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Relativistic Quantum Fields 2 Mark Hindmarsh University of Sussex [email protected] http://www.pact.cpes.susx.ac.uk/users/markh/Teaching/ Spring Term 2002 Monday 4pm Arundel 1A Tuesday 11:30pm Arundel 1A Friday 11:30am Arundel 1A PDF version of the lecture notes available: http://www.pact.cpes.susx.ac.uk/users/markh/RQF2/rqf2.pdf Aims and Learning Outcomes Aim: to enable 1st year particle physics graduate students to understand quantum field theory as used in current research in particle physics. Learning outcomes: to be able to give an account of the quantisation pro- cedures for fermions and gauge fields; to understand the gauge principle in particle physics; to be able to compute 1-loop Feynman diagrams in renor- malisable field theories; to understand the theortical basis of the Standard Model of particle physics. Skills: students will improve their problem-solving; and gain necessary subject-specific concepts (gauge field theory, renormalisation, symmetry and symmetry-breaking). Syllabus Fermions (Ryder 4.3, 6.7). Dirac equation in Lagrangian formulation; Canonical quantisation; Grassman variables; Fermionic path integral; Feynman rules. 1
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Page 1: Relativistic Quantum Fields 2

Relativistic Quantum Fields 2

Mark HindmarshUniversity of Sussex

[email protected]

http://www.pact.cpes.susx.ac.uk/users/markh/Teaching/

Spring Term 2002Monday 4pm Arundel 1A

Tuesday 11:30pm Arundel 1AFriday 11:30am Arundel 1A

PDF version of the lecture notes available:http://www.pact.cpes.susx.ac.uk/users/markh/RQF2/rqf2.pdf

Aims and Learning Outcomes

Aim: to enable 1st year particle physics graduate students to understandquantum field theory as used in current research in particle physics.

Learning outcomes: to be able to give an account of the quantisation pro-cedures for fermions and gauge fields; to understand the gauge principle inparticle physics; to be able to compute 1-loop Feynman diagrams in renor-malisable field theories; to understand the theortical basis of the StandardModel of particle physics.

Skills: students will improve their problem-solving; and gain necessarysubject-specific concepts (gauge field theory, renormalisation, symmetry andsymmetry-breaking).

Syllabus

• Fermions (Ryder 4.3, 6.7). Dirac equation in Lagrangian formulation;Canonical quantisation; Grassman variables; Fermionic path integral;Feynman rules.

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• Gauge field theory (Ryder 4.4, Cheng & Li Ch 8). Internal symme-tries; Gauge symmetry 1: Abelian; The electromagnetic field; Gaugesymmetry 2: non-Abelian; Non-Abelian gauge field theory; Sponta-neous symmetry breaking 1: Abelian; Spontaneous symmetry breaking2: SU(2).

• Quantum gauge theory (Cheng & Li Ch 9, Ryder Ch7). Path-integralquantisation; Fade’ev-Popov procedure, ghosts; Feynman rules in co-variant gauge.

• Electroweak theory (Cheng & Li Ch 11). 1-family SU(2)×U(1) La-grangian; Higgs mechanism; Mass spectrum; Family replication.

• QCD (Cheng & Li Ch 10). QCD Lagrangian and symmetries; Asymp-totic freedom; Anomalies

Teaching methods

There will be 3 lectures a week throughout the term. Problem sheets willbe given every two weeks.

Assessment

Problem sheets will count for 100% of the total mark for the course, with a25% weighting for a take-home exam at the end of the course.

Reading list

* Gauge theory of elementary particle physics, T.P. Cheng & L.F. Li(O.U.P., Oxford, 1984).

* Quantum Field Theory, L. Ryder (C.U.P., Cambridge, 1984).

An introduction to quantum field theory, Michael E. Peskin, Daniel V.Schroeder (Addison-Wesley, Reading, Mass; 1995).

Quantum field theory: a modern introduction, Michio Kaku (O.U.P.,Oxford, 1993)

Quantum Field Theory, C. Itzykson and J.-B. Zuber (McGraw-Hill,New York, 1980).

There are quite a few errors in Peskin and Schroeder which are supposed tobe corrected in the next printing of the book. Meanwhile, a list can be foundat

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http://www.slac.stanford.edu/~mpeskin/QFT.html

The other books are complementary in some way. Kaku is very completebut rather rushed. Itzykson & Zuber is compendious but lacks a coherentdevelopment of the subject.

Prerequisites

Relativistic Quantum Fields 1, Further Quantum Mechanics.

Course Lecturer

Mark Hindmarsh, CPES, Arundel 213.Telephone: 8934E-mail: [email protected] web: http://www.pact.cpes.susx.ac.uk/users/markh/RQF2/Office hour: Tuesday 2–3pm, or by arrangement (email is best).

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Contents

1 Fermions 61.1 Plane wave solutions . . . . . . . . . . . . . . . . . . . . . . . 71.2 The interpretation of negative energy states . . . . . . . . . . 91.3 Quantising the spinor field . . . . . . . . . . . . . . . . . . . . 101.4 Conserved charge . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Gauge field theory 142.1 Internal symmetries . . . . . . . . . . . . . . . . . . . . . . . . 142.2 Gauge symmetry 1 . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Maxwell’s equations in covariant form . . . . . . . . . . . . . . 192.4 Lagrangian formulation . . . . . . . . . . . . . . . . . . . . . . 212.5 Gauges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.6 Coulomb gauge solutions . . . . . . . . . . . . . . . . . . . . . 232.7 Lorentz gauge solutions . . . . . . . . . . . . . . . . . . . . . . 242.8 Canonical quantisation in the Lorentz gauge . . . . . . . . . . 262.9 Feynman propagator in Lorentz gauge . . . . . . . . . . . . . 28

3 Quantum gauge theory 29

4 Electroweak theory 29

5 QCD 29

6 Renormalisation 29

A Pauli matrices 29

B Dirac matrices 30B.1 Standard representation of Dirac matrices . . . . . . . . . . . 30

C Identities for Dirac matrices 31

D Feynman rules 31

E Real scalar field 33

F Complex scalar field 33

G U(1) gauge field 35

H Spinor field with U(1) gauge symmetry 35

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I Non-Abelian gauge field 37

J Complex scalar field with non-abelian gauge symmetry 40

K Spinor field with non-abelian gauge symmetry 42

L Problem Sheets 44L.1 Problem Sheet 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 44L.2 Problem Sheet 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 46L.3 Problem Sheet 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 49L.4 Problem Sheet 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 52L.5 Problem Sheet 5 . . . . . . . . . . . . . . . . . . . . . . . . . . 54

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1 Fermions

Dirac knew of the problem with the probability interpretation of the Klein-Gordon equation, and traced it to the fact that it was second order in timederivatives. He therefore set out to find a relativistic wave equation withonly one time derivative. The requirement that the form of the equation beunchanged under Lorentz transformations, which mix up ∂/∂t and∇, meansthat the equation must be first order in spatial derivatives as well. HenceDirac proposed a relativistic free particle wave equation

i∂

∂tψ = −iα·∇ψ + βmψ. (1)

There clearly has to be something rather special about the objects α, β andψ in order that Lorentz covariance be preserved. In fact, α and β are 4×4Hermitean matrices, and ψ is an object with four components called a spinor.

Now, we want a wave equation with plane wave solutions which satisfythe relativistic energy-momentum relation E2 = p2 + m2. To see how thisproperty emerges from the Dirac equation, we act on both sides of (1) with(i∂t − iα · ∇), to obtain(

i∂

∂t− iα · ∇

)(i∂

∂t+ iα · ∇

)ψ = m

(i∂

∂t− iα · ∇

)βψ. (2)

At this point, it is more convenient to switch to index notation for the 3-vectors α and ∇, noting that α · ∇ = αi∂i. We expand out the brackets onthe left hand side of (2), and add and subtract −iβαi∂i on the right handside. The result is(− ∂2

∂t2+ αiαj∂i∂j

)ψ = m

[(i∂

∂t− iβαi∂iψ

)+ iβαi∂iψ + iαiβ∂iψ

]. (3)

It is now useful to define the anticommutator, represented by curly brackets:

A,B = AB +BA. (4)

(Sometimes you will also see the anticommutator written as [A,B]+). Werewrite the left hand side as (−∂2

t + 12αiαj∂i∂j + 1

2αjαi∂j∂i)ψ, which we are

entitled to do by renaming the indices on α and ∂, to obtain(− ∂2

∂t2+

1

2αi, αj∂i∂j

)ψ = m2β2ψ + iβ, αi∂iψ. (5)

In order to reproduce the relativistic relation between energy and momentum,we must end up with an equation like the Klein-Gordon equation. If αi andβ satisfy

αi, αj = δij, β, αi = 0, β2 = 1, (6)

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where 1 is the 4×4 identity matrix, ψ satisfies the equation(− ∂2

∂t2+∇2

)ψ = m2ψ. (7)

Hence, each of the four components of ψ satisfy the Klein-Gordon equation.We can make the Dirac equation more explicitly relativistic by defining

four new 4×4 matrices:

γ0 = β, γi = βαi. (8)

If we multiply both sides of the Dirac equation (1) by β, we obtain

(iγµ∂µ −m1)ψ = 0. (9)

The conditions (6) on αi and β are neatly unified into the matrix equation

γµ, γν = 2ηµν1. (10)

1.1 Plane wave solutions

We know that plane wave solutions exist, but what are they? We thereforelook for solutions of the form

ψ(x) = ue−ip·x, (11)

with u a constant 4-component spinor. Although it appears from (7) that uis arbitrary, this is not in fact true, as we shall see.

Substituting our plane wave ansatz (11) into the Dirac equation (9), wefind

(γµpµ −m)ue−ip·x = 0. (12)

At this point we shall introduce another piece of notation: we shall use the“slash” symbol 6 to denote contraction of a 4-vector with the Dirac gammamatrices:

aµγµ =6a. (13)

Thus u satisfies the equation

(6p−m)u = 0. (14)

Let us premultiply this equation by ( 6p+m), and use an important identity,

6a 6b = a · b 1, (15)

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to show that(6p+m)(6p−m)u = (p2 −m2)u = 0. (16)

This brings us back to the relation E2 = p2 +m2, although by a rather morerapid route. It is conventional to choose p0 ≡ E always to be positive, inwhich case we must also include solutions proportional to e+ip·x separately.We write them

ψ(x) = veip·x, (17)

which satisfy(6p+m)v = 0. (18)

Solutions with time dependence e−iEt (eiEt) are called positive (negative)energy solutions.

We can use the first equality in Eq. (16) to show that the following spinorsare solutions to the Dirac equations for u and v:

u±(p) = N(6p+m)

(χ±0

)v±(p) = −N(6p−m)

(0χ±

)

where N is a normalisation factor, and we have defined χ+ and χ− to spanthe space of 2-component spinors (which means they are spinor spanners):

χ+ =

(10

), χ− =

(01

). (19)

Any solution to the Dirac equation can be written as a superposition of thefour four-component basis spinors u± and v±.

To proceed further we need an explicit representation for the gammamatrices, and we will use the so-called “standard representation” given inAppendix B.1. It will turn out to be convenient to choose the normalisationfactor to be

N = 1/√

(E +m). (20)

The basis for the 4-component positive energy solutions can therefore bewritten

u± =1√

E +m

((E +m)χ±p · σχ±

), (21)

and the negative energy basis is

v± =1√

E +m

(p · σχ±

(E +m)χ±

). (22)

There are a couple of remarks that are worth making about the solutions (21)and (22). Firstly, when the 3-momentum is small, that is when |p| m,

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two of the components are much smaller than the other two. For the positiveenergy solutions, it is the lower two components that are negligible, while theopposite is true for the negative energy solutions. Secondly, the appearanceof two degrees of freedom for each set of solutions (that is, χ±) demands someexplanation. One can verify that χ± are eigenstates of the Pauli matrix σ3,with eigenvalues ±1:

σ3χ± = ±χ±. (23)

This degeneracy is actually a result of the fact that the particles describedby the Dirac equation (such as the electron) have an extra property, spin,which is a type of angular momentum intrinsic to the particle itself, unlikeordinary angular momentum which particles gain by virtue of rotation. Spinis quantised in half-integer units, with the spin of the electron being s = 1

2.

As with other types of quantised angular momentum, the value of the spinprojected along any axis (such as the z axis) takes the values s, s−1, . . . ,−s.Thus Dirac-type particles have only two spin states, which we conventionallycall “up” and “down”.

With the given normalisation convention, the spinors have the properties

u†AuB = 2EδAB, v†AvB = 2EδAB (24)

We define the adjoint spinors u and v by

uA = u†Aγ0, vA = v†Aγ

0. (25)

The adjoint spinors have the further properties

uAuB = 2mδAB, vAvB = −2mδAB. (26)

Note that this is a different normalisation convention than that used byGreiner and Reinhardt.

1.2 The interpretation of negative energy states

In the last section we called half of the solutions “negative energy” solutionswithout proper explanation. Of course, the explanation is that they appearto represent quantum states with negative energy, for if one acts with thequantum mechanical energy operator, one obtains a negative number:

i∂

∂t(veip·x) = −E(veip·x), (27)

with E = |(p2 +m2)12 |. This raises an ugly possibility: that an electron with

positive energy E1 could decay into one with negative energy −E2, radiating

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a photon with energy E1 +E2. There indeed seems to be nothing to stop theenergy of the electron from tumbling down towards minus infinity, releasingan infinite amount of energy in the process. This is clearly nonsense. Dirac’sway of circumventing this problem was to suppose that all the negative en-ergy states are filled already, and the Pauli exclusion principle prevents apositive energy electron from radiating a photon and occupying a filled nega-tive energy state. One can still imagine exciting one of these negative energyelectrons into a postive energy state, whereupon it becomes a real electron,leaving behind a hole. A hole is an absence of a negative electric charge,which has positive charge. Thus Dirac realised that his theory predictedthe existence of a positively charged spin 1

2particle with exactly the same

mass as the electron. Initially, Dirac identified this particle with the proton,hoping that somehow Coulomb interactions would provide a mass difference,but after others (including Weyl, Oppenheimer, and Tamm) showed that thiscouldn’t be right, he eventually had to abandon this conservative positionand predict a new particle now know as the positron. Its discovery in 1932by Carl Anderson in cosmic rays settled the issue.

However, this picture of a negative energy “sea” of electrons leaves muchto be desired. Bosons, particles with integer spin such as the photon (spin1) and the pion (spin 0) do not obey the Pauli Exclusion Principle andyet they still have negative energy states. Thus there can be no Dirac seapicture and we are left where we were before. It is only if we abandon simplewave mechanics and turn to quantum field theory that we find a satisfactoryresolution of this problem.

1.3 Quantising the spinor field

The Lagrangian density for a massive spinor field is

L = iψ 6∂ψ −mψψ, (28)

where we recall that in the standard representation of the Dirac gammamatrices the adjoint spinor ψ = ψ†γ0. Rather like the complex scalar field,one can obtain the field equation (which is the Dirac equation) by treating ψand ψ as independent quantities, and demand that the action be stationarywith respect to arbitrary variations in either. The variation of the actionwith respect to ψ gives

δS =∫d4x δψ

∂L∂ψ

= 0, (29)

(the Lagrangian is not a function of ∂µψ in this formulation). Hence∫d4x δψ(i 6∂ψ −mψ) = 0, (30)

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from which we derive the Dirac equation i 6∂ψ −mψ = 0. Similarly, we canvary with respect to ψ and obtain

δS =∫d4x (iψ 6∂δψ −mψδψ). (31)

Integrating by parts we find

δS =∫d4x (−i∂µψγµ −mψ)δψ +

∫dSµ iψγ

µδψ, (32)

where the last term is an integral over the space-time surface at spatialinfinity (|x| → ∞) with end-caps at |t| → ∞. As usual, we suppose that thevariations die away at infinity so that we can drop the surface term, so werecover the equation for the adjoint spinor i∂µψγ

µ +mψ = 0.The momentum conjugate to ψ is given by

π =∂L∂ψ

= iψγ0, (33)

which shows that in the standard reresentation ψ and iψ† are canonicallyconjugate variables. Thus we can find the Hamiltonian density:

H = πψ − L = −iψγk∂kψ +mψψ. (34)

The quantisation of the spinor field follows a familiar pattern. We first ofall suppose that ψ(x) is an operator satisfying some commutation relations,acting on some space of quantum states. We specify the equal time com-mutation relations, and then try to find the possible states. The problemwith the spinor field is that it is not obvious from the outset what commu-tation relations to impose on the field operator, and the obvious relation[ψ(t,x), iψ†(t,x′)] = iδ(x− x′) is not in fact correct.

To see why this is so, let us compute the zero-point energy of the field.Firstly, we expand the field in terms of its operator-valued Fourier coefficients:

ψ(x) =∑A=±

∫ d3p

(2π)3

1

2E

(cA(p)uA(p)e−ip·x + d†A(p)vA(p)eip·x

), (35)

where E2 = p2 + m2, and uA(p) and va(p) are 4-component spinors whichsatisfy

(6p−m)uA(p) = 0, (6p+m)vA(p) = 0. (36)

Similarly, the conjugate spinor has the expansion

ψ(x) =∑A=±

∫ d3p

(2π)3

1

2E

(c†A(p)uA(p)e+ip·x + dA(p)vA(p)e−ip·x

). (37)

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We can now compute the Hamiltonian, which is the spatial integral of theHamiltonian density H, or

H =∫d3x (−iψγi∂iψ +mψψ) (38)

The end result is

H =∑A

∫ d3p

(2π)3

1

2EE(c†A(p)cA(p)− dA(p)d†A(p)

). (39)

The vacuum state is traditionally defined as the state annihilated by allannihilation operators cA(p) and dA(p):

cA(p)|0〉 = 0, dA(p)|0〉 = 0. (40)

We expect an infinite zero point energy again, removed by normal ordering.However, H is not positive definite, and nor is :H: unless we define thecommutators of the operators such that

:dA(p)d†A(p): = −d†A(p)dA(p). (41)

This means that the fermionic creation and annihilation operators must an-ticommute, and we quantise the spinor field by anticommutation relations

dA(p), d†B(p′) = 2EδAB δ(p− p′), (42)

cA(p), c†B(p′) = 2EδAB δ(p− p′). (43)

All other anticommutator brackets vanish:

cA(p), cB(p′) = 0 = dA(p), dB(p′) (44)

c†A(p), c†B(p′) = 0 = d†A(p), d†B(p′) (45)

cA(p), d†B(p′) = 0 = c†A(p), dB(p′). (46)

There are four kinds of 1-particle states in this theory, created by the fouroperators c†A(p) and d†B(p). We write them

|p, A; 0〉 = c†A(p)|0〉, |0; p, A〉 = d†A(p)|0〉. (47)

As you can probably guess from the way the states are built up for thecomplex scalar field, c†A(p) creates a particle with momentum p and spinstate A, while d†A(p) creates an antiparticle.

Let us examine the 2-particle state

|p1, A1,p2, A2; 0〉 = c†A1(p1)c†A2

(p2)|0〉. (48)

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If we interchange the particles, we must interchange the creation operators,which gives the state a relative minus sign:

|p2, A2,p1, A1; 0〉 = c†A2(p2)c†A1

(p1)|0〉, (49)

= −c†A1(p1)c†A2

(p2)|0〉, (50)

= −|p1, A1,p2, A2; 0〉. (51)

Thus we see something that has to be taken for granted in ordinary quan-tum mechanics: the states of the spinor or Dirac field are antisymmetricunder particle interchange. We call these states fermionic, and the particlesfermions. Furthermore, suppose we take both momenta and both spin statesequal to p and A respectively: we then find

|p, A,p, A; 0〉 = −|p, A,p, A; 0〉 = 0. (52)

Hence two fermions can never be in the same state. This is precisely thePauli exclusion principle, which is fundamental to the understanding of thephysics of the atom. On the other hands, bosons, like the scalar particle weconsidered previously, have commuting field operators, and so any numberof them can be in the same state.

Armed with the anticommutation relations for the creation and annihila-tion operators, and the spinor completeness relations∑

A

uAa uAb = (γ · p)ab +mδab,

∑A

vAa vAb = (γ · p)ab −mδab, (53)

the fields can be shown to satisfy the equal time anticommutation relations

ψa(t,x), ψb(t,x′) = ihδabδ(x− x′), (54)

ψa(t,x), ψb(t,x′) = 0 = ψa(t,x), ψb(t,x

′) (55)

where the factor of h has been revived, and labels a and b have been put into keep track of the rows and columns of the spinors ψ and ψ.

1.4 Conserved charge

One can show straightforwardly from the Dirac equation (and its adjoint)that the 4-vector quantity

jµ(x) = ψ(x)γµψ(x) (56)

satisfies a continuity relation ∂ · j = 0. Hence there is a conserved charge

Q =∫d3x ψγ0ψ =

∫d3xψ†ψ. (57)

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By substituting the plane wave expansion for the Dirac field operator, onecan show that

Q =∑A

∫ d3p

(2π)3

1

2E

(c†A(p)cA(p)− d†A(p)dA(p)

). (58)

Applying this operator to to the single particle states, one can show that

Q|p, A; 0〉 = +|p, A; 0〉, Q|0,p, A〉 = −|0; p, A〉. (59)

Hence the states created by c†A(p) have equal and opposite charge to thosecreated by d†A(p), which fits in with their interpretation as particles andantiparticles.

One can also show that the existence of this conserved charge is a conse-quence of a symmetry of the Lagrangian, as we should expect from Noether’stheorem. This symmetry is

ψ(x)→ ψ′(x) = eiθψ(x), ψ(x)→ ψ′(x) = e−iθψ(x). (60)

2 Gauge field theory

2.1 Internal symmetries

Recall Noether’s theorem for a set of N real fields:For every transformation φa(x) → φ′a(x

′) which leaves the action S[φa]invariant, there is a conserved current. Defining the total variation in thefield, which is assumed infinitesimal,

δφa(x) = φ′a(x′)− φa(x), (61)

one can find an (infinitesimal) 4-vector fµ which satisfies a continuity equa-tion ∂µf

µ = 0, given by

fµ =∂L

∂(∂µφa)δφa(x)−

(∂L

∂(∂µφa)∂νφa(x)− Lδµν

)δxν . (62)

We studied in RQF1 the case of coordinate translation symmetry, x→ x′ =x+δx, with δxµ = εµ, a constant but arbitrary infinitesimal 4-vector. This re-sulted in four conserved currents, which were assembled into a rank-2 tensor,the energy-momentum tensor

θµν =∂L

∂(∂µφa)∂νφa(x)− Lδµν . (63)

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In this section we will study internal symmetries, which transform fieldsamongst themselves:

δφa = εTabφb, δxµ = 0 (64)

The non-infinitesimal matrix Tab is called the generator the symmetry.With this transformation it is straightforward to show that the following

non-infinitesimal 4-vector is conserved:

jµ =∂L

∂(∂µφa)Tabφb. (65)

We can extend these considerations to N complex scalar fields. For this weneed to recall that when differentiating with respect to a complex variablez, one treats z and z as independent. Given this fact, it follows that theinfinitesimal conserved current which follows from a symmetry of the actionS[φa, φa] is

fµ =∂L

∂(∂µφa)δφa(x) +

∂L∂(∂µφa)

δφa(x)

−(

∂L∂(∂µφa)

∂νφa(x) +∂L

∂(∂µφa)∂νφa(x)− Lδµν

)δxν . (66)

Internal symmetry transformations for complex fields take the form

δφa = εTabφb, δφa = εT †abφb, δxµ = 0. (67)

Example 1. Consider a theory with two real fields φ1 and φ2, and an action

S =∫d4x

(1

2∂µφa∂

µφa − V (φ)), (68)

where φ =√φ2

1 + φ22 is the magnitude of the 2-dimensional field vector ~φ. The

action is clearly symmetric under rotations in the 2-dimensional “internal”space of the field,

~φ→ ~φ′ = Ω~φ, (69)

where Ω is a 2×2 orthogonal matrix

Ω =

(cos θ sin θ− sin θ cos θ

). (70)

If the transformation is infinitesimal, we have

δ~φ ' θ

(0 1−1 0

). (71)

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Using the antisymmetric symbol εab, defined by ε12 = −ε21 = 1 and ε11 =ε22 = 0, we can rewrite the infinitesimal change in the field in componentform,

δφa = θεabφb. (72)

In order to evaluate jµ we require

∂L∂(∂µφa)

= ∂µφa. (73)

Hence the conserved current in this theory is

jµ = ∂µφaεabφb = (∂µφ1)φ2 − (∂µφ2)φ1. (74)

Note that the set of 2×2 orthogonal matrices form a group under matrixmultiplication. We have been considering those matrices which are contin-uously connected to the identity, and therefore have unit determinant. Thegroup of 2×2 orthogonal matrices with unit determinant is called SO(2).

Example 2. Let us consider N = 1 complex scalar fields φ with action

S =∫d4x

(∂µφ∂µφ− V (|φ|)

)(75)

where |φ| is the modulus of φ. This is clearly invariant under the transforma-tion φ → φ′ = e−iθφ, i.e. under multiplication by complex numbers of unitmodulus. If θ is infinitesimal, we have

δφ = −iθφ, δφ = iθφ. (76)

In order to calculate the conserved current, we need the differentials

∂L∂(∂µφa)

= ∂µφa,∂L

∂(∂µφa)= ∂µφa. (77)

It is now straightforward to show that

jµ = −i(∂µφ)φ+ i(∂µφ)φ (78)

is conserved. Note that the set of numbers of unit modulus form a groupunder ordinary complex multiplication, which is called U(1). Note also thatif one writes φ = (φ1 +iφ2)/

√2, with φ1 and φ2 real, one can recover both the

action and the conserved current of the SO(2) scalar field theory. This is justa consequence of the fact that the groups U(1) and SO(2) are isomorphic.

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Example 3. Consider now the spinor field, whose action is

S =∫d4x ψ(i 6∂ −m)ψ. (79)

Again, this has the obvious symmetry ψ → ψ′ = e−iθψ, which in infinitesimalform is δψ = −iθψ and δψ = iθψ. In order to evaluate the conserved currentwe need

∂L∂(∂µψa)

= iψγµ,∂L

∂(∂µφa)= 0, (80)

from which we can quickly infer that

jµ = ψγµψ (81)

is conserved. Note that one can show that the conserved charge Q = ψγ0ψis equal to the difference in the numbers of particles and antiparticles:

Q =∫ d 3p

2E

(c†A(p)cA(p)− d†A(p)dA(p)

), (82)

which demonstrates explicitly that particles and antiparticles have oppositecharges.

Example 4. Finally, let us consider a massless spinor field, with action

S =∫d4x iψ 6∂ψ. (83)

This also enjoys the symmetry ψ → ψ′ = e−iθψ, but has an extra chiralsymmetry ψ → ψ′ = e−iθ5γ

5ψ, where γ5 is defined in Appendix B. Note

that the adjoint spinor transforms as ψ → ψ′ = ψe−iθ5γ5, and so the mass

term ψψ is not invariant under this transformation. Similar considerationsto the previous examples show that the conserved current associated withthis symmetry is

jµ5 = ψγµγ5ψ. (84)

2.2 Gauge symmetry 1

If the parameters of a symmetry transformation are space-time dependent wecall it a local or gauge symmetry. Gauge theories have much richer structurethat their counterparts with global symmetries.

The prototypical gauge theory is electrodynamics. Let us consider “gaug-ing” the massive spinor field, i.e. demanding that the U(1) global symmetrydiscussed in the previous section becomes space-time dependent:

φ→ φ′ = e−iα(x)φ. (85)

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The simple spinor action is no longer invariant under this space-time depen-dent transformation, as one can easily show:

S → S ′ =∫d4x ψeiα(x)(i 6∂ −m)e−iα(x)ψ,

= S −∫d4x ψγµψ∂µα. (86)

In order to make the action symmetric under this transformation we canintroduce a 4-vector field Aµ with the transformation property

Aµ → A′µ = Aµ +1

e∂µα, (87)

where e is a constant. Then one can show that

(∂µ + ieAµ)ψ → (∂µ + ieA′µ)ψ′ = (∂µ + (ieAµ + i∂µα)e−iα(x)ψ

= e−iα(x)(∂µ + ieAµ)ψ. (88)

Introducing the notation Dµ = (∂µ + ieAµ), one sees that Dµψ transforms inthe same way as ψ under gauge transformations:

(Dµψ)→ (Dµψ)′ = e−iα(x)(Dµψ). (89)

We call this special object the gauge covariant derivative.Using the gauge covariant derivative, we can contruct a new action which

is invariant under gauge transformations:

S =∫d4x ψ(i 6D −m)ψ. (90)

We say that this action is gauge invariant.The dynamics of this theory is not very interesting as it stands. If we

want to treat Aµ as a dynamical field, and ask that the action be at anextremum with respect to its variations:

δS

δAµ(x)=

δ

δAµ(x)

∫d4x

[ψ(i 6∂ −m)ψ − eψγµψAµ

= −e∫d4y ψγµψ

δAν(y)

δAµ(x)

= −eψ(x)γµψ(x) = 0. (91)

Thus we see that the current of the global symmetry must vanish if thisaction is at an extremum. We need to supplement the action to obtain afield equation for Aµ, in which case ψ(x)γµψ(x) will act as a source.

In order to realise this project, we first need to study the formulation ofthe field equations for vector fields – otherwise known as Maxwell’s equa-tions.

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2.3 Maxwell’s equations in covariant form

This section recaps some important results, and introduces the formulationof the theory in an explicitly Lorentz covariant manner.

Firstly, we recall Maxwell’s equations in free space, writing them downin natural units, in which the permittivity and permeability of free space µ0

and ε0 are both unity:

Homogeneous Inhomogeneous∇·B = 0 ∇·E = ρ

∂tB +∇ ∧ E = 0 − ∂

∂tE +∇ ∧B = j

(92)

The homogeneous and inhomogeneous (i.e. having a source term on the righthand side) equations have differing status. The homogeneous equations implythe existence of potentials φ and A from which the physically measurablequantities E and B can be calculated. These potentials are specified onlyup to a gauge transformation, φ→ φ− Λ and A→ A +∇Λ, where Λ is anarbitrary function of space and time. The inhomogeneous equations implythat the source terms must obey a current conservation equation, ρ+∇·j = 0.To summarise:

Homogeneous Inhomogeneouspotentials φ,A current conservation

B = ∇ ∧A

E = −A−∇φφ → φ− ΛA → A +∇Λ

ρ+∇·j = 0

(93)

All these quantities can be assembled into explicitly Lorentz covariant ob-jects. The gauge potentials belong together in a 4-vector potential Aµ =(φ,A), while the charge density ρ and the current density j can be put to-gether into a 4-vector current density jµ = (ρ, j). Recall that putting quanti-ties together into 4-vectors is not just a matter of notation: it means that thequantities transform just like the space-time coordinates xµ under a Lorentztransformation.

The electric and magnetic fields E and B also belong together in a Lorentzcovariant object, as they are mixed up by Lorentz transformations (an ob-server moving through a magnetic field also sees an electric field). However,this object cannot be a 4-vector as there are a total of 6 components of theelectric and magnetic fields, when they are taken together. In fact, the objectis an antisymmetric tensor, the field strength tensor F µν , whose entries are

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as follows:

F µν =

0 −E1 −E2 −E3

E1 0 −B3 B2

E2 B3 0 −B1

E3 −B2 B1 0

. (94)

Note that E1 represents the first component of the electric field vector, whichin Cartesian coordinates is the x component. The reverse relations may bewritten

Ei = F i0, Bi = −1

2εijkF

jk, (95)

which introduces the Levi-Civita symbol εijk. The Levi-Civita symbol isdefined by

εijk =

+1 if i 6= j 6= k cyclic,−1 if i 6= j 6= k anticyclic,0 otherwise.

(96)

The field strength tensor has a neat expression in terms of the gauge potential4-vector Aµ:

F µν = ∂µAν − ∂νAµ. (97)

Let us check this explicitly for the components corresponding to the electricfield:

F i0 = ∂iA0 − ∂0Ai = ∂iA0 − Ai = −∂iφ− Ai = Ei, (98)

where we have used the expression for the electric field in terms of the gaugepotentials in equation (93).

When expressed in terms of 4-vectors and tensors, electromagnetism looksvery simple and beautiful. For example, the covariant expression of thecurrent conservation equation is simply

∂µjµ = 0. (99)

Maxwell’s equations become:

Homogeneous Inhomogeneous

∂λF µν + ∂µF νλ + ∂νF λµ = 0, ∂µFµν = jν .

(100)

The first of these expressions looks as if it contains many more equationsthan the 4 of the original homogeneous Maxwell equations. However, theantisymmetry of the field strength tensor (F µν = −F νµ) means that theexpression is trivial if any of the two indices are equal. Thus all the indicesλ, µ and ν must take different values, and the number of ways of choosingthree different numbers from a set of four is 4C3, which is equal to 4, preciselythe number of equations we started with.

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We can use the four-dimensional Levi-Civita tensor to re-express the ho-mogeneous equations more compactly. This tensor has four indices, and isdefined by

εµνρσ =

+1 if µ 6= ν 6= ρ 6= σ, symmetric,−1 if µ 6= ν 6= ρ 6= σ, antisymmetric,0 otherwise.

(101)

A symmetric permutation of the indices in one in which an even number ofindices are exchanged, while an antisymmetric permutation is one for whichan odd number of indices are exchanged. Thus, for example, ε0123 = ε1032 =+1, ε1023 = ε0132 = −1, but ε0012 = 0.

There is also a version with the indices raised:

εµνρσ = ηµαηνβηργησδεαβγδ, (102)

whose symmetric permutations take the value −1 and antisymmetric ones+1.

Using this totally antisymmetric (i.e. antisymmetric on the exchange ofany two indices) tensor, the homogeneous Maxwell equations become

εµνρσ∂ρF µν = 0. (103)

2.4 Lagrangian formulation

We recall that the correct action to reproduce the inhomogeneous Maxwellequations is

S =∫d4x

(−1

4FµνF

µν − jµAµ), (104)

where Fµν = ∂µAν − ∂νAµ. We firstly recall that the functional derivative ofa 4-vector field with respect to itself is

δAν(y)

δAµ(x)= δµν δ

4(y − x). (105)

To obtain the field equations we extremise the action, finding

δS

δAµ(x)= ∂νF

νµ(x)− jµ(x) = ∂2Aµ(x)− ∂µ∂ · A(x)− jµ(x) = 0. (106)

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2.5 Gauges

The physical quantities, the electric and magnetic fields, contained in thefield strength tensor Fµν , are invariant under a gauge transformation

Aµ(x)→ A′µ(x) = Aµ(x)− ∂µΛ(x), (107)

where Λ is an abitrary function of the spacetime coordinates xµ. Indeed,using (97) one finds

F µν → ∂µ(Aν−∂νΛ)−∂ν(Aµ−∂µΛ) = F µν− (∂µ∂ν−∂ν∂µ)Λ = F µν . (108)

The last step follows because we can take the partial derivatives in any order.There are two related problems with the theory as formulated above.

1. Gauge invariance means that there is no unique solution to the fieldequations for a given source jµ.

2. If we wish to find the Hamiltonian of the theory, as is necessary forcanonical quantisation, we must first find the canonical momentum,

πµ =∂L∂Aµ

= Fµ0. (109)

This means that π0 = 0, i.e. the canonical momentum associated withthe timelike component of the gauge field A0 is identically zero.

The solution to these problems begins with fixing the gauge: imposing aconstraint on the field to remove the freedom to make a gauge transformation.For each constraint, the field equations may be solved.

There are in principle an infinite number of constraints one could apply,but in practice three turn out to be convenient.

1. Temporal gauge A0 = 0.

2. Coulomb gauge ∇·A = 0.

3. Lorentz gauge ∂ · A = 0.

Only the last is obviously Lorentz invariant, and so is most useful forrelativistic field theory. The other two gauges are useful in other circum-stances: for example, temporal gauge is the easiest to use for solving thefield equations numerically on a lattice, while Coulomb gauge is useful fornon-relativistic applications in atomic physics.

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2.6 Coulomb gauge solutions

This is also called radiation gauge, and is defined by

∇·A = 0, (110)

which implies that the first of the inhomogeneous equations in (92) becomes

∇2φ = −ρ. (111)

This gauge is therefore very useful in solving electrostatics problems, and weshall use it later on when studying the Casimir effect. However, it is not sooften used in relativistic applications, as the gauge condition does not respectLorentz invariance.

The second of the inhomogeneous Maxwell equations becomes, on usingthe gauge condition,

− ∂

∂t

(−A−∇φ

)−∇2A = j, (112)

which we may write as (∂2

∂t2−∇2

)A = jT , (113)

where jT = j+∇φ. One can show, using the equation of current conservationand Eq. (111), that

∇·jT = 0. (114)

This of course is necessary for the consistency of Eq. (113), because if onetakes the divergence of the left hand side one gets zero as well.

In free space, that is, in the absence of charges and currents, we may takeφ = 0, and the equation for the vector potential becomes(

∂2

∂t2−∇2

)A = 0. (115)

One of the important discoveries of the last century was that this equation hasplane wave solutions, which carry energy and momentum: electromagneticwaves. A plane wave solution with a particular wavenumber k may be written

A(t,x) = ae−iωt+ik·x, (116)

where a is a constant 3-vectors, and ω = |k|.The gauge condition (110) imposes a condition on the vector a, for

∇·A = ik·a e−iωt+ik·x = 0, (117)

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which impliesk·a = 0. (118)

Hence the gauge potential A is orthogonal to the wave vector k. This meansthat both the electric and magnetic fields are also orthogonal: we say thatthe waves are transverse.

2.7 Lorentz gauge solutions

This is very often the gauge to choose if one is interested in wave propagationin electromagnetism. It is defined by

∂µAµ = 0, (119)

and is manifestly Lorentz invariant. In this gauge the equation of motion forthe gauge field is simplified, for

∂µFµν = ∂µ∂

µAν − ∂µ∂νAµ = ∂2Aν = jν . (120)

However, an irritating feature of this gauge is that it does not quite specifyAµ fully. One can still make a gauge transformation Aµ → Aµ − ∂µΛ whichsatisfies the Lorentz gauge condition (119), as long as the function Λ satisfies∂2Λ = 0. Such functions are called harmonic. In classical field theory this isnot too much of a problem, but in setting up the quantum theory of gaugefields care must be taken.

In free space, the field equation is (120)

∂2Aµ ≡(∂2

∂t2−∇2

)Aµ = 0. (121)

A particular solution isAµ(t,x) = aµe−ik·x, (122)

where k0 = ωk = |k|, and aµ is a constant 4-vector.The Lorentz gauge condition (119) gives ∂ · A = −ik · ae−ik·x = 0, which

impliesk · a = 0. (123)

Once again the field is orthogonal to the wavenumber k, but this time in the4-vector sense.

The general solution may be constructed from a superposition of planewave solutions

Aµ(x) =∫ d 3k

2ωk

(aµ(k)e−ik·x + a∗µ(k)eik·x

), (124)

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where the amplitudes aµ(k) must all satisfy k · a(k) = 0.It would be more convenient to be able to choose the amplitudes of the

components of the gauge field independently, and to this end we introduce aset of four basis 4-vectors, or polarisation vectors, εA

µ(k), with A = 0, 1, 2, 3,for each wavevector k. They must satisfy a completeness condition in orderfor them to be considered as basis vectors, and it is also convenient to makethem orthonormal, in a 4-vector sense.

εA∗µ(k)εB

ν(k)ηµν = ηAB (Orthonormality), (125)

εA∗µ(k)εB

ν(k)ηAB = ηµν (Completeness). (126)

A good choice is

ε0 = (1; 0),

ε1 = (0, n× k)/|n× k|,ε2 = (0,k× (n× k))/|k× (n× k)|,ε3 = (0; k),

(127)

where n = (0, 0, 1) is a unit 3-vector. We refer to these polarisation vec-tors as timelike (A = 0), transverse (A = 1, 2), and longitudinal (A = 3)respectively.

Using this basis we can write aµ = aAεAµ, and find that the gauge condi-

tion implies k · a = k0a0 − |k|a3 = 0, or

a0(k) = a3(k). (128)

Thus the general solution is now more easily written down as

Aµ(x) =∫ d 3k

2ωk

(aA(k)εA

µ(k)e−ik·x + aA(k)ε∗Aµ(k)eik·x

). (129)

There now appears to be, for each k, three independent degrees of freedom,down from the original four. However, one of the amplitudes, a3, does notaffect the physical value of the electric and magnetic fields. This is a result ofthe remaining freedom in the Lorentz gauge to make gauge transformationswith a harmonic function Λ(x), which may be decomposed into plane waves,according to

Λ(x) =∫ d 3k

2ωk

(λµ(k)e−ik·x + λµ(k)eik·x

). (130)

Hence under such a gauge transformation,

aµ(k)→ a′µ(k) = aµ(k)− ikµλ(k). (131)

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By multiplying with ε3µ we see that this is equivalent to shifting a3(k),

a3(k)→ a′3(k) = a3(k)− i|k|λ(k). (132)

Thus although the Lorentz gauge does not entirely remove the freedom tomake gauge transformations, it does confine the ambiguity in the solution toone and only one of the sets of constants in the solution. There are only twophysical amplitudes to choose, a1 and a2.

Let us end this section by calculating the Hamiltonian in the Lorentzgauge.

HLg =∫d3x

(πµA

µ − LLg

)= −1

2

∫d3x(AµA

µ +∇Aµ·∇Aµ). (133)

We already notice from this expression that the Hamiltonian is not positivedefinite, as the timelike component of the gauge field A0 appears with anegative sign. It turns out that we have already cured this problem in theclassical theory. To see this, let us substitute the plane wave expansion ofthe field operator, obtaining

HLg = −∫ d 3k

2ωk

ωkηABa∗A(k)aB(k). (134)

We saw earlier that the Lorentz gauge condition Eq. 119 implies a0 = a3,and so we find that

HLg =∑A=1,2

∫ d 3k

2ωk

ωka∗A(k)aA(k), (135)

which is clearly positive definite, and depends only on the physical tranverselypolarised components.

2.8 Canonical quantisation in the Lorentz gauge

In order to quantise this theory in the canonical manner, we must first iden-tify the canonical momentum conjugate to the degrees of freedom, imposecanonical commuation relations, identify the ladder operators and find theircommutation relations, and construct a Fock space of states. We should alsocalculate the Hamiltonian, and verify that it is positive definite.

As one might imagine, gauge invariance poses problems for the quantumtheory as well. Firstly, we have the problem mentioned in Section 2.5, that

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the canonical momentum vanishes. This turns out to be cured by fixing thegauge, as we can verify in the Lorentz gauge:

πµ =LLg

∂Aµ= −Aµ. (136)

Hence the equal time canonical commutation relations can be guessed to bethose for four independent real fields

[Aµ(t,x), πν(t,x′)] = iδµν δ

3(x− x′), (137)

[Aµ(t,x), Aν(t,x′)] = 0, [πµ(t,x), πν(t,x′)] = 0. (138)

By substituting the plane wave expansion for the field operator Aµ(x), onecan find (after some algebra) that

[aA(k), a∗B(k′)] = −ηAB2ωk δ3(k− k′), (139)

[aA(k), aB(k′)] = 0, [a∗A(k), a∗B(k′)] = 0. (140)

Hence we have four pairs of ladder operators, one pair for each polarisation.We define the vacuum state |0〉 by

aA(k)|0〉 = 0 ∀k, A. (141)

and we can construct excited states by acting with the raising operatorsa∗A(k). For example,

|k, A〉 = a∗A(k)|0〉. (142)

How do we impose the gauge constraint in the quantum theory? Firstly,it is clear that demanding ∂ ·A = 0 is too strong a condition on the operatorAµ, as it is inconsistent with the equal time canonical commutation relationsEq. 137. A weaker condition is to ask that matrix elements of the gaugecondition vanish. This we cannot do for every possible state, but we onlyaccept as physical states for which this is true: i.e.

〈φ′|∂ · A|φ〉 = 0 (143)

for any two physical states |φ〉 and |φ′〉. Any equivalent condition is todemand that the positive frequency part of the gauge condition gives zerowhen acting on a physical state:

∂ · A(+)|φ〉 = 0, (144)

where

∂ · A(+) = −∫ d 3k

2ωk

ik · ae−ik·x. (145)

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Hence physical states must satisfy

(a0(k)− a3(k))|φ〉 = 0, (146)

which is the equivalent of Eq. (128) in the quantum theory.Finally, we shall check that the Hamiltonian is positive definite for phys-

ical states. We must deal with the infinity in the vacuum energy by normalordering, and study the matrix elements of the operator

:HLg: = −∫ d 3k

2ωk

ωkηABa∗A(k)aB(k). (147)

The quantum version of the gauge condition Eq. (146) implies that for phys-ical states |φ〉 and |φ′〉,

〈φ′|:HLg:|φ〉 =∑A=1,2

∫ d 3k

2ωk

ωk〈φ′|a∗A(k)aA(k)|φ〉. (148)

In particular, the expectation value of the energy is clearly positive definitefor physical states.

2.9 Feynman propagator in Lorentz gauge

By now we should be accustomed to thinking of the Feynman propagator asa Greens function for solving the field equations with a small imaginary massterm,

(∂2 − iε)Aµ(x) = jµ(x). (149)

As there are four fields, the Green’s function is a 4× 4 object, and is definedby

(∂2 − iε)DFµν(x− x′) = −δµν δ4(x− x′). (150)

Hence the Feynman propagator for a gauge field in the Lorentz gauge is justfour copies of a massless scalar propagator. The Fourier space representationis

DFµν(x− x′) = ηµν

∫d 4k

e−ik·(x−x′)

k2 + iε. (151)

Strictly speaking, one should also show that this is related to the two pointfunction, i.e. the vacuum expectation value of the time-ordered product oftwo fields:

iDFµν(x− x′) = Gµν(x, x′) = 〈0|T [Aµ(x)Aν(x

′)]|0〉. (152)

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However, proving this relation will have to wait until we have studied thepath integral quantisation of the gauge field. Meanwhile, we will jump thegun and write down the Feynman rules for the electromagnetic field

Ingoingk, ε r εµ(k)

Outgoing r k, εε∗µ(k)

Internal rµ rν ∫d 4k

iηµνk2 + iε

3 Quantum gauge theory

4 Electroweak theory

5 QCD

6 Renormalisation

A Pauli matrices

The Pauli matrices σi (i = 1, 2, 3) are defined to be

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

). (153)

They are Hermitean, i.e.(σi)† = σi, (154)

and they square to unity:

(σ1)2 = (σ2)2 = (σ3)2 = 1. (155)

One can verify thatσ1σ2 = iσ3, (and cyclic). (156)

Lastly, any two different Pauli matrices anticommute, or

σ1σ2 + σ2σ1 = 0, (and cyclic). (157)

This set of relations can be neatly expressed using the 3 dimensional Levi-Civita symbol εijk:

[σi, σj] = 2iεijkσk, (158)

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and with the anticommutation relations

σi, σj = 2δij. (159)

One may thus write the product of any two Pauli matrices as

σiσj =1

2σi, σj+

1

2[σi, σj] = δij + iεijkσk. (160)

The Pauli matrices form a representation of the angular momentum al-gebra with (spin) angular momentum 1

2.

B Dirac matrices

Dirac matrices (in four space-time dimensions) are 4×4 matrices defined bythe anitcommutation relations

γµ, γν = 2ηµν14, (161)

where ηµν is the Minkowski metric, and 14 is the 4×4 identity matrix (whichis often dropped and left implicit in the equation).

Properties of the Dirac γ-matrices include:

(i). (γµ)† = γ0γµγ0.

One can also define another, linearly independent, γ-matrix

γ5 = iγ0γ1γ2γ3, (162)

which has the easily derivable properties

(γ5)2 = 1, (γ5)† = γ0γ5γ0, γµ, γ5 = 0. (163)

The eigenvector of this matrix are known as chirality eigenvectors. Eigenvec-tors with eigenvalue +1 are termed right-handed, with -1 left-handed. Onecan define projectors onto these eigenvectors:

PR =1

2(1 + γ5), PL =

1

2(1− γ5) (164)

B.1 Standard representation of Dirac matrices

γ0 =

(1 00 −1

), γi =

(0 σi

−σi 0

), γ5 =

(0 11 0

). (165)

Here, σi are the Pauli matrices, which are defined in Appendix A, and 0 and1 are the 2×2 zero and identity matrices respectively.

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C Identities for Dirac matrices

(i). tr(γµγν) = 4ηµν

(ii). tr(γµ1 . . . γµ2n−1) = 0, n a positive integer.

(iii). tr(γµγνγργσ) = 4(ηµνηρσ − ηµρηνσ + ηµσηνρ)

(iv). tr(6 a1 . . . 6 a2n) =∑2ni=2(−1)ia1 · ai tr(6 a1 . . .6ai . . . 6 a2n) (the hat denotes

that 6ai is removed from the trace)

(v). γµγµ = 41

(vi). γµ 6aγµ = −2 6a

(vii). γµ 6a 6bγµ = 4(a · b)1

(viii). γµ 6a 6b 6cγµ = −2 6c 6b 6a

(ix). γµ 6a 6b 6c 6dγµ = 2(6d 6a 6b 6c+ 6c 6b 6a 6d)

D Feynman rules

The rules which one derives from the LSZ reduction formula are for calcu-lating S-matrix elements Sαβ = 〈α|S|β〉. In momentum space one finds, fora real scalar field, that the following rules are useful:

Internal line (propagator) ∫d 4k

i

k2 −m2 + iε

Vertex

k1

k2

k4

k3

−iλ δ4(k1 + k2 + k3 + k4)Momentum is conserved for particle excitations in the vacuum, so when

one performs as many of the integrations using the vertex δ-functions asone can, one always finds a factor of δ4(K ′ − K), where K is the total 4-momentum in the initial state, and K ′ the total 4-momentum in the final

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state. Furthermore, there are also uninteresting contributions from graphswith no vertices.

It is therefore convenient to define matrix elements Mαβ through

iMαβ δ4(K ′ −K) = 〈α|(1− S)|β〉.

We can derived slightly simpler Feynman rules for these matrix elements,in which the propagators lose their associated integrations and the verticestheir δ-functions.

Consider a diagram with E external lines, V vertices, and I internal lines.For the simple φ4 theory we know that external lines end on a vertex, internallines have vertices at each end, and vertices must be attached to four lines.Hence

E + 2I = 4V.

There are I integrations and V δ-functions, and we know that there is alwaysone δ-function left over. Hence we can trivially perform V − 1 integrations,leaving

L = I − V + 1

integrations to perform.We call the number of free integrations the number of loops, as for simple

graphs it does indeed correspond to the number of closed loops in the graph.The following sections give the Feynman rules for working out the matrix

elements Mαβ for several theories of interest.

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E Real scalar field

S =∫d4x

(1

2∂µφ∂

µφ− 1

2m2φ2 − 1

4!φ4)

Ingoing line →k

1

Outgoing line → k

1

Internal line (propagator) →k i

k2 −m2 + iε

Vertex −iλ

F Complex scalar field

S =∫d4x

(∂µφ∂

µφ−m2φφ− 1

4λ(φφ)2

)

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Page 34: Relativistic Quantum Fields 2

Ingoing particle line →k

1

Ingoing antiparticle line →k

1

Outgoing particle line → k

1

Outgoing antiparticle line →k

1

Internal line (propagator) →k i

k2 −m2 + iε

Vertex −iλ

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G U(1) gauge field

S =∫d4x

(−1

4FµνF

µν − 1

2ξ(∂ · A)2

)

Ingoing line →k, ε

εµ(k)

Outgoing line → k, ε

ε∗µ(k)

Internal gauge line (propagator) →k

µ ν

iηµν + (ξ − 1)kµkν/k

2

k2 + iε

H Spinor field with U(1) gauge symmetry

S =∫d4x ψ(i 6D −m)ψ.

Here Dµψ = (∂µ + ieAµ)Ψ.

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Ingoing particle line →k, A

uA(p)

Ingoing antiparticle line →k, A

vA(p)

Outgoing particle line → k, A

uA(p)

Outgoing antiparticle line →k, A

vA(p)

Internal line (propagator) →p i

6p−m+ iε

3-point vertex

µ

k1

k2

−ieγµ

• Closed fermion loops are associated with a factor (−1).

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I Non-Abelian gauge field

S =∫d4x

(−1

4F aµνF

aµν − 1

2ξ(∂ · Aa)(∂ · Aa)− ca∂ · (Dc)a

)

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Ingoing line →k, ε, a

εµa(k)

Outgoing line → k, ε, a

ε∗µa (k)

Internal gauge line (propagator) →k

µ, a ν, b

iδabηµν + (ξ − 1)kµkν/k

2

k2 + iε

Internal ghost line (propagator) →k

a b

iδabk2 + iε

3-point gauge vertex

k1, µ, a

k2, ν, b

k3, ρ, c ig[(k1 − k2)ρηµν+(k1 − k2)µηνρ+(k1 − k2)νηρµ]

4-point gauge vertex µ, a

ν, b

σ, d

ρ, c

−ig2[feabfecd(ηµρηνσ − ηµσηνρ)+feacfedb(ηµσηρν − ηµνηρσ)

+feadfebc(ηµνησρ − ηµρησν)]

Gauge-ghost vertex

µ, a

b

k, c

gfabckµ

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• Closed ghost loops are associated with a factor (−1).

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J Complex scalar field with non-abelian gauge

symmetry

S =∫d4x

((DµΦ)†(DµΦ)−m2Φ†Φ− Vint(Φ,Φ

†)).

Here Φ is a column vector of Nf scalar fields φm, and DµΦ = (∂µ−igAaµT a)Φ.Equivalently, (Dµφ)m = (δmn∂µ−igAaµT amn)φn. Rules are given for Vint(Φ,Φ

†) =14λ(Φ†Φ)2.

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Ingoing particle line →k

1

Ingoing antiparticle line →k

1

Outgoing particle line → k

1

Outgoing antiparticle line →k

1

Internal line (propagator) ! →k iδmn

k2 −m2 + iε

3-point gauge-scalar vertex

" µ, a

k1,m

k2, n

igT amn(k1 + k2)µ

4-point gauge-scalar vertex

#µ, a

ν, b

m

n

ig2T a, T bmn

Scalar vertex

$m

n

q

p

−iλ(δmnδpq + δmqδnp)

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K Spinor field with non-abelian gauge sym-

metry

S =∫d4x Ψ(i 6D −m)Ψ.

Here Φ is a column vector of Nf spinor fields ψm, and DµΨ = (∂µ−igAaµT a)Ψ.Equivalently, (Dµψ)m = (δmn∂µ − igAaµT amn)ψn.

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Ingoing particle line %→k, A,m

umA(p)

Ingoing antiparticle line &→k, A,m

vmA(p)

Outgoing particle line ' → k, A,m

umA(p)

Outgoing antiparticle line ( →k, A,m

vmA(p)

Internal line (propagator) ) →p

m n

iδmn6p−m+ iε

3-point vertex

* µ, a

k1,m

k2, n

igT amnγµ

• Closed fermion loops are associated with a factor (−1).

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L Problem Sheets

L.1 Problem Sheet 1

1. (a) If we writeψ(x) = ue−ip·x + veip·x,

with p0 = E ≡ |(p2 + m2)12 |, show that the the following expres-

sions for u and v give solutions to the Dirac equation (i 6∂−m)ψ =0:

u±(p) = (E+m)−12 (6p+m)

(χ±0

)v±(p) = −(E+m)−

12 (6p−m)

(0χ±

)

where χ+ =

(10

), and χ− =

(01

).

(b) Show also that

u†AuB = 2EδAB, v†AvB = 2EδAB

where the indices A and B range over the spin states + and −,and that

uAuB = 2mδAB, vAvB = −2mδAB,

where uA = u†Aγ0, and vA = v†Aγ

0 are adjoint spinors.

2. The Lagrangian density for a spinor field may be written

L = ψ(i 6∂ −m)ψ.

(a) Given that the general expression for the canonical energy-momentumtensor is

θµν =∂L

∂(∂µψ)∂νψ + ∂µψ

∂L∂(∂µψ)

− Lδµν ,

evaluate the momentum density θi0 and the energy density θ00,

showing that the latter is equal to the Hamiltonian density H.

(b) Given the plane wave expansion of the spinor field

ψ(x) =∑A=±

∫ d 3p

2E

(cA(p)uA(p)e−ip·x + d†A(p)vA(p)eip·x

),

show that

H =∑A=±

∫ d 3p

2EE(c†A(p)cA(p)− dA(p)d†A(p)

).

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3. We define the matrix γ5 = iγ0γ1γ2γ3.

(a) Show that γ5 has the following properties:

(γ5)† = γ5, (γ5)2 = 1, γ5, γµ = 0,

where 1 is the 4× 4 identity matrix, and evaluate γ5 in the stan-dard representation. (You may find it useful to recall the relation(γµ)† = γ0γµγ0.)

(b) Eigenvectors of γ5 are called eigenvectors of chirality, and eigen-vectors with eigenvalue +1 are termed right-handed, with -1 left-handed. Show that ψR = 1

2(1 + γ5)ψ is right-handed, and that

ψL = 12(1−γ5)ψ is left-handed. Show also that ψψ = ψLψR+ψRψL

and ψγ5ψ = ψLψR − ψRψL. Note that ψL = (ψL)†γ0.

(c) Consider the current density

jµ = ψγµψ.

Show that this current is conserved if ψ is a solution to the Diracequation, and show that ψγµψ = ψRγ

µψR + ψLγµψL

(d) Using γ5, we may also define another current carried by spinors,called the axial-vector current :

j5µ(x) = ψ(x)γµγ5ψ(x).

Assuming that ψ(x) satisfies the Dirac equation, calculate ∂µj5µ,

and exhibit j5µ(x) in terms of chirality eigenstates ψR and ψL.Under what circumstances is the axial-vector current conserved?

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4. (a) Consider a function of a pair complex Grassmann variables f(θ, θ).

f(θ, θ) = a+ bθ + cθ + dθθ,

and define the following operators:

D =∂

∂θ+ iθ, D = − ∂

∂θ+ iθ.

i. Show that D, D = 2i

ii. Show that if Df(θ, θ) = 0, then f(θ, θ) = a+ cθ − iaθθ.(b) Now consider two pairs of complex Grassman variables, θa and θb,

with a and b taking the values 1, 2. Show that∫dθ2dθ1θaθb = εab, and

∫dθ2dθ1θaθb = εab,

where ε12 = −ε21 = 1 and ε00 = ε11 = 0.

(c) Given the function of two pairs of complex Grassman variables

f(θa, θb) = exp(−θaMabθb),

where M is a 2× 2 matrix, show that∫dθ2dθ2dθ1dθ1 f(θa, θb) = detM.

L.2 Problem Sheet 2

1. A complex scalar field theory has action

S =∫d4x (∂µφ)(∂µφ)− V (|φ|).

The field may be expanded

φ(x) =∫ d 3k

2ωk

(ake

−ik·x + b∗keik·x),

where ωk = (k2 +m2)12 . In the quantum field theory, the ladder oper-

ators ak, a∗k, bk, and b∗k satisfy the relations [ak, a∗k′ ] = 2ωk δ

3(k − k′),[bk, b

∗k′ ] = 2ωk δ

3(k− k′), with all other pairs of ladder operators com-muting.

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Page 47: Relativistic Quantum Fields 2

(a) As shown in the lectures, there is a U(1) symmetry leading to aconserved current jµ = iφ(∂µφ)− i(∂µφ)φ. Show that

:Q: =∫ d 3k

2ωk

(a∗kak − b∗kbk).

where Q =∫d3xj0(t,x) is the charge operator.

(b) Show that [:Q:, ak] = −ak, [:Q:, a∗k] = a∗k, [:Q:, bk] = bk, and[:Q:, b∗k] = −b∗k,

(c) Hence or otherwise show that the states a∗(k)|0〉 and b∗(k)|0〉(where |0〉 is the vacuum state) are eigenstates of the charge op-erator, and give their charges.

(d) Show that [:Q:, φ(x)] = −φ(x), and prove that of Q|0〉 = 0 thenthe vacuum expectation value of the field operator, 〈0|φ(x)|0〉,must vanish.

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2. Consider a massless spinor with action

S =∫d4x ψi 6∂ψ,

which is invariant under the two-parameter global symmetry transfor-mation ψ → ψ′ = e−iα−iβγ

5ψ.

(a) Suppose we wish to gauge these symmetries by allowing space-timedependent α and β. Find the transformation laws for the vectorfields Vµ and Aµ which are required to construct the covariantderivative

Dµψ = (∂µ − ieV Vµ − ieAAµγ5)ψ.

(b) Write down the gauge invariant action constructed from this co-variant derivative. Show that it can be re-expressed in terms ofthe chirality eigenspinors ψL and ψR as

Sgi =∫d4x

(ψLi 6D1ψL + ψRi 6D2ψR

),

where D1µ = ∂µ− ie1A

1µ and D2

µ = ∂µ− ie2A2µ, giving the coupling

constants e1 and e2 in terms of eA and eB, and the vector fieldsA1µ and A2

µ in terms of Vµ and Aµ.

3. The action for a free electromagnetic field is

S =∫d4x

(−1

4FµνF

µν).

(a) Show that

δFρσ(x)

δAµ(y)=

∂xρδ4(x− y)δµσ −

∂xσδ4(x− y)δµρ ,

and hence that δSδAµ(x)

= ∂νFνµ(x). Write down the field equation

for Aµ in the Lorentz gauge.

(b) Show also that the two actions

S1 =∫d4x

(−1

4FµνF

µν − 1

2(∂ · A)2

), S2 =

∫d4x

(−1

2∂µAν∂

µAν)

both give the field equation you wrote down in part (a). Explainbriefly how it is that apparently different actions give the sameequation.

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4. Consider an action for an SU(2) doublet spinor coupled to an SU(2)gauge field

S =∫d4x

(−1

4F aµνF

aµν + Ψi 6DΨ−mΨΨ).

where Dµ = ∂µ − igAaµτa, F aµν = ∂µA

aν − ∂νAaµ + gεabcAbµA

cν , and τa =

σa/2, where σa are the Pauli matrices.

(a) Show that the resulting field equations are

∂νFaνµ + gεabcAbνF

cνµ = −gΨγµτaΨ,

i 6DΨ−mΨ = 0.

(b) Show that F aµν = 1

2εµνρσF

aρσ satisfies the identity

∂νF aµν + gεabcAbνF c

µν ≡ 0.

You may find the identity εabcεcde = δadδbe − δaeδbd useful.

(c) (Optional) Defining a 4-vector “current” Kµ = AaνF aµν , show that

∂µKµ =1

2F aµνF a

µν

.

L.3 Problem Sheet 3

1. An action for a massive vector field is

S =∫d4x

(−1

4FµνF

µν +1

2m2AµA

µ).

(a) Show that the field equations resulting from extremising the actionis

[ηµν(∂2 +m2)− ∂µ∂ν ]Aν = 0,

and show that ∂ · A = 0 follows. Contrast this equation for ∂ · Awith the status of ∂ · A = 0 in the theory for a massless vectorfield. Justify the assertion that the massive vector field has threepolarisation states.

(b) By writing Aν(x) = aνe−ik·x, with aν a constant, show that themomentum space version of the operator in square brackets inpart (a) is [ηµν(−k2 +m2) + kµkν ].

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(c) The Feynman propagator for the massive vector field is defined inthe usual way by

[ηµν(∂2 +m2)− ∂µ∂ν ]Dν

ρ(x) = ηµρδ4(x).

Give an argument which shows that the propagator must have theform

Dµν(x) =∫δ4k [A(k2)ηµν +B(k2)kµkν ]e

−ik·x

and show that A(k2) = −(k2 −m2)−1, B(k2) = [m2(k2 −m2)]−1.

2. The action for a complex scalar field is

S =∫d4x

[(∂µφ)(∂µφ)− V (|φ|)

],

with V (|φ|) = −µ2φφ+ λ(φφ)2.

(a) By making the change of field variables φ = 1√2(v+ h(x) + ia(x)),

with v2 = µ2/λ, show that the quadratic (free) part of the actionis

S0 =∫d4x

(1

2∂µh∂

µh− 1

2m2 +

1

2∂µa∂

µa),

giving an expression for m in terms of the original parameters ofthe action.

(b) Suppose we introduce a spinor field ψ, so that the action gains anextra piece

Sψ =∫d4x

(ψi 6∂ψ − gφψLψR − gφψRψL

).

Show that the spinor field has a mass gv, and interacts with thefields h and a through the terms −ghψψ and −igaψγ5ψ

3. The Lagrangian for a complex doublet scalar Φ with an SU(2) gaugesymmetry is

L = (DµΦ)†(DµΦ)− V (Φ)− 1

2tr(FµνF

µν),

where Dµ = ∂µ − igAµ and −igFµν = [Dµ, Dν ].

(a) If we make the SU(2) tranformations Φ → Φ′ = UΦ and Aµ →A′µ = UAµU

−1 − ig∂µUU

−1, verify that the covariant derivativetransforms as

Dµ → D′µ = UDµU−1.

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(b) If the potential has the form V (Φ) = −µ2Φ†Φ + λ(Φ†Φ)2, showthat the Lagrangian has the U(1) global symmetry Φ → Φe−iβ.Suppose that we wish to gauge this symmetry: how should weamend the Lagragian? (You should display the transformationproperties of any extra fields you introduce).

(c) Suppose now that we have two SU(2) scalar doublets Φ1 and Φ2,transforming under the combined SU(2)×U(1) gauge symmetryas

Φ1 → Φ′1 = UΦ1e−iβ, Φ1 → Φ′2 = UΦ2e

iβ.

Verify that the following potential is gauge invariant:

V (Φ1,Φ2) =

−µ21Φ†1Φ1 − µ2

2Φ†2Φ2 + λ1(Φ†1Φ1)2 + λ2(Φ†2Φ2)2 + λ3(Φ†1Φ2)(Φ†2Φ1),

and show that the potential has an extremum at

Φ1 =

(0

v1/√

2

), Φ2 =

(0

v2/√

2

).

Does this theory have any other internal symmetries?

4. (a) It follows from the identity 1 =∫Dφδ[f(φ)− ω]

∣∣∣det δfδφ

∣∣∣ that

δ[f(φ)− ω] = δ[φ− φ]

∣∣∣∣∣detδf

δφ

∣∣∣∣∣−1

.

where φ is the solution to the equation f(φ)−ω = 0 (here assumedto be unique). Using the latter form of the identity, show that∫

Dφδ[eφ − ω]e−12

∫d4xφ2(x) = e−

∫d4x (lnω(x)+ 1

2ln2 ω(x)).

(b) The measure for the functional integration over a complex scalarfield is DφDφ =

∏x dφ(x)dφ(x). Introducing “polar” fields ρ(x)

and θ(x), where φ = ρiθ, changes the measure according toDφDφ =DρDθ| detM |, where

M(x, y) =

δφ(x)δρ(y)

δφ(x)δρ(y)

δφ(x)δθ(y)

δφ(x)δθ(y)

.Show that

DφDφ = DρDθe∫d4x ln ρ(x).

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L.4 Problem Sheet 4

1. The axial gauge is defined by

Ga(A) = tµAaµ = 0,

where tµ is a constant spacelike 4-vector.

(a) If αa are the parameters of an infinitesimal gauge transformationon Aaµ, find the operator

Mab(x, y) =δGa(A(x))

δαb(y),

and hence write down the complete action for a non-Abelian gaugefield Aaµ in the axial gauge, include gauge-fixing and ghost terms.

(b) Show that the momentum space propagator in the axial gauge is

Dµνab (k) =

δabk2 + iε

[−ηµν +

tµkν + kµtν

k · t+

(ξk2 − t2)

(t · k)2kµkν

].

2. (a) Show that

tr(γµγν) = 4ηµν , tr(γµγνγργσ) = 4(ηµνηρσ − ηµρηνσ + ηµσηνρ)

(b) Consider the following diagram in Quantum Electrodynamics (QED),which is a theory of a spinor field of mass m and charge e, coupledto an Abelian gauge field, in Feynman gauge.

µ ν

Show that applying the Feynman rules to this diagram results inthe expression

−iηµρk2 + iε

iΠρσ(k)−iησνk2 + iε

,

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Page 53: Relativistic Quantum Fields 2

where

iΠµν(k) = −4e2∫d 4q

[qµ(q + k)ν + (q + k)µqν − ηµν(q · (q + k)−m2)]

[(q + k)2 −m2 + iε][q2 −m2 + iε].

NB Πµν(k) is called the polarization tensor.

(c) It can be shown that kµΠµν(k) = kνΠµν(k) = 0. What does this

imply about the dependence of Πµν on k?

3. (a) Prove the following γ-matrix identities:

γµγµ = 41, γµγνγµ = −2γν .

(b) Consider the following diagram in QED:

Show that applying the Feynman rules (in Feynman gauge) to thisdiagram results in the expression

i

6p−m+ iεiΣ(p)

i

6p−m+ iε

where

iΣ(p) = −e2∫d 4q

4m− 2 6q[(p− q)2 + iε](q2 −m2 + iε)

.

4. Consider the following correction to the propagator of a non-Abeliangauge field:

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Page 54: Relativistic Quantum Fields 2

µ, a ν, b

(a) Find the expression for the contribution of this diagram to thepolarisation tensor Πµν

ab resulting from the application of the Feyn-man rules (in Feynman gauge) to this diagram.

(b) In a non-Abelian gauge theory, with no other fields, what otherdiagrams contribute to Πµν

ab ?

L.5 Problem Sheet 5

1. Consider a theory of a real scalar field φ and an N -component complexscalar field χa (a = 1, . . . , N), whose action is

S =∫d4x

(1

2∂φ · ∂φ+ ∂χa · ∂χa −

1

2m2φ2 −M2χaχa −

1

2gφ2χaχa

).

(a) Write down the generating functional for the n-point functions ofthis theory, and show that in the absence of external currents onecan perform the functional integral over χa to obtain

Z =∫Dφei

∫d4x ( 1

2∂φ·∂φ− 1

2m2φ2−V1(φ)),

where V1(φ) = −iN∫d 4k ln(k2 −M2 − 1

2gφ2 + iε).

(b) Define what is meant by a 1-particle irreducible (1PI) Feynmandiagram. Write down the Feynman diagram and its associatedintegral for the the 1PI function for 2n ingoing φ fields with zero4-momentum, Γ2n(0, . . . , 0), at one-loop order.

(c) Show that, up to a φ-independent (and possibly infinite) constant,

V1(φ) = i∞∑1

1

(2n!)φ2nΓ2n(0, . . . , 0).

Why can we ignore the constant?

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Page 55: Relativistic Quantum Fields 2

(d) Define m2(φ) = m2 + 12gφ2. Show that, using an upper cut-off

Λ m(φ) on the momentum integration,

V1(φ) = NΛ2

16π2m2(φ) +N

1

16π2m4(φ)

[ln(m2(φ)/Λ2

)− 1

2

].

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2. (a) Write down the covariant derivative terms for the fermion fieldsin the Glashow-Weinberg-Salam (GWS) model of electroweak in-teractions, in terms of their weak eigenstates, SU(2)×U(1) gaugefields Aaµ and A0

µ, and gauge couplings g and g′. Show that theinteraction terms can be written

Ls,i = g(WµJµ +W †

µJ†µ + ZµJ

µZ) + eAµJ

µem,

where Wµ = (A1µ − iA2

µ)/√

2, Zµ = cos θWA3µ − sin θWA

0µ, Aµ =sin θWA

3µ + cos θWA

0µ, and θW is the weak mixing angle. Be sure

to give full expressions for the currents Jµ, JµZ , and Jµem.

(b) Show that JµZ and Jµem maintain their form in the mass eigenstatebasis, i.e. that (unlike the charged current) these currents do notcouple fermions with different flavour indices.

(c) Let Φ be the scalar doublet of the GWS model, with Φ†Φ = (v +h(x))2/2 in the symmetry-broken phase, where h is the Higgs field.Show that the couplings of the Higgs to fermion mass eigenstatesare proportional to the masses of the fermions.

(d) Show also that there are Higgs-W interaction terms of the formshWW and hhWW , giving the coupling constants.

3. The bare interaction Lagrangian of QED is Li = e0ψ0γµψ0Aµ0 .

(a) Show that if we demand that the action be dimensionless in ddimensions, the mass dimension of the bare coupling constant e0

is ε/2, where ε = 4− d.

(b) Show that the photon self energy of Problem 4.2(b) becomes

iΠµν(k) =

−4e20

∫ 1

0dx

∫d dl

(1− 2d)ηµνl2 − 2x(1− x)pµpν + ηµν(m2

0 + x(1− x)k2)

(l2 + α)2,

where α = m20−x(1−x)k2. Note that in d dimensions, ηµνη

µν = d.

(c) Perform the integrals to show that iΠµν(k) = (ηµνk2−kµkν)iΠ(k2),where

Π(k2) = − e20

∫ 1

0dxx(1− x)

(2

ε− lnα− γ

).

(d) Given that the bare photon propagator can be written

iDµν(k) =−iηµν

k2(1− Π(k2)),

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show how the divergence in Π(k2) can be dealt with using thephoton wavefunction renormalisation constant Z3.

(e) Introducing an arbitrary renormalisation parameter µ to define adimensionless renormalised coupling e = Z3e0µ

−ε/2, find the QEDβ-function,

β = µde

in the limit ε → 0. What is peculiar about the relation betweenthe renormalised and bare couplings?

Formulae for Problem 3:

1

A1A2

=∫ 1

0dx

1

[xA1 + (1− x)A2]2

∫d dl

1

(l2 + α)n=

1

(4π)d/2Γ(n− d/2)

Γ(n)

(1

α

)n−d/2.

zΓ(z) = Γ(1 + z), limz→0

Γ(z) =1

z− γ

where γ ' 0.5772.

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