to prove that we’ve made a correct guess, so that

〈Ω|T φH(x1)...φH(xn)|Ω〉 =〈0|T φI(x1)...φI(xn)S|0〉

〈0|S|0〉. (22.13)

This result turns out to be true and is known as the Gell-Mann–––Low

theorem.2 Let’s be completely clear about what eqn 22.13 is saying.2The proof of the theorem can be foundin the book by P. Coleman. It wasfirst proved by Gell-Mann and Low in1951. Murray Gell-Mann (1929– ) re-ceived the 1969 Nobel Prize in Physicsand, borrowing a term from Finnegans

Wake, gave the name to ‘quarks’. Fran-cis Low (1921–2007) worked with Gell-Mann at Princeton, and spent most ofhis later career at MIT.

First examine the left-hand side:

• |Ω〉 is the interacting ground state, defined by the full Hamiltonianas H|Ω〉 = 0. φH(x) are Heisenberg field operators. They time-evolve according to the full Hamiltonian H.

Next consider the right-hand side:

• Since we’ve turned off the source term now, the Hamiltonian canbe split up into free (solvable) and interacting (usually unsolvable)parts H = H0 + H ′. The right-hand side is therefore writtenin terms of the interaction picture. |0〉 is the free ground state,defined such that H0|0〉 = 0. Quantities φI(x) are interactionpicture field operators. These time-evolve according to the freepart of the Hamiltonian H0.

22.3 How to calculate Green’s functions

with diagrams

We’ve seen how to get a Green’s functions from the generating func-tional of a theory. It’s simply a matter of differentiating. It’s natural tobe curious about what happens if we try to directly relate the Green’sfunctions to Feynman diagrams. The Gell-Mann–Low theorem is a quo-tient:

G(n)(x1, ..., xn) =〈0|T φI(x1)...φI(xn)S|0〉

〈0|S|0〉. (22.14)

22.3 How to calculate Green’s functions with diagrams 205

Presumably there might be some cancellation between the Feynman di-agrams thrown up in the numerator against those coming from the de-nominator. It turns out that this cancellation does indeed take placeand leaves us with a wonderfully simple result:

G(n) = 〈Ω|T φH(x1)...φH(xn)|Ω〉 =∑(

All connected diagramswith n external lines

).

(22.15)

Recall from Section 19.3 (see particu-larly Fig. 19.4) that a connected dia-gram is made up of a single piece oflinked lines and loops. A disconnected

diagram is composed of a number (> 1)of separate pieces.Why is this? If we start with the Gell-Mann–Low formula (eqn 22.14),

then the numerator factorizes into

〈0|T φI(x1)...φI(xn)S|0〉 =∑(

Connected diagramswith n external lines

)×exp

[∑(Connected

vacuum diagrams

)],

and the denominator is simply 〈0|S|0〉 = exp

[∑(Connected

vacuum diagrams

)],

so we are able to show that

G(n)(x1, ..., xn) = 〈Ω|T φH(x1)...φH(xn)|Ω〉

=

∑(Connected diagramswith n external lines

)×

exp

[∑(Connected

vacuum diagrams

)]

exp

[∑(Connected

vacuum diagrams

)] ,

and that proves eqn 22.15, the most important result of this chapter.

Example 22.1

Let’s see how this unfolds for the single-particle propagator G(2)(x1, x2). A typicalterm will result in a disconnected diagram made up of a connected diagram withexternal lines and several connected vacuum diagrams. The value of the full diagramis the product of all of the connected diagrams. We label each connected vacuumdiagram by some number i. There may be more than one copy of each connectedvacuum diagram: we’ll call the number of copies ni. Call the value of a connectedvacuum diagram Vi. Put all of this together and write the value of a full, disconnecteddiagram, representing one particular contraction, as

„Connected diagramwith external lines

«

×Y

i

1

ni!(Vi)

ni , (22.16)

where we need a factor 1ni!

because we can interchange the ni copies of the connecteddiagram i. The sum of all diagrams is then given by

X

Connected diagrams

with external lines

X

(all ni)


«

×Y

i

1

ni!(Vi)

ni , (22.17)

where, to generate all possible disconnected diagrams, we have summed over allpossible connected diagrams with external lines and over all ordered sets of integersni. To break this down we may factorize out the sum over connected diagramswith external lines:

»X


«–

×X

(allni)

Y

i

1

ni!(Vi)

ni . (22.18)


Then the rest of the expression may be factorized thus:

»X


«–

×

0

@X

n1

1

n1!(V1)n1

1

A

0

@X

n2

1

n2!(V2)n2

1

A ...

=

»X


«–

×Y

i

0

@X

ni

1

ni!(Vi)

ni

1

A

=

»X


«–

×Y

i

eVi

=

»X


«–

× eP

i Vi . (22.19)

Now, as long as we know the Feynman rules for a theory, we can for-get about the S-matrix and the interaction picture. We’ve reduced thewhole perturbation theory to one expression. To get the n-point Green’sfunction we just sum all Feynman diagrams with n external lines. Amaz-ingly simple.

22.4 More facts about diagrams

We’ve seen that G(2)(x1, x2), which is the amplitude for a particle tostart at x2 and propagate to x1, may be written as a sum of diagramsdescribing all the ways in which a particle can start at x2 and finish up atx1. We ask next, is there a way of expressing Z[J ] in terms of diagrams?Since Z[J ] describes a situation where we have sources (and sinks) ofparticles, but that we start at t = −∞ with no particles and end up att = ∞ with no particles, we might expect to be able to express Z[J ]as a sum of diagrams describing ways of starting and ending withoutparticles. This is indeed the case and we have

Z[J ] = 〈Ω(∞)|Ω(−∞)〉J = 〈0|S|0〉J (22.20)

=∑(

Disconnected vacuum andsource-to-source diagrams

).

The sum here is made up of all of the ways of starting and ending with noparticles, which corresponds to the sum of all disconnected vacuum andsource-to-source diagrams, of which there are potentially a very greatnumber. Fortunately there’s an even simpler way of stating this result,which is:

Z[J ] = e

P

0

@

Connected vacuum andsource-to-source diagrams

1

A

. (22.21)

This simplification is based on the linked-cluster theorem,3 which

3The linked-cluster theorem, whichequates the right-hand sides ofeqns 22.20 and 22.21, can be statedmore memorably as

X„

alldiagrams

«

= e

P

connecteddiagrams

!

. we now show follows from more combinatoric arguments, very similarto those used earlier. We call a vacuum or source-to-source diagram

22.4 More facts about diagrams 207

Vi. Now, in general, the amplitude for a general, possibly disconnecteddiagram is given by

∏i

(Vi)ni

ni!and the generating functional is given by

the sum

Z =∏

i

∞∑

ni=0

(Vi)ni

ni!= e

P

i Vi , (22.22)

proving the theorem. (Notice that all of the leg work was done in Ex-ample 22.1!)

Example 22.2

We can use the linked-cluster theorem to simplify the expression for the normalizedgenerating functional Z[J ] = Z[J ]/Z[0]. If we set J = 0, we turn off the sources andwe have

Z[0] = eP

(Connected vacuum diagrams). (22.23)

We write

Z[J ] = e

P

Connected source-to-source diagrams

!

eP

(Connected vacuum diagrams). (22.24)

The normalized generating functional is then

Z[J ] =Z[J ]

Z[0]= e

P

Connected source-to-source diagrams

!

, (22.25)

allowing us to give more meaning to the normalized functional Z[J ].

Notice that we derive most of the results on the way so that we never haveto use them. (One assumes this is a bit like learning knife fighting.) Theend result for a particular theory is that the propagator is representedby a sum of connected diagrams. Of course, the details of what thediagams look like come from a one-off, nitty-gritty calculation. We’ll seea few more examples of these in the following chapters. Finally we’llneed to do the integrals. This will lead into even stranger territory.

Chapter summary

• Green’s functions are derived from generating functionals. Thisapproach shows that G(n) is simply the sum of all diagrams withn external lines.

• The generating functional Z[J ] is the exponential of the sum of allconnected source-to-source and vacuum diagrams.


Exercises

(22.1) Consider the forced quantum oscillator

L =1

2mx(t)2 − 1

2mω2x(t)2 + f(t)x(t), (22.26)

where f(t) is a constant force f0 which acts only fora finite length of time: 0 ≤ t ≤ T . We can use theS-matrix to work out the generating functional forthis theory, by treating f(t)x(t) as an interactionterm.(a) Explain why we may write 〈0|S|0〉 = Z[f(t)] forthis theory.(b) By expanding the S-matrix and show that

Z[f(t)] = e(Dumbbell), where (Dumbbell) is theFeynman diagram shown in Fig. 22.2(a) whose am-plitude is

(Dumbbell)=(−i)2

2

Z

dtdt′f(t)〈0|T x(t)x(t′)|0〉f(t′).

(22.27)(c) Working in frequency space, show that theDumbbell may be written

(Dumbbell) = − 1

2m

Z

dν

2πf(−ν) i

ν2 − ω2 + iǫf(ν),

(22.28)where f(ν) = if0

ν

`

1 − eiνT´

.(d) What does the quantity |Z[f(t)]|2 describephysically? Using the identity 1

E+iǫ= P 1

E−iπδ(E),

where P is the principal part (see Appendix B),show that this quantity is given by

|Z[f(t)]|2 = exp

„

− 2f20

mω3sin2 ωT

2

«

. (22.29)

(a) (b)

f (t)

f (t′)

J(x)

J(y)

Fig. 22.2 (a) Dumbbell diagram for Exercise 22.1. (b)Dumbbell diagram for Exercise 22.2.

(22.2) Now consider scalar field theory in the presence ofa source field

L =1

2[∂µφ(x)]2 − m2

2φ(x)2 + gJ(x)φ(x). (22.30)

Notice the resemblance between this theory and theforced oscillator.(a) Show that the amplitude for the Dumbbell dia-gram in this theory [Fig. 22.2(b)] is given by

(−ig)2

2

Z

d4xd4y

Z

d4p

(2π)4J(x)

ie−ip·(x−y)

p2 −m2 + iǫJ(y).

(22.31)(b) Consider a source field describing two staticsources localized at x1 and x2:

J(x) = δ(3)(x − x1) + δ(3)(x − x2). (22.32)

Show that the interaction between the two sourcescan be written

(Dumbbell)12 ∝ −ig2

„Z

dx0dy0 dp0

2πe−ip0(x0−y0)

«

×Z

d3p

(2π)3eip·(x1−x2)

p2 −m2 + iǫ,

where the self-interaction of the sources is ignored.(c) Show further that

(Dumbbell)12 ∝ ig2

Z

dx0

Z

d3p

(2π)3eip·(x1−x2)

p2 +m2.

(22.33)(d) Finally, using the fact that e−iET = 〈0|S|0〉 =

e(Dumbbell), where T is the time over which thesources act (see Exercise 43.5 if in doubt), showthat the interaction energy E between two spatiallyseparate sources is given by

E ∝ −g2

Z

d3p

(2π)3eip·(x1−x2)

p2 +m2. (22.34)

The important feature here is that the energy is neg-ative. The two sources therefore attract each otherby exchanging a virtual particle of mass m.*(e) Compare the previous result to the correspond-ing one for a (spin-1) vector field dealt with inChapter 24. Is the interaction attractive or repul-sive? A graviton is a spin-2 object. What wouldyou expect for the interaction of sources in a gravi-tational field. See Feynman’s Lectures on Gravita-tion, Chapter 3, for a discussion.

Part VI

Path integrals

In this part we use a reformulation of quantum field theory which is dueto Richard Feynman, the so-called path integral approach. Quantitiesof interest in quantum field theory, such as the propagator, can be cal-culated using a functional integral, which is an integral over all possiblepaths of the particle from one spacetime point to another.

• We describe Feynman’s path integral approach in Chapter 23 anddemonstrate how to evaluate certain Gaussian integrals, using thesimple harmonic oscillator as an example.

• These ideas are then applied in Chapter 24 to working out func-tional integrals and thus calculating Green’s functions.

• There is a rather subtle connection between quantum field theoryand statistical physics, and this is explored in Chapter 25, wherethe concepts of imaginary time and the Wick rotation are intro-duced.

• An important concept is symmetry breaking and this is treatedin Chapter 26. This idea is introduced in the context of phasetransitions and we show how to describe symmetry breaking witha Lagrangian. Breaking a continuous symmetry leads to Goldstonemodes and we also show the consequence of symmetry breaking ina gauge theory, thereby introducing the Higgs mechanism.

• Chapter 27 describes coherent states which are eigenstates of theannihilation operator. These states do not have a definite numberof particles but they can be made to have a well-defined phase andare used to describe laser fields and superfluids.

• We introduce a new type of mathematical object in Chapter 28:Grassmann numbers. These anticommute and can be used to de-scribe fermions. We show how to construct a fermion path integralusing Grassmann numbers.

23Path integrals: I said to

him, ‘You’re crazy’

23.1 How to do quantum mechan-ics using path integrals 210

23.2 The Gaussian integral 213

23.3 The propagator for the sim-ple harmonic oscillator 217

Chapter summary 219

Exercises 220

Thirty-one years ago Dick Feynman told me about his ‘sumover histories’ version of quantum mechanics. ‘The electrondoes anything it likes’, he said. ‘It goes in any direction atany speed, forward and backward in time, however it likes,and then you add up the amplitudes and it gives you thewave-function.’ I said to him, ‘You’re crazy’. But he wasn’t.F. J. Dyson (1923– )

In this chapter, we introduce the path integral formulation of quantummechanics.

23.1 How to do quantum mechanics using

path integrals

Quantum mechanics is all about probability amplitudes. One way to doquantum mechanics is to ask a question like ‘what’s the amplitude thata particle starts at point qa at time ta and ends up at point qb at timetb?’ The amplitude for the process is, as we’ve discussed, known as thepropagator. Previously, we found the propagator using a conventionalapproach to quantum mechanics espoused by Schrodinger and Heisen-berg. In this chapter we’ll find the propagator using Richard Feynman’salternative approach. This will then be applied to quantum fields.

A

B

eiS 1/

heiS 2/

h

eiS 3/

h

eiS 4/

h

Fig. 23.1 Some possible paths withtheir amplitudes. A particle will takeall of them.

The propagator depends on the trajectory that, for whatever reason,the particle takes in getting from A to B. Some possible paths are shownin Fig. 23.1. In classical mechanics, we’d know what to do: write down aLagrangian L; plug that Lagrangian into the Euler–Lagrange equationand thus generate the equations of motion, which we’d then (at leasttry) to solve. The trajectory that we’d calculate would be the one thatminimizes the action

S =

∫ tB

tA

dt L[q(t)], (23.1)

just as we discussed in Chapter 1.How does this work in quantum mechanics? Richard Feynman’s sug-

gestion was that in getting from A to B a particle will take every pos-sible trajectory. We’ll say that again. A particle will take every singlepossible trajectory, forward and backward in time, zig-zagging, loop-ing, whatever. To get the quantum amplitude Feynman says that each

23.1 How to do quantum mechanics using path integrals 211

trajectory contributes a complex factor eiS/~, where S is the action de-scribing that trajectory, and we add up the contributions to get theamplitude. Crazy!

Let’s see if we can justify this approach of getting the amplitudeby summing over eiS/~ factors associated with each trajectory. Thisis also helpfully known as the sum over histories approach becausethe present wave function is obtained by summing over all possible pasttrajectories. The propagator amplitude G to get from spacetime pointA [≡ (ta, qa)] to point B [≡ (tb, qb)] may be written as 〈qb|U(tb, ta)|qa〉,where |q〉 is an eigenstate of position and U(tb, ta) is the time-evolution

operator U(tb, ta) = e−iH(tb−ta) (where we’ve returned to our usual unitsemploying ~ = 1). We have then

G = 〈qb|e−iH(tb−ta)|qa〉. (23.2)

To deal with this we split the trajectory into N infinitesimal steps oflength ∆t as shown in Fig. 23.2. This is known as time-slicing. Thefact that U(t) is a unitary operator allows us to time-slice. Unitaritylets us say U(tb − ta) = U(tb − tx)U(tx − ta) and so time-slicing U(t) =

e−iH(tb−ta) in this way yields

G = 〈qb|(e−iH∆t)N |qa〉= 〈qb|e−iH∆t . . . e−iH∆t . . . e−iH∆t|qa〉. (23.3)

Notice that each of these mini time-evolution operators moves the par-ticle for time ∆t between two closely spaced positions qn and qn+1 (say)as shown in Fig. 23.2.

q

t

ta

tb

∆t

qa

q1

q2

q3

q4

q5q6

q7

qb

Fig. 23.2 Time slicing a trajectoryfrom (ta, qa) to (tb, qb).

Example 23.1

Remember that in quantum mechanics we often make use of the completeness of aset of states. For the position states qn, we have

Rdqn|qn〉〈qn| = 1. The trick to

get any further with G is to insert a ‘resolution of the identity’ (also known as a fatunity) in between each mini time-evolution operator. We’ll need N − 1 fat unities todo this. Inserting these in (working from right to left) we obtain

G = 〈qb|e−iH∆t

»Z

dqN−1|qN−1〉〈qN−1|–

e−iH∆t . . .

. . . e−iH∆t

»Z

dqn+1|qn+1〉〈qn+1|–

e−iH∆t

»Z

dqn|qn〉〈qn|–

e−iH∆t . . .

. . . e−iH∆t

»Z

dq1|q1〉〈q1|–

e−iH∆t|qa〉. (23.4)

Notice what’s happening here. In each resolution we’re varying the qn’s indepen-dently and making new trajectories as we do so. In Fig. 23.2 this corresponds tomoving the blobs along each line independently to some new position to make a newtrajectory. We take a snapshot of the trajectory and then move the blobs again. Thepath integral is done by adding up all of the different trajectories that you make. Inthis way we add all of the possible paths between qa and qb. A little rearrangingreveals that we have

G =

Z

dq1dq2 . . . dqN−1 〈qb|e−iH∆t|qN−1〉 . . . 〈qn+1|e−iH∆t|qn〉 . . . 〈q1|e−iH∆t|qa〉.(23.5)

212 Path integrals: I said to him, ‘You’re crazy’

Between the two amplitudes at the end we have a string of mini-propagators Gn =

〈qn+1|e−iH∆t|qn〉. Let’s look at one of the typical mini-propagators:

Gn = 〈qn+1|e−iH∆t|qn〉

= 〈qn+1|e−i

»

p2

2m+V (q)

–

∆t|qn〉, (23.6)

where we’ve written H = p2/2m + V (q). To evaluate the mini-propagator we needto replace the operators by eigenvalues. The potential term acts1 on the position1There’s some sleight of hand

here. What we’re saying is that

e−i(T+V )∆t = e−iT∆te−iV∆t +O(∆t2), and since ∆t is assumedsmall, all is well. In general if twooperators have a commutation relation

[A, B] = C then eA+B = eAeBe−C/2.Here we take A = −iT∆t, B = −iV∆tand so C = −(∆t)2[T , V ] and thus

e−C/2 ≈ 1 +O[(∆t)2].

state as follows: e−iV (q)∆t|qn〉 = |qn〉e−iV (qn)∆t [we’ve moved the state |qn〉 pastthe potential operator V (q), replacing the operator with its eigenvalue V (qn)]. Wehave then

Gn = 〈qn+1|e−i p2

2m∆t|qn〉e−iV (qn)∆t. (23.7)

Next we need to know how p acts on a position state. Since |qn〉 isn’t an eigenstateof p (since [q, p] = i) we just expand |qn〉 in terms of momentum eigenstates in the

usual way |qn〉 =R dp

(2π)|p〉〈p|qn〉 =

Rdp|p〉e−ipqn/(2π)

12 . Replacing |qn〉 in this way

gives us

Gn =

Zdp

(2π)12

〈qn+1|e−i p2

2m∆t|p〉e−ipqne−iV (qn)∆t

=

Zdp

(2π)12

〈qn+1|p〉e−i p2

2m∆te−ipqne−iV (qn)∆t

=

Zdp

(2π)eipqn+1e−i p2

2m∆te−ipqne−iV (qn)∆t

=

Zdp

(2π)e−i p2

2m∆t+ip(qn+1−qn)e−iV (qn)∆t. (23.8)

We’ve now removed all of the operators in favour of numbers and are left with anintegral over p to do. This is a Gaussian integral which can be done exactly (see thenext section2). The result is2The result we need is that

R∞−∞ dx e−

ax2

2+bx =

q2πa

eb2

2a , with

a = i∆t/m and b = i(qn+1 − qn). Gn =

„ −im

2π∆t

« 12

eim2

(qn+1−qn)2

∆t e−iV (qn)∆t. (23.9)

Inserting back into eqn 23.5, and writing the factor (−im/2π∆t)12 = ξ−1, we find that

the propagator amplitude G can be written as a product of these mini-propagators

G =

N−1Y

n=1

Zdqn

ξe

im2

(qn+1−qn)2

(∆t)2∆t

e−iV (qn)∆t. (23.10)

Lastly we take the limit N → ∞, ∆t → 0. This corresponds to upping the contraston our grid in Fig. 23.2, turning the jagged trajectories into smooth curves. Taking

these limits turns(qn+1−qn)2

(∆t)2into q2 and

P

n∆t→R

dt.

We obtain our answer for the total amplitude

G =

∫D [q(t)] e

iR

dth

mq2

2 −V (q)i

, (23.11)

where we define the integration measure∫D[q(t)] as

∫D [q(t)] = lim

N→∞

N−1∏

n=1

∫dqnξ. (23.12)

Equation 23.11 is an example of a functional integral, a sum over allpossible trajectories encoded as a vast multiple integral over all time-sliced coordinates qn, and in fact as we letN → ∞ the functional integral


becomes an infinitely multiple integral. The curly D hides the guts ofthe functional integral: the instruction to integrate over all possibletrajectories.

Noticing that the Lagrangian is L = mq2/2 − V (q) and restoring ~,we recognize

G =

∫D[q(t)] e

i~

R

dt L[q(t)] =

∫D[q(t)] eiS/~, (23.13)

which is what we set out to justify. To recap: we calculate an amplitudeby adding up contributions eiS/~ from each possible trajectory, of whichthere are an infinite number.3

3Doing the sum relies on the shadybusiness of defining an integration mea-sure

RD[q(t)] which doesn’t look very

well behaved in the limit N → ∞.Although this is a legitimate concern,we’ll see that we’ll always be able tomanoeuvre ourselves out of the troublecaused by

RD[q(t)].

A

B

Fig. 23.3 The classical path is shownas a solid line, alternative paths allowedby quantum mechanics are shown asdashed lines.

What does the path integral tell us about quantum amplitudes? Al-though we add up all of the paths, each path is weighted by a phasefactor, and we expect that there’s some degree of cancellation betweenthe contribution from different paths. Also, as we approach the classicalworld (mathematically achieved by sending ~ → 0) we should recoverLagrange’s result that the trajectory the particle takes is the one forwhich the action is stationary, i.e. δS/δq(t) = 0. This can be under-stood by considering a particular trajectory with action S, typicallylarge compared to ~, so that its contribution is weighted by eiS/~ (thephase S/~ can be anywhere on the Argand clock). A close neighbouringtrajectory will have a different action S′, also large compared to ~, andif δS = (S′ − S) ≫ ~ it is likely to have a completely different phase.It doesn’t take much imagination to realize that a bunch of such tra-jectories will all have essentially random phases and will cancel, givingno net contribution. In contrast, trajectories sufficiently close to thestationary value that Lagrange would have predicted (some are shownin Fig. 23.3) all have similar actions and thus similar phases, giving astrong in-phase contribution. Therefore, the sum is dominated by thestationary result. The quantum mechanical effects, known as quantumcorrections or quantum fluctuations, are given, for the most part, bythe trajectories close to the classical path of least action.4

4An approach to path integral calcu-lations where we find the classical tra-jectory and then find the approximateform of the quantum corrections, isknown as the stationary phase approx-imation and is used in Chapter 50.

23.2 The Gaussian integral

We’ve outlined a new way to think about quantum mechanics, but it’snot worth much if we can’t actually do a path integral. In this sectionwe’ll build up a mathematical armoury to deal with evaluating pathintegrals. It turns out that it’s all related to a simple Gaussian integraland we’ll build it up via a series of successive steps of generalization.Step 0: The basic Gaussian integral around which this section is basedis given by the standard integral5

5Write I =R∞−∞ dx e−x

2and I =

R∞−∞ dy e−y

2, then multiply the two

equivalent integrals together to obtain

I2 =

Z ∞

x=−∞

Z ∞

y=−∞dxdy e−(x2+y2)

=

Z 2π

θ=0dθ

Z ∞

r=0dr re−r

2,

where in the last equality we’veswitched to polar coordinates. Theseare standard integrals, yielding I2 =(2π) × 1

2= π and hence I =

√π.

∫ ∞

−∞dx e−x

2

=√π. (23.14)

Step 1: The first generalization involves rescaling x2 → ax2/2 to get∫ ∞

−∞dx e−

ax2

2 =

√2π

a. (23.15)


Step 2: A very important variation, which turns out to be the usefulone for evaluating path integrals, is the integral:

∫ ∞

−∞dx e−

ax2

2 +bx =

√2π

ae

b2

2a , (23.16)

which is also not too difficult to prove, as shown below.

Example 23.2

Equation 23.16 is proved by ‘completing the square’. The maximum of P (x) =

−ax2

2+ bx is Pmax = b2

2aat x = b

a. We rewrite

P (x) − Pmax = −ax2

2+ bx− b2

2a

= −a2

„

x− b

a

«2

. (23.17)

If we now change variables y = (x− ba) so that dx = dy, then eqn 23.16 becomes

Z ∞

−∞dx e−

ax2

2+bx =

Z ∞

−∞dx e−

a2 (x− b

a )2+ b2

2a

= eb2

2a

Z ∞

∞dy e−

ay2

2 =

r2π

ae

b2

2a . (23.18)

There are some things to note about this equation for later use. Mostof them are obvious, but important.

• Since x was our integration variable it doesn’t appear on the right-hand side. We say that we have ‘integrated out’ x.

• We are left with an expression which has the structure

∫dx e−x

a2 x+bx =

√2π

ae

12 (b

1a b), (23.19)

where on the left-hand side a touches two x’s and b touches one,while on the right-hand side the two b’s are linked by the 1

a .

We’ll see this structure again a little later.Step 3: In order to be able to integrate over functions like q(t), wehave to consider an intermediate step: integrating over vectors. This isbecause a function can be thought of as an infinite dimensional vector.We start with an N -dimensional vector x, with components6 xj and

6For simplicity here we label ordinaryvectors in Euclidean space with indicesin the down position and assume a sumover repeated indices. We will returnto our usual vector notation in the nextchapter.

consider the integral

I =

∫dx1dx2 . . . dxN e−

12xiAijxj =

∫dNx e−

12x

TAx, (23.20)

where A is an N ×N , real, symmetric matrix.7

7This is nothing to worry about aswe’re used to doing multiple dimen-sional integrals over volumes or sur-faces. Remember that the tricky thingabout these integrals is the dependenceof one component on another. Thingsare a lot simpler if we can separate thecomponents out. That’s exactly whatwe’re going to do here.


Example 23.3

To evaluate the integral we make an orthogonal transformation A = OTDO, wherewe choose our matrices O in such a way as to make D a diagonal matrix. Our integralis then

I =

Z

dNx e−12xTOTDOx. (23.21)

Next make the transformation Ox = y, and since the matrix O is orthogonal, theJacobian of this transformation is 1 so

RdNx =

RdNy, and we get

I =

Z

dNy e−12yTDy. (23.22)

As D is diagonal, we have yTDy =P

iDii(yi)2, which allows us to separate out our

multidimensional integral.Z

dNy e−12yTDy =

Z

dy1 e−D11(y1)2

2

Z

dy2 e−D22(y2)2

2 . . .

Z

dyN e−DNN (yN )2

2

=

„2π

D11

« 12„

2π

D22

« 12

. . .

„2π

DNN

« 12

=NY

i

„2π

Dii

« 12

=

„(2π)N

detD

« 12

=

„(2π)N

detA

« 12

, (23.23)

where we’ve usedQ

iDii = detD = detA.

Step 4: We’re finally in a position to generalize our important result,eqn 23.19, to the multidimensional case. We want to evaluate the inte-gral

K =

∫dNx e−

12x

TAx+b

Tx, (23.24)

where b is an N -dimensional vector.

Example 23.4

By analogy with Example 23.2 we consider P (x) = − 12xTAx + bTx. We can easily

find its minimum (remembering that Aij is symmetric and using ∂xi∂xj

= δij):

∂P

∂xk= −1

2δikAijxj −

1

2xiAijδjk + bjδjk

= −Akjxj + bk

= −(Ax − b)k = 0. (23.25)

This gives us a minimum of Pmin = bTA−1b2

at x = A−1b. So again, just as ineqn 23.17, we can write

P (x) − Pmin = −1

2xTAx + bTx − bTA−1b

2

= −1

2(x − A−1b)TA(x − A−1b). (23.26)

This all looks familiar, of course. Again, we can perform a transformation andintegrate over our new variable y = (x − A−1b). We’ll obtain a factor of the form

e12bTA−1b emerging at the front.


The previous example leads us to

K =

∫dNx e−

12x

TAx+b

Tx =

((2π)N

detA

) 12

e12b

TA

−1b. (23.27)

Again we note that we’ve integrated out x, and while A touches two x’son the left-hand side, A−1 takes b to bT.Step 5: Functions can be perfectly well described as infinite dimensionalvectors, and so our next generalization is the functional integral

Q =

∫D[f(x)] e−

12

R

dxdy f(x)A(x,y)f(y)+R

dx b(x)f(x). (23.28)

To deal with this we’ll revert to the discrete world by turning the func-tions into N -dimensional vectors; we’ll write down the solution and thenupgrade the result back into the world of functions.

Example 23.5

We convert A(x, y) → A, f(x) → x and so on to obtain

Z

D[f(x)] e−12

R


dx b(x)f(x) →Z

dNx e−12xTAx+bTx

=

„(2π)N

detA

« 12

e12bTA−1b (and then returning to functions)

→„

(2π)N

detA(x, y)

«1/2

e12

R

dxdy b(x)A−1(x,y)b(y). (23.29)

We have the final result that

Q =

∫D[f(x)] e−

12

R


dx b(x)f(x)

= B [detA(x, y)]−1/2

e12

R

dxdy b(x)A−1(x,y)b(y), (23.30)

where we’ve introduced a constant B to soak up the factor of (2π)N

(which is a potential embarrassment in the limit N → ∞).The inverse function A−1(x, y) is that function that satisfies

∫dz A(x, z)A−1(z, y) = δ(x− y). (23.31)

That’s just another way of saying that A−1(x, y) is the Green’s functionfor the operator A(x, y). It’s been a long slog, but we can now do one sortof functional integral,8 as long as we can find the determinant detA(x, y)

8The key path integral equation(eqn 23.30), which we’ll use most fre-quently in later chapters, may be sum-marized as

Z

D[f ] e−12

R

fAf+R

bf = Ne12

R

bA−1b,

where N is a constant.

and find the Green’s function of A(x, y). In the next section, at longlast, we’ll finally do a functional integral!

23.3 The propagator for the simple harmonic oscillator 217

23.3 The propagator for the simple

harmonic oscillator

An illuminating example of the path integral approach is the simpleharmonic oscillator. We’re going to find the probability amplitude thatif, at ta = 0, we put a particle in the minimum of a harmonic potentiallocated at q = 0, when we come back at time tb = T the particle willstill be at q = 0, where we originally put it. We need, therefore, tocompute the propagator

G(qb = 0, tb = T, qa = 0, ta = 0)

=

⟨Particle ends atqb = 0, tb = T

∣∣∣∣Particle starts atqa = 0, ta = 0

⟩

=

∫ qb=0,tb=T

qa=0,ta=0

D[q(t)] eiR T0

dt L[q(t)]. (23.32)

Let’s do the integral. The Lagrangian for a simple harmonic oscillator

is L = mq(t)2

2 − mω20q(t)

2

2 , where ω0 = (K/m)1/2. The path integral istherefore

G =

∫D[q(t)] e

im2

R

dth

( dq(t)dt )

2−ω20q(t)

2i

. (23.33)

Unfortunately this is not in the form of the one integral we know howto do (that one looks like

∫Dq e−

12 qAq+bq). Can we massage the simple-

harmonic oscillator into this form? We can and the trick, which is basedon integrating by parts, is one that we’ll use repeatedly for path integrals.

Example 23.6

The action in the argument of the exponential part of the path integral is

S =

Z

dt

»mq(t)2

2− mω2

0q(t)2

2

–

, (23.34)

and the trick just involves integrating the first term, which represents the kineticenergy, by parts

I =

Z

dt

„∂q(t)

∂t

«2

=

Z

dt

„∂q(t)

∂t

«„∂q(t)

∂t

«

, (23.35)

giving

I =»

q(t)∂q(t)

∂t

–T

t=0

−Z

dt q(t)∂2

∂t2q(t). (23.36)

The boundary conditions (q = 0 at the beginning and end of the trajectory) wipeout the first term and the second term has the form ‘qCq’, where C is an operator.

The use of our neat trick gives the functional integral

G(0, T, 0, 0) =

∫D[q(t)] e

i2

R

dt q(t)Cq(t), (23.37)


with C = m(− ∂2

∂t2 − ω20

). We can do this integral using a slightly

modified form of the result in eqn 23.30. Setting A(x, y) = −iC andb(x) = 0 in eqn 23.30, we discretize the integral to obtain

G→∫

dq1dq2 . . . dqN ei2

P

ij qiCijqj =

((2πi)N

det C

) 12

. (23.38)

Again this fills us with a sense of dread since the result is quite frighten-ing with its factor (2πi)N that won’t look pretty as N → ∞. Undaunted,we’ll assume that when we pop back into the world of functions the an-swer will at least be that → G ∝ [det C(x, y)]−1/2. Admittedly, thedeterminant of the differential operator C(x, y) might also send a chilldown the spine. To give such an object meaning, we can find the eigen-values (and eigenfunctions) of the operator and use the fact that thedeterminant is given by the product of eigenvalues. Our answer is then

G(0, T, 0, 0) = B det

[m

(− ∂2

∂t2− ω2

0

)]− 12

, (23.39)

where B is some constant coming from the integration measure thatwe’ll worry about shortly.

Example 23.7

The next job is to work out the determinant. This we do with the trick describedabove that makes use of the fact that the determinant of a matrix is equal to theproduct of its eigenvalues. Finding the eigenfunctions of the operator C isn’t too

tough either. You can check that m“

− ∂2

∂t2− ω2

0

”

q(t) = λnq(t) has eigenfunctions

q(t) = c sin(unt), with eigenvalues λn = m(u2n − ω2

0), of which there are an infinitenumber. Remembering that our boundary conditions are that q = 0 at t = 0 andq = 0 at t = T we need un = nπ/T for everything to work. Putting this all together,we have an answer

G(0, T, 0, 0) = B

( ∞Y

n=1

m

»“nπ

T

”2− ω2

0

–)− 12

. (23.40)

At this stage we notice that the answer isn’t very useful in its present form. This isbecause the infinite product looks like it gives infinity whenever ω0T = nπ and wedon’t know what B is yet.

A sanity check is useful here. When there’s no spring constant ω0 = 0, there’s noharmonic potential and we have a free particle. In that case we have

Gω=0(0, T, 0, 0) = B

( ∞Y

n=1

m“nπ

T

”2)− 1

2

, (23.41)

but we already know (from Example 16.5) that for a free particle Gfree(0, T, 0, 0) =“

−im2πT

” 12. This will be enough to provide the proportionality constant B for our

problem.

23.3 The propagator for the simple harmonic oscillator 219

We multiply eqn 23.40 by Gfree/Gω0=0 = 1:

G(0, T, 0, 0) = G(0, T, 0, 0)Gfree(0, T, 0, 0)

Gω0=0(0, T, 0, 0)

=BnQ∞

n=1mh`nπT

´2 − ω20

io− 12

BnQ∞

n=1m`nπT

´2o− 1

2

„−im

2πT

« 12

=

( ∞Y

n=1

"

1 −„ω0T

nπ

«2#)− 1

2 „−im

2πT

« 12

, (23.42)

which removes all of the dangerous terms. Finally we use the identityQ∞n=1[1 −

(x/nπ)2]−1 = x/ sinx and write

G(0, T, 0, 0) =

„ω0T

sinω0T

« 12„−im

2πT

« 12

. (23.43)

The final answer is that

G(0, T, 0, 0) =

( −imω0

2π sinω0T

) 12

, (23.44)

and the probability density |G|2 derived from this is shown in Fig. 23.4.Interestingly, the probability density peaks strongly for T = nπ/ω0,where n is an integer, making it very likely that the particle will befound at the origin at these times. Perhaps this shouldn’t come as toomuch of a surprise: we know that classical particle oscillates back andforth in an harmonic potential. From the path integral point of view,the peaks in probability density result from the constructive interferenceof all of the possible particle paths that start and end at the origin in atime T ; while the minima result from a large degree of cancellation ofsuch path sections. It’s as if the paths at the origin go in and out offocus as a function of time!

0

2

4

6

8

10

2π|G

|2 /(m

ω0)

0 1 2 3 4

ω0T/π

Fig. 23.4 The (scaled) probabilitydensity |G(0, T, 0, 0)|2 as a function oftime for a particle that starts at the ori-gin of a harmonic oscillator potential tobe found there a time T later.

Chapter summary

• Feynman’s path integral approach involves finding amplitudes byadding up contributions from all possible trajectories. Each tra-jectory makes a contribution exp(iS/~).

• One type of functional integral has been evaluated and yields the

result∫D[f ]e−

12

R

fAf+R

bf =(

(2π)N

detA

) 12

e12

R

bA−1b.

• The simple harmonic oscillator provides an example showing theinfinities that have to be tamed in a typical calculation.


Exercises

(23.1) A back of the envelope derivation of the most im-portant result of the chapter.Start with a Lagrangian L = 1

2xAx + bx, where A

is an operator. We will be brash and treat A as anumber. Use the Euler–Lagrange equation to findx and show that the Lagrangian may be expressedequivalently as L = −b 1

2A b.

(23.2) The path integral derivation of Wick’s theorem.

(a) CalculateR∞−∞ dxx2e−

12ax2

.(b) Define

〈xn〉 =

R∞−∞ dxxne−

12ax2

R∞−∞ dx e−

12ax2

. (23.45)

Calculate (i) 〈x2〉, (ii) 〈x4〉 and (iii) 〈xn〉.(c) Make sense of the result for 〈xn〉 diagrammati-cally, representing each factor of 1

aas a line linking

factors of x.(d) Now consider the N -dimensional vector x andthe integral

K =

Z

dx1 . . . dxN e−12xTAx+bTx

=

„

(2π)N

detA

«12

e12bTA−1b. (23.46)

By differentiating with respect to components of thevector b, and then setting b = 0, show that

〈xixj〉 =

R

dx1 . . . dxN xixje− 1

2xTAx

R

dx1 . . . dxN e−12xTAx

= (A−1)ij .

(23.47)(e) Using these results, argue that

〈xixjxkxl〉 = (A−1)ij(A−1)kl + (A−1)ik(A

−1)jl

+(A−1)il(A−1)jk. (23.48)

(f) Write down an expression for the general case〈xi . . . xz〉.This is the basis of Wick’s theorem in the path in-tegral approach.

(23.3) Consider the forced quantum oscillator from Exer-cise 22.1

L =1

2mx(t)2 − 1

2mω2x(t)2 + f(t)x(t). (23.49)

Take f(t) to be a constant force f0 which acts onlyfor a finite length of time: 0 ≤ t ≤ T .(a) Show that the amplitude for a particle that is inthe ground state at t = 0 to be in the ground stateat t = T is given by

A = e−12

R

dt′dtf(t)G(t,t′)f(t′), (23.50)

where

G(t, t′) =θ(t− t′)e−iω(t−t′) + θ(t′ − t)eiω(t−t′)

2mω.

(23.51)(b) Carry out the integral (being careful of the lim-its) and show that the amplitude is given by

A = exp

»

if20

2mω2

„

T − sinωT

ω+ i

2

ωsin2 ωT

2

«–

.

(23.52)(c) What is the probability for the oscillator to stillbe in its ground state at t = T? Compare this withthe result of Exercise 22.1.(d) What is the physical meaning of the imaginarypart of the argument in the exponential?

24Field integrals

24.1 The functional integral forfields 221

24.2 Which field integrals shouldyou do? 222

24.3 The generating functional forscalar fields 223

Chapter summary 226

Exercises 226

Life is not a walk across a fieldRussian proverb, quoted by Boris Pasternak (1890–1960) inhis poem Hamlet

In this chapter we will apply the functional integrals of the last chapter toquantum fields. Remarkably, the generating functional Z[J ] for quantumfields can be written as a functional integral. In this chapter we’ll findthat integral. Using functional integrals can be seen as an alternativeto canonical quantization and the S-operator (as shown in Fig. 24.1)although it’s best to have the two approaches available since both havetheir strengths and weaknesses.

Lagrangian

Path integral

gives

Z[J ]

Green’s

functions

Physical

predictions

add a source current J

differentiate

Fig. 24.1 Using functional integrals todo quantum field theory.

24.1 The functional integral for fields

In the previous chapter we calculated a propagator amplitude for a par-ticle by doing a functional integral. This involved summing all possibletrajectories for a single particle travelling between two spacetime points.Our result was that

G(x, tx, y, ty) =

∫D[q(t)] ei

R tytx

dt L[q(t)]. (24.1)

However, we know that single-particle quantum mechanics isn’t enoughto describe reality: we need fields. We therefore want to find a functionalintegral that sums all possible field configurations that can exist betweenthe two spacetime points. The bad news is that this integral won’t yielda propagator. In the next example we’ll pursue a method of integratingover field configurations and in the following section we will show (thegood news) how this leads directly to the generating functional.

Example 24.1

The jump between integrating over particle trajectories to integrating over field con-figurations is quite straightforward. We can still write down the eiS/~ factor forfields, but now the action S is the integral of the Lagrangian density over space-time: S =

Rd4xL[φ(x)]. Note that the fields appearing here are classical Heisenberg

fields. Their time dependence is given by the full Hamiltonian and they haven’tbeen through a canonical quantization machine so have no commutation relations.A functional integral over fields may then be written as

Z

D [φ(x)] eiR∞−∞ d4xL[φ(x)]

. (24.2)

222 Field integrals

The difference with the path integral that we had before is that the dynamic variableis now the field φ(x) rather than the trajectory q(t) and the integral is therefore overfunctions of position and time rather than just the functions of time. Despite thisdifference, we do the integral in the same way as before: we discretize. Previously wejust sliced up time, now we slice up spacetime. We obtain a multiple integral over Nspacetime variables.

We know how to do the integral for the special, but important, case that theLagrangian can be worked into the form L[q] = i

2qTAq + ibTq. Based on the

observation thatZ

dx ei2ax2+ibx =

„2πi

a

« 12

e−ib2

a , (24.3)

we can read off that the answer is

Here’s the recipe to revert back fromcontinuous functions to a discrete lat-tice:

• The field φ(x) evaluated at aspacetime point x becomes anN -component vector qj , wherej = 1, . . . , N labels a point inspacetime.

• Differential operators becomereal symmetric matrices. Thingslike ∂φ

∂xbecome 1

l(qi − qj) =

Aijqj .

• Integrals become sums:R

dx be-comes l

P

j .

P =

Z

dN q ei2qTAq+ibTq =

„(2πi)N

detA

« 12

e−i2bTA−1b. (24.4)

Our important result for a calculable functional integral over fields is thatZ

D[φ(x)] ei2

R

d4xd4y φ(x)A(x,y)φ(y)+iR

d4x b(x)φ(x)

= B [detA(x, y)]−12 e−

i2

R

d4xd4y b(x)A−1(x,y)b(y), (24.5)

where we’re hidden the factor (2πi)N in the (divergent) constant B as we did before.

24.2 Which field integrals should you do?

We’re now at the stage where we’d like to have a go at doing a functionalintegral over some fields. Before getting too excited about all of this, weneed to decide which integrals are worth doing. What are we trying tolearn? Which fields do we start at, and which do we end with?

We want access to all of the Green’s functions for a theory, since thesecontain all of the physics. If we only want to work out one thing, thenit should be the generating functional Z[J ]. We saw before that thisfunctional contained all of the Green’s functions and therefore all of theinformation for the theory. Remember that the generating functionalZ[J ] tells us the amplitude

Z[J ] =

⟨no particles no particlesat x0 = ∞ at y0 = −∞

⟩

J

, (24.6)

that is, in the presence of a source, we start with no particles and endwith no particles.1 The remarkable answer turns out to be that the1Also remember that our goal is ac-

tually the Green’s functions. Toget to these we can take func-tional derivatives of the generat-ing functional inG(n)(x1, ..., xn) =

1Z[J]

δnZ[J]δJ(x1)...δJ(xn)

˛˛˛J=0

(eqn 21.10).

generating functional can be written in terms of the functional integralas2

2Note that this is the unnormalizedgenerating functional. Recall that toobtain the normalized generating func-tional we can simply divide through byZ[J = 0] since Z[J ] = Z[J ]/Z[0].

Z[J ] =

∫D[φ(x)] ei

R

d4x L[φ(x)]+J(x)φ(x). (24.7)

In order to go about calculating Z[J ], we start by splitting the La-grangian up into two parts L = L0 + LI. As usual, L0 is the free part:it’s the part that can be cast in the quadratic form 1

2φ(x)Aφ(x) and istherefore solvable by our one known field integral. The other term LI isthe interaction part: that is, the part which isn’t solvable by canonical

24.3 The generating functional for scalar fields 223

quantization. Written in terms of functional integrals, the generatingfunctional is therefore

Z[J ] =

∫D[φ(x)] ei

R

d4x (L0+LI+Jφ). (24.8)

Now let’s try to do the integral.

24.3 The generating functional for scalar

fields

We’ll calculate a generating functional for the free scalar field. Followingthe custom that quantities involving free fields are given a subscript ‘0’this will be denoted Z0[J ]. The free Lagrangian is given by

L0 =1

2(∂µφ)2 − m2

2φ2. (24.9)

The generating functional Z0[J ] for the free scalar field isNote that in eqn 24.10 and later we’vesuppressed the spacetime variable x toavoid unnecessary clutter.Z0[J ] =

∫Dφ e

i2

R

d4x(∂µφ)2−m2φ2+iR

d4x Jφ. (24.10)

As we’ve said above, we can do the integral if we massage Z0[J ] intothe form of eqn 24.4. We again invoke the neat trick of doing by partsthe integral of the kinetic energy part of the Lagrangian density in theargument of the exponential (e

i2

R

d4x (∂µφ)2):

∫d4x (∂µφ)2 =

[φ(∂µφ)]∞−∞ −∫

d4xφ∂2φ, (24.11)

where we assume that the first term disappears at the boundary, that isthe field dies off as we head out to spacetime infinity. This trick enablesus to make the replacement

∫d4x (∂µφ)2 → −

∫d4xφ∂2φ and allows us

to rewrite the functional integral as

Z0[J ] =

∫Dφ e

i2

R

d4xφ[−(∂2+m2)]φ+iR

d4x Jφ, (24.12)

which is in the form of the one field integral that we know how to do.3

3We will see that this is telling us thata Lagrangian that can be massagedinto this form may be quantized usingfunctional integration. The family ofquadratic and bilinear Lagrangians forwhich this is true are exactly those thatmay be diagonalized by canonical quan-tization and are conventionally callednon-interacting.

We have that A = −(∂2 +m2) and our answer is that

Z0[J ] = B [detA(x, y)]− 1

2 e−12

R

d4xd4y J(x)[iA−1(x,y)]J(y), (24.13)

where B is a (potentially infinite) constant determined by the integrationmeasure. The determinant also looks worryingly infinite!

Normalization will save us here. The normalized generating functional

is Z0[J ] = Z[J ]/Z[J = 0], where Z0[J = 0] = B det[A(x, y)

]− 12

. This

removes two divergent quantities (B and the determinant) at a stroke.

Recall that the normalization ensuresthat the amplitude to start and endwith no particles, in the absence of asource, is unity.

224 Field integrals

Finally, we recognize A(x, y) = −(∂2 + m2) as an operator givingrise to the Klein–Gordon equation of motion and note that its inversedetermines the free scalar propagator via

iA−1(x, y) = ∆(x, y) =

∫d4p

(2π)4ie−ip·(x−y)

p2 −m2 + iǫ. (24.14)

Example 24.2

To check this, we recall that the inverse of an operator satisfies AA−1 = I, which inthe continuum limit becomes

−(∂2x +m2)A−1 = δ(4)(x− y), (24.15)

where the ∂2x operator acts on the x variable. Next we note that we defined ∆(x, y)

via−(∂2

x +m2)∆(x, y) = iδ(4)(x− y), (24.16)

so we can identify iA−1 = ∆(x, y), that is, the Green’s function of the equation ofmotion, also known as the Feynman propagator.

Putting all of this together gives us a normalized generating functionalfor the free scalar field of

Notice that the field φ doesn’t appearon the right-hand side of eqn (24.17).We say that we’ve ‘integrated out’ thefield φ(x). Z0[J ] =

∫Dφ e

i2

R

d4xφ−(∂2+m2)φ+iR

d4x Jφ

∫Dφ e

i2

R

d4xφ−(∂2+m2)φ

= e−12

R

d4xd4y J(x)∆(x−y)J(y). (24.17)

Example 24.3

We now justify that eqn 24.17 is the same generating functional for the free scalarfield that we defined in Chapter 22. The Chapter 22 definition of the generatingfunctional may be written

Z[J ] =∞X

n=0

in

n!

Z

d4x1...d4xn J(x1)...J(xn)〈Ω|T φH(x1)...φH(xn)|Ω〉. (24.18)

Note that for a non-interacting field we have φH = eiH0tφe−iH0t = φI and so we canuse Wick’s theorem (which applies to freely evolving interaction picture fields only)to expand any time-ordered products. They fall apart into a sum over products offree propagators. For example, the n = 4 term involves a product

〈0|T φI(x1)φI(x2)φI(x3)φI(x4)|0〉 =∆(x1 − x2)∆(x3 − x4) + ∆(x1 − x3)∆(x2 − x4)

+ ∆(x1 − x4)∆(x2 − x3).

Renaming variables, we see that each of the three terms will therefore contribute afactor

R[J(x1)∆(x1 − x2)J(x2)]2. This enables us to write the fourth-order term as

i4

4!

Z

d4x1...d4x4 J(x1)...J(x4)〈Ω|T φH(x1)...φH(x4)|Ω〉

=1

2!

„

−1

2

Z

d4x1d4x2 J(x1)∆(x1 − x2)J(x2)

«2

. (24.19)

Repeating this process for other values of n generates the corresponding terms in theexponential and the answer indeed reduces to eqn 24.17.

24.3 The generating functional for scalar fields 225

The point of finding the generating function is to give us access to prop-agators. Specifically we have for free fields that the propagator is given,in terms of the normalized generating functional, by

G(n)0 (x1, ..., xn) =

1

inδnZ0[J ]

δJ(x1)...δJ(xn)

∣∣∣∣J=0

=1

in1

Z0[J = 0]

δnZ0[J ]

δJ(x1)...δJ(xn)

∣∣∣∣J=0

. (24.20)

We’ll evaluate this in two different ways for the single-particle propagatorG0(x, y). Differentiating the expression for the functional integral Z0[J ]with respect to the J ’s gives us

G0(x, y) =

∫Dφφ(x)φ(y)ei

R

d4xL0[φ]

∫Dφ ei

R

d4xL0[φ], (24.21)

while differentiating the expression for the normalized generating func-tional Z0[J ] = e−

12

R

d4xd4y J(x)∆(x,y)J(y) gives us the expected answerG0(x, y) = ∆(x, y). Thus the Green’s function, which we wrote down as〈Ω|T φH(x)φH(y)|Ω〉, is obtained by integrating two fields weighted bya factor ei

R

d4xL[φ] which tells us how the amplitudes of different fieldtrajectories are distributed.4 4This is just as we have in classical

statistical physics, where the analogousexpression for the average of two fieldsis

〈φ1φ2〉t =

P

ij φiφje−βH(φi,φj)

P

ij e−βH(φi,φj)

(24.22)where the sum is over all configurationsof φi and φj weighted by the Boltz-mann factor and divided by the normal-izing partition function.

Example 24.4

We’ll apply what we’ve learnt to the interesting case of the free massive vector fieldtheory from Chapter 13 described by

L = −1

4(FµνF

µν) +1

2m2AµA

µ

= −1

2(∂µAν∂

µAν − ∂µAν∂νAµ) +

1

2m2AµAµ (24.23)

(see Exercise 13.4 if in doubt). The only complication with this is the differentcomponents of the Aµ field. Note that in this case the result of the one functionalintegral that we can do is written

Z0[J ] =

RDA e

i2

R

d4xAµKµνAν+i

R

d4x JµAµ

RDA e

i2

R

d4xAµKµνAν= e−

12

R

d4xd4y Jµ(x)[iK−1µν ]Jν(y).

(24.24)We can, therefore, only do our functional integral if the Lagrangian is in the form12AµKµνAν , where Kµν is a differential operator (with an inverse). Just as before,

the way to achieve this is, as usual, to integrate L by parts. Of the three terms inthe Lagrangian, the final one is already in the correct form. Integrating the first one,and disregarding the boundary term, we get 1

2Aµ∂2Aµ = 1

2Aµ∂2gµνAν . The second

term yields − 12Aµ∂µ∂νAν . The result5 is that 5You are invited to confirm this result

in Exercise 24.1.1

2AµKµνA

ν =1

2Aµˆ(∂2 +m2)gµν − ∂µ∂ν

Ãν . (24.25)

We therefore do the integral

Z0[J ] =

RDA e

i2

R

d4xAµ[(∂2+m2)gµν−∂µ∂ν ]Aν+iR

d4x JµAµ

RDA e

i2

R

d4xAµ[(∂2+m2)gµν−∂µ∂ν ]Aν

= e−12

R

d4xd4y Jµ(x)[iK−1µν ]Jν(y), (24.26)

226 Field integrals

where iK−1µν is the free particle propagator for the theory G0µν(x, y). This already

tells us something very interesting. Namely, that we have access to the free prop-agator for a theory if we can work out iK−1

µν . This is often an easier method than

working out, for example, G0µν(x, y) = 〈0|TAµ(x)A†ν(y)|0〉 as we would have had to

have done here using a mode expansion if we were still using canonical quantization.Following this through, we can therefore obtain the free particle propagator from thefollowing equation (with the indices all included):

ˆ(∂2 +m2)gµν − ∂µ∂ν

˜G0νλ(x, y) = igµλδ

(4)(x− y), (24.27)

and the easiest way of doing this is to take a Fourier transform, yielding

[−(p2 −m2)gµν + pµpν ]G0νλ(p) = igµλ , (24.28)

whose solution may be shown to be

G0νλ(p) =−i`gνλ − pνpλ/m

2´

p2 −m2, (24.29)

which (with the addition of iǫ in the denominator) is the massive vector field propaga-tor. Interestingly, this has the form of the scalar propagator which we met previously,but this time with the transverse projection tensor on top, containing all of the in-dices.

Chapter summary

• The normalized generating functional for free scalar fields isZ0[J ] = e−

12

R

d4xd4y J(x)∆(x−y)J(y).

Exercises

(24.1) (a) Verify eqn 24.25.(b) Show that eqn 24.29 solves eqn 24.28.

(24.2) Consider the φ4 Lagrangian L = 12(∂µφ)2− m2

2φ2−

g8φ4. Shift the Lagrangian by adding a new field σ

via

L′ = L +1

2g

“

σ − g

2φ2”2

. (24.30)

(a) By performing a functional integral over thefield σ, show that σ doesn’t change the dynamics ofthe theory.(b) Confirm that σ has no effect on the dynamics ofthe theory by finding the Euler–Lagrange equationfor σ and showing that is has no time derivatives.This means that it can only provide a constraint andmay be eliminated, much like the component A0 inthe massive vector field in Chapter 13.(c) Contrast the Feynman diagrams of φ4 theory

with those of the shifted theory involving σ.The introduction of a new field to shift the La-grangian in this way is known as a Hubbard–Stratonovich transformation and is very useful inremoving the inconvenient φ4 term from a theory.

(24.3) In this problem we motivate the perturbation expan-sion for the functional integral.We want to do the integral

Z(J) =

Z

dx e−12Ax2− λ

4!x4+Jx. (24.31)

(a) Show that this integral may be recast as

Z(J) =

" ∞X

n=0

1

n!

„

− λ

4!

∂4

∂J4

«n#

Z

dx e−12Ax2+Jx,

(24.32)

Exercises 227

which has the advantage that we’ve effectively re-moved the interaction from inside the integral.(b) Re-exponentiate the series in the first squarebracket to give a differential operator acting on anintegral and then do the integral to obtain the result

Z(J) =

»

e− λ

4!∂4

∂J4

–

"

„

2π

A

« 12

eJ2

2A

#

. (24.33)

We now have a perturbation series that can be ex-panded by operating on the free generating function

Z0(J) =`

2πA

´ 12 e

J2

2A .

(24.4) By analogy with the previous problem, the gener-ating functional for φ4 theory is given by

Z[J ] =

»

e− iλ

4!

R

d4z δ4

δJ(z)4

–

Z0[J ], (24.34)

where Z0[J ] is the normalized generating functionalfor the free scalar field. In this problem we investi-gate the first-order term in the expansion of Z[J ],given by

Z1[J ]=»

− iλ4!

R

d4z δ4

δJ(z)4

–

e−12

R

d4xd4y J(x)∆(x−y)J(y).

(24.35)

We will act on Z0[J ] = e−12

R

J∆J with δδJ(z)

fourtimes. This will rely on some functional differenti-ation from Chapter 1.(a) Verify that hitting Z0[J ] once gives us

δZ0[J ]

δJ(z)=

»

−Z

d4y∆(z − y)J(y)

–

Z0[J ]; (24.36)

(b) the second time:

δ2Z0[J ]

δJ(z)2=

"

−∆(z − z) (24.37)

+

Z

d4y∆(z − y)J(y)

ff2#

Z0[J ];

(c) the third:

δ3Z0[J ]

δJ(z)3=

»

3∆(z − z)

Z

d4y∆(z − y)J(y)

ff

−Z

d4y∆(z − y)J(y)

ff3#

Z0[J ]; (24.38)

(d) and finally the fourth:

δ4Z0[J ]

δJ(z)4=

ˆ

3∆(z − z)2 (24.39)

−6∆(z − z)

Z

d4y∆(z − y)J(y)

ff2

+

Z

d4y∆(z − y)J(y)

ff4#

Z0[J ].

(e) Expand the brackets, multiply by −iλ/4! andintegrate

R

d4z to end up with the final result thatZ1[J ] is given by

Z1[J ] = −iλ

"

1

8

Z

d4z∆(z − z)2 (24.40)

−1

4

(

Z

d4zd4y1d4y2∆(z − z)

×∆(z − y1)J(y1)∆(z − y2)J(y2)

)

+1

4!

(

Z

d4zd4y1d4y2d

4y3d4y4 ∆(z − y1)

×J(y1)∆(z − y2)J(y2)∆(z − y3)J(y3)

×∆(z − y4)J(y4)

)#

Z0[J ].

We see that this has given us terms in J of orderO(J) = 0, 2 and 4.(f) Interpret this result by drawing Feynman dia-grams for each term. This procedure can be used towork out the Feynman rules for a theory. For moredetail see Ryder, Chapter 6.

25 Statistical field theory

25.1 Wick rotation and Euclideanspace 229

25.2 The partition function 231

25.3 Perturbation theory andFeynman rules 233

Chapter summary 236

Exercises 236

It is the mark of a truly intelligent person to be moved bystatistics.George Bernard Shaw (1856–1950)

We’ve seen how taking the derivatives of the generating functionalZ[J ] =

∫Dφ ei

R

d4x(L[φ]+Jφ) allows us access to the all-importantGreen’s functions of a quantum field theory via

〈Ω|T φ(x1)...φ(xn)|Ω〉 =1

in1

Z[J = 0]

δnZ[J ]

δJ(x1)...δJ(xn)

∣∣∣∣J=0

. (25.1)

This is very similar indeed to the state of affairs in statistical mechanicsthat we examined in Chapter 21. There we found that we had access toall of the interesting correlation functions via derivatives of the partitionfunction

〈φi1 ...φin〉t =1

Z(J = 0)

∂nZ(J)

∂Ji1 ...∂Jin

∣∣∣∣J=0

. (25.2)

In fact, our motivation for searching for a quantum field generating func-tional was simply the ease with which calculations in statistical physicscan be made with a partition function. The similarity between the twodisciplines is not just a superficial resemblance but reflects somethingmuch more fundamental. The crux of the matter is that the Green’sfunctions of quantum field theory are related to time-evolution opera-tors, which (roughly speaking) look like a complex exponential e−iEt/~,with E an energy and t the time. The probabilities (and density ma-trices) of statistical physics are based on the Boltzmann factor, given(roughly) by the real exponential e−E/kBT . The fundamental elementsof the two theories are therefore both exponentials, albeit one is real andthe other is complex. This similarity can be exploited by making thesubstitution

i t

~→ 1

kBT, (25.3)

and the claim is that this will map a quantum field theory on to astatistical field theory. The advantage of this mapping is that it allows usto transfer results from one subject directly to the other, and vice versa.Within the framework of the mapping, inverse temperature1 β = 1/kBT

1Usually we employ units where kB =~ = 1 so that β = 1/T = it. in statistical physics behaves like imaginary time in a quantum field

theory. It turns out that the concept of imaginary time is a remarkablypowerful one with uses across quantum and statistical field theory. Weexamine the imaginary time formalism in the next section.

25.1 Wick rotation and Euclidean space 229

25.1 Wick rotation and Euclidean space

Imaginary time, assigned the symbol τ = it, is defined, not simply bymultiplying time by i, but by making a rotation in the complex planeby an angle2 −π/2 (Fig 25.1). This transformation is known as a Wick 2Positive angles correspond to anti-

clockwise rotations in the complexplane.

rotation and leads to a very nice feature for spacetime four-vectors. Byrotating the time-like part of our vectors we essentially remove the an-noyance of the Minkowski metric (+,−,−,−) and replace it (to withinan overall minus sign) by the more civilized (+,+,+,+). This meansthat all four components of our vectors are treated equally, in the sameway that all spatial components are treated equally when doing Eu-clidean geometry in three dimensions. We refer to this four-dimensionalspace as Euclidean space. Euclid of Alexandria was a Greek math-

ematician who lived around 300 BC.He is best known for his Elements, thetextbook which laid the foundationsfor geometry and has influenced gen-erations mathematicians for more thantwo thousand years.

p0

Re

Im

Fig. 25.1 The Wick rotation involves arotation of −π/2 in the complex plane.We also show the poles of the Feynmanpropagator, demonstrating that Wickrotating of the contour of integrationfrom the real axis (integrating from−∞ to ∞ as we do in calculating Feyn-man amplitudes) to the imaginary axis(integrating −i∞ to i∞) is permissibleas we don’t cross any singularities.

Example 25.1

Let’s work through the consequences of this transformation. To make the Wickrotation to Euclidean space we will rotate the time coordinate by −π/2, definingx0 = −iτ . This means that

x2 = x20 − |x|2 = −(τ2 + |x|2) = −x2

E, (25.4)

where x2E = τ2 + x2. Note that it also implies that

d4x = −id4xE = −idτ d3x. (25.5)

We still want things like Fourier transforms to work in Euclidean space. The re-quirement that the exponential factor eip·x will not blow up requires us to rotate theenergy part of the momentum vector p0 by π/2 in the complex plane. We thereforedefine p0 = iω, giving

p2 = p20 − |p|2 = −(ω2 + |p|)2 = −p2E. (25.6)

We can also write the element of four-volume in momentum space as

d4p = id4pE = idω d3p, (25.7)

and the inner product between p and x as

p · x = p0x0 − p · x = ωτ − p · x. (25.8)

We also consider the derivatives. We have that ∂0 = i∂τ and

∂2 = ∂20 − ∇

2 = −(∂2τ + ∇

2) = −∂2E, (25.9)

where ∂2E = ∂2

τ + ∇2. Finally we need to note that

(∂µφ)2 = −ˆ(∂τφ)2 + (∇φ)2

˜= −(∂Eµφ)2. (25.10)

To summarize the previous example:

The rules of Euclidean space

p0 = iω p2 = −(ω2 + p2) = −p2E d4p = idω d3p = id4pE

x0 = −iτ x2 = −(τ2 + x2) = −x2E d4x = −idτ d3x = −id4xE

∂0 = i∂τ ∂2 = −(∂2τ + ∇

2) = −∂2E (∂µφ)2 = −

[(∂τφ)2 + (∇φ)2

](25.11)


Example 25.2

Our goal is to see the effect of the Wick rotation on the path integral. We start withthe φ4 theory with sources for which we have

iS = i

Z

d4x

»1

2(∂µφ)2 − m2

2φ2 − λ

4!φ4 + Jφ

–

, (25.12)

Making the rotation results in the expression33Note that we often write

LE[φ] = 12(∂τφ)2 + U [φ], where

U [φ]h

= 12(∇φ)2 + m2

2φ2 + λ

4!φ4i

is

the potential energy density.

iS = −Z

d4xE

»1

2(∂τφ)2 +

1

2(∇φ)2 +

m2

2φ2 +

λ

4!φ4 − Jφ

–

. (25.13)

Defining

SE =

Z

d4xE (LE[φ] − Jφ) =

Z

d4xE

»1

2(∂τφ)2 +

1

2(∇φ)2 +

m2

2φ2 +

λ

4!φ4 − Jφ

–

,

(25.14)we write a Euclidean path integral

Z[J ] =

Z

Dφ e−SE =

Z

Dφ e−R

d4xELE[φ]−J(x)φ(x). (25.15)

This is a useful mathematical object as it stands since it makes the path integralsomewhat more respectable. As we shall see in the next example, it removes theambiguity in defining the propagator which forced us to introduce those factors of iǫ.

Example 25.3

This Wick rotation may be carried out on some of our most useful equations toexpress them in Euclidean space. The Feynman propagator for free scalar fields inMinkowski space is

∆(x) =

Zd4p

(2π)4ie−ip·x

p2 −m2 + iǫ, (25.16)

which is the solution to the equation

(∂2 +m2)∆(x− y) = −iδ(4)(x− y). (25.17)

In Euclidean space4 this becomes4We drop the subscript ‘E’ on positionand momentum variables in Euclideanspace from this point. ∆E(x) =

Zd4p

(2π)4e−ip·x

p2 +m2. (25.18)

In the last equation we’ve dropped the iǫ, since it is no longer required in this equationwhich has no poles (see also Fig. 25.1). This is the solution to the equation

(−∂2E +m2)∆E(x− y) = δ(4)(x− y). (25.19)

That this is the case may be shown by transforming from Minkowski to Euclideanspace. We make the changes

∂2 → −∂2E, ∆(x) → ∆E(x). (25.20)

To see how the delta function transforms note that its definition is

δ(x0)δ(3)(x) =

Zd4p

(2π)4e−ip·x → i

Zdωd3p

(2π)4e−ip·x = iδ(τ)δ(3)(x),(25.21)

which may be substituted to give the desired result.

25.2 The partition function 231

25.2 The partition function

We will now attempt to express the partition function as a path integraland will find that it involves the Euclidean generating functional. The

partition function is given by Z = Tr[e−βH

]=∑λ〈λ|e−βH |λ〉. We

know how to express a similar object as a path integral. In Chapter 23we found

〈qB|e−iHtB |qA〉 =

∫D[q(t)] ei

R tB0 dtL[q(t)], (25.22)

which tells us that to evolve the state at A into that at B we add upall of the possible trajectories that start at A at t = 0 and end at B att = tB. Let’s cook up a similar description of

∑λ〈λ|e−βH |λ〉.

First set |qA〉 = |qB〉 = |λ〉, sum over λ and make the replacement

tB → −iβ. This turns the left-hand side into the desired∑λ〈λ|e−βH |λ〉,

where the consequences of the sum will be discussed below. Now for theright-hand side. The replacement tB → −iβ provides the integral in theexponent with the new limits

∫ t=−iβ

t=0. To give this meaning we make the

Wick rotation t→ −iτ and we obtain

∑

λ

〈λ|e−βH |λ〉 =∑

λ

∫D[q(τ)] e−

R β0

dτLE[q(τ)], (25.23)

where LE[q(τ)] = m2

(dqdτ

)2

+ V (q) is the Euclidean version of the La-

grangian for a single particle.The question is, how does the sum on the left affect the trajectories

over which we’ll need to integrate? If we think of e−βH as an evolutionoperator, then we’re taking a state |λ〉, evolving τ from 0 to β andrequiring that we end up at the same state. This means that we musthave q(τ = 0) = q(τ = β). But that’s not all. If we imagine τ asextending from −∞ to ∞ then the boundary conditions are periodic:after a period β, the state returns to where it started. The sum on theleft means that we sum over all configurations subject to these periodicboundary conditions. A quotable conclusion is that ‘statistical physicstakes place in periodic, imaginary time’.

ReψImψ

t

Reψ

Imψ

τ

β

τ

3β

2β

β

0 φ

(a)

(b)

(c)

Fig. 25.2 (a) The trajectory of a com-plex scalar field ψ in time t, where−∞ ≤ t ≤ ∞. (b) In statistical physicsthe field is defined only from imagi-nary time τ = 0 to τ = β. For Bosefields like ψ(τ) or the real scalar fieldφ(τ) we impose the boundary conditionφ(0) = φ(β) to give a picture of periodicboundary conditions shown in (c).

This approach can be simply extended to fields. If H is the Hamil-tonian describing what’s happening to fields in three-dimensional spacethen we have a partition function given by the field integral

Z = Tr[e−βH

]=

∫

PBC

Dφ e−R β0

dτR

d3xLE[φ(x)], (25.24)

where the periodic boundary conditions (illustrated in Fig. 25.2 anddenoted ‘PBC’) are now φ(0,x) = φ(β,x) and LE[φ] is the EuclideanLagrangian density.5 Identifying dτ d3x with d4xE, we see that this is

5Note, however, that these boundaryconditions are appropriate for Bosefields only. For Fermi fields we mustimpose the antiperiodic boundary con-dition that ψ(0,x) = −ψ(β,x).

the Wick rotated version of the functional integral.We made the replacement x0 → −iτ and so in momentum space we

must make the replacement p0 → iω. In fact, to guarantee the periodicboundary conditions are encoded in our functional integral we choose towork with the Fourier transforms of the fields. Since statistical physics


involves systems in boxes and explicit factors of volume we will put thesystem in such a box of volume V which discretizes momentum space.The Fourier representation is then given by6

6The rather awkward normalization ischosen to make the field φn(p) di-mensionless. This is done with mal-ice aforethought in order to guaran-tee that the expressions which will re-sult make sense in describing a statisti-cal system. An important expressionrequired to manipulate these fields isR

dτd3x eiωnτ e−ip·x = βVδp,0δωn,0.

φ(τ,x) =

√β

V∑

n,p

e−iωnτ+ip·xφn(p), (25.25)

where the frequencies (quantized by the boundary conditions) must sat-isfy ωn = 2πn

β with n an integer to ensure the periodicity.7 The ωn’s are7This is another condition only appro-priate for Bose fields. To impose theantiperiodic boundary conditions forFermi fields we require that the Mat-

subara frequencies obey ωn =2π(n+1)

β.

known as Matsubara frequencies and in our calculations the integral

Takeo Matsubara (1921– )

over imaginary time will become a sum over these frequencies.We may now state the rules of statistical field theory required to find

the partition function.

The rules of statistical field theory

To find the partition function for scalar field theory follow the followingrecipe

• Write down the Euclidean Lagrangian LE = 12 (∂τφ)2 + U [φ].

• The partition function is given by the functional integral of theEuclidean action, with periodic boundary conditions (PBCs)

Z[J ] =

∫

PBC

Dφ e−R

dτd3x (LE[φ]−Jφ), (25.26)

where φ is a function of τ and x.

• Impose the PBCs with the expansion

φ(τ,x) =

√β

V∑

n,p

e−iωnτ+ip·xφn(p), (25.27)

where ω = 2πn/β.

• Note that we now use∫

dτ d3x eiωnτ−ip·x = βVδp,0δωn,0. (25.28)

Example 25.4

Let’s examine the familiar example of free scalar field theory in more detail. TheEuclidean Lagrangian is given by

LE =1

2(∂τφ)2 +

1

2(∇φ)2 +

m2

2φ2. (25.29)

This leads to a partition function

Z[J ] =

Z

PBCDφ e

−R

dτd3x

»

12(∂τφ)2+ 1

2(∇φ)2+ m2

2φ2−Jφ

–

. (25.30)

25.3 Perturbation theory and Feynman rules 233

We will calculate some thermodynamic properties of this theory. We therefore set J =0 and insert the Fourier expansion of the fields into the argument of the exponential.We obtain an argument of

−β2V

Z

dτd3xX

n,m,p,q

φm(q)e−iωmτ+iq·x [(−iωn)(−iωm)+(−ip)·(−iq)+m2]φn(p)e−iωnτ+ip·x

=−β2V

X

n,m,p,q

φm(q)(−ωnωm − p · q +m2)φn(p)βVδp,−qδωm,−ωn

= − β2

2

X

n,p

φn(−p)(ω2n + p2 +m2)φn(p).

(25.31)

This results in a partition function

Z[J = 0] =

Z

PBCDφ e−

β2

2

P

np φn(−p)(ω2n+p2+m2)φn(p). (25.32)

This is, of course, in the form of our favourite (and only) tractable functional integral.The solution (for J = 0) is that

Z[J = 0] = N detˆβ2(ω2

n + p2 +m2)˜− 1

2 . (25.33)

This is diagonal in p, allowing us to write ln[Z(J = 0)] as the expression− 1

2

P

n,p lnˆβ2(ω2

n + p2 +m2)˜+const. The sum over n can be done (using a couple

of tricks described in the book by Kapusta and Gale) which allows the replacementX

n

lnˆβ2(ω2

n + E2p)˜

= βEp + 2 ln“

1 − e−βEp

”

+ const., (25.34)

where Ep =p

p2 +m2. To obtain the thermodynamic limit we make our usual

replacement for summing over closely spaces levels in phase space,P

p → VR d3p

(2π)3,

leading to a result

lnZ = VZ

d3p

(2π)3

»

−1

2βEp − ln

“

1 − e−βEp

”–

. (25.35)

We then have a Helmholtz energy F = Vβ

R d3p(2π)3

ˆ12βEp + ln

`1 − e−βEp

´˜.

Notice that the zero-point energy leads to a ground-state energy given by E0 =

− ∂∂β

lnZ = V2

R d3p(2π)3

Ep [which is just like the 12

~ω from (n + 12

~ω)] and this is

associated with a pressure P0 = − ∂F∂V = E0

V . Our meaningful energies and pressureswill be expressed relative to these vacuum values. Finally we obtain an energy,expressed relative to the (arbitrary) ground state energy of

E − E0 = VZ

d3p

(2π)3Ep

eβEp − 1, (25.36)

and a pressure

P − P0 =1

β

Zd3p

(2π)3ln“

1 − e−βEp

”

. (25.37)

Although we have not learnt anything we couldn’t have extracted with simplermethods it is, as usual, heartening that the more sophisticated machinery pro-duces the same results. The power of statistical field theory comes in extract-ing predictions from unsolvable theories such as φ4 theory, to which we now turn.

25.3 Perturbation theory and Feynman

rules

The joy of statistical field theory is that the rules we have already learntwhich allow us to encode perturbation theory in terms of Feynman dia-grams carry over wholesale to statistical field theory. This is not magic;


we have made every effort to construct things this way! We found beforethat the generating functional could be expressed in terms of Feynman

diagrams as Z[J ] = e

P

0

@

Connected source-to-sourceand vacuum diagrams

1

A

. Since the re-sults of statistical field theory depend on lnZ rather than Z we use the

neat result that lnZ[J ] =∑(

Connected source-to-sourceand vacuum diagrams

). Remem-

ber that the sources are included in order that we can extract Green’sfunctions. If we are just interested in the thermodynamic equilibriumproperties we can turn off the sources and we have the key result

lnZ[J = 0] =∑

(Connected vacuum diagrams) . (25.38)

In statistical physics we often find ourselves in a situation where theaverage value of the operator 〈φ(x)〉t 6= 0. An example is the orderedmagnet encountered in the next chapter. In that case, more useful indetermining the fluctuations in the fields is the connected correlationfunction

Gc(x, y) = 〈φ(x)φ(y)〉t − 〈φ(x)〉t〈φ(y)〉t. (25.39)

This may also be derived8 directly from lnZ too:8See Exercise 25.2.

Gc(x, y) = 〈φ(x)φ(y)〉t − 〈φ(x)〉t〈φ(y)〉t =∂2

∂J(x)∂J(y)lnZ[J ]

∣∣∣∣J=0

=∑(

All connected diagrams with twoexternal legs

). (25.40)

Since there are far fewer connected diagrams than disconnected dia-grams, this is a wonderfully labour-saving simplification. This also ex-plains why Gc(x, y) is known as the connected correlation function. No-tice that this is the same result we had in eqn 22.15, where we insistedon normalizing the generating functional.

We already know how to draw diagrams, so the question we mustask is how the calculations encoded by these diagrams differ from thosewe’ve seen before. Luckily, all of the Feynman rules and momentumspace tricks we’ve built up can be employed with little modification.In the case of scalar field theory, the propagator in momentum spacei/(p2

0 −p2 −m2 + iǫ) becomes the Euclidean space version 1/(ω2n + p2 +

m2). Integrals over all momenta will now involve a sum over Matsubarafrequencies in place of the integral over p0. The conversion rule is that

d4p

(2π)4→ 1

β

∑

n

∫d3p

(2π)3. (25.41)

As long as we also have (2π)δ(p0n − p0

m) → βδωn,ωmthen all is well.

As an example, we now state the Feynman rules for translating dia-grams into thermodynamic quantities for φ4 statistical field theory:

25.3 Perturbation theory and Feynman rules 235

The Feynman rules for φ4 theory for T 6= 0

To calculate the amplitude of a diagram in statistical field theory

• Each internal propagator line contributes1

ω2n + p2 +m2

.

• A factor −λ results from each vertex.

• Integrate over all unconstrained internal momenta with a mea-

sure 1β

∑n

∫d3p

(2π)3 .

• Divide by the symmetry factor.

• An overall energy-momentum conserving delta function(2π)3δ(3)(pin − pout)βδωn,ωm

is understood for each diagram.For vacuum diagrams, this results in a factor β(2π)3δ3(0) → βVin the thermodynamic limit.

(a)

(b)

Fig. 25.3 (a) The double-bubble di-agram is the first-order correction tolnZ[J = 0] in φ4 theory. (b) The line-bubble diagram is the first-order correc-tion to the propagator.

For practice, we now evaluate a few diagrams in statistical field theory.

Example 25.5

The first-order correction to lnZ[J = 0] is given by the double-bubble diagram shownin Fig. 25.3(a). The symmetry factor for this one is D = 8, so it makes a contributionto lnZ of

lnZ(1) = −λVβ8

"

1

β

X

n

Zd3p

(2π)31

ω2n + p2 +m2

#2

. (25.42)

The first-order correction to the propagator G(1)(k, q) is the line-bubble shown inFig. 25.3(b). The symmetry factor is D = 2 and the amplitude for this diagram isgiven by9

9Here the incoming line carries four-momentum q = (ωl,q) and outgoingline four-momentum k = (ωm,k).

G(1)(k, q) =1

ω2m + k2 +m2

(

−λ2

"

1

β

X

n

Zd3p

(2π)31

ω2n + p2 +m2

#)

× 1

ω2l + q2 +m2

(2π)3δ(3)(k− q)βδωm,ωl , (25.43)

Example 25.6

We finish this chapter by examining some of the consequences of these corrections onthe thermodynamics of the scalar field with φ4 interactions. The loop integral andMatsubara sum in brackets in eqns 25.42 and 25.43 can be done.10 In fact, it may 10See Kapusta and Gale, Chapter 3.be shown that it falls apart into two pieces:

Zd4p

(2π)41

(p0)2 + E2p

+

Zd3p

(2π)31

Ep

„1

eβEp − 1

«

. (25.44)

The first term is independent of temperature and will not concern us here. Thesecond term provides the first-order thermodynamic correction to lnZ and to theparticle energies.

The resulting contribution to the temperature-dependent part of lnZ is

lnZ(1) = −λβV8

»Zd3p

(2π)31

Ep

„1

eβEp − 1

«–2

. (25.45)


This will give a contribution to the pressure that varies11 as T 4 in the limit that11See Exercise 25.3.m→ 0.

Turning to the line-bubble correction to the propagator: we note that in Chap-ter 16 it was suggested that the stuff that fits between the two external lines of thepropagator, known as the self-energy, gives a correction to the particle dispersionEp . (In Chapter 33 we will see that this is true.) For our present purposes we claimthat the correction to Ep is provided by the part in curly brackets in eqn 25.43. Inparticular, it is straightforward to show that in the limit that m → 0 the correctionto Ep varies as λT 2.

Chapter summary

• Statistical field theory is obtained from quantum field theory bya Wick rotation, so that (temperature)−1 behaves like imaginarytime.

• The diagrammatic techniques developed for quantum fields canthen be used to solve problems in statistical physics.

Exercises

(25.1) (a) What are the dimensions of the field in φ4 the-ory? Show that the normalization in eqn 25.25makes the field φn(p) dimensionless.(b) What normalization would be required for non-relativistic particles? What about Dirac particles?(c) Investigate the consequences of choosing a dif-ferent normalization for φ4 theory.

(25.2) Verify that Gc(x, y) may be obtained by differenti-ating lnZ[J ] as suggested in the text.

(25.3) Verify the T dependence of the corrections to φ4

theory claimed in the chapter, in the limit thatm→ 0.

(25.4) (a) After Wick rotating the Schrodinger equationand investigating the consequences, show that wave

functions evolve according to ψ(τ) = e−Hτψ(0)and the Heisenberg equation of motion becomes∂AH∂τ

=h

H, AH

i

.

(b) For a system described by a Hamiltonian H =ωc†c show that the operators evolve according toc(τ) = e−ωτ c and c†(τ) = eωτ c†.

(25.5) Using a Wick rotation ofR

dE2πG(E)e−iE(tx−ty) =

R

dE2π

ie−iE(tx−ty)/(E − Ep + iǫ), suggest a prop-agator for describing non-relativistic electrons atnonzero temperature. By convention the propaga-

tor for T 6= 0 is defined as minus this quantity,preventing a proliferation of minus signs.

(25.6) Define the imaginary time free propagator for thequantum oscillator as

G(τ) = −〈T x(τ)x(0)〉t, (25.46)

where T is the imaginary time ordering symbol andthe average is a thermal one. Vitally important hereis that 〈a†a〉t = 〈n〉t = (eβω−1)−1 rather than zero!(a) Show that

G(τ) = − 1

2mω

ˆ

θ(τ)(〈n〉t + 1)e−ωτ + θ(τ)〈n〉teωτ

+ θ(−τ)(〈n〉t + 1)eωτ + θ(−τ)〈n〉te−ωτ˜

.(25.47)

(b) Using the definition

G(iνn) =

Z β

0

dτ G(τ)eiνnτ , (25.48)

where νn is a Matsubara frequency satisfying νn =2πnβ

, show that the propagator may be expressed as

G(iνn) =1

2mω

»

1

iνn − ω− 1

iνn + ω

–

=1/m

(iνn)2 − ω2.

(25.49)

26Broken symmetry

26.1 Landau theory 237

26.2 Breaking symmetry with aLagrangian 239

26.3 Breaking a continuous sym-metry: Goldstone modes 240

26.4 Breaking a symmetry in agauge theory 242

26.5 Order in reduced dimensions244

Chapter summary 245

Exercises 245

On the earth the broken arcs; in the heaven, a perfect round.Robert Browning (1812–1889)

In this chapter we turn to the profoundly important topic of brokensymmetry. It’s hard to overestimate the influence this concept has hadon condensed matter and particle physics and on quantum field theoryin general. We will start by discussing the arena where these ideas wereborn: Lev Landau’s theory of the statistical physics of phase transitions.We then turn to classical field theory and will see how broken symmetrymanifests itself in a system described by a Lagrangian.

26.1 Landau theory

Our discussion of Landau’s theory of phase transitions begins with asimple observation about magnets. We imagine a magnet whose spinscan point either up or down. It’s well known that at high temperatureseach spin is equally likely to be found up or down. The magnetization,that is, the spatial average of the magnetic moment, is zero. This systemis shown in Fig. 26.1(a). This system has a symmetry: turn all of thespins through 180 [as in Fig. 26.1(b)] and the magnetization is still zero.Of course, each individual moment is pointing in the opposite direction,but the number pointing in the upwards direction is still half of the totaland the magnetization is still zero. The symmetry here is a global one:we rotate all of the spins through the same angle, here 180.

(a) (b)

Fig. 26.1 The magnet at T > Tc. Themagnetization, or average moment, iszero in (a) and, after rotation of eachspin through 180, is still zero in (b).

(a) (b)

Fig. 26.2 For temperatures T < Tc

the spins align along a single direction.The magnetization in (a) is different in(b), where each spin has been rotatedby 180.

It is found experimentally that upon cooling the systems through acritical temperature Tc, the system undergoes a phase transition andthe magnetization M becomes nonzero as all of the spins line up along asingle direction, as shown in Fig. 26.2(a). The direction along which thespins align could either be all in the up direction or the down direction.If we rotate each spin of the aligned system through 180 [Fig. 26.2(b)]M is obviously reversed. We say that the system has broken (or lowered)its symmetry in the ordered phase.

One puzzling feature of this story is the reason why the system choseto point all of the spins in the one direction rather than the other.After all, there’s nothing in the Hamiltonian which describes the systemthat distinguishes between up and down. The original symmetry of thesystem appears to have spontaneously broken. The same thing happensin the Euler strut shown in Fig. 26.3. A weight is balanced on top ofan elastic strut. If the weight is large enough the strut will buckle. The

238 Broken symmetry

buckling can be either to the left or to the right. There is nothing inthe underlying physics of the strut and weight that allows one to predictwhich way it will go.1 The result is that the ground state does not have

1Of course, in real life, some very smallperturbation or fluctuation will havetipped the system towards its choice ofground state.

the symmetry of the Hamiltonian describing the system. For the magnetthis Hamiltonian could be H = −J∑i S

zi S

zi+1, where the up direction

is along z. This has no bias either favouring the spins pointing up orfavouring them pointing down.

Lev Landau had a scheme for thinking about phase transitions involv-ing a breaking of symmetry.2 Since equilibrium in thermodynamic sys-

2Landau’s scheme is a form of mean-field theory, where the magnetization isdescribed by a uniform field M . Wewill meet mean field theories through-out this book, most notably in Chap-ter 43.

tems relies on both minimizing the internal energy U and maximizing theentropy S of a system, Landau considered the free energy F = U − TS.To find the equilibrium state of the system we need to minimize F . Thefree energy is a function of an order parameter, namely some fielddescribing the system whose thermal average is zero in the T > Tc un-broken symmetry state and nonzero in the T < Tc broken symmetrystate. For a magnet, the order parameter is simply the magnetizationfield M . With joyful and deliberate ignorance of any microscopic de-scription, Landau wrote F (M) as a power series

F = F0 + aM2 + bM4 + . . . , (26.1)

where a and b are parameters which are independent of M , but may inprinciple depend on the temperature. This free energy has the symmetryM → −M , that is, reversing the magnetization doesn’t affect the en-ergy. If a is positive we have the free energy shown in Fig. 26.4(a), whichis minimized at M = 0, which is the correct prediction for the high-temperature regime. If, however the parameter a is negative then wehave the free energy shown in Fig. 26.4(b), which is known, by some, asLifshitz’s buttocks. This has two minima at nonzero values of magnetiza-tion. These minima correspond to the spins all aligning up (M = +M0)or all aligning down (M = −M0). The previous minimum M = 0 is nowat a position of metastable equilibrium.3 If we take a = a0(T −Tc) (with

3This means that any slight perturba-tion of the system which has been pre-pared in a M = 0 state will propel itinto one, or other, of the new minimathat have emerged at ±M0.

a0 a constant) and b to be T -independent, then clearly a is positive forT > Tc and negative for T < Tc, and so we predict a phase transition ata temperatures T = Tc.

(a) (b) (c)

Fig. 26.3 The Euler strut. (a) Thishas a vertical axis of symmetry and canbuckle either (b) one way or (c) theother, in each case breaking symmetry.

(a)

M

F (b)

M

F

Fig. 26.4 The Landau free energy for(a) T > Tc with a minimum at M =0 and (b) for T < Tc with minima at±M0.

Example 26.1

We can find the minima of F straightforwardly for T < Tc. We set

∂F

∂M= 2aM + 4bM3 = 0, (26.2)

from which we conclude that the minima occur at

M20 = − a

2b. (26.3)

For T < Tc we have two minima at M = ±M0. The system will chooseone of these as its new ground state and the symmetry will be broken.When a symmetry is broken in a many-particle system there are fourfeatures4 that one should look out for.

4This classification is due to PhilipAnderson and described in detail inhis classic Basic Notions of CondensedMatter Physics.

26.2 Breaking symmetry with a Lagrangian 239

• Phase transitions We saw that in Landau’s example, the pa-rameter a in the free energy was temperature dependent. At atemperature Tc, at which a changes sign, a phase transition takesplace. The transition separates two distinct states of different sym-metry. The low-temperature phase has lost some symmetry, moreprecisely it is missing a symmetry element.5 5In fact, you don’t need to regard the

parameters in a model as being depen-dent on anything: in the Lagrangianexamples below we won’t, and then thephase transition is not such a usefulconcept. However, the link betweenphase transitions and symmetry break-ing is a very powerful one that uni-fies our understanding of an enormousrange of phenomena in Nature.

• New excitations Our philosophy has been that every particle isan excitation of the vacuum of a system. When a symmetry is bro-ken we end up with a new vacuum (e.g. a vacuum withM = −M0).The fact that the vacuum is different means that the particle spec-trum should be expected to be different to that of the unbrokensymmetry state (such as M = 0 in our example). We will seethat new particles known as Goldstone modes can emerge uponsymmetry breaking.6 6See Section 26.3.

• Rigidity Any attempt to deform the field in the broken symmetrystate results in new forces emerging. Examples of rigidity includephase stiffness in superconductors, spin stiffness in magnets andthe mechanical strength of crystalline solids.

• Defects These result from the fact that the symmetry may bebroken in different ways in different regions of the system, and aretopological in nature. An example is a domain wall in a ferromag-net. These are described in Chapter 29.

26.2 Breaking symmetry with a

Lagrangian

Let us now apply the arguments developed for magnets to our La-grangian treatment of (classical) field theory.7 Here instead of follow-

7Classical field theory will be enoughto show us the main features of brokensymmetry. See S. Coleman and Peskinand Schroeder for more details on thequantum mechanics of broken symme-try.

ing the ground state magnetization, we’re interested in the ground statevalue of φ(x) or (returning briefly to quantum mechanics) the expecta-tion value of the field operator 〈Ω|φ(x)|Ω〉, which is the order parameterin field theory. We start with a scalar field theory

L =1

2(∂µφ)2 − U(φ), (26.4)

where we’ve split off all of the potential energy terms and called them8

8More strictly U(φ) is the potential en-ergy density. It should not be confusedwith the internal energy, which is alsoconventionally called U .

U(φ). In the familiar case of φ4 theory we have that

(a)

φ

U(φ) (b)

φ

U(φ)

Fig. 26.5 Equation 26.5 for (a) µ2 > 0with a minimum at φ = 0 and (b) forµ2 < 0 with minima at ±

p6µ2/λ.

U(φ) =µ2

2φ2 +

λ

4!φ4. (26.5)

This function resembles the free energy in our magnet example above.It admits the global symmetry φ(x) → −φ(x). Assuming µ2 is positivewe have a potential with a minimum at φ = 0 as shown in Fig. 26.5(a).The minimum corresponds to a quantum mechanical ground state, alsoknown as the vacuum, of 〈Ω|φ(x)|Ω〉 = 0. As we’ve seen earlier, theexcitations of this field are phions with mass m = µ.

240 Broken symmetry

Now we turn to a very interesting possibility: what if we swap the

sign in front of µ2? In that case we have U(φ) = −µ2

2 φ2 + λ

4!φ4 and

again have a potential that looks like Fig. 26.5(b). The minima9 of the9From ∂U/∂φ = 0 we find

0 = −µ2φ+λ

3!φ3,

from which we deduce stationary pointsat (0,±

p6µ2/λ). The second deriva-

tive,∂2U

∂φ2= −µ2 +

λφ2

2,

is negative (and equals −µ2) for φ = 0and is positive (and equals +2µ2) forthe minima at φ0 = ±

p6µ2/λ.

potential now occur at

φ0 = ±(

6µ2

λ

) 12

, (26.6)

so we have the choice of two new vacua. The system will spontaneouslychoose one and the symmetry φ0 → −φ0 of the ground state is broken.

Next we ask if the new vacuum has the same excitations (a.k.a. par-ticles) as those of the unbroken symmetry state.

Example 26.2

To do this we choose a vacuum (for definiteness let’s choose +φ0) and we expand thefield around this new minimum. The Taylor expansion gives us

U(φ− φ0) = U(φ0) +

„∂U

∂φ

«

φ0

(φ− φ0) +1

2!

„∂2U

∂φ2

«

φ0

(φ− φ0)2 + . . .

= U(φ0) + µ2(φ− φ0)2 + . . . , (26.7)

and since U(φ0) is an ignorable constant, we can write the Lagrangian in terms ofφ′ = φ− φ0 as

L =1

2(∂φ′)2 − µ2φ′2 +O(φ′3). (26.8)

By comparison with our original scalar field theory,10 we find from10For the ordinary scalar field La-grangian, the factor multiplying theterm quadratic in field is m2/2, withm the mass of the particle excitationsin the theory.

eqn 26.8 that the mass of the phion excitations in the broken symmetrystate is now no longer m = µ but has become m =

√2µ.

An important point to note is that the Lagrangian does not ‘breaksymmetry’ itself. Indeed our Lagrangian, just like the Landau free en-ergy for the magnet, is invariant with respect to the transformationφ → −φ. The breaking of symmetry is a property of the ground state(or vacuum) of the system.1111In the quantum mechanical version

we say that the ground state does nothave the symmetry of the Hamiltonian.

To conclude, the effect of spontaneous symmetry breaking is (i) for φor (again, briefly returning to quantum mechanics) the vacuum expec-tation value 〈Ω|φ(x)|Ω〉 to acquire a non-vanishing, constant amplitude

φ0, in this case φ0 =(

6µ2

λ

)1/2

; (ii) the particle excitations of the theory

are the same as before, but now have a mass√

2µ.

26.3 Breaking a continuous symmetry:

Goldstone modes

The scalar field theory had a discrete global symmetry (φ→ −φ). Moredramatic and interesting still is the effect of symmetry breaking for a

26.3 Breaking a continuous symmetry: Goldstone modes 241

theory with a continuous global symmetry. A two-component field withthe sign of its mass term flipped looks like

L =1

2

[(∂µφ1)

2 + (∂µφ2)2]+µ2

2

(φ2

1 + φ22

)− λ

4!

(φ2

1 + φ22

)2. (26.9)

This has a global SO(2) symmetry: it is symmetric with respect torotations in the (internal) φ1(x)-φ2(x) plane (where the same rotationtakes place at all points in spacetime). The potential U(φ1, φ2) for thistheory is sketched in Fig. 26.6(b) and looks a lot like the bottom of awine bottle [as shown in Fig. 26.6(a)]. There are an infinite number ofpotential minima, and these are found to lie on a circle [Fig. 15.4(c)]

whose equation can easily be found12 to be φ21 + φ2

2 = 6µ2

λ .

12To quickly show this, you can write

U(x) = −µ2

2x+ λ

4!x2 where x = φ2

1+φ22,

and then ∂U/∂x = 0 yields the result.

(a)

(b)

(c)

φ1

φ2

Fig. 26.6 (a) The potential for theSO(2) symmetry breaking looks likethe bottom of a punted wine bottle.(b) There is a maximum at the pointφ1 = φ2 = 0, but surrounding thisthere is a set of minima which lie ona circle. (c) The circle of minima areshown on a φ1-φ2 plot (this is thereforeviewing the surface sketched in (b) from‘above’). The symmetry can then bebroken by choosing a particular pointin the circle of minima and setting thisto be the ground state. We can then ex-amine small deviations away from thatpoint.

Fig. 26.7 Breaking symmetry at

the position (φ1, φ2) =

„

+

q6µ2

λ, 0

«

.

There are two possible excitations: theparticle can oscillate up and down inthe φ1 direction (a massive excitation)or trundle round in the gutter (a mass-less excitation).

Let’s break the symmetry by supposing the system chooses a par-ticular vacuum. Each is as good as any other, so for computational

simplicity, let’s choose (φ1, φ2) =

(+√

6µ2

λ , 0

)and expand around this

broken symmetry ground state using variables φ′1 = φ1 −√

6µ2

λ and

φ′2 = φ2. This is illustrated in Fig. 26.7.

Example 26.3

With the potential U(φ1, φ2) given by

U(φ1, φ2) = −µ2

2(φ2

1 + φ22) +

λ

4!(φ2

1 + φ22)2, (26.10)

Taylor expanding around the minimum (φ1, φ2) =

„q6µ2

λ, 0

«

gives ∂2U/∂φ21 = 2µ2

and ∂2U/∂φ22 = 0 and so we obtain (ignoring constant terms and truncating the

expansion at order φ2)

L =1

2

ˆ(∂φ′1)2 + (∂φ′2)2

˜− µ2(φ′1)2 +O(φ′3). (26.11)

Equation 26.11 shows that the particles of the φ′1 field now have massm =

√2µ, but that there is no quadratic term for the φ′2 term: thus the

particles of the φ′2 field are completely massless! This seemingly magicalresult is less surprising if we look again at the potential landscape shownin Fig. 26.7. We can see immediately that excitations in the φ1 directionwill cost energy since such excitations have to climb up the walls of thepotential. However, a small movement in the φ2 direction corresponds torolling round the gutter. Such motion has no opposing force as it involvesno change in potential energy and therefore costs no energy, exactly whatwe would expect for a massless excitation. In other words, we still haveparticles in the φ2 field, it’s just that their dispersion starts at the origin,rather than from some offset determined by the mass. In the languageof condensed matter physics one would say that the excitations in theφ2 field are gapless, rather than massless, but it amounts to the samething.

242 Broken symmetry

This vanishing of the mass is a manifestation of Goldstone’s

theorem, which says that breaking a continuous symmetry alwaysresults in a massless excitation, known as a Goldstone mode. The

massless particle associated with this excitation is known as aGoldstone boson.

Jeffrey Goldstone (1933– ). Thebosons had previously been discoveredby Yoichiro Nambu (1921– ) in the con-text of superconductivity.

Example 26.4

Perhaps more illuminating is the complex scalar field theory. In this theory we’llbreak global U(1) symmetry. Of course U(1) is isomorphic to SO(2), so we won’thave anything too different from the previous example. The U(1) theory, with apositive mass term, reads

L = (∂µψ)†(∂µψ) + µ2ψ†ψ − λ(ψ†ψ)2. (26.12)

This theory enjoys a global U(1) symmetry: a transformation ψ → ψeiα has no effecton the Lagrangian. We switch to ‘polar coordinates’13 via ψ(x) = (x)eiθ(x) and this13See chapter 12, although note that

there we were dealing with a non-relativistic theory and wrote Ψ(x) =pρ(x)eiθ(x). Here, in dealing with

the relativistic theory we write ψ(x) =(x)eiθ(x) and so

Rd3x |(x)|2 [rather

thanR

d3x ρ(x)] gives the number den-sity of particles.

theory readsL = (∂µ)

2 + 2(∂µθ)2 + µ22 − λ4, (26.13)

where the two fields are now called (x) and θ(x). The symmetry transformation forthis version is → and θ → θ + α.

Let’s break this symmetry. The potential part is U = −µ22 +λ4 whose minima

are on a circle of radius =qµ2

2λ. We set 0 =

pµ2/2λ and (arbitrarily, but with

convenience in mind) choose a minimum by selecting θ0 = 0, then expand aroundthe minimum with ′ = − 0 and θ′ = θ − θ0, to obtain

L =“µ2

2λ

”

(∂µθ′)2 (θ′-field terms)

+(∂µ′)2 − 2µ2′2 − 4“µ2λ2

” 12′3 − λ′4 (′-field terms)

+

»

′2 +“

2µ2

λ

” 12′–

(∂µθ′)2 + . . . (interaction terms) ,

(26.14)

where constant terms have been dropped and terms have been arranged accordingto the contributions of the two individual fields and then the interactions betweenthem. As might be expected, there is no term in θ′2 and so the θ′-field is massless.As before, this is because it costs nothing to roll in the gutter. The mass of the′-field has been shifted to m =

√2µ.

26.4 Breaking a symmetry in a gauge

theory

An extraordinary feature that emerges upon breaking a symmetry in agauge theory was spotted by Anderson, Higgs, Kibble, Guralnik, En-glert and Brout and is famously known as the Higgs mechanism.14 As

14Philip Anderson (1923–), FrancoisEnglert (1932– ), Robert Brout (1928–2011), Peter Higgs (1929– ), Tom Kib-ble (1932– ), Gerald Guralnik (1936– )and Carl Hagan (1937– ). Higgs andEnglert shared the 2013 Nobel Prizein physics for their discovery (Englertand Brout worked together, but Broutdied before the prize was awarded).Anderson (who won the Nobel prizein 1977 for other work) had a non-relativistic version in 1962 which wasinspired by superconductivity, and in1964 all the others (Brout and En-glert published first, closely followed byHiggs) produced relativistic treatmentsof what we now call the Higgs mecha-nism (though only Higgs mentioned themassive boson).

we’ve said previously (see Chapter 14), a Lagrangian which has a localsymmetry contains gauge fields. The simplest example is the gaugedcomplex scalar field theory, whose Lagrangian (with a flipped signed forthe mass term) is given by

L = (∂µψ† − iqAµψ†)(∂µψ + iqAµψ) + µ2ψ†ψ − λ(ψ†ψ)2 − 1

4FµνF

µν ,

(26.15)

26.4 Breaking a symmetry in a gauge theory 243

which is symmetric under the local transformation ψ → ψeiα(x) as longas we also transform Aµ → Aµ − 1

q∂µα(x). This Lagrangian describes aworld made up of two sorts of oppositely charged, massive scalar particleswith energies Ep = (p2 + µ2)

12 and two sorts of transversely polarized

photons with energies Ep = |p|, which are, of course, massless.Now we’ll see what happens when we break symmetry in a system

with a local symmetry. Working in polars, the ground state has thefield ψ(x) = (x)eiθ(x) and takes on a unique phase angle θ(x) = θ0 forall x. We are no longer permitted to change the phase of the groundstate at different values of x (local symmetry) or indeed change thephase of the ground state for the entire system (global symmetry). Thebroken symmetry ground state therefore breaks global symmetry and,as a consequence, local symmetry too.

Next we turn our attention to the excitations. Remember that theLagrangian is gauge invariant so we are able to make gauge transforma-tions to simplify the physics as much as possible. In the next examplewe show that the gauge transformation we make to elucidate the par-ticle spectrum of the broken symmetry system reveals something quiteexciting.

Example 26.5

In polar coordinates, the second bracketed term in eqn 26.15 can be written

∂µψ + iqAµψ = (∂µ)eiθ + i(∂µθ + qAµ)eiθ. (26.16)

Thus we notice that Aµ enters the theory as

Aµ +1

q∂µθ ≡ Cµ, (26.17)

and so we will replace Aµ by Cµ, thereby simplifying the term (∂µψ†−iqAµψ†)(∂µψ+iqAµψ) = (∂µ)2 + 2q2CµCµ. The replacement is a gauge invariant one sinceFµν = ∂µAν − ∂νAµ = ∂µCν − ∂νCµ. The Lagrangian in eqn 26.15 can now bewritten in terms of and Cµ fields as

L = (∂µ)2 + 2q2C2 + µ22 − λ4 − 1

4FµνFµν , (26.18)

where we use the shorthand C2 ≡ CµCµ.Now to break the symmetry! The minima of the potential are again on the circle

at =qµ2

2λ. We choose the symmetry to be broken with 0 =

qµ2

2λand θ0 = 0.

The expansion that reveals the excitations above the new ground state is most easilydone by expanding in terms of a field χ, defined as χ√

2= − 0. Ignoring constant

terms we obtain

L =1

2(∂µχ)2 − µ2χ2 −

√λµχ3 − λ

4χ4

−1

4FµνF

µν +M2

2C2

+q2„µ2

λ

« 12

χC2 +1

2q2χ2C2 + . . . , (26.19)

where M = q

qµ2

λ.

244 Broken symmetry

As usual, we have the contribution that derives from the radial field (herecalled χ), which has a mass

√2µ. However, there is a big surprise here

in the second line of eqn 26.19: the theory also now contains a massivevector field Cµ, whose particles have mass M . But what of the θ field?It has completely disappeared! The excitations of the θ field, which weremassless in the global symmetry breaking version, have disappeared andhave been replaced by those of a massive vector field15 Cµ(x). As Sidney15Recall from Chapter 13 that the La-

grangian for a massive vector field is

L = − 14FµνFµν + m2

2AµAµ.

Coleman put it, it’s as if the massless photon field Aµ(x) has eaten theGoldstone bosons from the θ field and grown massive, changing its nameto Cµ(x). The theory now describes one sort of scalar particle, which isan excitation in the χ(x) field with energy Ep = (p2 + 2µ2)

12 , and three

sorts of massive vector particle, which are excitations in the Cµ(x) field

with energies Ep =[p2 +

(q2µ2

λ

)] 12

.

One might now worry that the loss of the Goldstone bosons meansthat we have lost some of the precious degrees of freedom in the mathe-matical description of the fields. This worry is unfounded because uponsymmetry breaking we have the change(

2 × massive scalar particles2 × massless photon particles

)→(

1 × massive scalar particles3 × massive vector particles

),

(26.20)that is, four types of particle are excitable before and after we breaksymmetry.

So where did the Goldstone boson go? The answer becomes clear whenwe reexamine what we did to remove it: we made the change of variablesAµ + 1

q∂µθ = Cµ, which is a gauge transformation! If the Goldstoneparticle can be removed with a simple gauge transformation then itcan’t have been there in the first place. In other words, it must be pure

gauge.16 Another way of describing the Higgs mechanism, therefore, is16The notion of a pure gauge will ap-pear again in our treatment of topolog-ical effects in Chapter 29.

the removal by gauge transformation of all Goldstone modes.This feature of symmetry breaking, where the Goldstone mode may

be removed and turned massive through combination with a gauge field,has become one of the most important in modern physics. Historicallyit told particle physicists, puzzled by that apparent lack of Goldstoneparticles in Nature, that symmetry breaking allows another way. Wewill return to this undeniably important topic in Chapter 46.

26.5 Order in reduced dimensions

Field theory allows us to ask questions about systems which have less (ormore!) than the ordinary three spatial dimensions of our world. One fa-mous result is the Coleman–––Mermin–––Wagner theorem, which tells

Sidney Coleman (1937–2007)N. David Mermin (1935– )Herbert Wagner (1935– )

us that symmetry breaking is impossible in two or fewer spatial dimen-sions in systems with an order parameter with a continuous symmetry.

We’ll show that this is the case for complex scalar field theory withspontaneous symmetry breaking.17 The idea of the proof is to ask what

17We use the theory of a complex scalarfield ψ(x) since there a continuous U(1)internal symmetry is broken, leading tothe emergence of a Goldstone mode.The presence of the Goldstone mode iscrucial here.

the fluctuations look like at one point in space. We get this by evaluatingthe Euclidean (two-point) correlation function at the origin. We want18

18Strictly we should write the integra-

tion measure asRDψDψ†, but we will

abbreviate it here to save on clutter.

Exercises 245

G(0, 0) = 〈|ψ(0)|2〉 =1

Z

∫Dψ |ψ(0)|2e−SE[φ]

= limx→0

1

Z

∫Dψ ψ(x)ψ†(0)e−SE[φ]. (26.21)

We can evaluate this since we know that the propagator is given ind-dimensional Euclidean space by

G(x, y)=〈ψ(x)ψ†(y)〉= 1

Z

∫Dψ ψ(x)ψ†(y)e−SE[φ] =

∫ddp

(2π)de−ip·(x−y)

p2 +m2.

(26.22)If the system breaks symmetry then the propagating particles will beGoldstone particles. These are massless, so give rise to a fluctuationwith m = 0 and result in a correlation function

G(0, 0) =

∫ddp

(2π)d1

p2, (26.23)

whose integrand contains a singularity at p = 0. For d > 2 the singular-ity is integrable and the fluctuations in the field are well behaved. Ford ≤ 2 the singularity is not integrable and the fluctuations diverge. Thedivergent influence of Goldstone bosons breaks up the order in the sys-tem. We conclude that symmetry breaking is impossible in our complexscalar field theory for d ≤ 2.

A simple example of the Coleman–Mermin–Wagner theorem can be foundin the behaviour of a ferromagnet.Magnons (quantized spin waves) arebosons and cost energy Ep = αp2 (inthe low p limit) where α > 0 is a con-stant. Thus the number of magnons ex-cited at temperature T is proportionalto Z

d|p| |p|d−1

eβEp − 1.

The integrand is proportional to |p|d−3

at low momentum (assuming T 6= 0),and therefore diverges when d ≤ 2.Thus long-range order in a ferromag-net is wiped out by spin waves at allnonzero temperatures in two or fewerdimensions. This argument uses noquantum field theory, and holds onlyin this specific case (the analogous ar-gument is more elaborate for antifer-romagnets, see the book by Auerbach)but it is apparent that the mathemat-ical origin is the same as that used forG(0, 0) in eqn 26.23, namely that an in-tegral of this form diverges when d ≤ 2.

Chapter summary

• In many phase transitions, symmetry is broken below Tc and theground state possesses only a subset of the symmetry of the Hamil-tonian.

• Breaking a continuous symmetry results in a massless excitation,known as a Goldstone mode or Goldstone boson. These are notstable in two or fewer dimensions and symmetry breaking is thennot possible (Coleman–Mermin–Wagner theorem).

• In a gauge theory, the Goldstone mode can be removed and be-comes massive through combination with a gauge field (the Higgsmechanism).

Exercises

(26.1) In this problem we prove that if the vacuum is in-variant under an internal symmetry then a multi-plet structure (that is, a different field with the same

energy) will emerge.Consider creation operators for two types of parti-

246 Broken symmetry

cles φ†A and φ†

B with the propertyh

QN, φ†A

i

= φ†B, (26.24)

for some generator of a symmetry group, such that[QN, H] = 0. We impose the condition

QN|0〉 = 0. (26.25)

(a) Show that eiαQN |0〉 = |0〉.(b) Show also that

EA = EB. (26.26)

where Hφ†A|0〉 = EAφ

†A|0〉 and Hφ†

B|0〉 = EBφ†B|0〉.

Note that symmetry breaking occurs in the con-trasting case that QN|0〉 6= 0.

(26.2) The Fabri–Picasso theorem demonstrates that if aLagrangian is invariant under an internal symmetrywith charge operator QN =

R

d3x J0, then there aretwo possibilities: either(i) (as above) we have QN|0〉 = 0; or(ii) QN|0〉 has an infinite norm.Here we prove this fact.(a) By using the translation operator eip·a show that

〈0|J0N(x)QN|0〉 = 〈0|J0

N(0)QN|0〉. (26.27)

(b) By considering the matrix element 〈0|QNQN|0〉along with the result of (a), show that eitherQN|0〉 = 0 or QN|0〉 has an infinite norm.Note that if QN|0〉 has an infinite norm, then it doesnot exist in the same space as |0〉. This is the casein spontaneous symmetry breakdown. In this case,the state QN|0〉 is not zero, but another possiblevacuum state. See Aitchison and Hey, Chapter 17for more details.

(26.3) We’ll work through a famous proof of Goldstone’stheorem. This can be found in many books. Startwith a theory with a continuous symmetry withcharge QN, where QN|0〉 6= 0. Consider a field φ(y)which is not invariant under QN such that

h

QN, φ(y)i

= ψ(y), (26.28)

where ψ is some other field. We are going to exam-ine 〈0|ψ(0)|0〉, which we assume takes on a nonzerovalue when symmetry is broken.(a) Show that

∂

∂x0〈0|ψ(0)|0〉 = −

Z

dS · 〈0|h

JN(x), φ(0)i

|0〉,(26.29)

where JN is the space-like part of the Noether cur-rent. Hint: You should assume ∂µJ

µN = 0.

Since the commutator in the previous equationcontains local operators potentially separated by

a large space-like interval, we conclude that〈0|ψ(0)|0〉 is independent of time.(b) Insert a resolution of the identity to show

Z

d3xX

n

h

〈0|J0N(0)|n〉〈n|φ(0)|0〉e−ipn·x

− 〈0|φ(0)|n〉〈n|J0N(0)|0〉eipn·x

i

6= 0. (26.30)

(c) Evaluate the spatial integral to verify

X

n

(2π)3δ(3)(pn)h

〈0|J0N(0)|n〉〈n|ψ(0)|0〉 eip0n·x0

− 〈0|ψ(0)|n〉〈n|J0N(0)|0〉e−ip0n·x0

i

6= 0. (26.31)

From (a) we conclude that this expression is inde-pendent of x0.(d) Argue that for |n〉 = |0〉, eqn. 26.31 vanishes.(e) Argue that if |n〉 describes a massive particlestate then matrix element 〈n|J0

N|0〉 must vanish.(f) Argue that if we want 〈n|J0

N|0〉 6= 0 then we re-quire p0

n → 0 as pn → 0.(g) This proves Goldstone’s theorem: the stateslinked to the ground state via the Noether currentare massless Goldstone modes. Convince yourselfthat this is so! Again, see Aitchison and Hey formore details.

(26.4) Consider another gauge theory: complex scalarelectrodynamics with a symmetry breaking massterm, whose Lagrangian is

L = −1

4FµνF

µν + |Dµψ|2 − V (ψ), (26.32)

where V (ψ) = −m2(ψ†ψ) + λ2(ψ†ψ)2 and Dµ =

∂µ + iqAµ.(a) Take the broken symmetry ground state to be

ψ0 =“

m2

λ

” 12. Expand about this using ψ =

ψ0 + 1√2(φ1 + iφ2) and show that the potential of

the broken symmetry theory is

U(φ1, φ2) = − 1

2λm4 +m2φ2

1 +O(φ3i ), (26.33)

exactly as in the case without electromagnetism.(b) Consider the kinetic energy term. Show that,on symmetry breaking, this becomes

|Dµψ|2 =1

2(∂µφ1)

2 +1

2(∂µφ2)

2 (26.34)

+√

2qψ0Aµ∂µφ2 + q2ψ2

0AµAµ + . . .

Interpret this result in terms of massive photons.This example shows that the massive gauge parti-cle in a broken symmetry system may be found byexpanding the covariant derivative.

27Coherent states

27.1 Coherent states of the har-monic oscillator 247

27.2 What do coherent states looklike? 249

27.3 Number, phase and thephase operator 250

27.4 Examples of coherent states252

Chapter summary 253

Exercises 253

’Tis all in pieces, all coherence goneJohn Donne (1572–1631)

There exists a class of problems in which a macroscopic number ofquanta all pile into the same momentum state. Examples include thelaser, where we put a large number of photons in the same wave-vectorstate into a cavity, and the superfluid, where a macroscopic number ofBose particles sit in a zero-momentum state. The best way to describesuch systems uses a set of basis states discovered by Schrodinger andchristened coherent states by Roy Glauber. A coherent state is an

Roy Glauber (1925– )eigenstate of an annihilation operator. In real space coherent states areGaussian wavepackets. As the occupation of the coherent state becomeslarge, the wavepacket becomes more and more narrow, increasingly re-sembling an object with classical properties. In this chapter we examinethe properties of these states. Having built up quantum field theory froman analogy between identical, non-interacting particles and harmonic os-cillators we start with the coherent states of the harmonic oscillator.

27.1 Coherent states of the harmonic

oscillator

The harmonic oscillator has a Hamiltonian H = ω(a†a+ 1

2

). Recall

from Chapter 2 that this was obtained from the Hamiltonian1 H =

1In this chapter we use reduced vari-

ables P 2 = p2

2mωand Q2 = 1

2mωx2,

and hence

Q = 1√2(a+ a†), P = − i√

2(a− a†).

ω(P 2 + Q2) by the introduction of creation and annihilation operatorsa and a†. This produced energy eigenstates |n〉 containing n quantaand with energy eigenvalue (n + 1

2 )ω. Thus 〈n|(P 2 + Q2)|n〉 = n +12 , but P and Q are individually not good quantum numbers.2 For a 2In fact 〈Q〉 = 〈P 〉 = 0 for energy

eigenstates.classical harmonic oscillator, the position and momentum behave likeQ = Q0 cos(ωt − ϕ) and P = −P0 sin(ωt − ϕ), so that both quantitiesoscillate but there is a definite 90 phase relationship between them.3 A

3This feature is not captured in ourquantum mechanical treatment of theharmonic oscillator in Chapter 2, sinceenergy eigenstates |n〉 are, by defini-tion, stationary states and individually

do not possess any signature of any-thing oscillating at angular frequencyω. Coherent states will achieve thisby being combinations of energy eigen-states. Coherent combination will giverise to interference, and this will recoverwavelike behaviour.

quantum mechanical state that possesses this phase relationship is whatwe are after, and so we are led to consider eigenstates of the operatorcombination Q+iP in which this phase relationship is hard wired. Sincea = (Q + iP )/

√2, we thus seek eigenstates of a. We therefore define a

coherent state |α〉 by

a|α〉 = α|α〉, (27.1)

where a is the annihilation operator and α is the eigenvalue (which will

248 Coherent states

be a complex number).4 Of course, a coherent state will be a sum of many4There is no restriction on α being realsince a is manifestly not Hermitian. energy eigenstates and so we may write |α〉 =

∑∞n=0 cn|n〉. Substituting

this into eqn 27.1 leads to the recursion relation cn+1 = αcn/√n+ 1,

and hence55In eqn 27.2 we have used |n〉 =(a†)n√n!

|0〉 in the second line.

|α〉 = c0

(|0〉 +

α√1!|1〉 +

α2

√2!|2〉 +

α3

√3!|3〉 + ...

)

= c0

(1 +

α

1!a† +

α2

2!(a†)2 +

α3

3!(a†)3 + ...

)|0〉, (27.2)

and normalization 〈α|α〉 = 1 fixes c0 = e−|α|2/2. Thus we can write thecoherent state compactly as

|α〉 = e−|α|2

2 eαa† |0〉. (27.3)

The amplitude to find |α〉 in the state |n〉 is cn = 〈n|α〉 = e−|α|2

2 αn/√n!,

corresponding to a probability density Pn = |cn|2 = e−|α|2 |α|2n/n! andthis represents a Poisson distribution (see Fig. 27.1). In the next exam-ple, we establish some basic properties of this probability distribution ofcoherent states.

n

Pn

Fig. 27.1 The Poisson distribution

Pn = e−|α|2 |α|2n/n! = e−〈n〉〈n〉n/n!plotted for |α|2 = 〈n〉 = 4. Note thata coherent state consists of a sum ofharmonic oscillator states: the quantityPn represents the probability of findingthis coherent state in the nth harmonicoscillator state.

Example 27.1

We don’t know the number of quanta in a coherent state but we can work out theaverage number6

6In fact this whole example can be donevery quickly if you use a property ofa Poisson distribution which is that itsmean and variance are the same. In thiscase, the mean 〈n〉 is |α|2, and hence sois the variance, meaning that ∆n, thesquare root of the variance, is simply|α|.

〈n〉 = 〈α|a†a|α〉 = |α|2. (27.4)

We can calculate the uncertainty ∆n in the number of quanta in a coherent statefrom ∆n =

p〈n2〉 − 〈n〉2. Thus by working out

〈n2〉 = 〈α|a†aa†a|α〉 = |α|4 + |α|2, (27.5)

we find that the uncertainty ∆n is

∆n = |α|, (27.6)

and therefore the fractional uncertainty is

∆n

〈n〉 =|α||α|2 =

1

|α| . (27.7)

We conclude that although we don’t know exactly how many quanta are in a coherentstate, the fractional uncertainty in the number tends to zero as α (and hence theaverage occupation) tends to infinity.

Example 27.2

In this example we see how coherent states respond to an operator U(θ) = e−iθn,where θ is a real number and n is our usual number operator. Let’s practise first onthe state |n〉 (where n|n〉 = n|n〉). This is easily done when we realize that U(θ) canbe written as a power series and hence

U(θ)|n〉 =

„

1 + (−iθn) +(−iθn)2

2!+ · · ·

«

|n〉 = e−iθn|n〉, (27.8)

27.2 What do coherent states look like? 249

since any power of the operator n on the state |n〉 just gives that power of the numbern. This now allows us to work out

U(θ)|α〉 = e−|α|2

2

∞X

n=0

(αe−iθ)n√n!

|n〉 = |αe−iθ〉. (27.9)

Thus the action of U(θ) on the coherent state |α〉 is to turn it into another coherentstate with an eigenvalue multiplied by e−iθ.

This last result tells us that the time-dependence of the coherent state|α〉 under the Hamiltonian H = (n+ 1

2 )ω is given by7 7Equation 27.10 should be contrastedwith the analogous result for energyeigenstates of the harmonic oscillator.Equation 27.8 gives

|n(t)〉 = e−i(n+ 12)ωt|n(0)〉,

so in this case the only time-dependence in the state is found in theuninteresting phase factor.

|α(t)〉 = e−iHt|α(0)〉 = e−iωt/2|α(0)e−iωt〉. (27.10)

The exponential prefactor e−iωt/2 is a simple phase factor originatingfrom the zero-point energy, but the main message should be read fromthe state |α(0)e−iωt〉: coherent states remain coherent over time and theeigenvalue α executes circular motion around the Argand diagram at anangular speed ω. It looks as if we have recovered the coherent oscillatorybehaviour we were after. But is there a way to visualize these states?

27.2 What do coherent states look like?

To see what a state looks like as a function of some coordinate Q weproject it along Q to form a wave function, via the amplitude 〈Q|α〉.

Example 27.3

Let |Q〉 and |P 〉 be the eigenstates of Q and P . We write a|α〉 = 1√2

“

Q+ iP”

|α〉 =

α|α〉 and act on the right with a state 〈Q| to obtain

1√2〈Q|

“

Q+ iP”

|α〉 = α〈Q|α〉. (27.11)

Since Q|Q〉 = Q|Q〉 and P = −i ∂∂Q

, we can make the replacements 〈Q|Q = Q〈Q|(since position is real) and 〈Q|iP = ∂

∂Q〈Q|, from which we obtain the differential

equation for the wave function:

∂

∂Q〈Q|α〉 = −

“

Q−√

2α”

〈Q|α〉. (27.12)

This equation has the normalized solution

〈Q|α〉 =1

π14

e−(Q−√

2α)2/2, (27.13)

so the wave function is a Gaussian. This one is displaced by√

2α = 〈Q〉+ i〈P 〉. Youcan also show that the wave function in momentum space is given by

〈P |α〉 =1

π14

e−(P+i√

2α)2/2. (27.14)

Using the position and momentum wave functions we can work out the uncertaintyin position and momentum. This is most easily done via the definitions

(∆Q)2 = 〈Q2〉 − 〈Q〉2, (∆P )2 = 〈P 2〉 − 〈P 〉2. (27.15)

250 Coherent states

We find (Exercise 27.3) that

(∆P )2(∆Q)2 =1

4. (27.16)

Comparing this with the uncertainty principle ∆Q∆P ≥ 1/2 shows that the coherentstates have the minimum possible value of uncertainty. This is the sense in whichcoherent states are the closest quantum mechanics allows us to come to classicalobjects which are localized in position and momentum space.

Finally we ask what these states do as a function of time. We haveshown that the state |α(0)〉 evolves to |α(t)〉 (apart from an uninterestingphase factor) where α(t) = α(0)e−iωt. An immediate consequence ofthis is that the expectation value of the position and momentum ofour state may be calculated. Since we have 〈a〉 = 〈α|a|α〉 = α and〈a†〉 = 〈α|a†|α〉 = α∗, we also have

〈Q〉 =α+ α∗√

2and 〈P 〉 = −i

α− α∗√

2, (27.17)

and we find that (writing α(0) = α0eiϕ)

〈Q〉 =√

2α0 cos(ωt− ϕ) and 〈P 〉 = −√

2α0 sin(ωt− ϕ), (27.18)

meaning that the expectation values execute simple harmonic motion(see Fig. 27.2). These results are exactly as we expected for the classicalharmonic oscillator at the beginning of the chapter.

Q

P

“blob”

Fig. 27.2 A coherent state can be pic-tured as a ‘blob’ orbiting the origin in aplot of P against Q. Its mean positionis given by 〈Q〉 =

√2Reα and 〈P 〉 =√

2Imα, where α(t) = α(0)e−iωt. Itis a blob, rather than a point, becausethere is uncertainty in both Q and P ,given by ∆Q = ∆P = 1√

2(see Exer-

cise 27.3).

27.3 Number, phase and the phase

operator

Because α is a complex number, we can write α = |α|eiθ and obtain

|α〉 = e−|α|2

2

(1 +

|α|eiθa†

1!+

|α|2e2iθ(a†)2

2!+

|α|3e3iθ(a†)3

3!+ ...

)|0〉.

(27.19)If we differentiate with respect to the phase θ we find that

−i∂

∂θ|α〉 = n|α〉. (27.20)

This implies that just as we have conjugate variables momentum andposition linked via p = −i ∂∂x , we identify conjugate variables n and θ.However, a closer examination of the phases of states shows that somecare is needed in their interpretation. The definition of an operator thatmeasures the phase of a system turns out to be rather problematical.One possibility suggested by Dirac is to make a phase operator via

a = eiφ√n,

a† =√ne−iφ. (27.21)

These already look rather odd in that it’s unclear how to interpret thesquare root of the number operator. We’ll examine the consequences ofthese definitions.

27.3 Number, phase and the phase operator 251

Example 27.4

The exponential operators can be rearranged to read

eiφ = a(n)−12 ,

e−iφ = (n)−12 a†. (27.22)

Since we know that a†a = n, then we must also have

e−iφeiφ = n− 12 nn− 1

2 = 1. (27.23)

However, if we try to find eiφe−iφ we encounter a problem since

eiφe−iφ = (n)−1(n+ 1) 6= 1, (27.24)

which means that the exponential operator eiφ is not unitary. Since these exponentialoperators are not unitary, the phase operator φ cannot be Hermitian8 and so cannot 8Remember that U = eiH , where U is a

unitary operator and H is a Hermitianoperator.

be used to describe an observable. There’s another problem. The phase operatorsyield a commutation relation (exercise)

h

n, eiφi

= −eiφ, (27.25)

which implies9 that we have 9If [A, B] = c then [A, eλB ] = λceλB .

We take A = n, B = φ and λ = i. Itfollows that c = i.

h

n, φi

= i. (27.26)

This, in turn, gives us a number-phase uncertainty relation

∆n∆φ ≥ 1

2. (27.27)

This is a problem if taken literally, since if we take ∆n to be small we must have ∆φgreater than 2π. Dirac was so troubled by these inconsistencies that he left theseoperators out of the third edition of his seminal textbook The Principles of QuantumMechanics.

One possible way forward was suggested by Peter Carruthers in the Peter Carruthers (1935–1997) is per-haps best known for his leadership ofthe theoretical division at Los AlamosNational Laboratory from 1973–1980,overseeing a notable period of expan-sion.

1960s.10 Here we regard the exponentials e±iφ as a whole as the fun-

10Actually by three students in hisclass. His problem set contained thefollowing: Apparently no one has in-vestigated whether quantities φ and

√n

‘defined’ by equations (1) really exist.A bonus will be given for an answer tothis question. The bonus turned outto be a bottle of beer. See M.M. Ni-eto, arXiv:hep-th/9304036v1 for moredetails.

damental operators. For things to work out we need the exponentiatedphase operators to obey the relations

eiφ|n〉 = |n− 1〉,e−iφ|n〉 = |n+ 1〉. (27.28)

We may then form Hermitian operators from the exponentials via thedefinitions

cosφ =1

2

(eiφ + e−iφ

),

sinφ =1

2i

(eiφ − e−iφ

). (27.29)

These operators, which do not commute with each other, have the prop-erties

[n, cosφ

]= i sinφ,

[n, sinφ

]= −i cosφ. (27.30)

252 Coherent states

and uncertainty relations

∆n∆cosφ ≥ 1

2|〈sinφ〉|,

∆n∆sinφ ≥ 1

2|〈cosφ〉|. (27.31)

Carruthers’ prescription then gives us a meaningful phase operator toplay with, as long as we agree only to ask questions about sines andcosines of the phase. In the next section we examine some examples ofcoherent states.

27.4 Examples of coherent states

Example 27.5

A box containing an electromagnetic field (or in the jargon, a cavity containingphotons) allows various normal modes. If we put a fixed number of photons into thebox and have them all in a single mode labelled by a wave vector k, then they willbe in an eigenstate of nk , which we will write as |nk〉. Since nk |nk〉 = nk |nk〉, weknow precisely the number of photons (i.e. we know nk) and we have absolutely nouncertainty in this number (∆nk = 0). However, we have complete uncertainty inthe phase of this state.

If instead we put photons, still all with the same wave vector k, into a coherentstate |αk〉, we are now uncertain about how many photons we have (∆nk = |αk |),by the uncertainty in phase11 ∆ cosφk → 0 as |αk | → ∞. As we put more photons11It is possible to show that

∆ cosφk =sin θk

2|αk |.

A proof is given in the book by Loudon.

into this state (〈nk〉 = |αk |2) then |αk | grows and the state becomes more phasecoherent. This situation is a good model for what occurs inside a laser.

Example 27.6

Another important use of the coherent states is in describing the matter fields ofsuperfluids and superconductors.12 These phenomena are examined in detail a little12Superfluids are described in Chap-

ter 42 and superconductors in Chap-ter 44.

later, but here we introduce a method of describing superfluids in terms of coherentstates. Like the laser, the superfluid also contains a large number of quanta in thesame quantum state. The particles are interacting bosons and are described by acoherent state

|ψ〉 = |αp0αp1

...〉. (27.32)

The coherent state |ψ〉 is an eigenstate of the field annihilation operator Ψ(x) givenby

Ψ(x) =1√VX

p

apeip·x , (27.33)

so that

Ψ(x)|ψ〉 = ψ(x)|ψ〉. (27.34)

The eigenvalue ψ(x) can be written simply as

ψ(x) =1√VX

p

αpeip·x , (27.35)

Exercises 253

and is often called a macroscopic wave function, for reasons which will becomeapparent.13

13Note that the expectation value

〈ψ|Ψ(x)|ψ〉 = ψ(x) = 1√VP

p αpeip·x ,which is significant since foroccupation-number states (withfixed numbers of particles) we wouldexpect

〈np1np2

...|Ψ(x)|np1np2

...〉= 1√

VP

p〈np1np2

...|ap |np1np2

...〉eip·x

= 0,

since 〈npi|api

|npi〉 = 0 for all i. So

clearly the interactions between bosons,which lead to a coherent state |ψ〉, aredoing something very special. In fact, itwill transpire that ψ(x) is the order pa-rameter for the ordered, coherent stateof bosons, known as a superfluid anddiscussed in Chapter 42.

In the superfluid a macroscopic number of particles reside in the p = 0 state. Wethen have ψ(x) = α0/

√V, a simple complex number which we can write as

√n0 eiθ0 .

Thus the expected number density of bosons is |α0|2/V = n0 and since n0 is verylarge (we have macroscopically occupied the coherent state), the uncertainty in thephase is vanishingly small.

We conclude that the macroscopic wave function ψ0(x) for a state where the p

state is coherently occupied is best written

ψ0(x) =√n0eiθ0 , (27.36)

where θ0 is a constant field. We can therefore think of superfluid order as a sponta-neous symmetry breaking of phase symmetry, resulting in the phase becoming fixedto the same value everywhere. This is examined in more detail in Chapter 42.

Chapter summary

• A coherent state is an eigenstate of the annihilation operator witheigenvalue α, a complex number.

• An occupation-number state contains a fixed number of particlesbut the phase is not well defined. A coherent state does not containa fixed number of particles, but that number is Poisson distributedwith mean |α|2. Its phase however can be well-defined, and itsuncertainty decreases as |α| increases.

• Coherent states can be used to describe lasers and superfluids.

Exercises

(27.1) Verify that

〈α|β〉 = eα∗β− |α|2

2− |β|2

2 . (27.37)

Compare this expression to the analogous onefor energy eigenstates of the harmonic oscillator,〈m|n〉 = δmn. Why is there a difference? Use theseresults to show that 〈n|a|n〉 = 0, but 〈α|a|α〉 = α.Again, why the difference?

(27.2) Show that

α =〈Q〉 + i〈P 〉√

2. (27.38)

(27.3) (a) Show that 〈Q〉 =√

2Reα, 〈P 〉 =√

2Imα,〈Q2〉 = 1

2+ 2(Reα)2 and 〈P 2〉 = 1

2+ 2(Imα)2.

(b) Show that ∆Q = ∆P = 1√2.

(c) Using the previous results, show that the uncer-tainty relation for coherent states is indeed

(∆P )2(∆Q)2 =1

4. (27.39)

(27.4) Verify that the momentum space coherent state isgiven by

〈P |α〉 =1

π14

e−(P+i√

2α)2 . (27.40)

(27.5) Show that“

Q− 〈Q〉”

|α〉 = −i“

P − 〈P 〉”

|α〉. (27.41)

One way of stating this result in words is that, forcoherent states, the uncertainty is balanced betweenposition and momentum.

254 Coherent states

(27.6) Show thath

n, eiφi

= −eiφ. (27.42)

(27.7) Once again consider the forced quantum oscillator

L =1

2mx(t)2 − 1

2mω2x(t)2 + f(t)x(t), (27.43)

where f(t) is the force which you should assumeacts only for a finite length of time: 0 ≤ t ≤ T .The evolution of 〈x(t)〉 will be given by the sumof the time dependence in the absence of the forceadded to the response:

x(t) = x0(t) +

Z

dt′χ(t− t′)f(t′). (27.44)

(a) Show that x(t) is given by

x(t) =

„

1

2mω

« 12 “

ae−iωt + a†eiωt”

(27.45)

+θ(t− t′)

Z

dt′if(t′)

2mω

“

e−iω(t−t′) − eiω(t−t′)”

.

Hint: Use the definition of the susceptibility in Ex-ercise 21.2.(b) For times t > T verify that this may be written

x(t) =

„

1

2mω

« 12

"

a+i

(2mω)12

f(ω)

!

e−iωt

+

a† − i

(2mω)12

f(−ω)

!

eiωt

#

.(27.46)

(c) Argue that the Hamiltonian must take the form

H = ω

a† − i

(2mω)12

f(−ω)

!

a+i

(2mω)12

f(ω)

!

.

(27.47)(d) Show that this Hamiltonian is diagonalized bythe coherent state |α〉 which is the eigenstate of thea operator and find α.(e) Show that the number of quanta emitted by thesource is

〈n〉 =|f(ω)|22mω

. (27.48)

and find the energy imparted to the system by thesource.(f) Find an expression for |α〉 in terms of the un-forced ground state |0〉.

(27.8) Consider (yet again) the forced quantum oscillatorfrom the previous question with the forcing part ex-pressed as an interaction.(a) Show that

−i

Z

dtHI(t)|0〉 =if(ω)

(2mω)12

|1〉. (27.49)

(b) By considering the S-matrix, show that the am-plitude for the source to create a state containing asingle quantum is given by

A1 =if(ω)

(2mω)12

e(Dumbbell), (27.50)

where (Dumbbell) denotes the Feynman diagram inFig. 22.2.(c) Extend the previous result to show that the am-plitude for the source emitting n quanta is given by

An =1√n!

if(ω)

(2mω)12

!n

e(Dumbbell). (27.51)

(d) Show that the probability of the source emittingn quanta is given by

Pn =|α|2nn!

e−|α|2 , (27.52)

where |α| = |f(ω)|(2mω)

12

. Comparing with the previous

question, explain why this result was inevitable.

(27.9) A coherent state is an eigenstate of the annihila-tion operator a. Why do we never bother with theeigenstates of the creation operator a†?

28Grassmann numbers:

coherent states and the

path integral for fermions

28.1 Grassmann numbers 255

28.2 Coherent states for fermions257

28.3 The path integral forfermions 257

Chapter summary 258

Exercises 258

Thus far, our discussion of fields and coherent states has concentratedexclusively on bosons. We would also like to describe fermions with pathintegrals and coherent states. This turns out to be a non-trivial matter.1

1For this reason, Chapter 28 can beskipped on a first reading. Grass-mann numbers will only be used in Sec-tion 38.2.

Consider the path integral, for example. In calculating Z[J ] for Bosefields we have argued that we integrate over the classical action in thepresence of sources Z[J ] =

∫Dφ e−i

R

d4x(L+Jφ). Unlike the canonicalformalism, the path integral formulation doesn’t deal with operatorsand their commutation relations; just classical fields. (The quantum be-haviour comes about through the constructive interference of the classi-cal paths.) This formulation works for Bose fields because the ordinarynumbers outputted by the functions over which we integrate commute,just like the quantum operators of the canonical formalism.

As we discussed in Chapter 3 the operator-valued fields that representfermions in the canonical approach anticommute. This is the propertythat, for example

ψ(x)ψ(y) = −ψ(y)ψ(x), (28.1)

where ψ(x) is an operator that annihilates a fermion at spacetime pointx. In order to be able to describe fermions with a path integral we needa new sort of number: a number that anticommutes.

Even before the advent of quantum mechanics, the properties of suchnumbers had already been examined by Hermann Grassmann. Although Hermann Grassmann (1809–1877)

Grassmann’s work was largely neglected during most of his life, theanticommuting numbers he invented are now known as Grassmann

numbers. In this chapter we will examine the properties of Grassmannnumbers. This will allow us to write down a coherent state for fermionsand will allow us to write a path integral for fermions. The latter willbe useful in our description of the quantum field of the electron.

28.1 Grassmann numbers

Grassmann numbers anticommute. This means that two Grassmannnumbers η and ζ have the property that

ηζ = −ζη. (28.2)

256 Grassmann numbers: coherent states and the path integral for fermions

Immediately following from this is that η2 = ηη = −ηη = 0. In words:the square of a Grassmann number is zero. This simplifies the algebraof Grassmann numbers. Because η2 = 0 we can’t have any terms in theTaylor expansion of a function with more than a single power of η, andso the most general function of η is then given by

f(η) = a+ bη, (28.3)

where a and b are real numbers.22Real numbers commute with Grass-mann numbers. We’d like to be able to do calculus with Grassmann numbers. We

define the act of differentiating via

∂∂ηη = 1, ∂

∂ηa = 0. (28.4)

Note also that this derivative operator itself anticommutes, so if ζ isanother Grassmann number we have ∂

∂η ζη = −ζ ∂∂ηη = −ζ.

Example 28.1

Differentiating the most general function of a Grassmann variable, we have∂f(η)

∂η= b. (28.5)

The most general function of two Grassmann variables is written

g(η, ζ) = a+ bη + cζ + eηζ. (28.6)

We can differentiate this to show that∂∂ηg(η, ζ) = b+ eζ, ∂

∂ζg(η, ζ) = c− eη. (28.7)

We would also like to integrate Grassmann numbers. We define integra-tion over Grassmann variables via

∫dη η = 1,

∫dη = 0. (28.8)

(Note again that the integration operator dη is to be regarded as aGrassmann number.) As a result of the definitions, integration of ourmost general function f(η), given in eqn 28.3, yields

∫dη f(η) =

∫dη a+

∫dη bη = b. (28.9)

Notice that this is the same result that we obtained by differentiating.In fact it is a general result that integration and differentiation of Grass-mann numbers yield exactly the same result!

Example 28.2

Let’s check this assertion by integrating the function g(η, ζ). Integrating with respectto η, we have Z

dη (a+ bη + cζ + eηζ) = b+ eζ, (28.10)

while integrating with respect to ζ gives usZ

dζ (a+ bη + cζ + eηζ) = c− eη, (28.11)

which is identical to what we had in eqn 28.7.

28.2 Coherent states for fermions 257

28.2 Coherent states for fermions

One use of Grassmann numbers is in defining a coherent state that de-scribed fermions. Taking the creation operator for a fermion to be c† wedefine a fermionic coherent state as |η〉 = e−ηc

† |0〉, where η, as usual, isa Grassmann number. We can immediately simplify this using η2 = 0and write

In order for things to work out, we re-quire that Grassmann numbers like ηanticommute with fermion operators.For example, we require η, c = 0.|η〉 = |0〉 − η|1〉. (28.12)

Example 28.3

We will check that a coherent state defined this way is the eigenstate of the fermionannihilation operator c. We have

c|η〉 = c|0〉 − cη|1〉 = 0 + ηc|1〉 = η|0〉. (28.13)

Butη|η〉 = η|0〉 − η2|1〉 = η|0〉, (28.14)

and so c|η〉 = η|η〉 as required.

We can also define a state 〈η|c† = 〈η|η, where

〈η| = 〈0| − 〈1|η = 〈0| + η〈1|. (28.15)

Note that η is not the complex conjugate of η and 〈η| is not the adjoint of|η〉. With these definitions it follows3 that the value of an inner product 3See Exercise 28.1.

is〈ζ|η〉 = eζη. (28.16)

Finally, for these coherent states to be useful, we require that they forma complete set via the completeness relation

∫dηdη eηη|η〉〈η| = 1. (28.17)

28.3 The path integral for fermions

We start by evaluating the Gaussian integral for coherent states4

4Had these been ordinary c-numbers,we would haveZ

dzdz∗ e−z∗az =

π

a, (Re a > 0),

where dzdz∗ represents the indepen-dent integration over real and imagi-nary parts of z. While the factor of πis neither here nor there, the importantpoint to note is that the a comes out inthe numerator for the Grassmann ver-sion.

∫dηdη eηaη =

∫dηdη (1 + ηaη) =

∫dη aη = a. (28.18)

This carries over to the case of Grassmann-valued (N -component) vec-tors η = (η1, η2, ..., ηN ) and η = (η1, η2, ..., ηN ) and we find5 that for a

5See Exercise 28.3.

matrix A: ∫dNηdN η eηAη = detA, (28.19)

where dNηdN η = dη1dη1dη2dη2...dηNdηN .Finally, we can work out the most important integral of all. Starting

with Grassmann numbers ψ and η, we will want to integrate an equationof the form ∫

dψdψ eψKψ+ηψ+ψη. (28.20)

258 Grassmann numbers: coherent states and the path integral for fermions

We play the usual trick of completing the square6 and we find6Recall that we originallysolved the path integral problem

I =R

dx e−ax2

2+bx by completing the

square, writing

−ax2

2+ bx = −a

2

„

x− b

a

«2

+b2

2a,

allowing us to conclude that I =q

2πa

eb2

2a .

ψKψ + ηψ + ψη =(ψ + ηK−1

)K(ψ +K−1η

)− ηK−1η, (28.21)

which enables us to conclude that∫

dψdψ eψKψ+ηψ+ψη = Ce−ηK−1η, (28.22)

where the number C (which will be proportional to a determinant forthe vector case) will later be removed by normalization. (Rememberthat most of our problems only require part of the solution: the part inthe exponential is important and the multiplying prefactor far less so.)

We can immediately extend this result to the functional integral weneed. Our goal is to find the generating functional for fermions

Z[η, η] =

∫DψDψ ei

R

d4x[L(ψ,ψ)+η(x)ψ(x)+η(x)ψ(x)]. (28.23)

If necessary we massage the Lagrangian L(ψ, ψ) into the formψ(x)Kψ(x), with K a differential operator, allowing us to read off theanswer that the normalized7 generating functional is given by:7Recall that to normalize we define

Z[η, η] = Z[η, η]/Z[0, 0].

Z[η, η] = e−iη(x)K−1(x,y)η(y). (28.24)

As before, we can then read off the propagator as iK−1 and the quanti-zation process is complete!

Chapter summary

• Grassmann numbers anticommute and can be used to describefermions.

• The fermion coherent state can be written |η〉 = |0〉 − η|1〉. Suchstates can be used to construct the fermion path integral.

Exercises

(28.1) From the definitions of the coherent states |η〉 and〈ζ|, prove that 〈ζ|η〉 = eζη.

(28.2) Find (a)R

dηdη ηη and (b)R

dηdη ηη.

(28.3) Show thatR

d2ηd2η eηAη = detA, where η =„

η1η2

«

, η =`

η1 η2´

, d2ηd2η = dη1dη1dη2dη2,

and A is a 2 × 2 matrix.

(28.4) Prove thatZ

dηdη eηη|η〉〈η| = |0〉〈0| + |1〉〈1|. (28.25)

and may therefore be taken as a resolution of theidentity.

(28.5) Verify eqn 28.21.

Part VII

Topological ideas

In many cases in quantum field theory we are concerned with workingout some integral over spacetime coordinates and therefore the precisegeometry of spacetime is important. However, there are some problemswhere the problem is entirely insensitive to the shape of spacetime, thatis it has no way of sensing the spacetime metric. In this case, the theorywill give an answer which is some kind of invariant which neverthelessdoes depend on some global property of the topology of the spacetime.

• In Chapter 29 we introduce some simple topological ideas and usethem to describe simple topological objects which can exist in fieldtheories. These include kinks and vortices.

• Topological field theories are discussed in Chapter 30 and we pro-vide some concrete examples of applications of these ideas by dis-cussing anyons, fractional statistics and Chern–Simons theories.

29 Topological objects

29.1 What is topology? 260

29.2 Kinks 262

29.3 Vortices 264

Chapter summary 266

Exercises 266

You better shape upOlivia Newton-John (1948– ), Grease1

1Newton-John’s grandfather was MaxBorn (1882–1970), one of the found-ing fathers of quantum mechanics. Thequoted lyric is by John Farrar.

In this chapter we meet a new class of excitation in the field knownas a topological object. These are configurations of the field whichmay exist in systems that show spontaneous symmetry breaking. Asdiscussed in Chapter 26, when a symmetry is broken we should watchout for objects known as defects that reflect the possibility that a fieldcan break symmetry in different ways in different regions of space. Theobjects we will encounter in this chapter are examples of such defects.We will discuss the two simplest topological objects: kinks and vortices.Both of these may be thought of as time-independent lumps of energywhich are held together by interactions. An important additional featureof vortices is that they’re not stable alone and we must add gauge fieldsin order for them to be realized. As always, the presence of gauge fieldsprovides the theory with some particularly interesting properties. Webegin with a crash course on basic topology.

∼= ∼=

Fig. 29.1 A cup can be continuouslydeformed into a doughnut (a torus).

29.1 What is topology?

The doughnut and the coffee mug are topologically equivalent becauseone can be continuously deformed into the other (Fig. 29.1). We imaginethat everything is made of some kind of mouldable clay and so all objectscan be pressed and prodded so that their shape changes. However,you are not allowed to puncture or heal, to add or remove holes, sothat neighbouring points in the object remain neighbouring after thedeformation. Topological spaces are given mathematical names: thereal line is denoted R, so the plane is R2 and the n-dimensional versionis Rn. One can define a product space in an obvious way, and soR × R = R2 [Fig. 29.2(a)]. Similarly R × R × · · · × R = Rn. A segmentof a R joined head to tail gives a circle, denoted S1. The sphere isS2. (Note that this is a two-dimensional space, because by ‘sphere’mathematicians mean the surface of a ball, not its interior.) The spherecannot be embedded in R2 (you can’t fully represent a sphere on a pieceof paper without cutting it2) but can be embedded in R3. (Similarly S1

2This is why projections of the Earth’ssurface on maps usually involve a cutthrough one of the major oceans.

can’t be embedded in R.) The torus T 2 (back to the doughnut again)can be constructed using the product T 2 = S1 × S1 [Fig. 29.2(b)].

∼=

∼=

×

×

S1 S1

T 2

R RR

2

(a)

(b)

Fig. 29.2 A product space is con-structed by attaching a copy of onespace to every single point in anotherspace. (a) A product space of the realline (R) with itself produces the two-dimensional plane R2. (b) The productof two circles (S1) produces the torusT 2. A cup can be continuously de-formed into a doughnut (a torus). Once we have a topological space, we can define a path through the

space as a mapping f(x) of the real-line segment [a, b] (shorthand for

29.1 What is topology? 261

a ≤ x ≤ b) to the topological space. If f(a) = f(b), the path becomes aloop. Paths can be continuously deformed into each other in an obviousway and one finds that in a particular topological space the set of possibleloops can be divided up into a number of classes and these can be mappedonto a group (called the fundamental group and given the symbol π1).For example, in Rn all loops are contractible (i.e. can be continuouslydeformed to a point) and so there is only one class of loop and π1 is atrivial group just consisting of the identity element. It’s more interestingif the space has a hole in it, so that the loops can be divided into classes,each one characterized by the number of times the loop winds round thehole. This integer is known as the winding number. Thus π1 = Z, theset of integers. Further examples are π1(S

1) = Z and π1(T2) = Z × Z

(see Fig. 29.3).

Fig. 29.3 Loops on a surface of atorus T 2 can be contractible (contin-uously deformable to a point), such asC, or non-contractible such as A and B(each of these winds round once). Anyloop can be described using two wind-ing numbers, the number of times theloop winds round like A and the num-ber of times like B. In topological par-lance: π1(T 2) = Z × Z.

The crucial point about topological arguments is that they do not relyon any notion of the geometrical structure of space. Topological spacesare continuously deformable and therefore a topological description re-flects the underlying ‘knottiness’ of the problem.

Example 29.1

A thin solenoid (radius R, aligned along ez) inserted into the two-slit experiment(see Fig. 29.4) gives rise to a magnetic vector potential outside the solenoid equal toAθ = BR2/2r and inside Aθ = Br/2. The magnetic field B = ∇× A is then equalto zero outside and (0, 0, B) inside the solenoid.

electron

s

electrons

scre

en

B

Fig. 29.4 The Aharonov–Bohm effect.Electrons diffract through two slits buttheir paths enclose a solenoid. Thefield inside the solenoid is B, but theelectrons only pass through regions inwhich there is no magnetic field.

Even though B = 0 outside the solenoid, the electronic wave function can stillbe affected. This is because a plane wave ψ ∝ eip·r/~ and in the presence of A themomentum p → p − qA and so over a trajectory ψ acquires an extra phase factorei∆α where ∆α = − q

~

RA·dr. This might not seem to matter because if you perform

a gauge transformation A → A + ∇χ then ∆α = − q~

R∇χ · dr and so ∆α can be

anything you want, all you need to do is to pick a particular form for χ. However,when we consider the two-slit experiment, we are interested in the phase difference∆δ between the two paths which is given by

∆δ = ∆α1 − ∆α2 = − q

~

Z

a→1→bA · dr +

q

~

Z

a→2→bA · dr

=q

~

I

A · dr, (29.1)

where the final integral is carried out anticlockwise around the trajectories. However,we may use Stokes’ theorem to write

∆δ =q

~

Z

∇× A · dS =q

~

Z

B · dS =q

~Φ, (29.2)

where Φ is the flux through the solenoid. This leads to a shift in the interferencepattern on the screen and is observed experimentally. Gauge transformations makeno difference here because

H∇χ · dr = 0. Remarkably we expect a nonzero gauge-

independent phase difference ∆δ even though electrons only pass through regionswithout magnetic field. The effect is known as the Aharonov–––Bohm effect and Yakir Aharonov (1932– )

David Bohm (1917–1992)is topological. The electronic wave function is defined on the plane R2 minus theorigin (where we place the flux), i.e. a sheet with a hole in it. Electromagnetism hasU(1) symmetry, which has the same topology as S1 (the phase can be defined on acircle in the Argand diagram). Thus to define a phase everywhere on a sheet witha hole in it amounts to mapping S1 on to a path around the hole. These fall intodisjoint classes (labelled by an integer winding number, because π1(S1) = Z) andthese cannot be deformed into each other.


29.2 Kinks

Consider the Lagrangian L = 12 (∂µφ)2 − U(φ) in (1 + 1)-dimensional

spacetime, with U(φ) = λ4 (v2−φ2)2. This is just our symmetry breaking

Lagrangian for φ4 theory as can be quickly seen by expanding the bracketand setting v2 = m2

λ . The potential, shown in Fig. 29.5, has two minimawith U(φ) = 0 at φ(x) = ±v. The vacuum of the theory is to be foundat φ(x) = v or φ(x) = −v. We know that one way to work out what sortof particles live in this potential is to expand about one of the vacua.The excitations are then predicted to be particles of mass m = (2λv2)

12 .

The point of this chapter is to tell you that objects other than particlescan live in the potential too. These are stationary configurations of thefield φ(x) whose energy density 1

2 (∂iφ)2 +U(φ) goes to zero at x = ±∞,but which do something non-trivial in between.

φ

U

−a a

Fig. 29.5 The double well potential.

x

φ(x)

φ(x) = v

Fig. 29.6 The uniform ground state ofthe field in a system with broken sym-metry.

x

φ(x)

v

−vl

Fig. 29.7 The kink solution.

Let’s examine a field which has zero energy density at x = ±∞. Sucha field must be stationary and occupy a zero of the potential at ±∞. Aboring way of achieving this is for φ(x) = v as shown in Fig. 29.6. Theconfiguration φ(x) = −v is similar. This is just symmetry breaking:the system falls into a ground state of φ = v or φ = −v. However,there is a more interesting configuration of the field. What if the fieldtakes φ(−∞) = −v and φ(∞) = v? Then we must have somethinglike the object shown in Fig. 29.7. This configuration is known as akink and involves the curious state of affairs of half of the field living inone vacuum and half in the other. Notice that the field has broken thesymmetry in two different ways. At x→ ∞ the field takes on the valueφ = v, whereas for x → −∞ it takes on value φ = −v. We might saythat the fields at ±∞ live in difference vacua. To find out whether thekink is stable, we may evaluate its total energy. If the energy is finitethen we’re in luck as such configurations are allowed to exist.

Example 29.2

We will evaluate the energy of a kink. We take the kink to be stationary in time, so∂0φ = 0. We want to work out E =

Rdxˆ12(∂1φ)2 + U(φ)

˜. From the equations of

motion we have that ∂2φ∂x2 = ∂U

∂φ, which may be integrated to give 1

2

“∂φ∂x

”2= U(φ).

We therefore have an expression for the energy given by

E =

Z

dx

"

1

2

„∂φ

∂x

«2

+ U(φ)

#

=

Z ∞

−∞dx 2U(φ) =

Z v

−vdφ

dx

dφ2U(φ)

=

Z v

−vdφ [2U(φ)]

12 . (29.3)

Plugging in [2U(φ)]12 = (λ/2)1/2(v2 − φ2), we obtain the energy3

3Notice that the kink has energy in-versely proportional to the couplingconstant. This is significant since itimplies that an expansion in powersof the coupling constant, of the sortwe make in perturbation theory, wouldnever predict such a field configuration.This is one of the hallmarks of topo-logical objects: they are fundamentallynon-perturbative.

E =1√2

4m3

3λ. (29.4)

This is finite and the kink is therefore allowed to exist!

29.2 Kinks 263

The kink has a finite energy and, by construction, is a time-independentfield configuration. It is also of a finite size l (see Fig. 29.8), the extentof which is determined by a balance of kinetic and potential energy. Thekinetic energy’s contribution

∫dx 1

2 (∂1φ)2 ≈ l× (v/l)2 tries to smear thekink out, but the potential contribution

∫dxU ≈ λv4l tries to limit the

region over which the field changes. The minimum energy is then to befound when the size of the kink is l ≈ (λv2)−

12 = 1/m, where m is the

mass of the conventional particles of the theory.x

E(x)

l

Fig. 29.8 The energy density of a kinkis localized in an interval l.

Example 29.3

The domain wall in a ferromagnet has much in common with the kink describedhere. For the ferromagnet the vacua correspond to all spins pointing up (φ = v) ordown (φ = −v). The kink corresponds a region where the fields change from up todown over a finite distance. This separates two magnetic domains. A domain is aregion where the symmetry is broken in a particular way. It is, for example, part ofthe Universe where the vacuum corresponds to all spins pointing up. Domain wallstherefore separate regions with different vacua.

Domain walls are real and their existence may be detected in magnets througha variety of means. In a magnet, domains allow a magnetic system of finite spatialextent to save energy by reducing the field density in free space. However, the domainwall might be expected to exist on the more general grounds discussed here.

Also interesting is an idea of Nambu’s that in reaching its final state the Universehas broken various symmetries. This, he suggests, might imply that domains withdifferent vacua may exists across the Universe.

We have a field configuration with finite energy above the ground statewhich is of finite spatial size. The theory is translationally invariant:the kink’s centre can be anywhere. The theory is Lorentz invariant: thekink can be boosted to an arbitrary velocity. The kink is, therefore,very much like a particle. In addition, the kink is stable: any attemptto remove the kink involves lifting a (semi)infinite length of field fromone of the potential minima to another. This costs an infinite amountof energy and is therefore impossible. A more mathematical statementof this is that the kink is not deformable. We cannot hold the endsof the kink tightly and then remove the part that crosses the axis inbetween. The one way of annihilating a kink would be to have an anti-kink: an object whose field crosses the axis in the opposite direction sothat φ(−∞) = v and φ(∞) = −v, as shown in Fig. 29.9. The kink andantikink pair is deformable. If we hold the ends of the field tightly wesee that we can continuously change the part that crosses the axis toremove the parts that cross the axis. We end with a field lying in thevacuum at φ = −v.

x

φ(x)

v

−vkink antikink

Fig. 29.9 A kink and antikink.

This stability of kinks is encoded in a description of its charge. Acharge, remember, is designed to be a conserved quantity. A single kinkshould carry a topological kink-charge of QT = 1; an antikink a kink-charge of QT = −1. In order to come up with a sensible recipe forfinding the kink-charge, we define the kink-current4 as 4Note that this is different to the

Noether currents that we have consid-ered so far. Its existence owes nothingto a symmetry of the Lagrangian.

JµT =1

2vεµν∂νφ, (29.5)


where εµν is the antisymmetry symbol, whose signs are determined byknowing that ε01 = 1. This leads to an expression for the charge of akink of

QT =

∫ ∞

−∞dxJ0 =

1

2ν

∫ ∞

−∞dx

∂φ

∂x

=1

2v[φ(∞) − φ(−∞)] = 1, (29.6)

whereas an antikink has QT = −1. Notice that the phions of the theorydo not have a kink-charge since QT = φ(∞) − φ(−∞) = 0. We call thekink-charge a topological charge, which explains the subscript. Itsexistence is independent of the geometry of spacetime. That geometryis encoded in gµν , which is an instruction book telling us what watchesand measuring rods measure in spacetime. In contrast the topologicalcharge has its indices summed with the antisymmetric symbol5 εµν . The

5Remember that our convention is notto treat this as a tensor and so εµν =εµν

dependence on εµν rather than gµν turns out to be a general feature oftopological objects and is examined in more detail in the next chapter.6

6To summarize the argument so far, thetheory

L =1

2(∂µφ)2 − U(φ),

with

U(φ) =λ

4(v2 − φ2)2,

allows the existence of kinks. These aretopological objects which behave ratherlike particles. The field at x = ∞ andx = −∞ live in different vacua.

29.3 Vortices

(a) (b)

(c) (d)

Fig. 29.10 Some vortices with windingnumber n equal to (a) 1; (b) 1; (c) −1;(d) 2.

Now we move to (2 + 1)-dimensional spacetime and examine the sameLagrangian. Our potential now resembles a Mexican hat, whose minimadescribe a circle in the φ1-φ2 plane. It will be convenient to work in polar(internal) coordinates (r, θ) and write φ(x) = φ1(x)+iφ2(x) ≡ r(x)eiθ(x).In the case of the kink we found a topological field whose two endslived in different vacua at spatial |x| → ∞. By analogy we ask what acontinuous field would look like whose elements live in different vacuaat spatial infinity. Some examples of such configurations are shown inFig. 29.10. These objects are known as vortices. They are described bya field whose form at infinity is written

φ(x) = Kei[nθ(x)+ϕ] (|x| → ∞), (29.7)

where θ(x) = tan−1(x2/x1) and ϕ is a constant, arbitrary phase.7 Notice

7The phase ϕ can be chosen to globallyalter the directions at infinity. So, forexample, Fig. 29.10(a) corresponds toϕ = 0, while Fig. 29.10(b) correspondsto ϕ = π/2. This can also be absorbedinto the complex amplitude K.

how this equation relates the direction in which the field φ points in theinternal complex plane to the angle θ that tells us where we are incoordinate space. Here n is known as the winding number; it tells ushow many times the field winds around as we travel around the circleat spatial infinity.

Example 29.4

We may evaluate the energy of a vortex. For a static field configuration we have aHamiltonian

H =1

2∇φ† · ∇φ+ U(φ), (29.8)

and we take U(φ) =`|K|2 − φ†φ

´2, so that U(φ) = 0 on the boundary at infinity.

The gradient of the vortex in cylindrical (spatial) coordinates is

∇φ =1

r(inKeinθ)eθ. (29.9)

29.3 Vortices 265

The core of the vortex looks frighteningly singular, but we will assume it is ultimatelywell behaved and call the core energy Ecore(a), where a is the core size. We evaluatethe energy of the rest of the configuration at distances larger than r = a and find

E = Ecore +

Z ∞

adrdθ rH = Ecore + πn2|K|2

Z ∞

adr

1

r. (29.10)

This is a logarithmically divergent energy. We conclude that a single vortex is not astable object.

The single vortex is unstable. This is understandable if we look at theform of the field at large distances (Figs. 29.11 and 29.12). It is stillswirly as r → ∞. Actually this is predicted by a general theorem due toG.H. Derrick, which says that time-independent topological objects areimpossible for our theory in more than one spatial dimension.

x

y

Fig. 29.11 Fields at large distancesfor an n = 1 vortex.

Fig. 29.12 The vortex field configura-tion for an n = 1 vortex.

To stabilize the vortex we must tamper with the Lagrangian. The wayto do this is to add a gauge field. Recall that adding a gauge field (knownas ‘gauging the theory’) is done by introducing a covariant derivative asfollows:

Dµφ = ∂µφ+ iqAµφ. (29.11)

In order to cancel the divergence of the vortex we’re going to choose ourgauge field in such a way that it cancels the divergent part of the vortexenergy, which resulted from the derivative of the field configuration.We therefore want the covariant derivative Dµφ to vanish at infinity.A gauge field that does this is one whose limit as r → ∞ tends toA(r, θ) = 1

q∇(nθ).

Example 29.5

The components of the proposed gauge field at infinity are8

8Recall that A = Ai = −Ai in follow-ing the sign changes here.

Ar → 0, Aθ → − nqr

as r → ∞. (29.12)

As r → ∞ the components of the covariant derivative become

Drφ = 0, Dθφ = 1r∂φ∂θ

+ iqAθφ = 0 as r → ∞. (29.13)

This shows that the covariant derivative (which represents the ‘kinetic energy’ ofthe field) vanishes at infinity. Since it was this term that diverged for the ungaugedtheory, we see that A has saved the day.

The gauge field should also be expected to make a contribution to theLagrangian of the form − 1

4FµνFµν . However, this is not the case here.

Usually we have a gauge field which can be changed by a gauge trans-formation which adds a contribution ∂µχ. In our case we notice that atinfinity A = ∇χ, where χ = 1

qnθ. That is, our gauge field is entirelyformed from a gauge transformation and is therefore another exampleof the pure gauge which we first saw in Chapter 26. This means thatFµν = 0 and the gauge field cannot result in a nonzero − 1

4F2 contribu-

tion to the Lagrangian. This is also a relief, since such a contributionmight also threaten to blow up, ruining the stability of the vortex.


What then is the contribution of the gauge field to the physics? Theanswer is found by carrying out the integral

∮A · dl around a circle at

infinity. From Stokes’ theorem we know what this means physically: theintegral is

∮A · dl =

∫B · dS = Φ, that is the magnetic flux through

the circle. Carrying out this integral we obtain

Φ =

∮A · dl =

∮Aθr dθ =

∫dθ

n

q=

2πn

q. (29.14)

We see that vortices carry quantized magnetic flux!The next step in this line of argument to examine a topological object

in (3+1)-dimensional spacetime. Remarkably, this object turns out tobe a form of a magnetic monopole and is examined in Chapter 49. In thenext chapter we turn our attention to another aspect of topology anddescribe a field theory that can be legitimately described as topological.

Chapter summary

• Topology relates the structure of objects that are preserved un-der continuous deformations involving stretching but no cutting orgluing.

• Topological objects include kinks and vortices and can be charac-terized by a topological charge.

Exercises

(29.1) Show that the antikink has kink-charge QT = −1and that two kinks and three antikinks have a totalcharge of QT = −1.

(29.2) This is the first time we have used the antisymmet-ric symbol εµν in anger. We regard the up and downversions as identical since we are not using it as atensor. Let’s examine its properties. Take ε01 = 1and ε012 = 1 and show the following:(a) εijεij = 2.(b) εijεin = δjn.(c) εijεmn = δjnδim − δinδjm.(d) εijkεimn = δjmδkn − δjnδkm.

(29.3) A simple model of a domain wall in a magnetsays that the interaction energy of two neighbour-ing (classical) spins of magnitude S is given byE = −JS2 cos θ where θ is the relative angle be-tween the spins and J is a constant.(a) Show that the energy of a domain wall sepa-

rating a domain of spins up and a domain of spinsdown costs energy

U =1

2JS2 π

2

N, (29.15)

where N is the number of spins in the wall.(b) Show that this picture will not lead to a stablewall.(c) In reality spin–orbit effects lead to an energycost for not having spins pointing along certain spa-tial directions. In this case these lead to a contri-bution to the energy of NK

2. Show that the wall is

now stable and find its size.

(29.4) (a) Draw vortices with winding numbers n = −2and n = 3.(b) Draw the fields resulting from vortices withn = 1 and n = −1 sharing the same region of space.This situation is discussed in Chapter 34.

30Topological field theory

30.1 Fractional statistics a laWilczek: the strange case ofanyons 267

30.2 Chern–––Simons theory 269

30.3 Fractional statistics fromChern–––Simons theory 271

Chapter summary 272

Exercises 272

Imagine the perplexity of a man outside time and space, whohas lost his watch, his measuring rod and his tuning fork.Alfred Jarry (1873–1907), Exploits and Opinions of DoctorFaustroll, Pataphysician

So far we have met topological objects that can exist in scalar field the-ories with a broken symmetry. In this chapter we will take the studyof topology further and cook up an inherently topological field the-

ory. This theory will apply to the special case of (2+1)-dimensionalspacetime. We will see that physics in the flatland of (2+1) dimensionshas some remarkable properties whose origin may be traced back totopological considerations.

We start our discussion by examining the sorts of particle statisticsthat apply in two spatial dimensions. We will see that flatland supportsanyons: particles that are neither bosons nor fermions. We will thengo on to formulate a field theory that not only captures anyon statistics,but also describes some curious new electrodynamics.

30.1 Fractional statistics a la Wilczek: the

strange case of anyons

We are quite used to the idea of bosons and fermions as entities distin-guished by their behaviour upon exchange of identical particles. Specif-ically, changing the labels on two identical particles results in a changeof the wave function according to the rule

ψ(x1, x2) = ±ψ(x2, x1), (30.1)

where the + sign applies to bosons and the − sign to fermions.For the case of two-dimensional space, the exchange of particles needs

more careful attention. In fact, we need to think more precisely aboutwhat the exchange of particles actually involves. We do not simplymake the two particles vanish and then reappear in different positions.Rather we imagine the process in which particles are moved aroundeach other in real space. Let’s identify two moving-around processes.We start with two identical particles at positions x1 and x2 and identifytwo fundamentally distinct ways of exchanging them. The result ofprocesses of type A is to move x1 → x1 and x2 → x2. Some examplesare shown in Fig. 30.1(a) and (b). In this sort of process the particles

268 Topological field theory

end up where they were originally, although they may move around eachother. The result of type B processes is to move x1 → x2 and x2 → x1

[Fig. 30.1(c) and (d)]. Again particles may move around each otherseveral times before settling at their final positions [as in Fig. 30.1(d)].This is particle exchange, albeit in a more tangible form than the usualmagic trick method.

We ask what the relative quantum mechanical phase difference is be-tween processes of type A and type B. The key parameter to consideris the angle that one particle is moved around the other. This is wheretopology comes in: given a set of particle paths, we may smoothly dis-tort the paths of the particles, but we may not change the number oftimes particles wrap around each other without introducing singularitiesin the particle paths. Processes of type A involves rotating particle 2around particle 1 by angle φ = 2πp, where the winding number p takesan integer value (including zero). Processes of type B involves rotationsof φ = π(2p + 1). Each value of p describes a topologically distinctprocess.

We suppose that these topologically distinct processes make a mul-tiplicative contribution to the total wave function (or, in field theory,to the path integral) of Φ(φ) which we expect to be pure phase. If wecarry out a sequential string of these processes then we require that theangles add, whilst the wave functions should multiply. That is, we needΦ(φ1 + φ2) = Φ(φ1)Φ(φ2) which implies that Φ(φ) = eiηφ where η is aparameter which, crucially, doesn’t need to be an integer.

Fig. 30.1 Examples of ways of ex-changing particles. (a) and (b) are typeA processes, where the particles end upat the same positions. In case (b) oneparticle loops around the other once inthe exchanging process. (c) and (d) aretype B processes where the particles ex-change positions. In case (d) one parti-cle loops once around the other duringthe exchange.

Now we compare our real life exchange to the old-fashioned definitionin eqn 30.1. If we carry out the exchange (x1, x2) → (x2, x1) [shown inFig. 30.1(c)] then the formal definition of exchange embodied in eqn 30.1tells us that the wave function should be identical (for bosons) or pickup a minus sign (fermions). However, the realistic version of exchangemerely tells us that φ = π, resulting in a phase factor of Φ(φ) = eiηπ.The two versions of exchange are only identical for the special cases that(a) we have η = even integer, when we recover the expected exchangebehaviour for bosons or (b) we have η = odd integer, when we recoverfermion exchange. However, this analysis shows that there are manymore possible values of η in two dimensions. It doesn’t have to bean integer! We are therefore not tied simply to bosons and fermions,the freedom to choose η means we can have any exchange statistics.Particles with such statistics were dubbed anyons by Frank Wilczek,who has made important contributions to elucidating the study of these

Frank Wilczek (1951– ) was a co-recipient of the Nobel Prize in Physicsin 2004 for their discovery of asymp-totic freedom. His popular book The

Lightness of Being (2008) is a very read-able account of modern ideas in particlephysics.

particles and their ‘fractional statistics’.Notice that having only two spatial dimensions to play with was vital

to the argument. In three spatial dimensions all of the type A processesare topologically identical since they are all deformable into paths wherethe particles don’t move. Similarly all of the type B processes are topo-logically identical and may be reduced to a simple exchange of particles.This reduction occurs because the extra dimension allows us to movethe paths past each other in the third dimension, shrinking all loops tozero.

30.2 Chern–––Simons theory 269

30.2 Chern–––Simons theory

We now turn to the problem of writing down a Lagrangian that can bedescribed as topological. What do we mean by this? Our usual La-grangians have relied on the use of the metric tensor gµν , which is anobject that tells us how to measure displacements in space and time.It provides a set of instructions for taking the scalar product of vectorsthrough a contraction of indices. In contrast, in a topological theorythe contraction of indices is carried out through the use [in (2+1) di-mensions] of the antisymmetric symbol εµνλ. Topological theories aretherefore blind to the details of the watches and measuring rods thatare contained in gµν . Their content is dependent purely on the topol-ogy of the manifold1 in which we’re working. The first consequence of 1A manifold is a topological space

that locally resembles Euclidean space.this feature is a shock when we try to derive the Hamiltonian from atopological Lagrangian: the Hamiltonian vanishes!

Example 30.1

To see this we note that a general method of finding the energy-momentum tensorTµν is to vary the action with respect to the components of the metric. For an actionStop derived by integrating a topological Lagrangian we have

Tµν =−2√− det g

δStop

δgµν. (30.2)

Since topological theories don’t include gµν at all then we must have Tµν = 0. Sincethe Hamiltonian is the 00th component of this tensor then we conclude that H = 0.

A consequence of this is that all of the states of a topological Hamiltonian haveenergy E = 0. So there is a significant degeneracy in the system. Unfortunately itis rather difficult to determine how many states we have since this depends on thetopology of the manifold.

We will now describe a topological term that may be part of a La-grangian. This is known as a Chern–––Simons Lagrangian and applies Chern–Simons theories were first intro-

duced into field theory by Edward Wit-ten (1951– ) based on work in differen-tial geometry. They are named afterShiing-Shen Chern (1911–2004), math-ematician, and James Harris Simons(1938– ), mathematician, hedge fundmanager and philanthropist.

to the case of (2+1)-dimensional spacetime. It is written2

2Notice that the Chern–Simons theorywill work in (2+1)-dimensional space-time but not in (3+1) dimensions. Trywriting it in (3+1) dimensions andyou’ll see that the indices won’t matchup.

L = −1

2κεµνλaµ∂νaλ, (30.3)

where aµ is a U(1) gauge field and κ is a constant. Chern–Simons termsare easy to spot since they have the form L = εa∂a. We will see thatChern–Simons theory contrasts in many respects with the other U(1)gauge theory with which we’re familiar: electromagnetism.

Example 30.2

The Chern–Simons Lagrangian gives rise to a gauge invariant action. Actually thegauge invariance is slightly more subtle than the usual version seen in electromag-netism. To show this, we make the gauge transformation aµ → aµ + ∂µχ, and weobtain

εµνλaµ∂νaλ → εµνλaµ∂νaλ + εµνλ∂µχ∂νaλ. (30.4)


Here we have used the fact that εµνλ∂µ∂νχ = 0, which follows from the antisymmetryof εµνλ. The resulting change in the action (that is, the second term in the aboveequation) may be written as a total derivative

δS =

Z

d3x εµνλ∂µ(χ∂νaλ). (30.5)

This vanishes under the assumption that we may drop boundary terms. We concludethat the Chern–Simons Lagrangian is gauge invariant, as long as we discard theboundary terms.

The next thing to note is that if we plug this term into the Euler–Lagrange equations we obtain an equation of motion for the aµ fieldsas

κεµνλ∂νaλ = 0, (30.6)

or defining a field strength tensor fµν = ∂µaν−∂νaµ we have an equationof motion 1

2κεµνλfνλ = 0 or

fµν = 0. (30.7)

This looks very unexciting! Unlike electromagnetism, for which, in theabsence of sources, we have an equation of motion ∂µF

µν = 0 whichsupports plane waves, the Chern–Simons theory seems to have no dy-namics of its own. Nevertheless, the Chern–Simons theory will turn outto be interesting, but we just need to couple it to another field.

Therefore, we now couple our Chern–Simons field aµ to a source Jµ

which is the conserved current of some other field and write

L = −1

2κεµνλaµ∂νaλ + aµJ

µ, (30.8)

and this leads to an equation of motion

Jµ = κεµνλ∂νaλ =1

2κεµνλfνλ. (30.9)

Clearly this expression for current isn’t a consequence of Noether’s the-orem; rather, it occurs as a result of our coupling a source to the Chern–Simons gauge field by hand. The Chern–Simons term therefore providesa constraint on the current.

In order to understand the properties of the current we now point outan odd, but undeniably relevant feature of (2+1)-dimensional spacetime.This feature stems from the fact that, in two spatial dimensions, thecross product results in a pseudoscalar, rather than a pseudovector. Theresult is that the magnetic field in (2+1)-dimensional spacetime is apseudoscalar: B = εij∂iA

j .

Example 30.3

To see this we note that in three dimensions we have the definition of the crossproduct in component form V i = εijkAjBk. In two dimensions we lose an indexfrom our antisymmetric symbol and we have S = εijAiBj . We run out of indices,so the result of the cross product is a pseudoscalar as claimed.

30.3 Fractional statistics from Chern–––Simons theory 271

Now turning to the components of the current Jµ = (ρ,J) predictedfrom eqn 30.9, we find

ρ = κ(∂1a2 − ∂2a1), J1 = κ(−∂0a2 + ∂2a0), J2 = κ(∂0a1 − ∂1a0).(30.10)

Although the field aµ does not describe electromagnetism, we may usethe language of electromagnetism and call b = ∂2a1 − ∂1a2 the aµ mag-netic field b and call ∂0ai − ∂ia0 the ith component of the aµ electricfield ei. We therefore have

ρ = −κb, J i = −κεijej , (30.11)

where ε12 = 1. The second of these equations tells us that an aµ electricfield in the y-direction gives a source current in the x-direction. Themeaning of the first equation is made clear if we integrate over the two-dimensional space to find that the charge Q of the source field is relatedto the flux of the aµ magnetic field by

Q = −κ∫

d2x b, (30.12)

or (Charge ofsource field

)∝(

Flux ofaµ field

). (30.13)

Thus we conclude the following:

Chern–Simons theory ties aµ-flux to the charge of the source field.

30.3 Fractional statistics from

Chern–––Simons theory

The fact that external source charge couples to aµ-flux has an influenceon the statistics we find on the exchange of particles. Recall that in(2+1) dimensions moving one particle completely around another givesrise to a phase contribution e2πiη, where η is an odd integer for fermionsand an even integer for bosons.

Now recall the Aharonov–Bohm effect from Chapter 29. Moving onecharged particle completely around a source of flux gives rise to a phasecontribution eiqΦ, where q is the particle charge and Φ is the flux. Usingour Chern–Simons result that q = −κΦ, we obtain a prediction of the

Aharonov–Bohm phase of e−i q2

κ . However, particle exchange (whichtells us about statistics) is slightly different to this in that it involvesmoving one particle around the other through an angle π. Interpretingthe above Aharonov–Bohm process as two lots of particle exchange (andtherefore contributing a phase 2πη) we obtain a result for the statisticsof our particles,3 namely that

3In fact, there’s another tricky factor oftwo here. One is tempted to say thatsince all particles carry both charge andflux then moving one charge around aflux also moves a flux around a charge,doubling the effect compared to whatwe have written. A more careful treat-ment confirms our naive picture of asingle charge moving around a singleflux and gives eqn 30.14. See Wen,Chapter 7 for further discussion.2πη = −q

2

κ, (30.14)


and hence η = − q2

2πκ is not necessarily an integer. This implies that theparticles exhibit fractional statistics.

This chapter has explained how to construct a topological Lagrangian.It is based on gauge fields; it supports excitations with fractional statis-tics; its charged excitations carry flux. In Chapter 45 we’ll examine aphase of matter which is described by a Chern–Simons theory: this isthe fractional quantum Hall fluid.

Chapter summary

• Topological field theories such as Chern–Simons theory are builtfrom products of gauge theories summed together using the anti-symmetric symbol εµνλ in place of a metric tensor. These theoriesprovide constraints on other fields to which they are coupled.

• Chern–Simons theory attaches flux to the charge of the source field.

• Chern–Simons theories may produce fractional excitations.

Exercises

(30.1) (i) Verify that εµνλ∂µ∂νχ = 0.(ii) Verify eqn 30.4.

(30.2) Show that the Chern–Simons term won’t work in(3+1)-dimensional spacetime. Suggest a form for(4+1)-dimensional spacetime.

(30.3) Derive the equations of motion for the Chern–Simons fields coupled to a source in eqn 30.9 usingthe Euler–Lagrange equations.

(30.4) The propagator for Chern–Simons theory may befound in a manner very similar to that used inelectromagnetism in Chapter 39. Starting withChern–Simons theory we add a gauge fixing term− 1

2ξ(∂µa

µ)2 to obtain

L = −κ2εµνλaµ∂νaλ −

1

2ξ(∂µa

µ)2 + Jµaµ. (30.15)

(a) Show that the equations of motion for the fieldaµ are given by

Jµ =

»

κεµνλ∂ν −1

ξ∂µ∂λ

–

aλ = Mµλaλ. (30.16)

(b) Show that the inverse of the equations of motionis given, in momentum space, by the matrix

(M−1)λσ = − iελνσpν

κp2+ ξ

pλpσp4

, (30.17)

which provides the propagator for the theory viai(M−1)λσ.(c) Show that the Chern–Simons term thereforeleads to a term in the Lagrangian that may be writ-ten

L = −1

2Jµ„

εµνλ∂ν

κ∂2

«

Jλ. (30.18)

This is known as the Hopf term.

(30.5) Consider Chern–Simons electromagnetism de-scribed by a Lagrangian

L = −1

4fµνf

µν − κ

2εµνλaµ∂νaλ − 1

2ξ(∂µa

µ)2,

(30.19)where fµν = ∂µaν − ∂νaµ.(a) Find the equations of motion for this theory.(b) Show that the inverse of the equations of motionis given by

(M−1)λσ =p2gλσ − pλpσ + iκελνσp

ν

p2(p2 − κ2)+ ξ

pλpσp4

.

(30.20)What is the significance of the pole in the propaga-tor at p2 = κ2?

Part VIII

Renormalization: taming

the infinite

This part is structured as follows:

• Interactions in quantum field theories change the properties of theparticles, producing dressed particles, also known as quasiparticles.These are examined in Chapter 31 which also gives an outline ofLandau’s Fermi liquid picture.

• Quantum field theory has some uncomfortable divergences andChapter 32 shows how these can be tamed using renormalizationusing additional counterterms.

• These ideas are applied to deriving propagators and making Feyn-man diagrams in Chapter 33. This allows us to write down aGreen’s function called the self-energy which describes the changein the mass of a particle due to the interactions, and also a vertexfunction which describes screening.

• The renormalization group is introduced in Chapter 34. Examplestreated include asymptotic freedom, Anderson localization, andthe Kosterlitz–Thouless transition.

• Chapter 35 treats the ferromagnetic transition as a further exam-ple of the renormalization group procedure and shows how criticalexponents can be extracted.

31Renormalization,

quasiparticles and the

Fermi surface

31.1 Recap: interacting and non-interacting theories 274

31.2 Quasiparticles 276

31.3 The propagator for a dressedparticle 277

31.4 Elementary quasiparticles ina metal 279

31.5 The Landau Fermi liquid 280

Chapter summary 284

Exercises 284

This chapter focuses on how the behaviour of particles is affected byinteractions between them. We will see that the passage from a non-interacting theory to an interacting theory involves not only a propensityfor particles to scatter from each other but, more dramatically, changesin the properties of particles themselves, such as their mass and charge.We will say that the values of these properties are renormalized. Fur-thermore, we will find that interactions even change what we mean by a‘particle’. The process of renormalization may be imagined as a particledressing itself in interactions.

A rather familiar example of interactions changing the apparent prop-erties of particles is the screening of a positive charge embedded in ametal. The positive charge builds up a cloud of electron density aroundit. Distant electrons do not feel its full charge because the nearby elec-tron density screens the charge, reducing its apparent value. A secondexample is the effective mass of an electron in a crystal. The fact that acrystal contains immobile ion cores which interact electrostatically withthe electron means that the apparent mass of the electron is not that ofthe bare electron in a vacuum, but instead an effective mass m∗.

We will discover that analogous processes occur in all interacting quan-tum field theories. Of course the difference between condensed matterphysics and a ‘fundamental’ theory such as quantum electrodynamics isthat, for the former, we may measure the mass and charge of electronsboth inside the crystal and in vacuo. In contrast, we cannot remove elec-trons from the vacuum of quantum electrodynamics to see the differencein their properties caused by interactions. However, the framework inwhich we understand the fate of interacting particles is the same in bothcases and it is to this set of ideas which we now turn.

31.1 Recap: interacting and

non-interacting theories

Our simplest field theories describe particles that don’t interact. We callthese free or non-interacting theories.1

1The simple scalar field theory with La-

grangian L = 12(∂µφ)2 − m2

2φ2 is an

example of a free field theory.

31.1 Recap: interacting and non-interacting theories 275

Example 31.1

Here’s a recap of the properties of a non-interacting theory. For a theory describedby a free Hamiltonian H0, we define freely evolving field operators and a vacuumground state. These may be defined through

H0|0〉 = 0, p|0〉 = 0, 〈0|φ(x)|0〉 = 0. (31.1)

Particles are excitations of the vacuum. Creation and annihilation operators cre-ate single-particle modes, one at a time, labelled by their momenta. An example isa†p |0〉 = |p〉. The amplitude for a field operator to create a (relativistically normal-ized) single particle with momentum p is given by

〈p|φ†(x)|0〉 = eip·x. (31.2)

Finally, to talk about particles we need a propagator. This describes a single particle,inserted into the system at y and removed at x, and which, for a non-interactingtheory, will not affect or be affected by any other particles in the system. TheFeynman propagator describing the evolution of a free, single particle is given by

G0(p) =i

p2 −m2 + iǫ. (31.3)

If we didn’t know the mass of the particle before (from its dispersion relation E2p −

p2 = m2) then we could read it off the propagator as the pole of the Green’s function.

T

λ

0

1

H = H0

H = H0 + H ′

Fig. 31.1 Adiabatic turning on of H′.The full Hamiltonian is given by H =H0+λ(T )H′. This thought experimentallows us to examine what happens toparticles in an interacting theory.

So much for the world of non-interacting particles. What happens whenparticles interact? Consider a Hamiltonian written as a sum of non-interacting and interacting parts

H = H0 + λH ′. (31.4)

In order to examine the interacting system, we will imagine what hap-pens to a free theory described by H0 when we slowly turn on an inter-action H ′. To do this we multiply H ′ by a function λ(T ) which startsvanishingly small and slowly approaches unity as we turn up some pa-rameter T , as shown in Fig. 31.1. As λ grows, our system graduallybecomes a bubbling cauldron of interactions: the new eigenstates of thefull Hamiltonian H (called |pλ〉) are different to those of H0 (called |p〉).The interacting ground state |Ω〉 is different from the non-interactingground state |0〉 and the dispersion of the particles is also altered. Ifwe put a test particle into this system it will interact with particles andantiparticles, pulling them out of the vacuum. These particles and an-tiparticles may also interact with still more particles and antiparticles,doing all manner of complicated things. Our original particle may belost in all of this havoc and it might be doubted that we can even iden-tify single particles in this system any more. We may have to abandonour quantum field theory which, after all, is based around the notion ofcreating and annihilating single particles. We are led to ask: are theresuch things as single-particle excitations in an interacting system? Thepoint of this chapter is to show that there are! They are called dressed

particles or quasiparticles.

276 Renormalization, quasiparticles and the Fermi surface

Quasiparticles are the excitations in interacting systems. Theyresemble free particles, but have different masses and interact

differently with their environment. The process of free particlesturning into dressed particles is called renormalization.

31.2 Quasiparticles

We imagine that as the interaction is turned on the particle begins itstransformation from a single non-interacting particle into a quasiparticle.We say that the particle becomes dressed by interactions and becomesa quasiparticle. The process of renormalization is therefore embodied inthe equation

(Quasiparticle) = (real particle) + (interactions), (31.5)

where the word quasiparticle may be substituted by ‘dressed particle’ or‘renormalized particle’, depending on the context.

Let’s examine what happens to a free particle as the interaction isturned on. How do we know if it’s possible for an interacting quasi-particle to exist with the same momentum once we’ve turned on theinteraction? To answer this, we act on the interacting vacuum with theparticle creation operator φ†(x) ∝

∫pa†peip·x + ape−ip·x that we had for

the non-interacting system. As we turn on the interaction the groundstate is no longer |0〉, it’s now |Ω〉 (where H|Ω〉 = 0). The excited statesare called |pλ〉, where λ tells us that we have interactions. The key tounderstanding dressed particles is that the state a†p|Ω〉 created by ourfield operator won’t generally contain one single-particle excitation. Infact, our operator a†p is a bull in a china shop: there’s nothing to stopthe free field operator producing several excitations all at once!

To make a single-particle excitation with momentum p in the freetheory we acted on the free ground state with an operator a†p and said|p〉 = a†p|0〉. To make a single particle with momentum p in the inter-acting theory we might act on the interacting ground state |Ω〉 with adifferent operator |pλ〉 = q†p|Ω〉. This all makes sense, in that we needto have the right tools for the job, that is, the right operators for theground states, and we make the expected number of particles (i.e. one).Unfortunately we don’t know what q†p is and so can’t use it on the sys-tem. All we have is the single-particle creation operators a†p for thenon-interacting states.

But what happens if we use the wrong operator on the interactingground state? If we apply the operators for the non-interacting theorya†p to the interacting ground state |Ω〉 we make a superposition of states.One might be the state we intend to make |pλ〉, but others will bemultiparticle states whose momenta each add up to p. That is

a†p|Ω〉 = |pλ〉〈pλ|a†p|Ω〉 +∑

(multiparticle parts). (31.6)

31.3 The propagator for a dressed particle 277

For example, a†p|Ω〉, might contain a multiparticle state2 with two par- 2Note that, in a metal, these multipar-ticle parts may be thought of in termsof the emission of electron–hole pairs.See Chapter 43.

ticles and an antiparticle like this: q†p1q†p2

qp1+p2−p|Ω〉.In fact, it can be more complicated still, since we may only approxi-

mately create a state |pλ〉 (along with multiparticle parts) and in actualfact create an ultimately unstable, narrow wave packet of width Γp re-sembling this state. In particle physics this is known as a resonance.In this case the energy of the particle may have an imaginary part Γp.A complex energy Ep + iΓp will cause a particle to time evolve accord-ing to eiEpte−Γpt or, in words, the particle is unstable to decay and hasa lifetime3 (2Γp)−1. After a time that’s long compared to (2Γp)−1 we 3Where the factor of two comes from

considering the square modulus of theparticle wave function.

can’t expect there to be a quasiparticle any more.With all of these particles flying about and possibly decaying how can

we say whether we have a single particle? We call the amplitude for

making the desired single-particle part the quasiparticle weight Z12p ,

where Z12p = 〈pλ|a†p|Ω〉, and where |pλ〉 is a relativistically normalized

momentum state. We say that we have a quasiparticle if Z12p 6= 0. In the

interacting theory, we can state that the amplitude in position space forcreating a single particle with a momentum p is

〈pλ|φ†(x)|Ω〉 = Z12p eip·xe−Γpt. (31.7)

In fact, to meaningfully describe a particle as a quasiparticle we willrequire that Ep > Γp, that is, its energy must be greater than its decayrate. Comparing with the analogous result obtained for non-interactingscalar field theory we conclude that, when Γp is small, we can describeparticles in an interacting theory by multiplying4 our single-particle wave 4In the jargon, one says ‘renormalizing’

rather than ‘multiplying’.functions by the factor Z12p .

It’s important to note that the energies of interacting single particlesare different from those of free particles. In the (relativistic) scalarfield theory we’re describing5 the interactions change the mass of the 5In which the free particle dispersion is

given by Ep = (p2 +m2)12 .single-particle excitation from m (often called the bare mass) to mλ,

giving an energy for the state |pλ〉 of (p2 +m2λ)

12 . Since mλ is the mass

of a single particle measured in our experiments, it’s often called thephysical mass and given the symbol mP. The following table showsquantites for both non-interacting and interacting theories.

Non-interacting: |p〉 = a†p|0〉〈p|φ(x)|0〉 = eip·x m

Interacting: |pλ〉 = q†p|Ω〉〈pλ|φ(x)|Ω〉 = Z12p eip·xe−Γpt mP

31.3 The propagator for a dressed particle

For serious calculations we need to identify a propagator for the inter-acting system via our definition

G(x, y) = 〈Ω|T φH(x)φ†H(y)|Ω〉. (31.8)


It can be shown, without the aid of perturbation theory, what the gen-eral form for the interacting propagator G should be.6 The propagator,

6This can be achieved simply by slidinga resolution of the identity between thefield operators in eqn 31.8. This reso-lution will be made up from the exacteigenstates |pλ〉 of the full HamiltonianH. For details of the calculation, seePeskin and Schroeder, Chapter 7.

which is built out of free operators ap and a†p, will not only describe thepropagation of single-particle states, but also describe the fate of themultiparticle states, and takes the form

G(p) =iZp

p2 −m2P + iΓp

+

(multiparticle

parts

), (31.9)

where, as before, Z12p = 〈pλ|φ(0)|Ω〉 is the amplitude for creating a single-

particle state |pλ〉 out of the interacting vacuum and mP is the mass ofthe state |pλ〉. Notice the role of Γp here: it is the inverse quasiparticlelifetime in the place of the usual infinitesimal factor ǫ.In most cases involving long-lived ele-

mentary particles Zp is independent ofthe three-momentum p and is simplydenoted Z. As we shall see, a depen-dence on p is more relevant for metalsin solids.

Example 31.2

One way to write the full propagator that makes contact with experiment uses thespectral density function ρ(M2). For example, for scalar field theory we rewritethe propagator in terms of the free propagator7 ∆(x, y,M2) which is now made a7Recall that the free scalar field propa-

gator ∆(x, y,M2) can be written in mo-mentum space as

∆(p,M2) =1

p2 −M2 + iǫ.

function of particle mass as well as the spacetime points x and y:

G(x, y) =

Z ∞

0

dM2

(2π)ρ(M2)∆(x, y,M2). (31.10)

Suppose we have a spectral density function

ρ(M2) = (2π)δ(M2 −m2P)Z +

„multiparticle

parts

«

. (31.11)

This is shown in Fig. 31.2(a). It is made up of a single-particle peak at massM ≡ mP:the physical mass of a single particle in the interacting theory. It also contains amultiparticle continuum which starts around 4m2

P, which is the threshold for makingtwo real particles in the centre of mass frame. This sort of spectral function is theone usually seen in particle physics problems where the creation of more than oneparticle is due to pair production. Plugging in this spectral density we obtain

(a)

(b)

M2

ρ(M2)

m2p

M2

ρ(M2)

m2p ∼ 4m2

p

Fig. 31.2 (a) The spectral functionshowing a single-particle peak. (b) Thespectral function showing a quasiparti-cle peak with lifetime (2Γp)−1.

G(p) =iZ

p2 −m2P + iǫ

+

Z ∞

≈4m2p

dM2

2πρ(M2)

i

p2 −M2 + iǫ. (31.12)

Notice that the lifetime of the particle is ǫ−1, which is effectively infinite since theparticle pole is infinitesimally close to the real axis. We expect these particles to belong-lived and stable.

Now suppose we have the spectral density in Fig. 31.2(b), which has a quasiparticlepeak with width 2Γp . This leads to a propagator

G(p) =iZp

p2 −m2P + iΓp

+

„multiparticle

parts

«

, (31.13)

where the quasiparticle now has a finite lifetime (2Γp)−1 (and we allow the possibilitythat Z is a function of p). We see that our requirement that Ep > Γp for a meaningfulquasiparticle tells us that the width of the peak in the spectral weight should benarrow compared to the distance of the peak from the origin. A spectral weightwith a broad quasiparticle peak is often seen in condensed matter physics wherethere is usually a large number of very closely spaced energy levels. As a resultthe quasiparticle is a wave packet spread over (potentially very many) levels whicheventually decays away.

In addition to the quasiparticle having a finite lifetime it can also be shown thatthe multiparticle part usually undergoes some degree of destructive interference anddecays after a time that is fairly small compared to Γ−1

p , leaving the lone quasiparticleamongst the ruins!

31.4 Elementary quasiparticles in a metal 279

Once we obtain the full propagator G(p) we can immediately read offthe mass energy of the excited states mP from the position of the polein the function, the residue at that pole gives us iZp, the lifetime isdetermined from Γp and, of course, we can use G(p) to calculate otherquantities too, such as scattering amplitudes.

31.4 Elementary quasiparticles in a metal

The concept of dressed particles is especially important in condensedmatter physics. Although, in reality, the condensed matter physicistusually deals with a system made up from a large number of very stronglyinteracting, real particles (such as electrons in a metal), renormalizationinvolves changing our point of view and describing the system in termsof a small number of weakly interacting or non-interacting, fictitiousparticle-like excitations. Despite the fact that the ground state of asystem, such as a crystal for example, isn’t an empty box8 we regard

8Far from it, it usually comprises N ≈1023 particles of one sort or another.

it as a vacuum, in that it contains no excitations. A weakly excitedstate of the whole system (that is, one at low temperature) can then bedescribed in terms of a small number of these excitations. These areoften also called the elementary excitations of the system.9

9The approach of describing a systemin terms of small numbers of excitationsworks in condensed matter physics be-cause we are most often interested inthe behaviour of a solid at low temper-ature and this behaviour is dominatedby the occupation of only those energylevels of the system very close to theground state.

A distinction can be made between two types of elementary excitation.The first are called collective excitations and correspond to particle-like excitations that occur through motion of all of the constituent partsof the underlying system. An example is the chain of coupled masses weconsidered in Chapter 2. There we described the excited states of thesystem, which involved the motion of all of the masses, as particle-likeentities called phonons.10

10Notice that if we turn off the inter-actions in the system (in this case thesprings between the masses) then theseexcitations cease to exist. Another ex-ample of a collective excitation is theplasmon, which is a collective excita-tion of all of the electrons in a metal.We will examine these in Chapter 43.

(a) (b) (c)

Fig. 31.3 (a) The non-interacting elec-tron gas with a single electron abovethe Fermi level. (b) The particle exci-tations in the non-interacting electrongas. (c) The quasiparticles in the inter-acting electron gas [identical to (b) butwith renormalized masses].

The second sort of elementary excitation is the quasiparticle, a single-particle excitation from the non-interacting system dressed in interac-tions. An example is the single-particle excitation of an electron gas. Ina non-interacting electron gas [Fig. 31.3(a)] we have single-particle exci-tations above the ground state called electrons and holes11 [Fig. 31.3(b)].

11Remember that in the picture dis-cussed here, excitations only existabove the ground state. The groundstate here consists simply of electronstates filled up with electrons all theway to the Fermi level. In the non-interacting system an electron abovethe Fermi level is an excitation, whichbecomes a quasiparticle upon turningon the interactions. An electron be-low the Fermi sea is not an excita-tion and does not become a quasipar-ticle. A contrasting philosophy, whereall electrons become quasiparticles, isLandau’s Fermi liquid theory, which isexamined later in the chapter.

As we turn on the interactions these excited electrons and holes becomequasielectrons and quasiholes [Fig. 31.3(c)] .

Example 31.3

Let’s look at the case of electron quasiparticles in more detail. We start with thenon-interacting version of a metal: the Fermi gas at T = 0. This comprises a boxof N electrons, stacked up in momentum eigenstates |p〉 up to the Fermi level |pF〉.(For simplicity we regard the electrons as spinless and, in that case, the momentum

states may contain either 1 or 0 electrons.) The momentum distribution n(0)p for the

Fermi gas is shown in Fig. 31.4(a). This is the ground state of this non-interacting

system, which we call |0〉. The particle dispersion is given by E(0)p = p2/2me, where

me is the free electron mass. Note that near the Fermi level the dispersion may be

written E(0)p = vF(|p| − pF), where

vF = ∇pEp ||p|=pF =pF

me, (31.14)


is known as the Fermi velocity12 [see Fig. 31.5(b)].12The Fermi surface is spherical here sothe Fermi velocity only depends on themagnitude of pF and we write |pF| =pF.

We now turn on interactions between the electrons. The interacting ground stateof the system is |Ω〉, the eigenstates of the system are now called |pλ〉, and the energiesof these states are now Ep . Near the Fermi level we again have Ep = vF(|p|−pF), butnow vF = pF/m

∗, where we call m∗ the effective mass. Following from our assertionthat we may describe the weakly excited states of a system as quasiparticles, weexpect that those states that lie close to the Fermi level pF will have nonzero values

of Z12p and may therefore be described as containing quasiparticles. As will be shown

in Chapter 43 the free propagator for particles in a metal is given by

G0(p) =iθ(|p| − pF)

E − E(0)p + iǫ

+iθ(pF − |p|)E − E

(0)p − iǫ

. (31.15)

The interacting version becomes

G(p) =iZpθ(|p| − pF)

E − Ep + iΓp

+iZpθ(pF − |p|)E − Ep − iΓp

+

„Multiparticle

parts

«

. (31.16)

It will be useful in what follows to have the propagator in three-momentum space andthe time domain. This is achieved by doing a Fourier transform (see Appendix B)with the result

G(p, t) = Zp

h

θ(t)θ(|p| − pF)e−iEpte−Γpt − θ(−t)θ(pF − |p|)e−iEpteΓpti

+

„Multiparticle

parts

«

. (31.17)

|p|

np

pF

|p|

np

pF

Fig. 31.4 (a) The momentum distri-bution of the Fermi gas. (b) The mo-mentum distribution of the interactingFermi system. A discontinuity of sizeZpF exists at the Fermi surface.

|p|

Ep

pF

|p|

Ep

pF

Fig. 31.5 (a) The dispersion of thefree particles in a Fermi gas. (b) Nearthe Fermi momentum the dispersion isgiven by Ep = (|p|−pF)vF, where vF =pF/m

∗ and m∗ is the effective mass ofthe particles.

We may use the result of the last example to test whether we really havequasiparticles in a real-life metal by making a direct measurement of thequasiparticle weight Zp. To measure Zp we make a measurement of themomentum distribution np of the weakly excited system (that is, whenit contains some quasiparticles), given by np = 〈a†pap〉 which may berelated to the propagator via (see Exercise 31.1.)

np = −limt→0−G(p, t), (31.18)

giving us a prediction

np = Zpθ(pF − |p|) −(

Multiparticleparts

). (31.19)

Assuming that the multiparticle contribution varies smoothly, we con-clude that the interacting metal should have a discontinuity in its mo-mentum distribution np at its Fermi surface of size Z|p|=pF , as shownin Fig. 31.4(b). This prediction can be tested by performing Comptonscattering experiments on metals. The results are found to be in goodagreement with the theoretical predictions.

31.5 The Landau Fermi liquidPhilip Anderson is arguably the mostinfluential condensed matter physicistsince Landau. Not only is he the dis-coverer/creator of a variety of theoret-ical descriptions of electronic and mag-netic phenomena (among many otherthings); his elucidation of the structureunderlying the subject pervades currentcondensed matter physics. (Even thename ‘condensed matter physics’ is oneof his contributions!)

In general, I believe that the attitude is at least justifiablethat the instructive use of quasi-particles by the founders ofsolid state physics and by Landau’s school is hardly less rig-orous than the sophisticated many-body theory, and perhapsit is more foolproof, because it has less appearance of beingrigorous.P. W. Anderson (1923– ), Concepts in Solids


A rather different way of understanding metals in terms of fictitious par-ticles was formulated by Lev Landau13 and is known as Fermi liquid

13Lev Landau (1908–1968) was a So-viet physicist and one of the great-est scientists of the twentieth centurywho made contributions in many ar-eas, including phase transitions, mag-netism, superconductivity, superfluid-ity, plasma physics and neutrinos. Hewas also the model for the principalcharacter, Viktor Shtrum, in VassilyGrossman’s novel Life and Fate.

theory. It has been so useful in understanding metals that it is lit-tle exaggeration to claim that it has become the standard model of themetal.14 Landau’s theory is phenomenological, but appealingly intuitive

14The concept of a Fermi liquid is ahigh successful phenomenological the-ory of interacting electrons. It is a bitof a departure from our main themeof field theories of interacting systems,but it is too useful to omit and more ad-vanced treatments than the one givenhere demonstrate that the machinery ofdiagrammatic perturbation theory canindeed be employed to derive importantresults in Fermi liquid theory.

since it involves a description of a strongly interacting metal as beingalmost identical to the non-interacting Fermi gas! Moreover Fermi liq-uid theory allows the complete description of an interacting system interms of a small number of parameters, while avoiding all of the com-plexities of perturbation theory. Central to the model is the fact thatLandau describes an interacting metal in terms of quasiparticles; butthese are different to the field theory quasiparticles described thus farin this chapter. We will call the former Landau quasiparticles to avoidconfusion.

In Landau’s picture we pay particular attention to the process of ‘turn-ing on’ the interaction between the non-interacting electrons of the Fermigas. Landau assumed that if we very slowly turn on the interaction thesystem evolves continuously from Fermi gas to Fermi liquid, with eachsingle-particle momentum eigenstate of the gas evolving into a single-particle momentum eigenstate of the liquid. This adiabatic turning onmay be shown in ordinary quantum mechanics to lead to a very smallamplitude for transitions out of the level, providing the density of finalstates is small, as it is for electrons within the Fermi sea as a result ofthe Pauli principle.

Vital to Landau’s Fermi liquid concept is the notion that all of thethe states |p〉 occupied by an electron in the gas with n

(0)p = 1 be-

come single-particle eigenstates |p〉 in the liquid occupied by a Landauquasiparticle. The Landau quasiparticles in the interacting ground statetherefore also have momentum distribution n

(0)p . We say that there is a

one-to-one correspondence between the free particles and Landau quasi-particles. On turning on the interaction the Landau quasiparticles takeon an effective mass m∗ which parametrizes the change in energy of theeigenstates due to the effect of the field from the other quasiparticles.

| 〉| 〉 | 〉

| 〉

| 〉| 〉 | 〉

| 〉

Fig. 31.6 Turning on interactionschanges the states but the energy lev-els shouldn’t cross during the turningon process. Thus we can have (a) butnot (b).

Despite the change in energy, there should be no ambiguity in theidentity of an occupied eigenstate after turning on the interaction andtherefore the energy levels of the states shouldn’t cross during the turn-ing on process. That is to say, upon turning on the interaction we musthave the process shown in Fig. 31.6(a) rather than that in Fig. 31.6(b).We might ask whether this non-crossing of levels is a realistic propo-sition. Recall that when two eigenstates are related by a symmetrytransformation which leaves the Hamiltonian invariant we obtain a de-generacy in energy. If we turn on a perturbing potential V which leadsto a nonzero matrix element δ between the two states we have a Hamil-

tonian H =

(E δδ E

)which leads to a splitting in energy of 2δ. It’s

as if the energy levels repel each other by virtue of a matrix element δexisting between them, which prevents them from ever crossing. Theconsequence of this for the metal is that, as we turn on the interaction,


the levels repel each other when they get too close and never cross. Thisis shown in Fig. 31.7. However, this repulsion does not occur when thematrix element between two levels is zero. This is the case when thelevels have different symmetries and then there is nothing stopping thelevels from crossing.15 Since in the absence of a phase transition the slow15This is exactly the case at a phase

transition when, as we’ve seen, the sys-tem finds a new ground state witha lower symmetry than its previousground state. The requirement that thelevels don’t cross is therefore a require-ment that the evolution of the systemdoesn’t involve a change in thermody-namic phase.

turning-on of interactions provides a one-to-one correspondence betweenLandau quasiparticles and free particles, the theory is said to possessadiabatic continuity.

non

-int

erac

ting

inte

ract

ing

λturning on

Energy

Fig. 31.7 Evolution of single-particleenergy levels as interactions are turnedon. Notice how they repel when theydraw close to each other and nevercross.

Notice how different the Landau quasiparticles are to our field the-ory idea of quasiparticles. In field theory the ground state containsno quasiparticles; in Fermi liquid theory it contains as many Landauquasiparticles as we had electrons in the Fermi gas, guaranteeing chargeconservation, provided Landau quasiparticles carry the same charge aselectrons. Landau quasiparticles do not, therefore, rely on the renor-malization constant Zp to guarantee their existence as our field theoryquasiparticles do.

An insight comes when, instead of starting with a ground state Fermigas, we start with a ground state Fermi gas with an additional electronadded with |p′| > pF and, again, turn on the interaction. We would hopethat the excited electron evolves into a quasiparticle in a momentumeigenstate |p′〉. However, unlike the electrons within the Fermi sea, theextra electron potentially has rather a lot of phase space to explore as weturn on the interactions. Through interactions with the other electronsit may therefore have a considerable amplitude to scatter out of its state|p′〉 and so the excited quasiparticle will have a finite lifetime (2Γp′)−1

against scattering into other states and decaying.16 The phase space for

16Notice that this sets a limit on thespeed at which we may turn on theinteractions. Although we would liketo turn them on infinitely slowly, anyslower than Γ−1

p′ and our excited quasi-particle loses its momentum and, con-sequently, its identity. For a generalquantum liquid to exist it must be pos-sible to select a turning on time thatallows the ground state to evolve adia-batically, but also retains the low-lyingquasiparticles.

this scattering, restricted by the Pauli principle, is such that those statesnear the Fermi momentum have the smallest probabilities for scatteringand therefore survive for longest in quasiparticle momentum states. Thisis examined in the exercises with the result that the decay rate varies asΓp ∼ (|p| − pF)2. We conclude that really it is meaningful to speak ofthe properties of a Landau quasiparticle only near the Fermi surface ina metal where Γp is small.

Finally, if we lift a Landau quasiparticle out of a level below pF inthe ground state and promote it to a level with momentum ∆p abovepF, we simultaneously create an unoccupied quasiparticle state or hole17

17Holes will be examined in Chap-ter 43.

with momentum ∆p below pF. Since ∆p = |p| − pF, the quasielectronsand quasiholes will both have decay rates Γp ∝ (|p| − pF)2, so it is onlymeaningful to discuss their properties if their momenta are close to pF.

To summarize the philosophy of the Fermi liquid:

The Fermi liquid ground state contains Landau quasiparticles stackedup to the Fermi level. Landau quasiparticles have the same charge asordinary electrons but their mass takes on a renormalized value m∗.A quasiparticle excitation outside the Fermi sea (or, equivalently, aquasihole within the sea) will have a decay rate Γp ∝ (|p| − pF)2.

A cartoon of the Landau Fermi liquid theory is shown in Fig. 31.8.


Where the Fermi liquid and quantum field theories coincide is in thedescription of the weakly excited states of the metal. The excited Landauquasiparticles have exactly the same properties as our quantum fieldtheory quasiparticles and the argument about their lifetime applies toboth. Near the Fermi surface the energy of an excited quasiparticle maybe written Ep ≈ pF

m∗ (|p| − pF) and its decay rate Γp ∝ (|p| − pF)2 Wesee that for small (|p| − pF) the real part of the quasiparticle energywill be larger than the imaginary part, or Ep > Γp, and a meaningfulquasiparticle exists in the metal.

(a) (b)

Fig. 31.8 A cartoon of the LandauFermi liquid. (a) The non-interactingFermi gas with a single electron pro-moted into an excited state (indicatedas a dotted line). (b) In the LandauFermi liquid the interactions are turnedon so that the electrons become quasi-particles and the holes become quasi-holes. There is a one-to-one corre-spondence between the quasiparticlesand quasiholes in the interacting liquidand the electrons and holes in the non-interacting picture.

We now complete our description of the Landau Fermi liquid witha discussion of the total energy of a weakly excited state of a metal.In the spirit of many of Landau’s arguments, in which any ignoranceof microscopic details is no barrier to formulating a meaningful powerseries expansion, we write the energy of the Fermi liquid in the limit ofa low density of quasiparticles as

E = Eg +∑

p

(E(0)p − µ) δnp +

1

2

∑

pp′

fpp′ δnp δnp′ + · · · , (31.20)

where δnp = np−n(0)p is the difference between a distribution with exci-

tations and the ground state distribution18 and Eg =∑

pE(0)p n

(0)p is the

18This implies that it is neither np nor

n(0)p that is the crucial quantity, but

rather it is their difference δnp , whichtells us the number of excitations in theexcited state. This is fortunate since weknow neither np nor n

(0)p with much ac-

curacy, but we can find δnp .

energy of the ground state. The term linear in δnp describes the excita-

tion of isolated quasiparticles of energy E(0)p −µ = ∂E/∂np. The second-

order coefficient19 fpp′ = ∂2E/∂np∂np′ describes the contribution from

19The partial differentials in the ex-

pressions for E(0)p −µ and fpp′ are eval-

uated about the ground state configu-rations, i.e. a frozen Fermi sea.

quasiparticle–quasiparticle scattering (i.e. from interactions), and canbe described via an interaction Hamiltonian HI = 1

2

∑pp′ fpp′ δnp δnp′ .

Example 31.4

To understand the scattering process in more detail, one should include spin andthis amounts to replacing fpp′ by fpσ;p′σ′ . One can show20 that for spin-conserving

20Further details and a fuller accountmay be found in the book by P. Cole-man.

interactions this quantity can be written as

fpσ;p′σ′ = f s(cos θ) + fa(cos θ)σ · σ′, (31.21)

where cos θ = p·p′

|p||p′| . The functions f s and fa can be expanded in terms of Legendre

polynomials so that

fs,a(cos θ) =1

g(EF)

∞X

ℓ=0

Pℓ(cos θ)Fs,aℓ , (31.22)

where F sℓ and F a

ℓ are Landau parameters and g(EF) is the density of states atthe Fermi level.21 These expressions allow some of the key properties of the Landau 21See Chapter 43.Fermi liquid to be deduced. For example, one can show that the effective mass ism∗ = m(1+F s

1) and the spin susceptibility is χ = µ0µ2Bg(EF)/(1+F a

0 ). Many of thepredictions of Landau’s Fermi liquid theory may be shown (after much hard graft!)to coincide with those of quantum field theory and, owing to its comparative ease ofuse, it is still regarded as the best way of understanding the results of experiment onnumerous condensed matter systems.

We have seen in this chapter that renormalization is an essential stepin understanding what happens to particles in an interacting quantum


field theory. In fact, as shown in Fig. 31.9, we could summarize the stepsto make a working theory as: (i) write a Lagrangian; (ii) quantize thefree part; (iii) derive Feynman rules for interactions; (iv) renormalize.However, renormalization is often presented as a method of eliminatingnonsensical infinities from theories. The elimination of these infinities isthe subject of the next chapter. It should be remembered throughoutthat renormalization is necessary for all interacting theories, indepen-dent of whether or not we have infinities.

Lagrangian

CanonicalQuantization

Dyson’sexpansionfor S

Path integral

gives Z[J ]

Feynmanrules

Renormalize

Green’sfunctions andpredictions

add a source

current Jfree

part

inte

racting

part

infer

infer

quanti

zati

on

pro

pagato

rs

Fig. 31.9 The process of doing quan-tum field theory includes the necessarystep of renormalizing.

Chapter summary

• Turning on the interactions in a theory changes the properties ofthe particles. We say that they dress themselves in interactions.

• The interacting propagator is

G(p) =iZp

p2 −m2P + iǫ

+ (multiparticle parts),

where Zp is the quasiparticle weight

• In Landau’s Fermi liquid theory there is a one-to-one correspon-dence between electrons in the non-interacting theory and quasi-particles in the interacting theory. The effect of interactions is torenormalize physical quantities, such as the effective mass, via theLandau parameters F s

ℓ and F aℓ .

Exercises

(31.1) Verify that, for fermions,

np = − limt→0−

G(p, t). (31.23)

(31.2) Consider an electron with energy E1 ≥ EF scatter-ing with an electron with energy E2 ≤ EF at T = 0.In order for this to occur we must have final elec-tron state E3 ≥ EF and E4 ≥ EF.(a) Show that this implies that the lifetime of anelectron with E1 = EF is infinite.(b) If E1 is a little different to EF, why does thescattering rate go as (E1 − EF)2?(c) For T 6= 0 argue that we expect a scattering

rate 1τ

= a(E1 −EF)2 + b(kBT )2, where a and b areconstants.

(31.3) (a) Using the relation

1

x0 + iǫ=

Px0

− iπδ(x0), (31.24)

(discussed in Appendix B) identify the delta func-tion part of the propagator

iZp

E−Ep+iǫ.

(b) For the propagatoriZp

E−Ep+iΓp, show that the

full width at half maximum of the peak is given by2Γp .

32Renormalization: the

problem and its solution

32.1 The problem is divergences285

32.2 The solution is counterterms287

32.3 How to tame an integral 288

32.4 What counterterms mean290

32.5 Making renormalizationeven simpler 292

32.6 Which theories are renor-malizable? 293

Chapter summary 294

Exercises 294

It turns out that the obvious generalization of this idioticallysimple manipulation gets rid of all of the infinities for anyfield theory with polynomial interactions, to any order inperturbation theory.Sidney Coleman (1937–2007)

In the last chapter, we saw how interactions dress a particle and changeits propagator. The dressed particles were quite different to the bare par-ticles that are revealed after canonical quantization of the free theory.This might mean that our perturbation theory, which involved expan-sions in the bare mass m and coupling constant λ, is at risk. It is: itturns out that we’ve been doing our perturbation theory about the wrongmasses and coupling constants! The solution to this problem involvesmaking a shift so that we’re expanding in physical masses and couplingconstants. The procedure also has a bonus. It solves a big problem inour perturbation theory: the problem of divergent amplitudes.

In this chapter we approach the problem by starting with the diver-gences in our perturbation theory. We will look at why we obtain infini-ties in our calculations and suggest an idiotically simple mathematicalworkaround. The physical content of this workaround will turn out toshift the parameters of the theory from the wrong ones (m and λ) tothe right ones (mP and λP).

32.1 The problem is divergences

We’ve looked at calculating amplitudes and propagators for several in-teracting theories. Perhaps you’re curious (or suspicious) as to why wehaven’t done the integrals at the end of the calculations. The reason isthat many of them diverge. Understanding the source of the divergencesrelies purely on the following facts (with a finite and positive):

∫ ∞

a

dxxn =

[xn+1

n+ 1

]∞

a

diverges for n ≥ 0, (32.1)

∫ ∞

a

dx

x= [lnx]

∞a diverges, (32.2)

∫ ∞

a

dx

xm=

[x−m+1

−m+ 1

]∞

a

=a−m+1

m− 1for m > 1. (32.3)

286 Renormalization: the problem and its solution

In the first integral the divergence arises because there are more powersof x in the numerator than the denominator. The second integral, wherethe number of powers of x is the same on top and bottom, is called log-

arithmically divergent, for rather obvious reasons. The last exampleis convergent.

Let us return to φ4 theory, which is described by a Lagrangian

L =1

2(∂µφ)2 − m2

2φ2 − λ

4!φ4. (32.4)

We ended our analysis of this theory in Chapter 19 with a perturbationexpansion encoded in Feynman diagrams. Consider the amplitude fortwo-particle scattering, which is equal to the sum of all connected, am-putated Feynman diagrams with four external legs. Up to second orderin the expansion (i.e. drawing all diagrams with one or two interactionvertices) we obtain the diagrams shown in Fig. 32.1. The main messageof this section is that the diagrams with loops in them [(b),(c) and (d)]are divergent. This is to say, with the upper limit of the integral setto infinity we get infinite and therefore nonsensical answers. This is adisaster.

(a) (b)

(c) (d)

p3 p4

p1 p2

−iλ

p3 p4

p1 p2

q

p3 p4

p1 p2

qp3 p4

p1p2

q

p1 + p2 − q

p1 − p3 − q

p1 − p4 − q

s

t u

Fig. 32.1 The expansion of four-point diagrams for two-particle scatter-ing up to second order in the interactionstrength. Diagrams (b)–(d) define theprocesses labelled s, t and u, respec-tively.

Example 32.1

Let’s see how this unfolds. We’ve encountered the first diagram before. The result is

iMa = −iλ. (32.5)

For the subsequent diagrams, the integral we’ll need is

Z Λ

0

d4q

(2π)4i

q2 −m2 + iǫ

i

(p− q)2 −m2 + iǫ= −4ia ln

„Λ

p

«

, (32.6)

where a is some numerical constant whose exact value isn’t important to us. Noticethat the limits of this integral are zero and Λ. The parameter Λ is a large momentumcut-off, and we want to send Λ → ∞. By counting powers we see that there are four

powers of momentum on top and four below, telling us that the integral ∼R d4q

q4is

logarithmically divergent. In fact, for the diagrams in Figs. 32.1(b)–(d) we obtainthe amplitudes:

iMb = iaλ2˘ln Λ2 − ln

ˆ(p1 + p2)2

˜¯= iaλ2

˘ln Λ2 − ln s

¯, (32.7)

iMc = iaλ2˘ln Λ2 − ln

ˆ(p1 − p3)2

˜¯= iaλ2

˘ln Λ2 − ln t

¯, (32.8)

iMd = iaλ2˘ln Λ2 − ln

ˆ(p1 − p4)2

˜¯= iaλ2

˘ln Λ2 − lnu

¯. (32.9)

All of these tend to ∞ as Λ → ∞. The total amplitude for two-particle scattering isgiven by the sum of these diagrams, giving

iM = = i (Ma + Mb + Mc + Md)

= −iλ+ iaλ2˘3 ln Λ2 − ln s− ln t− lnu

¯, (32.10)

and so we need to find a way to tame the iaλ2˘3 ln Λ2

¯∝ ln Λ term.

32.2 The solution is counterterms 287

We can avoid the divergence in these integrals if we make the pragmaticchoice to only integrate up to a large, but finite, value of the cut-off Λ.This amounts to deliberately ignoring any fine-scale details in the fieldsbelow a length scale ≈ 1/Λ. This is frequently what is done in condensedmatter physics, where we usually ignore all detail smaller than the size ofan atom. If we apply our theory to fundamental particles it’s not quiteso obvious why there should be a smallest scale but it might representsome graininess in spacetime.1 1Actually, the idea of a length scale like

this emerges very naturally from a wayof looking at the Universe called therenormalization group, which we dis-cuss in Chapter 34.

For now, let’s see if we can live (uncomfortably) with Λ 6= ∞. In-troducing finite Λ has an immediate consequence: the amplitudes wecalculate using Feynman diagrams will depend on Λ. This is a seriousproblem: if a prediction for something we want to measure depends onΛ then we have an arbitrary constant which will need to be given somevalue. One strategy to avoid this is to add terms to the Lagrangian that,when we do the perturbation expansion up to some order (second ordersay), will remove the dependence of amplitudes on Λ. Doing this willclean up the theory up to second order. Although we should still expectthe higher order terms to diverge (i.e. blow up as Λ → ∞), we’ll at leasthave healed the theory to the extent that we can obtain a prediction ofphysical behaviour that is valid up to second order.

32.2 The solution is counterterms

The terms that are added to the Lagrangian to remove the dependenceon Λ, and hence the divergences, are called counterterms. We sawthat for φ4 theory at second order the divergent part of the amplitudelooked like 6iaλ2 ln Λ; we therefore change the Lagrangian by adding acounterterm −(6aλ2 ln Λ)φ4/4! thus:

L → L−(

6aλ2 ln Λ

4!

)φ4. (32.11)

We now have a Lagrangian

L =1

2(∂µφ)2 − m2

2φ2 − λ

4!φ4 +

C(2)

4!φ4, (32.12)

where the counterterm C(2) = −6aλ2 ln Λ and we write the superscript(2) to remind ourselves that we’re only getting rid of divergences atsecond order.

Now we start again with our new Lagrangian. We canonically quan-tize, use the Dyson expansion and figure out the Feynman rules and thensum the new diagrams. With the counterterm, the Dyson expansion ofthe S-operator is now just

S = T e−iR

d4x 14! (λφ

4−C(2)φ4). (32.13)

Every order of the expansion therefore also includes a contribution fromthe counterterm, whose Feynman diagram is shown in Fig. 32.2. Since


the counterterm is proportional to φ4 it has the same behaviour as theλ4!φ

4 term. It therefore represents a vertex which, instead of carrying afactor −iλ carries a factor iC(2).

Fig. 32.2 The C-counterterm.

We calculate amplitudes to second order in λ and (we only need to)include one diagram involving a counterterm and we obtain

iM(2) = −iλ+ iaλ2[3 ln Λ2 − ln s− ln t− lnu

]+ iC(2), (32.14)

and substituting C(2) = −6aλ2 ln Λ yields

iM(2) = −iλ− iaλ2 (ln s+ ln t+ lnu) . (32.15)

This amplitude depends on the momenta of the incoming and outgoingparticles (which is no problem) but crucially it doesn’t depend on Λ.We’ve done what we set out to do! Predictions for the theory will nownot depend on an arbitrary cut-off Λ, and hence won’t yield infinitywhen we send Λ → ∞.

32.3 How to tame an integral

We now have a plan, which is to start calculating using the Lagrangianand then add a counterterm when we see a divergence. We then workout the coefficient C(n) in front of the counterterm and all is set. Quiteapart from the as yet unexamined physical consequences of this extraterm, we might worry that this process will turn out to be ultimatelyfutile. What if you calculate to second order as above, but at third ordernot only do we need a counterterm C(3)φ4 but also a new countertermD(3)φ6 to swallow up some extra divergent term that emerges? What ifit gets even worse and, at 27th order in the expansion, we need to add513 new types of counterterm to cancel all of the divergences and then,at 28th order, we find we need 752 new counterterms? A theory mightnever stop absorbing new types of counterterm!

Amazingly, it often turns out that we only need to add a small numberof types of counterterm (three, for example, in QED) and that thesecancel divergences to all orders of perturbation theory! We still need toknow the right constants (like C(n)) for each order of perturbation theoryin order to cancel the divergences, but these will turn out to be numbersthat we can calculate. Theories that have the property that only a finitenumber of counterterms are needed to cancel all divergences are said tobe renormalizable theories. This is the miracle of renormalization.

p p

k

q

p− k − q

Fig. 32.3 The Saturn diagram.

Example 32.2

To get a feel for this we’ll look at the Saturn diagram, shown in Fig. 32.3. Writtenout in full the amplitude is

iM =(−iλ)2

6

Z Λ

0

d4q

(2π)4d4k

(2π)4i

q2 −m2 + iǫ

i

k2 −m2 + iǫ

i

(p− q − k)2 −m2 + iǫ.

(32.16)

32.3 How to tame an integral 289

These integrals are getting worse! If we’re going to systematically renormalize anytheory we’ll need a system. The system involves writing the integral as a polynomial.Usefully there’s a key fact that will help us: Feynman diagrams can be expanded as aTaylor series in the external momentum. The polynomial representing the diagramwill have divergent coefficients, reflecting the divergence of the integral. We’ll pickour counterterms to cancel the divergences in the coefficients. Happily, this resultsin counterterms of the right type to cancel the divergences in a particular integral.

The integral in eqn 32.16 may be written in a series about p = 0 as

I = α+ βp2 + γp4 + . . . , (32.17)

where we don’t have any odd terms because of the symmetry φ ≡ −φ in the originalLagrangian.

Here’s how to find the counterterms:

• The Saturn diagram integral is quadratically divergent. It has eight powers ofmomentum on top and six on the bottom. If we set the external momentump equal to zero in our expansion then I = α and we conclude that α di-verges quadratically. This tells us that we need one counterterm that divergesquadratically.

• If we differentiate the Saturn integral twice with respect to p then it becomeslogarithmically divergent, since we have reduced the number of momenta ontop by two. If we differentiate the series expansion I twice and set p = 0 weget I′′ = 2β and we conclude that β diverges logarithmically. We thereforeneed another counterterm which is logarithmically dependent on the cut-offwhich will have a Feynman rule that it appears multiplied by p2.

• Differentiating the integral two more times makes it convergent so we don’tneed any more counterterms.

We need two counterterms to cancel the divergences in the Saturn integral I: onewith coefficient A quadratically dependent on Λ and one with coefficient B which islogarithmically dependent. How do we make sure that the B counterterm will givea Feynman amplitude that is multiplied by p2 and that A will not be multiplied byanything? The answer is to introduce counterterms into the Lagrangian that looklike

Lct =A

2φ2 +

B

2(∂µφ)2, (32.18)

remembering that in momentum space [∂µφ(x)]2 → φ(−p)p2φ(p) we haveB(∂µφ)2 → p2B and A→ A. Adding the counterterms gives the full Lagrangian forrenormalized φ4 theory:

L =1

2(∂µφ)2 − m2

2φ2 − λ

4!φ4 +

A

2φ2 +

B

2(∂µφ)2 +

C

4!φ4. (32.19)

Notice that all of the counterterms have the same form as the terms in the originalequation; they just have different coefficients.

These counterterms all have Feynman diagrams and rules attached to them. Wesaw before that the C(n)-counterterm diagram corresponds to a rule iC(n). The B(n)

counterterm makes a contribution i(p2B(n)), where p is the momentum along the lineand the A counterterm contributes iA(n). The A and B terms are usually representedby the single diagram shown in Fig. 32.4, which has a rule i(B(n)p2 +A(n)).

Fig. 32.4 The A and B countertermdiagram

To summarize, we have the Feynman rules for renormalized φ4 theorygiven in the box. These eliminate all infinities.


Feynman rules for renormalized φ4 theory

• A factor ip2−m2+iǫ for each propagator.

• A factor −iλ for each interaction.

• Add sufficient counterterm diagrams to cancel all infinities.

• A factor i(B(n)p2 +A(n)

)for each counterterm propagator,

where n is the order of the diagram.

• A factor iC(n) for each interaction counterterm, where n is theorder of the diagram.

• All other rules regarding integrating, symmetry factors and over-all energy momentum conserving delta functions are identical asfor the case of unrenormalized perturbation theory.

32.4 What counterterms mean

We now turn to the question of the physics behind the counterterms.We’ve added terms to the Lagrangian which surely alters the physicswe’re studying. What we are going to find is that the counterterms shiftthe parameters from fictional ones to the real-life ones, simultaneouslyremoving infinities and forcing our theory to describe real life. To recap:we started with an interacting theory,

L =1

2(∂µφ)2 − m2

2φ2 − λ

4!φ4, (32.20)

with the mass and coupling constant m and λ respectively. We’ll calleqn 32.20 the unrenormalized Lagrangian. The excitations made by thefield operators when the coupling λ = 0 have mass m. However, whenwe turn on the coupling this Lagrangian gives rise to revolting infinitiesunless we cut off the integrals at some momentum Λ.

To fix the divergence encountered as Λ → ∞ we find we need to renor-malize the theory. As described in the previous chapter, renormalizationinvolves accepting that we’re dealing with dressed particles, which havea mass mP and coupling λP. These particles only involve a fractionZ

12 of the fields. This realization doesn’t remove the infinities, however.

To do that we need to include counterterms in our Lagrangian. Therenormalized Lagrangian is

L′ =1

2(∂µφr)

2 − m2P

2φ2

r −λP

4!φ4

r

+B

2(∂µφr)

2 +A

2φ2

r +C

4!φ4

r , (32.21)

where we have rescaled the fields using2 φ =√Zφr(x) and called the2Note that φr(x) are known as renor-

malized fields. We assume Z is inde-pendent of three-momentum here.

mass mP and coupling constant λP respectively. This theory won’t giveus nonsensical, infinite results for Λ → ∞.

32.4 What counterterms mean 291

Example 32.3

To see the consequence of renormalization, we now collect the coefficients of therenormalized Lagrangian

L =1 +B

2(∂µφr)

2 − (m2P −A)

2φ2

r − (λP − C)

4!φ4

r , (32.22)

which suggests that the counterterms represent shifts in the parameters in the La-grangian. Writing −A = δm2, −C = δλ and B = δZ we have

L =1 + δZ

2Z(∂µφ)2 − (m2

P + δm2)

2Zφ2 − (λP + δλ)

4!Z2φ4, (32.23)

where we’ve restored the original, unrenormalized fields. From this, we read off thatwe can relate our original and renormalized Lagrangians through

φ =√Zφr, Z = 1 + δZ, m2 =

(m2P+δm2)

Z, λ =

(λP+δλ)

Z2 . (32.24)

This tells us that renormalization is simply an exercise in shifting parameters.To truly see what’s going on it may make more sense now to replay the argument

backwards. We start with the unrenormalized Lagrangian and shift from parametersm and λ to parameters mP and λP. We start by renormalizing the fields to obtain

L =1

2(∂µφ)2 − m2

2φ2 − λ

4!φ4

=Z

2(∂µφr)

2 − Zm2

2φ2

r − Z2λ

4!φ4

r . (32.25)

We then use the shifts

Z = 1 + δZ, Zm2 = m2P + δm2, Z2λ = λP + δλ, (32.26)

giving us

L =1

2(∂µφr)

2 − m2P

2φ2

r − λP

4!φ4

r +δZ

2(∂µφr)

2 − δ2m2φ2

r − δλ

4!φ4

r , (32.27)

allowing us to identify the counterterms A, B and C as the shifts in the parameters−δm2, δZ and −δλ.

We started the whole venture of quantum field theory by writing downa simple Lagrangian with parameters we thought would tell us aboutthe masses and couplings of real particles in Nature. It turns out wewere wrong. We were doing the wrong perturbation theory expanding aseries in terms of the wrong constant mass m and wrong coupling λ. Wetherefore had asked a nonsensical question and got some infinite (andtherefore nonsensical) answers. Actually we should have been expandingour series in terms of a mass mP and coupling λP. The price we payfor making this shift from wrong to right variables is counterterms. In asense, we’re not really ‘adding counterterms’ at all; we’re really taking abare Lagrangian and making a shift of variables from wrong ones to theright ones and the shifts required are the counterterms. Renormalizationis not, therefore, an exercise in hiding infinities, it’s an exercise in makinga theory describe real life. For φ4 theory, the counterterms are divergent,meaning that these shifts may be infinite! This implies that m and λ,the bare mass and coupling are infinite quantities which are shifted byinfinite amounts upon dressing themselves by interactions.3 3This explains why it’s often claimed

that the bare charge of the electron isinfinite. The spontaneous appearanceof electron-positron pairs screens thecharge leading to the finite charge weencounter in Nature.

There’s one last, but very important, point to make. What numberdo we take for mP and λP? What are the right masses and couplingwe should expand around? The answer is that they’re the ones Na-ture’s given us! For this reason, mP and λP are known as the ‘physical’parameters, in contrast to the ‘bare’ parameters m and λ.


32.5 Making renormalization even simpler

We’d like a method to avoid having to analyse each diagram separatelyas we did for the Saturn diagram above. Is there a method which can,once and for all, tell us how divergent a diagram or class of diagrams is?There is and it’s based on simple dimensional analysis.

Define the superficial degree of divergence4 of an integral arising from4The clumsy name, ‘superficial’ degreeof divergence, arises because the actualdegree of divergence can be complicatedin gauge theories and some pathologicalcases, see Peskin and Schroeder, page316.

a Feynman diagram in φ4 theory to be

D =

(Powers of momentum

in numerator

)−(

Powers of momentumin denominator

).

(32.28)If D > 0 the integrals diverge, if D = 0 they logarithmically diverge andif D < 0 they don’t diverge. For a diagram with BE external lines, thedegree of divergence in φ4 is given by D = 4 −BE.

Example 32.4

We can prove the above theorem for predicting whether diagrams diverge. In additionto the previously defined symbols

• V is the number of vertices,

• L is the number of loops,

• BI is the number of internal lines.

Each loop brings with it an integral with four powers of momentum. Each internalline brings a propagator which brings −2 powers of momentum. Therefore

D = 4L− 2BI. (32.29)

To get the number of loops: L is the number of momenta we integrate over andeach internal line gives a momentum. The number of loops is less than BI though.Momentum-conserving delta functions eat up integrals and there are V of them (onefor each vertex, since we must conserve momentum at every vertex). However, oneof the delta functions conserves momentum for the entire diagram, so doesn’t eat anintegral. The number of loops L is therefore equal to the number of internal linesBI, minus (V − 1) the number of vertices, with one removed for overall momentumconservation:

L = BI − (V − 1). (32.30)

Each vertex has four lines coming out of it. Each external line comes out of onevertex and each internal line connects two vertices. Therefore

4V = BE + 2BI. (32.31)

Inserting eqn 32.30 and eqn 32.31 into eqn 32.29 we get D = 4−BE and the theoremis proved.

The great significance of this theorem is that it shows that the threecounterterms we have identified are the only ones that will ever appear.The fact that D = 4−BE means that the only diagrams that diverge (i.e.with D ≥ 0) have ≤ 4 external legs. There are no allowable diagramswith one or three legs and we’ve considered the divergences for the twoand four leg diagrams. We’ll never need any more.

32.6 Which theories are renormalizable? 293

32.6 Which theories are renormalizable?

Renormalizable theories are those in which a finite number of countert-erms cancel all divergences. Unfortunately, there are theories for whichthis property does not hold true. An example is Fermi’s theory of weakinteractions between fermions, whose Lagrangian is given by

L = ψ(ˆp−m)ψ +G(ψψ)2, (32.32)

where ψ is the fermion field.5 The superficial degree of divergence of this 5The meaning of the cross through themomentum operator p and the bar overψ will be explained in Chapter 36.

theory is given by

D = 4 − 3

2FE + 2V, (32.33)

where FE is the number of external Fermi lines in the diagram. This hasthe unfortunate feature that D depends on V , the number of interactionvertices. If we consider fermion–fermion scattering, for which FE =4 then for V > 1 we have divergent diagrams, which become moredivergent as V gets larger. We would need new counterterms at everyorder of perturbation theory.

In fact, whether a theory is renormalizable or not may be read offsimply from the dimensions of the interaction coupling constant. Therules are

• A super-renormalizable theory has only a finite number of superfi-cially divergent diagrams. Its coupling constant has positive massdimension.

• A renormalizable theory has a finite number of superficially diver-gent diagrams, however, divergences occur at all orders of pertur-bation theory. Its coupling constant is dimensionless.

• A non-renormalizable theory has diagrams that are all divergentat a sufficiently high order of perturbation theory. These theorieshave a negative mass dimension.

In our units the action S =∫

d4xL must be dimensionless since eiS

appears in the path integral. We therefore need L to have dimensions[Length]−4. Also note that, using these units, mass and energy havedimensions [Length]−1. Our φ4 theory has a Lagrangian L = 1

2 (∂µφ)2 −m2

2 φ2− λ

4!φ4 from which we conclude that [φ] = [Mass] = [Length]−1 and

λ is dimensionless. The φ4 theory is therefore renormalizable. On theother hand, looking at the mass term in Fermi’s theory we can concludethat [ψ] = [Mass]

32 and that, therefore, [G] = [Mass]−2. The theory is

non-renormalizable.The reason for the rule may be seen if we ask about the momentum

dependence of scattering in Fermi’s theory for small external momenta.The lowest order contribution will vary as M1 ∼ G. The next orderis G2 and so that the units agree with M1 we require M2 ∼ G2p2

or ∼ G2Λ2, when we integrate. This is divergent when we send Λ →∞. Things only get worse as we go to higher orders. This increase inpowers of momentum will occur for any theory with a coupling which has


negative mass dimension. The disaster clearly doesn’t occur for theorieswith dimensionless couplings, where no extra momentum dependence isrequired. For super-renormalizable theories, we need momentum factorsraised to negative powers, improving convergence at large momenta.

Finally we note that this argument also allows us to see that non-renormalizable interactions in Fermi’s theory only cause trouble whenΛ is larger than ≈ G−1. That is, if we’re interested in physics at muchlower energies than G−1 then we never explore momenta between G−1

and infinity and so the presence of the non-renormalizable term in theLagrangian wouldn’t cause us trouble. It may be that all of the theoriesof Nature that, at one time, we believed to be renormalizable (includingquantum electrodynamics and the electroweak theory) actually containnon-renormalizable terms in their true Lagrangians. It’s just that all ofour experience is limited to energies where we don’t notice them sincetheir coupling constants Gi are so small that our experiments cannotreach the energies of G−1

i . In this way of looking at the world, our theo-ries of Nature are low-energy, effective theories, which will eventuallybreak down at high enough energies. The true theory of Nature maynot, therefore, be a quantum field theory at all...

Chapter summary

• Quantum field theory contains divergences and the solution to theproblem is renormalization. Counterterms can be added to theLagrangian to remove the divergences and these counterterms havethe effect of shifting parameters in the original Lagrangian.

Exercises

(32.1) Consider a theory described by the LagrangianL = 1

2(∂µφ)2 − (m2/2)φ2 − (g/3!)φ3.

(a) Write a renormalized Lagrangian and determinethe relationships between the bare and renormal-ized parameters.(b) Working in (5 + 1)-dimensional spacetime, de-termine the superficial degree of divergence of thetwo diagrams shown in Fig. 32.5.(c) Suggest Feynman rules for the counterterms inthe renormalized theory.

(a) (b)

Fig. 32.5 Two divergent diagrams occurring in φ3 the-ory.

33Renormalization in action:

propagators and Feynman

diagrams

33.1 How interactions change thepropagator in perturbationtheory 295

33.2 The role of counterterms:renormalization conditions

297

33.3 The vertex function 298

Chapter summary 300

Exercises 300

Now we are going to see yet another way in which the wholeworld is built out of recursion ... We are going to see thatparticles are - in a certain sense which can only be definedrigorously in relativistic quantum mechanics - nested insideeach other in a way which can be described recursively, per-haps even by a sort of ‘grammar’.Douglas Hofstadter (1945– ), Godel, Escher, Bach

Some of the most useful tools in quantum field theory are propagatorsand Feynman diagrams. We’ve seen in the previous two chapters howrenormalization affects single particles and Lagrangians. Now we’ll ex-amine how renormalization affects Feynman diagrams and propagators.This approach provides a particularly vivid picture of how a particledresses itself in interactions. The main features of the propagator ap-proach to renormalization is that the dressing-up process is implementedwith two new Green’s functions.1 The first results from the single particle 1The beauty of describing renormaliza-

tion in terms of Green’s functions isthat each of these new Green’s func-tions may be written as a sum of Feyn-man diagrams.

interacting with the vacuum as it propagates through spacetime. Theseinteractions gives rise to a Green’s function we call the self-energy.The self-energy describes the changes to the particle’s mass caused byinteractions. The second new Green’s function comes from the virtualfluctuations screening the interactions between particles. The screeningis described by the vertex function Γ. This screening changes the The notation is unfortunate, but please

try not to confuse the vertex functionΓ and the quasiparticle decay rate Γp .

coupling constant of the theory.

33.1 How interactions change the

propagator in perturbation theory

We will discuss the usual example of a scalar field theory with φ4 inter-actions, described by a Lagrangian

L =1

2(∂µφ)2 − m2

2φ2 − λ

4!φ4, (33.1)

where we’ve written the theory in terms of bare fields and parameters.When there are no interactions present, the amplitude for a particle

with momentum p to propagate between two points is given by G0(p) =

296 Renormalization in action: propagators and Feynman diagrams

i/(p2−m2+iǫ). We saw previously (without using perturbation theory atall) that when you turn on interactions particles become dressed particlesdescribed by a propagator

G(p) =iZp

p2 −m2P + iΓp

+

(Multiparticle

parts

), (33.2)

where Z1/2p = 〈pλ|φ(0)|Ω〉 tells you how much particle we have in the

momentum state |pλ〉, Γp is the particle decay rate and mP is the phys-ical particle’s exact mass. Notice that the pole is found at p2 = m2

P

and the residue of the pole is iZp. We now take a propagator for non-interacting particles with bare mass m and examine how it changes whenwe add an interaction to our theory as a perturbation. Since we knowthat the exact answer is given by eqn 33.2, we know how to extract theexact mass mP of the particle from any propagator we calculate withperturbation theory: we simply look for the pole. We also know how tofind the quasiparticle weight Zp: we find the residue at the pole. Wetherefore have a very important rule:

The physical mass of a particle in an interacting theory is given bythe position of the pole in the propagator. The quasiparticle weight

is found from the residue at the pole.

In perturbation theory the propagator can be found by adding up Feyn-man diagrams. That is

G(p) =∑(

All connected Feynman diagramswith two external legs

). (33.3)

One helpful way to think of the propagator is to picture it as being madeup of the two external legs and everything we can slot in between thosetwo legs. What we can slot in is called the self-energy, and we describethat using what is called a 1-part irreducible diagram (or 1PI di-

agram for short), which is defined as a connected diagram which can’tbe disconnected by cutting one internal propagator line. Some examplesof 1PI diagrams in φ4 theory are shown in Fig. 33.1(a) and some exam-ples of diagrams that aren’t 1PI are shown in Fig. 33.1(b). These latterdiagrams contain cutlines, and so fall apart into disconnected pieces ifyou attack one of the cutlines with a pair of scissors. In contrast, theconnectedness of an 1PI diagram is immune to a single scissor attack onone of their internal lines. They essentially represent the smallest non-trivial Feynman diagram and a basic building block of more complexdiagrams. They are, if you like, the guts of the self-energy.

(a) (b)

Fig. 33.1 (a) Some 1PI diagrams fromφ4 theory. They can’t be turned intotwo meaningful diagrams by cuttingone propagator line. (b) Diagrams thataren’t 1PI. All can be turned into legit-imate 1PI diagrams by cutting at somepoint along the horizontal line.

The sum of 1PI diagrams with two external legs forms a Green’s func-tion known as the 1PI self-energy Σ(p), given by2

2Note that the 1PI self-energy does notinclude an overall energy-momentumconserving δ-function. We also ampu-tate the external lines, that is, we don’tinclude propagators for these lines.

−iΣ(p) =∑(

All amputated 1PI diagramswith two external lines

). (33.4)

Some contributions to −iΣ(p) in φ4 theory are shown in Fig. 33.2. In

= +

+ +

+ +

+ · · ·

1PI

Fig. 33.2 Some contributions to the1PI self-energy −iΣ(p) in φ4 theory.

the following example, we will explore how the propagator G(p) can beexpressed in terms of the 1PI self-energy Σ(p).

33.2 The role of counterterms: renormalization conditions 297

Example 33.1

The definition of Σ leads to one of the neatest tricks in perturbation theory. We canwrite the interacting propagator as a series of encounters with an interaction repre-sented by Σ. This is the story of how the propagator dresses itself: it’s the amplitudefor propagating with no interactions, added to the amplitude for propagation beinginterrupted by one interaction with Σ, added to the amplitude for two interactionswith Σ, added to... The series, shown in Fig. 33.3, is written as follows:

G(p) =i

p2 −m2+

i

p2 −m2

h

−iΣ(p)i i

p2 −m2

+i

p2 −m2

h

−iΣ(p)i i

p2 −m2

h

−iΣ(p)i i

p2 −m2+ . . . (33.5)

We now treat this as a geometric series and sum:

G(p) =i

p2 −m2

1 +h

−iΣ(p)i i

p2 −m2

+h

−iΣ(p)i i

p2 −m2

h

−iΣ(p)i i

p2 −m2+ . . .

ff

=i

p2 −m2

8<

:

1

1 − Σ(p)

p2−m2

9=

;

=i

p2 −m2 − Σ(p) + iǫ, (33.6)

where, in the last line, we’ve reinserted the iǫ from the free propagator. The sum can,more amusingly, be done in terms of diagrams as shown in Fig. 33.4. Equation 33.6 is

= +

+ + · · ·

1PI

1PI 1PI

Fig. 33.3 Diagrammatic version ofeqn 33.5.

= +

+

+ · · ·

=( )

1−

=

( )−1 −

1

1PI

1PI

1PI 1PI

1PI

=i

p2 −m2 + iǫ− Σ(p)

Fig. 33.4 Diagrammatic version ofeqn 33.6

another instance of Dyson’s equation, which we met in Chapter 16. The form of theequation makes it look a lot like the free propagator, albeit with an extra self-energyterm in the denominator.

To find the mass-energy mP of the physical particles we just follow our renormal-ization rule: we look for the position of the pole. It is found using the equation

p2 −m2 − Re“

Σ(p)”

= 0, (33.7)

when p2 = m2P. That is to say that the physical mass of the particles is given by

m2P = m2 + Re

“

Σ(p2 = m2P)”

, (33.8)

which tells us that the real part of the self-energy tells us the shift in energy causedby interactions. The latter equation is known as a renormalization condition.3 3It allows us, at the level of Feynman

diagrams and perturbation theory, torenormalize a theory. By the same to-ken, the decay rate of the physical par-

ticle is found from Im“

Σ(p2 = m2P)”

.

33.2 The role of counterterms:

renormalization conditions

But what of counterterms in this way of looking at things? Thus farwe’ve summed all of the 1PI self-energy diagrams to make an objectΣ(p). In the absence of counterterms, the role of the self-energy Σ(p) isto shift the mass from m to mP. However, we have ignored the fact thatmany of the diagrams that contribute to Σ(p) will be divergent, makingthe shift in mass seem dangerously infinite! One way to interpret this isto say that m must be infinite and requires an infinite shift to bring itdown to the physical value mP. However all this talk of infinity should


make us nervous and we would be better to remove divergences as wedid in the previous chapter.

To remove the divergences in Σ(p) we should really sum all of the self-energy diagrams along with counterterms in order to cancel infinities.Recall that the effect of including counterterms was to shift the massof the theory from the unphysical fictional value of m to the measuredand reliable value mP. Using the 1PI self-energy Σ(p), we won’t need toshift the mass at all! That is, when we include counterterms we requirethe renormalization condition

Re(Σ(p2 = m2

P))

= 0. (33.9)

We see that there are two ways of doing business. We may use therenormalization condition eqn 33.8, which neglects counterterms withthe result that the shifts from fictional to real masses are potentiallyinfinite. On the other hand we can use the renormalization condition ineqn 33.9 which allows us to start with the mass set to the correct valueand we ensure that it doesn’t change as we do our perturbation theory.4

4In the theory of the electron gas,where we do not encounter divergencesto the extent that we do in some othertheories, the first option is the logicalone. In quantum electrodynamics thesecond option is the one to choose.

33.3 The vertex function

−iΓ

Fig. 33.5 In φ4 theory the vertexfunction is contained in the Greensfunction shown here, correspondingto 〈0|T φ4S|0〉/〈0|S|0〉, or equivalently

G(p1)G(p2)h

−iΓ(p1, p2, p3)i

G(p3)G(p4).

Diagrammatically, the vertex part −iΓ,the central blob in the diagram above,may be extracted by amputating theexternal propagator legs.

The next thing to do is to work out how to renormalize the couplingconstant. Physically, this is changed because vacuum fluctuations screenthe interactions between two particles. The interaction vertex in φ4

theory has the Feynman rule

(Vertex) = (2π)4δ(4)(p4 + p3 − p2 − p1)(−iλ), (33.10)

that is, we get a factor of −iλ for every vertex and the δ-function ensuresthat momentum is conserved. The vertex is plugged in between fourexternal lines to tell us about two particle scattering.

We define the vertex function Γ for φ4 theory by5

5We could also define a 1PI vertex func-tion, which is useful for many applica-tions, but we won’t need it here.

−iΓ =∑(

All connected four-point diagramswith external legs amputated

). (33.11)

This is shown diagrammatically in Fig. 33.5. (Note that −iΓ does notinclude an overall energy-momentum conserving δ-function.) Like theφ4 interaction vertex, the vertex function may be plugged in betweenfour propagators. It is designed to tell us how interactions (involvingvirtual particles) affect how real particles interact with each other. Someexamples of contributions to −iΓ are shown in Fig. 33.6.

= + +

+ + · · ·

Fig. 33.6 Some contributions to the

vertex function −iΓ in φ4 theory. Thevertex function is represented as ashaded blob with four external legs.

Example 33.2

For two-particle scattering Γ can be obtained from the amplitude for the two-particlescattering process. To first order in the coupling, the four-point vertex function isgiven by a single interaction vertex: −iΓ = −iλ. To second order in the couplingwe eliminated divergences using a counterterm (see Chapter 32) and saw that thetwo-particle scattering amplitude was given by

−iΓ(p1, p2, p3) = −iλ− iaλ2 (ln s+ ln t+ lnu) , (33.12)

33.3 The vertex function 299

that is, it depends on the momenta of the incoming and outgoing particles[parametrized by s = (p1 + p2)2, t = (p1 − p3)2 and u = (p2 − p3)2].

The last example demonstrates that the exact vertex functionΓ(p1, p2, p3) is a function of the incoming and outgoing momenta. (Ofcourse momentum conservation means we only need give three of thefour momenta in the problem.) In order to define a renormalizationcondition for Γ we need to choose the values of these momenta at whichto fix Γ to some value. However, unlike mass which is unambiguouslydefined, there is no unique definition of the coupling constant. We aretherefore free to choose a renormalization condition at our convenience.We finally write the definition of the physical coupling constant:

−iλP = −iΓ(p1, p2, p3) =∑

Amputated, connected diagrams withfour external legs

and momenta p1, p2 and p3

,

(33.13)where the values of p1, p2 and p3 that we choose are called the renor-malization point. The sign is chosen so that to first order λP = λ.

In Chapter 41 we will use the analogous

Green’s functions Σ and Γ to assess howelectromagnetic interactions change theproperties of photons and electrons inquantum electrodynamics. In Chap-ter 43 we will examine the renormaliza-tion of the theory of metals in terms ofthe self-energy of electrons and the pho-tons that mediate the Coulomb forcethat acts between them.Example 33.3

We can use the definition of λP to calculate the amplitude for two-particle scatteringup to second order in unrenormalized perturbation theory (i.e. where we haven’t usedcounterterms to remove the cut-off). If the physical coupling constant λP is definedto be that measured at a renormalization point s0, t0, u0 then, to second order, wehave that

−iλP = −iΓ(2)(s0, t0, u0) = −iλ+ iaλ2`3 ln Λ2 − ln s0 − ln t0 − lnu0

´. (33.14)

The two-particle scattering amplitude is given, for particles with momenta s, t andu, by

iM = −iλ+ iaλ2`3 ln Λ2 − ln s− ln t− lnu

´. (33.15)

At the level of approximation to which we’re working we may use eqn 33.14 to write

−iλ = −iλP − iaλ2P

`3 ln Λ2 − ln s0 − ln t0 − lnu0

´+O(λ3), (33.16)

which allows us to eliminate λ (and Λ) from eqn 33.15 to give the answer

iM = −iλP − iaλ2P

»

ln

„s

s0

«

+ ln

„t

t0

«

+ ln

„u

u0

«–

+O(λ3). (33.17)

We see that we obtain the amplitude in terms of the physical coupling constant λP,where momenta are measured relative to the renormalization point. This argumentmay be repeated using renormalized perturbation theory (i.e. with the use of coun-terterms, which eliminate all ln Λ terms) to obtain an identical answer, but withoutthe troubling business of carrying around the (potentially infinite) Λ-dependent term.


Chapter summary

• The self-energy Σ can be related to the sum of all 1-part irreducible(1PI) diagrams with two external legs.

• In φ4 theory, the vertex function Γ can be related to the sum of allfour-point diagrams with external legs amputated.

Exercises

(33.1) To find the quasiparticle weight we can expand theself-energy in a Taylor series. We’ll do this aboutthe interesting point p2 = m2

P.

Σ(p2) ≈ Σ(m2P) + (p2 −m2

P)dΣ(p2)

dp2

˛

˛

˛

˛

p2=m2P

+ . . .

(33.18)(a) Using this expansion show that

G(p) ≈ i

(p2 −m2P)

»

1 − dΣ(p2)

dp2

˛

˛

˛

p2=m2P

– . (33.19)

(b) Use this to show, to the order to which we’reworking, that

Z ≈ 1 +dΣ(p2)

dp2

˛

˛

˛

˛

p2=m2P

. (33.20)

This gives us a scheme for calculating Z, the quasi-particle weight, in terms of the quasiparticle self-energy.

(33.2) (a) Draw diagrams showing the contributions to theself-energy −iΣ(p) of psions for ψ†ψφ theory up tofourth order in the interaction.(b) Draw the contributions to the vertex function−iΓ in ψ†ψφ theory up to third order in the inter-action.Here the vertex function includes all amputated in-sertions with one psion line, one antipsion line andone phion line.

(33.3) Repeat the argument in Example 33.3 using renor-malized perturbation theory (i.e. including the useof counterterms to remove the momentum cut-offΛ).

(33.4) Consider the action derived for a one-dimensionallattice in (1 + 1)-dimensional spacetime

S=1

2

X

p

Z

dω

(2π)φ−p(−ω)

`

ω2−ω20+ω2

0 cos pa´

φp(ω).

(33.21)We will treat the final term in the bracket as a per-turbation.(a) Identify the free propagator and the interactionvertex.(b) Write the full propagator as a sum to infinityinvolving the free propagator and interaction term.(c) Show that, on carrying out the sum, the ex-pected full propagator is recovered.

(33.5) Consider a model of an atom made up of two energylevels (E2 > E1 > 0) occupied by a single fermionand described by a Hamiltonian

H = E1c†1c1 + E2c

†2c2 + V (c†1c2 + c†2c1). (33.22)

(a) Treat the non-diagonal term as a perturbationand find the free propagators 〈0|T c1(t)c†1(t′)|0〉 and〈0|T c2(t)c†2(t′)|0〉 for electrons in level |1〉 and |2〉respectively.(b) Now consider the interaction term. Find theFeynman rule for the processes that contribute tothe self-energy and invent appropriate Feynman di-agrams to describe an electron in the system.Hint: the potential here is time independent, so theprocess to consider involves the electron changingstates and immediately changing back.(c) Find the self-energy of the electron in each stateand show that the interaction causes a shift in theenergy of the particle in level |1〉 by an amount

∆E1 = − V 2

E2 − E1. (33.23)

Exercises 301

(d) Compare this result with (i) the exact solutionand (ii) with second-order perturbation theory.*(e) Harder In the time-dependent version of theproblem the interaction part is given by

H ′ =

Z

d3q

(2π)3V (ωq)(aqe−iωqt+a†qeiωqt)(c†1c2+c

†2c1).

(33.24)By evaluating the self-energy of an electron in eachof the levels, show that the decay rates of the levels

are given by

2Γ1 = 2πV (ω0)2〈n(ω0)〉

Z

d3q

(2π)3δ(q0 − ω0),

2Γ2 = 2πV (ω0)2[1 + 〈n(ω0)〉]

Z

d3q

(2π)3δ(q0 − ω0),

where ω0 = E2 − E1.

34 The renormalization group

34.1 The problem 302

34.2 Flows in parameter space304

34.3 The renormalization groupmethod 305

34.4 Application 1:asymptotic freedom 307

34.5 Application 2:Anderson localization 308

34.6 Application 3:the Kosterlitz–––Thoulesstransition 309

Chapter summary 312

Exercises 312

Everything is in motion.Everything flows.Everything is vibrating.William Hazlitt (1778–1830)

A question we keep asking in physics is ‘when is a theory valid’. In thecontext of quantum field theory an answer is provided by the philosophyof the renormalization group.1 This is a way of looking at the world

1The renormalization group is a bit ofa misnomer as it is not really a group.The name arises from the study ofhow a system behaves under rescalingtransformations and such transforma-tions do of course form a group. How-ever, the ‘blurring’ that occurs whenwe rescale and then integrate up to acut-off, thereby removing the fine struc-ture (and this is the very essence ofthe renormalization group procedure)is not invertible (the fine details arelost and you can’t put them back).Thus the transformations consisting ofrescaling and integrating up to a cut-offdo not form a mathematical group be-cause the inverse transformation doesnot exist.

pioneered by several physicists, but most notably Kenneth Wilson. It

Kenneth Wilson (1936–2013) wasawarded the Nobel Prize in Physics in1982 in recognition of ‘his theory forcritical phenomena in connection withphase transitions’, a theory that grewout of his work on the renormalizationgroup.

allows us to make sense of why a renormalized quantum field theorydescribes Nature.

34.1 The problem

Theories are described by Lagrangians. These are the sum of severalterms, each of which is some combination of fields and their derivatives,multiplied by a so-called coupling constant. For φ4 theory, we have theLagrangian

L =1

2(∂µφ)2 − m2

2φ2 − λ

4!φ4, (34.1)

where m and λ are the coupling constants. Calculations that start witha Lagrangian like this soon encounter disaster in the form of divergent(i.e. infinite) integrals. That is, we take our Lagrangian, canonicallyquantize the free part, treat the interacting part as a perturbation todefine an S-operator using Dyson’s equation and then start working outamplitudes in the form of S-matrix elements which lead to Feynmanrules. If a Feynman diagram involves a loop, we often get a divergence.To cure a divergence, we can introduce a finite cut-off in momentum Λin such a way that we integrate up to Λ rather than infinity. In theprevious chapters our strategy was then to renormalize, that is to saythat we remove the mention of Λ from things that we calculate. Thisinvolves expressing amplitudes in terms of physical coupling constants,which are the ones we measure at some agreed point in momentum space.Our definitions of coupling constants depend on this point in momentumspace.

Kenneth Wilson had another strategy: rather than hiding the cut-off,we live with it. In Wilson’s view, to properly define the theory, we needto admit that we do our integrals up to a maximum momentum Λ, whichwe are entirely free to choose. The arbitrary choice of Λ immediately

34.1 The problem 303

raises the question: how do we know what value to choose? One answeris that we want the length scale Λ−1 to be far smaller than the length2 2Remember that in units where ~ =

c = 1, mass m, energy E and momen-tum p have inverse units to length.

scale p−1 of any of the physics in which we’re interested. Consider, forexample, a gas of atoms in a box. If we’re interested in sound waves,which are oscillations in the number density of the constituents of thegas involving many millions of atoms, then the scale in which we’reinterested in p−1 is of the order of centimetres and we could take Λ−1

to be a few microns. If we’re interested in the electron cloud of an atomin the gas, then we’re interested in a length scale p−1 of the order ofthe size of an atom and we could take Λ−1 to be the size of an atomicnucleus. The size of Λ−1 that we choose therefore changes depending onthe scale p−1 of the physics that we’re analysing.

The renormalization group is a machine that tells us how the predic-tions of the theory change as we alter the scale of interest. It gives us anequation telling us how a coupling ‘constant’ changes with the length ormomentum scale in which we’re interested. To summarize this chapter:

The goal of renormalization group analysis is to discover how thecoupling constants change with the scale of interest.

Example 34.1

In general, this goal may be achieved by examining how your Lagrangian behaveswhen the cut-off Λ is systematically varied. This approach will be outlined later. It is,however, sometimes possible to leapfrog the systematic method to get an equation forthe coupling constant in terms of the scale of interest. We’ll follow such an approachhere first. Recall how we defined the renormalized coupling constant in φ4 theory.We said that λP was the measured scattering amplitude when the particles hadsquared momenta s0, t0 and u0, which is to say, the coupling constant is defined ata particular point in momentum space. How do we choose this point? A pragmatistwould say that you choose it to be a convenient energy scale for the problem you’readdressing. If we’re dealing with electrons then we could set s0 = t0 = u0 = µ2 whereµ is close to the electron mass. If we’re dealing with pions, we’d use µ close to thepion mass and so on. However, this ‘idiot-proof’ scheme could easily be challengedif an ingenious idiot started thinking about electron physics but set µ equal to thepion mass.

As we saw in Chapter 33 the physical coupling constant is given to second orderby

−iλP(s0, t0, u0) = −iλ+ iaλ2`3 lnΛ2 − ln s0 − ln t0 − lnu0

´. (34.2)

This allowed us to write the two-particle scattering amplitude to second order:

iM = −iλP(µ) + ia[λP(µ)]2»

ln

„µ2

s

«

+ ln

„µ2

t

«

+ ln

„µ2

u

«–

, (34.3)

which is to say that it’s given by the physical coupling constant added to a logarithmiccorrection. This correction is small since we chose µ2 to be of order s0, t0 and u0.

If, on the other hand, we idiotically choose a renormalization point µ′2 to be muchlarger than s0, t0 and u0, then the logarithmic term could be very large compared tothe first term. Then we would have

iM = −iλP(µ′) + ia[λP(µ′)]2»

ln

„µ′2

s

«

+ ln

„µ′2

t

«

+ ln

„µ′2

u

«–

, (34.4)

where the second term is large compared to the first. Subtracting, we can find anexpression for λP(µ′) in terms of λP(µ):

λP(µ′) = λP(µ) + 6a[λP(µ)]2 ln

„µ′

µ

«

+O(λ3P), (34.5)


which is expressed in differential form thus:

µd

dµλP(µ) = 6aλP(µ)2 +O(λ3

P). (34.6)

So in this case, we have a simple equation that tells us how the coupling λ changeswith the momentum scale µ.

If we are interested in high energies, we can examine how λ behaves as µ, rep-resenting the scale of interest, gets larger and larger, that is, as µ → ∞. We callthis the ultraviolet behaviour. Equally well, we might be interested in the physicsat very large length scale, which would involve asking how λ changed as µ → 0.This is known as the infrared behaviour, since it corresponds to the limit of small

Remember that [Mass]−1 ≡[Length]−1.

energies.

34.2 Flows in parameter space

A theory is characterized by gi, the set of coupling constants thatmeasure the strength of the various interactions. (For φ4 theory wehave g2 = m and g4 = λ.) Each coupling constant is a function of thecut-off Λ, so we can write the ith coupling constant as gi(Λ). We aregoing to examine what happens as we look at different length scales,which is achieved by changing the value of the cut-off. If we reduce thecut-off by a factor b, then we want to see how the coupling constantschange and so we will look at the transformation

gi(Λ) → gi(Λ/b). (34.7)

If we are interested in the limit of large length scales (corresponding

g1

g2

Fig. 34.1 The renormalization groupflow in a configuration space defined

by two coupling constants g1 and g2.

to what is measured in condensed matter physics) then we choose thedirection of positive flow by fixing b > 1. This is known as coarse

graining.3 A theory can be described as a point in the multidimensional

3Alternatively there is nothing stop-ping us examining the theory at shortlength scales in which case we could setb < 1.

space formed by all of the gi, i.e. by the point (g1, g2, · · · ) in thatspace. The rescaling of the cut-off causes the point to travel through thismultidimensional space along a renormalization group trajectory,and this is called a renormalization group flow (see Fig. 34.1). Wewill sometimes find fixed points in this space; these remain invariantunder coarse graining and correspond to scale-invariant Lagrangians.We will see that such Lagrangians are very important in determiningthe physics described by the theory.

Example 34.2

A very simple example of renormalization group flow occurs for a theory when thereis only a single coupling constant g. A graph with only one axis is hard to illustrate(much like the problem with the sound of one hand clapping) so to illustrate the flowit is conventional to plot the function

β(g) =dg

d ln b, (34.8)

which is known as a Gell-Mann–Low equation4 or simply a flow equation.5

4This is, of course, a different ‘Gell-Mann–Low equation’ to that describedin Chapter 22. See M. Gell-Mann andF. Low, Phys. Rev. 84, 350 (1951)

5Sometimes one can write

β(g) =dg

db

instead. Since b divides Λ, the rescalingis more usefully modelled by looking athow g changes with ln b. But, as weshall see in this chapter, different defi-nitions can, and are, made.

g

β

g

β

gg∗

β

gg∗

β

Fig. 34.2 Four examples of flows ina configuration space defined by a sin-gle coupling constant g, visualized byplotting β(g) = dg

db. Arrows show the

direction that the length scale ℓ→ ∞.

Note that:

34.3 The renormalization group method 305

• If β is always positive then g will be blasted out to infinity, as shown inFig. 34.2(a). If β is always negative then g will be sucked backwards to g = 0as shown in Fig. 34.2(b).

• If at position g = g∗ we have β(g∗) = 0 then g is stuck and will remain at g∗

for evermore. We call g∗ a fixed point.

• Even if a fixed point exists the system may never get stuck there. If thevelocity function β looks like Fig. 34.2(c) then if g > g∗ the system will endup at ∞, while g < g∗ will be drawn back to g = 0: here g∗ is known as arepulsive fixed point.

• If the function β(g) looks like Fig. 34.2(d) then the system will be attractedtowards g∗ regardless of its initial condition. There are no prizes for guessingthat under these circumstances g∗ is known as an attractive fixed point.

34.3 The renormalization group method

Now that we know what it’s useful for, we’d like to know how the renor-malization group method works in more detail. We start with the func-tional integral, written in Euclidean space

Z(Λ) =

∫

Λ

Dφ e−R

ddxL[φ], (34.9)

where Λ is an instruction to integrate over configurations of φ(x) con-taining Fourier components up to Λ.

Fig. 34.3 Removing the largest Fouriercomponents results in a loss of detail.

We will use the renormalization group method to examine the physicsas we increase the length scale of interest. The method involves threesteps. In step I we’re going to remove the largest Fourier components ofmomentum. This is achieved by doing that part of the functional integralthat involves the largest Fourier components in momentum space. Theremoval of these components is equivalent to losing one’s glasses. Acertain level of fine detail is lost and we have a simpler looking field(Fig. 34.3). We then compare the theory to what we started with. Sincewe now seem to be comparing apples (the original integral) and oranges(the partly integrated integral) we need to scale up the partly integratedintegral so that it is defined over the same momentum range as theoriginal. To do this involves (step II) relabelling the momenta andthen we (step III) scale the size of the fields so that the Lagrangianresembles what we had before. The beauty of the procedure is that, insuccessful cases, it will not change the form of the Lagrangian, only thecoupling constants. We then imagine repeating the procedure over andover again (losing successive pairs of spare glasses) and extract how thecouplings change as we partially integrate the functional integral.

Λ/bΛ

Λ/b

Λ

(a)

(b)

(c)

Fig. 34.4 The renormalization groupprocess in momentum space. (a) Thefield φ(x) before the process. (b) Theresult of integrating out the largest mo-mentum states (step I) is to remove thehigh spatial frequency components. (c)The field after rescaling (steps II andIII). Note that the region in the dashedbox is identical to the entire trace in(b).

|p|

0

Λb

Λ

φs

φf

Fig. 34.5 Components in momentumspace are divided up into slowly varyingand fast varying, with any componentwith |p| > Λ being ignored.

We now demonstrate the procedure, which is illustrated in Fig. 34.4.First we need to set the scene. We choose to work in momentumspace, where we divide the fields φ(p) into those parts that vary slowlywith p [called φs(p)] and those parts [called φf(p)] that vary quickly(see Fig. 34.5). We define the fast Fourier components as those in themomentum shell Λ/b ≤ |p| ≤ Λ, where b > 1 is a scaling factor, which


will be central to our analysis. The slow Fourier components of thefields are then those with momenta6 0 ≤ p ≤ Λ/b. The functional6We therefore have

φ(p) = φf(p) for Λ/b ≤ p ≤ Λ,

φ(p) = φs(p) for 0 ≤ p ≤ Λ/b.

integral becomes

Z(Λ) =

∫

Λ/b

Dφs e−R

ddxL[φs]

∫ Λ

Λ/b

Dφf e−R

ddxLI[φs,φf ]. (34.10)

We’re now ready to start the renormalization group procedure.Step I: This is the difficult step. We need to integrate over the quickly

varying fields. This is usually impossible, so we need an approximationscheme involving perturbation theory. Assuming we can do this, we’llobtain an action in which the remaining slow fields and coupling con-stants are different to what they were before. If we can (somehow) dothe integral over the fast varying part φf we define the answer to be

∫ Λ

Λ/b

Dφf e−R

ddxLI[φs,φf ] = e−R

ddx δL[φs], (34.11)

leading to the result for our integral:

Z(Λ) =

∫

Λ/b

Dφs e−R

ddx (L[φs]+δL[φs]), (34.12)

which only features the slow fields.Step II: We relabel the momenta. We do this by expressing our

theory in terms of scaled momenta p′ = pb. Since b > 1 this has theeffect of stretching the momentum space out over the range we hadoriginally, as shown in Fig. 34.6. This is seen by examining the limit ofthe integral, which is p′/b = Λ/b or p′ = Λ.

0

Λb

Λ

0

Λ

p p′ = bp

Fig. 34.6 Rescaling the momentum.

Step III: Finally we scale the fields. We select the term that weimagine will be most important in determining the scaling behaviourand require it to be invariant with the scaling of step II. We use thisto rescale the field φ(p) = φ(p′/b) = bd−dφ φ′(p′), where dφ is chosen toleave the important term invariant.7

7The subject is notorious for its arcaneterminology, and dφ is known as theanomalous dimension.

Now that the procedure has been carried out once, we repeat it aninfinite number of times. In fact, we only need to carry out the procedureonce and then infer the consequence of repeating it. That is, we do theapproximate integral which tells us how all of the coupling constantschange with scale. We then imagine a continuous flow of the couplingconstant caused by repeating the renormalization group method againand again.

By removing the fastest varying fields we are looking at the physicsat lower momentum, or equivalently at longer length scale. We cantherefore ask how the physics changes as we send the length scale ofinterest ℓ to infinity. One way to do this is to set b = eℓ and search forflow equations of the form

dgidℓ

= βi(g1, g2, ..., gN ). (34.13)

This is another form of the Gell-Mann–Low equation.8

8Sending ℓ to infinity corresponds tolooking at the infrared behaviour of thecouplings (since large length is the sameas small energy). We could send ℓ→ 0,if we want the ultraviolet behaviour.

34.4 Application 1: asymptotic freedom 307

The renormalization group is most famously applied to phase transi-tions, an example we will defer until the next chapter. Here we focuson three examples which illustrate other features of the renormalizationgroup.

34.4 Application 1: asymptotic freedom

Perturbation theory is carried out in terms of series expansions in cou-pling constants. The renormalization group teaches us that couplingconstants change depending on the region in energy-momentum spacein which we’re working. If we work in a region of momentum spacewhere the coupling constants are small, then perturbation theory willwork well. If, however, we work in a region where the coupling con-stants are large then we can expect perturbation theory to fail to yieldsensible results. The coupling ‘constant’ in quantum electrodynamicsis the electronic charge. As we’ll see in Chapter 41, to first order, theβ function describing the evolution of e is written9 µdβ

dµ = |e|3/12π2,

9Notice again that slightly differentdefinitions of the flow are used depend-ing on context. This is nothing toworry about since the idea is always thesame: we want to know how the cou-pling changes with scale.

where µ is the energy scale of interest. The important thing to noteis that the right-hand side is positive. This means that as we flow tolarge energy-momentum transfer (by increasing the mass-energy scaleµ), the coupling constant gets larger and larger. Eventually, when themomentum involved in the processes we’re calculating implies a verylarge mass-energy scale, perturbation theory ceases to be sufficient tocapture the physics. Of course, running this process backwards (cor-responding to examining processes involving small momentum transferand large length scale) the coupling gets smaller and smaller and per-turbation theory captures the physics better and better. This confirmsthe expected result that perturbation theory works best for processesinvolving small momenta and energy and gets worse as the energy getslarger. It also confirms the physical picture that electric charge is largestclose to a charged source (that is at small length scales), and dies awayat large distances.10

10This occurs because of screeningby virtual charged particle–antiparticlepairs. See Section 41.1 and Fig. 41.6.

A surprising result was discovered for a class of theories, which in-cludes the famous non-abelian Yang–Mills theory that will be examinedin Chapter 46. In this case, β is negative! Therefore, if we flow to largemomentum transfer, the coupling constants become small and perturba-tion theory at low orders is a very good approximation. However, nowif we reverse the process, allowing the momentum transfer to becomesmall, then g gets very large and perturbation theory no longer givesgood results. Theories of this sort are said to have asymptotic free-

dom:11 they are most strongly interacting at low energy scales, but at

11Frank Wilczek now regrets coiningthis term, which was first suggestedby Sidney Coleman, and wishes hehad used ‘charge without charge’ in-stead. The idea is that closer and closerto a quark the effective colour charge‘asymptotically’ approaches zero. Hav-ing zero colour charge means complete‘freedom’ from its interaction. Playingthe argument in reverse, at large dis-tances away from the intrinsically weakcolour charge of the quark, the cloudof virtual particles give rise to anti-screening and this greatly enhancesthe effective colour charge and resultsin the strong interaction being strong.

large scales they are weakly interacting (and therefore act like nearlyfree theories).12

12One might think of masses connectedby springs as showing a sort of asymp-totic freedom. At small length scalesthey interact weakly; as the length scaleis increased and the springs becomestretched they interact more strongly.This is why you cannot isolate quarksand gluons as free particles, but onlyfind them in bound states inside mesons(quark, antiquark) or baryons (threequarks).

The importance of asymptotic freedom is that it allows us to explain(among other things) the physics of the strong interaction. Under thenormal (i.e. low energy) circumstances of much of the Universe, thequarks in a proton are very tightly bound. However, it was found exper-


imentally in deep inelastic scattering experiments that quarks impactedby very highly energetic electrons behave as if they’re free (or, at leastthat the coupling between quarks is very small). This is entirely thecircumstance described by theories that show asymptotic freedom. Thestrong interaction is described by a theory known as quantum chromo-dynamics, which is a non-abelian gauge theory with an internal SU(3)symmetry which gives rise to a conserved charge known as colour.1313The virtual gluon excitations in the

vacuum carry colour charge and serveto reinforce the field, rather than screenit. This effect is sometimes called anti-screening (see above) but is essentiallyanother way of describing asymptoticfreedom.

The theory shows asymptotic freedom. We introduce non-abelian gaugetheories in Chapter 46.

34.5 Application 2: Anderson localization

Crystalline metals conduct electricity, but what happens when you in-troduce more and more disorder? Philip Anderson realized in 1958 thatbeyond a critical amount of disorder, the diffusive motion that resultsfrom impurity scattering would completely stop and the electrons wouldbecome localized, residing in bound states rather than in the delocalizedband states of the pure metal. This process is called Anderson local-

ization. Thus we can imagine making a metal more and more impureby implanting specks of dirt in it and at some point we expect to crossfrom metallic behaviour to insulating behaviour, as Anderson localiza-tion sets in. This tipping point in behaviour is known as the mobilityedge.

Let’s think about the scaling properties of conductivity, since that willbe important in a renormalization group treatment. A three-dimensionalmetal of volume L3 conducts electrons. It has a small resistance R anda large conductance G = 1/R. How does the conductance vary withthe length of a piece of metal? As examined in the exercises14 G(L) =

14See Exercise 34.1.

σL, where σ is the conductivity. In fact, for a d-dimensional solid,G(L) ∝ Ld−2. In contrast, an insulator has a small conductance, givenby G(L) ∝ e−L/ξ, where ξ is a length scale determined by microscopicconsiderations. What we really want to know is that if we start witha material with some fixed amount of dirt in it, do we expect it to bea conductor or an insulator? The renormalization group provides theanswer! For this problem, David Thouless showed that the coupling

David Thouless (1934– )

constant of interest turns out to be the conductance G. We define ourβ function as15

15This is another example of how theβ function is defined in various ways tosuit the problem in hand.

β(g) =d ln g

d lnL=L

g

dg

dL, (34.14)

where the dimensionless conductance is defined as g = ~G(L)/e2. Dif-ferentiating our expressions, we obtain a prediction for the scaling in themetallic and insulating regions for a d-dimensional solid:

β

ggc

d = 3

d = 2

d = 1

Fig. 34.7 The renormalization groupflows for Anderson localization. Arrowsshow the flow for long length scale.

β ≈

(d− 2) metallic (large g)ln g insulating (small g).

(34.15)

Patching these limits together we get the flow shown in Fig. 34.7. Noticethe fixed point at gc, which will turn out to represent the mobility edge.

34.6 Application 3: the Kosterlitz–––Thouless transition 309

The flow diagram predicts the behaviour of the system as we look atlarger and larger length scales. This is useful as the limit of large lengthscales is what our measurements of electrical resistance probe.

A material containing a particular amount of dirt will be an insulatoror metal, depending on whether the flows takes it to the left (small g)or right (larger g) respectively.16 We see that, for d = 3, we have the

16Note that a flow implying g → ∞doesn’t mean that we will measure thisbehaviour in a real system. We wouldexpect some other piece of physics, notincluded in our Lagrangian will cut offsuch flows eventually.

behaviour we predicted in our naive picture above: if the material has aconductance above that at the mobility edge (the fixed point in the flowknown as gc) the material is destined to be a metal. If it has so muchdirt that g < gc it will be an insulator. The mobility edge is representedby a repulsive fixed point. The surprise comes when we look in d = 2and d = 1 dimensions. In these cases there is no fixed point and hence,no mobility edge! If the system has any dirt in it at all it must flow tog = 0 and will be an insulator. This is a profound statement!

34.6 Application 3: the

Kosterlitz–––Thouless transitionJ. Michael Kosterlitz. The transi-tion was simultaneously discovered byVadim L. Berezinskii in the SovietUnion.

For any theory of a magnet in two spatial dimensions with a continu-ous symmetry, the Coleman–Mermin–Wagner theorem (of Chapter 25)predicts that there can be no symmetry breaking phase transition toan ordered magnetic state. However, a phase transition of a differentsort is predicted by a remarkable renormalization group analysis. Thisis a topological phase transition, which separates two phases of mattercontaining rather different topological objects.

We work in two spatial dimensions, and examine vortices, which arethe topological objects that can exist in (2+1)-dimensional spacetime.As discussed in Chapter 29, lone vortices are unstable17 at T = 0. An

17To recap: this is because the spinfield is swirly at spatial infinity and thiscosts the system an infinite amount ofenergy for a single vortex.

example of such a vortex is shown in Fig. 34.8. However, it’s a differentstory at high temperatures. This is because, rather than minimizing theenergy to find the thermodynamic equilibrium state of the system, wemust minimize the free energy F = U −TS, and so at high temperaturethere is a need to maximize entropy. The presence of a vortex gives thesystem some entropy, since its centre may be placed at any one of a largenumber of lattice sites and this provides a way of minimizing the freeenergy. The entropic term in F is weighted by the temperature, makingthe entropic advantage of vortices dominant at high T . We are led tothe conclusion that thermodynamics favours the proliferation of vorticesat high temperature.

+

Fig. 34.8 A vortex has infinite energy,due to the swirliness at spatial infinity.The question is, what happens at low temperatures? This is the ques-

tion that Kosterlitz, Thouless and Berezinskii addressed. Typically ofcondensed matter applications of the renormalization group, we ask whathappens to the system of vortices as we look at successively longer lengthscales. The renormalization group analysis of the two-dimensional vor-tex problem therefore involves asking what happens to the importantcoupling constants as we send the length scale of interest to infinity.


Example 34.3

We model this system with a complex scalar field given in polar coordinates φ(x) =eiθ(x). At a point in space x we see the field as an arrow of unit length pointing alonga direction given by an angle θ(x). Such a picture is known as the two-dimensionalXY model18 since the arrows are constrained to the two-dimensional plane, where

18See also Chapters 42 and 44 for moreon the XY model.

their direction may vary continuously. The Euclidean action of a gas of vortices isgiven by1919The reason for writing the action this

way is to prevent the integral from di-verging at the centre of the vortex,where the field looks to wind an infi-nite amount.

S = Score(a) +K

2

Z L

ad2x |∇φ(x)|2, (34.16)

where the first term represents the energy of the core region of the vortex, whichhas size a and the second term represents the energy cost of having a spin fieldthat varies with position outside the core in a system with sides of length L. Theconstant K = J/T where J is known as the stiffness of the spin system and T isthe temperature. We may plug in our ansatz for the field at a vortex, given byφ(x) = eiθ(x), where θ(x) + ϕ = tan−1(x2/x1) for |x| → ∞, just as we had inChapter 29. We integrate to find an action20

20See Exercise 34.2.

S = Score(a) + πK ln(L/a). (34.17)

What are the coupling constants that we change as the system flows tolarger length scales?

• The spin stiffness J is one of the coupling constants and tells ushow much energy cost we pay for having the spins describe somenon-uniform configuration. This is the parameter that causes asingle vortex to cost infinite energy.21 At high temperature, where

21More generally J arises from therigidity of the system. Rigidity is ageneral feature of ordered systems, re-flecting the energy cost associated withdeforming the order.

spins are presumably disordered, we do not expect any contribu-tion of the spin stiffness to the action determining the physics ofthe system over long distances. At low temperature J will be de-cisive in determining the configuration of spins. We thus need toconsider the ratio K = J/T which reflects the balance betweenJ and T and therefore we use K (actually K−1) as one of ourvariables.

• The other coupling reflects the energy cost of a vortex core dividedby the temperature. It is known as the fugacity and is given by y =e−S

core(a) (note that the Euclidean action is effectively a ratio ofenergy and temperature in statistical mechanics, see Chapter 25).The fugacity effectively tells us how the system feels the presence ofthe vortices. A small fugacity tells us that we have a configurationwith small, energetic and compact vortex cores [with large Score(a)]and that the spins don’t swirl much at infinity; a large fugacity tellsus that we have less action in the cores and a significant energycost from swirling fields at large distances owing to the existenceof free vortices. The energy cost here is, as usual for statisticalmechanics problems, measured with respect to the energy scaleset by kBT .

y

K−1

π2

low T phase

hig

hT

phas

e

Fig. 34.9 Flow diagram for theKosterlitz–Thouless transition. Thehorizontal axis is K−1 = T/J .

The behaviour of these two couplings predicted by the renormaliza-tion group22 is shown in Fig. 34.9. Following the arrows (which show

22See Altland and Simons, Section 8.6for details of the calculation.

34.6 Application 3: the Kosterlitz–––Thouless transition 311

the flow to large length scales) we see that most of the trajectories pointto K−1, y → ∞. This is the expected behaviour in the high-temperaturephase. The fact that y → ∞ tells us that we have free vortices presentmaking a non-uniform texture of spins at infinity. In this phase entropywins out and J is ultimately an unimportant consideration in determin-ing the thermodynamics, and so K = J/T flows to zero.

The high-temperature phase, however, is not the only fate of the sys-tem. We see that in the bottom-left of the plot there are flows that leadto fixed points on the K−1 axis, with y = 0. This is the low-temperaturephase of the system, where stable configurations of vortices are possible,even at T = 0. The fact that y flows to zero means that the vortex con-figurations must be quite compact, so that the cores look less and lesssignificant at larger and larger length scales and that the spin textureat infinity must be uniform. We also have that K 6= 0, meaning thatthe system has rigidity. Since rigidity is one of the inevitable side-effectsof magnetic order, this tells us that the low-temperature phase involvessome (possibly exotic) type of ordering.

+ −

(a) (b)

Fig. 34.10 (a) A vortex. (b) An an-tivortex.

−+

−+

− +

−+

−+

− +

−+

Fig. 34.11 Schematic of the low-temperature phase, viewed at long dis-tance. The +s represent vortices, the−s antivortices. They form a gas ofdipole-like pairs.

How can this be? The key to understanding the low-temperaturephase is the fact that it involves bound states of vortices and antivortices.As shown in Fig. 34.10, an antivortex has spins that wind in the oppositedirection to a vortex. A picture of the low-temperature phase is shownin Fig. 34.11 where we see that bound pairs of vortices and antivorticesare the stable field configurations. The stability of these pairs at lowtemperature is evident if we look at the fields at a large distance froma vortex–antivortex pair as shown in Fig. 34.12. The clockwise swirlof the vortex at infinity is cancelled by the anticlockwise swirl of theantivortex, making y → 0. This means that in a system with nonzeroJ , the pair costs a finite amount of energy.

−

+

Fig. 34.12 A vortex dipole made froma vortex and antivortex in the same re-gion of space. Notice that the fieldsmay now be made uniform at infinity.

We can even come up with an estimate of the transition temperaturebetween the low and high temperature phases. Consider the stabilityof the high-temperature phase, containing a single vortex. The energyof the vortex is given (from eqn 34.17) by πJ ln(L/a). The size of theentire system is L and the size of a vortex core is a. We may thereforefit a vortex into the system in L2/a2 ways, giving an entropy kB ln Ω =kB ln(L/a)2. The free energy is then F = (πJ − 2kBT ) ln(L/a). Theonly way of preventing this free energy from becoming very large andpositive is for 2kBT > πJ . Below T ≈ πJ/2kB the system is unstableto the formation of bound vortex–antivortex pairs. Of course, this isidentical to the fixed point in the renormalization-group flow diagramof Fig. 34.9 at K−1 = π

2 (apart, of course, from the factor kB that wasset to one in our hair-shirt quantum field treatment). The transition isoften also called the vortex unbinding transition, describing the breakupof the bound states on warming through the transition region.


Chapter summary

• The renormalization group is a method for studying how couplingconstants change with rescaling.

• This method provides insight into many physical phenomena,including asymptotic freedom, Anderson localization, and theKosterlitz–Thouless transition.

Exercises

(34.1) We will show using dimensional analysis that theconductance of a d-dimensional solid varies asG(L) ∝ L(d−2).(a) For d = 3 consider a cubic sample with sides oflength L and show G = σL, where σ is the conduc-tivity.(b) For d = 2 consider a flat plate of thickness aand other sides of length L and show G = σa.(c) For d = 1 show G = σa/L.

(34.2) By considering the ansatz in the text verify that theaction of a vortex is given by

S = Score(a) + πK ln(L/a). (34.18)

If instead the phase winds n times round a vortex,show that

S = Score(a) + πKn2 ln(L/a). (34.19)

35Ferromagnetism: a

renormalization group

tutorial

35.1 Background: critical phe-nomena and scaling 313

35.2 The ferromagnetic transitionand critical phenomena 315

Chapter summary 320

Exercises 320

In this chapter we are going to study how the coupling constants in aEuclidean φ4 theory vary with increasing length scale. This will tell usabout the physics of phase transitions. To make things concrete, we willdiscuss the magnetic transition separating a paramagnetic phase and aferromagnetic phase.1 Perhaps the most wonderful feature of a phase

1We have already discussed the mean-field treatment of this in Chapter 26.In this chapter we will include the pos-sibility of fluctuations in the fields.

transition is universality. This is the notion that some of the keyproperties of a phase transition depend only on the dimensionality of asystem and dimensionality of its order parameter field. The details ofthe microscopic interactions between the constituent parts, whether thetransition is ferromagnetic or superconducting, or whether it involvespolymers or the early Universe, are all irrelevant. Thus focussing on theferromagnetic transition will nevertheless give results that are applicableto any system described by the same Lagrangian.

35.1 Background: critical phenomena and

scaling

Magnetic phase transitions may be observed with several experimen-tal techniques. Some thermodynamic properties of a magnet in whichwe might be interested include the magnetization M (which is the or-der parameter of the ferromagnet system); the magnetic susceptibilityχ = limB→0

µ0MB , which tells us how much magnetization we can achieve

through applying a field B; and the heat capacity C, which tells us howthe energy of the system varies with temperature. To these thermody-namic properties we can add the correlation length ξ, which tells ushow well correlated the fluctuations are in space (Fig. 35.1).

ξ

Fig. 35.1 The correlation length ina paramagnet showing a ferromagneticfluctuation of size ξ.

The correlation length is related to the correlation functions, whichare the Green’s functions of Euclidean space. As described in Chap-ter 25, a slight modification of the usual correlation function G(x, y) =〈φ(x)φ(y)〉t is useful when studying phase transitions. This is the con-

nected correlation function Gc(x, y), given by

Gc(x, y) = 〈φ(x)φ(y)〉t − 〈φ(x)〉t〈φ(y)〉t, (35.1)

314 Ferromagnetism: a renormalization group tutorial

and differs from G(x, y) by the subtraction of a product of thermallyaveraged fields 〈φ(x)〉t〈φ(y)〉t. At temperatures above the transitiontemperature Tc we have 〈φ(x)〉t = 0 and the definition is identical tothe usual form of the correlation function. At temperatures below Tc,subtracting off the mean-fields allows us to just see the fluctuations awayfrom average behaviour. For example, in a magnet above Tc, despitethe spins being disordered they may save energy by pointing in thesame direction, and will tend to do so within the correlation length ξ.The correlation between spins at position x and position y decreaseswith increasing |x− y| because thermal fluctuations have a tendency torandomize the spins. Thus we find that2

2Closer to the disturbance the correla-tions are found to decay more slowly, asdescribed below.

Gc(x, y) ∼ e−|x−y|/ξ, (35.2)

for |x − y| ≫ ξ. The subtraction of the average fields in our definitionof Gc means that this is also the behaviour below Tc and represents thefluctuations away from perfect order.

At temperatures close to the ordering temperature Tc it is found exper-imentally that the physical properties of a magnet described above fol-low power law behaviour3 as a function of temperature, applied magnetic3Unlike, for example, exponential func-

tions such as e−x/ξ, power laws are freefrom any dependence on a length scale.This scale-free property is a hallmark ofcritical behaviour, where the character-istic fluctuations occur over all lengthscales.

field or distance. This motivates the definition of critical exponents

to describe the phase transition. Defining the reduced temperature ast = (T − Tc)/Tc, the critical exponents α, β, γ, δ, ν, ξ and η are relatedto physical properties as follows:

• Heat capacity: C ∼ |t|−α,

• Magnetization: M ∼ (−t)β , for B → 0, T < Tc,

• Magnetic susceptibility: χ ∼ |t|−γ ,• Field dependence of χ at T = Tc: χ ∼ |B|1/δ,• Correlation length: ξ ∼ |t|−ν ,• The correlation function G(r) behaves like

G(r) ∼

1|r|d−2+η |r| ≪ ξ

e−|r|ξ |r| ≫ ξ,

(35.3)

where r is distance and d is the dimensionality of the system.

A renormalization group analysis of the ferromagnet relies on varyingthe length scale at which we examine the physics. Well before the as-cent of the renormalization group, Benjamin Widom suggested that thecritical behaviour that is observed experimentally may be explained ifthe free energy4 of the system has a particular form under such a scaling4In these calculations, one studies the

reduced free energy f (dividing Fby both volume V = Ld and temper-ature T ) so that f = F/(LdT ). Wealso define reduced temperature t =|T −Tc|/Tc and reduced field h = B/T .

of the length scale. The idea is that when rescaling lengths by a factorb (so that a length L→ bL), we should expect

f(t, h) = b−df(t′, h′), (35.4)

where the factor b−d is because of the L−d factor in f . Widom assumedthat t′ = bytt and h′ = byhh, where yh and yt are exponents yet to be

35.2 The ferromagnetic transition and critical phenomena 315

determined. Thus the Widom hypothesis is

f(t, h) = b−df(bytt, byhh). (35.5)

We may use the fact that b is an arbitrary parameter and choose a valueof b such that bytt = 1. This reduces the number of free parameters byone, and we obtain

f(t, h) = td/ytf(1, h/tyh/yt), (35.6)

which is Widom’s hypothesis, written in reduced variables. Using thisresult, and also expressions relating the free energy to physical propertiessuch as magnetization, heat capacity, magnetic susceptibility, etc., onecan show that the critical exponents can be expressed in terms of they’s as follows:5 5See Exercise 35.1.

α = 2 − dyt, β = d−yh

yt, γ = 2yh−d

yt,

δ = yh

d−yh, ν = 1

yt, η = 2 + d− 2yh.

Our task is now to use the renormalization group to find yt and yh.We find these quantities by noting how the quantities t and h flow onrenormalization. Since we have the predictions t → bytt and h → byhh,we should be able to simply read off the values of yt and yh. If we cando this then we have access to all of the critical exponents, which wecan compare with experiment.

35.2 The ferromagnetic transition and

critical phenomena

Let’s feed a theory of the ferromagnet through the renormalization groupmachine and extract some physics. A continuum field theory of theferromagnet is given by the Landau–Ginzburg model. This model Vitaly Ginzburg (1916–2009)

is equivalent to (time-independent) d-dimensional Euclidean φ4 theory,which has an action6 6We can choose d to suit whatever di-

mensionality of system we need. Or-dinarily we imagine materials havingd = 3, but it is possible in condensedmatter physics to find materials whicheffectively have d = 2 and d = 1.

SE =

∫ddxLE =

∫ddx

[1

2(∇φ)2 +

m2

2φ2 +

λ

4!φ4

]. (35.7)

Example 35.1

We may see what the Landau–Ginzburg model predicts in mean-field theory wherewe take φ to be uniform across the whole system.7 We then have a free energy 7Note that this approach recreates the

Landau theory of phase transitionsused in Chapter 26.F =

m2

2Φ2 +

λ

4!Φ4, (35.8)

where Φ =R

ddxφ(x). If λ = 0 then the minimum free energy occurs at M = 0 andwe have a paramagnet. If λ > 0 and m2 = a(T − Tc) < 0 then we have minima

at M = [6a(Tc − T )/λ]12 6= 0, giving a ferromagnet. However, mean-field theory

is blind to fluctuations. We will show with the more sophisticated approach of therenormalization group that even if m2 < 0 and λ > 0 a phase transition is notguaranteed due to the influence of these wobbles in the field.


As a warm-up exercise we’ll set the coupling λ = 0 and carry out therenormalization group procedure for the free theory.8

8In this context, the λ = 0 free theoryis called the Gaussian model.

Example 35.2

We choose to work in momentum space, so we write the action in terms of fields inmomentum space in the usual manner:

SE =1

2

Zddp

(2π)dφ(−p)

`p2 +m2

´φ(p). (35.9)

To set the scene, we split the fields up into slow and fast via φ = φf + φs where

φ(p) = φf(p) for Λ/b ≤ p ≤ Λ,

φ(p) = φs(p) for 0 ≤ p ≤ Λ/b.(35.10)

The action in eqn 35.9 is diagonal in p, which means that slow and fast fields willnot be mixed up in the calculation of S. Let’s go through the steps with this freetheory.Step I involves integrating over fast momenta. This just results in a constant con-tribution to the action, which we are free to ignore. We therefore have the integral

Z =

Z

Λ/bDφs e−SE[φs] =

Z

Λ/bDφs e

− 12

R d4p

(2π)d φs(p)(p2+m2)φs(p)

. (35.11)

Step II: We scale the momenta with p′/b = p to obtain an action

SE =b−d

2

Z Λ

0

ddp′

(2π)dφs(−p′/b)

`b−2p′2 +m2

´φs(p

′/b). (35.12)

Step III: We scale the fields via φs(p′/b) = bd−dφ φ′(p′) to obtain

SE =b2(d−dφ)b−d

2

Z Λ

0

ddp′

(2π)dφ′(−p′)

`b−2p′2 +m2

´φ′(p′). (35.13)

Fixing the constant dφ so that the gradient term φ(−p)p2φ(p) is invariant, we haved− 2dφ − 2 = 0 or dφ = (d− 2)/2 and

SE =1

2

Z Λ

0

ddp′

(2π)dφ′(−p′)

`p′2 +m2b2

´φ′(p′). (35.14)

The result of our renormalization procedure is that after scaling, we find that wehave a mass term m′2 = m2b2. That is, since b > 1, the coupling constant m2 getslarger with each rescaling.

Since b > 1, the fact that m′2 = m2b2 means that m2 is a relevant variableand φ2 is known as a relevant operator. This means that it gets larger with eachrenormalization. (In contrast a variable that becomes smaller at each rescaling isknown as an irrelevant variable.) Identifying m2 from our action with the reducedtemperature t, we expect that each rescaling changes m2 by a factor byt . From thisanalysis we can conclude that yt = 2 when λ = 0.

In order to find yh we need to consider the application of an external magneticfield. Still holding λ = 0 we write an action that includes the influence of the field as

SE =

Z

ddx

»1

2(∇φ)2 +

m2

2φ2 − hφ

–

. (35.15)

On rescaling the field-dependent term will become −hbd−dφ φ′(p′). Substituting for

dφ as before we have h′ = hbd+22 , and so yh = (d+ 2)/2. This allows us to write the

exponents predicted from this λ = 0 theory.


Now let’s turn on the interaction. Since we don’t know how to integratethe functional integral for the φ4 theory it should come as little surprisethat neither can we do the integral over the fast modes for φ4 theory. Wetherefore have to treat LI = λ

4!φ4 as a perturbation and do the integral

approximately. Our previous method for working out approximations tothe generating functional Z[J ] using Feynman diagrams may be adaptedto get an approximate answer for the result of removing the fast modes.The only difference in calculating the Feynman diagrams for the renor-malization group analysis is that we integrate over internal momentaonly over the fast modes, up to the cut-off.

In order to work out the lowest order corrections to the couplings weconsider diagrammatic corrections to the theory in order of how manyloops they contain. The lowest order diagrammatic corrections to thetheory therefore involve diagrams with a single loop that correct theself-energy (which determines m2) and the phion–phion vertex function(which determines λ). These one-loop diagrams are shown in Fig. 35.2.

Fig. 35.2 One-loop Feynman dia-grams in φ4 theory.

Example 35.3

We give the results of the important computations here.

The first diagram in Fig. 35.2 corresponds to an amplitude −λ2

R ddp(2π)d

1p2+m2 ,

which yields a contribution δS(2) to the quadratic part of the action given9 by

9Although both of our diagrams havesymmetry factor 2, there is a subtletyhere in working out the prefactors re-flecting the fact that there is more thanone equivalent diagram for each loop.As discussed in Binney, Dowrick, Fisherand Newman, the correct prefactor fora diagram with p loops is 2p!

2p2p.

δS(2) =λ

2

Z Λ

Λ/b

ddp

(2π)d1

p2 +m2

Z

ddxφ2

s

2. (35.16)

Doing I: the integral over fast modes, II: rescaling the momentum and III: rescalingthe fields, we obtain a contribution to the quadratic term

δS(2) = b2»

λΩd

2(d− 2)(1 − b2−d) − m2λΩd

2(d− 4)(1 − b4−d)

– Z

ddxφ2

s

2, (35.17)

which we give in terms of Ωd, which is the volume of a unit sphere in d dimensions.

Measured in units of 2π this volume is given by Ωd = 1(2π)d

2πd/2

Γ(d/2), where Γ(z) =

R∞0 dt e−ttz−1 is the Gamma function. The usual procedure in these calculations is to

work in four-dimensional spacetime where the integrals converge and Ωd=4 = 1/8π.To see the effects in other dimensions we set d = 4 − ǫ, and expand in the smallparameter ǫ. To examine the real-life case of three dimensions we then set ǫ = 1,well outside the presumed range of applicability of this approach. Amazingly, thisprocedure describes reality very well!

Turning to the second diagram in Fig. 35.2, it turns out that we only need toevaluate the amplitude for zero external momentum,10 which means both internal

10See Altland and Simons, Section 8.4for a discussion.

propagators carry the same momentum. We then obtain a contribution δS(4) to thequartic term in the action

δS(4) =3λ2

2

Z Λ

Λ/b

ddq

(2π)d1

(q2 +m2)2

Z

ddxφ4

s

4!

= b4−d3λ2Ωd

2(d− 4)(1 − b4−d)

Z

ddxφ4

s

4!. (35.18)

The result of this expansion is that we must shift the variables according to

m′2 = b2»

m2 +λΩd

2(d− 2)(1 − b2−d) − m2λΩd

2(d− 4)(1 − b4−d)

–

,

λ′ = b4−d»

λ− 3λ2Ωd

2(d− 4)(1 − b4−d)

–

. (35.19)


Now to extract the physics. We’re looking at a system of condensed mat-ter, so we’re interested in a scaling that makes the length scale longer.We therefore set b = el. Since, as stated in the example, the inte-grals only converge for d = 4, we set d = 4 − ǫ and expand in theparameter ǫ. Evaluating the expressions to leading order in ǫ, usingΩ4−ǫ ≈ Ω4 = 1/8π2, we obtain the Gell-Mann–Low equations

dm′2

dl= 2m2 +

λ

16π2(1 −m2),

dλ′

dl= ǫλ− 3λ2

16π2. (35.20)

As shown in Fig. 35.3, these equations have fixed points at (m2, λ) =

(0, 0) and(− ǫ

6 ,16π2

3 ǫ). By expanding the flow around the fixed points

we can use this information to predict the behaviour of the ferromagnet.

Example 35.4

To see the behaviour near the fixed point we employ a first-order expansion to comeup with matrix equations predicting the flow. A Taylor expansion of f(x) about afixed point x∗ yields f(x) = f(x∗)+f ′(x∗)(x−x∗), but since f(x∗) = 0, by definition,we simply need to evaluate the derivatives at the fixed points. Our linearized flowequations take the form

„dm′2/dldλ′/dl

«

=

0

@

∂∂m2

“dm′2

dl

”∂∂λ

“dm′2

dl

”

∂∂m2

“dλ′

dl

”∂∂λ

“dλ′

dl

”

1

A

„m2 −m∗2

λ− λ∗

«

, (35.21)

with the elements evaluated at each of the fixed points.

β

λ16πǫ

30

Fig. 35.3 The β function for the be-haviour of the coupling constant λ.

For our ferromagnet, the matrices for the two sets of fixed points are

W1 =

„2 1

16π2

0 ǫ

«

, W2 =

„2 − ǫ

31

16π2

0 −ǫ

«

. (35.22)

Let’s examine how these equations predict the behaviour of λ. For the fixed pointat (0, 0) (corresponding to W1) we have λ′ = ǫλl. If ǫ > 0 (that is, we are examiningthe physics in less than four dimensions since d = 4 − ǫ) then λ increases as we lookat larger scales. In that case λ is a relevant variable and will be important to thephysics. Moving on to the other fixed point (−ǫ/6, 16π2ǫ/3), we see that close toλ∗ = 16πǫ/3 we have λ′ = −ǫ(λ− λ∗)l. In this case λ is relevant (i.e. gets larger onrescaling) for λ < λ∗ but irrelevant (i.e. gets smaller on rescaling) for λ > λ∗. Wewill see the influence of this on the flow of the system next.

λ

m2W1

W2

A

B

C

D

FM PM

Fig. 35.4 Renormalization group flowfor a ferromagnet. FM labels ‘ferro-magnet’, PM labels ‘paramagnet’. This information, along with an analogous analysis of how m2 behaves,

allows us to draw the phase diagram of the ferromagnet, as shown1111The predictions of the previous ex-ample are seen in the flows close to W2,where we see λ decrease or increase de-pending, respectively, on whether westart above (flow B) or below (flow C)the value λ∗ = 16πǫ/3. Flows A andD are clearly more complex and rely onthe magnitude and behaviour of m2.

in Fig. 35.4 for ǫ > 0. The fixed point at W1 is repulsive and thatat W2 is mixed: some flows are directed towards it, some away fromit. Let’s return to the physics of the system: at high temperatures wehave a disordered, paramagnetic phase; at low temperatures, we suspectfrom our mean-field analysis that there exists an ordered, ferromagneticphase. Relating this to our model, we writem2 = a(T−Tc), so flows thatsend m2 to large, positive values predict that the system will end in theparamagnetic phase. Conversely, flows that send m2 to negative values


predict that the system becomes ferromagnetic. By examining the flowswe ask what influence the fluctuations have on the phase transition.Recall that our mean-field analysis makes a prediction that any systemthat starts with a negative m2 will be a ferromagnet and any systemthat starts with a positive m2 will be paramagnetic.

The result of our renormalization group analysis is that flows startingwith positive values of m2 are always driven towards the paramagneticphase, whatever the value of λ, in agreement with the mean-field treat-ment. However, starting with a negative value of m2 doesn’t guaranteea ferromagnetic ground state! The fact that we have fluctuations, whosestrength is given by λ, may force the system to be a paramagnet. Thisis the case for the flow B in Fig. 35.4, where starting with a significantenough λ is enough to drag the system towards paramagnetism. On theother hand, flows starting with sufficiently small λ and sufficiently neg-ative m2 (such as paths C and D) eventually flow off to the left, pushingthe system towards ferromagnetism.

Example 35.5

So much for our qualitative analysis of the phase diagram. Let’s extract yt andyh and make some quantitative predictions regarding the thermodynamics of theferromagnet. One of the uses of the linearized flow matrices is that they may beused to read off the scaling constant yt. Since we have set b = el, we have thateach rescaling goes as m′2 = elytm2. Close to the fixed point, where our linearizedequations are valid this implies that ∆m′2 = yt(m2 − m∗2)∆l. We can thereforeread off from the matrix describing the flow around our nonzero fixed point thatyt = 2 − ǫ

3. The one-loop correction doesn’t change yh, which continues to be given

by yh = (d+ 2)/2 = (6 − ǫ)/2 just as it was from the analysis of the λ = 0 case.With these two pieces of information, we obtain all six of the critical parameters:

α = ǫ6, β = 1

2− ǫ

6, γ = 1 + ǫ

6, δ = 3 + ǫ, ν = 1

2+ ǫ

12, η = 0. (35.23)

These results can be used to make an estimate of the critical exponents of the three-dimensional Ising model, as shown in the following table.12 12For a discussion of the comparison

see the books by Kaku, Zinn-Justinand Binney et al. See also Peskin andSchroeder, Chapter 13.α β γ δ ν η

ǫ = 0 (d = 4, mean field) 0 12

1 3 12

0ǫ = 1 O(ǫ) 0.167 0.333 1.167 4 0.583 0ǫ = 1 O(ǫ2) 0.077 0.340 1.244 4.462 0.626 0.0193D Ising 0.110 0.327 1.237 4.789 0.630 0.036

2D Ising (exact) 0 18

74

15 1 14

The first line of this table gives the mean-field results, exact for d = 4 and so canbe obtained from eqn 35.23 by setting ǫ = 0. The second line gives a crude, butsurprisingly accurate, estimate for d = 3 by setting ǫ = 1. One can derive expressions

for the critical exponents to second order in ǫ [e.g. β = 12− ǫ

6+ ǫ2

162+ O(ǫ3)] and

these yield the estimates in the third line of the table. These approach the currentlyaccepted values13 (fourth line) which agree well with experimental values. In the last

13Taken from A. Pelissetto and E. Vi-cari, Phys. Rep. 368, 549 (2002),which reviews a large number of theo-retical studies using various techniquesand also summarises experimental re-sults.

line we list the exact results for the two-dimensional Ising model for comparison.


Chapter summary

• This chapter has provided a renormalization group analysis of theferromagnetic transition following Widom’s hypothesis for the scal-ing of the free energy.

Exercises

(35.1) (a) Using Widom’s hypothesis in the form f(t, h) =

tdyt f(h/t

yhyt ), show that critical parameters are

given in terms of the scaling parameters as follows

α = 2 − dyt, β = d−yh

yt, γ = 2yh−d

yt. (35.24)

(b) We now redefine the scaling function slightly so

that it reads f(t, h) = hd

yh g(h/tyhyt ) where g(z) =

z− d

yh f(z). Use this form of the function to showthat δ = yh

d−yh.

(c) From these results prove the following relations:

α+ 2β + γ = 2 Rushbrooke’s law,α+ β(δ + 1) = 2 Griffith’s law.

(35.25)

(d) Assuming the correlation function behaves as

Gc(x, t) = f“

xt2−α

d

”

/xd−2+η, argue that we re-

quire νd = 2 − α (known as Josephson’s law) anduse this to show ν = 1

yt.

(e) Finally use Fisher’s law (2 − η)ν = γ to find η.

(35.2) (a) Starting from eqns 35.19, verify the Gell-Mann–cLow equations in eqn 35.20.Hint: let b = 1 + ρ with ρ small. Then ln b ≈ ρand one may expand to order ρ to capture the flowequations.(b) Verify the positions of the fixed points.(c) Verify the matrices in 35.22 and the behaviourof the flow near the fixed points.

(35.3) What is the behaviour of λ for the case of dimen-sions d > 4?

Part IX

Putting a spin on QFT

This part extends our treatment of quantum field theory to spinors andallows us to treat the electron spin. These ideas set us on a path thatinevitably arrives at quantum electrodynamics.

• The Dirac equation for an electron is introduced in Chapter 36.We describe the γ-matrices necessary for formulating this and alsothe concepts of chirality and helicity. The basis states are calledspinors and we describe two types: Dirac spinors and Weyl spinors.

• We show how spinors transform under rotations and boosts inChapter 37. We also demonstrate that the parity operator turns aleft-handed spinor into a right-handed spinor and vice versa.

• In Chapter 38 we apply the machinery of quantization to the Diracequation and derive the Noether current. This allows us to writedown the fermion propagator.

• Arguably the most important achievement of quantum field theoryis quantum electrodynamics (QED), which describes the interac-tion between electrons and photons, and this theory is presentedin outline in Chapter 39.

• Three examples of ideas presented in the previous chapter arefleshed out in Chapter 40 via three examples. These are Ruther-ford scattering, the Mott formula and Compton scattering. Webriefly discuss the useful feature of crossing symmetry.

• Some important consequences of QED arise when you consider pro-cesses involving electron–positron loops. These include the renor-malization of the electron charge due to vacuum polarization andthe deviation of the g-factor from g = 2. These ideas are exploredin Chapter 41.

36 The Dirac equation

36.1 The Dirac equation 322

36.2 Massless particles: left- andright-handed wave functions

323

36.3 Dirac and Weyl spinors 327

36.4 Basis states for superposi-tions 330

36.5 The non-relativistic limit ofthe Dirac equation 332

Chapter summary 334

Exercises 334

I have an equation, do you have one too?Paul Dirac (1902–1984), on meeting Richard Feynman

Our story so far: as a single-particle equation of motion, the Klein–Gordon equation has some serious problems. The first is negative energystates and the second is negative probabilities. Paul Dirac sees that theroot of these problems lies in the fact that the Klein–Gordon equationcontains a second-order time derivative as part of ∂2. While staring intothe fire in the senior common room of St John’s College, Cambridge,Dirac realized that there’s a way to construct an equation of motionthat’s first order in the time derivative.

After seeing 2001: A Space Odyssey,Paul Dirac was so impressed that, twodays later, he attended three consecu-tive screenings of the film in the cinema.His favourite television programme wasthe Cher show.

In this chapter we examine Dirac’s equation. We will treat it as anequation of motion for single particles and examine the form of thesolutions and how features of these solutions describe the physical prop-erties of real particles. Although our experience with the Klein–Gordonequation has taught us that, ultimately, a single-particle treatment ofrelativistic quantum mechanics will not provide an adequate descriptionof Nature, the experience gained with the solutions of Dirac’s equationwill carry over to the world of Dirac fields1 whose excitations are the1This is the subject examined in the

two chapters following this one fermions we observe in our Universe.

36.1 The Dirac equation

In order to avoid all of the problems with the Klein–Gordon equation,you might wonder whether we could take the square root of the gutsof that equation (∂2 + m2) and then apply it to a wave function. Anaive guess might be that if we could treat things like ∂2 like ordinaryalgebraic symbols then we could write

(∂2 +m2) = (√∂2 + im)(

√∂2 − im). (36.1)

Unfortunately,√∂2 isn’t defined. How do we take the square root of the

∂2 operator? Dirac found the answer was to define a new four-vectorγµ, whose components have the properties

(γ0)2 = 1, (γ1)2 = −1, (γ2)2 = −1, (γ3)2 = −1. (36.2)

Furthermore, these components aren’t ordinary numbers; they anticom-mute. That is to say that for µ 6= ν they have the intriguing property

36.2 Massless particles: left- and right-handed wave functions 323

γµγν + γνγµ = γµ, γν = 0. Taken with eqn 36.2, the properties ofthese γ’s may be summarized in the compact form2 2Equation 36.3 defines what is known

as a Clifford algebra, named afterWilliam Kingdon Clifford (1845–1879).See Penrose, Chapter 11 for more de-tails.

γµ, γν = 2gµν . (36.3)

We use the gammas to cope with the square root in√∂2, via a strange

looking new symbol a = γµaµ, which is pronounced ‘a-slash’. We canthen say that

∂2 = (γµ∂µ)

2 =

(γ0 ∂

∂t+ γ1 ∂

∂x+ γ2 ∂

∂y+ γ3 ∂

∂z

)2

. (36.4)

Expanding this bracket and using the all-important anticommutationrelations we find that ∂2 = ∂2. Substituting ∂2 for ∂2 in the Klein–Gordon equation does indeed allow us to factorize

(∂2 +m2) = (∂2 +m2) = (∂ − im)(∂ + im). (36.5)

Now let’s just take the plus-sign bracket and treat it as an operatoracting on the wave function ψ(x),

(∂ + im)ψ(x) = 0, (36.6)

and this may be tidied up to give us the famous Dirac equation:

(iγµ∂µ −m)ψ(x) = 0. (36.7)

We note first that the Dirac equation may also be written as (p−m)ψ =0, where p = γµ(i∂µ) and we recognize the energy-momentum operator3 3Recall that pµ = i∂µ = i( ∂

∂t,∇).

pµ = i∂µ.We’ll soon see that the dispersion relation of the particles described

by the Dirac equation of motion continues to be E2p = (p2 +m2), which

still admits negative energy states.4 4It does successfully eliminate the neg-ative probability problem. Of course,we’ve already found a way to deal withthe negative energy states of the Klein–Gordon equation: Feynman’s prescrip-tion of interpreting them as positive en-ergy antiparticles, so the negative en-ergy states are not a deal-breaker. Infact, the Dirac equation will turn outto provide the equation of motion fora type of field whose excitations arefermions such as the electron. As such,the results from this temporary forayinto wave mechanics should be treatedas provisional.

To get to Dirac’s equation, we have had to introduce the mysterious,anticommuting γµ objects. These will have a profound influence on thephysics described by the Dirac equation because the simplest way tosatisfy the commutation relations is to represent the γ’s as four-by-fourmatrices. This means that the wave function ψ(x) described by theDirac equation is a special four-component wave function. Why fourcomponents? What could they represent? To make progress, we’ll usean explicit form for the matrices and examine some illuminating specialcases.

36.2 Massless particles: left- and

right-handed wave functions

There is no unique way of writing the γµ matrices and so we have achoice about how to represent them. One set of matrices that satisfies


the commutation relations is

γ0 =

0 0 1 00 0 0 11 0 0 00 1 0 0

, γ1 =

0 0 0 10 0 1 00 −1 0 0−1 0 0 0

,

γ2 =

0 0 0 −i0 0 i 00 i 0 0−i 0 0 0

, γ3 =

0 0 1 00 0 0 −1−1 0 0 00 1 0 0

.

(36.8)

Writing the γ’s in this form is known as the chiral representation,55This is sometimes called the Weylrepresentation. for reasons that will become apparent. Writing out lots of four-by-

four matrices is a tedious business, so to save on writing it’s useful towrite these four-by-four matrices out as two-by-two matrices, where eachelement is itself a two-by-two matrix.

Example 36.1

There are several ways to simplify the notation. Defining I =

„1 00 1

«

, we notice

that we can express eqn 36.8 in a 2 × 2 form:

γ0 =

„0 II 0

«

, γ1 =

„0 σx

−σx 0

«

,

γ2 =

„0 σy

−σy 0

«

, γ3 =

„0 σz

−σz 0

«

,(36.9)

where the Pauli matrices have their usual definitions.It turns out that γµ transforms as a four-vector. Motivated by this, we write

γµ = (γ0, γ1, γ2, γ3) = (γ0,γ) and then

γ0 =

„0 II 0

«

, γ =

„0 σ

−σ 0

«

, (36.10)

where σ = (σ1, σ2, σ3). Finally, in a desire to write as little as possible, we defineσµ = (I,σ) and σµ = (I,−σ) to obtain66If, at any point in the future, any

of this notation seems overly compact,then it may sometimes be helpful towrite things out in full.

γµ =

„0 σµ

σµ 0

«

. (36.11)

Substituting these γ’s into the Dirac equation gives

(γ0p0 − γ · p −m)ψ(x) = 0, (36.12)

where p0 = i∂0 and p = −i∇. We’ll make use of a handy two-by-twomatrix form for the equations, where the four-component wave function

ψ is written

(ψL

ψR

), where ψL and ψR are, themselves, two-component

column matrices.7 In this form the Dirac equation reads7The meaning of these subscripts willbe explained shortly. [(

0 p0

p0 0

)−(

0 σ · p−σ · p 0

)−(m 00 m

)](ψL

ψR

)= 0,

(36.13)

36.2 Massless particles: left- and right-handed wave functions 325

which can be rewritten as

(p0 − σ · p)ψR = mψL,

(p0 + σ · p)ψL = mψR. (36.14)

Now let’s further simplify things by considering the special case of amassless particle (which obviously has m = 0). We then obtain

(p0 − σ · p)ψR = 0,

(p0 + σ · p)ψL = 0, (36.15)

which implies that a four-component eigenstate ψ =

(ψL

ψR

)splits into

two two-component pieces, ψL and ψR, that don’t get mixed up by theequations of motion. It’s as if the Dirac equation is telling us thatwe have two sorts of massless Dirac particles in Nature: left-handed

particles described by wave functions ψL and right-handed particles

described by wave functions ψR. In the chiral representation, left-handedfunctions live in the upper two slots of the Dirac wave function and right-handed functions live in the lower slots.

Example 36.2

We need to be slightly more precise with our definition of left- and right-handed wavefunctions. To do this we define the chirality operator as

γ5 = iγ0γ1γ2γ3

=

„−I 00 I

«

, (36.16)

where the final equality holds in the chiral representation only. A wave function„ψL

0

«

is an eigenstate of γ5 with eigenvalue −1. An entirely right-handed wave

function

„0ψR

«

is then an eigenstate with eigenvalue +1. The Dirac equation

therefore predicts the existence of two types of Dirac wave function in the Universe:left-handed functions with chirality −1 and right-handed functions with chirality +1.

If we want to extract the left- and right-handed parts of a wave function we maydefine projection operators via

PL = 1−γ5

2, PR = 1+γ5

2, (36.17)

which will project out the desired part.

We conclude that since ψL and ψR are eigenstates of the massless Diracequation, then a free and massless left-handed particle may never changeinto a right-handed particle. Moreover, using the fact that, for masslessparticles, the eigenvalue of p0 is Ep = |p|, our eqns 36.15 give us

σ · p|p| ψR = ψR, (36.18)

σ · p|p| ψL = −ψL. (36.19)


This means that the massless states ψR and ψL, defined as being chiralityeigenstates, are also eigenstates of the helicity operator

h =σ · p|p| , (36.20)

with eigenvalues +1 and −1 respectively.From this discussion it looks rather like chirality and helicity, while

being represented by different operators, tell us the same thing. How-ever, one needs to be careful here: helicity and chirality are identicalfor massless particles, but are, in general, not the same. Clearly helic-ity tells us whether the particle’s spin and momentum are parallel orantiparallel. Helicity therefore depends on the frame in which it is mea-sured. In the case of massive particles it is possible to describe a particlewith positive helicity and then to boost to a frame where the particle’smomentum is reversed but its spin is unchanged, thereby reversing itshelicity.

If we now consider massive particles again then the equations of mo-tion revert back to eqn 36.14. This shows us that massive Dirac particlesinvolve both left- and right-handed wave functions coupled by the par-ticle’s mass. We can think of massive Dirac particles oscillating backand forth in time between left- and right-handed at a rate determinedby their mass. This is most easily seen by looking at massive Diracparticles at rest, where we have i∂0ψR = mψL and i∂0ψL = mψR.

Example 36.3

We’ll find the dispersion relation for massive Dirac particles. Assuming that ψL andψR are eigenstates of momentum and energy, we may eliminate ψL from eqns. 36.14,to give us

(p0 + σ · p)(p0 − σ · p)ψR = m2ψR. (36.21)

Multiplying this out givesˆ(p0)2 − p2

˜= m2, (36.22)

which demonstrates that Ep = ±(p2 +m2)12 and we still have negative energy states

as claimed above. Another way of saying this is that the eigenvalue of p0 could betaken to be positive or negative and we still get the same dispersion.

Finally we turn to antiparticle solutions of the Dirac equation. These arethe solutions which appear to have negative values of energy.8 Writing8Of course we know that this is a bad

way of thinking about them since phys-ical antiparticles have positive energy.We apologize for the brief lapse intobad habits.

energies as E = −|p0|, we find, for massless antiparticles, that

(−|p0| − σ · p)ψR = 0(−|p0| + σ · p)ψL = 0

antiparticles, (36.23)

which impliesσ·p|p| ψR = −ψRσ·p|p| ψL = ψL

antiparticles. (36.24)

This is to say that right-handed antiparticles have negative helicity andleft-handed antiparticles have positive helicity. Note that this is theother way round compared to the particle case.


To summarize our progress so far. We’ve discovered that the four-component solutions to the Dirac wave function have left-handed partsin the upper two slots and right-handed parts in the lower two slots andcan be written

ψ(x) =

(ψL(x)ψR(x)

). (36.25)

The four-component wave functions are known as Dirac spinors andare quite different to other objects that we have so far encounteredsuch as scalars and four-vectors. The two-component objects ψL(x) andψR(x) are themselves known as Weyl spinors. Spinors are best defined Hermann Weyl (1885–1955)

in terms of their transformation properties, which we turn to in the nextchapter. In the meantime we will examine them further as solutions toDirac’s equation.

36.3 Dirac and Weyl spinors

We now investigate what’s hidden away in ψL(x) and ψR(x). Let’s workon the positive and negative energy solutions separately. Since physicalparticles and antiparticles really have positive energies, we’ll conform toconvention and call these the positive and negative frequency solutions.The positive frequency solutions will describe Fermi particles, while thenegative frequency solutions will describe Fermi antiparticles. It’s easyto show that a set of solutions to the Dirac equation may be written

Particles: u(p)e−ip·x =

(uL(p)uR(p)

)e−ip·x,

Antiparticles: v(p)eip·x =

(vL(p)vR(p)

)eip·x.

(36.26)

Note that the negative frequency solutions are defined such that p0 =E > 0 so that antiparticles will have positive energies. The four-component objects9 u(p) and v(p) are momentum space Dirac spinors 9Just as a polarization vector ǫµ(p) car-

ries the spin information for the pho-ton, the spinors u(p) and v(p) carry thespin information of the fermion.

and themselves satisfy the momentum space Dirac equation:

(p−m)u(p) = 0,

(−p−m)v(p) = 0. (36.27)

The left-handed [uL(p) and vL(p)] and right-handed [uR(p) and vR(p)]parts are momentum space Weyl spinors.

We start by considering the particle solution at rest where pµ = (m, 0).In this case the Dirac eqn 36.27 becomes

(−m mm −m

)(uL(p0)uR(p0)

)= 0, (36.28)

with solutions

u(p0) ≡(uL(p0)uR(p0)

)=

√m

(ξξ

), (36.29)


where ξ =

(ξ1ξ2

)is a two-component column vector that we will choose

to normalize10 according to ξ†ξ = 1. Note that ξ is also often called a10The factor√m is included in

eqn 36.29 for later convenience. It re-sults in uu = 2m where u ≡ u†γ0 isdefined later (eqn 36.41). For mass-less particles one includes a factor

pEp

instead. See Peskin and Schroeder,Chapter 3, for more details.

spinor. The argument may be repeated for antiparticle solutions withthe result

−(m mm m

)(vL(p0)vR(p0)

)= 0, (36.30)

leading to

v(p0) =√m

(η−η

), (36.31)

where we write η =

(η1η2

)for the antiparticle spinor. Notice that the

solutions to the Dirac equation have two degrees of freedom. These arethe two components of ξ (for the positive frequency solutions) or η (forthe negative frequency solutions). It turns out that it is ξ and η thattell us about the spin of the particle.

We know the form of the operator for the spin of a particle at restis given by S = 1

2σ. The object on which this operator acts is η or ξ.

A particle with spin up along the z-axis therefore has ξ =

(10

)and

one with spin down along the z-axis has ξ =

(01

). We can use these

states as a basis to represent arbitrary spin states.What about antiparticles? We define an antiparticle at rest with

physical spin up along the z-direction to have η =

(01

), so that Szη =

− 12η. An antiparticle with physical spin down along z has Sz = 1

2 and

η =

(10

)so that Szη = 1

2η. Admittedly this is rather confusing! The

properties of particles and antiparticles are summarized in the table atthe end of the chapter.

Example 36.4

A positive helicity antiparticle with momentum directed along the positive z-axis

must have ση = +η. It must also, therefore, have η =

„10

«

, which corresponds to

physical spin down.

What about the spinor for a particle in the general case of particles thataren’t at rest? We quote the result here (and will justify it in the nextchapter). For a particle or antiparticle with momentum pµ the spinorsare given by

u(p) =

( √p · σ ξ√p · σ ξ

), v(p) =

( √p · σ η

−√p · σ η

). (36.32)

In eqn 36.32 we are again using the no-tation σ = (I,σ) and σ = (I,−σ) andwe understand that in taking the squareroot of a matrix we take the positiveroot of each eigenvalue.


Example 36.5

We will evaluate the spinor for a particle with pµ = (Ep , 0, 0, |p|) and ξ =

„01

«

.

Evaluating the left-handed part uL(p) =√p · σξ which lives in the upper two slots

of the spinor, we have

√p · σ =

„Ep − |p| 0

0 Ep + |p|

« 12„

01

«

=

„ pEp − |p| 0

0pEp + |p|

«„01

«

,

(36.33)

which gives uL(p) =√p · σξ =

pEp + |p|

„01

«

.

The analogous result for the right-handed, lower slots of u(p) follows by a similar

argument with the result that uR(p) =√p · σξ =

pEp − |p|

„01

«

.

We obtain the final result for the spinor

u(p) =

0

BB@

pEp + |p|

„01

«

pEp − |p|

„01

«

1

CCA. (36.34)

The general expression for the spinors in eqn 36.32 allows us to explorethe helicity and chirality of particles and antiparticles in more detail inthe next example.

Example 36.6

We will look at particles and antiparticles in the ultra-relativistic limit, obtained bygiving the particles a big boost. Consider a particle with spin up along the z-axis andmomentum |p| along +z. This particle, shown in Fig. 36.1(a), has helicity h = 1,corresponding to physical spin parallel to the momentum. This is described by aspinor

u(p) =

„uL(p)uR(p)

«

=

0

BB@

pEp − p · σ

„10

«

pEp + p · σ

„10

«

1

CCA. (36.35)

This leads us to

u(p) =

0

BB@

pEp − |p|

„10

«

pEp + |p|

„10

«

1

CCA

big−−−−→boost

p2Ep

0

BB@

0010

1

CCA, (36.36)

which has uL = 0 and uR =p

2Ep

„10

«

.

e− p(a)

e− p

(b)

e+ p

(c)

e+ p

(d)

Fig. 36.1 Helicity states of particlesand antiparticles. (a) A highly rela-tivistic electron with spin up along thez-axis and momentum along z, whichhas h = 1. (b) The same with spindown, giving helicity h = −1. (c) Ahighly relativistic positron with phys-ical spin up along the z-axis and mo-mentum along z, giving a helicity h =−1 (d) The same with physical spindown along z and h = 1.

We can carry out the same analysis for a negative helicity particle [Fig. 36.1(b)],which has spin down along the z-axis and momentum along +z. This has a spinor

u(p) =

0

BB@

√E − p · σ

„01

«

√E + p · σ

„01

«

1

CCA

=

0

BB@

pE + |p|

„01

«

pE − |p|

„01

«

1

CCA


p2Ep

0

BB@

0100

1

CCA,

(36.37)

which has uL =p

2Ep

„01

«

and uR = 0.


Now for the antiparticle states. Again we take momentum to be along +z. An

antiparticle with physical spin up along the z-axis [Fig. 36.1(c)] has

„01

«

. It has

a helicity of +1 (opposite to what a particle would have) and has a spinor

v(p) =

0

BB@

√E − p · σ

„01

«

−√E + p · σ

„01

«

1

CCA

=

0

BB@

0pE + |p|

0

−pE − |p|

1

CCA


p2Ep

0

BB@

0100

1

CCA.

(36.38)

This has vL =p

2Ep

„01

«

and vR = 0.

Finally, an antiparticle with spin down along the z-axis and momentum along +z[Fig. 36.1(d)] has helicity −1 and a spinor

v(p) =

0

BB@

√E − p · σ

„10

«

−√E + p · σ

„10

«

1

CCA

=

0

BBBB@

pE − |p|

0

−pE + |p|0

1

CCCCA


−p

2Ep

0

BB@

0010

1

CCA,

(36.39)

which has vL = 0 and vR = −p

2Ep

„10

«

.

Notice that highly relativistic particles with positive helicity have right-handedspinors, while relativistic antiparticles with positive helicity have left-handed spinors.

Imagine that there was an interaction that only coupled to left-handedparticles. In the ultra-relativistic limit, where chirality and helicity be-come identical, we’ve demonstrated that only negative helicity particlesand positive helicity antiparticles would interact. Amazingly, this is in-deed the case for the weak interaction, which describes a field Wµ(x)which only couples to left-handed Dirac particles!11 This means that

11Neutrinos only exist as left-handedparticles. They are (at least approx-imately) massless, which means thatonly negative helicity neutrinos andpositive helicity antineutrinos have everbeen observed. only a left-handed fermion may emit a W± particle. For a massive elec-

tron, which oscillates between being left- and right-handed, it may onlyemit a W− particle (turning into a neutrino as it does so) when it isleft-handed as shown in Fig. 36.2.

t

R

L

R

L

R

L

R

L

R

L

W−

R

L

R

L

R

L

R

L

R

L

(a) (b)

Fig. 36.2 Parity violation in the weakinteraction. The electron oscillates be-tween left- and right-handed chiralities.It may only emit a W− particle whenit is left-handed.

To summarize this section: the information content of a spinor tells usabout the spin of a particle through the object ξ (or η for an antiparticle)which has two degrees of freedom. If we know ξ and the momentum ofthe particle then all four components of the spinor (two left-handed partsand two right-handed parts) are completely determined.

36.4 Basis states for superpositions

We turn to the orthogonality of the Dirac spinors. This property is usefulif we are to use the spinors as a basis in order to express arbitrary wavefunctions as linear superpositions which contain spinor parts. Spinorscarry information about the spin of the particle, so we want one spinorto be orthogonal to another if they describe different spin states. (Theorthogonality of momentum states will be handled, as usual, by thee−ip·x factors.)

Now, it’s straightforward to show12 that12See Exercise 36.3.

u†(p)u(p) = 2Epξ†ξ, v†(p)v(p) = 2Epη

†η. (36.40)

36.4 Basis states for superpositions 331

However this isn’t going to be very useful. The reason is that theseexpressions are not Lorentz invariant: the combinations u†u and v†vare proportional to Ep, which is frame-dependent. To make a Lorentzinvariant quantity we define the adjoint or bar of a spinor as

u(p) = u†(p)γ0. (36.41)

This is useful as a Lorentz invariant quantity can now be formed via

u(p)u(p) = 2mξ†ξ. (36.42)

Now for the orthogonality: let’s also choose a basis in which to describe

the spin of a particle. We usually choose the basis ξ1 =

(10

)and

ξ2 =

(01

)such that ξs†ξr = δsr. The result of all of this is that we

can now write an expression for a Dirac spinor which is suitable for useas a basis state as

us(p) =

( √p · σξs√p · σξs

), (36.43)

where s = 1 or 2 and we have

us(p)ur(p) = 2mξs†ξr = 2mδsr. (36.44)

We are now able to express an arbitrary particle solution of the Diracequation as an integral over momentum and a sum over spin states asfollows:

ψ−(x) =

∫d3p

(2π)32

1

(2Ep)12

2∑

s=1

aspus(p)e−ip·x, (36.45)

where asp is an amplitude here.In the case of the antiparticle spinor we pick up a minus sign:

vs(p)vr(p) = −2mηs†ηr = −2mδsr, (36.46)

and the antiparticle wave function is written

ψ+(x) =

∫d3p

(2π)32

1

(2Ep)12

2∑

s=1

b∗spvs(p)eip·x, (36.47)

where the components have amplitude b∗sp. Finally we note that the u’sand v’s are orthogonal:

ur(p)vs(p) = vr(p)us(p) = 0. (36.48)

We summarize all of our definitions in the master table below.

Spinor physical physical normalizationspin up spin down

particle

( √p · σξ√p · σξ

) (10

) (01

)us(p)ur(p) = 2mδsr

antiparticle

( √p · ση

−√p · ση

) (01

) (10

)vs(p)vr(p) = −2mδsr


Example 36.7

One of the most useful expressions in subsequent calculations is the sum

2X

s=1

us(p)us(p) =X

s

„ √p · σξs√p · σξs

«`ξs†

√p · σ ξs†

√p · σ

´. (36.49)

We use the fact that

2X

s=1

ξsξs† =

„10

«`

1 0´

+

„01

«`

0 1´

=

„1 00 1

«

, (36.50)

giving us a four-by-four matrix

2X

s=1

us(p)us(p) =

„ √p · σ√p · σ √

p · σ√p · σ√p · σ√p · σ √

p · σ√p · σ

«

=

„m p · σp · σ m

«

. (36.51)

Finally we use our γ matrices to write this four-by-four object asX

s

us(p)us(p) = γ · p+m. (36.52)

It’s left as an exercise to prove the analogous expression for antiparticlesX

s

vs(p)vs(p) = γ · p−m. (36.53)

So why care about these spin sums?We’ll see a little later that when we ac-tually descend from our ivory tower tocalculate something that people actu-ally measure a certain amount of av-eraging is going to be required. Thisis because the incoming and outgoingpolarizations of our particles are usu-ally unknown. In that case the usualinstruction is to average over initialstates and sum over final states. Thiswill lead to sums over the spin stateswhere relaxations like

P

s us(p)us(p) =

p+m come in very useful.

36.5 The non-relativistic limit of the Dirac

equation

It is often stated that the Dirac equation is required to describe quantummechanical spin. This isn’t quite true as there’s another equation thatdoes this job perfectly well at low energy: the Pauli equation given by

Hψ =(σ · p)2

2mψ. (36.54)

The Pauli equation emerges naturally as the non-relativistic limit of theDirac equation.

Example 36.8

As in Chapter 12, we extract the non-relativistic limit by factorizing out the largemass contribution through a wave function ψL(t,x) = φL(t,x)e−imt. From Dirac’sequation (p−m)ψ = 0 we obtain

„−m E − σ · p

E + σ · p −m

«„φL(t,x)e−imt

φR(t,x)e−imt

«

= 0, (36.55)

where E = i∂0. Noting that Eφa(t,x)e−imt = e−imt(m + E)φa(t,x) (where a = Lor R) we obtain

−mφL + (m+ E − σ · p)φR = 0,

(m+ E + σ · p)φL −mφR = 0. (36.56)

36.5 The non-relativistic limit of the Dirac equation 333

Rewriting the second of these equations we produce

φR =

1 +E

m+

σ · pm

!

φL, (36.57)

allowing us to eliminate φR from the first equation, so that

2E +E2

m− (σ · p)2

m

!

φL = 0. (36.58)

Recognizing that the E2 term may be neglected in comparison with the other termsat low energy, we end up with Pauli’s equation

EφL =(σ · p)2

2mφL. (36.59)

The same equation applies for φR too, as can easily be shown.13 13See Exercise 36.8.

Example 36.9

The previous example gives us enough ammunition to show that the Pauli equation(and by extension, the Dirac equation) implies that the g-factor of the electron isg = 2. We may include the electromagnetic interaction in the Pauli equation bymaking the replacements14 p → p− qA and E → E − qA0. The result is 14This is equivalent to the replacement

∂µ → Dµ that we examined in Chap-ter 14.(E − qA0)φ =

[σ · (p− qA)]2

2mφ. (36.60)

Multiplying out the right-hand side and using the identity σiσj = δij + iεijkσk wefind15 15This is made slightly tricky by the

fact that p and A don’t commute. Ifin doubt, write p = −i∇ and remem-ber that the differential operator actson everything to its right.

(E − qA0)φ =(p− qA)2

2mφ− q

2mσ · Bφ, (36.61)

where we have identified B = ∇ × A. The expected interaction Hamiltonian of amagnetic moment with an external B-field may be written

H = −µ · B ≡ −g q

2mS · B. (36.62)

Comparing this with − q2m

B · σ from eqn 36.61 and identifying the spin operator as

S = 12σ we conclude that g = 2.

The prediction that g = 2 (rather than g = 1) was historically an important onein confirming the validity of the Dirac equation. However, it is found experimentallythat g = 2.002319304.... The reason for this discrepancy is that it is not appropriateto treat Dirac’s equation as the equation of motion for a single particle as we havedone here. In fact, the Dirac equation is correctly interpreted as providing the equa-tion of motion for fermionic quantum fields. As we shall see in Chapter 41 this willlead to a prediction that agrees with experiment to a truly remarkable extent.

In this chapter we’ve encountered a new object, the spinor, to add tothe scalar and vector objects that we’ve considered before. We have,however, treated the Dirac equation as a single-particle equation. Weknow that in reality particles are excitations in quantum fields. In orderto be able to define spinor fields we need to know the all-importanttransformation properties of these spinor fields. In the next chapter weturn to the Lorentz transformation properties of these objects.


Chapter summary

• The Dirac equation (p−m)ψ = 0 results from an attempt to takethe square root of the Klein–Gordon equation. The four compo-nents are required to represent spin and reflect symmetry underthe parity operation.

• The operator ‘p-slash’ is p = γµ(i∂µ) where γµ are the gamma

matrices. In the chiral representation γµ =

(0 σµ

σµ 0

)where

σµ = (I,σ), σµ = (I,−σ) and γ5 = iγ0γ1γ2γ3.

• Massless solutions of the Dirac equation are also the eigenstates ofthe chirality operator, and are known as left- and right-handed so-lutions. Left- or right-handed states with mass are not eigenstatesof the Dirac equation because the mass term couples them. Wecan think of Dirac particles as oscillating in time between beingleft- and right-handed.

• There are positive and negative energy eigenstates of the Diracequation. These give rise to particles and antiparticles.

• The non-relativistic limit of the Dirac equation yields the Pauliequation and predicts g = 2.

Exercises

(36.1) An illustration of the reason for anticommutationand spin(a) Show that the Dirac equation can be recast inthe form

i∂ψ

∂t= HDψ, (36.63)

where HD = α · p + βm and find α and β in termsof the γ matrices.(b) Evaluate H2

D and show that for a Klein–Gordondispersion to result we must have:(i) that the αi and β objects all anticommute witheach other; and(ii) (αi)2 = (β)2 = 1.This provides some justification for the anticommu-tation relations we imposed on the γs.(c) Prove the following commutation relations

(i)h

H, Lii

= i(p × α)i where L = x × p.

(ii)h

H, Sii

= −i(p × α)i where S = 12Σ and we

define Σ = i2γ × γ.

This tells us that the Dirac Hamiltonian doesn’tcommute with the orbital angular momentum op-erator L, mandating the existence of another formof angular momentum (i.e. the spin S) so that Hcommutes with the sum J = L + S.

(36.2) (a) Using the facts that γi† = −γi and γ0† = γ0,take the Hermitian conjugate of the Dirac equationto show that we obtain the non-covariant form:

−i∂0ψ†γ0 + i∂iψ

†γi −mψ† = 0. (36.64)

(b) Show that to restore covariance we may multi-ply through by γ0 which results in

i∂µψγµ +mψ = 0. (36.65)

(36.3) (a) Verify eqn 36.40.(b) Verify eqn 36.42.(c) Verify eqn 36.46.

(36.4) The γµ matrices and spinors don’t just have oneform. Any form that obeys the anticommutation

Exercises 335

relations will also work. Consider, for example, theunitary transformation matrix

U =1√2

„

1 −11 1

«

. (36.66)

Use this to transform the Dirac wave functionsψ → Uψ and show that for the resulting wave func-tion to satisfy the Dirac equation the γ matriceshave the explicit form

γ0 =

„

1 00 −1

«

, γi =

„

0 σi

−σi 0

«

. (36.67)

This is known as the standard or Dirac–Pauli rep-resentation.

(36.5) Show that in (1+1)-dimensional spacetime, the re-quirement γµ, γν = 2gµν is satisfied by γ0 = σ2

and γ1 = iσ1.

(36.6) Prove the following identities:(a)

˘

γµ, γ5¯

= 0.(b) (γµ)† = γ0γµγ0.

(c) (γ5)† = γ5.(d) (γ5)2 = 1.(e) (γ5γµ)† = γ0γ5γµγ0.

(36.7) (a) Show that

iu(p′)σµν(p′ − p)νu(p) (36.68)

=1

2u(p′)

h

−γµ(p′ −m) + (m− p)γ

µi

u(p),

where σµν = i2

[γµ, γν ] .(b) Show γµp = 2pµ − pγ

µ.(c) Use these two results to prove the Gordon

identity

u(p′)γµu(p) = u(p′)

»

p′µ + pµ

2m+

iσµνqν2m

–

u(p),

(36.69)where qµ = (p′ − p)µ. This identity will be used inChapter 41.

(36.8) Take the non-relativistic limit of the Dirac equationand find the equation of motion for φR.

37 How to transform a spinor

37.1 Spinors aren’t vectors 336

37.2 Rotating spinors 337

37.3 Boosting spinors 337

37.4 Why are there four compo-nents in the Dirac equation?

339

Chapter summary 340

Exercises 340

‘Do you ever run across a fellow that even you can’t under-stand?’‘Yes,’ says he.‘This will make a great reading for the boys down at the of-fice,’ says I. ‘Do you mind releasing to me who he is?’‘Weyl,’ says he.The interview came to a sudden end just then, for the doctorpulled out his watch and I dodged and jumped for the door.But he let loose a smile as we parted and I knew that all thetime he had been talking to me he was solving some problemthat no one else could touch. But if that fellow ProfessorWeyl ever lectures in this town again I sure am going to takea try at understanding him! A fellow ought to test his intel-ligence once in a while.Roundy interviews Professor Dirac (1929), Wisconsin StateJournal1

1TheWisconsin State Journal’s humor-ous writer ‘Roundy’ began his article asfollows: I’ve been hearing about a fel-

low they have up at the U. this spring

– a mathematical physicist, or some-

thing, they call him – who is pushing

Sir Isaac Newton, Einstein and all the

others off the front page. So I thought

I better go up and interview him for

the benefit of the State Journal readers,

same as I do all the other top notchers.

The article seems almost too good to betrue, and unfortunately it seems thatit is in fact a fake. Joseph Coughlin(nicknamed Roundy because of his ro-tund appearance) certainly existed but,as detailed in Graham Farmelo’s biog-raphy of Dirac The Strangest Man, thearticle is most probably a spoof. Dirachowever kept a copy of it.

The humble electron cannot be described by either the scalar or vectorfield theories that we have discussed thus far. The electron, and indeedall spin-1

2 fermions, are excitations in a spinor field. The differentsorts of field are classified by how they transform under rotations andLorentz transformations. We know how scalars transform: by definitionthey don’t change. We also know how to Lorentz transform and rotatevectors. In this chapter we discuss how to transform a spinor.

37.1 Spinors aren’t vectors

Recall that a general Lorentz transformation and rotation can be writtenas a single matrix:

D(θ,φ) = e−iJ·θ+iK·φ, (37.1)

where θ gives angle of rotation and φ gives the velocity boost.2 This

2Recall that we define tanφi = βi.

general transformation may be applied to scalars, vectors or tensors bypicking the appropriate form of the rotation generator J and boost gen-erator K, subject to their all-important commutation relations.3 We’ve

3The commutation relations are

ˆJi, Jj

˜= iεijkJk,

ˆJi,Kj

˜= iεijkKk,

ˆKi,Kj

˜= −iεijkJk.

seen how rotations and boosts affect scalars and vectors, but how dothey affect spinors? We have a problem in that our previous explicitexpressions for K are only geared for vectors, which are objects whoserotations are described by the group SO(3). We will see that spinorsare rotated by the elements of a different group.

37.2 Rotating spinors 337

37.2 Rotating spinors

As Eugene Wigner proved, the rotations of spinors are generated by theelements of the group SU(2). This is the group of two-by-two unitarymatrices with unit determinant. The generators are the Pauli spin matri-ces whose commutation relations are [σi, σj ] = 2iεijkσk. This comes asno surprise since all of us learn at our mother’s knee that two-componentS = 1/2 spin states are acted on by an angular momentum operatorJ = σ

2 and the angular momentum operator is the generator of rota-tions. The set of matrices that rotates two-component Weyl spinors istherefore

D(θ) = e−i2 σ·θ. (37.2)

These operators rotate the left-handed Weyl spinors ψL(p) and theright-handed Weyl spinors ψR(p) in exactly the same way, so the four-component Dirac spinor is rotated by the matrix

(ψL

ψR

)→(

e−i2 σ·θ 0

0 e−i2 σ·θ

)(ψL

ψR

). (37.3)

Example 37.1

With this knowledge we can calculate the 2 × 2 matrix that rotates Weyl spinorsaround a particular axis. We’ll try the rotation by an angle θ1 about the x-direction,whose matrix is4 4Recall eqn 15.23.

D(θ1) = e−i2σ1θ1 = I cos

θ1

2− iσ1 sin

θ1

2. (37.4)

To rotate four-component Dirac spinors we simply act on both the left- and right-handed spinors with this 2 × 2 matrix.

37.3 Boosting spinors

Next we examine how to boost spinors by finding an explicit form forthe boost generators K. Since we know that spinors are rotated by thePauli matrices via J = 1

2σ and we know the commutation relations ofthe generators K with themselves and J , we guess that K ∝ σ. Wefind that two possible choices of proportionality constants satisfy thecommutation relations. It is easily checked that both

K = ± iσ

2(37.5)

will work. It seems, therefore, that there must be two sorts of spinors:those boosted by K = + iσ

2 and those boosted by K = − iσ2 and the two

choices of sign correspond to two distinct representations of the group.At rest, these two sorts of spinors should be indistinguishable. This isindeed correct and the two sorts of spinors are exactly the left- and right-handed Weyl spinors, ψL and ψR, that fell out of the Dirac equation inthe previous chapter.


Example 37.2

To show that the Weyl spinors are transformed by different representations of thegroup we should be able to show that each representation obeys the Lie algebra ofthe generators of SU(2), which is [Ci, Cj ] = iεijkCk. This is possible if we define

J± =J ± iK

2, (37.6)

since then the commutation relations readh

Ji+, Jj+

i

= iεijkJk+, (37.7)h

Ji−, Jj−

i

= iεijkJk−. (37.8)

Our new generators J+ and J− now have independent Lie algebras and each sep-arately generates the rotations and boosts for a type of Weyl spinor. The simplestnon-trivial case of this algebra corresponds to one or other of the new generatorsbeing zero, which gives the rules

J = −iK for J+ = 0,J = iK for J− = 0.

(37.9)

We then have generators J = σ/2,K = iσ/2 for the Weyl spinor ψL and generatorsJ = σ/2,K = −iσ/2 for the Weyl spinor ψR, leading to our final answer for thetransformations appropriate for Dirac spinors:

„ψL

ψR

«

→„

DL(θ,φ) 00 DR(θ,φ)

«„ψL

ψR

«

, (37.10)

where

DL(θ,φ) = eσ2·(−iθ−φ),

DR(θ,φ) = eσ2·(−iθ+φ). (37.11)

After all of this formalism, we’re left with the answer to our question ofhow to Lorentz boost a Dirac spinor. The Dirac wave function, which ismade up of two spinors stacked on top of each other, is boosted accordingto ψ → eiK·φψ, or

(ψL

ψR

)→(

e−12 σ·φ 0

0 e12 σ·φ

)(ψL

ψR

). (37.12)

Example 37.3

We can get some practice by boosting the spinors along the z-direction.

„ψL

ψR

«

→

e−12σ3φ3

0

0 e12σ3φ3

!„ψL

ψR

«

. (37.13)

The elements of the boost matrix are given by

e±12σ3φ3

= 1 ± σ3φ3

2+

1

2!

„σ3φ3

2

«2

± . . .

= I cosh(φ3/2) ± σ3 sinh(φ3/2). (37.14)

So the boost is„

ψL

ψR

«

→

I cosh φ3

2− σ3 sinh φ3

20

0 I cosh φ3

2+ σ3 sinh φ3

2

!„ψL

ψR

«

.

(37.15)

37.4 Why are there four components in the Dirac equation? 339

Reverting to 4 × 4 matrices we see that the components are boosted by a matrix0

BBBBB@

e−φ3

2 0 0 0

0 eφ3

2 0 0

0 0 eφ3

2 0

0 0 0 e−φ3

2

1

CCCCCA

(37.16)

Because e±x = coshx± sinhx we can write e±φ3

2 =p

coshφ3 ± sinhφ3. Identifyingcoshφ3 = γ = E/m and sinhφ3 = βzγ = |p|/m we see that a boost along z isenacted by a matrix

1√m

0

BB@

pE − |p| 0 0 0

0pE + |p| 0 0

0 0pE + |p| 0

0 0 0pE − |p|

1

CCA. (37.17)

We found previously that the Dirac wave function for a particle at rest is given by„uL

uR

«

=√m

„ξξ

«

. Putting everything together we deduce that a boost in the

z-direction has the following effect0

BB@

u1Lu2Lu1Ru2R

1

CCA

=√m

0

BB@

ξ1

ξ2

ξ1

ξ2

1

CCA

→

0

BB@

pE − |p| ξ1

pE + |p| ξ2

pE + |p| ξ1

pE − |p| ξ2

1

CCA. (37.18)

This provides some justification for the expression

u(p) =

„ √p · σξ√p · σξ

«

, (37.19)

as we claimed in the previous chapter.

37.4 Why are there four components in

the Dirac equation?

The Dirac equation is written in terms of a four-component Dirac spinorψ. These Dirac spinors have two degrees of freedom: the two componentsof ξ (or η). So why do we have four components in the Dirac equation?

We found that for the case of massless particles the Dirac spinorsfell apart into two: left-handed Weyl spinors and right-handed Weylspinors. Each of these spinors carried the same spin information in ξ (orη). The question is, then, why does the world need left- and right-handedfields. We will answer this question by asking what fundamental opera-tion turns left-handed spinors into right-handed spinors? The answer isparity.

Many of the properties of fermions are invariant with respect to parityand so we require that Dirac particles have the same properties in aparity reversed world. (The weak interaction violates parity, but weignore that for now.) The parity operation P does the following: itmaps x → −x and v → −v and so it results in the changes to thegenerators:

P−1KP = −K, P−1JP = +J (37.20)


(recall that we say that K is a vector and J is a pseudovector). Theparity operation therefore turns a left-handed spinor into a right-handedspinor

PψL → ψR, PψR → ψL. (37.21)

An explicit form for the parity operator, suitable for use on Dirac spinors,

is P ≡ γ0 =

(0 II 0

).

As we have stressed, the Dirac wave functions are not actually single-particle wave functions: they represent fields. In the next chapter wewill quantize these fields to make them operator-valued, so that when weinput a position in spacetime we output a field operator. Dirac particlesare excitations in these quantum fields.

Chapter summary

• We have derived the transformations for rotating and boostingspinors.

• Parity, enacted with the γ0 matrix, swaps left- and right-handedparts of the wave function.

Exercises

(37.1) (a) Verify eqn 37.4.(b) Show that the rotation matrix

D(θ3) = e−i2σ3θ3 =

e−iθ3

2 0

0 eiθ3

2

!

. (37.22)

(37.2) Show that both choices of sign K = ± iσ2

obey thenecessary commutation relations.

(37.3) A one-line derivation of the Dirac equation.(a) Given that the left- and right-handed parts ofthe Dirac spinor for a fermion at rest are identical,explain why we may write (γ0 − 1)u(p0) = 0.(b) Prove that eiK ·φγ0e−iK ·φ = p/m.(c) Use the result proven in (b) to boost

(γ0 − 1)u(p0) = 0,

and show that you recover the Dirac equation.

38The quantum Dirac field

38.1 Canonical quantization andNoether current 341

38.2 The fermion propagator 343

38.3 Feynman rules and scatter-ing 345

38.4 Local symmetry and a gaugetheory for fermions 346

Chapter summary 347

Exercises 347

After examining the Dirac equation as an equation of motion for singleparticles and investigating the transformation properties of the spinorsolutions we now turn to quantum fields that obey Dirac’s equation astheir equation of motion. We will approach the problem with a varietyof the weapons we have developed in previous chapters. We start withcanonical quantization. We will write down a Lagrangian density whoseequations of motion are the Dirac equation. We will then quantize usingthe rules. The quantized field that we will produce is of fundamentalimportance since its excitations describe fermions such as the quarksand leptons of particle physics and various excitations in solids such asquasielectrons in metals. In order to carry out the sort of calculationsthat can make contact with experiments, such as scattering, we will needa propagator and Feynman rules. We will obtain the propagator with thepath integral. Finally the properties of Noether’s current will motivateus to write down the Lagrangian for perhaps the most successful of allquantum field theories: quantum electrodynamics.

38.1 Canonical quantization and Noether

current

To quantize a field we start by writing down a Lagrangian density de-scribing classical fields. It seems rather wrong-headed to write down aclassical equation for a field that describes fermions. Although we canmake a model of bosonic excitations like phonons out of masses andsprings there is no such obvious classical analogue for fermions. Un-daunted, we will write down the Lagrangian by choosing a form thatwhen inserted into the Euler–Lagrange equation outputs Dirac’s equa-tion. The Lagrangian that leads to the Dirac equation is given by

L = ψ(i∂ −m)ψ = ψ(iγµ∂µ −m)ψ, (38.1)

where ψ = ψ†γ0. Just as free scalar field theory describes fields whoseequation of motion is the Klein–Gordon equation, this Lagrangian de-scribes fermion matter fields whose equation of motion is the Dirac equa-tion.

We can use this Lagrangian to canonically quantize the Dirac field.This field has two components, just like the complex scalar field.The momentum conjugate to ψ is easily calculated using the ordinary

342 The quantum Dirac field

method. We find that

Πµψ =

δLδ(∂µψ)

= iψγµ, (38.2)

and Πµ

ψ= 0. We obtain a momentum field

Π0ψ = iψγ0 = iψ†. (38.3)

The Hamiltonian follows in the usual fashion from H = Π0ψ∂0ψ − L,

yieldingH = ψ†(−iγ0γ · ∇ +mγ0)ψ. (38.4)

However, we also know another piece of information, namely the Diracequation itself which simplifies things here. The Dirac equation tells usthat iγ0∂0ψ = (−iγ ·∇+m)ψ. This allows us to rewrite the Hamiltonianas

H = ψ†i∂0ψ. (38.5)

This looks much simpler! Next we make the fields quantum mechanicalby imposing equal-time anticommutation relations:11We also put hats on ψ and ψ† to make

the point that they are now operators.ψa(t,x), ψ†

b(t,y) = δ(3)(x − y)δab, (38.6)

ψa(t,x), ψb(t,y) = ψ†a(t,x), ψ†

b(t,y) = 0, (38.7)

where a and b label the spinor components of the four-component fieldoperators ψ(x). The mode expansions of these quantum fields are givenby

ψ(x) =

∫d3p

(2π)32

1

(2Ep)12

2∑

s=1

(us(p)aspe−ip·x + vs(p)b†speip·x

),

ˆψ(x) =

∫d3p

(2π)32

1

(2Ep)12

2∑

s=1

(us(p)a†speip·x + vs(p)bspe−ip·x

). (38.8)

The creation and annihilation operators must themselves obey the anti-commutation relations

asp, a†rq = bsp, b†rq = δ(3)(p − q)δsr. (38.9)

Now it simply remains to insert the mode expansion into the Hamilto-nian and normal order to obtain a quantized field theory.

Example 38.1

We obtain

H =

Z

d3xd3pd3q

(2π)3(2Ep2Eq )12

X

s,r

“

us(p)a†speip·x + vs(p)bspe−ip·x”

×γ0Eq

“

ur(q)arqe−iq·x − vr(q)b†rqeiq·x”

=X

s,r

Zd3p

2

“

us†(p)a†spus(p)asp − vs†(p)bspvs(p)b†sp

”

. (38.10)

Lastly, we can use the properties of the normalized spinors to do the spin sum, sincewe know that us†ur = 2Epδsr.

38.2 The fermion propagator 343

The (normal ordered) Hamiltonian becomes

H =

∫d3p

2∑

s=1

Ep

(a†spasp + b†spbsp

). (38.11)

We see that the energy is given by the sum of the energies of the particlesand antiparticles, exactly as we should expect.

The Lagrangian also has a global U(1) symmetry, meaning that theLagrangian is invariant with respect to the change ψ(x) → ψ(x)eiα, sowe can use Noether’s theorem to find the conserved current and charge.

Example 38.2

With ψ → ψeiα we also have ψ → ψe−iα. Since the transformation is global we alsocan write ∂µψ → eiα∂µψ. Putting this all together we get

L = ψ(i∂ −m)ψ

→ ψe−iα(i∂ −m)ψeiα = ψ(i∂ −m)ψ. (38.12)

The use of Noether’s theorem goes through similarly to the case for thecomplex scalar field, except that we have

Πµψ = ψiγµ, Πµ

ψ= 0, (38.13)

and Dψ = iψ. So, upon normal ordering and swapping signs to givethe conventional number current direction, we obtain a Noether currentoperator

JµNc = ˆψγµψ. (38.14)

We can use this to work out the conserved U(1) charge which, uponsubstitution of the mode expansion, yields up

QNc =

∫d3p

∑

s

(a†spasp − b†spbsp

). (38.15)

The conserved charge is given by the number of particles minus thenumber of antiparticles, as expected. Once again, this is just like thecomplex scalar field case.


Dψ = iψ Dψ† = −iψ†

Πµψ = iψγµ Πµψ

= 0

DL = 0 Wµ = 0J0Nc = ψ†ψ JµNc = ψγµψ

QNc =R

d3pP2s=1(n

(a)sp − n

(b)sp )

38.2 The fermion propagator

The propagator for fermions can be worked out by calculatingG0(x, y) =

〈0|T ψ(x) ˆψ(y)|0〉. However, we can also use the path integral. For See Exercise 38.4.

fermions we treat the classical fields ψ and ψ that appear in the pathintegral as Grassmann numbers.2 In terms of the fields ψ and ψ, the 2See Chapter 28 for an introduction to

Grassmann numbers.generating functional is written

Z[η, η] =

∫DψDψ e

iR

d4xh

ψ(x)“

i∂−m”

ψ(x)+η(x)ψ(x)+ψ(x)η(x)i

, (38.16)


where we notice that both ψ(x) and ψ(x) must be coupled to separatesource fields η(x) and η(x) respectively. The leg-work for this calculationwas carried out in Chapter 28 and so we note that the result is

Z[η, η] = Ce−iR

d4xd4y η(x)(i∂−m)−1η(y), (38.17)

where C is proportional to a determinant. As usual, we normalize by di-viding through by Z[0, 0] to obtain the normalized generating functional

Z[η, η] = e−R

d4xd4y η(x)iS(x−y)η(y), (38.18)

where S(x − y) = (i∂ −m)−1. The propagator is then the solution ofthe equation

(i∂ −m)iS(x) = iδ(4)(x). (38.19)

The solution for the fermion propagator may be read off from this3 as3Note that there is a difference of a mi-nus sign in the equations defining thescalar and fermion propagators in termsof the Green’s function of the equationsof motion.

G0(x, y) = iS(x− y) =

∫d4p

(2π)4ie−ip·(x−y)

p−m+ iǫ. (38.20)

It’s important to note that the fermion propagator G0(x, y) is a four-by-four matrix. This is revealed in momentum space if we multiply topand bottom by (p+m) to give

G0(p) =i

p−m=

i(p+m)

p2 −m2 + iǫ

=i

p2 −m2 + iǫ

(m p0 − p · σ

p0 + p · σ m

), (38.21)

where the latter equation is given in chiral representation. An interpre-

Fig. 38.1 The fermion propagator in-terpreted in terms of left- and right-handed spinors.

tation of the four-by-four fermion propagator is revealed if we considerthe roles of the left- and right-handed parts of the Dirac spinor. The

fermion propagator G0 is formed from the combination ψ(x) ˆψ(y) which,

after substituting ψ =

(ψL

ψR

), has the form

(ψL(x)ψ†

R(y) ψL(x)ψ†L(y)

ψR(x)ψ†R(y) ψR(x)ψ†

L(y)

). (38.22)

The top-left slot deals with a right-handed part entering and a left-handed part leaving. In contrast, the top-right slots tells us abouta left-handed part entering and a left-handed part leaving. Recallingthat a particle propagator is formed from a series G = G0 + G0V G0 +G0V G0V G0 +G0V G0V G0V G0 + . . ., we may multiply the propagatormatrices and obtain the matrix represented in Fig. 38.1. We see thatthe top-right slot still describes a left-handed part entering and a left-handed part leaving, but now expresses this as a superposition of all ofthe possible oscillations between left- and right-handed parts that resultin the left-handed part leaving. The other slots of the matrix representthe other combinations of chiralities entering and leaving. In this sense

38.3 Feynman rules and scattering 345

we may think of a massive Dirac particle propagating while oscillatingbetween left- and right-handed.

With this in mind, our problem is solved since we have found thepropagator for free fermions. This enables us to start solving scatteringproblems through the use of Feynman diagrams and it is the Feynmanrules for fermions that we turn to next.

38.3 Feynman rules and scattering(a)

(b)

(c)

(d)

(e)p

p

p

p

= vs(p)

= us(p)

= vs(p)

= us(p)

=i

p−m + iǫ

Fig. 38.2 Feynman rules for fermions.

The generating functional Z[η, η] can be used to find the Feynman rulesfor fermions. We won’t do this, but simply list the rules in momentumspace (see also Fig. 38.2).

Feynman rules for fermions

• A factor i/(p−m+ iǫ) for each internal fermion line.

• A factor us(p) for an incoming fermion with momentum p andspin s.

• A factor vs(p) for an incoming antifermion with momentum pand spin s.

• A us(p) for an outgoing fermion with momentum p and spin s.

• A vs(p) for an outgoing fermion with momentum p and spin s.

• Trace over the matrix product arising from a fermion loop.

• Add minus sign factors reflecting fermion statistics, the mostcommon of which is a factor −1 for every fermion loop.

We will now get some practice with the use of these rules.

(a)

(b)

p

p′

k

k′

us(p) ur(k)

us′(p′) ur′(k′)

p

p′

k

k′

us(p) ur(k)

us′(p′) ur′(k′)

q = p′ − p

q = p′ − k

Fig. 38.3 (a) Electron–electron t-channel exchange. (b) Electron–electron u-channel exchange.

Example 38.3

Let’s use our rules to work out some Feynman diagrams for a theory of scalar phionsinteracting with electrons, via the interaction LI = −gψψφ. This is very much likethe example of the complex scalar field interaction in Chapter 11, except that weneed to consider the complication of the spinor polarizations.

We can start with electron–electron scattering mediated by the t-channel exchangeof a virtual phion with mass mφ, as shown in Fig. 38.3(a). Applying the rules, weobtain an amplitude

iM1 = (−ig)2us′(p′)us(p)

i

(p′ − p)2 −m2φ

ur′(k′)ur(k), (38.23)

where we have ignored an overall energy-momentum conserving delta function. Nextwe consider electron–electron u-channel exchange scattering shown in Fig 38.3(b).Swapping the lines over corresponds to swapping operators in the S-matrix element,so this diagram comes with an extra minus sign. Applying the rules, we obtain

iM2 = −(−ig)2ur′(k′)us(p)

i

(p′ − k)2 −m2φ

us′(p′)ur(k). (38.24)

We’ll get a little further by considering the non-relativistic limit for distinguish-able particles. This means that only the t-channel diagram contributes. For non-relativistic particles, we have the approximate properties:

p ≈ (me,p) p′ ≈ (me,p′) k ≈ (me,k) k′ ≈ (me,k′) (38.25)


and, for spinors, that us(p) =√me

„ξs

ξs

«

. The spinor products become

us′(p)us(p) = 2meξ

s′†ξs = 2meδs′s, (38.26)

showing that the scattering can’t change the spin. Putting these contributions to-gether yields a scattering amplitude44This may be compared to the version

in Chapter 20, which only differs innormalization and the absence of spinconserving Kronecker deltas. Here, asbefore, the Yukawa potential is attrac-tive. In the next chapter we will re-cover the expected result that the elec-tromagnetic interaction scattering likecharges is repulsive.

iM =4ig2m2

e

(p− p′)2 +m2φ

δs′sδr′r. (38.27)

We will see more examples of scattering in the next two chapters.

38.4 Local symmetry and a gauge theory

for fermions

Our final job in this chapter will be to conjure up a gauge theory forfermions. Why do this? We know that electrons interact with elec-tromagnetic fields (since they carry electric charge), and so we there-fore plan to extend the gauge field Lagrangian of electromagnetismL = − 1

4FµνFµν to include fermions and an interaction term telling

us how photons and electrons interact. The gauge principle that weused in Chapter 14 allows us to do this since the act of promoting theDirac Lagrangian to a gauge theory by making it locally U(1) symmetricwill provide us with an interaction term through the minimal couplingprescription.

As we’ve seen, the Dirac Lagrangian is invariant with respect toglobal U(1) transformations, which is to say that the internal trans-formation ψ → ψeiα leaves the Lagrangian invariant. The result of thisis a conserved number current JNc. To make a gauge theory we willpromote the global U(1) symmetry to a local symmetry exactly as wehave done before in Chapter 14. We simply replace our transformationwith ψ → eiα(x) and introduce a gauge field via a covariant derivativeDµ = ∂µ + iqAµ(x). In order for the gauge field to ensure a local sym-metry, it must itself transform as Aµ(x) → Aµ(x) − 1

q∂µα(x).The Dirac Lagrangian, fixed up to have a local U(1) invariance, is

then given by55We writeD = γµDµ = γµ(∂µ+iqAµ).

L = ψ(iD −m)ψ. (38.28)

Inserting our covariant derivative, we have L = ψ (iγµ[∂µ + iqAµ] −m)ψor

L = ψ(i∂ −m

)ψ − qψAψ. (38.29)

Lo and behold, we are told how the gauge field Aµ(x) interacts withour fermion fields ψ(x) and ψ(x): it’s via the interaction term LI =−qψγµAµψ. This method of determining the form of the interactionmerely by considering the consequences of the substitution pµ → pµ −qAµ is another example of the minimal coupling prescription we saw inChapter 14.6

6 It is also worth noting that the elec-tromagnetic field tensor Fµν may begenerated by evaluating the commuta-tor [Dµ, Dν ] = iqFµν , as examined inExercise 46.3.

Exercises 347

In classical electromagnetism the conserved currents Jµem are thesources of the electromagnetic fields. These fields interact with the elec-tric charges q, telling them how to move. This is encoded in an inter-action between the electromagnetic current density Jµem(x) and the fieldAµ(x) giving a contribution to the Lagrangian LI = −Jµem(x)Aµ(x),where Jµem is a conserved current obeying ∂µJ

µem = 0. Our minimal

coupling prescription motivates us to make the step of identifying thefermion Noether current JµNc(x) = ψγµψ with the electromagnetic cur-rent via Jµem = qJµNc, where q is the electromagnetic charge. This con-strains the source currents that may be coupled to the electromagneticfield and guarantees that the source current is a conserved quantity.This philosophy is illustrated in Fig. 38.4.

Global

symmetry

Local

symmetry

Conserved

charge

upgradewithgaugefield

covariant derivativeminimal coupling

Fig. 38.4 The minimal coupling pre-scription in QED.

We end this chapter with a landmark equation. We will write downperhaps the most successful theory in modern physics: Quantum elec-trodynamics or QED. The QED Lagrangian includes the contributionsfrom the gauge field of electromagnetism (eqn 5.49) and from the lo-cally gauge invariant Dirac Lagrangian describing fermions and theirinteractions (eqn 38.28) giving

L = −1

4FµνF

µν + ψ (iγµ∂µ −m)ψ − qψγµAµψ. (38.30)

In the next chapter we will examine the predictions of this theory.

Chapter summary

• The Lagrangian that leads to the Dirac equation is L = ψ(i∂−m)ψwhere ψ = ψ†γ0 and can be quantized. The Noether current isJµNc = ψγµψ.

• The fermion propagator is i/(p−m).

• In order to have local gauge invariance, minimal coupling tells usthat we must add to L a term LI= − qψγµAµψ.

Exercises

(38.1) Try to canonically quantize the Dirac Lagrangianusing commutation relations rather than anticom-mutation relations for the fields. You should get aminus sign in front of the b†p bp term in the Hamil-tonian. Explain why this is a catastrophe.See Peskin and Schroeder, Chapter 3 for help.

(38.2) Verify eqn. (38.15).

(38.3) Show that the massless Dirac Lagrangian is invari-ant with respect to the global chiral U(1) transfor-

mation ψ(x) → eiαγ5

ψ(x). Show that the Noethercurrent is JµNc(x) = ψ(x)γµγ5ψ(x).

(38.4) Calculate the free fermion propagator by insertingthe mode expansion into the equation G0(x, y) =

〈0|T ψ(x) ˆψ(y)|0〉. You will need to use the spin sum-mation rules from Chapter 36.

39A rough guide to quantum

electrodynamics

39.1 Quantum light and the pho-ton propagator 348

39.2 Feynman rules and a firstQED process 349

39.3 Gauge invariance in QED351

Chapter summary 353

Exercises 353

Quantum electrodynamics (QED) is the quantum field theory that de-scribes the interaction of light and matter. Arguably one of the greatestintellectual achievements ever made, its discovery was due to severalpeople but notably Tomonaga, Schwinger and Feynman, who shared

Sin-Itiro Tomonaga (1906-1979).

the Nobel Prize in 1965.Since QED describes an interaction it will come as little surprise that

this is not a solvable theory and we must rely on perturbation theoryand Feynman diagrams to understand the results of experiments. In thischapter we take the view that the Feynman diagrams are the theory andwill therefore concentrate on finding the propagator for the photon andthe Feynman rules for the photon–fermion interaction. One complicationis that electrodynamics is a gauge invariant theory so we must expendsome effort in ensuring that our dealings with the photon preserve thisinvariance. We will find that this can be done and, on the way, calculateour first amplitude in QED!

39.1 Quantum light and the photon

propagator

Our task is to find the propagator for photons. We might guess that it’sgiven by the m→ 0 limit of the propagator for the massive vector fieldwe found1 in Chapter 24, which is to say

1See eqn 24.29. Note that here wefollow the convention, introduced inChapter 17, that the free photon prop-agator is called D0µν(k).

D0µν(k) = limm→0

i(−gµν + kµkν/m2)

k2 −m2 + iǫ. (39.1)

However there’s a problem in taking this limit: the (longitudinally pro-jecting) term kµkν/m

2 threatens to blow up for photons on the massshell, which have k2 = m2 → 0. We will show later that this disasterdoes not occur and that gauge invariance allows the term kµkν/m

2 tobe removed. Accepting this step for now, we are left with the photonpropagator, given by

D0µν(k) =−igµνk2 + iǫ

. (39.2)

As usual the propagator describes the contribution from a virtual par-ticle, in this case the photon. By virtual we mean that the energyEk 6= |k|, where k is the three-momentum of the photon.2

2All photons that we detect actually in-teract with electrons in detectors suchas the eye. They must all then, in somesense, be virtual! How can this be?We know that particles that are off-shell have the range over which theycan propagate limited by the extentto which they’re off-shell. If we seephotons that have travelled from dis-tant stars they have to be pretty closeto being on-shell. We’ve seen beforethat when a particle is on-shell we hitthe pole of the particle’s propagator.Therefore photons from Andromeda,visible on a moonless night, must be soclose to the pole that there can’t be anyobservable effects from being off-shell.

39.2 Feynman rules and a first QED process 349

It’s worth checking that our propagator correctly predicts the mech-anism through which photons mediate the interaction between charges.Let’s consider the fate of two electric charges interacting via the ex-change of a photon, as shown in Fig. 39.1. Since we’re interested inthe photons we’ll take the charges as forming electric currents Jµ. Thishas the advantage that the fermion and vertex parts of the amplitudeare contained3 in the current Jµ, leaving us free to concentrate on the

3This assertion is justified later in thechapter.

physics of the photon.

µ ν

k

Jµa = qψaγµψa Jνb = qψbγ

νψb

Fig. 39.1 The exchange of a virtualphoton.

We therefore consider the amplitude

A = Jµa

(−igµνk2

)Jνb , (39.3)

where we’ve left out the iǫ, since its presence won’t be important forwhat follows.

Example 39.1

If we work in a frame where kµ = (k0, 0, 0, k3), then the amplitude looks like

A = −i`J0aJ

0b − Ja · Jb

´

(k0)2 − (k3)2. (39.4)

Current conservation implies that kµJµ = 0, so we have k0J0 − k3J3 = 0, whichallows us to eliminate the component J3 to yield

A = iJ0aJ

0b

(k3)2+ i

J1aJ

1b + J2

aJ2b

(k0)2 − (k3)2. (39.5)

Now for some interpretation.4 The first term is in terms of J0, the charge density. 4Note that in the general case, where

kµ = (k0,k), the first term in eqn 39.5will be iJ0

aJ0b/k

2.

If we (inverse) Fourier transform this quantity we obtain an instantaneously actingCoulomb potential, which is repulsive between like charges

Zd4k

(2π)4e−ik·x J

0aJ

0b

k2∝ q2

4π|r| δ(ta − tb). (39.6)

Don’t worry about the instantaneousness of the term. It only looks unphysically in-stantaneous because we’ve split up the propagator in a non-covariant manner. More-over, this is the term which dominates in the non-relativistic regime. This Coulombinteraction is of course the basis of much of condensed matter physics.

We argued above that the photons we observe are those very close to the pole. Forour case of photons propagating along the z- (or 3-) direction, we look at the residueof the second term and we see that there seem to be two sorts of photon: those thatcouple J1 currents and those that couple J2 currents. These are the two physicaltransverse photon polarizations.

39.2 Feynman rules and a first QED

process

We now turn to the interaction of the photon with a fermion. We sawin the previous chapter5 that the result is an interaction term which 5Remember that here, as elsewhere in

the book, q = Q|e|.expressed in Hamiltonian form is

HI = q¯ψγµψAµ. (39.7)

350 A rough guide to quantum electrodynamics

The diagrammatic version of this interaction is shown in Fig. 39.2(a).This one simple vertex, whose translation in words is ‘fermions can emitor absorb virtual photons’, governs the entire theory.

We now have the ingredients to state the Feynman rules for quantumelectrodynamics:

Feynman rules for QED

• Use all of the rules given previously for fermions.

• The interaction vertex contributes −iqγµ, where q(= Q|e|) is thecharge [Fig. 39.2(a)].

• Every internal photon line contributes D0µν(k) = −igµν/(k2+iǫ)

[Fig. 39.2(b)].

• Incoming external photon lines contribute a polarization vectorǫµλ(p) [Fig. 39.2(c)].

• Outgoing external photon lines contribute ǫ∗µλ(p) [Fig. 39.2(d)].

(a)

(b)

(c)

(d)

p

p

k

p

p′

p′ − p

= ǫ∗νλ(p)

= ǫµλ(p)

= − igµν

k2 + iǫ

= −iqγµ

ν

µ

νµ

µ

Fig. 39.2 The QED Feynman rules.

To get the feel of the rules, we will examine the case of e+e− → µ+µ−,shown in the Feynman diagram in Fig. 39.3. Using the Feynman rules

e− e+

µ− µ+

q

pp′

k k′

Fig. 39.3 A first QED process:e+e− → µ+µ−.

for fermions and photons, we obtain an invariant amplitude of

iM = vs′

(p′)(−iQ|e|γµ)us(p)(−igµν

q2

)ur(k)(−iQ|e|γν)vr′(k′), (39.8)

where we’ve written the charge as Q|e|. Note that the sign of the chargedoesn’t matter for many calculations and we will drop Q in what follows.

Example 39.2

We’ll get some more practice manipulating spinors in these calculations by consid-ering this process in the ultra-relativistic limit. We imagine the kinematics for the

Our treatment here follows Peskin andSchroeder.

process to be those depicted in Fig. 39.4. Remember that in the ultra-relativisticlimit particles in chirality eigenstates will simultaneously be in helicity eigenstates.

We’ll start with a right-handed electron with initial momentum along +z. Arelativistic, right-handed electron always has helicity h = +1 and so must have a

spinor ξ =

„10

«

, corresponding to a physical spin-up along z. We will collide

this electron with a left-handed positron with initial momentum along −z. A left-

handed, highly relativistic positron6 has h = +1, so its spinor is given by ξ =

„01

«

,

6Remember that relativistic, left-handed fermions have h = −1 whileleft-handed antifermions have h = +1.

corresponding to physical spin-up along z. Taking the limit of very large momenta7

7See Chapter 36 if in doubt. We alsoassume here that E = Ep ≈ Ep′ .

we have spinors for the incoming electrons given by

u(p) =√

2E

0

BB@

0010

1

CCA

v(p′) =√

2E

0

BB@

000−1

1

CCA. (39.9)

To calculate our matrix element we need to evaluate products looking like uγµu =u†γ0γµu. It’s therefore useful to know that

γ0γµ =

„0 11 0

«„0 σµ

σµ 0

«

=

„σµ 00 σµ

«

. (39.10)

39.3 Gauge invariance in QED 351

Plugging in our spinors and using this result gives us the result for the incominginteraction vertex that

v(p′)γµu(p) = 2E(0, 0, 0,−1)

„σµ 00 σµ

«

0

BB@

000−1

1

CCA

= −2E

0

BB@

01i0

1

CCA. (39.11)

e−

µ+

θ e+

µ− x

z

p = (E,Ez )

p′ = (E,−Ez )

Fig. 39.4 Kinematics for our firstQED process.

So far, so good. Next we need to evaluate the outgoing part describing the muons. Auseful insight here is that the quantity v(p′)γµu(p) can be thought of as a four-vectordescribing the spin and momenta of the incoming electron states. We will take theinner product with a similar vector describing the outgoing muon states. We noticethat the muon vector u(k)γνv(k′) is almost the same four-vector as v(p′)γµu(p) whichwe evaluated above for the incoming electrons. The difference is that it’s complex-conjugated and rotated by an angle θ in the x-z plane. The complex conjugation ismade easy with the identity8 [u(p)γµu(k)]∗ = u(k)γµu(p) and we certainly know how 8See Exercise 39.3.to rotate a four-vector. Conjugating and rotating the result in eqn 39.11 thereforegives us

u(k)γµv(k′) = [v(k′)γµu(k)]∗ (Complex conjugation)= [−2E(0, cos θ, i,− sin θ]∗ (Rotation about y by angle θ)= −2E(0, cos θ,−i,− sin θ).

(39.12)Putting the amplitude together via the dot product vγµugµν uγνv and including thephoton propagator yields up the invariant amplitude

iM = −i4e2E2

q2(1 + cos θ). (39.13)

Noting that q2 = (2E)2, we obtain the simple result that M(e−Re+L → µ−Rµ

+L ) =

−e2(1 + cos θ).This can be repeated for other combinations of chirality with the result that

M(e−Re+L → µ−Rµ

+L ) = M(e−L e

+R → µ−L µ

+R) = −e2(1 + cos θ),

M(e−Re+L → µ−L µ

+R) = M(e−L e

+R → µ−Rµ

+L ) = −e2(1 − cos θ), (39.14)

and all other combinations yield an amplitude of zero.

39.3 Gauge invariance in QED

Gauge invariance

ψ(x) → eiα(x)ψ(x)Aµ → Aµ − 1

q∂µα

Local EM chargeconservation∂µJ

µem = 0

Ward identitykµMµ(k) = 0

Minimalcoupling Noether’s

theorem

Feynman diagrams

Fig. 39.5 Gauge invariance, the Wardidentity and conservation of electro-magnetic current are all intimatelylinked.

kµ · = 0µ

p

p′

k

S-matrix element

Fig. 39.6 The Ward identity. Dot thevector kµ with the contribution to theS-matrix shown and you get zero.

Gauge invariance is central to our notion of QED. Global U(1) symmetryof the Dirac Lagrangian guarantees the conservation of fermion number.Local U(1) symmetry [guaranteed by the addition of the electromagneticgauge field Aµ(x)] allows us to identify the conserved fermion currentwith the conserved electromagnetic current and fixes the form of theQED interaction. Gauge invariance therefore leads to the conservationof electromagnetic current in QED. This is certainly central to the theoryand must be maintained in our Feynman diagrams. Gauge invariance isindeed guaranteed in Feynman diagrams by the Ward identity. The John Ward (1924–2000). It has been

said that his advances were used by oth-ers ‘often without knowing it, and gen-erally without quoting him’ (M. Dun-hill, The Merton Record, 1995).

symbiotic relationship between the concepts is shown in Fig. 39.5. TheWard identity comes in many forms9 and the simplified one with which

9It is a special case of the Ward–Takahashi identities. See Peskin andSchroeder for its derivation.

we’ll be concerned is shown diagrammatically in Fig. 39.6. It says thatif a sum of diagram parts contributing to an S-matrix element may bewritten Mµ(k, p1, p2, ...), where µ labels the vertex to which a photonline is attached, and that the photon line carries momentum k and theexternal lines are on mass shell, then

kµMµ(k, p1, p2, ...) = 0. (39.15)


This effectively kills the term kµkν/k2 in the numerator of the photon

propagator, since it always combines kµ with the sort of vertex shownin Fig. 39.6.

We began our discussion of the photon propagator by arguing thatthe term kµkν/k

2 disappears from the numerator of our propagator. Weconclude here that its disappearance is caused by gauge invariance andis encoded in Feynman diagrams by the Ward identity. A diagrammaticexample of the Ward identity is examined in the exercises and a justifi-cation of the physics may be found in the next example.

Example 39.3

To see how gauge invariance leads to the disappearance of this term we naivelycouple a general source Jµ (not the conserved current of electromagnetism) to amassive vector field Aµ. We will calculate the amplitude for starting with no vectormesons and ending with one, represented by the half-dumbbell diagram shown inFig 39.7. We can write the amplitude for creating a massive vector meson with any

Fig. 39.7 The half-dumbbelldiagram.

polarization as

A ∝X

λ

ǫ∗λµ(k)Jµ(k), (39.16)

where the sum is over the polarizations. Assuming a basis of linear polarizations (i.e.so ǫ∗λµ = ǫλµ), this leads to a probability

P = |A|2 ∝X

λ

Jµ(k)Jν†(k)ǫλµ(k)ǫλν(k)

= (−gµν + kµkν/m2)Jµ(k)Jν†(k), (39.17)

where we’ve used the result forP

λ ǫλµ(k)ǫλν(k) = −PTµν discussed in Chapter 13.

Next we ask what this tells us about the source emitting a photon, by attemptingto take a limit m → 0. We immediately see that the kµkν/m term blows up asm → 0, which is the same problem we had with trying to apply the massive vectorpropagator to photons.

However, this won’t cause a problem if kµJµ(k) = 0, since this kills the trouble-some term. Notice that this is current conservation ∂µJµ = 0 written in momentumspace! We conclude that the disappearance of the troublesome kµkν/m2 term isdue to electromagnetic current conservation, which is itself attributable to gaugeinvariance.

k

p

p′

(p′ − k)

ℓ p

ℓ′ p′

(ℓ′ − k) (p′ − k)

k

q

pk

p′

(p + k)

p

k

p′

(p + k)

ℓ

q

ℓ′

(ℓ− k)

µ

ηλ

ν

(a)

(b)

(c)

(d)

µ

νλ

η

Fig. 39.8 Illustration of gauge in-variance in the diagrammatic languageused in Exercise 39.4.

Looked at in another way, the disappearance of the kµkν/k2 term allows

us to add any multiple of this quantity back into the propagator. Thatis to say, vast helpings of a term that has no effect on the physics maybe added for your convenience. A more general version of the photonpropagator can therefore be given by

D0µν(k) =−i(gµν + (1 − ξ)kµkν/k

2)

k2 + iǫ. (39.18)

The simple choice ξ = 1 is known as Feynman gauge, while ξ = 0 isknown as Landau gauge. As examined in the exercises, exactly this prop-agator follows from fixing the gauge in which we work in the Lagrangianitself. The freedom to choose ξ is therefore another manifestation ofgauge invariance.

Exercises 353

We have written down the Feynman rules for QED and examined afirst process. In the next chapter we go further and look at the main (andmeasurable) application of QED calculations: calculating the scatteringamplitudes for some fundamental processes.

Chapter summary

• The photon propagator has been derived and every internal photonline contributes a factor −igµν/(k

2+iǫ), though alternative choicescan be made in different gauges.

• We have presented the Feynman rules for QED.

Exercises

(39.1) We will use the gauge fixing technique to find theQED propagator.(a) Show that the equations of motion for theelectromagnetic field Aµ(x) written in momentumspace are given by

`

k2gµν − kµkν´

Aµ(k) = 0. (39.19)

(b) We will attempt to find the inverse of theseequations of motion. Explain why this must be ofthe form (M−1)νσ = C(k)gνσ +H(k)kνkσ.(c) Show that defining the inverse viaMµν(M−1)νσ = gµσ demands the impossible condi-tion

−k2C(k)gµσ + C(k)kµkσ = gµσ . (39.20)

(d) We may fix this by adding a term to the La-grangian that ensures we work in Lorentz gauge∂µA

µ = 0. We therefore add a gauge fixing termand obtain a Lagrangian

L = −1

4FµνFµν −

1

2ξ(∂µA

µ)2, (39.21)

where ξ is an arbitrary parameter. The point hereis that the gauge fixing term will force the theoryto be well behaved, but will contain the parameterξ which (we claim) will not enter into any measur-able quantity. Show that the equations of motion,written in momentum space, are now given by„

−k2gµν + kµkν − 1

ξkµkν

«

Aν(k) = 0. (39.22)

(e) Finally, verify that the photon propagator, de-fined as i(M−1)µν , is given by

D0µν(k) = −i

gµν − (1 − ξ)kµkν

k2

k2 + iǫ

!

. (39.23)

(39.2) Consider the reaction e+e− → µ+µ− again. Mostexperiments are done by firing unpolarized beams ofelectrons and positrons at each other. Since muondetectors are usually not sensitive to muon polar-ization we actually need to throw away the spin in-formation. What we want is an average over initialelectron spin states and a sum over the final muonspin states.(a) Explain why this is the right thing to do andshow that this prescription leads to a probability

1

2

X

s

1

2

X

s′

X

r

X

r′

|M(s, s′ → r, r′)|2. (39.24)

(b) Show that we obtain a scattering proba-bility for unpolarized electrons and muons of14

P

spins |M|2 = e4(1 + cos2 θ).We will revisit this next chapter.

(39.3) Verify [u(p)γµu(k)]∗ = u(k)γµu(p).

(39.4) We will examine how the term kµkν/k2 is killed

in a simple situation. Consider the Feynman dia-gram shown in Fig. 39.8(a) which contributes, viathe S-matrix to the second-order correction to theamplitude for fermion–fermion scattering.


(a) Using the Feynman rules write down the ampli-tude for the whole process.(b) Consider the bold part, shown in Fig. 39.8(b).This features a virtual electron propagator, with aphoton line hitting it at the end labelled ν. Showthat the amplitude Apart for this part of the dia-gram is

Apart = u(p′)(−ieγη)

„

i

p+ k −me

«

(−ieγν)

×„−igµν + ikµkν/m

2γ

k2 + iǫ

«

u(p), (39.25)

where we’ve called the photon mass mγ and thefermion mass me.(c) Now just consider the dangerous part, which isproportional to 1/m2

γ . Show that the guts of thisboil down to something proportional to

Aguts = u(p′)γηkµkν/m

2γ

p+ k −me

γνu(p). (39.26)

(d) Show that this may be further simplified to

Aguts =1

m2γ

u(p′)γηkµu(p)

k2 + iǫ. (39.27)

Hint: rewrite the denominator as k = (p + k −me)−(p−me) and use Dirac’s equation in the form

pu(p) = meu(p).(e) Now turn to the process shown in Fig. 39.8(c)which it is also necessary to consider when calcu-lating the second-order correction to the fermion–fermion scattering amplitude. Just consider thepart shown in Fig 39.8(d) (and note the change inlabelling).(f) Show that the dangerous guts of this part aregiven by

Bguts = u(p′)γνkµkν/m

2γ

p′ − k −me

γηu(p), (39.28)

and that this may be reduced to

Bguts = − 1

m2γ

u(p′)kµγηu(p)

k2 + iǫ. (39.29)

The point, then, is that we must consider both ofthese processes and the sum of the dangerous gutsAguts +Bguts = 0, which means that the dangerouskµkν/m

2γ term disappears.

This argument is discussed in more depth in Zee,Chapter II.7.

40QED scattering: three

famous cross-sections

40.1 Example 1: Rutherford scat-tering 355

40.2 Example 2: Spin sums andthe Mott formula 356

40.3 Example 3: Compton scat-tering 357

40.4 Crossing symmetry 358

Chapter summary 359

Exercises 359

In this chapter we examine three famous scattering examples in QED: (1)Rutherford scattering, which led to the discovery of the atom; (2) Mottscattering, which is the relativistic version of Rutherford scattering and(3) Compton scattering, which demonstrates the particle-like propertiesof light. These examples will introduce a number of useful tricks whichare employed freely in more advanced applications.

40.1 Example 1: Rutherford scattering

i|e|γµAµcl(q)

Aµcl(q)

us(p)

us′(p′)

p′

p

−iQ|e|γµ

(a)

(b)

Fig. 40.1 (a) Scattering from a clas-sical potential. (b) The Feynman rulefor a Q = −1 electron to scatter fromthe potential.

We’ll examine the scattering of an electron from a heavy, charged objectsuch as a point-like nucleus. This scattering of distinguishable particlesis described by a Feynman diagram shown in Fig. 40.1(a). The nucleusis very heavy compared to the electron, so we treat the the process asinvolving an electron passing through a classical potential Aµcl(x). TheFeynman rule for the vertex describing the electron interacting withthe potential is −iQ|e|γµAµcl(q) [Fig. 40.1(b)], that is, the electron (withQ = −1) interacts via the Fourier transform of the potential Aµcl(x).This is easy to understand: the interaction between an electron anda potential is HI = Q|e|ψγµψAµcl. The ψ brings an electron in andψ takes is out, so the interaction vertex is proportional to what’s left,which is Q|e|γµAµcl. For a static, positive electromagnetic potential we

have A0cl(r) = Z|e|

4π|r| , and Acl(r) = 0, so the potential we are after is

the Fourier transform of the Coulomb interaction A0cl(q) = Z|e|

q2 . Theamplitude for the diagram in Fig. 40.1(a) for a (Q = −1) electron to bescattered by a nuclear potential is given by

iM = iZe2

q2u(p′)γ0u(p). (40.1)

p = mv |p ′| cos θ

|p ′| sin θθ

p′

Fig. 40.2 Kinematics for Rutherfordscattering.

Example 40.1

The kinematics of the scattering is shown in Fig. 40.2. Calculating the transferredmomentum gives us1 q2 = −|q|2 = −4p2 sin2 θ

2. In this example we will work in the 1See Exercise 40.1.

non-relativistic limit, for which we have that u(p′)γ0u(p) = 2mξ†ξ = 2m. PuttingRecall that in the non-relativistic limit

we have u(p) ≈ √m

„ξξ

«

.

everything together we obtain an amplitude

|M|2 =4Z2e4m2

16p4 sin4(θ/2). (40.2)

356 QED scattering: three famous cross-sections

Finally, we can relate our amplitude to something measurable. For scattering froma potential, the differential cross-section2 is given by2See, for example, Peskin and

Schroeder, Chapters 4 and 5.dσ

dΩ=

|M|2(4π)2

, (40.3)

and so plugging in our expression for |M|2, we obtain

dσ

dΩ=

Z2α2

4m2v4 sin4(θ/2), (40.4)

which is Ernest Rutherford’s celebrated result.An alleged scientific discovery has no

merit unless it can be explained to a

barmaid.

Ernest Rutherford (1871–1937)

40.2 Example 2: Spin sums and the Mott

formula

If we work through the previous example again, but this time withoutinvoking the non-relativistic approximation, we obtain the relativisti-cally correct version of Rutherford’s result, which is known as the Mottformula. The complication with working with relativistic particles willbe the manipulation of spinors.

We collect here some useful tricks forevaluating expressions in QED.

Spinor trick 1:

[u(f)Γu(i)]∗ =h

u(i)γ0Γ†γ0u(f)i

Spinor trick 2:X

s

us(p)us(p) = p+m

Trace trick 0:

Tr(I) = 4

Trace trick 1:

Tr(odd number of γ matrices) = 0

Trace trick 2:

Tr(γµγν) = 4gµν

Trace trick 3:

Tr(γµγνγργσ)

= 4(gµνgρσ − gµρgνσ + gµσgνρ)

Trace trick 4:

γµaγµ = −2a

Photon trick:X

polarizations

ǫµ(p)ǫ∗ν(p) → −gµν

See Peskin and Schroeder, Chapter 5,for the origin of the photon trick.

Example 40.2

The scattering amplitude is still given by iM = iZe2

q2us

′(p′)γ0us(p), but here we’ve

included the spin indices. We could work out the spinor part by brute force. How-ever, this isn’t usually necessary in real life. Experiments involve scattering initiallyunpolarized fermions and detecting the products of the scattering in detectors whicharen’t sensitive to polarization. We therefore average over the initial spin state andsum over the final spin states. We therefore want 1

2

P

s′,s |M|2, whose spinor partis given by

1

2

X

s′,s

|us′ (p′)γ0us(p)|2 =1

2

X

s′,s

h

us′(p′)γ0us(p)

i h

us′(p′)γ0us(p)

i∗. (40.5)

The calculations of these spin sums are quite straightforward, but do rely on theknowledge of a number of tricks (see sidenote). We will draw freely from themthroughout this chapter.

Using spinor trick 1, we have |u(f)Γu(i)|2 =û(f)γ0u(i)

˜ û(i)γ0u(f)

˜. Using

this, and writing out indices, our spinor term becomes

1

2

X

s,s′

us′(p′)αγ0

αβus(p)β u

s(p)ηγ0ηλu

s′ (p′)λ, (40.6)

where the explicit inclusion of subscripts allows us to rearrange the elements behindthe sum sign (and we assume a sum over repeated indices, as usual). Next we usespinor trick 2 which says

P

s [us(p)us(p)]αβ = [p+m]αβ . This is used to perform

the sum over the spin polarizations s and s′ as follows3

3Manipulations used in eqn 40.7 are:

Line 1: us(p)β us(p)η = [p+m]βη .

Line 2: Rearranging and using spinortrick 2 on primed variables.

Line 3: Identifying a trace.

12

P

s,s′ |us′(p′)γ0us(p)|2 = 1

2

P

s′ us′ (p′)αγ0

αβ [p+m]βη γ0ηλu

s′ (p′)λ= 1

2γ0αβ [p+m]βη γ

0ηλ [p

′ +m]λα= 1

2Trˆγ0(p+m)γ0(p

′ +m)˜.

(40.7)

To recap, we’ve boiled down the sum over spinors into the trace of a number ofmatrices.

40.3 Example 3: Compton scattering 357

Next we note that the vector a = γµaµ and we use some fun properties of thetraces of products of γ matrices. Trace trick 1 says that traces of odd numbers of γsvanish which means we may write:

1

2Trˆγ0(p+m)γ0(p

′ +m)˜

=1

2

˘Trˆγ0

pγ0p′˜+m2Tr

ˆ(γ0)2

˜¯. (40.8)

It’s fairly easy to work out Tr(γ0pγ

0p′) by plugging in explicit forms of the matrices,

or alternatively we can use the trace trick 3, to obtain 4(Ep′Ep +p′ ·p). Trace trick 0gives us 4m2 for the last term. Remembering that the scattering is elastic, we obtain

1

2

X

s,s′

|us′ (p′)γ0us(p)|2 = 2(E2p + p · p′ +m2). (40.9)

Referring back to the kinematics of the scattering, it’s straightforward to show that2(E2

p +p2 cos θ+m2) = 4E2p(1−β sin2 θ/2), which combined with |q|2 = 4p2 sin2 θ/2

allows us to complete the problem. We finally obtain

1

2

X

s,s′

|M|2 =Ze4

4p2β2 sin4 θ/2(1 − β2 sin2 θ/2). (40.10)

We can utilize our cross-section equation once more to obtain

dσ

dΩ=

Zα2

4p2β2 sin4 θ/2(1 − β2 sin2 θ/2), (40.11)

which is Mott’s formula.

I myself am neither an experimental-

ist nor a real mathematician; my the-

ory stops at the Schrodinger equation.

What I’ve done in this subject is to look

at the evidence, do calculations on the

back of an envelope and say to the the-

oretician ‘if you apply your techniques

to this problem, this is how it will come

out’ and to the experimentalists just

the same thing.

Nevill Mott (1905–1996)

40.3 Example 3: Compton scattering

Compton scattering is the process e−+γ → e−+γ. The two lowest order

Arthur Compton (1892–1962)With his wife and two sons, Dr. Comp-

ton lives in Chicago in a big brick house

filled with souvenirs of their world

tour. He does not know the taste of

hard liquor, almost never smokes, al-

ways offers a cigaret to women visi-

tors. He plays such a bang-up game

of tennis that he sometimes has a hard

time finding worthy opponents. Several

times a month he puts in an evening

of mandolin-playing with three friends.

When his graduate students have fin-

ished an examination, he likes to dine

them and take them to the theatre.

Time Magazine (1936).

diagrams (which are second order in the interaction term) are shown inFig. 40.3.

p′

pk

k′

p + k

p′

pk

k′

p− k′

(a)(b)

Fig. 40.3 Compton scattering. (a)The s-channel contribution. (b) The u-channel contribution.

Example 40.3

As practice in translating Feynman diagrams, we may translate the Compton dia-grams into amplitudes. The first diagram yields (upon dropping Qs again)

iMs-channel = u(p′)(i|e|γν)ǫ∗νλ′ (k′)

i

(p+ k) −m+ iǫǫµλ(k)(i|e|γµ)u(p), (40.12)

and the second diagram gives

iMu-channel = u(p′)(i|e|γν)ǫνλ(k)i

(p− k′) −m+ iǫǫ∗µλ′ (k

′)(i|e|γµ)u(p). (40.13)

These amplitudes may be evaluated to give the Klein–Nishina formula. However, the

Yoshio Nishina (1890–1951).

full derivation of the Klein–Nishina formula is rather lengthy and can be found inmany books,4 so we’ll evaluate the s-channel contribution from the first diagram for

4Peskin and Schroeder give, as usual,an especially clear derivation.

the special case of a highly relativistic particle. (Recall that here ‘relativistic’ meanswe can ignore the mass of the electron.) We therefore have that p2 = 0 and (asusual) k2 = 0. We’re interested in |M|2 averaged over initial spin states and photonpolarizations and summed over final ones (we pick up a factor 1/4 in doing this).We therefore want (employing spinor trick 15 and dropping the photon polarization

5We have the identity

γ0`γνγλγµ

´†γ0 = γµγλγν , and

so γ0 (γνǫ∗νpǫµγµ)† γ0 = γµǫ∗µpǫνγ

ν .

labels):

1

4

X

s,s′

X

polarizations

e4

(p+ k)2

h

us′(p′)γνǫ∗ν(k

′)(p+ k)ǫµ(k)γµus(p)i

×h

us(p)γσǫ∗σ(k)(p+ k)ǫρ(k′)γρus

′(p′)

i

. (40.14)

358 QED scattering: three famous cross-sections

Freely employing our array of tricks we haveSuccessive lines of eqn 40.15 use the fol-lowing ideas:

Line 1: Photon trick.

Line 2: Contracting indices and spinortrick 2.

Line 3: Using p2 = p2 = 0 and trace

trick 4.

Line 4: Trace trick 3.

Line 5: u = (p′ − k)2 = −2p′ · k,s = (p+ k)2 = 2p · k.

1

4

X

|Ms-channel|2 =

e2

4s2

X

s,s′

“

gµσgνρus′ (p′)γν(p+ k)γ

µus(p)

×us(p)γσ(p+ k)γρus

′(p′)

”

=e2

4s2Trh

γν(p+ k)γµpγµ(p+ k)γνp

′i

=e2

s2Trh

kpkp′i

=4e2

s22(p · k)(p′ · k) = −2e2u

s. (40.15)

It turns out that the second diagram contributes − 2e4su and there is

no interference term between the two. The electron spin and photonpolarization averaged, squared amplitude for Compton scattering is then

1

4

∑

s,s′

∑

polarizations

|M|2 = −2e4(us

+s

u

). (40.16)

40.4 Crossing symmetry

As a final illustration of the utility of Feynman diagrams in describingscattering, we briefly note another very useful feature. Feynman’s inter-pretation of antiparticles as particles travelling backward in time maybe used to manipulate Feynman diagrams, allowing us access to sev-eral more physical amplitudes from the evaluation of a single Feynmanamplitude.

φ

p

φ

k = −p

Fig. 40.4 Crossing symmetry. A Feyn-man diagram with an incoming par-ticle may be manipulated to describea new process involving an outgoingantiparticle with opposite charge andfour-momentum.

The principle is shown in Fig. 40.4, where we see that if we start witha diagram with an incoming particle φ then we may flip the leg to createa valid process involving an outgoing antiparticle φ with opposite chargeand momentum. This process, known as a crossing, has the wonderfulproperty that

M (φ(p) + ...→ ...) = M(...→ ...+ ¯φ(k)

), p = −k. (40.17)

Example 40.4

Consider the s-channel process e−e+ → µ−µ+ examined in Chapter 39 and shownin Fig. 40.5. If we ‘cross’ the diagram by reversing the momentum direction of the

e− e+

µ− µ+

pApB

pD pC

Fig. 40.5 The s-channel processe−e+ → µ−µ+.

e− µ−

e− µ−

kA kC

kB kD

Fig. 40.6 The t-channel processe−µ− → e−µ− found by crossing theprevious diagram.

incoming e+ and outgoing µ+ (changing the sign of their charge as we do so), thenupon flipping the diagram over we obtain Fig. 40.6 describing the process e−µ− →e−µ−. The amplitude for the resulting diagram is the same as the M we calculatedin Chapter 39 if we make the replacements: pA → kA, pB → −kB, pC → kC andpD → −kD. Note that we have turned an s-channel process into a t-channel processwith our crossing, reflected by the fact that instead of reversing individual momentumwe may simply replace s→ t in the amplitude.

Exercises 359

Crossing symmetries are clearly a very useful feature of the formalismbecause calculating |M|2 for one process immediately gives us accessto |M|2 for all processes related by crossings. As we have only brieflytouched on this topic here, we recommend you consult one of the stan-dard texts before deploying crossings in anger. Peskin and Schroederand Halzen and Martin both contain several examples.

Chapter summary

• This chapter has illustrated three simple scattering examples fromQED: (1) Rutherford scattering; (2) the Mott formula; (3) Comp-ton scattering.

• Crossing symmetry makes a Feynman amplitude for one processdescribe another process.

Exercises

(40.1) Verify that the transferred momentum in theRutherford calculation is given by q2 = 4p2 sin2 θ

2.

(40.2) Verify the Trace Tricks used in this chapter.

(40.3) (a) Calculate Tr(γ0

pγ0

p′) using explicit forms of the

matrices.(b) Repeat the calculation using Trace Trick 3.

(40.4) Consider the e−e+ → µ−µ+ problem from the pre-vious chapter. Using the tricks considered in thischapter show that

1

4

X

spins

|M|2 =e4

4q4Trh

(p′ −me)γ

µ(p+me)γνi

×Trh

(k +mµ)γµ(k′ −mµ)γν

i

.(40.18)

(b) Taking all particles to be highly relativistic,show that

1

4

X

spins

|M|2 =8e4

q4ˆ

(p · k)(p′ · k′) + (p · k′)(p′ · k)˜

.

(40.19)(c) Finally, use the kinematics discussed in Chap-ter 39 to show 1

4

P

spins |M|2 = e4(1 + cos2 θ).See Peskin and Schroeder, Chapter 5 for help.

(40.5) Show that eqn 40.19, which describes the processe−e+ → µ−µ+, may be written in terms of Man-delstam variables as

1

4

X

spins

|M|2 =8e4

s2

"

„

t

2

«2

+“u

2

”2#

. (40.20)

Use a crossing to determine the analogous equationfor the process e−µ− → e−µ−.

(40.6) Møller scattering is the process e−e− → e−e−.(a) Two diagrams contribute to the Feynman am-plitude for this process. Identify these and showthat

M = −e2

tu(k)γµu(p)u(k′)γµu(p

′)

+e2

uu(k′)γµu(p)u(k)γµu(p

′).(40.21)

(b) Show further that, in the ultra-relativistic limit,

1

4

X

spins

|M|2 =e4

4

1

t2Trˆ

kγµ

pγν˜Tr

h

k′γµp

′γνi

+1

u2Trh

k′γµpγ

νi

Trh

kγµp′γνi

− 1

tuTrh

kγµ

pγνk

′γµp′γνi

− 1

tuTrh

k′γµpγ

νkγµp

′γνi

ff

.

(c) Use a crossing symmetry to turn the previousequation into one describing the process e−e+ →e−e+, which is known as Bhabha scattering.

41The renormalization of

QED and two great results

41.1 Renormalizing the photonpropagator: dielectric vac-uum 361

41.2 The renormalization groupand the electric charge 364

41.3 Vertex corrections and theelectron g-factor 365

Chapter summary 368

Exercises 368

His laboratory is his ball-point pen.Caption to a photograph of Julian Schwinger (1918–1994)(holding a pen)

α

2πEngraved on Julian Schwinger’s tombstone

Some really stunning consequences of QED are seen when we examineprocesses that contain electron–positron loops. The integrals we en-counter when considering these processes contain divergences that wemust tame with renormalization. In this chapter we will demonstratetwo of the great results that follow from renormalizing QED. These are(1) the fact that the electronic charge is not a constant and (2) the factthat the g-factor of the electron is not quite g = 2.

(a)

(b)

(c)

(d)

(e)

(f)

1PI

1PI

Fig. 41.1 Feynman diagrams forrenormalized QED.

To renormalize QED we calculate three Green’s functions:

(a) Electron 1PI self-energy −iΣ(p),(b) Photon 1PI self-energy iΠµν(q),

(c) Interaction vertex function −iQ|e|Γµ(p, p′).(41.1)

These functions, each made of an infinite sum of Feynman diagrams,are themselves shown diagrammatically in Fig. 41.1(a–c). In calcu-lating each of these Green’s functions we will encounter contributionswhich lead to divergences. These divergences are fixed through the in-troduction of counterterms. QED is a renormalizable theory and onlythree counterterm diagrams, each corresponding to one of the Green’sfunctions above, are required. The QED counterterms are shown inFig. 41.1(d–f). They have Feynman rules:

(d) Electron self-energy (Bird on a wire) i (pB +A) ,(e) Photon self-energy (Bird on a wave) i(gµνq2 − qµqν)C,(f) Interaction vertex (Electrocuted bird) iQ|e|γµD.

(41.2)In order to renormalize the theory we impose conditions on each ofthe Green’s functions which ensure that we are expanding in the correctmasses and coupling constants. For QED the renormalization conditionsare:

−iΣ(p = m) = 0 fixes electron mass to m,

iΠµν(q = 0) = 0 fixes photon mass to 0,

−iQ|e|Γµ(p′ − p = 0) = iQ|e|γµ fixes charge to Q|e|.(41.3)

41.1 Renormalizing the photon propagator: dielectric vacuum 361

We will begin our tour of the consequences of renormalizing QED by ex-amining the case of the photon propagator and show how renormalizingit explains the dielectric properties of the vacuum.

41.1 Renormalizing the photon

propagator: dielectric vacuum

A dielectric is an insulator which may be polarized with an appliedelectric field. This is encoded in the dielectric constant of a material ǫ,which tells us how much the dielectric alters the electrostatic potentialof a charge. Renormalized QED predicts that the interacting vacuumis a dielectric! The dielectric properties of the vacuum arise becauseinteractions dress the bare photon in a cloud of electron–positron pairs.To see how this works we will examine the photon’s self-energy. As inChapter 33 we define the 1PI photon self-energy which is given by

iΠµν(q) =∑(

All 1PI diagrams that can beinserted between photon lines

). (41.4)

(This differs in sign from the 1PI self-energy Green’s functions encoun-tered previously.) Some examples of contributions to iΠµν(q) are shownin Fig. 41.2. Notice that these involve the photon removing electron–positron pairs from the vacuum in various ways and then returning them.

= + +

+ + + · · ·

1PI

Fig. 41.2 Some contributions to the

Green’s function iΠµν(q).Looked at in terms of pictures, it’s clear how the photon self-energy

will renormalize the photon propagator. As shown in Fig. 41.3, we sim-ply sum the contribution of the self-energy to infinity. The version inequations is made a little trickier by the presence of the indices, whichreflect the vector-like nature of the photon field.1 1Remember that the massless nature of

the photon field means that photons arenot vector particles.

= + + + · · ·

= 1

( )−1 − 1PI q2

(a)

(b)

1PI

1PI

1PI

Fig. 41.3 Renormalization of the pho-ton propagator.

Example 41.1

In equations the sum is

Dµν(q) = D0µν(q) + D0µλ(q)[iΠλη(q)]D0ην(q) + . . .

=−igµν

q2+

„−igµλ

q2

«

iΠλη(q)

„−igην

q2

«

+ . . .

=−igµν

q2+

„−i

q2

«

iΠµν(q)

„−i

q2

«

+

„−i

q2

«

iΠηµ(q)

„−i

q2

«

iΠην(q)

„−i

q2

«

+ . . .

(41.5)

Summing the series we obtain

Dµν(q) = − igµν

q2+

1

q2Πµη(q)D

ην (q), (41.6)

or “

q2gµη − Πµη(q)”

Dην (q) = −igµν . (41.7)

Things are simplified rather by the use of the Ward identity, which guarantees thatqνΠµν(q) = 0. As a result, we can write the photon self-energy as

Πµν(q) = (q2gµν − qµqν)Π(q). (41.8)

Substitution of this new form yields

q2[1 − Π(q)]Dµν(q) + Π(q)qµqηDην (q) = −igµν , (41.9)

362 The renormalization of QED and two great results

whose solution is

Dµν(q) =−igµν

q2[1 − Π(q)]+

iqµqνΠ(q)

q4[1 − Π(q)]. (41.10)

In practical calculations the qµqν part never contributes to any observable amplitude.This is because Dµν(q) is always coupled to conserved currents and we know thatqµJ

µem = 0.

The resulting renormalized photon propagator is given by

Dµν(q) =−igµν

q2[1 − Π(q)]. (41.11)

Note that we would expect that Π(q), which contains loops, will divergeunless we use counterterms in our calculation of Π(q2).

Example 41.2

Let’s see how to use the counterterm. Instead of summing over all 1PI inser-tions, we’ll limit our calculation at a single electron–positron loop, which we’ll calliπµν(q) = i(gµνq2 − qµqν)π(q). To prevent the loop causing any trouble we include

Fig. 41.4 The lowest order correctionto the photon propagator and the coun-terterm that prevents its divergence.

a second-order counterterm which makes a contribution i(gµνq2 − qµqν)C(2). Thetotal insertion in the photon line is shown in Fig. 41.4. Summing this insertion to allorders, we obtain

Dµν(q) =−igµν

q2˘1 −

ˆπ(q) + C(2)

˜¯ . (41.12)

The renormalization condition is that the photon self-energy Π(q) vanishes at q2 = 0,implying that π(q = 0) = −C(2) and the propagator becomes

Dµν(q) =−igµν

q2 1 − [π(q) − π(0)] . (41.13)

This will be enough to remove the divergences from the calculation of Dµν(q) tosecond order.

The bare photon propagates while tearing electron–positron pairs fromthe vacuum. These processes have the effect of modifying the photon’samplitude for propagation between two points. More than that, weknow that virtual photons (and remember that all photons are, to someextent, virtual photons) couple to a fermion line at both ends of theirtrajectory. If we write the vertex Feynman rule as −iQ|e0|γµ, we canbundle up the factors of |e0| with the photon propagator, so that wehave2 (considering Q = 1 for simplicity):2Using the shorthand that e0 is the

unrenormalized charge and assumingthat the counterterm contribution is ac-counted for by Π(0). Dµν(q) =

−ie20gµν

q2

1 −[Π(q) − Π(0)

] =−igµνq2

e20

1 −[Π(q) − Π(0)

]

.

(41.14)Written the second way, the term in the brackets may be considered themomentum-dependent electric charge e(q). We could say that the real-life electric charge, renormalized by the interactions with the vacuum, is

given by |e| = |e0|/

1 −[Π(q) − Π(0)

]. The electric charge therefore

41.1 Renormalizing the photon propagator: dielectric vacuum 363

depends on the momentum transferred by the virtual photon that me-diates the electromagnetic force! A physical explanation for this featureof the dielectric properties of the vacuum is provided by the concept ofscreening, as we discuss below.

Example 41.3

In what follows we will require an expression for the one-loop amplitude in Fig. 41.5,

q

q

pp + q

µ

ν

Fig. 41.5 The photon propagator witha single electron–hole loop insertion.

which may be translated from diagram to equation, giving

iπ(q) = (−1)(−ie0)2Z

d4p

(2π)4Tr

„

γµi

p−mγν

i

p+q −m

«

. (41.15)

This integral can be done. With the inclusion of the counterterm it is found thatπ(q) − π(0), the one-loop contribution to Π(q), is given by

π(q) − π(0) = − e202π2

Z 1

0dx (x− x2) ln

„m2

m2 − (x− x2)q2

«

. (41.16)

We’ve seen that the renormalized propagator is given, to second orderin the interaction, by

Dµν(q) =−igµνq2

e201 − [π(q) − π(0)]

≈ −igµνe20

q2[1 + π(q) − π(0)] . (41.17)

To see the consequence of this on electrostatics, we need the static ver-sion of the propagator, obtained by taking q2 = −q2 thus:

π(−q) − π(0) = − e202π2

∫ 1

0

dx (x− x2) ln

(m2

m2 + (x− x2)q2

). (41.18)

Expanding this in the limit q2 ≪ m gives us

Dµν(q) ≈ igµνe20

q2

(1 +

α

15π

q2

m2,

), (41.19)

where α = e20/4π.Recall from Chapter 20 our (Born approximation) interpretation of

the momentum space propagator as proportional to a matrix elementof a potential V (r) via D00(q) ∝ −

∫d3r eiq·rV (r). Doing the inverse

Fourier transform of D00(q) yields a potential due to the electron of

V (r) = −α

|r| +4α2

15m2δ(3)(r)

. (41.20)

The first term is Coulomb’s potential for a point charge. The secondterm is the correction due to the screening of the electron charge thatis provided by the virtual electron–hole pairs. The screening effect isvisualized in Fig. 41.6, where we represent the pairs as effective dipoles.The screening is known as vacuum polarization.

−

−+

−+

−+

− +− +

−+

−+ −

+

−+

−+

−+

−+

−+−+

− +

−+

Fig. 41.6 Screening in QED.


Importantly, this effect can be measured! It causes a shift in the hy-drogen energy levels of ∆E = − 4α2

15m2 |ψ(0)|2, where ψ(x) is the hydrogenwave function (which is only nonzero at the origin for l = 0 levels). Thiseffect causes a shift in the 2S 1

2to 2P 1

2transition in hydrogen of −27 MHz

which, despite making up only a small part of the famous Lamb shift(of +1057 MHz), has indeed been experimentally verified.3 The effect of3Willis E. Lamb, Jr (1913–2008) mea-

sured the energy level shift in hydro-gen in 1947. [See W. E. Lamb and R.C. Retherford, Phys. Rev. 72, 241(1947)]. Historically, this measurementwas of immense importance in convinc-ing physicists to take corrections due tovirtual photons seriously. Shortly afterthe announcement of the experimentalresult, Hans Bethe (1906–2005) made anon-relativistic calculation of the shiftcaused by the self-energy of an elec-tron in the electromagnetic field of anucleus using an early form of renor-malization. The agreement with ex-periment was good and the fully rela-tivistic explanation soon followed froma number of physicists. See Schweberfor the history and Weinberg, Chapter14, for the full calculation of the Lambshift.

vacuum polarization is also very important in the quantum field theoryof metals and will be discussed further in Chapter 43.

41.2 The renormalization group and the

electric charge

Using what we know about vacuum polarization along with the renor-malization group (of Chapter 34) we can work out how the electric chargechanges, depending on the size of the momentum that the photon is car-rying. That is, how the QED coupling depends on the energy scale. Todo this we need to calculate the β-function, given here by β = µd|e|

dµ ,which tells us how the renormalized charge e depends on the choice ofenergy scale µ.

Example 41.4

Using our equation for the renormalized electric charge we can expand for small π(q)to obtain

e2(µ) =e20

1 − (π(µ) − π(0))≈ e20[1 + (π(µ) − π(0))] +O(e4), (41.21)

or

|e| ≈ |e0|»

1 +1

2(π(µ) − π(0))

–

. (41.22)

The β-function is then given by

β = µd|e|dµ

=1

2µ|e0|

dπ(µ)

dµ. (41.23)

Now we use eqn 41.16 which states that

π(µ) − π(0) = − e202π2

Z 1

0dx (x− x2) ln

»m

m2 − (x− x2)µ2

–

. (41.24)

In the large energy scale limit (µ≫ m) differentiation gets us

dπ(µ)

dµ=

e20π2µ

Z 1

0dx (x− x2) =

e206π2µ

. (41.25)

Note that, to the order to which we’re working, it’s permissible to swap e0 for e onthe right-hand side in what follows.

We obtain a β-function

β = µd|e|dµ

= +|e|312π2

, (41.26)

where we accentuate the + sign, since that’s the most important part.This shows that the electromagnetic coupling increases with increasing

41.3 Vertex corrections and the electron g-factor 365

energy scale. Another way of saying this is that it increases with de-creasing length scale. The electron appears to be more strongly chargedas you get closer to it. This makes sense in the context of the last sec-tion, where we interpreted vacuum polarization as a screening process.As you get closer to the electron you penetrate the cloud of screeningpairs and the charge seems to increase.

41.3 Vertex corrections and the electron

g-factor

Other people publish to show you how to do it, but JulianSchwinger publishes to show you that only he can do it.Anon, quoted from S. Schweber, QED and the Men whoMade It

p

p′

p′ − p

Fig. 41.7 The vertex Green’s function−iQ|e|Γµ(p, p′).

The magnetic moment operator of the electron may be written

µ = g

(Q|e|2m

)S, (41.27)

where Q = −1 for the electron. The Dirac equation predicts4 that the

4See Example 36.9.

electron g-factor of the electron is exactly 2. In this section we will makethe link between renormalized QED and the g-factor. This will lead toJulian Schwinger’s famous prediction that the first-order correction tothe QED interaction vertex causes a shift from g = 2 to g = 2 + α/π.That is to say that the fact that electrons can emit virtual photonschanges the form of the electron–photon interaction in a measurablemanner.

The Green’s function for the electron–photon vertex can be written−iQ|e|Γµ(p, p′) and its diagram is shown in Fig. 41.7. It is defined as5

5The overal sign is chosen becausethe first-order interaction is given by−iQ|e|γµ so to first-order Γµ = γµ.

−iQ|e|Γµ(p, p′) =∑

All amputated insertions with oneincoming fermion line, one outgoingfermion line and one photon line.

.

(41.28)Some examples are shown in Fig. 41.8. The vertex function can be

=

+

+

+ + · · ·

Fig. 41.8 Contributions to the Green’s

function −iQ|e|Γµ. The contribution tofirst order is simply given by the inter-action vertex −i|e|γµ.

read as describing an off mass-shell photon decaying into an electron–positron pair. Since the photon has6 JP = 1−, there are two possible

6In this notation we list the total angu-lar momentum J and the intrinsic par-ity P of the photon.

configurations of the pair. It can have L = 0 and S = 1, or L = 2 andS = 1. This fact is reflected in the form in which we write the vertexfunction as a sum of two terms.

Γµ(p, p′) = γµF1(q) +iσµνqν

2mF2(q), (41.29)

where qµ = p′µ − pµ and where σµν = i2 [γµ, γν ]. The function F1(q) is

known as the Dirac form factor7 and F2(q) is known as the Pauli form7In fact, F1(q) may be thought of asa Fourier transform of the charge dis-tribution and has the property thatF1(0) = 1 at all orders of perturbationtheory.

factor. For vanishing q we need only consider the first-order contributionto Γµ = γµ, which is the first diagram in Fig. 41.8. We read off that forq → 0 we have F1(0) = 1 and F2(0) = 0.


We will now sketch out the determination of the g-factor of the elec-tron in a manner intended ‘to show you how to do it’. Since the calcula-tion is quite lengthy and involved, we will break it down into a numberof steps.

Step I: Consider the diagram in Fig. 41.9 showing the interactionvertex coupled to a classical source Acl

µ (x), giving a Feynman rule con-

tribution of −iQ|e|ΓµAclµ (q), where qµ is the momentum of the photon.

Coupling this to the spinor current of the incoming and outgoing elec-tron, we have an amplitude

iM = −iQ|e|u(p′)Γµ(p, p′)u(p)Aclµ (q), (41.30)

where q = p′ − p and an energy momentum conserving delta functionhas been suppressed. Inserting eqn 41.29 for Γµ yields

iM = −iQ|e|u(p′)[γµF1(q) +

iσµνqν2m

F2(q)

]u(p)Acl

µ (q). (41.31)

p

p′

q

Aclµ(q)

Fig. 41.9 The vertex function coupledto a classical source.

Step II: The next step makes the u′γµu in the first term much morecomplicated and uses an identity, known as the Gordon decomposition(proved in Exercise 36.7), given by

u(p′)γµu(p) = u(p′)

[p′µ + pµ

2m+

iσµνqν2m

]u(p). (41.32)

The reason for this step is that is eliminates γµ, pushing the spin-independent part into the first term and the spin-dependent part intothe second. Substituting this into our Feynman amplitude yields

iM = −iQ|e|u(p′)(p′µ + pµ

2m

)u(p)Acl

µ (q)F1(q)

+Q|e|2m

[u(p′)σµνqνu(p)] Aclµ (q) F1(q) + F2(q) . (41.33)

We now have an expression for the amplitude which is separated intothe spin-independent first term and a spin-dependent second term. Theinformation on the g-factor is to be found in the second term.

Step III: Now we take the non-relativistic limit. This involves writing

u(p) ≈ √m

(ξξ

). We also need to use the explicit forms for the

components of σµν

σ0i =

(−iσi 0

0 iσi

)and σij =

(εijkσk 0

0 εijkσk

). (41.34)

Using these we find that u′σ0iu = 0 and, more importantly, thatu′σiju = 2m

[ξ′†σkξ

]εijk.

Putting this all together we have, for the spin-dependent part of theamplitude of our diagram in the limit q → 0, that

Spin-dependentamplitude

= (2m)Q|e|2m

[ξ′†σkξ

]εijkqjAcli(q) F1(0) + F2(0)

= (2m)iQ|e|2m

[ξ′†σkξ

]Bk(q) 1 + F2(0) , (41.35)

41.3 Vertex corrections and the electron g-factor 367

where we’ve assumed Acl(x) = (0,Acl(x)) and used the fact that, inmomentum space, the components of the magnetic field are given byiεijkqiAclj(q) = Bk(q). We conclude that the spin-dependent part ofthe amplitude is proportional to Q|e|1 + F2(0)〈S〉 · B, where the ex-pectation value of the spin operator is 〈S〉 = ξ′†σξ/2.

We now have an amplitude, so we may extract a scattering potentialusing the Born approximation.

iM = −i〈f |V (x)|i〉. (41.36)

Since we are still using the relativistic normalization, we need to correctby dividing through by twice the energy of the electron that’s scattered(= 2m). We obtain a real-space potential

V (x) = −Q|e|m

1 + F2(0)〈S〉 · B(x). (41.37)

Comparing this to the expected expression for the potential energy of amagnetic moment µ in a magnetic field, namely

V (x) = −〈µ〉 · B(x) = −[g

(Q|e|2m

)〈S〉]· B(x), (41.38)

we may read off that g = 2 [1 + F2(0)].So g is not exactly 2! There’s a correction determined by F2(0). The

correction can be calculated by evaluating the contributions to Γµ.

p

p′

q

k

k′ =

k+q

p− k

Fig. 41.10 The third-order diagramwhich contributes to F2(0).

Step IV: Let’s look at the contributions to −iQ|e|Γµ, order by order.To first order we have the bare vertex whose contribution (for Q = −1)is i|e|γµ, so the first-order contribution to Γµ is γµ implying F1(0) = 1and F2(0) = 0 as we said earlier. There are no second-order diagrams,but to third order the only diagram that contributes to F2(q) is thatshown in Fig. 41.10.

Example 41.5

From the labelled diagram, shown in Fig. 41.10, we can write down the amplitudefor the vertex as

Γµdiagram

=

Zd4k

(2π)4−igνρ

(p− k)2 + iǫu(p′)(i|e|γν) i

k′ −m+ iǫγµ

i

k −m+ iǫ(i|e|γρ)u(p).

(41.39)This is divergent, so we must also include a third-order counterterm. With theinclusion of the counterterm in Fig. 41.1(e) and two pages or so of Dirac algebra8 8See Peskin and Schroeder, which is

particularly informative on this sub-ject.

one can derive a result for the Pauli form factor which is F2(0) = e2/8π2.

The result: We evaluate g = 2[1 + F2(0)] with F2(0) = e2

8π2 and findthe famous result for the shift in the g-factor: Recall that in the units used here, the

fine structure constant is given by α =e2/4π. Because α = 1/137.036... wehave g = 2.0023... In SI units, α =e2/4πǫ0~c.

g = 2(1 +

α

2π+ . . .

). (41.40)

This was also Julian Schwinger’s result and one for which he justifiablygained a spontaneous round of applause when he revealed, in a talk at


the New York APS meeting in 1948, that his theory agreed precisely withthe crucial experiments. Of course, higher order corrections to this andother QED processes may be evaluated and have led to QED becomingthe most stringently tested theory in physics.9

9See Peskin and Schroeder, page 196,for a discussion of the remarkable agree-ment of QED with modern experimen-tal tests.

Chapter summary

• Renormalizing the photon propagator shows that the electriccharge is given by |e| = |e0|/[1 − Π(q)]. The electric charge isscreened by virtual electron–positron pairs (known as vacuum po-larization) and this gives rise to the Lamb shift.

• The Dirac theory predicts g = 2, but including the electron–photonvertex −iQ|e|Γ(p, p′) (for an off mass-shell photon decaying into anelectron–positron pair) yields g = 2

(1 + α

2π + · · ·)

which agreesextremely well with experiment.

Exercises

(41.1) Verify

Dµν(q) =−igµν

q2[1 − Π(q)]+

iqµqνΠ(q)

q4[1 − Π(q)], (41.41)

is the solution to eqn 41.9.

(41.2) (a) Verify

Dµν(q) ≈ igµνe20

q2

„

1 +α

15π

q2

m2

«

. (41.42)

(b) Take the inverse Fourier transform and showyou obtain the form in the text.

(41.3) The electron 1PI self-energy is defined as

−iΣ =X

„

All 1PI insertions with incomingand outgoing electron lines

«

.

(41.43)Draw contributions to −iΣ in QED up to fourthorder in the interaction.

(41.4) The Schwinger model deals with quantum electro-dynamics in (2+1)-dimensional spacetime and pre-dicts a one-loop vacuum polarization

−iΠµν(p) = − ie2

πp2(pµpν − p2gµν). (41.44)

Show that the photon takes on a mass as a resultand determine its size.

(41.5) Verify eqn 41.25.

(41.6) (a) Show that

σ0i =

„

−iσi 00 iσi

«

(41.45)

and

σij =

„

εijkσk 0

0 εijkσk

«

. (41.46)

(b) Show that uσiju = 2mξ†σkξεijk.

Part X

Some applications from the

world of condensed matter

Quantum field theory finds many applications in the field of condensedmatter physics which is concerned with the properties of many-particlesystems whose fundamental excitations can be regarded as particles. Wehave already seen that lattice vibrations can be treated as particles calledphonons. In the case of metallic systems which contain large numbersof electrons, the interactions between electrons can be very importantin determining the properties.

• Our first condensed matter physics example is the superfluid, andin Chapter 42 we show how an approximation due to Bogoliubovallows weakly interacting Bose systems to be treated. The quasi-particles are called bogolons and the ground state of the systemcan be treated as a coherent state. The Noether current is shownto arise from a gradient in the phase of this coherent state.

• Interactions between electrons in a metal are treated in Chapter 43and we introduce the Cooper, Hartree and Fock terms in the per-turbation expansion for the ground state energy shift. We explainhow to model the excitations and study them using propagatorsand we introduce the random phase approximation.

• The BCS theory of superconductivity is presented in Chapter 44and we show how to write down the BCS coherent state, derivethe quasiparticles and explore the effect of broken symmetry. Thisallows us to explore the Higgs mechanism in a superconductor.

• The fractional quantum Hall effect is introduced in Chapter 45 andwe explain how quasiparticles carrying fractional charge emerge inthis theory.

A note on notation: In this part we conform to the conventions of thecondensed matter literature by denoting the number of particles by Nand define the number density of particles n = N/V, where V is thevolume. For consistency we write number operators and occupationnumbers for the state with momentum p as Np and Np respectively.

42 Superfluids

42.1 Bogoliubov’s hunting license370

42.2 Bogoliubov’s transformation372

42.3 Superfluids and fields 374

42.4 The current in a superfluid377

Chapter summary 379

Exercises 379

There is no doubt a special place in hell being reserved forme at this very moment for this mean trick, for the task isimpossible.Robert Laughlin (1950– ), on asking his students to deducesuperfluidity from first principles

A gas of weakly interacting, non-relativistic bosons is a problem wherethe methods of quantum field theory allow us to understand the groundstate, the excitations and the breaking of symmetry. Not only is thisan illustration of many of the methods we’ve discussed; it also appliesto one of the most fascinating phenomena in condensed matter physics:superfluidity. A superfluid is a state of matter where momentum flowswithout dissipation. When 4He is cooled below 2.17 K it becomes asuperfluid and may flow through capillaries without resistance. It alsoexhibits the fountain effect whereby a beaker of superfluid spontaneouslyempties itself.

The dispersion of superfluid helium has been measured with inelasticneutron scattering and is shown in Fig. 42.1. At low momentum thedispersion is linear, at large momentum the energy goes as p2/2m asexpected for free particles. In the middle is a large minimum. We willdiscover that our gas of weakly interacting bosons captures the low andhigh momentum behaviour of the superfluid dispersion. We will alsofind that the form of the field that we obtain on spontaneous symmetrybreaking necessarily gives us a superflow of momentum.

0

0.5

1

1.5

Ep

(meV

)

0 1 2 3

|p|/h (A−1)

Fig. 42.1 The measured dispersionof superfluid helium. Data taken fromA.B.D. Woods and R.A. Cowley, Can.J. Phys. 49, 177 (1971).

42.1 Bogoliubov’s hunting license

The problem revolves around the Hamiltonian for non-relativistic Boseparticles interacting with a momentum-independent potential g. TheHamiltonian describing this state of affairs is

H =1

2m

∫d3x∇φ†(x) · ∇φ(x)

+g

2

∫d3xd3y φ†(x)φ†(y)φ(y)φ(x)δ(3)(x − y)

=∑

p

p2

2ma†pap +

g

2V∑

kpq

a†p−qa†k+qakap, (42.1)

where the second form follows from putting the system in a box andinserting the mode expansion φ(x) = 1√

V∑

p apeip·x. As usual, we call

42.1 Bogoliubov’s hunting license 371

the interacting ground state of this Hamiltonian |Ω〉. This is intendedto describe a very large number of interacting particles. For a macro-scopic sample of matter in a laboratory this number might be around1023. Unfortunately, the interaction part of the Hamiltonian, given byHI = g

2V∑

kpq a†p−qa

†k+qakap, is too complicated to solve, so we need

an approximation. Bogoliubov came up with one by thinking about the

Nikolay Bogoliubov (1909–1982)

low-energy behaviour of the system.1 Bogoliubov’s assumption (based

1For ‘low-energy’, you might read ‘low-temperature’ if you’re imagining an ex-periment in a laboratory.

upon Bose–Einstein condensation) is that, if the system has very littleenergy, the number of particles N0 in the p = 0 ground state of theHamiltonian will be macroscopically large. The consequence of this isthat instead of the exact relation ap|Ω〉 =

√N0|N0 − 1〉 (where |N0 − 1〉

is a state achieved by removing one zero momentum particle from theground state), we can take ap=0|Ω〉 ≈

√N0|Ω〉.

Bogoliubov used this as an approximation2 to allow the replacement 2In his words, this approximation pro-vides a ‘hunting license’ to search fornew phenomena.

of operators ap=0 and a†p=0 with the number√N0. He then broke down

the sum over momentum states∑

kpq a†p−qa

†k+qakap by considering, in

turn, the terms in the sum in which particular momentum subscripts arezero, and replacing that operator with the number

√N0. Since terms

with odd numbers of operators will always give zero expectation value,we need only consider the cases of two or four of the indices being zero.The method then relies on sending the momentum index of pairs, or allfour, of the operators to zero. Then we replace the operators a0 and a†0with

√N0.

Example 42.1

We take turns setting pairs of momentum subscripts to zero. Replacing pairs ofsubscripts yields six terms:

p = 0, k = 0, a†−q a†q a0a0 → N0a

†−q a

†q ,

p = 0, k + q = 0, a†−q a†0ak a0 → N0a

†kak ,

p = 0, p− q = 0, a†0a†kak a0 → N0a

†kak ,

k = 0, k + q = 0, a†p a†0a0ap → N0a

†p ap ,

k = 0, p− q = 0, a†0a†q a0ap → N0a

†p ap ,

k + q = 0, p− q = 0, a†0a†0a−q aq → N0a−q aq . (42.2)

Lastly, we replaces all four of the subscripts to get

a†0a†0a0a0 = N2

0 . (42.3)

With this approximation we obtain an approximate interaction Hamil-tonian

HI ≈g

2V

N2

0 + 4N0

∑

p6=0

a†pap +N0

∑

p6=0

(a†pa†−p + apa−p)

. (42.4)

Now we must prepare ourselves for a shock. This seemingly quite reason-able approximation of saying that there are a large number of particles

372 Superfluids

in the lowest momentum state has a dramatic consequence: it breaks asymmetry. Our original Hamiltonian is invariant under the global U(1)transformation φ(x) → φ(x)eiα, or equivalently ap → apeiα. Looking

at HI, we see that the terms a†pa†−p and apa−p both change if we make

the latter transformation.3 We have therefore lost the U(1) symmetry3Note that this global transformationdoesn’t depend on the momentum. and its conserved quantity, which (remembering back to our discussion

of global U(1) symmetry) was particle number.44This result is perhaps unsurprising inthat our ‘hunting license’ approxima-tion is to equate |N−1〉 with |Ω〉, whichshows a slightly cavalier attitude to par-ticle number.

Since we’ve now lost the conservation of particle number, we don’tknow what value to assign to N0. To remedy this we set the totalnumber of particles N equal to the number in the p = 0 ground state,plus all the rest, that is:

N = N0 +∑

p6=0

a†pap. (42.5)

The result of these manipulations is to yield an effective Hamiltonian

H =∑

p6=0

(p2

2m+ ng

)a†pap +

1

2

∑

p6=0

ng(a†pa†−p + apa−p), (42.6)

where n = N/V and we’ve dropped the constant 12gn

2 term.The next problem is that the potential term in the Hamiltonian isn’t

diagonal. That is, it isn’t expressed in terms of number operators of theform b†q bq. This may spell trouble unless there’s a way to turn these non-diagonal objects into diagonal ones. That is, we want to turn objectslike a†pa

†−p and apa−p into number operators.

42.2 Bogoliubov’s transformation

The way to turn these non-diagonal bilinears into number operators isto transform to a new set of operators. The procedure that does thisis a Bogoliubov transformation which defines a new set of operatorsαp and α†

p via

(ap

a†−p

)=

(up −vp

−vp up

)(αp

α†−p

), (42.7)

and we will use these operators to diagonalize the Hamiltonian. Thequantities up and vp obey the following rules:

u2p − v2

p = 1, u∗p = up, v∗p = vp, (42.8)

which are cunningly designed so that the new operators αp and α†p obey

the same commutation relations as ap and a†p:

[αp, α

†q

]= δpq, [αp, αq] =

[α†

p, α†q

]= 0. (42.9)

The physical significance of the new operators is that they describe a newsort of excitation: a new bosonic quasiparticle we will call a bogolon.

42.2 Bogoliubov’s transformation 373

Example 42.2

We can show how the bogolon operator diagonalizes the Hamiltonian. The Hamilto-nian can be written in matrix form as

H =X

p

“

a†p a−p

”„ ǫp12ng

12ng 0

«„ap

a†−p

«

, (42.10)

where ǫp = p2

2m+ ng. We’re going to make the transformation given in eqn 42.7

and fix the constants up and vp to make the resulting Hamiltonian matrix diagonal,which is to say

H =X

p

“

α†p α−p

”„ D11 00 D22

«„αp

α†−p

«

. (42.11)

Assuming that we’re successful in diagonalizing, how do we get the excitation energy,that is the constant in front of α†

p αp? The trick is to look at the diagonalized formwhich is

H =X

p

h

D11α†p αp +D22α−p α

†−p

i

=X

p

h

D11α†p αp +D22(1 + α†

−p α−p)i

. (42.12)

The sum over −p’s can be re-indexed to be a sum over positive p’s and we see thatthe constant in front of α†

p αp is D11 + D22, otherwise known as the trace of thediagonal matrix. The trace is given by

Ep = ǫp(u2p + v2p) − 2ngupvp . (42.13)

The condition for the off-diagonal elements to be zero is that

2upvp

u2p + v2p

=ng

ǫp. (42.14)

Substituting the condition allows us to eliminate up and vp and we obtain5 the 5You are invited to verify this in Exer-cise 42.1.answer given in eqn 42.15.

The diagonalized Hamiltonian is given by

H =∑

p

Epα†pαp, where Ep =

√p2

2m

(p2

2m+ 2ng

). (42.15)

The point of diagonalizing the Hamiltonian is to tell us about the quasi-particle excitations from the ground state. These are the bogolons, whichrepresent the motion of a large number of the original interacting bosons.The operator α†

p creates a single bogolon. These bogolons don’t interactwith each other so the problem is solved.

|p|

Epp2

2m

(ngm

)12 |p|

Fig. 42.2 The dispersion predicted byBogoliubov’s model. At low momen-tum the energy is linear in |p|, at largemomentum it is quadratic.

Turning to the bogolon dispersion, shown in Fig. 42.2, we have thatfor small momenta the dispersion is linear and looks like

Ep =(ngm

) 12 |p|. (42.16)

This linear dispersion occurs in another important problem: that ofphonons. In a vibrating system phonon quasiparticles are collective ex-citations of a large number of atoms. The small-momentum bogolons

374 Superfluids

are similar, and they may be thought of as representing the compli-cated, collective motion of a large number of bosons. Loosely speaking,the original bosons have a number density whose behaviour is wave-like and these waves are quantized into bogolons. At large momentumthe dispersion becomes Ep = p2/2m telling us that large-momentumbogolons behave6 like free particles with a mass m.6We say that the particles are ballistic.

Finally we note that Bogoliubov’s treatment does not predict the mini-mum seen in experimentally in Fig. 42.1. This owes its existence in liquidhelium to the fact that the interactions in that system are very strong,not weak, as considered here. The minimum corresponds to excitationsknown as rotons, which represent a back-flow of Bose particles, ratherlike the motion of particles in a smoke ring.77See Feynman, Statistical Physics.

42.3 Superfluids and fields

The argument presented above, in terms of creation and annihilation op-erators, represents a quick and illuminating way of solving Bogoliubov’sproblem. An alternative approach is to attack the problem using theLagrangian and the machinery of spontaneous symmetry breaking thatwe developed in Chapter 26.

We started our treatment with Bogoliubov’s approximation ap=0|Ω〉 ≈√N0|Ω〉. Actually we know that something very similar to this relation

holds exactly, by definition, for a coherent state. If |Ω〉 is a coherent state,we have ap=0|Ω〉 =

√N0 eiθ0 |Ω〉. Note also that if this coherent state is

macroscopically occupied then the uncertainty in the phase ∆ cos θ tendsto zero, so that the phase θ0 is a well defined quantity. Thus Bogoliubov’s‘approximation’ holds exactly for a macroscopically occupied coherentstate.

We can also understand how the occurrence of such a state necessarilyimplies broken symmetry. In real space (and ignoring time variation fornow) the annihilation field Φ(x) = 1√

V∑

p apeip·x has a coherent state

eigenstate |φ(x)〉 defined such that

Φ(x)|φ(x)〉 =√ρ(x)eiθ(x)|φ(x)〉, (42.17)

where ρ(x) is a number density of particles. Substituting the modeexpansion shows that Φ(x) has the property, for the coherent groundstate |Ω〉, that

〈Ω|Φ(x)|Ω〉 =1√V∑

p

〈Ω|ap|Ω〉eip·x

=1√V∑

p

√Npeiθpeip·x, (42.18)

but since, for |Ω〉, only the p = 0 state is occupied with any sizeableprobability, this term dominates the sum and we have

〈Ω|Φ(x)|Ω〉 =√n eiθ0 , (42.19)

42.3 Superfluids and fields 375

where√n =

√Np=0/V. An important point here is that the field Φ(x)

has developed a ground state with a nonzero vacuum expectation value,which is a tell-tale sign of a broken symmetry state. In fact we take〈Ω|Φ(x)|Ω〉 as the order parameter of the system. We also read off fromeqn 42.17 that in the ground state |Ω〉 the number operator expectationvalue is

√ρ(x) =

√n across the sample and that the state’s phase is

θ(x) = θ0. We stress that while there is still uncertainty in the numbern in a macroscopically occupied coherent state, the uncertainty in thephase θ0 is zero. Thus, the breaking of symmetry involves fixing thephase to θ(x) = θ0 across the entire system. Clearly it is this picking ofa unique phase that breaks the U(1) symmetry of the system.8

8The breaking of symmetry is oftenrepresented pictorially by drawing aset of compass needles which depictθ(x): a phase angle which may be dif-ferent at every point in space. Thebreaking of symmetry, and formationof a superfluid phase, corresponds toall of the needles lining up, as shownin Fig. 42.3(a). This picture of a super-fluid is known as the XY model, wherethe name is motivated by the freedomof the needles to point in the x-y planein the normal state.

How do we end up with a Lagrangian that describes non-relativisticbosons which is unstable to symmetry breaking? A system containinga fixed number of non-interacting, non-relativistic bosons has a Hamil-tonian H0 = (Ep − µ)a†pap, with Ep = p2/2m and where µ is thechemical potential. Translating back into position space fields we have aLagrangian for the non-relativistic system, including chemical potential,given by

L = iΦ†∂0Φ − 1

2m∇Φ† · ∇Φ + µΦ†Φ. (42.20)

This is similar to the form we had in Chapter 12. The interaction isincluded by subtracting a term g

2 (Φ†Φ)2.

Fig. 42.3 (a) Order in the superfluidcorresponds to a uniform phase angleacross the entire system. This is repre-sented on a lattice here. (b) A gradientin the phase ∇θ(x) results in a currentin the superfluid.

Example 42.3

The full, interacting Lagrangian is

L = iΦ†∂0Φ − 1

2m∇Φ† · ∇Φ + µΦ†Φ − g

2(Φ†Φ)2, (42.21)

which, with its positive mass-like term µΦ†Φ, is unstable to spontaneous symme-try breaking. The Lagrangian may be helpfully rewritten (to within an ignorableconstant) as

L = iΦ†∂0Φ − 1

2m∇Φ† · ∇Φ − g

2

“

n− Φ†Φ”2, (42.22)

and remembering that the number density of particles is given by Φ†Φ, we interpretn = µ/g as the boson density of the ground state.

The Lagrangian for non-relativistic bosons with a short ranged repulsiveinteraction is therefore given by

L = iΦ†∂0Φ − 1

2m∇Φ† · ∇Φ − g

2(n− Φ†Φ)2, (42.23)

which is unstable to symmetry-breaking. To see the consequences ofbroken symmetry we will use the polar coordinates of Chapter 12, sothat the field looks like Φ(x) =

√ρ(x)eiθ(x), and so is described by an

amplitude-field part ρ(x) and a phase-field part θ(x). Our Lagrangianbecomes9

9Note that Φ† =√ρ e−iθ and so

∂0Φ = i∂0θ√ρ eiθ +

1

2√ρeiθ∂0ρ,

resulting in

Φ†∂0Φ = iρ∂0θ +1

2∂0ρ.

L = −ρ∂0θ −1

2m

[1

4ρ(∇ρ)2 + ρ(∇θ)2

]− g

2(n− ρ)2, (42.24)

376 Superfluids

where we have dropped the term i∂0ρ/2 as it is a total derivative andshould therefore give a vanishing contribution to the action since ourfields are defined to vanish at ±∞.

This Lagrangian has a Mexican hat potential (Fig. 42.4) which has aminimum at ρ = n and is invariant with respect to global U(1) trans-formations of phase. That is, we can swap θ(x) → θ(x) + α and theLagrangian is the same as long as α is the same at every point in space-time. The possible ground states of the theory are therefore to be foundat Φ(x) =

√ρ(x)eiθ(x) =

√neiθ0 , where θ0 is the same at every point

in space, but may itself take any value. We break the global U(1) sym-metry of the theory by fixing the phase of the ground state to be ata particular value, say θ0 = 0 as shown in Fig. 42.4(c). Breaking thiscontinuous U(1) phase symmetry will result in the emergence of a Gold-stone mode, i.e. an excitation which costs vanishingly little energy toexcite at low momentum.

(a)

(b)

(c)

ReΦ

ImΦ

U(|Φ|)

ReΦ

ImΦ

ReΦ

ImΦ

θ

Fig. 42.4 (a) The Mexican hat poten-tial in the superfluid problem. There isone of these at every point x in space.(b) The potential viewed from above ascontours, with a minimum at ρ = n.Each value of θ is equivalent in energy.(c) Upon symmetry breaking the vac-uum state picks a unique value of θ(here θ = 0) at every point across thesystem.

To find the Goldstone mode we use a method much beloved of con-densed matter physicists. We are going to remove the modes that costa lot of energy.10 This will leave behind a theory that only includes

10Particle physicists sometimes callthis integrating out the high-energyparticles.

the low-energy excitations, which (we hope) will include the Goldstonemodes. Staring at the form of the potential, we see that excitations thatinvolve climbing the wall of the potential by changing ρ will cost moreenergy than those involving rolling around in the gutter (changing θ).We therefore examine the excitations that arise from departures fromthe ground state by considering an excited state field

√ρ(x) =

√n+ h.

This species of excited state just involve climbing the wall slightly fromthe ground state. We use these excited states to expand the Lagrangian,which yield (on dropping11 the total time derivative −n∂0θ):

11This is a Berry phase term which re-veals some interesting physics in thepresence of vortices. See Wen, Sec-tion 3.6.

L = − 1

2m(∇h)2 − 2gnh2 −

(2√n∂0θ

)h− n

2m(∇θ)2 + . . . (42.25)

To remove the energetic states we may plug a Lagrangian into a pathintegral and integrate out the variable describing these states. We willuse this technique to remove the amplitude h from our Lagrangian.

Example 42.4

We’ll make use of two neat functional integral tricks here. Remember that whendealing with the Lagrangian in a path integral, we can make the integrate-by-partssubstitution to the quadratic parts:

(∂µφ)2 −m2φ2 → −φ`∂2 +m2

´φ, (42.26)

which for our equation in terms of h requires

− 1

2m(∇h)2 − 2gnh2 → −h

„

− 1

2m∇

2 + 2gn

«

h. (42.27)

Next we use the one functional integral we can do, i.e.RDφ e

i2

R

φKφ+iR

Jφ =

e−12

R

J(iK−1)J , which allows us to make the replacement

1

2φˆ−(∂2 +m2)

˜φ+ Jφ→ 1

2J(x)

1

∂2 +m2J(y). (42.28)

42.4 The current in a superfluid 377

Setting φ = −√

2h and J =√

2n∂0θ, we have

h

„

− 1

2m∇

2 + 2gn

«

h−`2√n∂0θ

´h→

√n∂0θ

1

(−1/2m)∇2 + 2gn

√n∂0θ. (42.29)

Our Lagrangian becomes

L = n∂0θ1

2gn− 12m∇

2 ∂0θ −n

2m(∇θ)2 + . . . (42.30)

To obtain the low-energy, small-momentum behaviour we treat the en-ergy density term (1/2m)∇2 as small compared to the potential densityterm 2gn and we find a Lagrangian purely describing the low-energyphysics

L =1

2g

(∂θ

∂t

)2

− n

2m(∇θ)

2. (42.31)

This is merely a wave equation Lagrangian with a linear dispersion anda wave speed c =

√gnm . We have found our Goldstone mode: the θ field

has no mass term, so the energy of excitations tends to zero as p → 0.Our manipulations have resulted in a low-energy dispersion

Ep =(ngm

) 12 |p|, (42.32)

just as we had before. We’ve shown through another method that break-ing the U(1) symmetry gives the linearly dispersing mode at low mo-mentum.

42.4 The current in a superfluid

Noether’s theorem: U(1) internalsymmetry for the low-energytheory

Dθ = 1Π0θ = 1

g∂0θ Πiθ = n

m∂iθ

DL = 0 Wµ = 0J0Nc = − 1

g∂0θ JNc = n

m∇θ

Let’s examine the fate of Noether’s symmetry current for the low-energytheory. Looking at the Lagrangian in eqn 42.31, we see that the U(1)translation θ(x) → θ(x) + α leaves the function invariant.12 This is 12See Chapter 12 for a reminder.

merely the U(1) symmetry that is spontaneously broken in the groundstate. Applying Noether’s theorem to this symmetry we have thatDθ(x) = 1 and

Π0θ = 1

g∂0θ, Πiθ = n

m∂iθ. (42.33)

As a result, we have the conserved currents

J0Nc(x) = − 1

g∂0θ(x), JNc(x) = nm∇θ(x). (42.34)

This final equation tells us that gradients in the phase in a superfluid[as shown in Fig. 42.3(b)] result in currents. The broken symmetryground state of the superfluid has a uniform phase and if we deform thephase we set up a current in the system along the gradient. Notice thatthe current JNc depends on the strength of the condensate (or vacuumexpectation value) given by n. In the absence of symmetry breaking,

378 Superfluids

this is zero, so it makes sense that this is indeed the current that resultsfrom symmetry breaking.

What does this tell us about the dissipationless flow of momentumwhich, after all, is one of the most striking aspects of a superfluid? Wewill use a beautiful argument based on topology that demonstrates thata moving superfluid can’t dissipate momentum. That is, it’s a runawaytrain that can’t slow down! The superfluid’s broken symmetry groundstate is a constant field Φ =

√neiθ0 . We imagine putting our superfluid

in a box of size L with periodic boundaries Φ(x = 0) = Φ(L). Now weboost the superfluid so it has a momentum P . This changes the groundstate Φ → eiPxΦ =

√nei(θ0+Px). Notice that the boost effectively twists

the phase of our field. The boost also twists the boundary conditions;instead of Φ(0) = Φ(L), they become Φ(0)eiPL = Φ(L). This is shownin Fig. 42.5(a). To satisfy the periodic boundary conditions, we musthave PL = 2πw, where w is an integer (known as the winding number,as discussed in Chapter 29).

(a)

(b)

(c)

w = 1

w = 1w = −1

Fig. 42.5 (a) The superfluid with aw = 1 phase twist. Any attempt tountwist the phase would tear the su-perfluid. (b) The same superfluid anda w = −1 vortex propagating to theright. (c) The propagation of the vor-tex unwinds the twist allowing the su-perfluid to dissipate momentum.

The nonzero momentum state has more energy than the ground state.If the superfluid can slow down it will lower its energy. Now we ask, canwe untwist the superfluid phase to remove this extra momentum? Theanswer is no, we can’t. The nonzero momentum state involves an integernumber of 2π rotations of the phase angle between the ends of the box.The angles at the ends of the box are locked to each other by the periodicboundary conditions, so removing the twists will lead to singularities inthe phase field, costing enormous amounts of energy.

The only way we could remove a twist is to create a vortex, a localizedphase twist. Such vortices carry integer winding numbers, so creatinga vortex which travels across the box will remove a twist as shown inFig. 42.5(b).13 The creation of a vortex is a tunnelling process with a fi-13We have considered annihilating our

broad phase twist in Fig. 42.5(a) witha vortex in Fig. 42.5(b). The broadnessof the phase twist and the localized na-ture of the vortex are irrelevant. Whatmatters is their topological character,namely that one has w = 1 and theother has w = −1 so that they annihi-late each other.

nite energy barrier. Ignoring such subtleties would lead us to believe thata moving superfluid is a perpetual motion machine. However, Natureis not so kind and allows a quantum tunnelling process which unwindsthe condensate. As a result, the superfluid can, over a very long period(and even in the absence of excitations) dissipate some momentum.

Our model of weakly interacting bosons captures the small and largemomentum behaviour of the dispersion, so explains at least some of thephysics of the excited superfluid. One might question whether this dis-persion is essential to the existence of the superfluid state. It is essentialsince it guarantees that the moving fluid can’t dissipate too much mo-mentum through its excitations. The reason is that a linear dispersiondoesn’t give rise to enough states for quasiparticles to scatter into anddissipate momentum. To see this consider the density of states g(E) fora three-dimensional fluid of bosons. For the linear dispersion of a su-perfluid, this varies as g(E) ∝ E2, whereas for the quadratic dispersionof a normal particle, this varies as g(E) ∝ E

12 . As a result, there are

fewer excited states available for a superfluid than a normal fluid at lowvalues of energy, reducing the phase space for scattering events that willreduce momentum.

Exercises 379

Example 42.5

It is also worth noting another beautiful argument by Landau that shows an upper

bound on the critical velocity for superfluid flow is given by (ng/m)12 , a quantity

which only exists due to the presence of interactions. In the simplest version of thisargument, we imagine the superfluid as a body of mass M moving with velocityv. The body creates an excitation, with momentum p and energy Ep , with theresult that the body’s velocity changes to v′. By conservation of momentum, wehave Mv = Mv′ + p, from which we see that the body ends up with kinetic energyMv′2

2= Mv2

2−v ·p+ p2

2M. The point is that the excitation cannot be created unless

we have Mv2

2> Mv′2

2+ Ep , from which we conclude that

Ep < v · p− p2

2M≈ v · p, (42.35)

if M is assumed large. In other words, we have an upper bound on the superflow

against creating excitations, given by Ep/|p| = (ng/m)12 . So while a non-interacting

Bose gas (which has g = 0) will undergo a transition into a Bose–Einstein condensateat low temperatures, such a state will not exhibit a superflow as the critical velocityis zero.

Chapter summary

• Bogoliubov’s approximation states that the number of particles inthe p = 0 ground state becomes macroscopically large, allowing thereplacement of operators ap=0 and a†p=0 with the number

√N0.

• The effective Hamiltonian in a non-relativistic Bose gas can bediagonalized using a Bogoliubov transformation, and the quasi-particles are called bogolons.

• The low-energy excitations in a superfluid are Goldstone modes,the broken symmetry is the phase θ and the superfluid current isJNc = n

m∇θ(x).

Exercises

(42.1) Verify eqn 42.15 using the method suggested in thetext. You may find it useful to make the substitu-tions

up = cosh θp ,

vp = sinh θp . (42.36)

(42.2) Verify the algebra leading to eqn 42.31.

(42.3) (a) Show that, in three dimensions, the density ofstates g(E) = dN(E)/dE for a superfluid goes asg(E) ∝ E2.

(b) Show that for a Fermi gas g(E) ∝ E12 .

43The many-body problem

and the metal

43.1 Mean-field theory 380

43.2 The Hartree–––Fock groundstate energy of a metal 383

43.3 Excitations in the mean-fieldapproximation 386

43.4 Electrons and holes 388

43.5 Finding the excitations withpropagators 389

43.6 Ground states and excita-tions 390

43.7 The random phase approxi-mation 393

Chapter summary 398

Exercises 398

The sentries report Zulus to the south west. Thousands ofthem.Colour Sergeant Bourne (Nigel Green), Zulu (1964)

The physics of large numbers of non-relativistic particles and their in-teractions is a branch of quantum field theory known as the many-bodyproblem. In this chapter we’ll examine the properties of a large numberof non-relativistic fermions confined in a box, which is the basis of thedescription of electrons in solids. As we’ve seen, non-relativistic fermionsscatter via the diagram shown in Fig. 43.1. The full formalism of Diracspinors isn’t required to understand this problem owing to the low en-ergies involved in metals and so we can treat the particles as interactingvia an instantaneous potential V (x−y) =

∑q eiq·(x−y)Vq. Rather than

starting with propagators and path integrals, we’ll gain some intuitionby treating the interaction with some very simple approximations.

p kq

p − q k + q

Fig. 43.1 The Coulomb vertex. Notethat in this chapter the transferred mo-mentum runs from left to right.

43.1 Mean-field theory

We start with the Hamiltonian

H = H0 + V , (43.1)

where H0 =∑

pp2

2m a†pap and

V =1

2

∑

pkq

Vqa†p−qa

†k+qakap, (43.2)

where the anticommuting operators a†p and ap create and destroy

fermions respectively. We take H0 as the dominant contribution andtreat V as a perturbation. As usual we call |0〉 the ground state of H0

with energy E0 and |Ω〉 the ground state of the full Hamiltonian of Hwith energy E.1

1It is worth noting that |0〉 doesn’tmean that we have no particles inthe ground state. Usually in con-densed matter physics we study sys-tems with a finite density of parti-cles at zero temperature. The non-interacting ground state for a metal isa gas of electrons stacked up in en-ergy up to the Fermi level pF. Soif we take spin into account then theground state of a metal may be written|0〉 =

Q

|p|<pF a†p↑a

†p↓|Empty box〉.

We need a way to deal with the perturbation provided by the com-plicated two-body potential V . A good first step is to ask what themean-field behaviour of the system is. Mean-field theory is a schemeof approximations that crops up all over physics and involves takinga system of many particles and asking how a single particle reacts tothe average behaviour of all of the others. In the context of many-body

43.1 Mean-field theory 381

theory, we will interpret the mean-field approximation as a method of re-placing pairs of operators with averages, or more correctly, with vacuumexpectation values (VEVs), and we will later link this to the conven-tional idea of mean-field theory. The mean-field correction ∆E to thenon-interacting ground state energy E0 is found by taking a VEV of theperturbation, which is to say we approximate the expectation value ofthe troublesome four-operator term taken with the unperturbed groundstate:2 〈0|a†p−qa

†k+qakap|0〉.

2This is exactly how we usually workout the energy shift caused by a pertur-bation in first-order perturbation the-ory, so represents nothing new. Nov-elty comes with the next step, wherethis VEV is boiled down to somethingmore useful using a version of Wick’stheorem.

Example 43.1

Wick’s theorem reduces the four operators in eqn (43.2) into products of contractionsof pairs of (normally ordered) operators.3 We therefore apply 3Since we’re dealing with fermion oper-

ators in this chapter then we need to re-member that swapping the order of twooperators when we bring them togetherand make a VEV, gives us a minus sign.

a†p−q a†k+q

ak ap →Y

N

»all paired

contractions

–

. (43.3)

Wick’s theorem on our string of four operators yields

a†p−q a†k+q

ak ap = Nh

a†p−q a†k+q

ak ap

i

+ a†p−q a†k+q

N [ak ap ] +Nh

a†p−q a†k+q

i

ak ap

+ a†p−q ap Nh

a†k+q

ak

i

+Nh

a†p−q ap

i

a†k+q

ak

− a†p−q akNh

a†k+q

ap

i

−Nh

a†p−q ak

i

a†k+q

ap

+ a†p−q a†k+q

ak ap + a†p−q ap a†k+q

ak − a†p−q ak a†k+q

ap . (43.4)

Everything within the normal ordering signs is normally ordered as written, so thesecan be dropped. Mean-field theory involves replacing the contractions with averagestaken over the ground state: 〈0|O|0〉.

The terms with uncontracted operators represent excitations, and sowe ignore them for now, since we’re concerned with the ground stateproperties. For the ground state energy shift 〈V 〉, we therefore onlyconsider the completely contracted terms and this procedure thereforeyields up the three terms:

1

2

∑

pkq

Vq

〈a

†p−qa

†k+q〉〈akap〉︸︷︷︸

leads to C0

+ 〈a†p−qap〉〈a†k+qak〉︸︷︷︸leads to D0

−〈a†p−qak〉〈a†k+qap〉︸︷︷︸leads to F0

=C0 +D0 + F0,(43.5)

where we’ve shortened 〈0|O|0〉 to 〈O〉 to save on clutter. We’ll call C0

the Cooper term,4 D0 is Hartree’s direct term and F0 Fock’s exchange

4The Cooper term will be importantwhen we come to deal with supercon-ductors in the next chapter. We’ll post-pone discussing it until then and, forthe moment, use the fact that undernormal circumstances, the combination〈0|a†n a†m |0〉 = 0 for any n and m, andso we expect no contribution from C0.

Douglas Hartree (1897–1956)

Vladimir Fock (1898–1974)term.

Let’s examine the Hartree term in more detail. It gives rise to acontribution to the ground state energy of

D0 =1

2

∑

pkq

Vq〈a†p−qap〉〈a†k+qak〉. (43.6)

382 The many-body problem and the metal

We know that 〈0|a†ras|0〉 = 0 unless r = s, so we can immediately fixq = 0 in eqn 43.6. Recall that the Vqs are Fourier components of the realspace potential V (r). A wave with wave vector q = 0 is a constant (thatis, it has an infinitely long wavelength), so the Hartree term evaluatesthe energy due to the constant part of the potential. Maybe this is whatwe would expect from a first-order guess! We obtain

D0 =1

2

∑

pk

Vq=0〈a†pap〉〈a†kak〉. (43.7)

Notice that we’ve reduced the VEVs of operator pairs to simple numberoperators and the Hartree term has become

D0 =1

2Vq=0

(∑

p

〈0|Np|0〉)2

. (43.8)

The mean-field treatment has a diagrammatic interpretation: we cantake the interaction diagram and join the legs of the operators that we’reaveraging (Fig. 43.2). This turns out to be equivalent to the more usualways of getting diagrams with the S-matrix or path integral approach.The Hartree term can therefore be represented in diagrammatic form asthe double-headed tadpole shown in Fig. 43.2(b).

(a)

(b)

(c)

p

k

p kq = 0

qap

a†p−q

ak

a†k+q

q = p − k

Fig. 43.2 (a) The basic interactiondiagram. (b) The double tadpole di-agram giving the Direct (or Hartree)contribution to the ground state energy.(c) The double oyster diagram givingthe Exchange (or Fock) contribution tothe ground state energy.

Another point to note is that for theories where there are no par-ticles in the ground state |0〉 the tadpole [whose head corresponds to〈0|a†pap|0〉, see Fig. 43.2(b)] gives zero. This isn’t the case for nuclear orelectronic matter however, where we have a finite number of fermions inthe ground state.

Example 43.2

Let’s see how the Hartree term looks in terms of position space fields. We examinethe product 〈a†p−q ap〉 We have that ap = 1√

VR

d3x ψ(x)e−ip·x , which enables us tosay

〈a†p−q ap〉 =1

V

Z

d3xd3x′ 〈ψ†(x)ψ(x′)〉ei(p−q)·xe−ip·x′. (43.9)

This in turn allows us to write the Hartree energy as

D0 =1

2V2

X

p,k,q

Z

d3xd3x′d3yd3y′ Vq 〈ψ†(x)ψ(x′)〉〈ψ†(y)ψ(y′)〉 (43.10)

×ei(p−q)·xe−ip·x′ei(k+q)·ye−ik·y′

.

Doing the sums over p and k yields Vδ(3)(x − x′) and Vδ(3)(y − y′) respectively,which eat up two of the space integrals. We’re left with

D0 =1

2

X

q

Z

d3xd3y Vq 〈ψ†(x)ψ(x)〉〈ψ†(y)ψ(y)〉e−iq·(x−y). (43.11)

Next we sum over q, which takes an inverse Fourier transform of the potential Vq ,and we obtain

D0 =1

2

Z

d3xd3y V (x− y)〈ψ†(x)ψ(x)〉〈ψ†(y)ψ(y)〉. (43.12)

This is just what one might have guessed classically for the energy of two chargedistributions ρ(x) and ρ(y) interacting via a potential V (x− y).

43.2 The Hartree–––Fock ground state energy of a metal 383

Of course, decompressing the notation here

〈ψ†(x)ψ(x)〉 = − limt′ → 0−

x′ → x

〈0|T ψ(t′,x′)ψ†(0,x)|0〉, (43.13)

is related to a propagator.5 The internal lines that form the double 5See also eqn 31.18.

tadpole’s two heads are just propagators, as we expect for a respectableFeynman diagram.

Next we examine the Fock term. This is represented by the doubleoyster diagram in Fig. 43.2(c) which corresponds to the equation

F0 = −1

2

∑

pkq

Vq〈a†p−qak〉〈a†k+qap〉. (43.14)

Example 43.3

Again, in order to make the VEVs nonzero, this gives us a condition on q. This timewe have that p− q = k, leading to a result

F0 = −1

2

X

pk

Vp−k〈a†k ak〉〈a†p ap〉

= −1

2

X

pk

Vp−k〈0|Nk |0〉〈0|Np |0〉, (43.15)

which is diagonal (that is, contains only number operators) but is clearly more com-plicated than what we had for the Hartree term. This is shown diagrammaticallyas the double oyster in Fig. 43.2(c). There is a lot of physics in the exchange termincluding, for example, magnetic ordering in metals, as examined in the exercises.

The position space version unfolds much as before, although now the momentumsums mix up x’s and y’s, giving us

F0 = −1

2

Z

d3xd3y V (x− y)〈ψ†(x)ψ(y)〉〈ψ†(y)ψ(x)〉. (43.16)

43.2 The Hartree–––Fock ground state

energy of a metal

We will use our mean-field results to work out an approximate valueof the total ground state energy of a metal. This will involve potentialenergy contributions from the Hartree term and the Fock term and thisis therefore known as the Hartree–Fock ground state energy.6

6The other contribution is from H0 andreflects fact that each electron has ki-netic energy E

(0)p = p2

2mand obeys the

Pauli principle. Our approach will beto add these kinetic energies up by not-ing that the sums are carried out upto pF and that states are so closelyspaced that we may make the replace-

mentP

|p|<pF → VR

|p|<pFd3p

(2π)3.

Example 43.4

Recall that we put non-interacting electrons into a box, with one spin-up and onespin-down electron in each momentum state. The electrons stack up in energy and theupper-most occupied energy level is called the Fermi level. The number of occupiedmomentum states is

X

|p|<pF

→ VZ

|p|<pF

d3p

(2π)3= V

Z pF

|p|=0(4π)

d|p|(2π)3

|p|2 =Vp3F6π2

, (43.17)


and with two spin states per momentum level, the total number of electron states isN = Vp3F/3π2. The electron density of the system n = N/V is related to the Fermi

momentum pF via pF = (3π2n)13 , and the Fermi energy is EF = p2F/(2m). The total

energy of the box of non-interacting electrons is given by

W0 = 2X

|p|<pF

E(0)p = 2V

Z

|p|<pF

d3p

(2π)3p2

2m=

3

5NEF, (43.18)

and so the kinetic energy per electron W0/N = 35EF. A popular unit for energy

among theoreticians is the Rydberg (Ry), equal to ~2/(2mea20) where a0 is the Bohr

radius. Writing the volume occupied per electron as 1/n = 43πr3 where r is the

average distance per particle, and using the dimensionless length rs = r/a0, we canexpress the kinetic energy per electron as

W0

N=

3

5EF =

3

5

„9π

4

«2/3 1 Ry

r2s≈ 2.21

r2s

Rydbergs

electron. (43.19)

Now for the potential energy. This is not something we can calculate ex-actly, but our mean-field approach tells us that to first order we shouldadd the Hartree term D0 and Fock term F0. However an immediatesimplification is possible. Metals may be modelled as charge-neutralboxes full of electrons interacting via the Coulomb force. Although,in reality, metals contain positive ion cores and mobile electrons, thislevel of detail is not always necessary. Instead, we use a model known77Named by John Bardeen.

as jellium in which the electrons are put in a box full of a homoge-nous, positively charged jelly in order to guarantee charge neutrality.This uniform positive charge exactly cancels the contribution from theHartree term which originates from the uniform electron distribution.We therefore only need consider the Fock contribution to the energygiven by F0 = − 1

2

∑pk Vp−k〈Np〉〈Nk〉. The ingredients of the sum are

the ground state occupation numbers

〈Np〉 =

1 |p| ≤ pF

0 |p| > pF,(43.20)

and the Fourier transform of the interaction potential Vp.

Example 43.5

The electrostatic potential energy between two electrons is given by8 V (x − y) =

8Here we return to SI units, since theseare often used in the many-body liter-ature.

e2

4πǫ0|x−y| . The Fourier transform can be most easily evaluated by working with

the screened potential energy V (r) = (e2/4πǫ0)e−λ|r|/|r| and then sending λ → 0.Thus9

9The integral was evaluated in Chap-ter 17.

Vq = limλ→0

1

Ve2

4πǫ0

Z

d3re−iq·re−λ|r|

|r| = limλ→0

e2

Vǫ0(q2 + λ2)=

e2

Vǫ0q2. (43.21)

43.2 The Hartree–––Fock ground state energy of a metal 385

Now to evaluate the Fock oyster term. It will be useful for what is tofollow if we arrange the terms as follows (including a factor of 2 for spin):

F0 = 2∑

|p|<pF

1

2

−

∑

|k|<pF

Vp−k

= 2∑

|p|<pF

1

2

−

∑

|k|<pF

e2

Vǫ01

|p − k|2

. (43.22)

The part in the square bracket corresponds to the oyster diagram partshown in Fig. 43.3 and is known as the Fock self-energy10 Σ

(F)p .

10We will see this again in the next sec-tion.

p

p

k

p − k

Fig. 43.3 The Fock contribution to

the self-energy Σ(F)p .

Example 43.6

The Fock self-energy may be evaluated as follows:11

11Use has been made of the integral

R

|k|<pFd3k

|p−k|2=R pF0 2πk2 dk

R π0

sin θ dθk2+p2−2kp cos θ

= 2πp

R pF0 k dk ln

˛˛˛k+pk−p

˛˛˛

= 2πpF F“

|p|pF

”

.Σ(F)p = − e

2

ǫ0

Z

|k|<pF

d3k

(2π)31

|p− k|2 = −pFπ

„e2

4πǫ0

«

F

„ |p|pF

«

, (43.23)

where

F (x) = 1 +1 − x2

2xln

˛˛˛˛

1 + x

1 − x

˛˛˛˛ . (43.24)

The function F (x) is shown in Fig. 43.4. It has an infinite slope at x = 1 correspond-ing, in our example, to the Fermi momentum.

Ep

|p|/pF

x

F (x)

0 1 2

1

2

(a)

(b)

Fig. 43.4 (a) The function F (x).(b) The energy of an electron in theHartree–Fock approximation Ep =p2

2m+ ΣF

p .

The Fock energy is given (finally) by

F0 = 2∑

|p|<pF

1

2Σ(F)

p = −V pF

π

(e2

4πǫ0

)∫

|p|<pF

d3p

(2π)3F

( |p|pF

)

= −3NpF

2π

(e2

4πǫ0

)∫ 1

0

dxx2F (x). (43.25)

The integral yields 1/2 and we obtain the potential energy per electron

F0

N= −3pF

4π

e2

4πǫ0= −0.916

rs

Rydbergs

electron. (43.26)

Putting this together with the kinetic energy yields the total Hartree–Fock energy per electron of jellium metal

EHF

N=

(2.21

r2s− 0.916

rs

)Rydbergs

electron. (43.27)

Although this is as far as our mean-field theory will get us, this equationhas the appearance of the start of a series in powers of rs and we expectadditional terms to add to this series. These terms were given the namecorrelation energy12 by Eugene Wigner and Frederick Seitz. The next

12Feynman suggested it could be calledthe stupidity energy.

Eugene Wigner (1902–1995) made fun-damental contributions to many areasof physics and mathematics. His sisterwas married to Dirac, who would oftenrefer to her as ‘Wigner’s sister’.

Frederick Seitz (1911–2008) has ar-guably the best claim to be the found-ing father of solid state physics.

two terms were calculated by Gell-Mann and Brueckner in 1957 with the

Keith Brueckner (1924– )

result

E

N=

(2.21

r2s− 0.916

rs− 0.094 + 0.0622 ln rs

)Rydbergs

electron. (43.28)


43.3 Excitations in the mean-field

approximation

Next we examine the changes the interactions can make to the dispersionrelations of single particles in the metal. For the moment we will returnto the simple minded picture of a box of fermions, which does not includejellium’s positive background.

The excitations of a quantum field theory are particles, which have acharacteristic dispersion Ep. To examine the excitations of a field theorywe want to manoeuvre our Hamiltonian into the form H =

∑pEpa

†pap.

So when we examine the fate of particles in mean-field theory we tryto massage the two-particle interaction, described by the four operatorsin eqn 43.2, into one that describes a single particle interacting with anexternal field. The external field here is caused by the effect of all of theother particles, which (finally) explains the logic of naming this proce-dure a ‘mean field’ approximation. Recall from Chapter 4 that a singleparticle in an external potential is described by an operator Vqa

†p+qap,

so we want to turn the tricky term a†p−qa†k+qakap into a sum of terms

that look like this. Actually we want to go further than this and haveincoming and outgoing electrons residing in the same momentum stategiving

∑p V

MFp a†pap, where V MF

p is the effective mean-field potentialthrough which the particle with momentum p passes. We do this witha recipe: we use Wick’s theorem to generate terms containing the VEVof two of the four operators and treat this VEV part as an effectivepotential.

Example 43.7

Returning to our Wick expansion of the operator we identify the terms with onecontraction and one pair of uncontracted operators. We obtain a sum of terms:

D = 〈a†k+q

ak〉a†p−q ap + 〈a†p−q ap〉a†k+qak ,

F = −〈a†k+q

ap〉a†p−q ak − 〈a†p−q ak〉a†k+qap ,

C = 〈a†k+q

a†p−q 〉ap ak + 〈ap ak〉a†k+qa†p−q . (43.29)

We will continue to ignore the Cooper terms for now. The two terms in the D sumare identical after a re-indexing of the sums, and the same is true of the two termsin the F sum. We end up with

D = 2〈a†k+q

ak〉a†p−q ap ,

F = −2〈a†p−q ak〉a†k+qap . (43.30)

We now have a Hamiltonian that reads

H =∑

p

p2

2ma†pap +

∑

qpk

Vq〈a†k+qak〉a†p−qap −∑

qpk

Vq〈a†p−qak〉a†k+qap.

(43.31)

43.3 Excitations in the mean-field approximation 387

(a)

(b)

(c)

= + + + + · · ·

+ + + + · · ·

p kq

p − q k + q

p kq

p − q k + q

p

pp − k

k

p

p

q = 0k

Fig. 43.5 (a) The tadpole, representing the first-order Hartree contribution to thesingle-particle energy. (b) The oyster giving the first-order Fock contribution to thesingle-particle energy. (c) The sum to infinity of oyster and tadpole diagrams.

Now we’re in a position to examine the simplified version of the two-particle interaction. First, let’s examine the direct term: exactly asbefore the VEV part fixes q = 0, giving a number operator and theexpression

Vdirect =∑

pk

Vq=0〈a†kak〉a†pap (43.32)

=∑

p

[Vq=0

∑

k

〈a†kak〉]a†pap.

Comparing with our equation for a single particle in an external poten-tial, we see that it looks as if each of the particles is sailing throughan effective potential V MF

p = Vq=0

∑k〈Nk〉. This is represented by the

tadpole diagram in Fig. 43.5(a), which is generated by joining up two ofthe four legs. Notice that there are two ways of forming this diagram,re-creating the factor of two we had from Wick’s expansion.

Next we examine the exchange term

Vexchange = −∑

qpk

Vq〈a†p−qak〉a†k+qap. (43.33)

In order to get a nonzero VEV, we just consider the terms with p = k+q.Then we obtain

Vexchange = −∑

p

[∑

k

Vp−k〈a†kak〉]a†pap. (43.34)


This corresponds to a single particle interacting with an effective poten-tial V MF

p =∑

k Vp−k〈Nk〉, which we represent as the oyster diagram inFig. 43.5(b), generated by joining two opposite legs from the interactionvertex.

Notice that the effective potentials we have derived are exactly thefirst-order contributions to the self-energy Σp of a particle. They areeach represented by a part of a Feynman diagram we can fit betweenthe two external legs. We can therefore write the energy of the particlesas a sum of kinetic and self-energy parts

Ep =p2

2m+ Σ(D)

p + Σ(F)p , (43.35)

where

Σ(D)p = Vq=0

∑k〈Nk〉, Σ

(F)p = −∑k Vp−k〈Nk〉. (43.36)

For the case of jellium metal, we note that the Hartree self-energy Σ(D)p

is cancelled by the positive background charge (just as previously) and

the Fock term becomes Σ(F)p from eqn 43.23, justifying its assignment

in the previous section. The energy of electrons in the Hartree–Fockapproximation is therefore given by

Ep =p2

2m− pF

π

(e2

4πǫ0

)F

( |p|pF

), (43.37)

which is shown in Fig. 43.4(b). It is worth stressing at this point that thisis not what is measured in a real metal! The Hartree–Fock approxima-tion doesn’t capture the physics very well and higher order correctionsare necessary. This requires the full machinery of quantum field theoryincluding propagators and Feynman rules and it is to this we now turn.

43.4 Electrons and holes

So far we have treated our metal as a box of electrons. For more seriouscalculations it is useful to distinguish two sorts of excitations. This isbecause a metal comprises electron states filled up to pF and so the‘vacuum’ for the metal is not empty. With this in mind we define newoperators

ap = θ(|p| − pF)cp + θ(pF − |p|)b†p,a†p = θ(|p| − pF)c†p + θ(pF − |p|)bp, (43.38)

with anticommutation relations

cp, c

†q

= δ(3)(p − q),

bp, b

†q

= δ(3)(p − q), (43.39)

with all other anticommutators vanishing. We say that the cp operatorsdescribe electrons, while the bp operators describe holes.13

13Notice that, defined this way, elec-tron operators only act for states abovethe Fermi surface |p| > pF while holeoperators only act for states below theFermi surface |p| < pF. This, in turn,means that we describe the system ashaving electron excitations above theFermi energy and hole excitations be-low the Fermi energy.

43.5 Finding the excitations with propagators 389

Defined in this way the kinetic energy of the system is given by

H0 =∑

|p|<pF

Ep +∑

|p|>pF

Epc†pcp −

∑

|p|<pF

Epb†pbp. (43.40)

The first term is the ground state energy of all of the states, filled up tothe Fermi level. The second term accounts for electron excitations. Thefinal term accounts for hole excitations. Note that, with these definitionsthe holes make a negative contribution to the energy. Although thereis a sense in which holes are antiparticles, they are not identical withpositrons whose contribution to the total energy is always positive.

To perform calculations we need an electron–hole field operator, whichis given by

ψ(x) =1√V∑

p

[θ(|p| − pF)cp + θ(pF − |p|)b†p

]e−ip·x. (43.41)

The use of this expansion to find theelectron–hole propagator is examinedin Exercise 43.4.

43.5 Finding the excitations with

propagators

Our rather ad hoc mean-field procedure will not get us much further.Fortunately, we can use the technology of propagators to formalize theprocedure, turning it into a perturbation series. This will have the ad-vantage of showing how the full expansion of the four-operator termwill go. Recall that, since we are dealing with an interacting theorywe should think of our electrons and holes as quasiparticles. In the fol-lowing example, we will recap the steps needed to make a perturbationexpansion and re-create all of our results so far in a flash.14 14See Mahan, Chapter 2, P. Coleman,

Chapter 8, or Abrikosov, Gorkov andDzyaloshinski, Chapter 2, for the fullstory. Note also that we return to rel-ativistic notation here, so that xµ =(t,x), pµ = (E,p), etc.Example 43.8

The interaction Hamiltonian is HI(x−y) = 12ψ†(x)ψ†(y)V (x−y)ψ(y)ψ(x)δ(x0−y0).

This interaction gives us an S-operator

S = e−i2

R

d4xd4y ψ†(x)ψ†(y)V (x−y)ψ(y)ψ(x)δ(x0−y0). (43.42)

The basic building block of the perturbation expansion is the Feynman propagator,which we can obtain as the Green’s function of the equation of motion. The equationof motion for non-relativistic electrons is the Schrodinger equation, so we want

„

Ep − id

dt

«

G0(p) = −iδ(t), (43.43)

with Ep = p2

2m. We obtain the propagator in momentum space as

G0(p) =i

E − Ep + iǫ, (43.44)

where we have added iǫ in the denominator to avoid integrals hitting the electronpole. This choice is discussed in Appendix B.


With the inclusion of holes, the free propagator for electrons in a metal is givenby

G0(p) =iθ(|p| − pF)

E − Ep + iǫ+

iθ(pF − |p|)E − Ep − iǫ

, (43.45)

which is a sum of electron and holes parts. This expression is similar to what we havefor scalar particles, although with a differing sign. Since, for a particular choice of p,we only need one of the terms in the expression for G0(p), a much simpler approachthat is commonly employed is to write the electron–hole free propagator as a singleexpression

G0(p) =i

E − Ep + iδp, (43.46)

where δp = +ǫ for |p| > pF and δp = −ǫ for |p| < pF.

Now that we have a propagator and an interaction, we’re ready to writedown the Feynman rules for a metal.

The Feynman rules for a metal

• A factor G0(p) = iE−Ep+iδp

for each internal line.

• A factor −iVq for each interaction vertex.

• A factor −1 for each closed fermion loop.

• Conserve energy-momentum at the vertices.

• Integrate over unconstrained energies and momenta with a mea-

sure V∫

d4p(2π)4 .

• A convergence factor eiE0+

for non-propagating lines.

(a) (b)p

p

Fig. 43.6 Non-propagating lines.

The final point is necessary for the two diagram parts shown in Fig. 43.6and stems from the instantaneous nature of the interaction, which ap-pears to allow lines to propagate instantaneously.15 We’re now ready to

15As shown in the examples below, itforces us to close the contour in our in-tegrals in a particular manner, guaran-teeing that we get a sensible answer.

go and calculate some things with the Feynman rules.

43.6 Ground states and excitations

Ground state energies may be evaluated from perturbation theory bysumming Feynman diagrams. As discussed in Exercise 43.5, this involves

finding the amplitude 〈0|S|0〉 = eP

(Connected vacuum diagrams), fromwhich it follows that

−iE

V =

∑(Connected

vacuum diagrams

)

VT . (43.47)

The lowest order vacuum diagrams are the double tadpole and oyster,

Fig. 43.7 Corrections to the groundstate energy of a metal.

just as we had with the mean-field approximation. The mean-field trick,which gave us the Hartree–Fock approximation, represents first-orderterms in the perturbation expansion of the ground state energy. Higherorder terms in the expansion give us diagrams such as those others shownin Fig. 43.7.

43.6 Ground states and excitations 391

What are the particle energies? These can be accessed through theinteracting propagator of the theory, which includes the effect of all self-energy contributions. Recall the exact result that

G(p) =iZp

E − Ep + iΓp

+

(multiparticle

parts

), (43.48)

so we should be able to read off the dispersion from the position ofthe pole in the denominator. To find the full propagator we use theprescription

G(p) =∑(

Connected diagrams withtwo external legs

). (43.49)

We will use Dyson’s formula to formally sum the effect of self-energyprocesses to infinity and renormalize the propagator. As before, wedefine the 1PI self-energy as

−iΣ(p) =∑(

Amputated 1PI diagramswith two external legs

). (43.50)

The sum to infinity, shown for the example of Hartree–Fock theory inFig. 43.5(c), yields Dyson’s equation for the propagator

G(p) =1

G0(p)−1 + iΣ(p)=

i

E − Ep − Σ(p) + iδp, (43.51)

so the energy of the excitations is E = Ep + Re[Σ(E,p)

]. Some of the

contributions to Σ(p) are shown in Fig. 43.8.

1PI = + + + + · · ·

Fig. 43.8 Contributions to the electron 1PI self-energy of the metal.

Example 43.9

Let’s examine the renormalization of the electron lines by first-order processes. Theonly self-energy terms containing one interaction wiggle are the tadpole and oysterdiagrams shown in Fig. 43.5(a) and (b). This is, of course, identical to the mean-fieldresult and demonstrates, again, that the mean-field approximation predicts the samecontribution as first-order perturbation theory, namely the Hartree–Fock contribu-tions. The joy of the propagator theory is that it tells us how to make sense of theself-energy contribution: we can use it to renormalize the propagator by summing toinfinity.


In terms of propagators the diagrams correspond to

−iΣ(Tadpole)p = (−1)X

k

ZdE

2π(−iVq=0)G0(k)eiE0+

,

−iΣ(Oyster)p =X

k

ZdE

2π(−iVp−k)G0(k)eiE0+

. (43.52)

It’s instructive to use the complex analysis of Appendix B to show the use of the

eiE0+factors. Let’s examine the integral

ZdE

2πG0(k)eiE0+

, (43.53)

whose pole structure is shown in Fig. 43.9(a). If we complete the contour in theupper half-plane [Fig. 43.9(b)] then we pick up all of the momentum state poles for|k| < pF, completing in the lower half-plane picks up the momentum state poles for

|k| > pF. The convergence factor eiE0+makes the choice for us! If we complete

in the lower half-plane (where imaginary E gets very large and negative) this factorwill diverge. On the other hand, it will vanish for a contour completed in the upperhalf-plane in the limit that the contour becomes very large. We therefore pick up thepoles in the upper half-plane for all |k| < pF, each contributing residue i. We find

ZdE

2πG0(k)eiE0+

=1

2π× (2πi)iNk , (43.54)

where Nk is the occupation number distribution which is unity for |k| < pF and 0otherwise. We obtain expressions for our diagrams of

−iΣ(Tadpole)p = (−1)X

k

(−iVq=0)(−Nk), (43.55)

and

−iΣ(Oyster)p =X

k

(−iVp−k)(−Nk). (43.56)

k0

k0(a) pF

(b)

Fig. 43.9 The integration in eqn 43.53can be carried out in the complex plane.

For the case of jellium metal, with its positive background charge, the tadpolecontribution is cancelled and we only need consider the oyster contribution. Weobtain for the oyster diagram

Σ(Oyster)p = −X

|k|<pF

e2

Vǫ01

|p− k|2 = − e2

ǫ0

Z

|k|<pF

d3k

(2π)31

|p− k|2 ,

= −pFπ

e2

4πǫ0F

„ |p|pF

«

, (43.57)

which is, of course, the same result as discussed earlier in this chapter.Finally, the renormalization of the electron line from the oyster contribution,

achieved with a sum to infinity, is shown in Fig. 43.10, showing that the quasiparticleenergy is shifted to p2/2m+ Σ(Oyster)p .

So far we have only re-created the same results as seen earlier. To seethe motivation for going further one can calculate the effective mass forHartree–Fock theory.16

16Close to the Fermi level, we may ex-pand the energy of an excitation as

Ep = EF+

„∂Ep

∂p

«˛˛˛˛|p|=pF

(|p|−pF)+. . .

which, for a non-interacting system, isgiven by

Ep =p2F2m

+pFm

(|p| − pF) + . . .

from which we identify m∗ =pF

“

∂Ep

∂|p|

” .

The result of computing the mass m∗ corresponding to the disper-sion Ep following from our expression for the Hartree–Fock energy(eqn 43.57) is that m∗ = 0 close to the Fermi energy, which is clearlywrong: electrons should have effective masses of order17 m. The prob-

17We ignore the heavy fermion materi-

als where m∗ ≈ 103m. See P. Coleman,Chapter 16, for a discussion of thesesystems.

lem here is that the Hartree–Fock theory is completely static. It treatsan electron as if it’s propagating in the static field of all of the others.In reality electrons will alter their configuration dynamically resulting intime-dependent correlations. These may be included in the calculation

43.7 The random phase approximation 393

= + + + + · · ·

=

=1

1−

( )−1−

Fig. 43.10 The summation of the Fock terms to infinity.

through the inclusion of more processes in the electron self-energy asshown in Fig. 43.8.

The next logical step would be to evaluate contributions to the self-energy that include two interaction wiggles, including the pair-bubblediagram in Fig. 43.11. However examining the amplitude for this dia-gram we find that it is divergent for small q. This looks like a problem!However, this divergence is avoided if we take a side-step and, ratherthan concentrating on corrections to the electron self-energy, we con-sider a correction analogous to the photon self-energy of Chapter 41.This will have the effect of summing up a class of diagrams that aresimilar to that shown in Fig. 43.11 which will remove the divergencein the electron self-energy. This summation is the subject of the nextsection. Fig. 43.11 The pair-bubble contribu-

tion to the self-energy.

43.7 The random phase approximation

For an electron gas with a high density of particles the most importantcorrection to the Hartree–Fock approximation is provided by the lowestorder correction to the wiggle in the interaction vertex. For historicalreasons18 this is known as the random phase approximation or RPA

18The randomness of a phase is unim-portant in our description of the RPA.The reason for the name comes froman alternative treatment where a termP

l eiq·xl is shown to be neglectable,

where xl labels an electron’s position.The physics of this approximation isthat if xl is distributed over a largerange then the random phases that re-sult will tend to cancel.

and was first formulated by David Bohm and David Pines.19 This correc-19David Pines (1924– )tion to the interaction vertex has much in common with the self-energy

correction to the photon propagator in QED and represents a changein the potential felt by the electrons due to the effect of shielding. Theidea is that the interaction wiggle creates an electron–hole pair, whichannihilates shortly afterwards.

Fig. 43.12 A first-order polarizationprocess showing a bubble in the inter-action line.

This state of affairs is known here, as in QED, as a polarization

process and is shown in Fig. 43.12. It is one example of a class of1PI self-energy processes that can be inserted into an interaction wigglewhose sum is a Green’s function denoted iΠ(q), which we define similarly


to its cousin in QED as

iΠ(q) =∑(

All 1PI diagrams that can be insertedinto an interaction wiggle

). (43.58)

Some of the terms contributing to iΠ(q) are shown in Fig. 43.13.

= + + + + · · ·1PI

Fig. 43.13 Some of the diagrams contributing to iΠ(q).

In order to fully take account of iΠ(q) we need to sum the contributionto infinity. Since we don’t have a photon propagator in this theory, theserenormalize the potential, changing the interaction strength from −iVq

to the effective potential −iVeff(q). The sum to infinity is simply carriedout in the usual manner:

−iVeff(q) = −iVq +[−iVq

] [iΠ(q)

] [−iVq

]

+[−iVq

] [iΠ(q)

] [−iVq

] [iΠ(q)

] [−iVq

]+ . . .

=−iVq

1 − VqΠ(q). (43.59)

The interaction potential Vq therefore becomes an effective potentialVeff(q) = Vq/[1 − VqΠ(q)], which is often written Veff = Vq/ǫ(q), wherethe permittivity ǫ(q) = 1 − VqΠ(q) is a function of energy q0 and wavevector q. So, just as in QED, the vacuum state of the metal takes ondielectric properties caused by electron–hole pairs buzzing in and out ofexistence. The physics of these dielectric properties lies in the screeningof the electrons from each other’s charge by the electron–hole pairs.Finally we note that since ǫ(q) is a function of energy, the inclusionof polarization diagrams will provide the all-important time-dependentcorrelations whose neglect resulted in a nonsensical effective mass in theprevious section.

Our treatment of Π(q) has so far been quite general. In the RPA weevaluate the effect of only the lowest order contribution to Π(q), that is,the electron–hole bubble shown in Fig. 43.12, whose amplitude we willcall π(q). (The sum to infinity of this interaction that renormalizes thephoton line in the RPA is shown in Fig 43.14.)


= + + + · · ·

==1

1− ( )−1−

Fig. 43.14 The sum of bubbles in the interaction line.

Example 43.10

Including a minus sign for the fermion loop, the amplitude of the bubble in Fig. 43.12is given by

iπ(q) = −VZ

d4q

(2π)4G0(p+ q)G0(p)

= −VZ

d3p

(2π)3

Z ∞

−∞

dp0

2π

i

p0 + q0 − Ep+q + iδp+q

i

p0 − Ep + iδp

= VZ

d3p

(2π)31

(Ep+q − iδp+q ) − (Ep − iδp) − q0

×Z ∞

−∞

dp0

2π

„1

p0 + q0 − Ep+q + iδp+q

− 1

p0 − Ep + iδp

«

, (43.60)

where, in the final line, we have rewritten the amplitude in a form which allowsus to do a contour integral (see Appendix B). The integrals over p0 have poles atp0 = −q0 + Ep+q − iδp+q and p0 = Ep − iδp respectively. The residues for theseare Fermi functions and so we pick up factors 2πiNp+q and 2πiNp respectively andobtain the result20 20Dropping the infinitesimal in the de-

nominator, since we won’t use it fur-ther.

π(q) = VZ

d3p

(2π)3Np+q −Np

Ep+q − Ep − q0. (43.61)

The integrand in eqn 43.61 is known as the Lindhard function. The evaluation ofthe integral involving the Lindhard function is carried out by P. Coleman, Chapter 8.The result is a rather complicated expression for ǫ(q). Analytically continuing toimaginary energies and doing the integral it is found that

π(iq0,q) = −1

2g(EF)F

„iq0

4EF,|q|2pF

«

, (43.62)

where g(EF) =dNp

dEp

˛˛˛EF

is the density of states at the Fermi level and

F(y, x) = 1+1

4x

»

1 −“

x− y

x

”2–

ln

˛˛˛˛˛

x− yx

+ 1

x− yx− 1

˛˛˛˛˛+

1

4x

»

1 −“

x+y

x

”2–

ln

˛˛˛˛˛

x+ yx

+ 1

x+ yx− 1

˛˛˛˛˛,

(43.63)is a dynamicized version of the function F (x) that featured in the evaluation ofthe Fock self-energy. This latter expression must be analytically continued back tothe result we want for real q0 (see Mahan, Chapter 5 for the details). However,for our purposes we will limit our attention to the important limit q0 → 0, whereF(iq0/4EF, |q|/2pF) reduces to F (|q|/2pF).

We will extract three great results from the RPA by examining threespecial cases of ǫ(q0, q).


Case I: In the static (q0 → 0) limit we have π(0, q) =− 1

2g(EF)F (|q|/2pF). Particularly illuminating is the static and smallq limit, where F (|q|/2qF) → 2 and limq→0 π(0, q) = −g(EF) = − 3N

2EF

(where we have included a factor of 2 to account for spin degeneracy)and so

limq→0

ǫ(0, q) = 1 +

(3ne2

2ǫ0EF

)1

|q|2 = 1 +q2TF

|q|2 , (43.64)

where qTF is known as the Thomas–Fermi wave vector. Notice that thepermittivity diverges as |q| → 0. Perversely this is good news! It saysthat a uniform electric field can’t penetrate a metal, which is somethingwe certainly expect.21 Plugging ǫ(0, q) into eqn 43.59 we see that the

21The order of the limits is crucial here.Here we have started with the staticlimit and examined what happens asq → 0. The other order, correspond-ing to considering a uniform field (i.e.q = 0) then taking the limit q0 → 0tells us about the response of a metal toac fields and hence its transport prop-erties.

static interaction potential is changed to an effective potential

|q |

Veff(0, |q |)

Fig. 43.15 The static effective poten-

tial lim|q|→0 Veff(0,q).

limq→0

Veff(0, |q|) =e2

ǫ0V1

|q|2 + q2TF

, (43.65)

which is shown in Fig. 43.15. Notice that this is well behaved as |q| → 0.We started this section by arguing that the RPA was going to tell usabout screening. The connection with screening arises because our resultfor Veff is the Fourier transform of the real-space interaction potential

lim|x|→∞

Veff(x) ∝ e2

|x|e−qTF|x|, (43.66)

which is, of course, also known as the Yukawa potential, perhaps thesimplest screened Coulomb potential. We conclude that the creation andannihilation of electron–hole pairs screens the electronic charge and inthe RPA approximation this screening results in the Coulomb potentialbeing replaced by an effective Yukawa potential.

|x |

Veff(0, |x |)

0

Fig. 43.16 The static effective poten-tial Veff(0, |x|) given by the wave vectorbehaviour close to pF.

Case II: In the static limit for nonzero q the infinite slope of F (x) atx = 1 means that ǫ(0, q) has a weak singularity at |q|/2pF = 1. Near thispoint π(0, q) ∝ (|q| − 2pF) ln(|q| − 2pF). When inserted into eqn 43.59and Fourier transformed this leads to an effective potential

Veff(0, |x|) ∝ cos(2pF|x|)|x|3 , (43.67)

shown in Fig. 43.16. This long-range, oscillatory potential reflects thefact that the metal as a whole reacts to the presence of each charge. Thesharpness of the Fermi surface causes this reaction to ‘ring’ in space, seenas the oscillations in the effective potential of an electron.22

22These are known as Friedel oscilla-tions after Jacques Friedel (1921– ).

Case III: Finally we examine a case with nonzero q0 in the limit|q| → 0.

Example 43.11

In this case we may expand23 the Lindhard function for small q:23Strictly one should consider the realpart of π(q) and so this derivationshould be regarded with caution!

Np+q −Np

Ep+q − Ep − q0≈ q · vp

q · vp − q0

„dNp

dEp

«

, (43.68)


where vp = ∇pEp . Expanding in powers of momentum, and including a factor 2for spin degeneracy, we have

lim|q|→0

π(q0,q) ≈ −2VZ

d3p

(2π)3

»q · vp

q0+

(q · vp)2

(q0)2

–„dNp

dEp

«

, (43.69)

= 2VZ

d|p|d(cos θ)dφ

(2π)3|p|2

"

|q||vp | cos θq0

+

„ |q||vp | cos θq0

«2#

m

|p| δ(|p| − pF),

where we’ve used the fact that“

dNp

dEp

”

= − m|p| δ(|p|−pF). The integration over angle

kills the first term and we obtain from the second that

lim|q|→0

π(q0,q) = g(EF)|vF|2

3

|q|2(q0)2

. (43.70)

Inserting into our expression for the permittivity yields

lim|q|→0

ǫ(q0, q) = 1 −ω2

p

(q0)2, (43.71)

where ω2p = ne2

mǫ0is known as the plasma frequency. The fact that

lim|q|→0 ǫ(q0, q) vanishes at q0 = ωp is telling us about the presence of

a new, long-wavelength, mode of excitation in the system. This is theplasma oscillation and corresponds to the entire system of electronsoscillating with respect to the positive background.

= +

+ · · ·++

Fig. 43.17 The oyster diagram within the RPA.

Finally we mention the reason we took the diversion into the RPA inthe first place: the correction to the excitation energy Ep. This is ob-tained from the RPA by inserting the renormalized interaction shown bythe shaded blob in Fig. 43.14 into our Feynman diagrams. For example,the corrected oyster diagram is shown in Fig. 43.17. The evaluation ofthe numbers is, as might be imagined, quite involved. It may be shownthat the quasiparticle energy is predicted to be

Ep =p2

2m− 0.166rs(ln rs + 0.203)

|p|pF

2m+ const., (43.72)

which leads to a nonzero effective mass.

plasmons

electron-holeexcitations

E

|p|

|p| |p| + 2pF

(a)

(b) (c)

0 2pF

Fig. 43.18 (a) The excitation spec-trum of a metal showing low-energyparticle–hole quasiparticle excitationsand the higher energy (collective) plas-mon mode. A quasiparticle excitationwith energy Ep may be created with:(b) a momentum transfer |p|, from oneside of the Fermi surface, and (c) ≈|p| + 2pF from the other (and all mo-menta in between) leading to the con-tinuum of states shown in the shadedregion.

The insights provided by the RPA allow us to draw a diagram of thedispersion relation of excitations of the metal, shown in Fig. 43.18. The


low-energy excitations are quasielectron–quasihole pairs, while at higherenergies we see the plasmon mode. Immediately noticeable is the factthat the low-energy excitations form a continuum of width 2pF. Thisarises because excitations involve promoting an electron from within aFermi sphere which has a diameter 2pF. For a given excitation energywe therefore have a range 2pF of possible momenta, with the limitscorresponding to forming the excitation from filled states on the nearestand furthest points on the Fermi sphere, as shown in Figs. 43.18(b)and (c).

Chapter summary

• A treatment of interactions in metals has been introduced at thelevel of the Hartree–Fock approximation. The Hartree term givesthe energy due to the constant part of the potential and is repre-sented by a tadpole diagram. The Fock term describes exchangeand is represented by an oyster diagram.

• The next correction to the Hartree–Fock approximation is theRPA, introducing a bubble into the interaction line due to virtualelectron–hole pairs and this allows us to treat screening.

Exercises

(43.1) Verify that − 34π

e2

4πǫ0pF = − 0.916

rs, where the right-

hand side is in units of Rydbergs per electron.

(43.2) (a) Using the Hartree–Fock expression for the quasi-particle excitation energy in eqn 43.37, show thatthe Hartree–Fock effective mass is given by

m

m∗ =e2m

2πpF

1

x2

„

1 + x2

xln

˛

˛

˛

˛

1 + x

1 − x

˛

˛

˛

˛

− 2

«

, (43.73)

with x = |p|/pF and show that this diverges atx = 1.(b) Using the RPA expression for the quasiparticleexcitation energy in eqn 43.72 show that the RPAeffective mass is given by

m

m∗ = 1 − 0.083rs(ln rs + 0.203). (43.74)

(43.3) Including spin leads to a many-body Hamiltonianfor electrons of

H =X

pσ

p2

2ma†pσapσ (43.75)

+1

2

X

pkqσσ′

Vq a†p−qσa

†k+qσ′ akσ′ apσ.

(a) Why does the kinetic energy term favour equalnumbers of spin-up and spin-down electrons?(b) Show that the Hartree contribution to the totalenergy is independent of the relative populations ofspin-up and spin-down electrons.(c) Show that the Fock term gives a negative con-tribution to the energy for electrons with like spinsand no contribution from the interaction of elec-trons with unlike spins.(d) Under what circumstances will this system bea ferromagnet?

Exercises 399

(43.4) From the definition of the free electron–hole prop-agator

G0(x, y) = 〈0|T ψ(x)ψ†(y)|0〉, (43.76)

along with the electron–hole mode expansion, showthat the electron–hole propagator is given byeqn 43.46.

(43.5) Use the fact (which was proved in Chap-ter 22.2) that 〈Ω(∞)|Ω(−∞)〉 = 〈0|S|0〉 =

exp

»

P

„

Connectedvacuum diagrams

«–

to show that

−iE

V =

P

„

Connectedvacuum diagrams

«

VT . (43.77)

Notice that dividing by the volume and time VT re-moves the factor of δ(4)(x = 0) that we showed inExercise 19.4 accompanies all vacuum diagrams.

(43.6) (a) Write down the amplitude for the pair-bubblediagram in Fig. 43.11 and show that it varies asR

dq|q|4 , which is divergent at small q.

(b) By considering the other diagrams containingtwo interaction wiggles, show that the pair-bubbleis the most divergent diagram at second order.(c) What is the most divergent diagram at third or-der?(d) Show that the RPA corresponds to a summa-tion of the most divergent diagrams in the electron

self-energy.See Mattuck for help.

(43.7) Consider an electron gas in (1+1)-dimensionalspacetime.(a) Show that the equation of motion for electronsnear the Fermi energy may be written

„

∂

∂t± vF

∂

∂x

«

ψ = 0, (43.78)

where the + sign is applied for electrons moving tothe right and the − for electrons moving to the left.(b) What is the Lagrangian for this system of left-and right-moving electrons?(c) Using the (1 + 1)-dimensional representation ofthe γ matrices from Exercise 36.5, γ0 = σ2 andγ1 = iσ1, show that the Lagrangian may be written

L = iψ†„

∂

∂t− vFσ3

∂

∂x

«

ψ, (43.79)

where ψ =

„

ψL

ψR

«

.

(d) Employing units where vF = 1, show that thismay be written the form of a massless Dirac La-grangian.This problem is discussed in more depth in Zee,Chapter V.5.

44 Superconductors

44.1 A model of a superconductor400

44.2 The ground state is made ofCooper pairs 402

44.3 Ground state energy 403

44.4 The quasiparticles are bo-golons 405

44.5 Broken symmetry 406

44.6 Field theory of a charged su-perfluid 407

Chapter summary 409

Exercises 409

I’ll use super-mathematics! One bean weighs 1/20 of anounce! The jar weighs 12 pounds! Allowing two pounds forthe jar, that makes 20 times 16 times 10 ... or 32,000 beans!Next request, please!Superman (1938–1992, 1993– )

Upon cooling, it is found that many metals undergo a transition toa superconducting state where all magnetic flux is expelled from theirinterior and their electrical resistance is zero. Experiments have probedthe excitation spectra of these materials, which are found to have anenergy gap ∆ per particle that separates the ground state from excited,quasiparticle states. In this chapter we will use quantum field theory todescribe this phase of matter.

44.1 A model of a superconductor

Like superfluidity, superconductivity involves a dissipationless flow ofparticles. In the late 1930s, Fritz London recognized that both effectsFritz London (1900–1954)

involve the condensation of particles into a macroscopically occupiedstate. We saw that in the case of superfluidity this went hand in handwith the spontaneous breaking of a global U(1) symmetry. This willalso turn out to be the case for superconductors. However, since theparticles in a superconductor are charged (they must be since they carryelectrical current), we are motivated to look for a theory which includeselectromagnetism. Electrodynamics is a gauge theory which is invariantwith respect to local U(1) transformations; it is the global breaking ofthis symmetry1 that leads to superconductivity and the expulsion of flux

1By ‘breaking global symmetry’ wemean that the ground state picks out aunique direction for the phases to pointfor the entire system.

and measured dispersion follow as consequences:

A superconductor is a state resulting from breaking global phasesymmetry in a system with local U(1) invariance.

Which particles should we be considering as the constituents of the su-perconducting state? Although originally it was assumed that thesewere the electrons of a metal, this is incorrect. The superconductor isbuilt, not from electrons, but from pairs of electrons. In 1949 HerbertFrohlich formulated the theory of the electron–phonon interaction2 and

Herbert Frolich (1905–1991)

2The electron–phonon interaction isquite similar to the electron–photon in-teraction in QED, with the interac-tion consisting of an electron emitting aphonon, rather than a photon as shownin Fig. 44.1.

this allowed the possibility of an attractive interaction between electronsmediated by phonons. This interaction between two electrons looks like

e

ephonon

Fig. 44.1 The electron–phonon inter-action vertex.

44.1 A model of a superconductor 401

a t-channel electron–electron scattering process as shown in Fig. 44.2.What makes an attractive interaction a realistic proposition is the vastdifference in time scales between the moving electrons and the move-ment of ion cores. Electrons move at the Fermi velocity and so the timescale over which they move through the crystal is E−1

F . The time scaleof the deformation of the lattice is set by the phonon Debye frequency3

Peter Debye (1884–1966)

3The Debye frequency ωD is the max-imum energy that a phonon is allowedto take. Any larger and the phononhas a wavelength so small that it beginsto explore length scales smaller thanthe distance between atoms and the no-tion of a phonon as a long-wavelength,collective excitation of a lattice breaksdown. Phonons move at around thespeed of sound while electrons veloci-ties are typically a factor of 102 larger.

as ω−1D . Since E−1

F ≪ ω−1D then the original electron has moved well

clear by the time the phonon finds the second electron. In the fieldtheory picture, this difference in time scales corresponds to an effectiveelectron–electron interaction that is strongly ‘retarded’, which is to say,strongly dependent on the phonon frequency.

p ↑

k ↑

−p ↓

−k ↓phonon

Fig. 44.2 The t-channel process in asuperconductor. The effective interac-tion is mediated by a virtual phonon.

Example 44.1

We can piece together the form of the effective electron–electron interaction. Byanalogy with QED, the electron–phonon interaction vertex will be that shown inFig. 44.1 with a coupling constant κ. The effective electron–electron interaction willbe given by the t-channel process, mediated by the phonon, which has a propagatorresembling the free propagator for the scalar field D0(q) ∝ (ω2 − ω2

q )−1, where ωq

is the dispersion of the phonons. This results in an effective interaction betweenelectrons given by

Veff(ω,q) ∝ κ2

ω2 − ω2q

. (44.1)

Taking the characteristic frequency of the phonons as the Debye frequency ωq ≈ ωD,we deduce that the interaction is attractive for ω < ωD. Leon Cooper showed more

Leon Cooper (1930– )

rigorously that an arbitrarily small electron–electron attraction would lead to pairingof electrons into bound states and that this pairing would occur between electronsacross the whole of the metallic Fermi surface, making the entire metal unstableto the formation of pairs! Specifically, Cooper showed that there is an attractiveinteraction between two electrons with equal and opposite momenta and oppositespins. This is mediated by phonons and occurs chiefly between electrons within anenergy ωD of the Fermi energy EF.

We will explore the consequences of the existence of these so-calledCooper pairs of electrons further by using an effective Hamiltoniangiven by4

4Comparing our notation from Chap-

ter 42, one can identify κ2 and g2V .

H =∑

pσ

εpc†pσ cpσ − κ2

∑

pk

c†k↑c†−k↓c−p↓cp↑, (44.2)

where σ(=↑, ↓) labels the electron spin. This Hamiltonian describesthe attractive potential energy between pairs of electrons, with oppositespins and momenta, whose energies will be taken in a shell of width ωD

at the Fermi surface.5

5Note that this form of the poten-tial was motivated by the mean-fieldapproach of Chapter 43. There wesaw that the mean-field Hamiltonianfor electrons in metals throws up termssuch as a contribution to the energy ofC0 = 〈c†p−q c

†k+q

〉〈ck cp〉, which we dis-carded for an ordinary metal since or-dinary single-electron states give zerofor each expectation value. This is notthe case for Cooper pairs of spin- 1

2elec-

trons.

This Hamiltonian in eqn 44.2 is the model that was used to solve theproblem of superconductivity by John Bardeen, Leon Cooper and Robert

John Bardeen (1908–1991) is the onlyperson to have been awarded the NobelPrize in physics twice.

Schrieffer, known colloquially as the BCS model. Perhaps the most

J. Robert Schrieffer (1931– )

significant feature of the BCS model is not the Hamiltonian itself, but themethod used to solve it. Schrieffer invented a trial many-particle state|ΨBCS〉, known as the BCS wave function, which ingeniously capturesthe crucial physics. The BCS wave function is a coherent state whosep = 0 mode is macroscopically occupied when the pairs condense to form

402 Superconductors

the superconducting ground state. It is used as a variational state withthe model Hamiltonian to find the ground state energy of the system.We therefore start by building the BCS coherent state function.

44.2 The ground state is made of Cooper

pairs

The building blocks of the superconductor are Cooper pairs. These arepairs of electrons with opposite spin and momenta. We can create aCooper pair from the vacuum |0〉 using the operator

P †p = c†p↑c

†−p↓. (44.3)

The BCS coherent state is built from the vacuum using this operator asfollows:

|ΨBCS〉 =∏

p

Cp eαpP†p |0〉. (44.4)

The superconducting state is often described in terms of Cooper pairs of

In eqn 44.4, the Cp ’s are normalizationconstants and αp is a complex numberwhich depends on p.

electrons in some sense ‘becoming’ bosons which subsequently undergoBose condensation. This can’t be quite right, since the commutationrelations for the pair operator aren’t those of a boson (see box). The

Commutation relations for Pp

[Pp , Pq ] = 0

[P †p , P

†q ] = 0h

Pp , P†q

i

= δpq (1 − Np↑ − N−q↓).pair operators do, however, commute with each other. Remember thatthe c†p operators create fermions which obey the exclusion principle and,in particular, we have

P †pP

†p = c†p↑c

†−p↓c

†p↑c

†−p↓ = 0. (44.5)

Example 44.2

The fact that (P †p )2 = 0 gives us enough information to simplify the BCS coherent

state. We notice that only the first two terms of the exponential can be nonzero(since they contain no powers and one power of P †

p respectively) and we have

|ΨBCS〉 =Y

p

CpeαpP†p |0〉 =

Y

p

Cp

“

1 + αp P†p

”

|0〉. (44.6)

We can also normalize the wave function:

1 = |Cp |2〈0|“

1 + α∗p Pp

”“

1 + αp P†p

”

|0〉 = |Cp |2(1 + |αp |2), (44.7)

giving Cp = (1 + |αp |2)−12 . Defining

up = 1

(1+|αp|2)12

, vp =αp

(1+|αp|2)12

, (44.8)

we have that |up |2 + |vp |2 = 1, we can write6

6It’s important to notice that |ΨBCS〉is very different from the usual situa-tion in a metal in which the wave func-tion is a single specific configuration ofempty and filled states. The product ofterms (up +vp P

†p ) ensures that |ΨBCS〉

contains a coherent sum containing allpossible configurations in which everypair state is filled or empty. The BCSstate combines all possibilities in a sin-gle wave function. |ΨBCS〉 =

Y

p

(up + vp P†p )|0〉. (44.9)

Finally we can give meaning to vp . If we evaluate the ground state expectation valueof the number operator for spin-up electrons, we obtain7

7See Exercise 44.3.

〈Np↑〉 = 〈ΨBCS|c†p↑cp↑|ΨBCS〉 = |vp |2, (44.10)

which tells us that |vp |2 gives the average pair occupation of a state labelled withmomentum p with spin up.8

8Similarly 〈Np↓〉 = |vp |2. Of course,

|up |2 = 1 − |vp |2 tells us the aver-age non-occupation of the state (since afermion state can either be unoccupiedor singly occupied).

44.3 Ground state energy 403

Written in terms of our single-fermion operators we have a simplifiedexpression for the coherent state given by

|ΨBCS〉 =∏

p

(up + vpc†p↑c

†−p↓)|0〉. (44.11)

The number operator N is given by

N =∑

pσ

c†pσ cpσ =∑

p

(Np↑ + Np↓

). (44.12)

Thus the total number of electrons in the superconducting state is foundby summing over spin-up and spin-down occupancies and so

〈ΨBCS|N |ΨBCS〉 =∑

p

(〈Np↑〉 + 〈Np↓〉

)= 2

∑

p

|vp|2. (44.13)

We now have enough ammunition to attack the variational problem offinding the ground state energy.

44.3 Ground state energy

We will now use the BCS ground state as a variational wave function forthe model Hamiltonian (eqn 44.2) describing the superconductor. Byminimizing the energy with respect to the parameters up and vp we willextract the properties of the ground state of the superconductor. Theexpectation value of the energy is

E = 〈ΨBCS|H|ΨBCS〉, (44.14)

which, upon substitution,9 gives us

9The first term is easy since it is thekinetic energy term in the BCS Hamil-tonian (eqn 44.2). It can be written

X

pσ

εpNpσ ,

and so we can use eqn 44.13 to reduceit to X

p

2εp |vp |2.

The second term is evaluated in the ex-ercises.

E =∑

p

2εp|vp|2 − κ2∑

pk

v∗pvku∗kup. (44.15)

We will minimize this subject to two constraints. The first fixes the totalparticle number via

N = 2∑

p

|vp|2, (44.16)

which we impose as a constraint using a Lagrange multiplier10 µ. We 10As will be obvious to those with ex-pertise in statistical mechanics, the La-grange multiplier µ is the chemical po-tential. The Lagrange multiplier Ep

will turn out to give the quasiparticleenergy.

also introduce a Lagrange multiplier Ep to enforce the constraint that|up|2 + |vp|2 = 1.

Example 44.3

Using the method of Lagrange multipliers, we write a function f using

f = E − µN +X

p

Ep(|up |2 + |vp |2 − 1), (44.17)

404 Superconductors

and look for solutions to ∂f/∂up = 0 and ∂f/∂vp = 0. This yields

∂E

∂up

− µ∂N

∂up

+ Epu∗p = 0,

∂E

∂vp

− µ∂N

∂vp

+ Epv∗p = 0. (44.18)

Doing the derivatives we find, after a little algebra (examined in Exercise 44.5), thematrix equation

„(εp − µ) ∆

∆∗ −(εp − µ)

«„u∗pv∗p

«

= Ep

„u∗pv∗p

«

, (44.19)

where we’ve written ∆ = κ2P

p u∗pvp . This matrix has eigenvalues

Ep = ±ˆ(εp − µ)2 + |∆|2

˜ 12 . (44.20)

Finding the eigenvectors is made easier if we define

cos 2θp =εp−µEp

, sin 2θp = ∆Ep

. (44.21)

Then we find eigenvectors

u∗p = cos θp , v∗p = sin θp , (44.22)

or

|up |2 =1

2

„

1 +εp − µ

Ep

«

,

|vp |2 =1

2

„

1 − εp − µ

Ep

«

. (44.23)

The behaviour of |up |2 and |vp |2 is shown in Fig. 44.3. We also find that u∗pvp =

cos θp sin θp = 12

sin 2θp = ∆/2Ep . We conclude that the BCS ground state is

|ΩBCS〉 =Q

p(cos θp + sin θp c†p↑c

†−p↓)|0〉.

Ep

εp − µ

εp − µ

|vp|2 |up|2

12

u∗ pv p

εp − µ

Fig. 44.3 (a) The function Ep (whichwe will later call the bogolon disper-sion). (b) The functions |up |2 and|vp |2 which show that well below µ thestate is predominantly electron-like andwell above it is predominantly hole-like.Near to µ it has a mixed electron-likeand hole-like character. (c) u∗pvp =∆/2Ep , which has a maximum whenthere are plenty of electron states andavailable states to scatter into.

We will see in the next section that ∆ is the energy gap between theground state and excited quasiparticle states. We may now eliminateup and vp to reveal the physics of the BCS state. Using u∗pvp = ∆/2Ep

with the definition of the gap ∆, we find our first result which is that

∆ = κ2∑

p

∆

2Ep

. (44.24)

Since the electrons that form the pairs only lie within ωD of the Fermienergy, we can restrict our sum to |(εp − µ)| < ωD. We then make thereplacement

∑p → g(εF)

∫dε, where g(εF) is the density of states at

the Fermi energy, and writing ε ≡ εp − µ, eqn 44.24 becomes

∆ = κ2 g(εF)︸︷︷︸Λ

∫ ωD

−ωD

dε∆

2(ε2 + ∆2)12

, (44.25)

where we have defined Λ (the effective coupling constant) using Λ =κ2g(εF). This integral gives

1

Λ= sinh−1

(ωD

∆

), (44.26)

44.4 The quasiparticles are bogolons 405

but since ∆ ≪ ωD we obtain

|∆| ≈ 2ωD e−1/Λ. (44.27)

This tells us how the energy gap that promotes superconductivity de-pends on the effective coupling and the Debye energy. Superconductivityis most robust if there is a large energy gap and so, if we want to find themost robust superconductor with the highest possible11 critical temper- 11This assumes BCS-like electron–

phonon pairing. A higher Tc may befound using other mechanisms.

ature Tc, we need a material with a large Debye frequency and a largecoupling constant Λ. The latter is achieved by having a large κ and alarge density of states at the Fermi energy.

44.4 The quasiparticles are bogolons

We now make good on our claim of the last section: we want to confirmthat ∆ is indeed the energy gap in the particle spectrum of the super-conducting ground state. To find the excited states we employ a similarapproach to that of the metal. We consider the model Hamiltonian withaveraged pairs of operators in the interaction part:

H =∑

pσ

(εp−µ)c†pσ cpσ−κ2∑

p

(〈c†p↑c

†−p↓〉c−p↓cp↑ + 〈c−p↓cp↑〉c†p↑c

†−p↓

),

(44.28)where the expectation values are taken with respect to the BCS groundstate |ΩBCS〉. Substitution of |ΩBCS〉 allows us to identify the quantityκ2∑

p〈ΩBCS|c−p↓cp↑|ΩBCS〉 as the energy gap12 ∆ = κ2∑

p u∗pvp, and 12In manipulating these expressions in-

volving the energy gap, note that if ∆is not real then, writing ∆ = |∆|eiφ,we may always make it real with theglobal transformation ci → eiφ/2ci,c†i → e−iφ/2c†i . In the next sectionwe will confirm that choosing a particu-lar value for ∆ corresponds to breakingsymmetry.

we have

H =∑

pσ

(εp − µ)c†pσ cpσ −∑

p

(∆∗c−p↓cp↑ + ∆c†p↑c

†−p↓

). (44.29)

We need to diagonalize this Hamiltonian, which contains cc and c†c†

bilinears. As in the last chapter, the Bogoliubov transformation givesus a method to do this.

Example 44.4

Written in matrix form we have the Hamiltonian13 13This looks a lot like eqn 44.19 butwith minus signs on the off-diagonal el-ements.H =

X

p

“

c†p↑ c−p↓

”„ εp − µ −∆−∆∗ −(εp − µ)

«

cp↑c†−p↓

!

. (44.30)

This may be diagonalized using the Bogoliubov procedure, whose details are exam-ined in Exercise 44.7, to give

H =X

p

“

b†p↑ b−p↓

”„ Ep 00 −Ep

«

bp↑b†−p↓

!

, (44.31)

where the operators b†pσ and bpσ create and destroy bogolon quasiparticle excitations.The bogolon operators obey anticommutation laws:

n

bp1σ1 , b†p2σ2

o

= δp1p2δσ1σ2 , (44.32)

n

b†p1σ1, b†p2σ2

o

= 0,n

bp1σ1 , bp2σ2

o

= 0.

406 Superconductors

The diagonalized Hamiltonian for the excitations in the superconductoris given by

H =∑

p

Ep(b†p↑bp↑ + b†−p↓b−p↓), (44.33)

where Ep =√

(εp − µ)2 + |∆|2. This is shown in Fig. 44.3. Notice thatthe bogolons have a dispersion relation that p starts at an energy ∆above the ground state energy. This is the energy gap:14 there are no

14This is key to superconductivity. Ex-citations will dissipate momentum andprevent the superflow of current. Thepresence of a gap prevents their cre-ation at low energies.

allowed quasiparticle excitations between Ep = −∆ and Ep = ∆.

44.5 Broken symmetry

Fig. 44.4 Order in the superconductorinvolves the lining up of phase angles,just as in the case of an uncharged su-perfluid.

As in the superfluid state examined earlier, superconductivity involvesthe macroscopic occupation of a coherent ground state |Ω〉. Recall thatthe order parameter for the superfluid was given by

〈Ω|Φ(x)|Ω〉 =1√V∑

p

〈Ω|ap|Ω〉eip·x

≈ 1√V〈Ω|ap=0|Ω〉 =

√neiθ0 . (44.34)

For the superconductor we have (by the same token)

〈ΩBCS|Ψ(x)|ΩBCS〉 =1√V∑

p

〈ΩBCS|c−p↓cp↑|ΩBCS〉ei(p−p)·x

=1√V∑

p

〈ΩBCS|c−p↓cp↑|ΩBCS〉

=√neiθ0 . (44.35)

From these expressions we see that we could also regard the supercon-ducting energy gap as the order parameter for the system since we have

∆ = κ2∑

p

〈ΩBCS|c−p↓cp↑|ΩBCS〉. (44.36)

Returning to the real space expectation value 〈ψ(x)|Ψ(x)|ψ(x)〉 =√ρ(x)eiθ(x) for a coherent state |ψ(x)〉, we see that for the supercon-

ducting ground state |ΩBCS〉 has the phase field θ(x) breaking symmetryby becoming a constant θ0. (As in the case of the superfluid we can vi-sualize this with an XY -model, as shown in Fig. 44.4.) As alluded tobefore, the difference between superfluidity and superconductivity comesfrom the fact that the phase symmetry that’s broken is a local one. Priorto breaking symmetry the superconductor enjoys local U(1) symmetry:we would alter the phase by θ(x) → θ(x) + α(x), where α(x) is differ-ent at all points in spacetime and the Lagrangian describing the systemwould remain invariant. A gauge field is needed in the Lagrangian toguarantee this. We have seen in Chapter 26 that the gauge field willconsume the Goldstone boson and become massive. This is the Higgsmechanism and occurs in superconductors! We will examine the Higgsphenomenon in superconductivity in the next section.

44.6 Field theory of a charged superfluid 407

44.6 Field theory of a charged superfluid

The superfluid Lagrangian is invariant with respect to the global trans-formation Ψ → eiαΨ. We now promote this global symmetry to a localone. As we have seen, this requires the introduction of a gauge field Aµthrough the replacement of the derivative ∂µ by the covariant deriva-tive Dµ = ∂µ + iqAµ. The coupling constant q is called the charge ofthe Ψ field, so by making a local phase transformation we are studyingthe physics of a charged superfluid. Starting with the non-relativisticsuperfluid Lagrangian (and denoting the coupling by g = 2Vκ2):

L = iΨ†∂0Ψ − 1

2m∇Ψ† · ∇Ψ − g

2(n− Ψ†Ψ)2, (44.37)

we gauge the theory by introducing the covariant derivative Dµ. Thiswill be easier in polar coordinates, so we introduce Ψ(x) =

√ρ(x)eiθ(x)

and DµΨ =[

12√ρ∂µρ+ i

√ρ (∂µθ + qAµ)

]eiθ to obtain the gauged La-

grangian

L = −ρ(∂0θ + qA0) −1

2m

[1

4ρ(∇ρ)2 + ρ (∇θ − qA)

2

]

−g2

(n− ρ)2 − 1

4FµνF

µν , (44.38)

where, as before, we drop the total time derivative term i∂0ρ/2. Notethat we have included the contribution of the gauge field via Fµν =∂µAν − ∂νAµ. The Lagrangian is now invariant with respect to thechange θ(x) → θ(x) + α(x) provided Aµ(x) → Aµ(x) − 1

q∂µα(x).As we’ve seen before, when we break the symmetry we’ll find that

the massless photon field Aµ(x) gobbles up the Goldstone mode θ(x)and acquires a mass. This is the Higgs mechanism, at play in a chargedsuperfluid.

Example 44.5

Here’s how it unfolds.15 We notice that we can cast our Lagrangian in terms of a 15See Example 42.4 for the simpler ver-sion of this argument appropriate forthe superfluid.

gauge invariant contribution Cµ(x) = 1q∂µθ(x) +Aµ(x), and the contribution of the

gauge field is Fµν = ∂µAν − ∂νAµ = ∂µCν − ∂νCµ. We have

L = −ρqC0 − 1

2m

»1

4ρ(∇ρ)2 + ρq2C2

–

− g

2(n− ρ)2 − 1

4FµνF

µν . (44.39)

Now break the symmetry of the ground state.16 We choose a ground state Ψ0(x) = 16Remember, you don’t break the sym-metry of a Lagrangian; that always re-tains the full symmetry. We simply ex-pand the Lagrangian around the brokensymmetry ground state.

√neiθ0 , where we’ll make our usual choice of θ0 = 0 as shown in Fig. 44.4. We

expand about the ground state by examining small displacements√ρ =

√n + h,

which yields

L = −nq2

2mC2 − 1

2m(∇h)2 − 2gnh2 − 2q

√nC0h− 1

4FµνF

µν , (44.40)

dropping the term −qnC0 as we did for superfluids. Already notice that the C-field

appears to have acquired a mass (nq2/m)12 . To complete our calculation we should

integrate out the energetic field h to obtain a low-energy field theory. Just as inChapter 42, we recognize that the h dependent part can be written in the form

−h„

− 1

2m∇

2 + 2gn

«

h− (2q√nC0)h. (44.41)

408 Superconductors

Again, just as before, we will integrate out the field h with our path integral via− 1

2φKφ+ Jφ→ 1

2JK−1J . Setting φ = −

√2h and J =

√2nqC0 we have

L = −nq2

2mC2 + q

√nC0

1

(−1/2m)∇2 + 2gnq√nC0 − 1

4FµνF

µν

=q2

2gC2

0 − nq2

2mC2 − 1

4FµνF

µν

=q2

2g

`C2

0 − v2C2´− 1

4FµνF

µν , (44.42)

where v2 = ng/m and we have neglected the term (1/2m)∇2 as small comparedto 2gn. This is a non-relativistic equation telling us that the fields in a chargedsuperfluid give rise to excitations which are spin-1 vector particles.1717See Chapter 13 if in doubt.

The consequence of the Higgs mechanism in a superconductor is the ex-pulsion of magnetic flux from the interior of the material. This is knownas the Meissner effect. Starting with the low-energy Lagrangian fromeqn 44.42 we recognize that we can eliminate the component C0 (since amassive vector field has only three degrees of freedom, see Chapter 13).We may therefore integrate out C0 (in the same manner as above) withthe result that

L = −nq2

2mC2 +

(Terms involvingderivatives of C

). (44.43)

Notice that the three massive fields described by this equation have massM2 = nq2/m.

Our low-energy Lagrangian in eqn 44.43 may be used to find thecurrent in the superconductor, which is given bySee Exercise 44.8.

J =nq

m(∇θ − qA) . (44.44)

We conclude that the current in the superconductor responds to gradi-ents in the phase field (just like the superfluid) and also to the appliedfield A.

Example 44.6

Upon taking the curl of the current, we obtain an equation describing the magneticinduction field B in the superconductor:1818This is one of the London equations;

the other is

∂J

∂t=nq2

mE,

which also follows from eqn 44.44 as-suming ∇θ has no time-dependence.

∇× J = −nq2

mB, (44.45)

which, combined with the static Maxwell equation ∇× B = J, yields

∇2B =

nq2

mB. (44.46)

In our units, the quantity nq2/m must have units (length)−2, so we define λ =(m/nq2)−1/2.

In one dimension eqn 44.46 now reads

d2B

dx2=

1

λ2B, (44.47)

Exercises 409

whose solutions have the form B(x) = B(0)e−x/λ. In other words, the field ina superconductor falls away as we enter the interior of the superconductor over alength scale determined by λ. Since λ tells us how far magnetic flux can penetratethe superconductor it is known as the penetration depth. Our treatment has allbeen aimed at describing a charged superfluid. A superconductor is quite similar. Ina superconductor we have Cooper pairs, so we conventionally set q = −2|e| and writens = 2n and m = 2m∗ for the superfluid density and effective mass respectively.This gives a penetration depth19 for a superconductor of 19In SI units, 1

λ2 = µ0nse2

m∗ .

1

λ2=nse2

m∗ . (44.48)

Chapter summary

• The BCS wave function is given by |ΨBCS〉 =∏

p(up +

vpc†p↑c

†−p↓)|0〉 and is a coherent state of Cooper pairs.

• The quasiparticle excitations are bogolons, with dispersion givenby Ep =

√(εp − µ)2 + |∆|2.

• The superconducting energy gap ∆ can be regarded as theorder parameter of a superconductor, and is given by ∆ =κ2∑

p〈ΩBCS|c−p↓cp↑|ΩBCS〉.• The Higgs mechanism in a superconductor leads to the London

equation, which can be written as J = nqm (∇θ − qA).

Exercises

(44.1) Show that the amplitude

〈ΨBCS|c†p↑c†−p↓|ΨBCS〉〈ΨBCS|c−p↓cp↑|ΨBCS〉

(44.49)is nonzero.

(44.2) Verify the commutatorh

Pp , P†q

i

= δpq(1 − Np↑ −N−q↓).

(44.3) Verify 〈ΨBCS|Np↑|ΨBCS〉 = |vp |2.(44.4) Verify eqn 44.15:

〈E〉 =X

p

2εp |vp |2 − κ2X

pk

v∗pvku∗kup . (44.50)

(44.5) Starting with the total energy and total particlenumber written in the form

E =X

p

εp

`

|vp |2 − |up |2 + 1´

− κ2X

pk

v∗pvku∗kup ,

N =X

p

`

|vp |2 − |up |2 + 1´

, (44.51)

derive eqn 44.19:„

(εp − µ) ∆∆∗ −(εp − µ)

«„

u∗p

v∗p

«

= Ep

„

u∗p

v∗p

«

.

(44.52)(b) Diagonalize the Hamiltonian and find |up |2 and|vp |2 using the method suggested in the text.

(44.6) (a) Define the Nambu spinor as ψp =

„

cp↑c†−p↓

«

and the energy gap as ∆ = ∆1 − i∆2. Show that

410 Superconductors

the BCS Hamiltonian can be written

H =X

p

ψ†php · τ ψp , (44.53)

where hp = (∆1,∆2, εp − µ) and τ is a vector madeup of the Pauli matrices.(b) Show that eqn 44.19 is identical to the Diracequation upon making the replacements:

ψL → v∗p ,ψR → u∗

p ,σ · p → εp − µ,m → ∆.

(44.54)

This is the basis of Nambu’s analogy.(c) When local U(1) symmetry is globally brokenin the superconductor ∆ takes on a nonzero value.What is the equivalent of this symmetry breakingin the Dirac problem?Nambu’s analogy is discussed further in Aitchisonand Hey, Chapter 18.

(44.7) We will find the excitations out of the BCS groundstate by diagonalizing eqn 44.30 using the Bogoli-

ubov procedure. The transformation to use in thiscase is

„

cp↑c†−p↓

«

=

„

u∗p vp

−v∗p up

«

bp↑b†−p↓

!

, (44.55)

and the algebra is simplified by using the substitu-tions

up = cos θp , (44.56)

vp = sin θp . (44.57)

(a) Check that the transformation maintains theanticommutation relations.(b) Verify that this does the job of diagonalizingthe Hamiltonian.(c) Confirm that the bogolons are really the excita-tions by checking bp↑|ΩBCS〉 = 0 and b−p↓|ΩBCS〉 =0, that is, that there are no bogolons in the BCSground state.

(44.8) Show that the current in a superconductor is givenby eqn 44.44.

45The fractional quantum

Hall fluid

45.1 Magnetic translations 411

45.2 Landau Levels 413

45.3 The integer quantum Hall ef-fect 415

45.4 The fractional quantum Halleffect 417

Chapter summary 421

Exercises 421

The more success the quantum theory has, the sillier it looks.Albert Einstein (1879–1955)

The application of a magnetic field to electrons in condensed matter hassurprising consequences, none more so than the integer and fractionalquantum Hall effects. The latter effect is a product of interactions, butthere are some interesting features even in the non-interacting case whichwe review first.

Throughout this chapter, it will be helpful to keep in mind that therelevant length scale in the problem is the magnetic length ℓB, givenby1 1For ease of notation we follow conven-

tion in condensed matter physics andwrite −e for the charge on the electronin this chapter.

ℓB =

(~

eB

) 12

. (45.1)

Note that2 the flux Φ through a circle with radius√

2ℓB is equal to 2The fact that we have to make the cir-

cle have radius√

2ℓB and not ℓB is an-noying, but we will have to live with it.As we shall see, sometimes the relevantlength scale is ℓB and sometimes

√2ℓB.

Φ = B(2πℓ2B) = h/e. This is double the value of the magnetic fluxquantum Φ0 = h/2e which is important in superconductivity (wherethe Cooper pairs have charge −2e) and plays the role of a flux quantumfor problems involved with unpaired electrons. It is often called theDirac flux quantum.

45.1 Magnetic translations

Problems involving a lattice in solid state physics make use of transla-tional symmetry. Recall from Chapter 9 that the translation operatorfor a translation by a is e−ip·a, where p is the momentum operator. Fora periodic system we would expect e−ip·a to commute with the Hamilto-

nian H. Writing U(a) = e−ip·a, we would then expect[U(a), H

]= 0 or

U−1(a)HU(a) = H. This expectation is not realized in the presence of amagnetic field where we find that U(a) does not commute with H. Thisis because H has become a function of the magnetic vector potential A

and this changes as you translate. We should therefore expect3 3Treating the magnetic vector poten-tial as a function, rather than a secondquantized field, we have U−1(a)A(r) =eip·aA(r) = A(r − a).

U−1(a)H(A)U(a) = H(U−1(a)A

). (45.2)

However, if the energy of the system is independent of global translationsthen this change in A can’t affect the magnetic field we would measurewith a magnetometer. We have, of course, met such shifts of the vectorpotential before: they simply represent gauge transformations.

412 The fractional quantum Hall fluid

Example 45.1

Consider4 a uniform field B. In the gauge A(r) = 12B × r we have4Recall from Chapter 14 that a gauge

transformation involves changes ψ →ψeiα and Aµ → Aµ − 1

q∂µα and we

sometimes write χ = α/q. Note alsothat the covariant derivativeDµ = ∂µ+iqAµ implies P = p−qA. In this chap-ter we will often deal with electrons,where q = −e.

U−1(a)A(r) = A(r − a) = A(r) − 1

2B × a. (45.3)

This is a gauge transformation Aµ → Aµ − 1q∂µα(x) with 1

q∇α(x) = − 1

2B × a or

α(r)

q= χ(r) = −1

2(B × a) · r. (45.4)

Along with a shift in Aµ, a gauge transformation also involves a shift of the matterfield ψ → ψeiα(x) which here is ψ → ψ exp

îe2

(B × a) · r˜.

To summarize: in the presence of a magnetic field, we have a slightlyunusual form of translational symmetry in that it is invariant under acombination of a translation with a gauge transformation.

We saw in Chapter 14 that the momentum operator pµ = i∂µ doesn’ttransform properly under gauge transformations, but the covariant ver-sion Pµ = iDµ = i(∂µ + iqAµ) does. It therefore comes as little surpriseto find that an upgraded translation operator that does commute withthe gauged Hamiltonian is given by

U(a) = e−iP ·a, (45.5)

where P = −i∇ − qA.To see some consequences of this formalism, we now return to con-

densed matter physics and consider a two-dimensional square lattice(with lattice constant a) of non-interacting electrons in the presence ofa magnetic field. The Hamiltonian is55Since we are discussing two-

dimensional space here we will revertto the ordinary notation of vectorcomponents where momentum alongthe x-direction is written px and so on.We will return to four-vector notationin Section 45.4.

H =1

2m(px + eAx)

2+

1

2m(py + eAy)

2+ V (x, y), (45.6)

where V (x, y) = V (x + a, y) = V (x, y + a) is a periodic potential.With a uniform magnetic field parallel to z we can write the magneticvector potential A = B

2 (−y, x, 0). In this case the magnetic translationoperators that translate us through one lattice spacing in the x- andy-directions are, respectively,

Ux = exp

[− ia

~

(px −

eBy

2

)],

Uy = exp

[− ia

~

(py +

eBx

2

)], (45.7)

where we have restored factors of ~. As advertised, these operatorscommute with H. However, they do not commute with each other and,in fact,6

6This shows that the magnetic trans-lation operators form a non-abeliangroup. See Chapter 46.

UxUy = e2πiφUyUx, (45.8)

where φ = ea2B/h is the dimensionless magnetic flux.7

7The proof of eqn 45.8 follows fromh

− ia~

“

px − eBy2

”

,− ia~

“

py + eBx2

”i

=

2πiφand the use of the identity

eAeB = eBeAe[A,B],

which holds ifh

A, Bi

commutes with A

and B. See Exercise 45.1.

45.2 Landau Levels 413

The quantity φ represents the number of flux quanta (h/e) passingthrough a square on the lattice (area a2, usually called a plaquette).Thus

φ =Φ

h/e, (45.9)

where Φ = Ba2 is the magnetic flux through one plaquette. If φ = p/qwhere p and q are integers then

[Uqx , Uy

]= 0, (45.10)

because (e2πiφ)p = 1. This means that a rectangle consisting of q ad-jacent plaquettes contains an integer number of flux quanta. Thus, thesystem has an effective periodicity equal to q times what would be thecase for B = 0.

Fig. 45.1 The Hofstadter butterfly.This idea is beautifully illustrated by the Hofstadter butterfly8

8For more discussion of the Hofstadterbutterfly see Douglas Hofstadter’s pa-per [Phys. Rev. B 14, 2239 (1976)]and his well-known book Godel, Es-cher, Bach, Chapter 5, which also fea-tures his take on renormalization.

shown in Fig. 45.1, which shows the energy level spectrum for a two-dimensional tight binding model (recall Example 4.8) with a magneticfield perpendicular to it. When B = 0 (and consequently φ = 0), theallowed energies satisfy −4t ≤ E ≤ 4t as you would expect from eqn 4.50.For φ = p/q, the allowed energies form q sub-bands (note that sometimesthe sub-bands touch, so the two sub-bands at φ = 1/2 meet at E = 0).The resulting band structure is fractal (and was published only a yearafter Benoit Mandelbrot coined the term). It has the extraordinary Benoit B. Mandelbrot (1924–2010)

chose his own middle initial, whichdoesn’t stand for anything. There’s anold joke that says that the ‘B’ in ‘BenoitB. Mandelbrot’ stands for ‘Benoit B.Mandelbrot’, thereby making the nameof the inventor of fractals a recursively-defined fractal.

feature that the spectrum depends on whether φ is rational or irrational(and can only be plotted for rational φ).

45.2 Landau Levels

In order to examine the quantum mechanics of charges in applied mag-netic fields we return to the problem of two-dimensional free electrons.The Hamiltonian for this system may be written

H =P 2x + P 2

y

2m, (45.11)

where

Px = px + eAx, Py = py + eAy. (45.12)

Classically, electrons with momentum p will orbit in circles of radiusp/eB in a magnetic field. Consequently, we write the position operatorsof an electron as

x = X + 1eB Py, y = Y − 1

eB Px, (45.13)

where (X,Y ) are the coordinates of the guiding centre of the orbit.We find that (X,Y ) and (Px, Py) are independent coordinates because

[X, Px

]=[X, Py

]=[Y , Px

]=[Y , Py

]= 0. (45.14)


However, we also find that[X, Y

]= iℓ2B,

[Px, Py

]= − i~2

ℓ2B. (45.15)

Thus the guiding centre cannot be determined more accurately than anarea ≈ ℓ2B, i.e. the area occupied by approximately one quantum of flux.

Example 45.2

Operators X and Y commute with H and hence dXdt

= dYdt

= 0. However,

dPx

dt=

1

i~

h

Px, Hi

= −ωcPy ,

dPy

dt=

1

i~

h

Py , Hi

= ωcPx, (45.16)

where ωc = eB/m is the cyclotron frequency. Thus (ignoring hats)

Px + ω2cPx = 0,

Py + ω2cPy = 0, (45.17)

and one can easily show that electrons perform circular orbits around (X,Y ).

As with many subjects in quantum field theory, the physics of this prob-lem may be described using the machinery of the simple harmonic os-cillator. To map this problem onto the simple harmonic oscillator weintroduce the operators

a = ℓB√2~

(Px − iPy

), a† = ℓB√

2~

(Px + iPy

),

b = 1√2ℓB

(X + iY

), b† = 1√

2ℓB

(X − iY

),

(45.18)

satisfying[a, a†

]=[b, b†

]= 1 and

[a, b]

=[a†, b

]= 0. The Hamiltonian

then assumes the expected form: H =(a†a+ 1

2

)~ωc and so there are

eigenstates |N〉, where N is the number of a-quanta in the oscillator.The energy levels of this system are called Landau levels.

However, note that we can write the states

|N,n〉 =1√N !n!

(a†)N (b†)n|0〉, (45.19)

and so, even though the energy depends only on N , we shouldn’t forgetabout n. This contributes to the degeneracy and, in fact, from this wewill now show that each electron occupies a real space area of πℓ2B.

Example 45.3

The density of states in momentum space is given by

g(k)dk =dk

(2π/L)2× 2, (45.20)

45.3 The integer quantum Hall effect 415

where we write k ≡ |k| = (k2x + k2

y)12 , the factor of 2 is for spin degeneracy and L is

the linear size of the system. The radius kN of the Nth Landau level in k-space isfound by writing

~2k2N

2m=

„

N +1

2

«

~ωc, (45.21)

and hence

k2N =

1 + 2N

ℓ2B. (45.22)

The area between two successive circles in k-space tells us how much momentumspace is taken up by each Landau level. It is given by

∆Ak = π(k2N+1 − k2

N ) =2π

ℓ2B, (45.23)

where the factor of 2 again accounts for spin. We conclude that, in each Landaulevel, the number of states9 is given by 9Thus the degeneracy of a Landau level

is 2eB/h = 1/(πℓ2B) per unit area of thesample.

2π

ℓ2B

1

(2π/L)2× 2 =

L2

πℓ2B= L2

„2eB

h

«

, (45.24)

which has the form:

Number of states in one Landau level =(Area taken up by all of them)

(Area taken up by one of them). (45.25)

So the effective real-space area occupied by one Landau state within a Landau levelis πℓ2B.

However, we also need to discuss the effect of spin splitting. Each Landau levelcontains electrons in both spin states and a magnetic field will cause these to havedifferent energies. The energy can be written

E = (N +1

2)~ωc ± 1

2g∗µBB, (45.26)

where g∗ is the effective g-factor. This effect will now double the number of levels(these are now spin-split half Landau levels) and each of these levels now has halfthe number of states in it.10 The effective area occupied by one state in a spin-split 10Thus the degeneracy of a spin-split

half Landau level is eB/h = 1/(2πℓ2B)per unit area of the sample.

half Landau level is then11 2πℓ2B.

11Note that 2πℓ2B also equals the areaoccupied by one Dirac flux quantum.

45.3 The integer quantum Hall effect

The Hall effect depends on the number density of carriers and so ourtreatment of the integer quantum Hall effect will begin by consideringhow many states are available for occupation by electrons in a two-dimensional metal12 when subject to a large magnetic field. 12Recall from Chapter 34 that we ex-

pect the conductance to vary as G(L) ∝L(d−2), where d is the dimensional-ity. For d = 2 we therefore ex-pect behaviour that is independent ofthe dimensions of the system, makingthe quantum Hall phenomena we willdescribe completely universal for two-dimensional systems.

Example 45.4

A reminder of the physics of the conventional Hall effect may be found in Exer-cise 45.4. In brief, for an isotropic material Ohm’s law is written

„JxJy

«

=

„σxx σxy−σxy σxx

«„ExEy

«

, (45.27)

and the Hall coefficient is defined as RH = σxy/B. In experiments we measure theresistivities ρxx and ρxy , which are given by

ρxx ≡ Ex

Jx=

σxx

σ2xx + σ2

xy

,

ρxy ≡ Ey

Jx=

σxy

σ2xx + σ2

xy

, (45.28)


and in the high magnetic field limit we find |σxy | ≫ |σxx| giving

ρxx ≈ σxx

σ2xy

, and ρxy ≈ 1

σxy= RHB. (45.29)

For a conventional system we expect ρxx to be a constant and ρxy to increase linearlywith field. As shown in Fig. 45.2, systems approximating a two-dimensional electrongas show something rather different with increasing field:

• Plateaux in ρxy occurring at values ρxy = 1νhe2

, with ν an integer [Fig 45.2(b)].

• Dramatic drops in ρxx which takes very small values when we have a plateauin ρxy [Fig 45.2(a)].

• Maxima in ρxx in the transition region between the plateaux.

This is known as the integer quantum Hall effect.1313See Singleton, Band Theory andElectronic Properties of Solids, formore detail on Hall effect physics.

Fig. 45.2 The quantum Hall effectin a GaAs-(Ga,Al)As heterojunction,shown in resistivity components (a) ρxxand (b) ρxy , measured at temperaturesbetween 0.03 and 1.5 K. (Figure fromJ. Singleton, Band Theory and Elec-tronic Properties of Solids, reprintedwith permission.) (c) The fractionalquantum Hall effect in a semiconduc-tor heterojunction, shown in resistivitycomponents ρxy (dotted line) and ρxx(solid line). Figure from P. Gee, D.Phil.Thesis, Clarendon Laboratory, Univer-sity of Oxford (1997).

0

25.8128

50

75

ρxy

(kΩ

)

0

2

4

ρxx

(kΩ/

)

0 5 10 15

B (T)

(c)

The first step in studying the quantum Hall effect is to confine an elec-tron gas into a two-dimensional region.14 A magnetic field is then applied14Experimentally, this is accomplished

by ensuring that electrons occupy thelowest sub-band in a quantum well in asemiconductor heterostructure.

perpendicular to this two-dimensional electron gas. As shown in Exam-ple 45.3 each spin-split Landau level has a number of states per unitarea of sample equal to 1/(2πℓ2B). If the number of electrons per unitarea is n2d, then we define the filling factor ν by

ν =(Areal density of electrons)

(Areal density of states in a (spin-split half) Landau level)

= n2d2πℓ2B =n2dh

Be. (45.30)

As a result, whenever the magnetic field takes a value

B =n2dh

e× 1

p, (45.31)

where p is an integer, then the filling factor ν will be an integer andhence ν (spin-split half) Landau levels15 will be completely filled.

15The filling factor ν counts the num-ber of spin-split half Landau levels.However, from now on we will simplyrefer to the spin-split levels as Landaulevels. Whether the electrons interactwith the magnetic field via their orbital(cyclotron) motion or via their spin an-gular momentum, the end result is thesame: orbits with discrete energy lev-els; hence we will call these energy lev-els Landau levels, and ν of them arefilled.

45.4 The fractional quantum Hall effect 417

This then has the interesting consequence that the system realizesa macroscopic quantum state in which there is an energy gap for thecreation of all charged excitations. The system is an incompressible

liquid, essentially because the energy of the system becomes indepen-dent of area when the Landau levels are full (i.e. integer ν). In this case,we can use the fact that the Hall coefficient RH = −1/(n2de) to findthat the Hall resistivity ρxy is16 16To prevent a minus sign here, we use

the convention that B < 0, so that(−e)B > 0. This convention is also fol-lowed in the discussion in Chapter 7 ofWen.

ρxy = RHB =B

n2de=

1

ν

h

e2. (45.32)

Thus we find plateaux in ρxy whenever ν Landau levels are full. Re-markably, the resistivity ρxy which is measured with a battery and somelengths of wire depends only on two fundamental constants of Nature.The quantity h/e2 = 25812.8 Ω has been measured with incredible pre-cision and is often called the resistance quantum.17 Equivalently, one 17It is also known as the von Klitz-

ing contant, after Klaus von Klitzing(1943– ) the discoverer of the quantumHall effect.

can write eqn 45.32 in terms of the Hall conductivity σxy as

σxy = νe2

h. (45.33)

Before feeling too pleased with ourselves we should remember that thisanalysis only tells us that ρxy is quantized at those values of magneticfield at which an integer number of Landau levels are completely filled,but does not explain the occurrence of the plateaux nor the vanishingof ρxx. In fact, these two features rely crucially on the fact that someof the states in a Landau level are localized via Anderson localization(described18 in Chapter 34 ). The idea is that we imagine starting at a 18See Singleton for a simple explana-

tion of how this works in this contextor, for more detail, the article by J.T.Chalker in Ecole des Houches: Topolog-ical Aspects of Low Dimensional Sys-tems, Eds. A. Comtet, T. Jolicoeur andS. Ouvry. (1998).

field where a single Landau level is exactly filled and then further increasethe magnetic field. This reduces the degeneracy of the Landau level,forcing some electrons into the next Landau level at a higher energy.These electrons will end up in the localized states of that upper leveland hence cannot contribute to the conductivity. As a result ρxx = 0,while ρxy remains at the quantized value corresponding to the completefilling of the lower level, resulting in a plateau.

45.4 The fractional quantum Hall effect

When ν = 1 only a single Landau level is occupied and this maximizesthe effect of electron–electron interactions. Although our treatment ofthe integer quantum Hall effect ignored these and treated the electrons asnon-interacting, this will no longer be sufficient. The intriguing physicsthat is revealed when ν < 1 is accessed by using magnetic fields withB > n2dh/e. In addition to the plateaux observed at integer values of ν,experiments in high-quality samples revealed further plateaux at frac-tional values of ν [see Fig. 45.2(c)]. These extra plateaux are due to theformation of a state of matter which can support fractional excitationsand which is descibed by a topological field theory of the sort we met inChapter 30.19

19Although we haven’t stressed topol-ogy in our treatment of the integerquantum Hall effect, it may also be un-derstood in these terms. See Altlandand Simons, Section 9.3 for the detailsof Pruisken’s field theory of the inte-ger quantum Hall effect and its link totopology.


We will assemble a quantum field theory that describes this fractional

quantum Hall (FQH) effect.20 This will be an effective theory in that20The original explanation of the FQHeffect was made by Robert Laughlinbased on a trial wave function ap-proach. We won’t pursue this methodhere. The interested reader should con-sult Wen, Chapter 7 for a more de-tailed description of the other possibleapproaches to describing the physics.

we won’t start with details of the electrons and holes found in a metal,rather with a set of fields that will model the ground state and the lowenergy or elementary excitations.

This is an important point. Nature doesn’t allow us the luxury ofbeing able to derive the properties of complex phases of matter from firstprinciples. In the same way that the rigidity of solids is not derivablefrom the physics of those atoms in a gas, so the FQH effect is a differentphase of matter from the two-dimensional electron gas from which itis created.21 The properties of the phase of matter that give rise to a21This is a form of emergence, where

the properties and excitations of aphase of matter arise upon the conden-sation of that phase.

fractional quantum Hall effect cannot, therefore, be derived from theelectron gas in zero magnetic field. So while most phases are separatedby a breaking of symmetry, as discussed in Chapter 26, this one is verydifferent. The FQH phase is topologically distinct from the electron gas.The phase transition between electron gas and FQH state is thereforetopological and cannot be described by a broken symmetry theory as wehad for the magnet, for example.

In the remainder of this chapter we will construct a quantum fieldtheory of the FQH fluid, which is a phase of matter showing a FQHeffect in its ground state. Specifically, we will explain the existence ofa subset of the possible FQH states: those with 1/ν equal to an oddinteger.22 We will also examine the elementary excitations of the FHQ22A slightly more general, but basically

similar, theory is needed to explain theothers. Again, see Wen, Chapter 7 forthe full story.

fluid, which are the quasiparticles of this phase.23

23We stress again that these quasipar-ticles can’t be expected to be elec-trons with a shielded charge and in-creased mass as we had for quasipar-ticles in QED and in metals. In fact,the FQH quasiparticles will turn outto have quite different properties fromelectrons and holes, including fractionalstatistics and even fractional electriccharge!

We begin by asking what we want from our theory. Obviously we wantthe theory to predict the fundamental facts of the FQH effect. Theseare:

• In units where ~ = 1, we want to predict a conductivity σxy =

Jx/Ey = e2

2πν with 1/ν taking odd integer values.

• The FQH fluid is incompressible with a charge density given by

ρ = −eJ0FQH = −e×

(Number of states per

Landau level

)× ν

L2

= −e2B

2πν, (45.34)

where we’ve used the fact that the number of states per unit areaL2 in a Landau level is 1/(2πℓ2B).

These two facts are embodied in an equation describing the electromag-netic FQH current2424Recall from Chapter 30 that in (2+1)

dimensions we may write the magneticfield as B = ∂1A2 − ∂2A2 and, follow-ing convention, we take B < 0 so that(−e)B > 0.

−eJµFQH = −νe2

2πεµνλ∂νAλ, (45.35)

where Aµ is the usual U(1) gauge field of electromagnetism. We will cookup a Lagrangian whose equation of motion, at the very least, predictsthis current.

The Lagrangian which will do this is of the Chern–Simons form and isthus a topological theory. It will describe another U(1) gauge field aµ,

45.4 The fractional quantum Hall effect 419

which couples to the electromagnetic field Aµ. We know from Chapter 30that in a Chern–Simons theory the field aµ gives rise to a current JµCS ∝εµνλ∂νaλ. We choose to adjust our normalization slightly25 here and 25We simply replace κ in Chapter 30

with 1/2π here.will define the Chern–Simons current as

JµCS =1

2πεµνλ∂νaλ. (45.36)

The key to constructing the theory is to say that the FQH current ineqn 45.35 will contribute to the total Chern–Simons current. Our taskis now to find a Lagrangian whose equation of motion tells us that theChern–Simons current JCS in eqn 45.36 is given by a contribution fromthe FQH current in eqn 45.35 added to a contribution from any sourcejµ of quasiparticles that we put in the system. This is achieved with theLagrangian

L = −1

2

s

2πεµνλaµ∂νaλ +

e

2πεµνλAµ∂νaλ + jµaµ, (45.37)

where s is a number. This Lagrangian is formed from three terms: I:a Chern–Simons term; II: a term where the electromagnetic gauge fieldAµ couples to the Chern–Simons current; III: a term where a source ofquasiparticles jµ couples to the topological gauge field aµ.

Example 45.5

We can check that this works by feeding the Lagrangian through the Euler–Lagrangeequation to extract the equation of motion. We find that

∂ν∂L

∂(∂νaµ)= 1

2s2πεµνλ∂νaλ − e

2πεµνλ∂νAλ,

∂L∂aµ

= − 12s2πεµνλ∂νaλ + jµ.

(45.38)We obtain s

2πεµνλ∂νaλ =

e

2πεµνλ∂νAλ + jµ, (45.39)

or

JµCS =e

2πsεµνλ∂νAλ +

1

sjµ. (45.40)

If we identify the dimensionless parameter s from the Lagrangian as being 1/ν, andtemporarily set the quasiparticle source jµ = 0 then this is the fractional quantumHall result we set out to predict.

Our low-energy Lagrangian would be of little use if it didn’t providesome further insight into the workings of the fractional quantum Hallfluid. We will obtain these by asking about the quasiparticle excitationsin the FQH ground state which are created by the conserved current jµ.Recall from Chapter 30 that coupling a current like jµ to the topologicalgauge field aµ constrains the properties of these particles, attaching theflux of the field aµ to each particle created by jµ and imbuing them wthfractional statistics. Specifically we will use the theory to show that:

• the FQH plateaux appear for half odd integer s;

• the FQH quasiparticles carry fractional electromagnetic charge;

• the FQH quasiparticles have fractional statistics.


Our source of FQH quasiparticles jµ will creates excitations with aµ-charge26 l, where l is constrained to be an integer. We will construct the

26Potential for confusion arises here asthere are two gauge fields in this prob-lem: (i) the U(1) gauge field of elec-tromagnetism Aµ, which gives rise tothe electric field E and magnetic fluxdensity B, whose excitations couple tothe charge −e; and (ii) the topologicalU(1) gauge field aµ, giving rise to anelectric field e and magnetic flux den-sity b, whose excitations couple to thecharge l. The FQH quasiparticles carryintegral aµ-charge l and also Aµ-chargeq, whose value we will determine. Asthe current jµ of FQH quasiparticles iscoupled to the topological field aµ weexpect, from Chapter 30, that they willalso carry flux of the b-field (which werefer to as aµ-flux).

source term jµaµ = la0δ(2)(x − x0), where we have chosen to localize

the quasiparticle source at position x0. The zeroth component of theequations of motion predicts a total electric charge density of

−eJ0CS =

e2

2πsB − el

sδ(2)(x − x0), (45.41)

where the first term on the right-hand side is the electric charge densityexpected from the FQH ground state and the second term is the extraelectric charge of the quasiparticle. This equation implies two things:firstly that the FQH quasiparticle have electric charge −el/s; secondly,returning to the properties of Chern–Simons theories from Chapter 30,we see that this excitation carries aµ-flux of 2πl/s. This is progress, butwe have yet to establish the values of s.

Next we find the statistics of the FQH quasiparticles. We picture twosuch excitations with charges l1 and l2. Move one in a circle around theother and we obtain an Aharonov–Bohm phase:

∆Φ = (aµ flux) × (quasiparticle charge)

= 2πl1s× l2. (45.42)

Now we set l1 = l2 = l to obtain identical quasiparticles. Using themethod of Chapter 30,27 we note that exchanging these particles means27In Chapter 30 we saw that the

Chern–Simons charge density is relatedto the flux density via ρ(= J0) = −κbwhere, for our quasiparticles ρ = l/sper unit area. Note also that κ = 1

2πand we take b < 0.

rotating one through an angle π around the other so that we expect aphase change πη, where η will determine the statistics of the quasipar-ticles. The corresponding phase change ∆Φ is half that in eqn 45.42,implying that we have statistics given by

η =l2

s. (45.43)

Our conclusions so far are that the excitations of the FQH ground stateare quasiparticles with aµ-charge of l, which carry an electric charge−el/s, carry aµ-flux of 2π/s, and cause a phase change on exchangegoverned by η = l2/s. Since we know that ν = 1/s, which experimentteaches us takes values such as 1/3, this suggests that the quasiparticlescarry a fraction of an electron charge and possess fractional statistics!However we need to complete the puzzle by establishing the necessityfor s to be an odd integer.

In order to complete the description it turns out that we must appealto the presence of electrons and holes in the metal and link these toour FQH quasiparticles. Following convention, we will call our FQHquasiparticle excitations quasielectrons and quasiholes since, despite thefact that we cannot adiabatically tune between the FQH quasielectronsand the quasielectrons of a metal, we hope that their properties canat least be related. To establish the relation, we note that a singleelectron has electric charge −el/s = −e, it therefore has l/s = 1 andcarries l = s units of aµ-charge. Its exchange property follows as η =

Exercises 421

s. However, we know that electrons are fermions and must thereforehave an exchange angle η of an odd integer. We therefore deduce thats = odd integer, which is exactly the property we wanted to explain.We may now conclude that the FQH quasiparticle has fractional electriccharge −el/s and fractional statistics governed by η = l2/s.

Example 45.6

Let’s examine the case of s = 3. Electrons with l = s = 3 units of aµ-charge formthe FQH fluid with filling factor ν = 1/3. The excitations of this liquid are FQHquasielectrons with aµ-charge l = −1 and quasiholes with l = 1. Quasielectronshave fractional electric charge −e/3, they carry aµ-flux of 2π/3 and have statisticsgoverned by η = 1/3.

Chapter summary

• In a magnetic field the Hamiltonian is invariant under a combina-tion of a translation and a gauge transformation. This leads to thephysics of the Hofstadter butterfly

• The integer quantum Hall effect follows from considering the oc-cupancy of Landau levels as a function of applied magnetic field ina two-dimensional electron gas.

• The fractional quantum Hall effect is caused by interactions andmay be explained using an effective, topological field theory.

Exercises

(45.1) Verify eqn 45.8 using the method suggested in thetext.

(45.2) Verify eqns 45.14 and 45.15.

(45.3) (a) Use the creation and annihilation operators de-fined in eqns 45.18 to diagonalize the Hamiltonianin eqn 45.11.(b) Evaluate X2 + Y 2 to show that each electronoccupies an area of 2πℓ2B.

(45.4) A reminder of the Hall effect.Consider driving a current of density Jx in the x-direction along a finite sized conductor, in the pres-ence of a magnetic field B directed along z. Con-

sider the equations of motion of a charge in therelaxation time approximation

mv = −mv

τ+ qE + qv × B, (45.44)

where τ is the relaxation time.(a) In the steady state, show that Ey/Ex = −ωcτ .(b) Verify that the Hall coefficient is given by

RH =EyJxB

= − 1

ne, (45.45)

where n is the electron density.

Part XI

Some applications from the

world of particle physics

Quantum field theory is foundational for particle physics, since particlesare simply excitations in quantum fields. In this part, we review severalimportant particle physics theories which use the ideas developed in thisbook.

• Our treatment of gauge theory in Chapter 14 assumed that theunderlying symmetry group is abelian (i.e. its elements commute).This is the case for U(1), the group that describes electromag-netism. In Chapter 46 we introduce non-abelian gauge theory andshow that the non-commutativity of group elements leads to a non-linear field tensor.

• A non-abelian gauge theory is at the heart of the electroweak the-ory due to Weinberg and Salam and this is presented in Chapter 47.Symmetry breaking of the Higgs field results in the emergence ofthe photon, the charged W+ and W− particles, and the neutralZ0 particle. The photon is massless, but the other three acquiremass and so the weak force becomes short-ranged.

• Majorana fermions, particles that are their own antiparticles, aredescribed in Chapter 48.

• Chapter 49 introduces two varieties of magnetic monopole: onedue to Dirac which provides a link between the quantization ofelectric charge and the quantization of the magnetic charge of amonopole; the other due to ’t Hooft and Polyakov which arisesfrom a non-abelian gauge theory.

• The final chapter, Chapter 50, introduces the concept of instan-tons, which allows us to treat problems involving tunnelling. Weask what would happen if the Universe were to tunnel from a falsevacuum into the true vacuum.

46 Non-abelian gauge theory

46.1 Abelian gauge theory revis-ited 424

46.2 Yang–––Mills theory 425

46.3 Interactions and dynamics ofWµ 428

46.4 Breaking symmetry with anon-abelian gauge theory

430

Chapter summary 432

Exercises 432

Physics is very muddled again at the moment; it is much toohard for me anyway, and I wish I were a movie comedianor something like that and had never heard anything aboutphysics.Wolfgang Pauli (1900–1958)

QED is a very successful field theory built on the gauge principle whichaccurately describes electromagnetism. Applying this principle involvedtaking a theory with a global (internal) U(1) symmetry and promotingthis to a local symmetry, necessitating the introduction of a gauge fieldAµ(x). Turning our attention now to particle physics, we would also liketo explain the weak and strong forces using fields. The weak and stronginteractions can also be described by gauge theories, but unfortunatelyneither are based on the simple case of local U(1) symmetry. In fact,both are based on local symmetry under more complicated groups oftransformations: for the weak interaction it is the group U(1) ⊗ SU(2)and for the strong interaction it is SU(3). Unlike U(1), these groups arenon-abelian.

A non-abelian group is formed from elements that don’t commute: aNiels Hendrik Abel (1802–1829), af-ter whom abelian groups are named,died aged 26 from tuberculosis. Non-abelian group theory was introduced byEvariste Galois (1811–1832) who waskilled in a duel aged 20, having spentthe entire night before writing down arevolutionary theory involving the useof groups to investigate the solubility ofalgebraic equations.

transformation by element a followed by one by element b has a differenteffect to the transformation by b followed by a. Non-abelian groupsare quite familiar from everyday life, the rotation group SO(3) beinga simple example. Figure 46.1 shows that a rotation R1 of a book byπ/2 about x followed by rotation R2 by π/2 about z leads to a differentfinal orientation of the book from the rotations R2 then R1. In this

x

y

z R2

x y

z R1

x y

z

x y

zR1

x

y

z R2

xy

z(a)

(b)

Fig. 46.1 A demonstration that the el-ements of the non-abelain group SO(3)don’t commute.

chapter we examine the features of a gauge theory with a local non-abelian symmetry. We will concentrate on the simplest possible non-abelian cases SU(2) and SO(3). We start by reviewing the importantproperties of an abelian gauge theory.

46.1 Abelian gauge theory revisited

In Chapter 14 we looked at the simplest possible gauge theory. Fun-damentally it is a theory with a local symmetry. It is symmetric withrespect to U(1) transformations which are different at every point inspace time. U(1) is abelian since two transformations, applied one afteranother, always lead to the same result, no matter the order in whichthey’re applied. As a reminder, here is the complex scalar field La-


grangian:

L = (Dµψ)†(Dµψ) −m2ψ†ψ − 1

4FµνF

µν , (46.1)

where Dµψ = (∂µψ+ iqAµψ). This is a locally gauge invariant quantityunder the simultaneous transformations

ψ → ψeiα(x), Aµ → Aµ − 1q∂µα(x). (46.2)

In order to make the jump to more complicated symmetries we will dealwith these in the form of infinitesimal transformations.

Example 46.1

We’ll replay the proofs that the Lagrangian is invariant with respect to local gaugetransformations using infinitesimal transformations.

ψ → (1 + iα)ψ,

∂µψ → ∂µψ + i(∂µα)ψ + iα(∂µψ),

Aµ → Aµ − 1

q∂µα(x). (46.3)

This works up to order O(α) because we have

ψ†ψ → (ψ† − iαψ†)(ψ + iαψ) = ψ†ψ +O(α2), (46.4)

and

Dµψ →»

∂µψ + i(∂µα)ψ + iα(∂µψ) + iq

„

Aµ − 1

q∂µα

«

(ψ + iαψ)

–

= [∂µψ + iqAµψ] + iα [∂µψ + iqAµψ] +O(α2)

= (1 + iα)Dµψ +O(α2). (46.5)

Equation 46.5 shows that Dµψ transforms exactly the same way as ψ. This, in turn,means that (Dµψ)†(Dµψ) → (Dµψ)†(Dµψ) + O(α2). So all of the terms involvingψ are invariant with respect to the transformations.

46.2 Yang–––Mills theory

In 1954 Chen-Ning Yang and Robert Mills asked if the theory of local Chen-Ning Yang (1922– )Robert Mills (1927–1999).A gauge theory based on the SU(N)group is called a Yang–––Mills the-ory in honour of Yang and Mills whoextended gauge theory to non-abeliangroups in 1954.

symmetry could be extended to more complicated groups of transfor-mations than U(1), such as SU(2) and SO(3). These groups are non-abelian. Although an SU(2) gauge theory doesn’t describe the weak orstrong interactions, it does display the properties of a more complicatednon-abelian theory and so we will consider it in this chapter.

We ask what needs to be done to make a theory with a global internalSU(2) symmetry also symmetric under local internal SU(2) transfor-mations. We’ll use exactly the same strategy we used to upgrade theglobal internal U(1) symmetry to a local one. We will first work out theconsequences of imposing a local symmetry and then we will add a fieldwhich cancels out any extra terms that arise.


Here’s the Dirac Lagrangian for two sorts of fermions, each with massm:

L = f(iγµ∂µ −m)f + g(iγµ∂µ −m)g, (46.6)

which we write more compactly as

L = Ψ(iγµ∂µ −m)Ψ, (46.7)

where

Ψ =

(fg

), Ψ =

(f g

). (46.8)

This Lagrangian is invariant under a global SU(2) transformation

Ψ → ei2 τ ·αΨ, Ψ → Ψe−

i2 τ ·α, (46.9)

where τ = (τx, τy, τz) are the Pauli matrices.1 The Noether currents

1We use τ for the Pauli matriceshere, rather than σ, to emphasize thatthey act on our states of two sorts offermions, Ψ =

`f g

´, rather than

on the spin of a single fermion, i.e. theyare Pauli ‘isospin’ matrices, rather thanPauli ‘spin’ matrices.

Noether’s theorem: SU(2) inter-nal symmetry

DaΨ = i2τaΨ (a: internal)

ΠµΨ = Ψiγµ (µ: vector)DL = 0 Wµ = 0

JµaNc = 12ΨγµτaΨ

QaNc =R

d3x 12Ψ†τaΨ

arising from this symmetry are given in the box. The conserved charge-like quantity for this theory is the isospin I =

∫d3x Ψ† τ

2 Ψ.

As examined in the exercises, Iz =12

Rd3x(f†f − g†g). We might say that

the Ψ-field carries I = 1/2 units ofisospin, with f -on particles correspond-ing to I3 = 1/2 and g-on particles toI3 = −1/2. Isospin will be importantin the next chapter.

To investigate further we’ll be working with infinitesimal transforma-tions, so in infinitesimal form we can write

Ψ →(

1 +i

2τ · α

)Ψ. (46.10)

Example 46.2

We can show that this is a global symmetry of our Lagrangian. As before we write

ΨΨ → Ψ

„

1 − i

2τ · α

«„

1 +i

2τ · α

«

Ψ

= ΨΨˆ1 +O(α2)

˜. (46.11)

Since we’re working with infinitesimals, we agree to jettison any term above firstorder. Since, for a global symmetry the derivative ∂µΨ transforms exactly as Ψ does,(since α has no dependence on spacetime) the entire Lagrangian is invariant.

Now for the pivotal point of this chapter. We upgrade this argument toa local transformation by letting α be a function of spacetime x. Thisis no problem for the mass term in the Lagrangian ΨmΨ, which is stillinvariant. The trouble comes from the derivative, since

∂µΨ → ∂µΨ +i

2[τ · α(x)] ∂µΨ +

i

2[τ · ∂µα(x)] Ψ, (46.12)

and the final term in eqn 46.12 prevents ∂µΨ transforming like Ψ andtherefore prevents the derivative term from being invariant.22Remember that the U(1) version of

this is ∂µψ → ∂µψ+iα(∂µψ)+i(∂µα)ψ. Just as we introduced the gauge field Aµ(x) for the U(1) case, weneed to introduce a new gauge field which will mop up the term in-volving ∂µα(x). This gauge field is called Wµ(x) and has three internalcomponents [that is (W 1

µ(x),W 2µ(x),W 3

µ(x))], each of which have fourcomponents in Minkowski spacetime (labelled with the index µ). Just


as before, the new field is combined with the derivative operator to makea covariant derivative. We therefore define a covariant derivative3 3Recall that for U(1) we have Dµ =

∂µ + iqAµ. The equation here is for-mulated analogously with, by conven-tion, a sign change reflecting the fact

that our chosen SU(2) rotation ei2

τ ·α

rotates in the opposite (internal) direc-tion to the U(1) rotation eiα.

Dµ = ∂µ − i

2gτ · Wµ(x), (46.13)

where g is the charge of the theory (it’s the quantity which will even-tually tell us how strongly the gauge field Wµ interacts with Ψ). Ifthis procedure is going to work, then DµΨ must transform in exactlythe same way as Ψ does. That is, the manner in which the gauge fieldtransforms must cancel out the extra term in the transformation of theordinary derivative.4 This will indeed be the case, but the fact that the 4Note that, written out ex-

plicitly, the covariant deriva-

tive reads DµΨ =

„∂µf∂µg

«

−

i2g

„W 3µ W 1

µ − iW 2µ

W 1µ + iW 2

µ −W 3µ

«„fg

«

.

transformation is non-abelian leads to a new complication.

Example 46.3

We want the derivative to transform in the same way as Ψ, which is to say that itshould transform as

DµΨ →„

1 +i

2τ · α(x)

«

DµΨ. (46.14)

This implies

DµΨ →„

1 +i

2τ · α(x)

«„

∂µΨ − i

2gτ · Wµ(x)Ψ

«

= ∂µΨ − i

2gτ · Wµ(x)Ψ +

i

2τ · α(x)∂µΨ

−„

i

2

«2

g [τ · α(x)] [τ · Wµ(x)Ψ] . (46.15)

So how do we get it to work? We suppose that the covariant derivative transformsas

Dµ → ∂µ − i

2gτ · Wµ(x) − i

2gτ · δWµ(x), (46.16)

where δWµ(x) is the change in the gauge field Wµ(x) with the transformation. Wealso know that the field transforms as Ψ →

ˆ1 + i

2τ · α(x)

˜Ψ. Combining these

transformations shows that

DµΨ →„

∂µ − i

2gτ · Wµ(x) − i

2gτ · δWµ(x)

«„

1 +i

2τ · α(x)

«

Ψ

= ∂µΨ +i

2τ · [∂µα(x)]Ψ +

i

2τ · α(x)∂µΨ

− i

2gτ · Wµ(x)Ψ −

„i

2

«2

g [τ · Wµ(x)] [τ · α(x)]Ψ

− i

2gτ · δWµ(x)Ψ, (46.17)

ignoring the αδWµ term as it’s second order.By comparing terms in eqns 46.15 and 46.17 we find that for our gauge field to

work we require

τ · δWµ(x) =1

gτ · [∂µα(x)] +

i

2[τ · α(x)] [τ · Wµ(x)] − [τ · Wµ(x)] [τ · α(x)] .

(46.18)Notice that this last term survives only because the transformation is non-abelianand the two transformations involve τ entering in different orders. Finally, we canuse the identity

(τ · a)(τ · b) = (a · b) + iτ · a× b, (46.19)

and conclude that

τ · δWµ(x) =1

gτ · [∂µα(x)] − τ · [α(x) × Wµ(x)] . (46.20)


The result of the preceding example is that the gauge field musttransform5 according to5For comparison, in the U(1) case this

transformation is Aµ → Aµ − 1q∂µα.

τ · Wµ → τ · Wµ +1

gτ · (∂µα) − τ · (α × Wµ). (46.21)

46.3 Interactions and dynamics of Wµ

Next we ask how the gauge field couples to the fermion particle field Ψ.Minimal coupling, which we first met in Chapter 14, tells us that thisis simply a matter of multiplying out the term in the Dirac Lagrangianinvolving the covariant derivative Dµ. We have

ΨiγµDµΨ = Ψiγµ∂µΨ +g

2Ψγµτ · WµΨ. (46.22)

Compare this to the U(1) version

ψiγµDµψ = ψiγµ∂µψ − qψγµAµψ, (46.23)

which told us that the excitations in the U(1) gauge field Aµ(x) mediatedinteractions between electrons by coupling to the charge q. Furthermore,the Feynman rule for the interaction was that a ΨΨAµ interaction vertexcontributed a factor −iqγµ. By analogy, the SU(2) version suggests thatthe Feynman rule for a ΨΨWµ vertex is the contribution of a factorproportional to igτγµ. After quantization, there will be excitations inthe Wµ gauge field which mediate the interactions between the fermionsof the theory. Explicitly, we expect there to be masslessW 1, W 2 andW 3

photon-like particles, each with two possible transverse polarizations.The particles couple to the isospin of the Ψ field with a strength set bythe charge g.

When we looked at the U(1) theory, an additional, but very impor-tant contribution to the Lagrangian was provided by the gauge fieldAµ(x) itself, whose dynamics gave us electromagnetism through the term− 1

4FµνFµν . We now ask what contribution the gauge field Wµ(x) makes

to the Lagrangian. An obvious guess would be that we should add a termof the form − 1

4Gµν ·Gµν where Gµν is a tensor analogous to Fµν in theelectromagnetic case. The obvious guess for the form of this tensor isGµν = ∂µWν − ∂νWµ, but since we will want Gµν to transform like Φ

and ∂µΦ, it turns out that the tensor we need is66This is a surprising result: the fieldstrength Gµν is a nonlinear functionof the gauge field Wµ. We thus havefound ourselves a nonlinear interactingtheory, even in the absence of matter!

Gµν = ∂µWν − ∂νWµ + g(Wµ × Wν). (46.24)

Our final expression for the Lagrangian of a locally invariant SU(2)theory is then given by

L = Ψ(iγµDµ −m)Ψ − 1

4Gµν · Gµν . (46.25)

The incredible thing about this Lagrangian is that the nonlinearity ofGµν allows a new sort of process to take place, namely, the direct in-teraction of particle excitations in one of the Wµ fields with other Wµ

46.3 Interactions and dynamics of Wµ 429

particles, as shown in the interaction vertex in Fig. 46.2, where we havethree Wµ particles interacting. (The Lagrangian also allows the inter-action of four Wµ particles.7) Note that photon–photon interactions are 7The terms leading to the interactions

of three and four Wµ particles can beseen by expanding Gµν · Gµν .

not possible in abelian electromagnetism as the photon carried no elec-tromagnetic charge of its own. In contrast, the isospin field Wµ carriesan isospin charge of one unit (I = 1) so remarkably it may act as asource of itself.8 8Since the algebra of isospin mirrors

that of ordinary spin, we say that theexcitations of the I = 1 field Wµ areparticles with Iz = 0 created (anddestroyed) by the field W 3

µ ; particleswith Iz = 1 created by the combi-nation 1√

2(W 1

µ + iW 2µ) [and destroyed

by 1√2(W 1

µ − iW 2µ)]; and particles with

Iz = −1 created by 1√2(W 1

µ−iW 2µ) [and

destroyed by 1√2(W 1

µ + iW 2µ)].

Fig. 46.2 The interaction of three Wparticles is possible in a gauge theorywith local SU(2) symmetry.

This self-interaction of the gauge field provides an explanation for afeature that makes Yang–Mills theories so interesting: the asymptoticfreedom described in Chapter 34. As we probe further away from anelectromagnetic charge (described by abelian U(1) electromagnetism)the charge is shielded by dipoles formed from virtual electron–positronpairs and the electric charge appears to get smaller. However, the non-abelian gauge field produces a marked difference as we probe furtheraway from a quark (where the strong interaction of colour charge isdescribed by a non-abelian SU(3) gauge theory). The screening effectwould be the same for the quark as for the electron, except that we havethe gauge field self-interaction which allows the gauge field quantum (agluon) to decay into a pair of gluons. As the gluons themselves carrycolour charge they lead to a proliferation of like colour charges aroundthe quark leading to an increased colour charge at large distances. Thisis asymptotic freedom: the interaction gets stronger as we probe furtherfrom the quark.

Example 46.4

A free, globally SO(3) invariant scalar field theory was introduced in Chapter 13. ItsLagrangian is

L =1

2∂µΦ · ∂µΦ − m2

2Φ · Φ, (46.26)

where Φ = (φ1, φ2, φ3). This theory is invariant with respect to the global (internal)rotation given by Φ(x) → Φ(x) − θ × Φ(x). Again, we upgrade this to a localtransformation by letting θ, the internal angle of rotation, be a function of spacetimex. That is, the transformation is now Φ(x) → Φ(x)−θ(x)×Φ(x). To make a locallyinvariant theory we need a gauge field, leading to the covariant derivative:

DµΦ = ∂µΦ − gΦ × Wµ. (46.27)

The correct Wµ to choose is one that transforms as

Wµ → Wµ +1

g∂µθ − (θ × Wµ), (46.28)

where g is the charge of the theory, and as for SU(2) we add a term − 14Gµν ·Gµν to

the Lagrangian in eqn 46.26, where for this theory Gµν = ∂µWν − ∂νWµ + g(Wµ ×Wν). This theory describes three massive scalar fields (the three components of Φ)and three massless gauge fields (the three internal components of Wµ).

This completes our discussion of the properties of a non-abelian gaugetheory.9 As a summary, the gauge fields that we have met so far are

9The next logical step would be toquantize the Yang–Mills Lagrangian.This can be done using the path inte-gral but requires some tricks that wewon’t pursue here. See Peskin andSchroeder, Chapter 16 for the details.listed in the margin on page 430.


46.4 Breaking symmetry with a

non-abelian gauge theory

We saw in Chapter 26 that spontaneous symmetry breaking in a gaugetheory led to the famous Higgs mechanism. Here we treat exactly thesame problem with the more complicated example of a non-abelianfield. This is useful, not only because it provides yet another exam-ple of the wonderful physics of symmetry breaking, but also because theelectroweak theory examined in the next chapter relies on exactly thisphysics.

(1) Abelian U(1) gauge theory forthe complex scalar field ψ:

L=(Dµψ)†(Dµψ)−m2ψ†ψ−1

4FµνF

µν

where

Dµψ = ∂µψ + iqAµ(x)ψ

Fµν = ∂µAν − ∂νAµ.

The symmetry is

ψ → ψeiα(x)

and the gauge field transforms as

Aµ → Aµ − 1

q∂µα.

(2) Non-abelian SU(2) gauge the-ory for fermions described by afield Ψ:

L = Ψ(iγµDµ −m)Ψ − 1

4Gµν · Gµν

where

DµΨ = ∂µΨ − i

2gτ · Wµ(x)Ψ

Gµν = ∂µWν − ∂νWµ

+g(Wµ × Wν).

The symmetry is

Ψ → Ψei2

τ ·α(x)


Wµ → Wµ +1

g∂µα− (α× Wµ).

(3) Non-abelian SO(3) gauge the-ory for real scalar fields Φ:

L=1

2DµΦ·DµΦ−m

2

2Φ·Φ−1

4Gµν ·Gµν

where

DµΦ = ∂µΦ − gΦ × Wµ

Gµν = ∂µWν − ∂νWµ

+g(Wµ × Wν).

The symmetry is

Φ → Φ − θ(x) × Φ


Wµ → Wµ +1

g∂µθ − (θ × Wµ).

We will examine the case of SO(3) gauge theory with an φ4-type in-teraction and, crucially, a positive, symmetry-breaking mass term, witha Lagrangian given by

L =1

2DµΦ ·DµΦ +

m2

2Φ · Φ − λ(Φ · Φ)2 − 1

4Gµν · Gµν . (46.29)

The potential possesses a spherical shell of minima at a radius (in in-

ternal Φ space) of |Φ0| =(m2

4λ

) 12

. Symmetry breaking involves picking

one of the infinite number of equivalent vacua, which breaks global (andlocal) symmetry. For simplicity, we chose the one that points along the

3-direction in isospin space, that is Φ0 =(m2

4λ

) 12

e3. For the excitations

above this ground state, we write

Φ =

Φ1(x)Φ2(x)

|Φ0| + χ(x)

, (46.30)

where χ(x) describes deviations in the 3-direction. Then, after a boutof tedious algebra, we find

L =1

2

[(∂µΦ1)

2 + (∂µΦ2)2 + (∂µχ)2

]− 4|Φ0|2λχ2

+g|Φ0|[(∂µΦ1)W

2µ − (∂µΦ2)W

1µ

]

+g2|Φ0|2

2

[(W 1

µ)2 + (W 2µ)2]− 1

4Gµν · Gµν + . . . , (46.31)

where we ignore higher order terms. The first line describes the dynamicsof the χ(x) scalar field, the final line the dynamics of the Wµ(x) vectorfield, but the terms that are mixed in components of Φ and Wµ aren’teasy to interpret. However, there is a useful trick we can use: we selecta gauge which ensures that the excited state field Φ points along the3-direction in isospace at every point in spacetime. (This is known asunitary gauge.) This removes all mention of the components Φ1 andΦ2 and means that the excited state field becomes simply

Φ(x) = [|Φ0| + χ(x)] e3. (46.32)

46.4 Breaking symmetry with a non-abelian gauge theory 431

Example 46.5

Since the term DµΦ = ∂µΦ − gΦ × Wµ mixes up the components of Φ, our choiceof unitary gauge yields

DµΦ1 = gW 2µ(|Φ0| + χ),

DµΦ2 = −gW 1µ(|Φ0| + χ),

DµΦ3 = ∂µχ, (46.33)

giving,

(DµΦ)2 = (∂µχ)2 + g2|Φ0|2ˆ(W 1

µ)2 + (W 2µ)2˜+ . . . (46.34)

This solves the problem of the mixing of components of Φ and Wµ fields.

The broken symmetry Lagrangian is now given by

L =1

2(∂µχ)2 − 4|Φ0|λχ2

+g2|Φ0|2

2

[(W 1

µ)2 + (W 2µ)2]− 1

4Gµν · Gµν + . . . , (46.35)

which is the sum of Lagrangians for a massive, single-component scalarfield χ and a massive vector field with two components W 1

µ and W 2µ .

Symmetry breaking has caused two massive scalar fields (Φ1 and Φ2) todisappear from our original Lagrangian and we are left with one, calledχ. The vector fields W 1

µ and W 2µ have grown massive, taking on a mass

g|Φ0| (see Fig. 46.3). The field W 3µ remains massless.

Fig. 46.3 The gauge fields W 1µ and

W 2µ eat the Φ Goldstone modes and

acquire mass when global symmetry isbroken in a non-abelian gauge theorywith local SU(2) or SO(3) symmetry.

In summary, symmetry breaking of the SO(3) gauge theory causes:

(3 massive scalar fields Φ

3 massless photon-like fields Wµ

)→

1 massive scalar field χ2 massive vector fields W 1

µ ,W2µ

1 massless photon-like field W 3µ

.

We may check that we haven’t lost any degrees of freedom. Massivescalar fields only have a single degree of freedom, while massless photon-like fields have two.10 This makes nine on the left. Noting that massive

10Remember the case of electromag-netism!

vector fields have three degrees of freedom we see that we also havenine on the right and all is well. Notice, finally, that we could describeelectromagnetism by taking W 3

µ(x) = Aµ(x). This immediately leads tothe question of whether electromagnetism in our Universe results frombreaking symmetry in an SO(3) or SU(2) non-abelian theory.

Is all this non-abelian gauge theory simply a mathematical exerciseto amuse the theoretically inclined? On the contrary, it seems that theUniverse has some non-abelian symmetries etched into it. In the nextchapter we will apply what we have learnt about symmetry breakdownin a non-abelian gauge theory to formulating the electroweak theoryof Steven Weinberg and Abdus Salam, and we will discover that thesymmetry group we break is a little more complicated than SO(3).


Chapter summary

• A non-abelian group is formed from elements which do not com-mute. The introduction of a non-abelian local symmetry leads to anon-abelian gauge field. The field tensor has an extra term whichmakes it nonlinear.

• The consequences of symmetry breaking were demonstrated forSO(3) theory.

Exercises

(46.1) Show that Iz = 12

R

d3x(f†f − g†g).

(46.2) (a) Write an SO(3) rotation about the z-axis in in-finitesimal form.(b) Show that the Lagrangian in eqn 46.26 is invari-ant with respect to global SO(3) transformations.(c) We want to arrange matters so that the theoryis invariant with respect to local SO(3) transforma-tion. As in the abelian case, the trouble comes fromthe derivative term. Show that

∂µΦ(x)→∂µΦ(x)− ∂µθ(x)×Φ(x)− θ(x)×∂µΦ(x).(46.36)

(d) Verify that DµΦ transforms in exactly the sameway as Φ.

(46.3) (a) Show that, for U(1) gauge theory, the commu-tator of the covariant derivatives gives

[Dµ, Dν ] = iq(∂µAν − ∂νAµ) = iqFµν . (46.37)

This suggests that we can extract the analogue ofFµν for non-abelian gauge theory by evaluating thecommutator. This turns out to be the case!(b) Show that the analogous commutator for SU(2)gauge theory is

[Dµ, Dν ] =i

2gτ · (∂µW

ν − ∂νW µ + gW µ × Wν)

=i

2gτ · Gµν .

(46.4) Verify eqn 46.31 and the simplified form eqn 46.35.

47The Weinberg–––Salam

model

47.1 The symmetries of Naturebefore symmetry breaking

434

47.2 Introducing the Higgs field437

47.3 Symmetry breaking theHiggs field 438

47.4 The origin of electron mass439

47.5 The photon and the gaugebosons 440

Chapter summary 443

Exercises 443

As theorists sometimes do, I fell in love with this idea. But asoften happens with love affairs, at first I was rather confusedabout its implications.Steven Weinberg (1933– ) on symmetry breaking

I remember travelling back to London on an AmericanAir Force (MATS) transport flight. Although I had beengranted, for the night, the status of a Brigadier or a FieldMarshal – I forget which – the plane was very uncomfortable;full of crying servicemen’s children – that is, the childrenwere crying, not the servicemen. I could not sleep. I keptreflecting on why Nature should violate left-right symmetryin the weak interactions.Abdus Salam (1926–1996)

The Universe as we know it is the result of a symmetry breaking phasetransition. In this chapter we will explain the features of a Universe inwhich:

• Electrons have mass, but neutrinos do not.1 1Of course neutrinos are not reallymassless, as shown by neutrino flavouroscillation experiments. Their mass is,however, undoubtedly small comparedto the electron mass and in this chapterthe Universe we describe is an approx-imation to ours.

• Electrons may occur with left- or right-handed chirality, neutrinosare only observed with left-handed chirality.

• The photon is massless.

The explanation of these fundamental properties emerges from a modelformulated by Abdus Salam and Steven Weinberg which unites the elec- Abdus Salam (1926–1996) is notable,

not only for his many achievements inquantum field theory, but for a tirelessadvocacy for the development of sciencein the third world.

Steven Weinberg’s (1933– ) influence onquantum field theory is difficult to over-state. See his three volume masterworkon the subject for a profound and dif-ferent take on the material.

tromagnetic and weak interactions. At the heart of the Weinberg–Salamelectroweak theory lies a proposition: we live in a broken symmetryUniverse and the particles that we observe in Nature are a result ofthe symmetry breaking.2 In order to make progress with this idea we

2Although we won’t consider muonsand tauons, these leptons may also beincluded in addition to the electron.

will write down a Lagrangian for the electroweak Universe before thissymmetry breaking phase transition took place. We will propose a setof local symmetries for the Lagrangian. As usual, this will require thatwe introduce gauge fields to force the systems to be locally symmetric.We will then examine the consequences of the spontaneous breaking ofsymmetry on the particle spectrum of the system. Let’s make a start!

434 The Weinberg–––Salam model

47.1 The symmetries of Nature before

symmetry breaking

Imagine the Universe a moment after the Big Bang. It is more symmet-rical than the Universe in which we live. Specifically, the lepton fieldsenjoy an internal SU(2)⊗U(1) symmetry.3 In around 10−12 seconds the3Strictly speaking this should be called

a U(2) symmetry. See Penrose for de-tails.

Universe will cool below a temperature 1016 K and undergo a symmetrybreaking phase transition. Before this occurs we will assess the state ofthe lepton fields of the Universe as described by the Weinberg–Salammodel.

Before symmetry breaking neutrinos and electrons are both masslessparticles. Electrons are possible in both left- and right-handed forms,but neutrinos only exist in left-handed form.4 Neutrinos and electrons4That this model doesn’t provide an

explanation for this feature of Naturehas led to much work over the last fewdecades regarding more general ‘super-symmetries’ that may be required todescribe the Universe more completely.

are fermions, and are therefore excitations in Fermi fields. We knowwhat to expect from their Lagrangian: it’s a Dirac Lagrangian for eachfield, without any mass terms:

L = νe(x)i∂νe(x) + eL(x)i∂eL(x) + eR(x)i∂eR(x), (47.1)

where we’ve written the right-handed electron field as eR(x) and so forth.The interesting thing about having several fields in a theory is the

possibility of internal symmetries. As we have seen, symmetries lead toconserved charges and one of our tasks in formulating the theory is todecide the values of the conserved charges that excitations in each ofour fields will carry. Below we will assign two mysterious charges to ourfields, Y and I, which we apparently pluck from the air on the groundsthat they lead to a successful theory that agrees with all experimentalobservations. It should be noted that these charges are related to themore familiar electromagnetic charge Q via the Gell-Mann–Nishijimarelation:

Kazuhiko Nishijima (1926–2009). Therelation was found by Nishijima andTadao Nakano, and independently byMurray Gell-Mann. Thus it is oftenalso known as the NNG rule.

Q = I3 +Y

2. (47.2)

The known values of Q for the particles in our Universe will constrainthe values of I and Y we assign.

We begin by writing the fields as components of a large column vector:

Ψ(x) =

νe(x)eL(x)eR(x)

. (47.3)

Next we ask what symmetries this theory has. The first is a local U(1)symmetry. The U(1) transformation causes the field to pick up a phasefactor eiβ(x). To ensure invariance we introduce a gauge field Bµ(x)which transforms as Bµ(x) → Bµ(x)+

1Y g′ ∂µβ(x), where Y is the charge

of the theory and the parameter g′ tells us how strongly particles willcouple to the hypercharge (just as we write q = Q|e| and |e| tells ushow strongly a photon couples to a single electronic charge). Rear-ranging the position of charge Y slightly in the equations and rescalingfor future convenience, we’ll rewrite the transformation as e

i2Y β(x) and

47.1 The symmetries of Nature before symmetry breaking 435

Bµ → Bµ + 1g′ ∂µβ. The former expression tells us that the strength of

the transformation at a point in spacetime depends on how much chargea field carries.5 Note carefully that the U(1) charge Y isn’t the electric

5The reason for doing this here is thatthe different components of the fieldare going to carry different amounts ofcharge. We will call Y a ‘U(1) charge’because it controls the coupling be-tween the fermions and the gauge fieldswe need to introduce to guarantee localU(1) symmetry.

charge; instead, Y is known as the weak hypercharge. We assign thefields hypercharges as shown in the table in the margin. Thus under

Field νe eL eR

Y −1 −1 −2

Table of weak hypercharge Y

U(1) transformations we have

νeeLeR

→

e−iβ(x)

2 0 0

0 e−iβ(x)

2 00 0 e−iβ(x)

νeeLeR

. (47.4)

To summarize so far: we may turn the internal dial labelled ‘U(1) weakhypercharge’ without changing the properties of the state of the earlyUniverse. The dial serves to multiply the fields by position-dependentphase factors, whose effects are cancelled by the gauge field Bµ(x).

The internal U(1) symmetry is not the only local symmetry possessedby the early Universe. There is also a local SU(2) symmetry. A SU(2)transformation results in the field components acquiring phases whosevalues depend on the details of the conserved charge6 I, known as weak 6See the previous chapter for further

details of how this arises.isospin, carried by each field component. The weak isospin I obeysthe usual rules of spin angular momentum so, for example, the thirdcomponent of isospin for a field with I = 1/2 has eigenvalues I3 = ±1/2.We assign isospin quantum numbers to particles as shown in the table inthe margin, with the result that the fields undergo the transformation:

(νeeL

)→ e

i2 τ ·α(x)

(νeeL

)eR → eR. (47.5)

For SU(2) transformations the gauge field needed to guarantee localinvariance is Wµ(x), which transforms according to τ · Wµ → τ · Wµ +1gτ · (∂µα)− τ · (α×Wµ), where g is the coupling to the weak isospin.7

7As in the previous chapter, the Wµ

field itself has I = 1. Note also thatthis field has Y = 0 units of weak hy-percharge.

Summarizing again: the dial labelled ‘SU(2) weak isospin’ may beturned at will in the early Universe without any physical consequence.This dial mixes up the massless fields νe and eL and multiplies them byposition dependent phases. Its effects are cancelled by the gauge fieldWµ(x). The transformation has no effect on eR.

Field νe eL eR

I 12

12

0

I312

− 12

0

Table of isospin quantumnumbers I3 and I

We have now described a set of local U(1) and SU(2) transformationsfor our set of field components. Together they make the theory locallyinvariant with respect to the enlarged group of transformations SU(2)⊗U(1). In order to make these symmetries of the Lagrangian we now needto replace derivatives ∂µ in the Lagrangian with covariant derivatives.We write

U(1) : DµΨ = ∂µΨ − i2g

′Y Bµ(x)Ψ,SU(2) : DµΨ = ∂µΨ − igIτ · Wµ(x)Ψ.

(47.6)

Since the original column vector (νe, eL, eR) is distinguished by the firsttwo (left-handed) components having the same Y charge and I chargewhile the right-handed components have different ones, we will rewrite


the fields as

Ψ =

(LR

), where L =

(νeeL

), R = eR. (47.7)

This prevents our having to keep writing Is and Y s in the Lagrangian.8

8The logic of this notation is that itputs the field with I3 = 1 in the upperslot of the L doublet and the one withI3 = −1 in the lower slot.

In terms of these new fields, the covariant derivatives are written:

DµL = ∂µL− i

2gτ · WµL+

i

2g′BµL, (47.8)

DµR = ∂µR+ ig′BµR. (47.9)

Our resulting, locally symmetric, Lagrangian is given by

L = Riγµ (∂µ + ig′Bµ)R+ Liγµ(∂µ − i

2gτ · Wµ +

i

2g′Bµ

)L

−1

4G(W )µν · G(W )µν − 1

4F (B)µν F (B)µν , (47.10)

where G(W )µν = ∂µWν − ∂νWµ + gWµ × Wν and F

(B)µν = ∂µBν − ∂νBµ,

which are the correct contributions from the respective gauge fields forSU(2) and U(1) theories.99Note that the addition of mass terms

to the Lagrangian in eqn 47.10 wouldviolate the local symmetry we have setup. As shown in eqn 47.26, a mass termlinks the left- and right-handed parts ofthe electron field via a term −m(eLeR+eReL) but, since the SU(2) transforma-tion affects the left- and right-handedparts of the field differently, this termcannot be admitted.

Example 47.1

Although apparently plucked from the air, the assignment of charges to the field com-ponents may be physically motivated as follows. We want our theory of electroweakinteractions to have both the weak interaction and electromagnetism correctly em-bedded within it. In terms of left- and right-handed components, we may write theweak isospin currents as

Jµ =1

2LγµτL, (47.11)

and the electromagnetic current as

Jµem = Q (eLγµeL + eRγ

µeR) . (47.12)

Note that, in terms of net charge transfer, Jµem and Jµ3 are neutral currents, whereasJµ1,2 are charged currents. Since Jµ3 does not involve eR, but Jµem does, we need toadd an extra current to the problem. This should ensure that we have a consistentgauge theory that has electromagnetism correctly embedded within it. The simplestsolution is to write

Jµem = Jµ3 +1

2JµY, (47.13)

where JµY is the weak hypercurrent and the factor 1/2 is included by convention.Equation 47.13 then immediately implies the Gell-Mann-Nishijima relation Q = I3 +Y2

. If we now expand out the expressions for Jµ3 and JµY we obtain

Jµ3 =1

2(νeγ

µνe − eLγµeL)

JµY = −νeγµνe − eLγµeL − 2eRγ

µeR, (47.14)

where we have taken Q = −1 for the electron field in the final line. The values ofQ, I3 and Y may then be read off from the coefficients of the νeγµνe, eLγ

µeL andeRγ

µeR terms in the expressions for Jµem, Jµ3 and JµY, in agreement with those listedin the tables above.

So far this seems merely an elaborate example of writing down a La-grangian with some potentially interesting local symmetries. The amaz-ing things start happening when we add a further field into the mix. Itis this field which, from behind the scenes, will determine the propertiesof our electroweak Universe.

47.2 Introducing the Higgs field 437

47.2 Introducing the Higgs field

We now introduce a massive, complex scalar field called the Higgs field

into the Universe. It is the interaction of this field with eL and eR whichwill make electrons massive particles. The Higgs field has four compo-nents which are conveniently arranged into a two-component vector10 as 10Again the upper slot contains the

fields with I3 = 1, while the I3 = −1fields sit in the lower one.

follows:

φ =

(φ+

φ0

)=

1√2

(φ3 + iφ4

φ1 + iφ2

). (47.15)

This arrangement works since we have

φ†φ = (φ+)∗φ+ + (φ0)∗φ0 =1

2(φ2

1 + φ22 + φ2

3 + φ24), (47.16)

giving something that looks like a magnitude.We need to know how the Higgs field transforms under U(1) and SU(2)

transformations. The Higgs field is defined to have weak hyperchargeY = +1 and weak isospin I = 1/2. It therefore transforms according to

(φ+

φ0

)→(

ei β2 0

0 ei β2

)(φ+

φ0

),

(φ+

φ0

)→ e

i2 τ ·α

(φ+

φ0

).

(47.17)This choice of charges requires a covariant derivative for the Higgs fieldof the form

Dµφ = ∂µφ− i

2gτ · Wµφ− i

2g′Bµφ. (47.18)

The Higgs field will give a contribution Lφ to the Lagrangian of

Lφ = (Dµφ)†(Dµφ) +m2

h

2φ†φ− λ

4(φ†φ)2. (47.19)

The positive mass term gives rise to the potential shown in Fig. 47.1,from which we see that we have set the Higgs field up for a fall: it’sunstable to symmetry breaking. Recall that an ordinary scalar field,with negative mass term, has a minimum in potential at φ(x) = 0 and so,classically at least, it takes the value zero in the vacuum. However, withits Mexican hat potential (Fig. 47.1), the ground state of the Higgs fieldis at (φ)0 6= 0, implying that the ground state of a Universe containingthe broken symmetry Higgs field, will be permeated by the uniform field(φ)0. As we shall see, it is the fact that electrons are moving through(and interacting with) this ether that gives them mass.

Re φ

Im φ

U(φ)

Fig. 47.1 The potential of the Higgsfield.

Of course, all things related to the Higgs field would be irrelevant toour discussion of electrons and neutrinos if we didn’t have an interactionbetween the Higgs field and the electron/neutrino field. This is given by

LI = −Ge(LφR+ Rφ†L), (47.20)


where Ge is the coupling strength. Putting everything together, theLagrangian for the Weinberg–Salam model is given by

L = LiγµDµL+ RiγµDµR+ (Dµφ)†(Dµφ)

+m2

h

2φ†φ− λ

4(φ†φ)2 −Ge(LφR+ Rφ†L)

−1

4G(W )µν · G(W )µν − 1

4F (B)µν F (B)µν , (47.21)

where the covariant derivatives are those appropriate for each field, asdiscussed above. To be clear, the massless fields described by this La-grangian are the left- and right-handed electron spinor fields, the left-handed neutrino spinor field, the massless Wµ gauge field and the mass-less Bµ gauge field.

The Higgs field is unstable to symmetry breaking. What will hap-pen? After 10−12 s have elapsed the ground state of the Universe willbreak symmetry, with the Higgs field choosing a unique state from theinfinity of possible ones in the gutter of the potential in Fig. 47.1. Ifthe Weinberg–Salam theory describes reality then, as this happens, theelectron should take on a mass, the neutrino should remain massless andthe massless photon of electromagnetism should emerge. This will turnout to be how things fall and, as a bonus, the existence of two moremassive particles, the W± and Z0 bosons, will be predicted.

47.3 Symmetry breaking the Higgs field

It is now a time t > 10−12 s after the Big Bang, the Universe has cooledbelow 1016 K and its ground state breaks the SU(2) ⊗ U(1) symmetrywe have been discussing. Just as in Chapters 26 and 46 we’re going toexamine the consequences of this broken symmetry.11 The minimum in11Since we have the non-abelian sym-

metry SU(2) then this will bear a re-semblance to the case examined in thelast chapter.

potential of the Higgs field described in eqn 47.19 is not at (φ)0 = 0,

but at (φ†φ)0 = v =(m2

h

λ

). We’ll choose to break the symmetry with a

new ground state at

(φ1)20 =

2m2h

λ , (φ2)0 = 0, (φ3)0 = 0, (φ4)0 = 0. (47.22)

Writing (φ1)0 =(

2m2h

λ

) 12

=√

2v we then have a ground state field

(φ)0 =

(φ+

φ0

)

0

=

(0v

). (47.23)

An important point here is that when we break a non-abelian symme-try, our choice of vacuum may be invariant under a subset of the originalsymmetry transformations, meaning that those symmetries aren’t bro-ken at all. The vacuum chosen here has the property that it is stillinvariant with respect to the local transformation U = ei( Y

2 +I3τ3)α(x)

which, using the Gell-Mann–Nishijima relation of eqn 47.2, we see isequivalent to the U(1) transformation eiQα(x). Invariance with respect

47.4 The origin of electron mass 439

to this transformation gives our broken symmetry Universe the gift ofMaxwell’s electromagnetism.

We could search for excitations from this ground state field in a verygeneral way by allowing every component to vary.12 However, just as in 12So, for example, we could write our

excited state field as

φ(x) =

1√2

[φ3(x) + iφ4(x)]

v + 1√2

[φ1(x) + iφ2(x)]

!

.

the last chapter, we are at liberty to perform a different gauge transfor-mation at each point in spacetime, and thus reduce the excited field tothe much simpler unitary gauge form

φ =

(0

v + h(x)√2

). (47.24)

We are now ready to examine the consequences of spontaneous sym-metry breaking for this theory. We will demonstrate that it predicts13 13Note that the masses of both the pro-

ton and the neutron (which contributemost to the masses of everyday objects)are dominated by the confinement en-ergy of quarks inside them. Thus thepopular notion of the Higgs field assome kind of unique giver of mass to ev-erything in the Universe is rather wideof the mark.

the existence of (i) massive electrons and massless neutrinos; (ii) mass-less photons; and (iii) massive W± and Z0 particles.

47.4 The origin of electron mass

The physics of the Weinberg–Salam model is revealed by picking out thechoice parts of the Lagrangian. Firstly we examine where we expect tofind a fermion mass in a Lagrangian theory written in terms of left- andright-handed fields. If we start with the Dirac Lagrangian L = ψ(p−m)ψand write ψ = ψL + ψR, we have

L = ψLpψL + ψLpψR + ψRpψL + ψRpψR

−ψLmψL − ψLmψR − ψRmψL − ψRmψR. (47.25)

In fact a number of these terms cancel.14 We are left with 14To show this we use the fact that theleft- and right-handed parts are projec-tions of the full field ψ, obtained via

ψL =

„1 − γ5

2

«

ψ

and

ψR =

„1 + γ5

2

«

ψ.

We commute the operators so that theprojection parts sit together. We findthat terms containing the momentumand mixed right- and left-handed fieldsalways contain the combination

(1 − γ5)(1 + γ5) = 1 − (γ5)2

and since (γ5)2 = 1 we conclude thatall mixed terms involving momentumcancel. Totally left-handed or right-handed terms involving the mass termsalso contain this combination, so welose those too.

L = ψLpψL + ψRpψR −m(ψLψR + ψRψL). (47.26)

This provides us with a clue for where to search for the masses of theelectrons in the Weinberg–Salam Lagrangian: we are looking for a termcontaining the combination (ψLψR + ψRψL), which we expect to be mul-tiplied by a scalar quantity, which will be the electron mass for whichwe’re searching. Just such a term is to be found in the interaction be-tween the Higgs field and the lepton field, given by

LI = −Ge

(L φR+ R φ†L

). (47.27)

We will insert our broken symmetry excited state (eqn 47.24) into thisequation and see what happens. The first term yields

L φR = eL

(v +

h(x)√2

)eR. (47.28)

Notice that the neutrino field has dropped out. This is good news sinceit implies that it will remain massless, just as we hope for a realistictheory. The second term gives

R φ†L = eR

(v +

h(x)√2

)eL. (47.29)


The neutrino field doesn’t feature here either and is destined to be mass-less. Putting the two halves together we obtain

Lint = −Ge(eLveR + eRveL) −Ge

(eLh(x)√

2eR + eR

h(x)√2eL

). (47.30)

The first term is in the predicted form of a scalar mass multiplied byψLψR + ψRψL. It predicts the electron mass to be me = Gev.

We conclude that, as claimed, the broken symmetry Higgs field, whichtakes the value

√2v throughout the vacuum, has provided a source of

quantum mechanical treacle that causes the electrons to acquire mass.The nature of the interaction between the Higgs field and the left- andright-handed electrons is represented in Fig. 47.2, which depicts the in-teraction in eqn 47.27.

v

v

v

v

eL

eR

eL

eR

eL

Fig. 47.2 The interaction of the left-and right-handed electron fields and theHiggs field from eqn 47.27.

47.5 The photon and the gauge bosons

Next we need to show that the theory predicts a massless photon. We’lldo this by looking at the place in the Lagrangian where the gauge fieldsinteract with the Higgs field. Minimal coupling tells us that this is theterm involving the covariant derivative of the Higgs field: (Dµφ)†(Dµφ).

Example 47.2

Inserting the broken symmetry version of the Higgs field φ, the covariant derivativeDµφ becomes

01√2∂µh(x)

!

−»

ig

2

„W 3µ W 1

µ − iW 2µ

W 1µ + iW 2

µ −W 3µ

«

+ig′

2Bµ

– 0

v +h(x)√

2

!

.

(47.31)Multiplying out, we obtain

Dµφ = − i

2

0

@gv(W 1

µ − iW 2µ) +

gh(x)√2

(W 1µ − iW 2

µ)

i√

2∂µh(x) + v(−gW 3µ + g′Bµ) +

h(x)√2

(−gW 3µ + g′Bµ)

1

A . (47.32)

Finally we multiply this by its adjoint to form the term in the Lagrangian

(Dµφ)†(Dµφ) =1

2[∂µh(x)]

2 +g2v2

4(W 1

µ)2 +g2v2

4(W 2

µ)2

+v2

4

ˆ(gW 3

µ − g′Bµ)2˜+ (higher order terms). (47.33)

Noting, as usual, that a boson mass always enters the Lagrangian inthe form 1

2 (mass)2 × (field)

2, we see that in breaking the symmetry of

the Higgs field we have developed massive W 1µ and W 2

µ vector particles,with masses M2

W = g2v2/2. The linear combination15 of gauge fields15Recall from Chapter 13 that such alinear combination of fields gives rise toa perfectly valid description of particlesfor a theory.

(gW 3µ − g′Bµ) has also grown massive, but this component has a mass

that depends on the relative sizes of the coupling constants g and g′.Notice that three Goldstone modes of the Higgs field have been con-

sumed with the result that three gauge fields have grown massive. The

47.5 The photon and the gauge bosons 441

excitations of these three fields will become the W+, the W− and the Z0

vector bosons in the next step. We therefore have three massive vectorparticles. Where’s the massless photon? To find it, we note the absenceof the orthogonal linear combination (g′W 3

µ + gBµ) and propose thatthis combination of fields is (proportional to) Aµ(x), the photon field ofelectromagnetism.

g

g′

θW

Fig. 47.3 Definition of the Weinbergangle θW in terms of g′ and g.

Since our particles depend, in a non-trivial manner, on the ratio ofg and g′, we may simplify things by drawing the right-angled triangleshown in Fig. 47.3, with the coupling constants g and g′ as the lengthsof the orthogonal sides and define the Weinberg angle tan θW = g′/g.We then define two new fields Zµ and Aµ by16 16The fields may be written equiva-

lently as

Zµ =(gW 3

µ − g′Bµ)

(g2 + g′2)12

,

Aµ =(g′W 3

µ + gBµ)

(g2 + g′2)12

. (47.34)

(ZµAµ

)=

(cos θW − sin θWsin θW cos θW,

)(W 3µ

Bµ

), (47.35)

where, as suggested by the notation, Aµ is to be interpreted as our oldfriend, the electromagnetic field. In terms of the new fields, our originalequation becomes

(Dµφ)†(Dµφ) =1

2[∂µh(x)]

2 +g2v2

4(W 1

µ)2 +g2v2

4(W 2

µ)2

+g2v2

4 cos2 θWZ2µ + (higher order terms), (47.36)

and from this equation we can read off17 that the mass of the Z0 particle

17The mass of the W boson is given by

M2W = g2v2/2. The term premultiply-

ing Z2µ in eqn 47.36 is g2v2

4 cos2 θW= 1

2M2

Z

and hence MZ = MW/ cos θW.

is MW/ cos θW.The fact that the Wµ and Zµ fields have taken on a mass means that

their propagators will have the formGµν

p2−M2+iǫ . Recall from our dis-cussion of the Yukawa force in Chapter 17 that this implies a potentialvarying as U(r) ∝ e−M |r|. We conclude that, compared to the elec-tromagnetic interaction which is mediated by the massless photon, theweak interaction is short-ranged, falling off over a distance 1/M .

We now have the ingredients for the non-interacting Lagrangian ofthe broken symmetry Universe (see margin on page 443). However, theonly way to observe the predictions of the theory is via the interactionsthat the particles enter into with each other. Our final task is thereforeto substitute our results back to see what the theory predicts for theseinteractions.

Wµ

W †µ

Zµ

Zµ

Zµ

Aµ

νe

eL

eL

νe

νe

νe

eL

eL

eR

eR

e

e

Fig. 47.4 The electroweak interactionvertices predicted by the Weinberg–Salam theory.

Example 47.3

As in the case of QED, minimal coupling allows us to see the interactions by consid-ering the covariant derivatives. Starting with

iRγµ(∂µ + ig′Bµ)R+ iLγµ„

∂µ − i

2gτ · Wµ +

i

2g′Bµ

«

L, (47.37)

and expanding out and using our expressions for Aµ, Zµ and Weinberg’s trigonom-etry, we may eliminate Bµ, W 3

µ and g′ and we obtain

ieγµ∂µe+ iνeγµ∂µνe − g sin θW eγµeAµ

+g

cos θW(sin2 θW eRγ

µeR − 1

2cos 2θW eLγ

µeL +1

2νeγ

µνe)Zµ

+g√2

h

νeγµeLW

†µ + eLγ

µνeWµ

i

, (47.38)


where e = eL + eR, likewise for ν and we define the field Wµ = (W 1µ + iW 2

µ)/√

2. Weexpect the quanta of the Wµ field to be two species of oppositely charged, massivevector particles: the W+ (the quanta created by the Wµ(x)-field, with I3 = 1) andthe W− (annihilated by Wµ(x), with I3 = −1). The Gell-Mann–Nishijima formula(eqn 47.2) tells us that W+ has Q = 1 units of electric charge, while the W− hasQ = −1 units. In contrast the Zµ(x) field is uncharged and we expect a single speciesof massive quantum Z0.

In eqn 47.38 we have an expression for some of the interactions18 in the18For a more complete description ofthe 18 interactions described by thetheory see Mandl and Shaw, Chap-ter 19.

electroweak theory in terms of the coupling g and the Weinberg angle θW.Some important points to note from our result is that Aµ only couples toelectrons and not to neutrinos, just as experiment shows the photon does.The vector field Wµ only couples left-handed electrons to neutrinos anddoesn’t touch right-handed ones. The vector field Zµ couples to left-and right-handed electrons with different strengths. Notice also thatthat parity operation P−1HIP, which exchanges left- and right-handedfields, is not a symmetry of those parts of the interaction involving theWµ(x) and Zµ(x) fields. This is the sense in which, famously, the weakinteraction is said to ‘violate parity’.

The vertices for these interaction are shown in Fig. 47.4. The moststartling of these involve the Wµ field and show that the emission or ab-sorption of a W± particle can result in the transformation of an electroninto a neutrino (and vice versa). Some examples of particle processesallowed by the weak interaction are shown in Fig. 47.5.

Note what the theory predicts for the electric charge. In QED thecharge is the coupling constant in front of the fermion–photon interactionterm: −qψγµψAµ. Taking q = |e| for a single electronic charge we readoff from eqn 47.38 that the electric charge is given by

|e| = g sin θW. (47.39)

W−W+

νeµ−

νµ e−

u d

µ+ νµ

Fig. 47.5 Some example electroweakprocesses. (a) µ−+νe → νµ + e−; (b)positive pion decay involving quarks uand d decaying into µ+ + νµ.

The Weinberg–Salam electroweak theory and its verification are twoof the greatest achievements of twentieth century science. The Z0 andW± particles were detected with masses MZ = 91.19 GeV and MW =80.40 GeV and correspond to a Weinberg angle (via cos θW = MW/MZ)of θW ∼ 28.19 The final prediction of the theory is the existence of a

19Historically, a value for the Weinbergangle had been estimated from ear-lier measurements of lepton scatteringcross-sections and provided a predic-tion for the masses of the W± and Z0

particles which were subsequently dis-covered at the predicted energies. SeeGary Taubes’ highly recommended No-bel Dreams for the frightening history.

scalar field excitation in the Higgs field, known as the Higgs boson.20

20Often incorrectly pronounced‘Boatswain’ by the UK media, possiblyas some kind of nautical throwback.

At the time of writing, experiments at the Large Hadron Collider havefound a particle, thought to be the Higgs, with a mass ≈ 125 GeV.

Thus in summary our current picture is of a symmetry-broken Higgsfield that permeates the Universe. Other fields are set to zero in thevacuum and fluctuate around that, giving rise to particle–antiparticleexcitations that wink in and out of existence; the Higgs field is nonzeroby default. With no Higgs field, the electron and electron neutrino wouldbe identical particles, and theW and Z particles, and in fact all standardmodel fermions, would be massless. That indeed was the situation inthe very early Universe before the Higgs field became symmetry broken.But immediately after the electroweak phase transition (that occurredperhaps 10−12 s after the Big Bang) the present broken-symmetry state

Exercises 443

emerged and the masses of these particles became simply a functionof how strongly they couple to the Higgs field. And couple they will,because the Higgs field is both nonzero and everywhere.

Chapter summary

• In the period t . 10−12s after the Big Bang the leptons of the earlyUniverse enjoy an internal SU(2) ⊗ U(1) local symmetry.

• The Weinberg–Salam electroweak theory describes the breaking ofthe symmetry of the Higgs field with the result that a nonzeroHiggs field permeates the vacuum, providing electrons with mass.

• This also results in the emergence of the photon and three-vectorparticles W+, W− and Z0. The excitations of the electromagneticfield Aµ do not acquire a mass, whereas the W± and Z0 particlesbecome massive.

Non-interacting electroweak La-grangian

L0 = e(i∂ −me)e+ νe(i∂)νe

−1

4FµνF

µν

−1

2W †µνW

µν +M2WW †

µWµ

−1

4ZµνZ

µν +1

2M2

ZZµZµ

+1

2(∂µh)

2 − 1

2m2

hh2

where Wµν = ∂µWµ− ∂νWµ and sim-ilarly for Zµν .

Exercises

(47.1) Show that the broken symmetry ground state

vacuum expectation value (φ)0 =

„

0v

«

is in-

variant with respect to the transformation U =

ei( Y2

+I3τ3)α(x).

(47.2) Verify eqn 47.26.

(47.3) By considering the covariant derivative of the Higgsfield, verify eqn 47.33.

(47.4) Verify eqn 47.38.

(47.5) Fermi’s theory of the weak interaction.(a) Read off the interaction vertices for weak in-teractions between electrons and neutrinos fromeqn 47.38.(b) Draw a Feynman diagram for negative muondecay µ− → e− + νe + νµ and show that, at lowmomentum, the process is described by an interac-

tion Hamiltonian

HI =g2

2νµγ

αµLgαβM2

W

eLγβνe. (47.40)

(c) Defining the current for the lepton species l asJαl = lγα(1 − γ5)νl, show that the interaction forthe muon decay can be recast as

HI =G√2Jα†µ Jeα, (47.41)

where G = g2

4√

2M2W

. This is known as the Fermi

current–current interaction.(d) For muon decay G ≈ 10−5m−2

p , where mp isthe proton mass. Comment on the ‘weakness’ ofthe weak interaction.

48 Majorana fermions

48.1 The Majorana solution 444

48.2 Field operators 446

48.3 Majorana mass and charge447

Chapter summary 450

Exercises 450

Tragically I was an only twinPeter Cook (1937–1995), title of an unrealized project.

Can a fermion particle be its own antiparticle? That is to say, if wetake the charge conjugate of the creation operator of a Fermi particlec†p, can it return the same operator C−1c†pC = c†p without creating aninconsistency in the theory? We know that particle excitations of thereal scalar field are identical to antiparticle excitations of the scalar field.We also know that the photon excitations in the electromagnetic fieldare identical to antiphotons. Scalar particles (spin-0), photons (spin-1)and gravitons (spin-2) can be described by real fields: φ = φ†. Sinceφ† creates a particle and φ an antiparticle, we expect these particlesto be identical to their antiparticles. The Dirac theory of the electron(Chapter 36) seems to require complex fields (though see below), and soit might seem at first glance as if fermion fields necessarily involve dis-tinct particle and antiparticle parts. This would explain why electronsand positrons (anti-electrons) are distinct particles, and also why eventhe electrically-neutral neutron is not identical to its antiparticle, theantineutron. However, this is not the whole story, and in this chapterwe examine the steps required to formulate a theory containing fermionswhose particles and antiparticles are identical or, if you prefer, symmet-ric.

MAJORANA

FERMION

(particle) (antiparticle)

=MAJORANA

FERMION

Fig. 48.1 A majorana fermion is itsown antiparticle.

48.1 The Majorana solution

Etorre Majorana came up with a theory in which a fermion particle wasEtorre Majorana (1906–[1938+x]) hada short, but brilliant, career in theoret-ical physics which was cut short whenhe disappeared whilst travelling by pas-senger ship from Palermo to Naples. Ithas been variously suggested that hecommitted suicide, was murdered, orescaped to Argentina.

its own antiparticle (see Fig. 48.1). Such particles are called Majorana

fermions and are unchanged by the act of charge conjugation.

Example 48.1

For charged particles we know that the charge conjugation operator swaps particlesfor antiparticles, so for a complex scalar field we have, for example: C−1a†pC = b†p .

Let’s start by taking a naive view and examining a mode expansion of a classicalscalar field (that is, the expansion of a function; not an operator):

φ(x) =

Zd3p

(2π)32

1

(2Ep)12

“

ape−ip·x + a∗peip·x”

. (48.1)

48.1 The Majorana solution 445

Notice that the second term is the complex conjugate of the first and that φ(x) isreal. Compare this with the case of a (classical) complex scalar field:

ψ(x) =

Zd3p

(2π)32

1

(2Ep)12

“

ape−ip·x + b∗peip·x”

. (48.2)

Here the complex conjugate of the first term yields something which is not identicalto the second term, i.e. the particle is different from the antiparticle.

Speaking very roughly, charge conjugation involves swapping each field for itscomplex conjugate and so a real field like φ(x) will describe particles that are theirown antiparticles. This will not be the case for the complex field ψ(x).

Motivated by the previous example, we will search for a real (ratherthan complex) solution to the Dirac equation. Such a solution followsif the Dirac equation itself is real. Whether the Dirac equation (i∂ −m)Ψ = 0 is real or complex depends on the choice of the γ matrices.Recall that there are many possible γs out there, constrained only bythe requirement that γµ, γν = 2gµν . If we can find a set of γs whichare purely imaginary, then the Dirac equation itself will be real andconsequently so will its solutions.

Majorana found such a set of purely imaginary γ matrices, given ineqn 48.3 in the margin. Using the Dirac equation given by these matrices

Majorana’s γ matrices:

γ0 =

„0 σ2

σ2 0

«

γ1 =

„iσ1 00 iσ1

«

γ2 =

„0 σ2

−σ2 0

«

γ3 =

„iσ3 00 iσ3

«

. (48.3)

we will find solutions ν(x) which will have the property that ν(x) =ν∗(x), which is known as the Majorana condition and reflects thefact that the solutions are identical to their complex, or more precisely,to their charge conjugates.

Having demonstrated the possibility of Majorana fermions existing asviable quantum mechanical entities, we would now like to make contactwith our previous approach and describe Majorana’s solution in thechiral representation, which involves stacking up pairs of two-componentWeyl spinors to make four-component Dirac, or in this case Majorana,spinors. We therefore need to consider how charge conjugation works forWeyl spinors. (Note that we’re treating our spinors as c-number wavefunctions for the moment; fields will follow shortly.)1 1Note also that, throughout the rest

of this chapter, we reserve the sym-bol ψ for two-component Weyl spinorsonly, which may be left-handed (ψL) orright-handed (ψR), while the symbolsΨ and ν denote four-component Diracand Majorana spinors respectively.

Example 48.2

There is a recipe for obtaining the charge conjugate ΨC(x) of a Dirac spinor Ψ(x) inchiral representation. It is

ΨC(x) = C−1Ψ(x)C = C0Ψ∗(x), (48.4)

with C0 = −iγ2. As an example, we could start with a Dirac spinor Ψ =

„ψL

0

«

containing only a left-handed Weyl spinor ψL and take the charge conjugate as fol-lows:

ΨC = −iγ2

„ψL

0

«∗=

„0 −iσ2

iσ2 0

«„ψ∗

L0

«

=

„0

iσ2ψ∗L

«

. (48.5)

This gives a component in the lower slot involving the original left-handed Weylspinor.2 You may verify that conjugating the charge twice returns the original spinor.

2Note that charge conjugation of the

right-handed spinor gives

„0ψR

«

→„

−iσ2ψ∗R

0

«

.


Here’s the solution: in the chiral basis a Majorana spinor may be builtout of a left-handed Weyl spinor and its charge conjugate as follows:

ν(x) =

(ψL(x)

0

)+

(0

iσ2ψ∗L(x)

)=

(ψL(x)

iσ2ψ∗L(x)

), (48.6)

that is ν(x) =

(ψL(x)ψL,C(x)

), where we write the charge conjugate of a

Weyl spinor as ψL,C = iσ2ψ∗L. This solution ν(x) obeys the all-important

property that νC(x) = ν(x), which is a more general expression of theMajorana condition above. Notice that a Majorana particle may bebuilt starting with only a left-handed Weyl spinor, putting its conjugatepart into the slot where the right-handed part usually lives.3 This is3We could also have started with a

right-handed spinor and then built a

Majorana fermion ν′ =

„−iσ2ψ∗

RψR

«

.

The point is that we only need tostart with either a left-handed spinoror right-handed spinor.

different to the ordinary Dirac fermion we met in Chapter 36, which hasindependent left- and right-handed parts.

48.2 Field operators

So far we’ve been working with wave functions. To make the the-ory respectable we should write things in terms of field operators.Like the wave functions, these fields must obey the Majorana con-dition, which expressed in terms of charge conjugation operators isν(x) = νC(x) = C0ν

∗(x). A slightly jarring abuse of notation hereis the complex conjugate of the field operator, since the complex con-jugate of a creation or annihilation operator is not, strictly speaking,defined. In taking the complex conjugate we are seeking to avoid thematrix transpose that goes along with Hermitian conjugation. There-fore a fussier, but more correct, way of writing the conjugation operationwould be ν†T. However, we’ll stick with the complex conjugate notation

where one should note that us∗(p) =

(us∗L (p)us∗R (p)

)and we freely abuse

the notation with the understanding that here a∗p ≡ a†p.

Example 48.3

For a Dirac particle, recall that the field annihilation operator is given by

Ψ(x) =

Zd3p

(2π)32

1

(2Ep)12

X

s

“

us(p)aspe−ip·x + vs(p)b†speip·x”

, (48.7)

whose charge conjugate is C−1Ψ(x)C = C0Ψ∗(x)

=

Zd3p

(2π)32

1

(2Ep)12

X

s

“

−iγ2us∗(p)a†speip·x − iγ2vs∗(p)bspe−ip·x”

. (48.8)

Moreover, using the explicit forms of us(p) and vs(p) it may be shown4 that

4Consider, for example, thecharge conjugation of the posi-tive helicity antiparticle spinorfrom Chapter 36: −iγ2vs∗(p) =

„0 −iσ2

iσ2 0

«

0

BB@

0pE + |p|

0

−pE − |p|

1

CCA

=

0

BB@

pE − |p|

0pE + |p|

0

1

CCA

= us(p),

i.e. a positive helicity particle spinor,as claimed.

−iγ2us∗(p) = vs(p) and −iγ2vs∗(p) = us(p) leading to a charge conjugate field

C−1Ψ(x)C =

Zd3p

(2π)32

1

(2Ep)12

X

s

“

vs(p)a†speip·x + us(p)bspe−ip·x”

. (48.9)

For what follows, note that this can be made the same as the original Dirac field if wewere to make the replacements a†sp → b†sp and bsp → asp , from which we concludethat the prescription indeed enacts charge conjugation on the Dirac field in that itreturns a Dirac field with particle operators exchanged for antiparticle operators.

48.3 Majorana mass and charge 447

Finally we will write down a Majorana field. We start by noting thatthe general form of a quantum field is

(Quantum Field) = (Particle part) + (Antiparticle part)

= (Particle part) + C−1(Particle part)C.

If particles and antiparticles are identical (which is what we’re after) thenC−1a†spC = a†sp and so C−1us(p)a†spC = −iγ2us∗(p)a†sp. To construct theMajorana field we therefore replace the antiparticle part of the Dirac fieldvs(p)b†sp with −iγ2us∗(p)a†sp. The result is the Majorana field

ν(x) =

∫d3p

(2π)32

1

(2Ep)12

∑

s

(us(p)aspe−ip·x − iγ2us∗(p)a†speip·x) .

(48.10)As may be checked, the Majorana field does indeed enjoy the Majoranacondition:

C−1ν(x)C =

∫d3p

(2π)32

1

(2Ep)12

∑

s

(−iγ2us∗(p)a†speip·x + us(p)aspe−ip·x)

= ν(x).(48.11)

Example 48.4

If we start with a left-handed (momentum space) Weyl spinor usL(p) the Majoranafield takes the form

ν(x) =

Zd3p

(2π)32

1

(2Ep)12

X

s

»„usL(p)

0

«

aspe−ip·x +

„0

iσ2us∗L (p)

«

a†speip·x–

.

(48.12)As in the case of wave functions, we see that Majorana fields may be built out ofWeyl spinors of only one handedness.

Although the argument of this chapter has been very formal, the point isa simple one: it is indeed quite possible to build a Majorana field whoseexcitations are Majorana fermions. We will now discuss the physics ofthese particles.

48.3 Majorana mass and charge

First we consider the mass of the particle excitations of the Majoranafield. Recall that Weyl spinors are necessarily massless. Massive Diracspinors must contain independent left- and right-handed parts and theparticles may be thought of as oscillating between the two. The massterm in the Dirac Lagrangian may be written in terms of Weyl spinorsas5 5See the previous chapter for an expla-

nation of this.mD(ΨLΨR + ΨRΨL), (48.13)


where we write ΨL =

(ψL

0

)and ΨR =

(0ψR

). Vitally, eqn 48.13 is

a Lorentz invariant quantity.66Notice that we don’t have terms

ΨLΨL and ΨRΨR which are notLorentz invariant and therefore are for-bidden.

As Majorana solutions may be written in terms of Weyl spinors ofa single chirality we might question whether it is possible to identify amassive Majorana field. This worry looks to be valid since, as shownin the exercises, a mass term in the Lagrangian of the form ν(x)ν(x)vanishes if we treat the fields in the Lagrangian as c-numbers, as we dofor the Dirac case. However, it turns out that we may define massiveMajorana fields, as long as they anticommute. In that case we mayhave massive Majorana fields built out of left-handed Weyl spinors only,with mass mL, or Majorana fields built from right-handed Weyl fieldsonly, with mass mR. Turning to the Lagrangian, we may write the mostgeneral, Lorentz invariant mass term for Majorana fields as77Note that the N objects in this equa-

tion are built from four-component Ma-jorana spinors ν, so themselves haveeight components.

−1

2

(NL,CMNL + NLMNL,C

), (48.14)

where

NL =

(νLνR,C

), NL,C =

(νL,CνR

), M =

(mL mD

mD mR

),

(48.15)

and where νL =

(ψL

0

), νL,C =

(0

iσ2ψ∗L

), νR =

(0ψR

)and νR,C =

(−iσ2ψ∗

R

0

). Here we have three types of mass: the Dirac mass mD

along with the right- and left-handed Majorana masses mR and mL.

Example 48.5

Working with Majorana fields ν(x) built from left-handed Weyl spinors, we set mD =mR = 0 and mL 6= 0. We may then write a Majorana Lagrangian as

L =1

2νi∂ν − 1

2(νL,CmLνL + νLmLνL,C), (48.16)

where ν = νL + νL,C. Varying this Lagrangian with respect to ψ†L, leads to the

Majorana equation for the field ψL, given by

iσµ∂µψL −mLiσ2ψ∗L = 0, (48.17)

which may be recast in the form of a Dirac equation

(i∂ −mL)ν(x) = 0. (48.18)

To complete this cavalcade of formalism, we briefly turn to the electro-magnetic charge of the Majorana field: it has none. This is, of course,necessary for the particle excitations of this field to be identical to theantiparticles. It can also be seen from the field equations by noting thatif ν = νL+νL,C, then if we make the transformation νL → νLeiα, we musthave νL,C → νL,Ce−iα (because νL and νL,C are related by complex con-jugation). It is therefore impossible to define a U(1) transformation for

48.3 Majorana mass and charge 449

Majorana spinors which simultaneously provides both upper and lowerslots with the same phase factor eiα. The Majorana equation cannot,therefore, be made invariant under local U(1) transformations. Put theother way: a particle carrying the conserved U(1) charge of electromag-netism cannot be a Majorana particle.

Having formulated this theory we now ask whether it’s of any use.Indeed, for many years it seemed to be an interesting solution in need ofa problem. However, in recent years, several such problems have arisen.

• The vast number of emergent quasiparticles in condensed matterphysics present us with an ideal playground for searching for ex-otic excitations such as Majorana fermions. In a semiconductor ora metal, electrons and holes look different because they are oppo-sitely charged, and so it does not seem possible that the particles(electrons) and antiparticles (holes) can be symmetrically related.A superconductor, on the other hand, blurs the distinction betweenelectrons and holes and so seems like a possible environment forrealizing Majorana fermions. One of the great things about su-perconductors is that they screen electric and magnetic fields, andso charge is not a good observable. The bogolon excitations of asuperconductor superpose electrons and holes and, under ratherspecial circumstances, it is possible that the quasiparticles of asuperconducting system are Majorana fermions. An example ofsuch a circumstance involves a superconductor in the presence ofvortices, which changes the equations of motion of the electronsand can lead to the trapping of electron–hole pairs which can bedescribed as Majorana fermions.

• At first sight neutrinos seem to be well described as solutions toWeyl’s equation. All neutrinos are left-handed massless particleswith negative helicity whereas all antineutrinos are left-handedmassless particles with positive helicity. However, the discoverythat neutrinos emitted from the Sun with one flavour may be de-tected with a different flavour suggests that these particles actuallypossess a nonzero, but small, mass. Whether this is a Dirac massis an open question and there exists the possibility that neutrinosare actually Majorana particles with Majorana mass.

• Consider the possibility that for every species of boson in the Uni-verse there exists a corresponding species of fermion (and viceversa) with the same mass. This would require the Lagrangiandescribing our Universe to exhibit a symmetry known as super-

symmetry. In a supersymmetric Universe we should expect theexistence of the ‘selectron’: a spin-0 particle with the mass of anelectron; and the ‘photino’: a spin-1/2 massless particle. If thephotino mirrors the properties of the photon then it must be itsown antiparticle. This implies that the photino is a Majoranafermion, as will be the ‘Higgsino’ and various types of ‘gaugino’.(Supersymmetry is a subject replete with wonderful terms, andhappily the supersymmetric partner of the W boson is called a


‘wino’.) Despite much effort, any direct evidence for supersym-metric particles has so far proved to be elusive.

Chapter summary

• A Majorana fermion is its own antiparticle and is a solution to aversion of the Dirac equation in which the γ-matrices are purelyimaginary.

• We have shown how to build Majorana fields from Weyl spinors ofa single handedness and we have written down a Lagrangian thatleads to their equation of motion.

Exercises

(48.1) Show that two operations of the charge conjuga-tion operator used in eqn 48.5 return the originalparticle.

(48.2) Show that −iγ2us∗(p) = vs(p) for a negative helic-ity particle spinor.

(48.3) (a) Show that, for a Majorana field ν built fromleft-handed Weyl spinors:

νν = −iψTLσ

2ψL + h.c., (48.19)

where h.c. denotes the Hermitian conjugate. Ex-plain why this must vanish if the fields are repre-sented by c-numbers.(b) Show that νν won’t vanish if the fields are rep-resented by anticommuting Grassmann numbers.

(48.4) Verify eqns 48.16 and 48.17.

(48.5) Verify that eqn 48.17 may be recast in the form ofa Dirac equation as claimed.

49Magnetic monopoles

49.1 Dirac’s monopole and theDirac string 451

49.2 The ’t Hooft–––Polyakovmonopole 453

Chapter summary 456

Exercises 456

One would be surprised if nature had made no use of it.Paul Dirac (1902–1984), on magnetic monopoles.

Do magnetic monopoles exist? We are taught at our Mother’s kneethat Maxwell’s equation ∇ · B = 0 insists that they do not. Howeverit is more the case that they have never been observed rather thanthat their existence is a physical impossibility. In fact there seems tobe nothing preventing the existence of the magnetic monopole and inthis chapter we will investigate the properties of these objects. We willpresent descriptions of two rather different sorts of monopole: thoseof Paul Dirac and those of Gerard ’t Hooft and Alexander Polyakov.The existence of Dirac’s monopoles are not mandated by any theory.However, we’ll see that if we really do live in a Universe that resultsfrom the sort of non-abelian gauge theories with symmetry breakingdiscussed in previous chapters, then the existence of ’t Hooft–Polyakovmagnetic monopoles is almost inevitable.

49.1 Dirac’s monopole and the Dirac

string

The discovery of a certain kind of magnetic monopole with a magneticcharge g wouldn’t be a disaster for Maxwell’s electromagnetism. Wecould include them in his equations, which would become rather moresymmetrical: Remember we are using Heaviside

units.

∇ · E = ρe, ∇ · B = ρm,

∇ × E = −Jm − ∂B

∂t, ∇ × B = Je +

∂E

∂t.

The extra terms involve ρm, the magnetic charge density, and Jm, themagnetic current.1 We would also need to adjust the Lorentz force law

1In conventional electromagnetismρm = 0 and Jm = 0.

which would become

F = q(E + v × B) + g (B − v × E) . (49.1)

This variety of magnetic monopole was first considered by Paul Diracand is therefore known as a Dirac monopole.

As a first step in our investigation of the properties of these objects,let’s recap the physics of electric monopoles, which certainly do exist!

452 Magnetic monopoles

They are the charges of classical electromagnetism (see Fig. 49.1). Foran electrical charge q at the origin, Maxwell’s equation yields

∇ · E = q δ(3)(r), (49.2)

whose solution, due to Monsieur Coulomb, takes the form

E =qr

4π|r|3 = −∇

(q

4π|r|

). (49.3)

The electric flux ΦE =∫

E ·dS through a spherical surface surroundingthe electric monopole is then just q.

Fig. 49.1 Monopoles (electric or mag-netic) are sources or sinks of field, be-cause ∇ · E = qδ(3)(r) and ∇ · B =gδ(3)(r).

Fig. 49.2 You can think of a monopoleas originating from an extremely longbar magnet.

AN AS

Dirac stringmonopole

θ r

Fig. 49.3 The vector field from amonopole can be described everywhereexcept along a line of singularities (theDirac string). AN is defined for θ < π.AS is defined for θ > 0.

If analogous magnetic monopoles exist then their behaviour should begoverned by similar equations. Instead of the Maxwell equation ∇ ·B =0, we should have for a magnetic monopole of charge g at the origin

∇ · B = g δ(3)(r), (49.4)

and a radial field equation (that is, the analogue of Coulomb’s law):

B =gr

4π|r|3 . (49.5)

The magnetic flux through a sphere surrounding the magnetic monopoleis given by ΦM = g.

Example 49.1

The problem with the Dirac monopole comes when you try and write down themagnetic vector potential A. There is no single function, without singularities, thatdoes the job everywhere; actually, this shouldn’t be a surprise. If you could find sucha function, then the flux ΦM out of a closed region S (of volume V ) would be

ΦM =

Z

S∇× A| z

B

· dS =

Z

∇ · (∇× A) dV ≡ 0, (49.6)

whereas this should come out as ΦM = g. One way of understanding this physicallyis to think of the monopole originating from an infinitely long and very thin solenoid,or permanent magnet, which has one end at the origin and the other end very faraway (somewhere out at infinity). This looks like a monopole at the origin (seeFig. 49.2), and the physical Universe can be modelled everywhere, except at a line ofsingularities where the solenoid or permanent magnet lies. To illustrate this problem(see Fig. 49.3), consider the function AN given by

AN =g

4πr

(1 − cos θ)

sin θeφ =

g

4πrtan

θ

2eφ (θ < π). (49.7)

This is well defined everywhere except at θ = π and taking the curl of it quickly2

2The expression for the curl of a vectorfunction only containing an azimuthal(eφ) component in spherical polars is

∇× A =1

r sin θ

∂

∂θ

Âφ sin θ

˜er

−1

r

∂

∂r

ˆrAφ

˜eθ.

brings back eqn 49.5. The function blows up at θ = π for all radii r, and so thismagnetic vector potential is singular along the whole of the negative z-axis. It’s asif the monopole carries around a tail of singularities. This tail is known as a Diracstring. Similarly, we can define another function

AS = − g

4πr

(1 + cos θ)

sin θeφ =

g

4πrcot

θ

2eφ (θ > 0), (49.8)

which does the same job but has a singularity at θ = 0, so although we’ve removedthe tail of singularities along the negative z-axis, we now have them along the positivez-axis: we’ve only managed to move the problem, not remove it. This illustrates thepoint that no single function will work everywhere.

49.2 The ’t Hooft–––Polyakov monopole 453

One interesting consequence of these results is that the two magnetic vector po-tential functions AN and AS only differ by3 3Use has been made of the expression

for grad in spherical polars:

∇f =∂f

∂rer +

1

r

∂f

∂θeθ +

1

r sin θ

∂f

∂φeφ.

AN − AS =g

2πr

1

sin θeφ = ∇

„gφ

2π

«

, (49.9)

which looks right since AN and AS correspond to the same field B and so theyshould differ only by a gradient of a scalar function.

The two functions AN and AS represent different choices of electromag-netic gauge. The wave function ψ of an electric charge q surroundingthe monopole can be written in either gauge, but the two expressionswill be related by a gauge transformation:

ψS(r) = ψN(r) exp

[−iq

~

(gφ

2π

)]. (49.10)

Thus as we take φ from 0 to 2π, the wave function will only be single-valued if

g = nh

q, (49.11)

where n is an integer.4 This equation tells us that if the electric charge q

4This is called Dirac’s quantizationcondition.

is quantized, then so is the magnetic charge g. The argument also worksin the other direction: if we were to discover quantized magnetic chargesthen this would explain the existence of quantized electric charge. Thisis Dirac’s formulation of monopoles. There is no necessity for them toexist, but if they are found, they provide a rationale for the quantizationof electric charge.

49.2 The ’t Hooft–––Polyakov monopole

So far we have been considering single particles: the electrically chargedparticle and, by analogy, the magnetically charged particle.5 Now we ask

5In fact, taking the Dirac monopole tobe a point particle is something of a ne-cessity since, like the electric monopoleit represents a singular solution to theMaxwell equations.

whether there is a quantum field theory for which a magnetic monopolerepresents a stable solution? There is: magnetic monopoles are topolog-ically stable solutions of a non-abelian gauge theory with spontaneoussymmetry breaking.6 As we will show, these field theory monopoles, pro-

6Since, in Chapter 47, we saw that sucha theory seems to describe the elec-troweak part of Nature, it seems likelythat magnetic monopoles might be re-alized in our Universe! There shouldalso, we hope, be a good reason whywe haven’t observed them yet.

posed in 1974 by ’t Hooft and Polyakov, are rather different entities toGerard ’t Hooft (1946– )Alexander Polyakov, (1945– )

the Dirac monopoles of the last section.

Example 49.2

The monopole solution is most easily seen if we review the topological objects ofChapter 29. There we started with a Lagrangian which was unstable to sponta-neous symmetry breaking. Then the topological solutions arose by considering time-independent solutions which sat in a different vacuum at different points in space:

• The simplest was the kink in (1+1) dimensions, which was a field existing indifferent vacuum states at ±∞.


• The second simplest was the vortex in (2+1) dimensions, where a field withbroken U(1) symmetry enjoyed different vacua at all points on a circle atspatial infinity.

Obviously the next step is to examine (3+1) dimensions and a configuration withbroken SO(3) symmetry which enjoys a different vacuum at all points at a sphericalshell at infinity.

The solution with this topological property in (3+1)-dimensional space-time is the monopole, which has the form as |r| → ∞ that

Φ(r) = A r|r| (|r| → ∞), (49.12)

with A a constant. The thing to notice here (as with the vortex) isthe distinction between internal and real space components. The vectorΦ lives in internal isospace but the radial vector r lives in real space.However, eqn 49.12 links these two together. In components we have,for example φ1(|r| → ∞) = A x

|r| , that is, along the spatial 1-direction

(also known as the x-direction) the field points along the internal 1-direction. Similarly along the y-direction the field points along theinternal 2-direction and so on. In this sense, the field points radiallyoutwards at infinity, which is why Polyakov called this the hedgehog

solution. This is represented in Fig. 49.4.

Fig. 49.4 The hedgehog solution.

Our task is now to find a theory for which the hedgehog is a solu-tion and then to show that it has the electromagnetic properties of amonopole. Let’s consider a gauge theory with local SO(3) symmetry,defined by a Lagrangian

L =1

2(DµΦ) · (DµΦ) +

m2

2Φ · Φ − λ(Φ · Φ)2 − 1

4Gµν · Gµν , (49.13)

where Gµν is the gauge field tensor and DµΦ is the covariant derivativegiven, respectively, by

Gµν = ∂µWν − ∂νWµ + qWµ × Wν ,

DµΦ = ∂µΦ − qΦ × Wµ, (49.14)

with q the SO(3) charge of the theory (which will be shown to be equiv-alent to electric charge a little later). Notice that the Lagrangian wehave written is unstable to spontaneous symmetry breaking so Φ is aHiggs field.7 As stated above, we seek a topological field configuration

7Recall what happens in such a the-ory. Owing to the positive mass term,the symmetry is spontaneously brokenand the Higgs field φ takes on a con-stant value in the ground state |Φ0| =“m2

4λ

” 12. The broken symmetry ground

states form a spherical shell S2 in Φ-space of radius |Φ0|. Previously wehave chosen to direct the Φ-field alongthe Φ3-direction. In that case W 3

µis equivalent to the usual electromag-netic field Aµ and we recover Maxwell’selectromagnetism. We therefore takethe charge of the theory to be q: theelectromagnetic charge. (Although ofcourse we know that our electromag-netism resulted from symmetry break-ing in a SU(2)×U(1) gauge theory, theSO(3) theory will be sufficient for ex-amining monopoles solutions.)

for this theory with a different ground state at each point on the spatialboundary of the theory, which is the spherical shell S2 at |r| → ∞. Theresulting hedgehog solution of the Lagrangian in eqn 49.14 is given by

Φ =(m2

4λ

) 12 r

|r| (|r| → ∞). (49.15)

As with all time-independent soliton-like solutions in more than onespatial dimension,8 Derrick’s theorem9 tells us that this object, taken8Recall that the vortex is one such ob-

ject.

9See Section 29.3.

alone, gives rise to an infinite energy. We therefore need to appeal tothe gauge field to stabilize the monopole and guarantee that its energy

49.2 The ’t Hooft–––Polyakov monopole 455

remains finite. For this to be true we require the gauge field Wµ for|r| → ∞ to be (see Exercise 49.1)

W ai = −εiab r

b

qr2,

W a0 = 0, (49.16)

where here a is the internal index and i is the real-space vector index.

Example 49.3

We have identified a topological object that can exist in (3+1)-dimensional spacetime,but we’ve yet to find its electromagnetic properties. This turns out to be a rathernon-trivial matter since we have three internal components of the field strength tensorGµν and, therefore, three different choices of magnetic field10 Bai = −εijkGajk (with 10Here, as elsewhere in this chap-

ter, a = 1, 2, 3 is the internal indexand i, j, k are vector indices. Com-pare these three choices with the sin-gle choice from conventional electro-magnetism: Bi = −εijkFjk (no sumimplied over repeated indices).

no sum implied over repeated indices). For the more usual uniform broken symmetryground state, where Φ points along the 3-direction, the choice is simple, the a = 3component of Gaµν reduces to the usual form Fµν . More generally for an arbitrarilyoriented ground state (i.e. one not necessarily pointing along the 3-direction) we have

Fµν =Φ

|Φ| · Gµν . (49.17)

For any situation other than the ground state of Φ we need a more complicatedexpression containing terms in DµΦ which vanish in the ground state.11 Fortunately, 11See Exercise 49.3.these complexities may be ignored since we only require the magnetic flux emergingfrom the hedgehog solution, for which we find

Fij = F ij = −εijk rk

q|r|3 . (49.18)

Finally we’re ready to see what the hedgehog field predicts for the magnetic field ofa monopole: it is

B =1

q

r

|r|3 , (49.19)

which is exactly the radial field expected for a monopole. It results in a magneticflux

ΦM =4π

q. (49.20)

The hedgehog solution is therefore confirmed as a magnetic monopole. Comparingthe flux from the hedgehog with the flux expected from a magnetic monopole Φ = g,we must have that 4π

q= g or, on restoring factors of ~,

g =2h

q. (49.21)

Comparing this with the result for a single Dirac monopole, for which g = h/q, we seethat the magnetic charge of the ’t Hooft–Polyakov monopole is twice the magnitudeof the Dirac monopole.

Notice how the ’t Hooft–Polyakov monopole and Dirac monopole arevery different beasts. Dirac monopoles are singular mathematical so-lutions of electrodynamics which necessitate the introduction of pointparticles as the sources of the magnetic flux. These particles have arbi-trary spin and mass. In contrast ’t Hooft–Polyakov monopoles are non-singular solutions arising from the interaction of a non-abelian gaugetheory and a scalar field. All of their properties, such as their mass, aredetermined by the original theory.12

12The fact that these monopoles are so-lutions to the sort of broken symme-try theory that describes our Universeleads us to question why we have notyet observed any of them. ’t Hooftestimated the mass of these beasts asMW/α ≈ 137MW ≈ 11 TeV, makingthem very heavy indeed and outside therange of our experiments. Their exis-tence therefore remains an open ques-tion.


Chapter summary

• Dirac’s magnetic monopole leads to a singularity in the magneticvector potential which cannot be removed (the so-called Diracstring). The quantization of magnetic charge of this monopoleis inextricably linked to the quantization of electric charge.

• The ’t Hooft–Polyakov monopole is a topological object in a non-abelian gauge theory.

Exercises

(49.1) Show that in order to kill the divergent derivative

of the monopole field Φ =“

m2

4λ

” 12 r

|r| at r → ∞ we

must introduce a covariant derivative with a gaugefield with components Wµ

W ai = −εiab r

b

qr2,

W a0 = 0. (49.22)

(49.2) Verify eqn 49.19.

(49.3) We can generalize the definition of the Maxwellelectromagnetic field Fµν so that it describes elec-tromagnetism in the presence of the topologi-cal monopole and reduces to ordinary electromag-netism under the normal circumstance of a brokensymmetry Universe with components Φ3 6= 0 and

Φ(1,2) = 0. We define the modified electromagneticfield tensor as

Fµν =Φ

|Φ| ·Gµν −Φ

q|Φ|3 · (DµΦ×DνΦ). (49.23)

(a) Check that this reduces to the ordinary form ofFµν as claimed.We may also define Aµ = 1

|Φ|Φ · Wµ, which, ina broken symmetry Universe without monopoles,picks out the part of the potential that gives ordi-nary electromagnetism.(b) Show that this leads to a cleaned up field equa-tion

Fµν = ∂µAν−∂νAµ−Φ

q|Φ|3 ·(∂µΦ×∂νΦ). (49.24)

(c) Starting with eqn 49.24 verify the algebra thatleads to the value of the B-field in eqn 49.19.

50Instantons, tunnelling and

the end of the world

50.1 Instantons in quantum parti-cle mechanics 458

50.2 A particle in a potential well459

50.3 A particle in a double well460

50.4 The fate of the false vacuum463

Chapter summary 466

Exercises 466

There is a theory which states that if ever anybody discov-ers exactly what the Universe is for and why it is here, itwill instantly disappear and be replaced by something evenmore bizarre and inexplicable. There is another theory whichstates that this has already happened.Douglas Adams (1952–2001)

In the true vacuum, the constants of nature, the masses andcouplings of the elementary particles, are all different fromwhat they were in the false vacuum, and thus the observer isno longer capable of functioning biologically, or even chemi-cally.Sidney Coleman (1937–2007)

One aspect of conventional quantum mechanics that we have not yetaddressed with quantum field theory is tunnelling. In this chapter wewill discuss a class of objects known as instantons which are foundin the path integral version of quantum mechanics. Their field theoryanalogues allow us to address the problem of tunnelling in quantum fieldtheory. In particular we will address the apocalyptic question of the endof the world! We know that when a system breaks a symmetry theresult is that the vacuum of a system is one of (potentially) a numberof equivalent vacua. This is presumably the case in our own Universe.But what would happen if the vacuum that the system adopts is notactually the lowest energy state? What if there is one with a slightlylower energy? If this is the case with our own Universe then we mightworry that the Universe will undergo a transition, via a tunnelling event,into the true vacuum with catastrophic consequences for those of us whohave grown dependent on the physics of the current (false) vacuum.1

1This problem was called the ‘The fateof the false vacuum’ by Sidney Cole-man and our approach to this problemfollows his treatment closely. See thesuperb lecture in his collection Aspectsof Symmetry for the full story.

We start by examining instantons. These are very similar to the kinkswe examined in Chapter 29 which exist as stable entities localized inspace. Instantons are constant energy solutions to quantum mechanicalequations of motion with a characteristic, stable structure localized intime. The name ‘instanton’, coined by ’t Hooft, reflects this particle-like existence in time, rather than space. Instantons are associated withcritical points (local maxima, minima or saddle points) of the action.

458 Instantons, tunnelling and the end of the world

50.1 Instantons in quantum particle

mechanics

To reveal the physics of the instanton we will temporarily leave behindfields and deal with single-particle quantum mechanics, albeit describedby the path integral and Green’s functions. We will examine a singleparticle in one spatial dimension described by a Lagrangian L = p2/2m−V (x). This gives rise to an action S =

∫ T/2−T/2 dt L and a path integral

G =∫D[x(t)] eiS/~ which gives us the Green’s function, or amplitude, for

any given scenario. (For simplicity we will henceforth assume the particledescribed here has unit mass.) We now make a switch to Euclideanspace,2 where we have a Euclidean action2In Chapter 25 we used the Wick ro-

tation, which turns Minkowski space-time into Euclidean spacetime via therotation taking x0 → −iτ , purely as ameans of relating quantum field theoryto statistical physics. In this chapter wewill see that working in Euclidean spacehas the advantage of leading to newand useful solutions to the equationsof motion of quantum particles. Notethat here we do not impose the peri-odic boundary conditions of statisticalphysics and so imaginary time stretchesfrom −∞ ≤ τ ≤ ∞. We will also callthe time interval over which the calcu-lations are carried out T in Minkowskispace and TE in Euclidean space.

SE =

∫ TE/2

−TE/2

dτ

[1

2

(dx

dτ

)2

+ V (x)

], (50.1)

and a path integral

G(xf , TE/2, xi,−TE/2) =

∫D[x(τ)] e−SE/~, (50.2)

where G is the Green’s function describing the amplitude for starting at(−TE/2, xi) and ending up at (TE/2, xf), and the integral is carried outover all trajectories that have this property.

Let’s suppose a stationary trajectory exists called x(τ). Applying theEuler–Lagrange equations to the Euclidean action leads to

d2x

dτ2=

dV (x)

dx, (50.3)

which differs from the usual equation of motion for a particle in a poten-tial by a minus sign. Doing our analysis in Euclidean space has resultedin equations of motion that seem to describe the motion of a particle in apotential −V (x). As a result, the constant of the motion correspondingto energy is given by

E =1

2

(dx

dτ

)2

− V (x). (50.4)

The solutions to the equations of motion that minimize the action haveconstant energy and so this equation for E enables us to identify themvery conveniently.

Example 50.1

In order to find the Green’s functions which describe amplitudes in this potential wewill use what is known as the stationary phase approximation whose guts aredescribed here. Since we know that the path integral will be dominated by thosepaths closest to the stationary trajectory3, we split the integral into3See Chapter 23.

G = e−(Stationary action)/~ × (quantum corrections). (50.5)

50.2 A particle in a potential well 459

The key to solving problems is then to find the stationary action for the situationwe are trying to describe and expanding around this point. The implementation ofthis approximation becomes clear if we look at a simple integral I =

Rdx e−f(x).

Expanding f(x) about a stationary point at x we have

f(x) = f(x) +1

2f ′′(x)(x− x)2 + ... (50.6)

The second term will present us with a Gaussian integral and so we obtain

I ≈ e−f(x)„

2π

f ′′(x)

« 12

. (50.7)

This result carries over to the case of the functional integral and we obtain an ex-pression for the Green’s function, for our Euclidean theory of a unit mass particle,given by the stationary phase approximation as

G(xf , TE/2, xi,−TE/2) = Ne−S[x]/~

»

det

„

− ∂2

∂τ2+ V ′′(0)

«–− 12

, (50.8)

which is accurate to order O(~), which will be sufficient for our purposes, and whereN supplies the normalization.

The conclusion from all of this scene-setting is simply that using theEuclidean action has the effect of flipping the sign of the potential. Thepoint of this is that there are stationary solutions which exist in theEuclidean world of upside-down potentials which may be analyticallycontinued back to Minkowski space. We therefore potentially gain accessto lots of new possibilities for the motion of particles to which we werepreviously ignorant. One of these is tunnelling, which would never bepredicted by calculating the quantum corrections to a stationary pathfound in Minkowski space.

In the next sections we will build some confidence in this approach byexamining the stationary solutions that follow from some simple upside-down potentials.

50.2 A particle in a potential well

Example 50.2

As a warm up exercise in getting used to the formalism, we will examine the case of aparticle in a potential well such as that shown in Fig. 50.1, which has V (x = 0) = 0.

In the Euclidean version of the story the well is turned upside down into the

x

V

Fig. 50.1 A potential well.

x

−V

Fig. 50.2 An upside-down potentialwell, inverted by the Wick rotation toEuclidean space.

potential hill shown in Fig. 50.2. If we impose boundary conditions that we wantxi = xf = 0 then clearly the only solution is for a particle to balance at the top of thehill. This involves the particle stopped at x = 0 for the entire interval TE and so hasaction S(x) = 0. Using our equation for the Green’s function we find the quantumamplitude that the particle starts at x = 0 at time τ = −TE/2 and is found thereagain at τ = TE/2 is given by

G(0, TE/2, 0,−TE/2) = N

»

det

„

− ∂2

∂τ2+ ω2

«–− 12

, (50.9)

where ω2 = V ′′(0). Recall that we’ve previously solved a related (exactly solvable)path integral problem: that of the simple harmonic oscillator potential. In that casewe concluded that (in Minkowski spacetime) we have

G(0, T/2, 0,−T/2) = 〈0|e−iHT/~ |0〉 = N

»

det

„

− ∂2

∂t2− ω2

0

«–− 12

, (50.10)


where ω0 = V ′′(0) and V (x) = 12ω2

0x2. The stationary phase version above is simply

an approximate generalization of this result with a sign change due to our workingin Euclidean space.

Returning to the more general potential and to Euclidean space we see that work-ing out G is made difficult by the determinant. However, we can argue the generalform of the answer. We know that in Euclidean space the Green’s function will look

like G = 〈0|e−HTE/~ |0〉 = e−E0TE/~ , and we expect that the ground state energyshould be something close to E0 ≈ 1

2~ω for small oscillations in a potential well,

suggesting that G ∝ e−ωTE/2. In fact, for large TE, it may be shown that

G(0, TE/2, 0,−TE/2) = N

»

det

„

− ∂2

∂τ2+ ω2

«–− 12

=“ ω

π~

” 12

e−ωTE

2 , (50.11)

and we read off the ground state energy as E0 = 12

~ω [1 +O(~)], as we expect.

We’ll follow the same procedure again for more interesting potentials.The procedure is to write down the potential and shift it to Euclideanspace, find the stationary configuration, work out its action and thenfind the approximate Green’s function using eqn 50.8.

50.3 A particle in a double well

x

V

−a a

Fig. 50.3 The double potential well.

x

−V−a a

Fig. 50.4 The double potential well,turned upside-down by the Euclideanrotation.

The instanton solution is found when we repeat the procedure outlinedabove for the double potential well. We define the well with minimaat ±a and take V (−x) = V (x). We also define V ′′(±a) = ω2. Thepotential for the double well is that shown in Fig. 50.3 and shiftingto Euclidean space turns it upside down, as shown in Fig. 50.4. Welook for solutions to the equations of motion with constant energy withboundary conditions that the particle itself is stationary at τ = ±TE/2.These are useful since we will eventually make TE very large and sothey will represent well defined solutions of finite energy. There aretwo obvious sets of possible solutions: one involves a particle fixed at aor at −a, just as we had in the previous section. However, there is amore interesting possibility: the particle starts at −a at −TE/2 and endsup at a at TE/2 as shown in Fig. 50.5. This is the instanton solution.Viewed with the potential flipped the right way up, the end points ofthis solution correspond to a particle tunnelling though the barrier.

x

−V

Fig. 50.5 The instanton solution.

Example 50.3

We may calculate the properties of the instanton. With the potential as defined inFig. 50.5 this solution has E = 0, so we have

dx

dτ= (2V )

12 . (50.12)

The action of an instanton is given by

S0 =

Z

dt

"

1

2

„dx

dτ

«2

+ V

#

=

Z

dτ

„dx

dτ

«2

=

Z a

−adx (2V )

12 . (50.13)

At large TE we have that x = ω(a− x) leading to

(a− x) ∝ e−ωτ , (50.14)

which tells us that the (temporal) size of an instanton is 1/ω.

50.3 A particle in a double well 461

The previous example demonstrates that the instanton has a well definedstructure in time of size τ ∼ 1/ω. It is shown in Fig. 50.6, where itstemporal extent is given by the width of the region where it crosses theaxis. Notice the similarity between this and the kink in Chapter 29.(The kink represented a well defined structure in space, with a spatialextent l ∼ 1/m.)

τ

φ(τ )

a

−a

Fig. 50.6 The instanton as a tightlybound structure in (imaginary) time.

In order to use the instanton to predict the properties of the doublewell we must allow the possibility of the existence of more than one tun-nelling event. This corresponds to examining more than one instanton.This is the subject of the next example.

Example 50.4

We will now examine the case of a dilute gas of instanton objects. Using the samedouble well, the ‘gas’ corresponds to a particle on top of one hill that occasionallyrolls down the hill and up to its neighbour, and then some time later rolls back downthe hill and up to where it started. Some time later it undergoes another rollingevent, and so on. The gas then comprises n rolls, constrained so that an instantona → −a must be followed by an anti-instanton −a → a. (Recall that we had thesame constraints on kinks and antikinks.) The gas is dilute in that each instantonhas a size (in time) far smaller than the gaps between instantons and anti-instantons.The gas is shown in Fig. 50.7. This may the strangest picture of a gas it’s possibleto draw!

τ

x

−TE/2 TE/2τ1τ2τ3

Fig. 50.7 A gas of instantons for thedouble well problem.

The action of n dilute instantons is given by nS0, where S0 is the action of oneinstanton that we calculated above. For a single well we had G(0, TE/2, 0,−TE/2) =

(ω/π~)12 e−ωTE/2. For the gas in the double well we need to consider this amplitude

multiplied by the contribution K from each of the n instantons:

“ ω

π~

” 12

e−ωTE/2Kn, (50.15)

where K is the amplitude for the occurrence of a single instanton. Since we don’tknow where each instanton is centred in time, we must integrate this quantity overthe centres of all of the instantons. This is conveniently carried out by noting that

Z TE/2

−TE/2dτ1

Z τ1

−TE/2dτ2...

Z τn−1

−TE/2dτn =

TnEn!. (50.16)

Putting all of this together we have, for example, to order O(~), an amplitude forthe particle to start and finish at x = −a of Note that ~ has been reintroduced for

clarity.

G(−a, TE/2,−a,−TE/2) =“ ω

π~

” 12

e−ωTE/2X

ne

(Ke−S0/~TE)ne

ne!, (50.17)

where ne means that n only includes even numbers, ensuring that, however large wemake n, the particle ends up where it started.4 These sorts of sums may be done and 4The same argument says that amnesia

may be avoided if one is hit over thehead an even number of times.

the general result is that

G

„

±a, TE

2,−a,−TE

2

«

=1

2

“ ω

π~

” 12

e−ωTE

2

»

exp(Ke−S0~ TE) ∓ exp(−Ke−

S0~ TE)

–

.

(50.18)We may read off the physics by comparing with e−ETE/~ as we did for the singlewell. In this case we find that we have two low-lying energy states with energiescorresponding to whether we finished at −a (summing over even n) or +a (summingover odd n). The energies of the two states are given by

E± =1

2~ω ± ~Ke−S0/~ . (50.19)

It’s worth pausing at this stage to take stock of what we’ve done by returning toMinkowski space and flipping the potential back the right way up.


The conventional quantum mechanics of the double well is instructive here. Wehave solutions that are localized in the left-hand well |−a〉 and in the right-hand well|a〉. We’ll call the energy for sitting in a potential minimum U0, but we allow theHamiltonian to permit barrier penetrating transitions between minima with matrixelement 〈−a|H|a〉 = 〈a|H| − a〉 = −V . The solutions are

1√2

“

| − a〉 + |a〉”

E = U0 − V,

1√2

“

| − a〉 − |a〉”

E = U0 + V.(50.20)

The system saves some energy through the wave function spreading out over the twowells. The symmetric combination lies lowest, separated by an energy 2V from theantisymmetric combination.

In the present case, we can identify U0 = 12

~ω as the energy for sitting in

the bottom of a well and V = ~Ke−S0/~ . This makes sense as the difference inenergies is proportional to the factor accounting for barrier penetration e−S0/~ .

x

V

a

Fig. 50.8 A potential with a barrier.

x

−Va

Fig. 50.9 The barrier potential turnedupside-down.

Example 50.5

We now examine tunnelling to freedom through a barrier as another example of theuse of instantons. We start with the potential shown in Fig. 50.8 and up-end it byworking in Euclidean space to obtain Fig. 50.9. The stationary solution of interest is adifferent instanton to those considered thus far. This instanton, shown in Fig. 50.10,involves the particle starting at xi = 0, travelling to x = a, where it bounces, reversesits direction and returns to its starting point at xf = 0.

x

−V

Fig. 50.10 The bounce solution: an-other example of an instanton.

This is rather similar to the previous problem of the instanton gas, except thatthe Green’s function is given by summing over all n, since any number of instantonsresult in the particle back where it started at x = 0. We may carry out the sum togive

G(0, TE/2, 0,−TE/2) =“ ω

π~

” 12

e−ωTE/2 exph

Ke−S0/~TE

i

, (50.21)

which we might think tells us of an energy eigenvalue of E0 = 12

~ω + ~Ke−S0/~ .However, the interpretation of this solution is slightly more subtle.

Let’s return to the right-way-up potential of Fig. 50.8. We’ve seen that the factor~Ke−S0/~ tells us about barrier penetration, but penetrating through the barrier inthis problem results in the particle rolling away to infinity. This suggests that thebound state in the potential is unstable. Additional evidence arises when we plot thetrajectory of the bouncing particle, which is shown in Fig. 50.11. The fact that thetrajectory has a maximum tells us that the wave function must have a node. Thismeans it’s not the lowest energy wave function and there must be a lower one with anegative energy. Since K is a function of the square-root of energy then this impliesthat K is imaginary. This fits with the idea of barrier penetration: we know thatunstable states have energies with imaginary parts which, when plugged into e−iEt

give e−iE0t−Γt where 2Γ is the decay rate of the state. What our instanton tells us isnot a correction to the energy of a bound state, but rather the imaginary part of theenergy, which gives is the decay rate. We conclude therefore that ImE0 ≈ ~|K|e−S0/~

and so the decay rate of the state is roughly given by

Γ ≈ ~|K|e−S0/~ . (50.22)τ

x

Fig. 50.11 The trajectory of the parti-cle during the bounce has a maximum,signalling a node in the wave function.

50.4 The fate of the false vacuum 463

50.4 The fate of the false vacuum

Finally we turn to fields. As stated in the introduction, our goal is todescribe what happens to a broken symmetry system in the case thatthe vacuum that the system has chosen (known as the false vacuum) liesslightly higher in energy than another minimum in the potential, whichwe call the true vacuum. A potential describing this state of affairs isshown in Fig. 50.12. To be more colourful, we imagine that the vacuumof the Universe in which we live is a false vacuum (at φ = φ+) and askwhether the Universe is about to tunnel through a potential maximuminto the real vacuum (at φ = φ−).

φ

U

φ−φ+

ε

Fig. 50.12 False vacuum potential.

By analogy with the case of the magnet in Chapter 26, the catas-trophic quantum collapse of the false vacuum would seem rather un-likely. In the magnet example the process of tunnelling to the truevacuum would rely on each electronic spin in the system simultaneouslytunnelling. We might have expected this to have a vanishingly smallprobability since there are so many particles. However, the possibilityof instanton solutions in the field allow a scenario where a cluster of spinsfinds the true vacuum and then an instanton causes the spins formingthe boundary of this ‘bubble’ to flip, causing the Universe to pass atipping-point beyond which there is sufficient volume of true vacuumto make this the preferred phase. Whether a change in phase of theUniverse actually occurs depends on the relative balance of the energysaving in realizing the true vacuum versus the cost of the domain wallbetween true and false vacua.

Example 50.6

Supercooled liquid presents us with a similar problem. In that case we cool a liquidwhich reaches a metastable state of liquidity, separated by a small energy from thetrue vacuum of solidity. A similar situation occurs for the superheated liquid, whichremains in a metastable state of liquidity rather than evaporating.5

5Although in this case, you should notethat there’s no difference in symmetrybetween a liquid and a gas, so the anal-ogy with symmetry breaking no longercarries.

r

E

Fig. 50.13 The energy of a bubble ofgas of radius r in a superheated liquid.

For the superheated liquid, thermal fluctuations often result in a small bubble ofgas appearing. The inside of the bubble is the true ground state and so results inan energy saving ǫ, per unit volume. On the other hand, the surface energy of thebubble costs energy σ per unit surface area. We might write the energy

E = −4

3πr3ǫ+ 4πr2σ, (50.23)

shown in Fig. 50.13. Small bubbles, with a small volume energy saving comparedto surface area cost will disappear. At some point a bubble will be large enoughthat it is energetically favourable for it to increase in size. By finding the stationarypoint in E we can work out how large a bubble must be to become ever-expanding.Extremizing, we find a maximum in E at

r =2σ

ǫ. (50.24)

Bubbles larger than this find it energetically favourable to expand. As this occurs,more and more space is taken up by the true vacuum, which eventually spreadsthroughout all space.


We examine the decay of the false vacuum by analogy with the particleproblems looked at so far in this chapter. The decay process is describedby a bouncing instanton, which gives a decay rate. Since in quantum fieldtheory we deal with a path integral6

∫Dφ e−

R

d4xLE , then the quantity6again setting ~ = 1.

we will find is Γ/V, a decay rate per unit volume, which we expect tobe given by

Γ/V = |K|e−S0 , (50.25)

where S0 is the action of the instanton bounce. Calculating the approxi-mate action of the bounce will be possible: since the minima of the falsevacuum potential are only separated by a small energy difference, theyclosely resemble the exact double well problem for which we know theaction.

Let’s now be specific. We consider a quantum field theory in four-dimensional Euclidean space. This has an action

S =

∫d4x

[1

2(∂µφ)2 + U(φ)

]. (50.26)

We cook up a symmetric, double-well potential Usym, which we thenmodify slightly to obtain the false vacuum potential U :

U = Usym + ε(φ− a)/2a. (50.27)

The potential has minima at φ± = ±a, although that at φ+ = a lies ata slightly higher energy, by an amount ε (Fig. 50.12).

We will follow the procedure we developed for identifying instantonsin particle mechanics. We may employ the results from the particle, aslong as we remember that the imaginary time τ in the Euclidean particleproblem has become a spacetime point r in four-dimensional Euclideanspace. The Euclidean potential is upside-down as shown in Fig. 50.14.The bouncing instanton clearly starts on the hill of the false vacuumand bounces from the point φc, where U(φ) cuts the φ-axis.

φ

−U

φ−

φ+φc

Fig. 50.14 False vacuum potential inEuclidean space.

We can also imagine this running backwards, with the particle startingat a point φc near the true vacuum at φ−. We choose to look at it thisway around because the bubble that will result in 4-space then resemblesthe bubble in the superheated liquid with true vacuum within and falsevacuum without (shown in Fig. 50.15). The particle will roll down thehill and come to a rest exactly on top of the hill at φ+ (the false vacuum).We’ll imagine this takes place very rapidly around some point in time rthat we’ll call r = R. Back in field language we ask what the hill-rolllooks like. In Euclidean space we have a large spherical bubble of radiusr = R. This has a thin wall representing the rolling process. (Thewall is thin because the roll takes place very quickly.) The bubble wallseparates the false vacuum on the outside of the bubble from the truevacuum on the inside. The wall of the bubble contains the instanton. IfR, the spacetime point where the instanton is centred, is greater thanthe critical radius Rc above which the bubble is self-sustaining, then theinstanton creates a bubble that will drag all of spacetime into the truevacuum.

Rc

Truevacuum

Falsevacuum

Fig. 50.15 The bubble in Euclideanspace.

50.4 The fate of the false vacuum 465

We will now construct an action S for the rolling process. As r in-creases we start inside the true vacuum bubble, move through the walland end up outside. The action is given for this spherically symmetricproblem by

S0 = 2π2

∫ ∞

0

dr r3

[1

2

(dφ

dr

)2

+ U

]. (50.28)

Unlike the previous examples, where only the instanton part contributedto the action, we need to be more careful here. Inside the bubble we arein the true vacuum where we have φ− and U = −ǫ, so only the secondterm in the integral contributes, yielding an action

Strue vacuum = −1

2π2R4ε. (50.29)

Over the thin surface of the bubble we have a contribution from theinstanton part:

Sinstanton = 2π2R3

∫dr

[1

2

(dφ

dr

)2

+ Usym

], (50.30)

where, as advertised, we choose to neglect the difference ε in energybetween the minima and so only need consider the symmetric doublewell potential Usym. We know that the instanton solution takes us from−a to a as r increases through R. By analogy with the particle problem,such an instanton has an action S1 =

∫ a−a dφ (2Usym)

12 . This gives us a

contribution to the action from the instanton of

Sinstanton = 2π2R3S1. (50.31)

Outside the bubble we are in the false vacuum where φ = φ+ and U = 0,which makes no contribution to the action. The total action is thereforegiven by

Stot = −1

2π2R4ε+ 2π2R3S1. (50.32)

We may now extremize the action to find the critical radius Rc at whichthe tunnelling destroys the false vacuum. We vary the total action withrespect to R and find a radius Rc = 3S1/ε. Substituting this back intothe expression for the total action of the bubble we predict a rate oftunnelling to annihilation of

Γ/V ≈ |K|e−Stot , (50.33)

where Stot = 27π2S41/2ε

3.

x

t

O

Bub

blelig

htco

neBub

blewall

|x|=

t Rc

Fig. 50.16 Spacetime diagram show-ing the growth of the bubble. An ob-server at O only knows of the bubblewhen its forward light-cone intersectsthe observer’s world-line. A time Rc/cseconds later the bubble wall intersectsthe observer’s world-line.

Now we swap back into the Minkowski space of our Universe, and askhow the Universe will end. An analytical continuation of these resultsback to Minkowski space shows that the shape of the bubble in Euclidean4-space is the same as that of the bubble in (3+1) dimensions. As Rincreases, the bubble expands, sweeping out a region in spacetime givenby the hyperboloid

|x|2 − (ct)2 = R2c . (50.34)


We might expect that since the initial bubble is created by quantummechanical fluctuations then Rc should be a very small number, perhapsof the order of 1 fm. This would mean, from eqn 50.34, that |x|2 ≈(ct)2 and the bubble expands at almost the speed of light. This, ratherinconveniently, gives an observer effectively no warning of the coming ofthe true vacuum. Looked at on the spacetime diagram in Fig. 50.16 wesee that the observer O doesn’t know about the existence of the bubbleuntil the light from the forward light-cone of the bubble’s outer wallmeets her world-line (see Fig. 50.17). We see that the true vacuum itselfcollides with her world-line a time R/c seconds later, and she ceasesto exist. This time (by assumption, ≈ 1 fm/c seconds) is orders ofmagnitude shorter than the time it takes a neuron to fire. In SidneyColeman’s words she ‘has literally nothing to worry about’.

Fig. 50.17 We have essentially nowarning of the coming of the true vac-uum. However, we can give some ad-vance notice of the imminent end of thisbook.

Chapter summary

• Instantons are kink-like solutions of the equation of motion asso-ciated with critical points in the action and allow tunnelling prob-lems to be treated.

• Instantons can be used to study the fate of the false vacuum ina symmetry broken system, such as the Universe, where the truevacuum state is only a tunnelling event away.

Exercises

(50.1) Verify the expression for the stationary phase ap-proximation in eqn 50.8.

(50.2) Verify the solution of the double well problem ineqns 50.20.

AFurther reading

Many of the ideas in the preceding essay have been takenfrom another...John Berger (1926– )

There are many other excellent books which introduce and describequantum field theory.1 Whether you are after the friendly approachabil- 1See the detailed list below, beginning

on page 470.ity of Mattuck or Zee, or perhaps the elegant but unforgiving grandeurof Weinberg, or something in between, there is something for everyone.We particularly like Peskin and Schroeder, and also Ryder, and either(or both) could serve as a useful complement to this book. Many of thearguments we have used are adapted from these books (Zee and Peskinand Schroeder deserve special mention here) and also from the legendarylecture course given by Sidney Coleman.2 Finally, for those eager to prac- 2These mid-1970s lectures are, at least

at the time of writing, viewable on theHarvard Physics website.

tise, a useful set of solved problems may be found in Radanovic.

Further reading by chapter:

Chapter 1: An accessible introduction to Lagrangian mechanics is givenin Feynman, Leighton and Sands. See Landau and Lifshitz (vol. I) for thefull story. Chapters 2 and 3: The simple harmonic oscillator and occupa-tion numbers are explained in most books on quantum field theory. Thebooks by Aitchison and Hey,3 P. Coleman and Feynman are especially 3Assume volume I of multivolume texts

unless stated otherwise.clear. Chapter 4: Several examples of non-relativistic applications of sec-ond quantization may be found in P. Coleman. Chapter 5: Classical fieldtheory is covered in Maggiore, Ryder and Itzykson and Zuber amongstmany others. Chapter 6: The Klein–Gordon equation and the problemswith its interpretation are explored in Aitchison and Hey. Chapter 7:The Lagrangians we describe are all examined in most books on quantumfield theory. Ryder, Huang, and Zee are good sources of information.Chapter 8: Time evolution operators and related issues are discussed inmost books on advanced quantum mechanics (Ziman, for example) andin the lectures by S. Coleman. Chapter 9: Transformations are describedvery clearly in Ryder. For a more sophisticated treatment see Weinberg.Chapter 10: An in-depth discussion of Noether’s theorem may be foundin Neuenschwander. Both Ryder and Huang provide accessible introduc-tions. Chapter 11: Our treatment of canonical quantization follows S.Coleman’s lectures. Aitchison and Hey, Bjorken and Drell4 (RQF) and

4We abbreviate Bjorken and Drell’sRelativistic Quantum Fields as hereRQF to avoid confusion with their Rel-ativistic Quantum Mechanics (RQM).

Schiff are other sources of clear information. Chapter 12: The propertiesof the complex scalar field are examined in nearly all books on quan-

468 Further reading

tum field theory [see Aitchison and Hey or Bjorken and Drell (RQF)].Our discussion of its non-relativistic properties follows Zee. Chapter 13:Internal symmetries and vector fields are discussed in Ryder, Zee andin Aitchison and Hey. Chapter 14: Gauge theory and its consequencesare very clearly described in Aitchison and Hey who also give a goodexposition of the difficulties inherent in the canonical quantization of theelectromagnetic field. The solution to these difficulties is discussed inPeskin and Schroeder. Chapter 15: Charge conjugation, parity and timeinversion are described in Aitchison and Hey and in Bjorken and Drell(RQF). A more sophisticated account is Weinberg. Relevant backgroundon the mathematics may be found in the books by Georgi and Nakahara.Chapter 16: Green’s functions are introduced in Mattuck in a similarway to that described here. An approach to perturbation theory basedon propagators may be found in Schiff. Classical Green’s functions aredescribed in Barton. Chapter 17: The Feynman propagator is discussedin Aitchison and Hey. The non-relativistic version is discussed in Mat-tuck. Chapter 18: Useful information on the S-matrix may be found inAitchison and Hey, and also Peskin and Schroeder. Weinberg providesthe philosophy. Chapter 19: The expansion of the S-matrix is carriedout for a simple toy theory in Aitchison and Hey, while the φ4 theory isdiscussed in Peskin and Schroeder and in Ryder. Chapter 20: Scatter-ing is covered by most books on quantum field theory and on particlephysics. An introduction to scattering for particle physics applicationsis given in Griffiths. Clear treatments from the field theory perspectiveappear in Peskin and Schroeder (who discuss ψ†ψφ theory) and also inWeinberg. Chapter 21: Statistical physics is described in Blundell andBlundell and at a more advanced level in Chaikin and Lubensky. Linearresponse theory is described in Binney, Newman, Dowrick and Fisher.The analogy between field theory and magnetism is described in moredetail in Peskin and Schroeder. See also Zinn-Justin. Chapter 22: Gen-erating functionals are described in Ryder, Peskin and Schroeder, andin Binney, Newman Dowrick and Fisher. Chapter 23: The path integralis described in Zee and Ryder. A different approach is described byits inventor in Feynman and Hibbs. Chapter 24: Functional integralsare clearly described in Zee, Peskin and Schroeder and in S. Coleman.Chapter 25: The Wick rotation is discussed in Peskin and Schroederand Altland and Simons. Kapusta and Gale is a good book on theapplications of statistical field theory and Mahan describes how thesetechniques are used in solid state physics. Chapter 26: Broken symme-try is discussed in Blundell and in more detail in Anderson. A lively,advanced, field-centred treatment may be found in S. Coleman. Chap-ter 27: Coherent states are introduced in Annett and in Loudon. Thehistory of the phase operator problem is described in the paper by Nieto,arXiv:hep-th/9304036v1. Altland and Simons and P. Coleman describecoherent states for field applications. Chapter 28: Grassmann variablesare treated in Chapter 44 of Srednicki, and in great detail by Negele andOrland. Chapter 29: Zee is a very clear source on topological effects andwe have followed his treatment in our presentation. Topological objects

469

are described in Zee and Ryder, with more advanced treatments in S.Coleman, Weinberg, Nakahara, and in Altland and Simons. Chapter 30:Our description of topological field theory follows Zee and the notes byDunne (arXiv:hep-th/9902115v1). It is examined in more advanced formin Wen. Chapter 31: Quasiparticles are described in Mattuck and P.Coleman. The Landau Fermi liquid is introduced in P. Coleman. Chap-ter 32: The philosophy of renormalization is made clear in Peskin andSchroeder and in Weinberg. Chapter 33: The use of renormalization inperturbation theory is stressed in Peskin and Schroeder. Chapter 34:Our presentation of the philosophy of the renormalization group is sim-ilar to that of Zee and of Peskin and Schroeder. Good sources on theuse of the method may be found in Altland and Simons, Peskin andSchroeder, Binney, Newman, Dowrick and Fisher and McComb. Chap-ter 35: More detail on ferromagnetism and the renormalization groupmay be found in Altland and Simons. Chapter 36: There are many dif-ferent approaches to introducing the Dirac equation. Aitchison and Heyand Ryder are good places to start. A more modern approach is foundin Maggiore. We have followed the approach in Penrose. Chapter 37:The transformation of spinors is discussed in Ryder and Maggiore andin more detail in Weinberg. Chapter 38: The quantum mechanics offermion fields is introduced very clearly in Peskin and Schroeder andin Bjorken and Drell (RQM and RQF). Chapter 39: Simple examplesfrom QED feature in Aitchison and Hey, Mandl and Shaw and Peskinand Schroeder. The history of the subject is explored in Schweber.Chapter 40: The examples of scattering in QED covered in this chapterare more fully explained in Peskin and Schroeder. See also Berestet-skii, Lifshitz and Pitaevskii for lots more examples. Chapter 41: Therenormalization of QED is covered clearly and in detail in Peskin andSchroeder. Chapter 42: Bolgoliubov’s treatment is described in Ziman.The Lagrangian treatment of a superfluid may be found in Zee and inmore detail in Wen. See also Fulde. Chapter 43: An introduction tothe field theory of metals may be found in P. Coleman. See the detailedtome by Mahan for the full story and also Giuliani and Vignale for morebackground. Chapter 44: Our treatment of superconductivity in termsof second quantized operators follows Annett. The Lagrangian formu-lation is described by Wen. See also Weinberg (vol. II) and Aitchisonand Hey (vol. II). Chapter 45: The field theory of the fractional quan-tum Hall effect is described straightforwardly in Zee and in more detailin Ezawa and Wen. Chapters 46 and 47: Non-abelian gauge theoryand the Weinberg–Salam model are described in Ryder, in Peskin andSchroeder and, of course, in Weinberg (vol. II). An approach motivatedby particle physics phenomenology is presented in Aitchison and Hey(vol. II). Chapter 48: Majorana fermions are described in Aitchison andHey (vol. II) and in a review by Pal (arXiv:1006.1718v2). Chapter 49:Magnetic monopoles are described in Ryder, Zee and by S. Coleman.Chapter 50: Our treatment of instantons is based very closely on thelecture in the collection by S. Coleman. Appendix B: Useful backgroundmay be found in Penrose. Our treatment follows that of Boas.

Further reading

470 Further reading

Bibliography

• A. A. Abrikosov, L. P. Gorkov and I. E. Dzyaloshinski, Methods of quantum

field theory in statistical physics, Dover, New York (1963).

• I. J. R. Aitchison and A. J. G. Hey, Gauge theory in particle physics, vol. I,3rd edition, IOP, Bristol (2003).

• I. J. R. Aitchison and A. J. G. Hey, Gauge theory in particle physics, vol. II,3rd edition, Taylor and Francis, New York (2004).

• A. Altland and B. D. Simons, Condensed matter field theory, CUP, Cambridge(2006).

• P. W. Anderson, Basic notions of condensed matter physics, Benjamin-Cumm-ings, Menlo Park (1984).

• J. F. Annett, Superconductivity, superfluids and condensates, OUP, Oxford(2004).

• A. Auerbach, Interacting electrons and quantum magnetism, Springer-Verlag,New York (1994).

• T. Banks, Modern quantum field theory, CUP, Cambridge (2008).

• G. Barton, Elements of Green’s function and propagation, OUP, Oxford(1989).

• V. B. Berestetskii, E. M. Lifshitz and L. P. Pitaevskii, Quantum electrodynam-

ics, 2nd edition, (volume IV of Landau and Lifshitz), Butterworth-Heinemann,Oxford (1982).

• J. J. Binney, N. J. Dowrick, A. J. Fisher and M. E. J. Newman, The theory

of critical phenomena, OUP, Oxford (1992).

• J. D. Bjorken and S. Drell, Relativistic quantum mechanics, Dover, New York(2012).

• J. D. Bjorken and S. Drell, Relativistic quantum fields, Dover, New York(2012).

• S. J. Blundell, Magnetism in condensed matter, OUP, Oxford (2001).

• S. J. Blundell and K. M. Blundell, Concepts in thermal physics, 2nd edition,OUP, Oxford (2010).

• M. L. Boas, Mathematical methods in the physical sciences, 2nd edition, Wiley,New York (1983).

• P. M. Chaikin and T. C. Lubensky, Principles of condensed matter physics,CUP, Cambridge (1995).

• P. Coleman, Introduction to many body physics, CUP, Cambridge, to be pub-lished. (See also http://www.physics.rutgers.edu/∼coleman/.)

• S. Coleman, Aspects of symmetry, CUP, Cambridge (1985).

• S. Coleman, Physics 253: Quantum field theory: lectures by Sidney Coleman.Lectures from 1975-1976 available to stream fromhttp://www.physics.harvard.edu/events/videos/Phys253.

• S. Doniach and E. H. Sondheimer, Green’s functions for solid state physicists,Imperial College Press, London (1998).

• Z. F. Ezawa, Quantum Hall effects, World Scientific, Singapore (2008).

• R. P. Feynman, Statistical mechanics, Westview, Boulder (1998).

• R. P. Feynman and A. R. Hibbs, Quantum mechanics and path integrals,Emended edition, Dover, New York (2005).

• R. P. Feynman, R. B. Leighton and M. L. Sands, Lectures in physics, vol. II,Addison-Wesley, Reading (1963).

• P. Fulde, Correlated electrons in quantum matter, World Scientific, Singapore(2012).

471

• H. Georgi, Lie algebras in particle physics, 2nd edition, Westview, Boulder(1999).

• G. F. Giuliani and G. Vignale, Quantum theory of the electron liquid, CUP,Cambridge (2005).

• D. J. Griffiths, Introduction to elementary particles, 2nd edition, Wiley VHC,Weinheim (2008).

• F. Halzen and A. D. Martin, Quarks and leptons, John Wiley and Sons, Hobo-ken (1984).

• K. Huang, Quantum field theory, Wiley VCH, Weinheim (2010).

• C. Itzykson and J.-B. Zuber, Quantum field theory, Dover, New York (1980).

• J. I. Kapusta and C. Gale, Finite-temperature field theory, CUP, Cambridge(2006).

• M. Kaku, Quantum field theory, OUP, Oxford (1993).

• L. P. Landau and E. M. Lifshitz, Mechanics (volume I of Landau and Lifshitz),Pergamon, Oxford (1976).

• E. M. Lifshitz and L. P. Pitaevskii, Statistical Physics, part 2 (volume IX ofLandau and Lifshitz), Pergamon, Oxford (1980).

• R. Loudon, The quantum theory of light, OUP, Oxford (2000).

• M. Maggiore, A modern introduction to quantum field theory, OUP, Oxford(2005).

• G. D. Mahan, Many-particle physics, Plenum, New York (1990).

• F. Mandl and G. Shaw, Quantum field theory, 2nd edition, Wiley, Chichester(2010).

• R. D. Mattuck, A guide to Feynman diagrams in the many-body problem,Dover, New York (1967).

• W. D. McComb, Renormalization methods, OUP, Oxford (2004).

• V. F. Mukhanov and S. Winitzki, Introduction to quantum effects in gravity,CUP, Cambridge (2007).

• M. Nakahara, Geometry, topology and physics, Adam Hilger, Bristol (1990).

• D. E. Neuenschwander, Emmy Noether’s wonderful theorem, Johns Hopkins,Baltimore (2011).

• J. W. Negele and H. Orland, Quantum many-particle systems, Addison Wes-ley, Reading (1988).

• R. Penrose, The road to reality, Vintage, London (2004).

• M. E. Peskin and D. V. Schroeder, An introduction to quantum field theory,Westview Press, Boulder (1995).

• V. Radovanovic, Problem book in quantum field theory, 2nd edition, Springer,Berlin (2008).

• L. H. Ryder, Quantum field theory, CUP, Cambridge (1985).

• J. J. Sakurai, Modern quantum mechanics, revised edition, Addison-Wesley,Reading (1994).

• L. I. Schiff, Quantum mechanics, 3rd edition, McGraw-Hill, New York (1968).

• S. S. Schweber, QED and the men who made it, Princeton University Press,New Jersey (1994).

• M. Srednicki, Quantum field theory, CUP, Cambridge (2007).

• R. Ticciati, Quantum field theory for mathematicians, CUP, Cambridge(1999).

• A. M. Tvselik, Quantum field theory in condensed matter physics, CUP, Cam-bridge (1995).

Bibliography

472 Further reading

• S. Weinberg, The quantum theory of fields, vol. I, CUP, Cambridge (1995).

• S. Weinberg, The quantum theory of fields, vol. II, CUP, Cambridge (1996).

• X.-G. Wen, Quantum field theory of many-body systems, OUP, Oxford (2004).

• A. Zee, Quantum field theory in a nutshell, CUP, Cambridge (2003).

• J. M. Ziman, Elements of advanced quantum mechanics, CUP, Cambridge(1969).

• J. Zinn-Justin, Quantum field theory and critical phenomena, OUP, Oxford(1989).

BUseful complex analysis

B.1 What is an analytic function?473

B.2 What is a pole? 474

B.3 How to find a residue 474

B.4 Three rules of contour inte-grals 475

B.5 What is a branch cut? 477

B.6 The principal value of an in-tegral 478

For every complex problem there is an answer that is clear,simple, and wrong.H. L. Mencken (1880–1956)

Throughout the book we have tried to keep the amount of complexanalysis to a minimum. This appendix provides a simple guide to someof the complex analysis commonly employed in quantum field theory.The guide is illustrated by examples drawn from the subject, includingthe most important function of a complex variable in quantum fieldtheory: the propagator.

B.1 What is an analytic function?

If a function is analytic in a region close to a point z, then it has aderivative at every point in that region. We define the derivative of acomplex number as

f ′(z) =df

dz= lim

∆z→0

∆f(z + ∆z) − f(z)

∆z. (B.1)

Importantly, for the function to be analytic, the derivative shouldn’tdepend on the way the interval in the complex plane ∆z is selected.

Example B.1

The function f(z) = z2 is analytic; g(z) = |z|2 is not. To see the first write

f ′(z) = lim∆z→0

(z + ∆z)2 − z2

∆z

= lim∆z→0

z2 + 2z + (∆z)2 − z2

∆z= 2z, (B.2)

just as for a normal derivative. Note that we didn’t have a choice of ∆z, this procedureworks whatever we choose.

On the other hand for g(z) = |z|2 we have

g′(z) = lim∆z→0

|z + ∆z|2 − |z2|∆z

. (B.3)

If ∆z = i∆y (with y real) then you will get a different derivative to the case of∆z = ∆x, with x real. See Boas for more details.

474 Useful complex analysis

B.2 What is a pole?

A pole is a type of singularity1 that behaves like the singularity of 1/zn1By singularity, we mean a point wherea mathematical object is not defined. at z = 0. Let f(z) be analytic between two circles C1 and C2. In the

region between them we can write f(z) as a so-called Laurent seriesexpanded about a point z0:

f(z) = a0+a1(z−z0)+a2(z−z0)2+. . .+b1

(z − z0)+

b2(z − z0)2

+. . . (B.4)

The part with b coefficients is known as the principal part of the se-ries. Don’t confuse this with the principal value of an integral, which isdifferent, and discussed below.

A result and some definitions:

• If all b’s are zero then f(z0) is analytic at z = z0.

• If all b’s after bn are zero then we say that we have a pole of ordern at z = z0. If n = 1 we have a simple pole.

• The coefficient b1 is called the residue of f(z) at z = z0.

An important example is the function

f(z) =α

z − β, (B.5)

which has residue b1 = α and all other ai and bi zero. This function hasa simple pole at z = β.

Example B.2

Let’s examine the pole structure of two of our propagators. The non-relativistic,

(a) E

Ep

−ε

(b) p0

Ep−Ep

−εε

Fig. B.1 (a) Position of the pole in

the complex E plane for G+0 (E). (b)

Positions of the poles in the complexp0 plane for the Feynman propagatorfor free scalar fields.

retarded, free electron propagator is given by

G+0 (E) =

i

E − Ep + iǫ. (B.6)

This has a first-order pole at Ep − iǫ. This is shown in Fig. B.1(a).The Feynman propagator for the free scalar field is usually written

∆(p) =i

p2 −m2 + iǫ; (B.7)

this is helpfully rewritten (see Chapter 17) as a function of the complex variable p0:

∆(p) =1

2Ep

»i

(p0) − Ep + iǫ− i

(p0) + Ep − iǫ

–

. (B.8)

The first (particle) part has a simple pole at p0 = Ep − iǫ. The second (antiparticle)part has a simple pole at −Ep + iǫ. This is shown in Fig. B.1(b).

B.3 How to find a residue

You can find a residue R(z0) at the pole z0 by writing a Laurent series.There are more direct methods too. For example, when we have a simplepole, we can write

R(z0) = limz→z0

(z − z0)f(z). (B.9)

B.4 Three rules of contour integrals 475

Example B.3

The residue of G(E) = iZE−Ep+iǫ

at the simple pole E = Ep − iǫ is given by

R(Ep − iǫ) = limE→Ep−iǫ

(E − Ep + iǫ)iZ

E − Ep + iǫ= iZ. (B.10)

B.4 Three rules of contour integrals

A contour C is a closed path in the complex plane with a finite numberof corners which doesn’t cross itself. Integrals around such contours havea number of useful properties.

Example B.4

We can get some practice with a contour integral by calculatingI

Cdz z2, (B.11)

where the contour is shown in Fig. B.2. We split the contour into two parts and startwith the straight line along the real axis. Take z = reiθ and this part becomes

(straight line) =

Z 1

r=−1dr r2 =

»r3

3

–1

−1

=2

3. (B.12)

Now for the semicircle, described by z = r0eiθ, where r0 = 1. We have dz = ir0eiθdθgiving

(semicircle) =

Z π

θ=0dθ ir30e3iθ =

»e3iθ

3

–π

0

= −2

3. (B.13)

Adding the contributions we have

(line) + (semicircle) = 0. (B.14)

z

1−1

Fig. B.2 An example of a contour inthe complex z plane.

There are three useful theorems for evaluating integrals taken aroundcontours. The first is Cauchy’s theorem:

Augustin-Louis Cauchy (1789–1857)

If f(z) is analytic on and inside C then∮

C

dz f(z) = 0. (B.15)

This is good news, since it says that if the region in a contour containsno poles then the integral gives zero. It also explains why the previousexample gives zero: z2 is analytic on and inside the contour.

The second theorem is known as Cauchy’s integral formula:


If f(z) is analytic on and inside a simple closed curve C, and the pointa is inside C, then the value of f(a) is given by

f(a) =1

2πi

∮

C

dzf(z)

z − a. (B.16)

The third is the residue theorem:If f(z) has singularities at points zi, then, for a closed curve enclosingthese points we have

∮

C

dz f(z) = 2πi∑

i

(Residue at f(zi)

inside C

), (B.17)

where the integral around C is performed in the anticlockwise direction.(You merely change the sign of the answer if you perform the integralin the clockwise direction.)

Often we want to do difficult integrals over real variables. These maybe turned into easier integrals if we form a contour in the complex planewhich includes the original domain of integration and use the rules givenabove. The art is in choosing the best contour to do the integral.

Example B.5

We can use Cauchy’s theorem along with the residue theorem to justify some of themore seemingly cavalier tricks employed in the discussion of propagators in Chap-ters 16 and 17. Let’s find the inverse Fourier transform of the retarded propagatorG+

0 (E), given by

G+0 (t− t′) =

Z ∞

−∞

dE

2π

ie−iE(t−t′)

E − Ep + iǫ, (B.18)

for which the integration path is along the real axis. To use our contour integral ruleswe must complete the contour by joining up this path with a further section of pathwhich will either be in the upper half of the complex E plane or in the lower half.

(a) E

(b) E

Fig. B.3 (a) The contour completed inthe lower half-plane enclosing the pole.Note that the direction here is clock-wise, earning us an extra minus sign.(b) The contour completed in the up-per half-plane. No poles are enclosedso the answer is given by Cauchy’s the-orem.

Suppose we take it in the lower half-plane. Then, as we take the limits along thereal axis to ±∞ the semicircular path gets larger and larger. This will make a large,negative imaginary contribution to E. Let’s call it −i|η|. The exponential will then

involve a contribution e−|η|(t−t′). If (t − t′) is positive then this contribution getssmaller, eventually vanishing as the contour becomes infinitely large.2 We conclude

2This disappearance of semicircularcontours in the limit of infinite radiusis a result of Jordan’s lemma. Thissays that the integral

R∞−∞ dz f(z)eiaz

along the infinite upper semicircle iszero, provided (i) a > 0, and (ii) f(z)is a well-behaved function satisfyinglimR→∞

˛˛f(Reiθ)

˛˛ = 0.

that, for the case t− t′ > 0, the integral above is equivalent to the contour shown inFig. B.3(a).

Let’s do that integral. The contour contains the pole at E = Ep − iǫ so we usethe residue theorem to say

I

C

dE

2π

ie−iE(t−t′)

E − Ep + iǫ= −2πi

„Residue atE = Ep − iǫ

«

, (B.19)

where the minus sign follows from our attempt to take the integral in the clockwisedirection. The residue at the pole is ie−iEp(t−t′)e−ǫ(t−t

′) and the answer is

G+(t− t′) = e−iEp(t−t′)e−ǫ(t−t′), (B.20)

which we stress applied for (t− t′) > 0.

B.5 What is a branch cut? 477

What if we tried to complete the contour in the upper half-plane? Then we wouldhave obtained a large, positive, imaginary contribution to the exponential resultingin a contribution e|η|(t−t

′) which blows up for t − t′ > 0. Such a badly behavedintegral is certainly not suitable for evaluating the Fourier transform. However, fort − t′ < 0 the semicircular contour has a vanishing contribution at infinity and weagain have the equivalence of the Fourier transform and the contour C′ shown inFig. B.3(b). Notice that C′ contains no poles, so Cauchy’s theorem says that theintegral is zero.

We conclude that G+(t− t′) = 0 for t− t′ < 0 and noting that ǫ is an infinitesimal

quantity, we may replace both (i) zero for t − t′ < 0 and (ii) eǫ(t−t′) for t − t′ > 0

with θ(t− t′) and conclude

G+(t− t′) = θ(t− t′)e−iEp(t−t′), (B.21)

just as we had in Chapter 16 without the need for adding damping factors by hand!

The meaning of the iǫ factors now becomes clear. These infinitesimalsposition the poles of the propagators in such a way as to ensure thecorrect causality relationships. Returning to the scalar field propagatorof Fig. B.1 we see that closing the contour in the lower half-plane picksup the positive energy pole, leading to a factor θ(x0 − y0), while closingthe contour in the upper half-plane picks up the negative energy poleand leads to the factor θ(y0 − x0). This motivates the definition of theFeynman propagator in Chapter 17.

B.5 What is a branch cut?

The function eiθ is multivalued. We are always at liberty to add wholenumbers of 2π to the argument in the exponential, i.e. eiθ = ei(θ+2πn)

where n is an integer. The fact that this function is multivalued meansthat care must be taken when taking roots and logarithms.

Example B.6

For the case of the logarithm we have, taking z = reiθ:

ln z = ln r + iθ. (B.22)

Clearly, for fixed r, ln z takes different values for θ and θ + 2πn, even though bothchoices correspond to the same z.

We therefore agree that we should only consider angles in some intervalin θ of size 2π, known as a branch of the function. In order to make thisclear in the complex plane, we can define lines that we will agree notto cross with any of our operations. These are known as branch cuts.The points from which these emerge are known as branch points.

For ln z the branch point is the origin and the branch cut may be takenalong the positive real axis as shown in Fig. B.4(a). It may also be takenalong the negative real axis, or indeed any convenient line. Crossing thebranch cut makes the function jump by 2πi.

(a) x

(b) |p|

Fig. B.4 (a) The branch cut alongthe positive, real axis. (b) The com-plex plane for eqn. B.23.


Recall from Chapter 8 the integral

−i

(2π)2|x|

∫ ∞

−∞d|p| |p|ei|p||x|e−it

√|p|2+m2

, (B.23)

which we consider in the complex |p| plane. The square root in thisequation

√p2 +m2 must be restricted to a single branch. The square

root vanishes for |p| = ±im. which are therefore the branch points.For convenience, we take the branch cuts to extend along the imaginaryaxis as shown in Fig. B.4(b). Notice that when we do the integral inChapter 8 we can’t cross the cuts with our contour, so we must directthe contour around the cuts.

Example B.7

Another occasion where we must consider a function with a branch cut is the fullpropagator discussed in Chapter 31, given by

G(p) =iZ

p2 −m2 + iǫ+

Z ∞

≈4m2

dM2

2πρ(M2)

i

p2 −M2 + iǫ, (B.24)

which has a pole structure shown in Fig. B.5. The second term in the expression tellsus to expect multiparticle contributions. These lead to a line of poles with infinites-imal separation between them. Such a line of poles is another way of describing abranch cut and so we draw a cut extending along the real axis from the branch pointgiven by the two-particle production threshold (p0)2 ≈ 4m2.

(p0)2

m2 4m2

Fig. B.5 The pole structure of the fullpropagator in eqn. B.24.

B.6 The principal value of an integral

The Cauchy principle value is a method of giving improper integralsa value. This is not complex analysis, but is often used in integralsinvolving the propagator. Suppose we want to evaluate the integral∫ ca

dxf(x) but we have the problem that the integrand f(x) diverges at

x = b (where a < b < c) and so both∫ ba

dx f(x) and∫ cb

dx f(x) will alsoblow up.3 In that case we may take the Cauchy principal value of the

3Specifically we require

Z b

adx f(x) = ±∞,

for a < b andZ c

bdx f(x) = ∓∞,

for c > b. (That is, one sign in front ofthe ∞ is plus and one minus.)

integral, denoted by P and defined by

P∫ c

a

dx f(x) = limǫ→0+

[∫ b−ǫ

a

dx f(x) +

∫ c

b+ǫ

dx f(x)

]. (B.25)

This gives the integral an unambiguous value, as demonstrated in theexample below.

Example B.8

The integral

I =

Z 10

0

dx

x− 2, (B.26)

B.6 The principal value of an integral 479

is not well defined as the integrand diverges at x = 2. You can see that the twointegrals

R 20

dxx−2

andR 102

dxx−2

both diverge, giving −∞ and +∞ respectively.To get around this we integrate from 0 up to 2 − ǫ and then from 2 + ǫ up to 10,

thereby cutting out the troublesome part of the problem. We obtain

I1 =R 2−ǫ0

dxx−3

= ln ǫ− ln 2, I2 =R 102+ǫ

dxx−3

= ln 10 − ln ǫ. (B.27)

We find that I = I1 + I2 = ln 5, independent of ǫ. We may therefore take the limitǫ→ 0 and obtain an unambiguous result. We conclude that

PZ 10

0

dx

x− 2= ln 5. (B.28)

The principal value arises in quantum field theory when we want to dointegrals of the form

∫ ba

dx f(x)x+iǫ , where f(x) is a complex-valued function

and a and b are real, obeying a < 0 < b. In this case we use the followingtheorem which says that

limǫ→0+

∫ b

a

dxf(x)

x± iǫ= P

∫ b

a

dxf(x)

x∓ iπf(0). (B.29)

Often we take f(x− x0) = δ(x− x0) and obtain the identity,4 4Sometimes called the Dirac relation inthe physics literature.

1

x0 ± iǫ=

Px0

∓ iπδ(x0), (B.30)

as used in Exercises 22.1 and 31.3.

Index

1-part irreducible (1PI) diagram, 2961PI self-energy, 296

abelian gauge theory, 424abelian group, 80, 424action, 14action, principle of least, 14active transformation, 80adiabatic continuity, 282Aharonov, Yakir, 261Aharonov–Bohm effect, 261, 271–272amputation, 192analytic function, 473Anderson localization, 308, 417Anderson, Philip, 242, 280annihilation operator, 23, 20–23anomalous dimension, 306anticommutator, 33

Dirac fields, 342fermion, 33

antiparticle, 61antiscreening, 307anyons, 267asymptotic freedom, 307axial gauge, 130axial vector, 136

Bardeen, John, 401BCS theory, 400, 400–406BCS wave function, 402–403Berezinskii, Vadim L., 309Bethe, Hans, 364Bhabha scattering, 359Bogoliubov transformation

in a superconductor, 405in a superfluid, 372

Bogoliubov, Nikolay, 371bogolon, 372

in a superconductor, 405in a superfluid, 374

Bohm, David, 261Born approximation, 363Born, Max, 194Bose, S. N., 28Bose–Einstein condensate, 379boson, 32branch cut, 477broken symmetry, 237, 237–245

false vacuum, 463for continuous symmetries, 240

Higgs field, 438in a gauge theory, 242Lagrangian formulation, 239non-abelian gauge theory, 430SO(2) symmetry, 240superconductor, 406superfluid, 374U(1) gauge theory, 242U(1) symmetry, 242

Brout, Robert, 242Brueckner, Keith, 385

canonical momentum, 50canonical quantization, 98, 98–108

fermion, 341for a complex scalar field, 109–111for a scalar field, 108

Carruthers, Peter, 251Casimir effect, 105Casimir, Hendrik, 105Cauchy principle value, 478Cauchy’s integral formula, 475Cauchy’s theorem, 475Cauchy, Augustin-Louis, 475charge

conserved, 94electric, 90in electromagnetism, 52Noether, 94renormalization, 364

charge conjugation, 135Chern, Shiing-Shen, 269Chern–Simons theory, 269, 269–272, 418chiral representation, 324chirality, 325Clifford algebra, 323Clifford, William, 323cluster decomposition principle, 181coarse graining, 304coherent states, 247, 247–253

fermion, 257harmonic oscillator, 247superfluid, 374

Coleman, Sidney, 126, 190, 244, 457Coleman–Mermin–Wagner theorem, 244collective excitations, 279complex scalar field, 55, 68

canonical quantization, 109–111Noether current, 111non-relativistic limit, 112

Compton wavelength, 76Compton, Arthur, 357condensed matter physics, 369, 369–422conjugate momentum, 56connected correlation function, 198, 313conservation law

connection with invariance, 91conserved charge, 94continuity equation, 54, 61continuum limit, 50contour integration, 475contraction, 172contravariant vector, 4Cooper pair, 401Cooper, Leon, 401correlation energy, 385correlation function, 198

connected, 234correlation length, 313Coulomb gauge, 130counterterms, 287, 287–294, 297–298

in QED, 360covariant, 3covariant derivative, 128covariant vector, 4CPT theorem, 139creation operator, 23, 20–23, 102critical exponents, 314, 319critical phenomena, 313cross-section, scattering, 193crossing symmetry, 358current

charged, 436in a superconductor, 407in a superfluid, 377in electromagnetism, 54in QED, 346neutral, 436Noether, see also Noether currentprobability, 61

cut-off, momentum space, 286–288,302–307

cutlines, 296

d’Alembert, Jean le Rond, 5d’Alembertian operator, 5Debye, Peter, 401Derrick’s theorem, 265determinant, 40differential cross-section, 193

Index 481

Dirac equation, 323, 322–334four components of, 339–340non-relativistic limit, 332

Dirac flux quantum, 411Dirac monopole, 451Dirac spinor, 327Dirac string, 452Dirac’s quantization condition, 453Dirac, Paul, 322dispersion

for weakly interacting bosons, 373in superfluid helium, 370massive scalar field, 66massless scalar field, 65of a metal, 397

divergence, 285logarithmic, 286superficial degree of, 292

domains, 263double cover, 141dressed particle, 275Dyson’s equation, 150Dyson’s expansion, 170Dyson, Freeman, 170

effective coupling constant, 404effective field theory, 294, 418Einstein summation convention, 4Einstein’s field equations, 95electromagnetic field tensor, 54electromagnetism, 7, 52, 52–57, 97, 125,

129Chern–Simons, 272gauge theory, 129massive, 120–123

electroweak theory, 433, 433–443elementary excitations, 279, 418emergence, 418energy-momentum tensor, 95, 97Englert, Francois, 242equal-time commutator, 99Euclid, 229Euclidean space, 229, 458Euler, Leonhard, 15Euler–Lagrange equation, 15, 91, 93

four-vector version, 16exchange symmetry, 28

fate of the false vacuum, 457Fermat’s principle, 10, 17Fermat, Pierre de, 10Fermi liquid theory, 281Fermi’s weak interaction theory, 293Fermi, Enrico, 28fermion, 32, 321–348

canonical quantization, 341equation of motion, 322Feynman rules, 345generating functional, 343Majorana, 444–450

Noether current, 341non-relativistic, 332, 380origin of mass, 439path integral, 343propagator, 343scattering, 345

ferromagnet, 237–238renormalization group analysis,

313–320Feynman diagram, 175

amputated, 192connected, 181, 192disconnected, 181double oyster, 382double tadpole, 382dumbbell, 208external line, 181general rules for drawing, 180internal line, 181link with generating functional, 206link with Green’s functions, 204–206oyster, 190, 387pair-bubble, 393reason for drawing, 182Saturn, 183, 288symmetry factors, 182–183tadpole, 190, 387two-particle scattering, 161vacuum diagram, 181

Feynman propagator, 156, 156–158, 173Feynman rules

fermion, 345metal, 390φ4 theory, 182, 185ψ†ψφ theory, 191QED, 350renormalized φ4 theory, 289statistical field theory, 235

Feynman, Richard, 61, 175field, 2, 76–77

complex scalar, 68, 109massive scalar, 65–67massive vector, 120–123massless scalar, 64–65quantum, 19

field operator, 37, 37–38, 98filling factor, 416first quantization, 19fixed point, 305flow equation, 304Fock space, 39Fock, Vladimir, 381four-point function, 162four-vector, 3four-vector inner product, 5Fourier transform, 6–7, 25Fourier, Joseph, 6Frolich, Herbert, 400fractional quantum Hall effect, 417,

417–421

fractional statistics, 267, 271, 420free propagator, 157Friedel oscillations, 396function, 11functional, 11, 11–13functional differentiation, 12functional integral, 212, see also path

integral, 221–226field version, 221scalar field, 223

fundamental group, 261

g-factorg = 2 from Dirac equation, 333QED correction, 365–368

γ matrices, 322–325Majorana basis, 445

gauge, 126axial, 130common gauges, 130Coulomb, 130Lorenz, 130unitary, 430Weyl, 130

gauge field, 128SO(3), 429SU(2), 428SU(2) ⊗ U(1), 434U(1), 128

gauge invariance, 126, 129QED, 351Ward identity, 351

gauge principle, 131gauge theory, 128, 126–134U(1), 346abelian, 424fermion, 346in a superconductor, 407non-abelian, 424–431SO(3), 430, 453SU(2), 425–428SU(2) ⊗ U(1), 434U(1), 351, 407, 424

gauge transformation, 126Gaussian integral, 213Gaussian model, 316Gell-Mann, Murray, 204Gell-Mann–Low equation, 304Gell-Mann–Low theorem, 204Gell-Mann–Nishijima relation, 434general linear group, 142generating function, 196generating functional, 198, 201, 201–207

fermion, 343from functional integral, 222–226in statistical field theory, 234link with Feynman diagrams, 206scalar field theory, 223

generator, 81Gibbs distribution, 196

482 Index

Ginzburg, Vitaly, 315Glauber, Roy, 247global symmetry, 90global transformation, 127Goldstone boson, 242Goldstone’s theorem, 242, 246Goldstone, Jeffrey, 242Gordon identity, 335, 366Gordon, Walter, 60Grassmann numbers, 255Grassmann, Hermann, 255Green’s function

two-point, 162Green’s function, 144, 144–153

from Feynman diagrams, 204–206from generating functional, 201–203link with S-matrix, 203–204

Green, George, 144group, 80

abelian, 80cyclic, 136Lie, 80representation of, 84table of groups, 142

guiding centre, 413Guralnik, Gerald, 242

Hagan, Carl, 242Hamilton’s equations, 51Hamilton’s principle of least action, 14Hamilton, W. R., 11Hamiltonian, 51, 50–59

BCS, 401electromagnetism, 57for a metal, 380for a superfluid, 370Hubbard, 47masses coupled by springs, 25particles a magnetic field, 412simple harmonic oscillator, 20tight-binding, 43–44

Hartree, Douglas, 381Hartree–Fock approximation, 383–388Heaviside, Oliver, 7Heaviside–Lorentz units, 7hedgehog solution, 454Heisenberg equation of motion, 74Heisenberg picture, 74Heisenberg, Werner, 74helicity, 326Hermite polynomial, 20Higgs field, 437Higgs mechanism, 242, 430–431

in a superconductor, 407in electroweak theory, 438

Higgs, Peter, 242Hilbert space, 39Hofstadter butterfly, 413hole, 388, 449Hubbard model, 46

Hubbard, John, 46

imaginary time, 228improper rotation, 139incoming line, 181incoming wave, 59incompressible liquid, 417infrared behaviour, 304insect, giant, 79instanton, 460, 457–466

in quantum field theory, 463in quantum particle mechanics, 458

integer quantum Hall effect, 415–417integration measure, 212interacting theory, 108, 154, 162interaction, 165

Coulomb, 176, 380electron–phonon, 400electron-electron, 380electron-phonon, 400electroweak, 441φ4 theory, 176ψ†ψφ theory, 176

interaction representation, 167, 167–168internal degree of freedom, 67internal line, 181internal symmetry, 111, 117, 117–120invariance

connection with conservation laws, 91invariant, 81, 90invariant amplitude, 192inversion, 136irrelevant variable, 316isospin, 117, 426

weak, 435

jellium, 384Jordan’s lemma, 476

Kibble, Tom, 242kink, 262Klein, Oskar, 60Klein–Gordon equation, 60, 59–63Kosterlitz, J. Michael, 309Kosterlitz–Thouless transition, 309Kramers doublet, 138Kramers’ theorem, 138Kramers, Hendrik, 138Kronecker delta, 5Kronecker, Leopold, 5

ladder operator, 20–23Lagrange, J. L., 11, 64Lagrangian, 14, 10–18, 50–59, 64–69,

92–94, 99charged superfluid, 407complex scalar field, 68electromagnetism, 125massive scalar field, 65massless scalar field, 64

φ4 theory, 67ψ†ψφ theory, 188QED, 347renormalized, 291SO(3) gauge theory, 430, 454SU(2) gauge theory, 428superfluid, 375Weinberg–Salam model, 438

Lagrangian density, 15, 64–69Lamb shift, 364Lamb, Willis, 364Landau Fermi liquid, 280Landau free energy, 237Landau levels, 414Landau parameters, 283Landau, Lev, 281Landau–Ginzburg model, 315laser, 252Laughlin, Robert, 370, 418least action, principle of, 14left-handed particle, 325Levi-Civita symbol, 6Levi-Civita, Tullio, 6Lie algebra, 84Lie group, 80Lie, Sophus, 80light cone, 75, 101Lindhard function, 395linked-cluster theorem, 206logarithmically divergent, 286London, Fritz, 400Lorentz covariant, 3Lorentz group, 88, 140, 142Lorentz invariance, 90Lorentz invariant, 3Lorentz transformation, 3, 86–88Lorentz, Hendrik, 7, 130Lorentz-invariant measure, 102Lorenz gauge, 130Lorenz Ludvig, 130Low, Francis, 204lowering operator, 22, 20–23

macroscopic wave function, 253magnetic flux quantization, 265magnetic length, 411magnetic monopole, 451, 451–456

Dirac, 451’t Hooft–Polyakov, 453

Majorana condition, 445Majorana equation, 448Majorana fermion, 444–450Majorana, Ettore, 444Mandelbrot, Benoit B., 413Mandelstam variable, 191Mandelstam, Stanley, 191manifold, 269mass

in scalar field theory, 65Majorana, 447

Index 483

origin of electron, 439physical, 290, 297

mass shell condition, 101massive electromagnetism, 120, see also

massive vector fieldmassive scalar field, 65–67massive vector field, 120–123

propagator, 225–226massless scalar field, 64–65Matsubara frequencies, 232Matsubara, Takeo, 232Maxwell’s equations, 7, 57

including magnetic monopoles, 451Maxwell, James Clerk, 7McWhirter, Norris, 90mean-field theory, 237, see also Landau

theory, 315, 380excitations, 386metal, 380

Meissner effect, 408Mermin, N. David, 244metal, 279, 380–398

effective potential, 394excitations, 386, 389Feynman rules, 390permittivity, 394random phase approximation, 393screening, 393–396

metric tensor, 5Mills, Robert, 425minimal coupling prescription, 131

in QED, 346mode expansion, 101, 106–108momentum density, 92, 99monopole, magnetic, 451Mott’s formula, 357Mott, Nevill, 357Møller scattering, 359

n-particle operator, 39Nambu spinor, 409Nambu’s analogy, 410Nambu, Yoichiro, 242negative energy states, 61–63neutrino, 433–443, 449Nishijima, Kazuhiko, 434Noether charge, 94, 111, 125Noether current, 93, 111

fermion, 341SO(3) internal symmetry, 119SU(2) symmetry, 426superfluid, 377U(1) symmetry, 112, 115

Noether’s theorem, 93, 92–97, 125Noether, Emmy, 90non-abelian gauge theory, 424, 424–431

broken symmetry, 430non-abelian group, 424non-interacting theory, 108, 154non-relativistic limit

complex scalar field, 112Dirac equation, 332

normal ordering, 44, 99, 104, 104–105,169, 171

number current, 112number density operator, 42number operator, 21, 105number, I am not a, 28number-phase uncertainty, 115, 251

O(1, 3), 140, 142O(3), 139O(n), 142occupation number representation, 24,

30, 28–37operator

annihilation, 20–23anticommuting, 255charge conjugation, 135chirality, 325creation, 20–23, 102field, 37–38, 99helicity, 326ladder, 20–23lowering, 20–23magnetic translation, 411number density, 42parity, 136, 339projection, 123raising, 20–23representation of, 84rotation, 79single-particle, 39time-evolution, 72–74time-reversal, 137translation, 79two-particle, 44unitary, 73, 80

operator valued field, 99order parameter, 238orthogonal group, 142oyster diagram, 190

parity, 136, 339particle exchange, 267–268particle physics, 423, 423–466partition function, 196, 231passive transformation, 80path integral, see also functional

integral, 210–219fermion, 257, 343

Pauli equation, 332Pauli exclusion principle, 28, 33Pauli, Wolfgang, 33penetration depth, 409periodic boundary conditions, 25

in statistical field theory, 231particle in a box, 28

permanent, 40phase operator, 250

phase transition, 237renormalization group treatment, 313

φ4 theoryferromagnetic transition, 315–319Feynman diagrams, 177Feynman rules, 182, 185Lagrangian, 67S-operator, 177thermodynamics, 232–236

phonon, 26, 25–27photon, 27, 133

in electroweak theory, 440propagator, 348renormalized propagator, 361

physical coupling constant, 291physical mass, 277, 291Pines, David, 393plaquette, 413plasma oscillation, 397Poincare group, 88, 89Poisson bracket, 51Poisson distribution, 248Poisson, Simeon, 51polar vector, 136polarization process, 361, 393polarization vector, 122pole, 474Polyakov, Alexander, 453principal value, 478principle of least time, 10probability current, 61Proca equation, 121Proca, Alexandru, 121product space, 260projection matrices, 124propagator, 144

Euclidean, 230fermion, 343Feynman, 156–158, 173field, 154–155four-point, 162free, 150, 157full, 150in the complex plane, 476massive vector field, 225–226metal, 390n-point, 162non-relativistic fermions, 389notation for, 156photon, 348renormalized, 277, 295–298renormalized photon, 361scalar field theory, 158–159simple harmonic oscillator, 217–219single-particle, 162two-point, 162

proper rotation, 139proper, orthochronous Lorentz subgroup,

140pseudoscalar, 136

484 Index

pseudovector, 136ψ†ψφ theory, 188

Feynman diagrams, 188Feynman rules, 191Lagrangian, 188scattering, 190

pure gauge, 244, 265

QED, 346–353Feynman rules, 350gauge invariance, 351renormalization, 360–368scattering, 349, 355–359

quantizationcanonical, 98–108first, 19second, 19–49

quantum charge, 135quantum field, 19

mode expansion, 106–108transformation of, 83, 85–86

quantum field theory, 1quantum fluctuations, 213quantum Hall effect

fractional, 417–421integer, 415–417

quantum statistics, 28quantum tunnelling, 457–466quasiparticle, 276–280

in a fractional quantum Hall fluid, 420in a metal, 279–280Landau, 281–284

quasiparticle weight, 277

raising operator, 22, 20–23random phase approximation, 393rapidity, 87reduced free energy, 314relativity

general, 95special, 3–6

relevant variable, 316renormalizable theories, 288renormalization, 276, 273–320

electric charge, 364physical coupling constant, 291physical mass, 291, 295propagator, 295–298QED, 360–368vertex, 298–299

renormalization group, 302, 302–312Anderson Localization, 308asymptotic freedom, 307for phase transitions, 313–320Kosterlitz–Thouless transition, 309QED, 364

renormalization group flow, 304renormalized fields, 290representation

Heisenberg, 74

interaction, 167–168occupation number, 28–37Schrodinger, 72–73

representation of a group, 84residue, 474residue theorem, 476resonance, 277response function, 199Ricci scalar, 95Ricci tensor, 95right-handed particle, 325rigidity, 310rotation, 82–83, 91

improper, 139proper, 139

Rutherford, Ernest, 356

s-channel process, 191S-matrix, 166, 165–187, 192

link with Green’s functions, 203–204S1 (circle), 141S2 (2-sphere), 141S3 (3-sphere), 141Sahl, Ibn, 10Salam, Abdus, 433Saturn diagram, 183scalar, 3scalar field, 55

canonical quantization, 108complex, 68free propagator, 158–159generating functional, 223massive, 65–67massless, 64–65

scattering, 186Bhabha, 359Compton, 357fermion, 345Mott formula, 356Møller, 359φ4 theory, 186prototypical experiment, 166ψ†ψφ theory, 190QED, 349, 355–359Rutherford, 355theory, 188–195

scattering cross-section, 193Schrieffer, J. Robert, 401Schrodinger picture, 72Schrodinger, Erwin, 72Schwinger, Julian, 175, 360screening, 363, 429

strong interaction, 307second quantization, 19, 19–49

of operators, 39–42second quantization trick, 25Seitz, Frederick, 385self-energy, 295, 296

metal, 391self-energy diagrams, 183

Simons, James Harris, 269simple harmonic oscillator, 19–28

path integral treatment, 217–219simply connected topological space, 140single-particle density matrix, 48single-particle operator, 39single-particle quantum mechanics

death of, 76singularity, 474SL(2,C), 141Slater determinant, 36Snell’s law, 10, 17Snellius, Willebrord, 10SO(1, 3), 142SO(3), 117, 139–141SO(n), 142SO+(1, 3), 140source current, 66, 201, 222source term, 197special linear group, 142special Lorentz group, 142special orthogonal group, 139, 142special relativity, 3–6special unitary group, 141, 142spectral density function, 278spermion, 190spin, 328spin splitting, 415spinor, 327, 327–330

boosts, 337Dirac, 327rotations, 337Weyl, 327

spinor field, 336stationary phase, 17stationary phase approximation, 458statistical field theory, 228–236statistical mechanics, 196–199string, waves on a, 15strong interaction, 307, 429structure constant, 84Stueckelberg, Ernst, 61stupidity energy, 385SU(2), 141SU(n), 142sum over histories, 211superconductor, 252, 400–409, 449

BCS theory, 400–406broken symmetry, 406current, 408energy gap, 403–405excitations, 405ground state, 402–405Higgs mechanism, 407

superfluid, 252, 370, 370–379broken symmetry, 374charged, 407excitations, 372, 377Landau’s argument, 379Noether current, 377

Index 485

supersymmetry, 449susceptibility, 199symmetry, 90–97

exchange, 28internal, 111

symmetry factors, 182–183

t-channel process, 191T -matrix, 192tadpole diagram, 190tensor, 6Thomas-Fermi wave vector, 396’t Hooft, Gerard, 453’t Hooft–Polyakov monopole, 453Thouless, David, 308tight-binding model, 43–44time reversal, 137time-advanced Green’s function, 146time-evolution operator, 73, 72–74time-ordered product, 169time-retarded Green’s function, 146time-slicing, 211Tomonaga, Sin-Itiro, 348topological charge, 264topological field theory, 267topological object, 260, 453, 457topological space, 140topology, 260transformation

continuous, 91discrete, 135

transition matrix, 192translation

spacetime, 79–82spatial, 90

time, 72–74, 91two-point function, 162two-point Green’s function, 162

U(1) symmetry, 111U(n), 142u-channel process, 191ultraviolet behaviour, 304unitary gauge, 430, 439unitary group, 142unitary matrix, 80unitary operator, 80universality, 313

vacuum diagram, 181vacuum expectation value (VEV), 171,

381vacuum polarization, 363vector, 3vector field, 55vertex

in a Feynman diagram, 181renormalization, 298–299renormalization in QED, 365

vertex function, 295, 298virtual particle, 160vortex, 264

unbinding transition, 309

W± particle, 330, 440Wagner, Herbert, 244Ward identity, 351Ward, John, 351weak hypercharge, 435weak interaction, 330, 433–443

weak isospin, 435Weinberg angle, 441Weinberg, Steven, 433Weinberg–Salam model, 433–443Weyl gauge, 130Weyl representation, 324Weyl spinor, 327Weyl, Hermann, 327Wheeler, John, 166Wick rotation, 229, 458Wick time-ordering symbol, 156Wick’s theorem, 172, 171–173, 184Wick, Gian-Carlo, 156Widom’s hypothesis, 314Wigner, Eugene, 385Wilczek, Frank, 268Wilson, Kenneth, 302winding number, 261, 264Witten, Edward, 269world, end of the, 466

Yang, Chen-Ning, 425Yang–Mills theory, 425, 425–428Yukawa interaction, 188Yukawa particle exchange, 160Yukawa potential, 194

in a metal, 396Yukawa theory, 194, see also ψ†ψφ

theoryYukawa, Hideki, 159

Z0 particle, 440Z2, 136zero-point energy, 20

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