8 Probing the Proton: Electron - proton scatteringvjm/Lectures/SHParticlePhysics2012_file… · 8 Probing the Proton: Electron - proton scattering Scattering of charged leptons by

8 Probing the Proton: Electron - proton scattering

Scattering of charged leptons by protons is an electromagnetic interaction. Electronbeams have been used to probe the structure of the proton (and neutron) since the1960s, with the most recent results coming from a high energy electron-proton collidercalled HERA at DESY in Hamburg.

These experiments provide direct evidence for the composite nature of protons andneutrons, and measure the distributions of the quarks and gluons inside the nucleon.

The results of e−p→ e

−p scattering depend strongly on the wavelength λ = E/c.

• At very low electron energies λ >> rp, where rp is the radius of the proton, thescattering is equivalent to that from a point-like spin-less object

• At low electron energies λ ∼ rp the scattering is equivalent to that from a extendedcharged object

• At high electron energies λ < rp: the wavelength is sufficiently short to resolvesub-structure. Scattering from constituent quarks

• At very high electron energies λ << rp: the proton appears to be a sea of quarksand gluons.

8.1 Form Factors

Extended object - like the proton - have a matter density ρ(r), normalised to unitvolume:

d

3r ρ(r) = 1. The Fourier Transform of ρ(r) is the form factor, F (q):

F (q) =

d

3r expiq · r ρ(r) ⇒ F (0) = 1 (8.1)

Cross section from extended objects are modified by the form factor:

dσ

dΩ

extended

≈dσ

dΩ

point−like

|F (q)|2 (8.2)

For e−p → e

−p scattering we need form factors are required: F1 to describe the distri-

bution of the electric charge F2 to describing the recoil of the proton

8.2 Elastic Scattering

The elastic scattering of a pointlike spin-1/2 electron by a pointlike spin 1/2 target isdescribed in the relativistic limit pe = Ee by the Mott formula:

dσ

dΩ

point

=α

2

4p2esin4 θ

2

cos2 θ

2−

q2

2m2p

sin2 θ

2

(8.3)

51

In this formula θ and pe are in the Lab frame. The no recoil limit corresponds to a verymassive target m

2p q

2.

In the non-relativistic limit pe me this reduces to Rutherford scattering:

dσ

dΩ

NR

=α

2

4m2ev4

esin4 θ

2

(8.4)

q

p(p2)

e−(p1)

p(p4)

e−(p3)

−ieKµ

ieγµ

2mpν = Q2 = −q

2 (8.5)

Note that ν > 0 by energy conservation, so Q2

> 0 and the mass squared of the virtualphoton is negative, q

2< 0!

8.3 Form Factors

Deviations from the point-like Mott scattering are described in the no recoil limit by aform factor F (q2), related to the finite size of a charge distribution inside the proton:

dσ

dΩ=

dσ

dΩ

point

|F (q2)|2 (8.6)

At low q2 the distances probed are large compared to the size of the proton, so the

scattering still appears point-like with F (0) = 1. As q2 gets larger the electron probes

deeper into the proton, and F (q2) is found to decrease. Mathematically, the form factoris the Fourier transform of the charge distribution inside the proton:

F (q2) =

ρ(x) exp(iq · x)d3

x (8.7)

The charge distribution ρ(x) is assumed to be spherically symmetric, and normalisedto one. It has a mean square radius < r

2>:

ρ(r)d3

r = 1 < r2

>=

r2ρ(r)d3

r (8.8)

The matrix element for elastic scattering is written in terms of electron and protoncurrents:

M(e−p→ e−p) =

e2

(p1 − p3)2(u3γ

µu1) (u4Kµu2) (8.9)

52

where the proton is treated as an extended structure, with a current operator Kµ that

is more complex than the point-like γµ:

Kµ = γ

µF1(q

2) +iκp

2mp

F2(q2)σµν

qν + qνF3(q

2) (8.10)

In this general form there are three form factors F1, F2 and F3 which are functions ofq2. However, electromagnetic current conservation δµ(u4K

µu2) = 0 implies that F3 = 0.

F1 is the electrostatic form factor, while F2 is associated with the recoil of the proton.

The anomalous magnetic moment of the proton is defined by:

µp =e(1 + κp)

2mp

κp = 1.79 (8.11)

The differential cross section for elastic electron-proton scattering is

dσ

dΩ

lab

=α

2

4E21 sin4 θ

2

E3

E1

F

21 −

κ2q2

4m2p

F22

cos2 θ

2−

q2

2m2p

(F1 + κF2)2 sin2 θ

2

(8.12)

where the two form factors F1 and F2 are functions of q2 which parameterise the struc-

ture of the proton. They have to be determined by experiment. For a point-like spin1/2 particle, F1 = 1, κ = 0, and the above equation reduces to the Mott scatteringresult.

It is common to use linear combinations of the form factors:

GE = F1 +κq

2

4m2p

F2 GM = F1 + κF2 (8.13)

which are referred to as the electric and magnetic form factors, respectively.

The differential cross section can be rewritten as:

dσ

dΩ

lab

=α

2

4E21 sin4 θ

2

E3

E1

G

2E

+ τG2M

1 + τcos2 θ

2+ 2τG

2M

sin2 θ

2

(8.14)

where we have used the abbreviation τ = Q2/4m2

p. The experimental data on the form

factors as a function of q2 are in good agreement with a “dipole” fit:

GE =GM

µp

=

β

2

β2 + Q2

2

β = 0.84GeV (8.15)

This corresponds to an exponential charge distribution:

ρ(r) = ρ0 exp(−r/r0) 1/r20 = 0.71GeV2

< r2

>= 0.81fm2 (8.16)

The whole of the above discussion can be repeated for electron-neutron scattering, withsimilar form factors for the neutron.

53

Note that even though the neutron has zero total charge, it has an anomalous magneticmoment:

µn =eκn

2mn

κn = −1.91 (8.17)

The anomalous magnetic moments are themselves evidence that the proton and neutron

are not pointlike Dirac fermions.

54

8.4 Deep Inelastic Scattering

During inelastic scattering the proton can break up into its constituent quarks whichthen form a hadronic jet. At high q

2 this is known as deep inelastic scattering (DIS).

q

p(p2)

e−(p1)

X

e−(p3)

ieγµ

The invariant mass, W , of the final state hadronic jet is:

W2 = m

2p+ 2mpν + q

2 (8.18)

Since W = mp the four-momentum and energy transfer, q2 and ν, are two independent

variables in DIS, and it is necessary to measure E1, E3 and θ in the Lab frame todetermine the full kinematics.

It is useful to introduce two dimensionless variables, a parton energy x, and a rapid-ity y, which replace ν and q

2

x =Q

2

2mpν=−q

2

2mpνy =

p2 · q

p2 · p1=

ν

E1(8.19)

It is an exercise to show that the allowed kinematic ranges of these variables are 0 ≤x ≤ 1 and 0 ≤ y ≤ 1.

8.5 Structure Functions

The matrix element squared for DIS can be factorised into lepton and hadron currents:

|M|2 =

e4

q2L

µν

e(Whadron)µν (8.20)

where the hadronic part Wµν

hadron describes the inelastic breakup of the proton. As inelastic scattering, there are two independent form factors in Whadron, which are knownas structure functions W1(ν, Q2) and W2(ν, Q2).

In the Lab frame, the doubly differential cross section for deep inelastic scatteringis:

dσ

dE3dΩ

lab

=α

2

4E21 sin4 θ

2

W2(ν, Q

2) cos2 θ

2+ 2W1(ν, Q

2) sin2 θ

2

(8.21)

55

Bjorken predicted that at high energy the structure functions should exhibit a propertycalled scaling:

mpW1(ν, Q2) → F1(x) νW2(ν, Q

2) → F2(x) (8.22)

where F1(x) and F2(x) are now functions of x alone.

Note that the structure functions F1(x) and F2(x) in DIS are different from the elastic

form factors F1(q2) and F2(q2)!

The experimental data for F2 are shown in figure 8.1. For intermediate regions ofx and Q

2 scaling holds, but at high Q2 and low x there is a significant amount of

scaling violation which we will discuss later.

Figure 8.1: Structure function F2 for large Q2 and small x, as measured at HERA using

collisions between 30 GeV electrons and 830 GeV protons.

56

8.6 The Parton Model

The parton model was proposed by Feynman in 1969, to describe deep inelastic scat-tering in terms of point-like constituents inside the nucleon known as partons with aneffective mass m < mp. Nowadays partons are identified as being quarks or gluons.

q

m

e−(p1)

m

e−(p3)

The parton model restores the elastic scattering relationship between q2 and ν, with m

replacing mp:

ν +q2

2m= 0 (8.23)

and the cross section for elastic electron-parton scattering is:

dσ

dΩ

lab

=Z2

α2

4E21 sin4 θ

2

E3

E1

cos2 θ

2−

q2

2m2sin2 θ

2

(8.24)

where Z is the charge of the parton (+2/3 or −1/3 for quarks).

Effectively the DIS structure functions for scattering off a single parton, have becomedelta functions:

2W1 =Q

2

2m2δ(ν −

Q2

2m) W2 = δ(ν −

Q2

2m) (8.25)

It can be seen that the parton energy variable:

x =Q

2

2mpν=

m

mp

(8.26)

is the fraction of the proton rest mass carried by the parton.

8.7 Parton Distribution Functions

We introduce the parton distribution functions, fi(x), defined as the probability thata parton of type i carries a fraction x of the proton mass. The structure functions canthen be written:

W1(x) =F1(x)

mp

F1(x) =1

2

i

Z2ifi(x) (8.27)

W2(x) =F2(x)

νF2(x) =

i

xZ2ifi(x) (8.28)

57

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x* f(

x, µ

=2 G

eV)

X

CT10.00 PDFs

g/5ud

ubardbar

sc

0

0.2

0.4

0.6

0.8

1

1.2

.9.8.7.6.5.4.3.2.1.0510-210-310-5

3*x5/

3 * f(

x, µ

=85

GeV

)

X

CT10.00 PDFs (area proportional to momemtum fraction)

gud

ubardbar

sc

The structure functions F1 and F2 satisfy the Callan-Gross relation:

2xF1(x) = F2(x) =

i

xZ2ifi(x) (8.29)

The valence quark distributions are written u(x) and d(x). In a proton they havenormalisations: 1

0

u(x)dx = 2

1

0

d(x)dx = 1 (8.30)

and the contribution of the valence quarks to F2 are:

Fp

2 (x) =4

9xu(x) +

1

9xd(x) (8.31)

For a neutron u and d are interchanged:

Fn

2 (x) =4

9xd(x) +

1

9xu(x) (8.32)

58

There is an additional sea of quarks and antiquarks at low x, which also contribute toF2, shown as u(x) = d(x) = s(x).

8.8 Quark, Antiquark & Gluon Fractions

The total number of valence quarks can be obtained from:

FCC

3 (νN) =u(x)− u(x) + d(x)− d(x)

(8.33)

1

0

FCC

3 (νN)dx = 3 (8.34)

The fraction of sea quarks is obtained from:

q(x)dxq(x)dx

=3R− 1

3−R≈ 0.1 (8.35)

where R is the ratio of the antineutrino to neutrino total cross-sections. For a pointlikespin 1/2 fermion, f :

R =σ(νf)

σ(νf)=

1

3(8.36)

whereas for an antifermion R = 3.

If we integrate the area under the electron DIS structure function F2(x) we measurethe total fraction carried by the valence and sea quarks.

1

0

Fp

2 (x)dx =4

9fu +

1

9fd

1

0

Fn

2 (x)dx =4

9fd +

1

9fu (8.37)

We obtain the surprising results:

1

0

Fp

2 (x)dx = 0.18

1

0

Fn

2 (x)dx = 0.12 (8.38)

fu = 0.36 fd = 0.18 (8.39)

The quarks constitute only 54% of the proton rest mass!The remainder is carried by gluons which must also be considered as partons.

Understanding the gluon component of the proton presents a problem, since the lowestorder scattering processes of electrons and neutrinos only couple to the quarks.

There are additional scattering processes that involve gluons either in the initial state(“hard scattering”) or in the final state (“gluon emission”), where there is an addi-tional hadronic jet from the gluon. These have different kinematics from the lowestorder scattering.

59

Figure 8.2: Gluon emission from a final state quark γ∗q → qg

Figure 8.3: Hard scattering off an initial state gluon γ∗g → qq

9 Quantum Chromodynamics

Quantum Chromodynamics or QCD is the theory of strong interactions between quarksand gluons. It is a quantum field theory similar to QED but with some crucial differ-ences.

9.1 Feynman rules for QCD amplitudes

The calculation of Feynman diagrams containing quarks and gluons has the followingchanges compared to QED:

• The coupling constant α becomes αs (where√

αs = gs).

• αs(q2) decreases rapidly as a function of q2.

At small q2 it is large, and QCD is a non-perturbative theory.

At large q2 QCD becomes perturbative like QED.

• A quark has one of three colour states (replacing electric charge).Antiquarks have anticolour states.

• A gluon propagator has one of eight colour-anticolour states.

• A quark-gluon vertex has a factor −igsλaγ

ν .

• As a consequence of the gluon colour states, gluons can self-interact in three orfour gluon vertices.

60

Figure 8.4: Evidence for gluon emission from transverse momentum squared of jets indeep inelastic scattering of muons. The difference between the solid and dashed linesis due to gluon emission.

q1q3

q2

a, µ b, ν

c, λa, µ

b, ν

d, ρ

c, λ

• There is a complicated factor for a three-gluon vertex:

− gSfabc [gµν(q1 − q2)λ + gνλ(q2 − q3)µ + gλµ(q3 − q1)ν ] (9.1)

where fabc are known as colour structure constants.

9.2 SU(3) Colour

Quarks carry one of three colour states, red r, green g, and blue b. Antiquarks carrythe anticolour states, r, g, and b. We define three quark eigenstates r, g, and b:

cr =

100

cg =

010

cb =

001

(9.2)

The colour states are related by an SU(3) symmetry group which defines rotationsin colour. Strong interactions are invariant under SU(3) colour.

The coupling is independent of the colour states of the quarks and gluons.

9.2.1 Gluon States

At a quark-gluon vertex a quark can either change colour or remain the same. A changeof colour requires that gluons also have colour states.

61

This differs from QED where the photon has no charge.

A gluon carries a colour and an anti-colour state cicj. Naively you would think thatthere are nine possible gluon states:

rb br rg gr bg gb rr bb gg

of which the last three are apparently colour neutral. The symmetry properties of theSU(3) group actually classify the states as 3 ⊗ 3 = 8 ⊕ 1. There is a colour-octet ofallowed gluon states which are colour antisymmetric:

G1 = 1√2

rb + br

G2 = 1√

2

rb− br

G4 = 1√2(rg + gr) G5 = 1√

2(rg − gr)

G6 = 1√2

bg + gb

G7 = 1√

2

bg − gb

G3 = 1√2

rr − bb

G8 = 1√

6

rr + bb− 2gg

(9.3)

and a colour-singlet, which is the only colour symmetric state:

G0 =1√

3

rr + gg + bb

(9.4)

The singlet state G0 is forbidden for gluons, because it would give rise to long-rangestrong interactions.

9.2.2 The λ matrices

There are eight λ matrices which are the generators of SU(3):

λ1 =

0 1 01 0 00 0 0

λ2 =

0 −i 0i 0 00 0 0

λ3 =

1 0 00 −1 00 0 0

λ4 =

0 0 10 0 01 0 0

λ5 =

0 0 −i

0 0 0i 0 0

λ6 =

0 0 00 0 10 1 0

λ7 =

0 0 00 0 −i

0 i 0

λ8 =

1√

3

1 0 00 1 00 0 −2

The λ matrices can be identified with the eight gluon states. The operators λ1,2,4,5,6,7

are the ones that change quark colour states. The diagonal operators λ3,8 are the ones

that do not change the colour states.

The SU(3) structure constants describe the commutators of the λ matrices:

λ

a, λ

b

= 2i

c

fabcλc (9.5)

62

Non-zero values are:

f123 = 1 f458 = f678 =

√3

2(9.6)

f147 = f165 = f246 = f257 = f345 = f376 =1

2(9.7)

The fabc are related to each other by the property that thay are antisymmetric underthe interchange of any pair of indices, e.g. f132 = f213 = f321 = −1 and f231 = f312 = 1from the value f123 = 1. The fabc not covered by permutations of the values given aboveare zero.

More familiar are the three Pauli matrices σ which are the generators of SU(2). They

can be seen as subsets of λ1,4,6 (σ1), λ

2,5,7 (σ2), and λ3 (σ3).

Note that the Pauli matrices have only one structure constant ijk = f123.

9.2.3 Non-Abelian Gauge Symmetry

A rotation in colour space is written as:

U = e−iαa·λa

(9.8)

where αa are the equivalent of “angles” in colour space. QCD amplitudes can be shownto be invariant under this non-Abelian gauge transformation.

The transformations of the quark and gluon states are:

q → (1 + iαaλa)q G

a

µ→ G

a

µ−

1

gs

∂µαa − fabcαbGc

µ(9.9)

9.3 Quark-antiquark scattering

g

p1, c1

p2, c2

p3, c3

p4, c4

gS

gS

The matrix element for quark-antiquark scattering is written:

M =u3c

†3

−igs

2λ

aγ

µ

[u1c1]

gµνδ

ab

q2

v2c

†2

−igs

2λ

bγ

ν

[v4c4] (9.10)

M =αs

4q2[u3γ

µu1] [v2γµv4] (c

†3λ

ac1)(c

†2λ

ac4) (9.11)

63

This looks very similar to the matrix element for electron-positron scattering, exceptthat α is replaced by αs and there is a colour factor cf :

cf =1

2(c†3λ

ac1)(c

†2λ

ac4) (9.12)

The calculation of the cf can be found in Halzen & Martin P. 67-69:

quark states gluon states cf

rr ↔ rr G7, G8 +2/3rr ↔ rr G7, G8 −2/3rb ↔ rb G8 −1/3rb ↔ br G1 +1rr ↔ bb G1 −1rb↔ rb G8 +1/3

9.4 Strong Coupling Constant αs

The coupling strength αs is large, which means that higher order diagrams are impor-tant. At low q

2 higher order amplitudes are larger than the lowest order diagrams, sothe sum of all diagrams does not converge, and QCD is non-perturbative. At highq2 the sum does converge, and QCD becomes perturbative. The coupling constant

αs can be renormalised at a scale µ in a similar way to α:

α(q2) =α(µ2)

1− α(µ2)3π

log

q2

µ2

(QED)

In QCD the renormalization is attributed both to the colour screening effect of virtualqq pairs and to anti-screening effects from gluons since they have colour-anticolourstates. This leads to a running of the strong coupling constant:

αs(q2) =

αs(µ2)

1 + βαs(µ2) ln

q2

µ2

β =(11Nc − 2Nf )

12π(QCD)

where Nc = 3 is the number of colours, and Nf ≤ 6 is the number of active quarkflavours which is a function of q

2.

From the positive sign of β it can be seen that the anti-screening effect of the gluonsdominates, and the strong coupling constant decreases rapidly as a function of q

2. Therunning of the coupling constant is usually written:

αs(q2) =

12π

(33− 2Nf )ln

q2

Λ2QCD

(9.13)

where ΛQCD = 217± 25 MeV is a reference scale which defines the onset of a stronglycoupled theory where αs ≈ 1.

There are many independent determinations of αs at different scales µ, which are sum-marized in the figures on the next page. The results are compared with each other byadjusting them to a common scale µ = MZ where αs = 0.1184(7).

64

Figure 9.1: Top: Running of the strong coupling constant αs(q2). Bottom: Measure-ments of αs(MZ), the equivalent strong coupling at q

2 = M2Z.

9.5 Confinement and Asymptotic Freedom

In deep inelastic scattering experiments at large q2, the quarks and gluons inside the

proton can be observed as partons. This property is known as asymptotic freedom.It is associated with the small value of αs(q2), which allows the strong interaction cor-rections from gluon emission and hard scattering to be calculated using a perturbativeexpansion of QCD. The perturbative QCD treatment of high q

2 strong interactions hasbeen well established over the past 20 years by experiments at high energy colliders.

In contrast, at low q2 the quarks and gluons are tightly bound into hadrons. This is

known as confinement. For large αs QCD is not a perturbative theory and differentmathematical methods have to be used to calculate the properties of hadronic systems.A rigorous numerical approach is provided by Lattice gauge theories.

The breakdown of the perturbation series is due to the colour states of the gluons and thecontributions from three-gluon and four-gluon vertices which do not have an analoguein QED. A pictorial way of thinking of this is as a colour flux tube, connecting thequarks in a hadron. Starting from the familiar dipole field between two charges, imaginethe colour field lines as being squeezed down into a tight line between the two quarks:

The additional energy density stored in the flux tube as the qq pair are pulled apartcan be parametrized by a string tension k, and the QCD potential can be written asthe sum of a Coulomb-like potential and the string potential energy:

Vqq(r) = −4

3

αs

r+ kr (9.14)

As a consequence of the positive term in the strong interaction potential, a qq paircannot be separated since infinite energy would be required. Instead the flux tube can

65

break creating an additional qq pair in the middle which combines with the original qq

pair to form two separate hadrons.

There are no free quarks or gluons!

66

10 QCD at Colliders

10.1 e+e− → Hadrons

In the electromagnetic process e+e− → qq, the flavour of the q and q must be the same.

This is known as associated production. The final state quark and antiquark each forma jet by a process known as fragmentation. The two jets follow the directions of thequarks, and are back-to-back in the center-of-mass system.

γ

e−(p1)

e+(p2)

q(p3)

q(p4)

There is an initial colour flux between the quark and antiquark, but this is broken bythe energy of the collision. The quarks then radiate gluons, which couple to quark-antiquark pairs, eventually forming a large number of bound states known as hadrons.The fragmentation process is characterised by the transverse momentum, pT , of thehadrons relative to the jet axis.

The cross section for e+e− → qq can be calculated using QED. The matrix element is

similar to e+e− → µ

+µ− apart from the final state charges and a colour factor, Nc = 3

σ(e+e−→ qq) = NcZ

2qσ(e+

e−→ µ

+µ−) (10.1)

Summing over all the active quark flavours gives the ratio:

R ≡σ(e+

e− → hadrons)

σ(e+e− → µ+µ−)= 3

q

Z2q

(10.2)

67

Note the importance of the factor 3. The measurement of the ratio R shows that there

are three colour states of quarks!6 41. Plots of cross sections and related quantities

! and R in e+e! Collisions

10-8

10-7

10-6

10-5

10-4

10-3

10-2

1 10 102

![m

b]

"

#

$

#!

J/%

%(2S)!

Z

10-1

1

10

10 2

10 3

1 10 102

R "

#

$

#!

J/% %(2S)

!

Z

!s [GeV]

Figure 41.6: World data on the total cross section of e+e! ! hadrons and the ratio R(s) = !(e+e! ! hadrons, s)/!(e+e! ! µ+µ!, s).!(e+e! ! hadrons, s) is the experimental cross section corrected for initial state radiation and electron-positron vertex loops, !(e+e! !µ+µ!, s) = 4"#2(s)/3s. Data errors are total below 2 GeV and statistical above 2 GeV. The curves are an educative guide: the broken one(green) is a naive quark-parton model prediction, and the solid one (red) is 3-loop pQCD prediction (see “Quantum Chromodynamics” section ofthis Review, Eq. (9.7) or, for more details, K. G. Chetyrkin et al., Nucl. Phys. B586, 56 (2000) (Erratum ibid. B634, 413 (2002)). Breit-Wignerparameterizations of J/$, $(2S), and !(nS), n = 1, 2, 3, 4 are also shown. The full list of references to the original data and the details ofthe R ratio extraction from them can be found in [arXiv:hep-ph/0312114]. Corresponding computer-readable data files are available athttp://pdg.lbl.gov/current/xsect/. (Courtesy of the COMPAS (Protvino) and HEPDATA (Durham) Groups, May 2010.) See full-colorversion on color pages at end of book.10.2 Gluon Jets

Sometimes one of the quarks radiates a hard gluon which carries a large fraction of thequark energy, e

+e− → qqg. In this process there are three jets, with one coming from

the gluon. This provided the first direct evidence for the gluon in 1979. The jets areno longer back-to-back, and the gluon jet has a different pT distribution from the quarkjets. The production rate of three-jet versus two-jet events is proportional to αs.

Including higher order processes, the ratio R is:

R = 3

q

Z2q

1 +

αs(q2)

π

(10.3)

This is used to measure the strong coupling constant as a function of q2.

10.3 Hadron Colliders

Proton-antiproton colliders were used to discover the W and Z bosons at CERN inthe 1980s, and the top quark at the Tevatron (Fermilab) in 1995. In 2010 the Large

68

69

Hadron Collider (LHC) began operation at CERN. This collides two proton beams withan initial CM energy of 7 TeV. This is half the eventual design energy, but is alreadythe world’s highest energy collider.

10.3.1 Parton Level Scattering

Figure 10.1: Stylized representation of a hadron-hadron collision from “Jet Physics atthe Tevatron” by A.Bhatti & D.Lincoln, arXiv:1002.1708.

• Each initial hadron provides one parton, which can be either a quark, antiquarkor gluon. At lower energy colliders the valence quarks dominate, whereas at theLHC most scattering is between low x gluons.

• The remnants of the two protons form backward and forward jets similar to thejet in electron-proton DIS. These remnants are known as the “underlying event”,and are usually regarded as background.

• The partons undergo “hard scattering”, a strong interaction with a large q2 trans-

fer. This leads to two or more quarks or gluons in the final state. The amplitudecan be calculated using perturbative QCD.

• The final state quarks and gluons form hadronic jets in a similar way to the quarks(and gluons) in e

+e− → qq(g).

• Additional gluons can be emitted from either the initial or final state quarks (andgluons). A distinction is made between “hard” gluons, and “soft collinear” gluons.

• Most collisions have no missing energy and no high mass final state particlesW, Z, H, b, t. These events are said to be “minimum biased”.

• A large transverse momentum (high pt) jet or isolated charged lepton is a signaturefor the production of a high mass final state particle.

70

• Large missing transverse energy (missing Et) is a signature for neutrinos or otherweakly interacting neutral particles.

10.3.2 Description of Jets

The hadrons that constitute a jet are summed to provide the kinematic informationabout the jet:

• Jet energy E =

iEi and momentum p =

ipi

The direction of p defines the jet axis.

• Jet invariant mass W2 = E

2 − |p|2

• Transverse energy flow within the jet |pT |2 =

i|pi − p|2

• Transverse jet momentum pt =

(p2x

+ p2y), relative to beam axis.

• “Pseudorapidity” y = ln [(E + pz)/(E − pz)]/2], related to cos θ.This goes to ∞ along the beam axis, and is zero at 90.

• Azimuthal angle φ = tan−1(py/px)

In the case of e+e− → qq(g) it is straightforward to decide which hadrons belong to

which jets. At a hadron collider a minimum bias event has at least two jets from thehard scatter, as well as two jets from the underlying event. With large numbers of jetsit is hard to decide which hadrons belong to which jets. What is done is to define aradial distance from the jet axis for each hadron:

R2i

= (yi − yjet)2 + (φi − φjet)

2 (10.4)

An initial estimate is made of yjet and φjet using a subset of the highest momentumhadrons. Then a cone radius R is defined containing all the i hadrons that make upthe jet. These are then used to recalculate yjet and φjet. The process can be iterateduntil it converges. The choice of radius for the cone is critical. A typical value is R = 0.7.

The cross-section for jet production as a function of pt and y can be measured andcompared to pertubative QCD calculations. These are usually carried out to next-to-leading order (NLO), with one additional hard gluon, and sometimes to next-to-next-to-leading order (NNLO). The measurements are used to constrain gluon and quarkparton density functions, and to determine the strong coupling constant αs(q2).

10.3.3 Jet Fragmentation

The initial stages of jet fragmentation involve the emission of soft collinear gluonsfrom the scattered partons. These can still be understood using perturbative QCD.Eventually the energy scales of the fragmentation become too low, and hadronisationinto bound states begins. This part is modelled in several different ways:

71

• PYTHIA breaks colour flux tubes between quarks and gluons.

• HERWIG clusters the quarks and gluons according to colour matching.

• SHERPA uses a cutoff in the transverse energy of a quark or gluon within a jetto stop the production of lower energy jet fragments.

The fragmentation models are compared to data and tuned appropriately. The mostrelevant jet parameter for this is the tranverse energy flow within the jet |pT |

2, althoughthe hadron multiplicity and individual momenta can also be looked at.

10.4 Production of Heavy Quarks

10.4.1 Discovery of the J/ψ

In November 1974, two groups simultaneously announced the discovery of a narrowJ/ψ resonance with a mass of 3100 MeV. At SLAC the resonance was observed ine+e− → qq, while at Brookhaven the inverse process was studied, e

+e− pair production

by a proton beam on a Beryllium target.

Figure 10.2: Discovery of J/ψ in 1974 at Brookhaven (left) and SLAC (right).

The J/ψ is identified as the lightest of the cc states, known as charmonium. Its widthΓ = 0.087 MeV, is much smaller than the experimental resolution, which is a surprise.

10.4.2 bb Production

The narrow bb bound state Υ(1S) with mass 9.5 GeV, was first identified in hadroncollisions at Fermilab in 1977. It is also produced in e

+e− collisions in a similar way to

72

the J/ψ.

The production of b quarks has a significant role at hadron colliders, because it ispossible to “tag” jets that come from b quarks. This is done by measuring detacheddecay vertices due to the finite lifetime of the b quark, τb = 1.5ps. Measurements of theproduction cross-section for b jets can then be compared with perturbative QCD calcu-lations. These are particularly useful for constraining the antiquark and gluon partondensity functions, since bb pairs are produced either by quark-antiquark annihilation orby gluon fusion.

An important use of b tagging is to identify jets that have been produced by the decay ofa heavy particle into a b quark. Examples of these are t→ bW , Z → bb, and eventuallyfor a light Higgs boson, H → bb.

10.4.3 The Top Quark

The top quark was not discovered until 1995, because of its unexpectedly large massmt = 172.0± 1.6 GeV. It was observed at the Tevatron(Fermilab) in proton-antiprotoncollisions at a centre-of-mass energy of 1.8 TeV. The main production mechanism isquark-antiquark annihilation giving tt, followed by the decays t→ bW :

t→ W+b t→ W

−b W

+→

+ν W

−→ qq (10.5)

There are no narrow tt resonances because of the short lifetime of the top quark. Thetotal cross-section for tt production at the Tevatron is 7.5± 0.5 pb. There is also singletop production via a W coupling to the valence quarks, W → tb, with a cross-section2.3± 0.6 pb.

Recently the CDF experiment at the Tevatron has measured a large forward-backwardasymmetry in tt production from proton-antiproton collisions. This is not expected inthe Standard Model! (see preprint arXiv:1101.0034)

73

11 Mesons and Baryons

11.1 Formation of Hadrons

As a result of the colour confinement mechanism of QCD only bound states of quarksare observed as free particles, known as hadrons. They are colour singlets, with gluonexchange between the quarks inside the hadrons, but no colour field outside.

Mesons are formed from a quark and an antiquark with colour and anticolour stateswith a symmetric colour wavefunction:

χc =1√

3(rr + gg + bb) (11.1)

Baryons are formed from three quarks, all with different colour states, with an anti-

symmetric colour wavefunction:

χc =1√

6(rgb− rbg + gbr − grb + brg − bgr) (11.2)

There may be other types of hadrons which are colour singlets: four quark states(qqqq), pentaquarks (qqqqq), hybrid mesons (qqg), or glueballs (gg, ggg). There is someexperimental evidence for these, but it is not yet convincing.

11.2 Isospin

Strong interactions are the same for u and d quarks. This is an SU(2) flavour sym-metry, known rather confusingly as isospin. The u and d quarks are assigned to anisospin doublet:

u : I =1

2, I3 = +

1

2d : I =

1

2, I3 = −

1

2(11.3)

The lowest lying meson states are the pions, which are pseudoscalars with spin J = 0(↑↓). They form an I= 1 triplet:

π+ [1, 1] = ud π

0 [1, 0] =1√

2(uu− dd) π

− [1,−1] = du (11.4)

There is also an I= 0 singlet, the eta meson:

η [0, 0] =1√

2(uu + dd) (11.5)

For baryons the lowest lying states are the J = 1/2 proton and neutron:

p [1/2, +1/2] = uud n [1/2,−1/2] = ddu (11.6)

Isospin symmetry means that protons, neutrons and pions all have the same stronginteractions.

74

11.3 SU(3) Flavour Symmetry

In 1961, before the discovery of quarks, Gell-Mann proposed a structure to classify thehadrons, which was known as the eightfold way. This classification is now understoodas an approximate SU(3) flavour symmetry between the three lightest quarks u, d, s.This symmetry is broken by the mass of the strange quark. The s quark is assigned astrangeness, S= 1 and I= 0, and the u and d quarks have S= 0 and I= 1/2.

The mathematics of SU(3) flavour is identical to that of SU(3) colour which was dis-cussed in lecture 8. The eight generators of the group are the λ

a matrices. The matricesλ

1, λ2, and λ

3 form the SU(2) isospin part of SU(3) flavour which deals with the u andd quarks. The two diagonal matrices, λ

3 and λ8 are related to eigenstates of isospin,

I3, and hypercharge, Y, respectively:

I3 =1

2λ

3 Y = S + B =1√

3λ8 (11.7)

where B= 1/3 is the baryon number of the quarks. The quarks charges can be writtenas:

Q = I3 +Y

2(11.8)

The quark and antiquark flavours can be represented as 2-dimensional SU(3) multipletsof isospin and hypercharge:

+- i1 2!!!!!

+-i4

5

!!!!!!+- i6 !!!!!

7

+- i1 2!!!!!

+-i6!!!!!

7

s

+- i4

5!!!!!!

-3d u

-1/3

2/3

-1/2 1/2I3

-2/3

-1/3I3

-2/3 s

du

2/3YY

-1/2 1/21/31/3

3

11.4 Hadron Multiplets

The SU(3) multiplet structure for the qq meson states is 3⊗ 3 = 8⊕1, i.e. it contains aflavour octet and a flavour singlet. The lowest lying J = 0 (↑↓) pseudoscalar and J = 1(↑↑) vector mesons are shown on the next pages.

Three of the nine J = 0 mesons are neutral with I3 = Y = 0. These states are π0,

η1 and η8, where η1 is the SU(3) singlet. The physically observed states, η and η, are

mixtures of η1 and η8. For the J = 1 mesons, the three neutral states that are observedare ρ

0, ω and φ, where the φ is purely an ss state.

According to SU(3) symmetry, the baryon states are classified as 3⊗3⊗3 = 10⊕8⊕8⊕1,i.e. the three light quarks form a decuplet, two octets and a singlet. The lowest lying

75

baryon octet with J=1/2 (↑↑↓) contains p, n, Λ, Σ+, Σ0, Σ−, Ξ0 and Ξ− states.

The decuplet has J = 3/2 (↑↑↑). It consists of ∆, Σ∗, Ξ∗ states and the Ω− whichis an sss state.

11.4.1 ∆++ and Proton Wavefunctions

The flavour wavefunctions for decuplet baryons are symmetric, e.g.:

1√

6(dus + uds + sud + sdu + dsu + usd) (11.9)

The ∆++(uuu) has symmetric (S) flavour, spin and orbital wavefunctions, and an an-tisymmetric (A) colour wavefunction. It is overall antisymmetric, as it must be for asystem of identical fermions.

The wavefunctions for the lowest baryon octet have a combined symmetry of the flavourand spin parts:

uud(↑↓↑ + ↓↑↑ −2 ↑↑↓) + udu(↓↑↑ + ↑↑↓ −2 ↑↓↑) + duu(↑↑↓ + ↑↓↑ −2 ↓↑↑) (11.10)

Hence the proton also has an overall antisymmetric wavefunction, ψ:

Hadron χc χf χS χL ψ

∆++ A S S S Ap A S S A

Note that there are no J=1/2 states uuu, ddd, sss because the flavour symmetric partwould have to be combined with an antisymmetric spin part.

11.4.2 Hadron Masses*

In the MS renormalization scheme of QCD the masses of the u and d quarks, mu andmd are only a few MeV, and the mass of the s quark, ms ≈ 100 MeV. In discussinghadron masses and magnetic moments we use constituent quark masses mu = md ≈

300 MeV and ms ≈ 500 MeV.

Gell-mann and Okubo introduced a semi-empirical mass relation for baryons:

M = M0 + YM1 + M2

I(I + 1)− Y2

/4

(11.11)

For the J = 3/2 baryon decuplet Y = B + S = 2 (I−1), and the formula reduces to:

M = M∆ + (md −ms)S (11.12)

For the J = 1/2 baryon octet the masses are related by:

3MΛ + MΣ = 2MN + 2MΞ (11.13)

76

K0(ds) K+(us)

π−

(du)π

0

η8 η1

π+

(ud)

K−(su) K0(sd)

Y,S+1

0

−1

I3

−1 −1/2 0 +1/2 +1

Pseudoscalar mesons

JPC = 0−+

π0 = (dd− uu)/

√2

η8 = (dd + uu− 2ss)/√

6

η1 = (dd + uu + ss)/√

3

K∗0(ds) K∗+(us)

ρ−

(du)ρ

0

ω φ

ρ+

(ud)

K∗−(su) K∗0

(sd)

Y,S+1

0

−1

I3

−1 −1/2 0 +1/2 +1

Vector mesons

JPC = 1−−

ρ0 = (dd− uu)/

√2

ω = (dd + uu)/√

2

φ = ss

77

S

−3

−2

−1

0 ∆−

ddd

∆0

ddu

∆+

duu

∆++

uuu

Σ−

dds

Σ0

dus

Σ+

uus

Ξ−

dss

Ξ0

uss

Ω−

sss

I3−3/2 −1 −1/2 0 +1/2 +1 +3/2

Y

−2

−1

0

+1

M

MeV/c2

1232

1385

1533

1672

S0

−1

−2

nddu

puud

Σ−

dds

Σ0

Λuds

Σ+

uus

Ξ−

dss

Ξ0

uss

Y+1

0

−1

I3−1 −1/2 0 +1/2 +1

M

MeV/c2

938.9

11931116

1318

78

These formulae are accurate to ≈ 1%.

The semi-empirical mass relation fails for mesons, but the mass differences betweenJ=0 and J=1 mesons and between J=1/2 and J=3/2 baryons can be understood ashyperfine splitting due to spin-spin coupling between quarks.

11.5 Resonances

The J=1 vector mesons decay to the J=0 pseudoscalar mesons through strong interac-tions in which a second quark-antiquark pair is produced, e.g. ρ

0 → π+π−. Similarily

the J=3/2 baryons (except for the Ω−), decay to the J=1/2 baryons through stronginteractions, e.g. ∆++ → π

+p. The decay widths for these processes are ≈ 100 MeV,

corresponding to lifetimes of ≈ 10−23s. Hadronic states that can decay through stronginteractions are known as resonances. They are observed as broad mass peaks in thecombinations of their daughter particles.

Figure 11.1: The ∆++ resonance observed in π+p scattering.

Figure 11.2: Quark line diagram of a ρ+ meson coupling to π

+π

0.

The Feynman diagram for resonance production can be drawn in reduced form showingjust the quark lines, which are continuous and do not change flavour. Since the processis a low energy strong interaction, there are lots of gluons coupling to the quark lineswhich form a colour flux between them.

79

11.6 Heavy Quark States

The c and b quarks can replace the lighter quarks to form heavy hadrons. However, thet quark is too short-lived to form observable hadrons.

The equivalent of the K and K∗ mesons are known as charm, D

(∗), and beauty, B(∗)

mesons. There are three possible D states (and antipartners):

D+ (cd) D

0 (cu) D+s

(cs) (11.14)

and four possible B states:

B+ (bu) B

0 (bd) B0s

(bs) B+c

(bc) (11.15)

Similarly there are heavy baryons such as Λc and Λb, where a heavy quark replaces oneof the light quarks.

The masses of heavy quark states are dominated by the heavy quark constituent massesmc ≈ 1.5 GeV and mb ≈ 4.6 GeV.

11.6.1 Charmonium and Bottomonium

The J/ψ meson is identified as the lowest 3S1 bound state of cc. The width of this stateis narrow because MJ/ψ < 2MD, so it cannot decay into a DD meson pair. It decaysto light hadrons by quark-antiquark annihilation into gluons. The quark line diagramis drawn as being disconnected:

c

c

Light hadrons

There is a complete spectroscopy of cc charmonium states with angular momentumstates equivalent to atomic spectroscopy. Only the states with M(cc) < 2MD havenarrow widths.

A similar spectroscopy is observed for bb bottomonium states.

80

(2S)

c(2S)

c(1S)

hadrons

hadrons hadrons

hadrons

radiative

hadronshadrons

c2(1P)

c0(1P)

(1S)J/

!JPC 0 " 1 0"" 1"" 1" 2""

c1(1P)

0

hc(1P)

, 0

hadrons

Figure 11.3: Spectroscopy of lightest charmonium states with M(cc) < 2MD. Notethat out of these states only the J/ψ and ψ(2S) are produced in e

+e− collisions. The

singlet ηc states, and the P-wave χc and hc states, are observed via electromagnetictransitions.

81

12 Weak Decays of Hadrons

12.1 Selection Rules for Decays

There are selection rules to decide if a hadronic decay is weak, electromagnetic or strong,based on which quantities are conserved:

• For strong interactions total isospin I, and its projection I3 are conserved. Stronginteractions always lead to hadronic final states.

• For electromagnetic decays I is not conserved, but I3 is. There are often photonsor charged lepton-antilepton pairs in the final state.

• For weak decays I and I3 are not conserved. They are the only ones that canchange quark flavour, e.g. ∆S = 1 when s→ u.

• Weak decays can lead to hadronic, semileptonic, or leptonic final states. A neu-trino in the final state is a clear signature of a weak interaction.

The lightest hadrons cannot decay to other hadrons via strong interactions. The π0,

η and Σ0 decay electromagnetically. The π±, K

±, K0, n, Λ, Σ±, Ξ and Ω− decay via

flavour-changing weak interactions, with long lifetimes.

12.2 Pion and Kaon Decays

12.2.1 Neutral Pion Decay π0 → γγ

The π0 meson is a (uu−dd) state which decays through the annihilation of its uu or dd

quarks into a pair of photons. This is an electromagnetic decays that conserves I3 andcharge conjugation symmetry C. The quantum numbers are [I, I3]π0 = [1, 0], Cπ0 = +1and Iγ = 0, Cγ = −1, so the allowed decay is to two photons. The decay to threephotons has not been observed, B(π0 → 3γ) < 3× 10−8.

The lifetime τπ0 = (8.4 ± 0.6) × 10−17s, is the shortest lifetime ever measured ex-

perimentally. The decay width is:

Γ(π0→ 2γ) = α

2N

2cg

2π0m

3π0 (12.1)

where Nc = 3 is the colour factor, and gπ0 = 92 MeV is the π0 decay constant, which

represents the probability that the quark-antiquark pair will meet inside the meson andannihilate.

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12.2.2 Charged Pion Decay π+ → µ

+νµ

W

d

u

νµ(p4)

µ+(p3)

fπ

Charged pions decay mainly to a muon and a neutrino:

π+→ µ

+νµ π

−→ µ

−νµ (12.2)

This occurs through the annihilation of the ud quark-antiquark into a charged W boson,which is described in terms of a pion decay constant, fπ. The matrix element for thedecay is:

M =VudGF√

2fπq

µ

uµγ

µ1

2(1− γ

5)uνµ

(12.3)

The matrix element squared in the rest frame of the pion is:

|M|2 = 4|Vud|

2G

2Ff

2πm

2µ[p3.p4] (12.4)

and the total decay rate is:

Γ =1

τπ

=|Vud|

2G

2F

8πf

2πmπm

2µ

1−

m2µ

m2π

2

(12.5)

From the charged pion lifetime, τπ+ = 26ns, the pion decay constant can be de-duced, fπ = 131 MeV. Note that this is very similar to the charged pion mass,mπ+ = 139.6 MeV.

The decay of a charged pion to an electron and a neutrino π+ → e

+νe, is helicity sup-

pressed. This can be seen from the factor m2µ/m

2π

in the decay rate. Replacing mµ withme reduces the decay rate by a factor ≈ 10−4. The suppression is associated with the he-licity states of neutrinos, which force the spin-zero π

+ to decay to a left-handed neutrinoand a left-handed µ

+, or the π− to decay to a right-handed antineutrino and a right-

handed µ−. The precise experimental measurement B(π+ → e

+νe) = 1.230(4) × 10−4,

is a test of the lepton universality of weak couplings.

12.2.3 Charged Kaon Decays

The charged Kaon has a mass of 494MeV and a lifetime τK+ = 12ns.Its main decay modes are:

• Purely leptonic decays, where the su annihilate, similar to π+ decay:

B(K+→ µ

+νµ) = 63.4% (12.6)

The Kaon decay constant fK = 160 MeV is 20% larger than fπ due to SU(3)flavour symmetry breaking.

83

• Semileptonic decays, where s → u and the W+ boson couples to either e

+νe or

µ+νµ:

B(K+→ π

0+ν) = 8.1% (12.7)

Semileptonic decays satisfy the ∆Q = ∆S rule, i.e. the change in strangeness isequal to the lepton charge, K

+ → + and K

− → −.

• Hadronic decays, where s→ u and the W+ boson couples to ud:

B(K+→ π

+π

0) = 21.1% (12.8)

B(K+→ π

+π

+π−) = 5.6% (12.9)

Hadronic weak decays prefer an isospin change ∆I= 1/2 to ∆I= 3/2. The originsof this rule are not well understood.

12.2.4 The Cabibbo Angle

The relationship between the couplings of the W boson to ud and us quarks is describedby the Cabibbo angle, θC = 12.7:

d

s

=

cos θC sin θC

− sin θC cos θC

d

s

(12.10)

and the vertex couplings which modify the usual weak coupling GF are:

Vud = cos θC = 0.974(1) Vus = sin θC = 0.220(3) (12.11)

A factor |Vus|2 appears in all the decay rates for charged Kaons.

12.3 Decays of Heavy Quarks

Hadrons with charm decay mainly by weak c → s transitions, while beauty hadronsdecay via weak b → c transitions. There are also suppressed c → d and b → u

couplings to the W . The lifetimes are quite short but measurable: τD0 = 410(2)fs andτB0 = 1.53(1)ps.

12.3.1 Semileptonic decays

Semileptonic decays of heavy quarks are used to determine the W couplings. D → Kν

and D → πν give |Vcs| = cos θC and |Vcd| = sin θC . More interestingly B → D(∗)

ν

is used to measure |Vcb| ≈ 10−2, and B → πν is used to measure |Vub| ≈ 10−3.The extraction of these couplings requires a knowledge of hadronic form factors, whichdescribe the B → D

(∗) or π transitions in terms of the overlap of initial and final statehadronic wavefunctions. These are usually calculated non-perturbatively using LatticeQCD.

84

12.3.2 B → τν and Ds → µν

The purely leptonic decays of heavy mesons have small branching fractions, B(Ds →

µν) = 6× 10−3 and B(B+ → τν) = 1.8× 10−4. The decay B → τν was first observedby the Belle experiment in 2006. It measures the combination f

2B|Vub|

2, where fB ≈

190MeV is the B decay constant, also obtained from Lattice QCD. The measuredbranching fraction is larger than expected from other determinations of |Vub|.

12.3.3 b→ sγ

The flavour-changing neutral current (FCNC) transitions b → s, b → d and c → u arenot allowed at first order in the Standard Model. They do occur as a second orderweak interaction through a “penguin” diagram containing a loop with a W boson anda t quark (or b quark in the case of c → u). The decay b → sγ was first observed bythe CLEO experiment in 1992. It has a branching fraction B(b → sγ) = 3.5 × 10−4.This is a strong constraint on new physics beyond the Standard Model, because newheavy particles could replace the W and t in the loop, giving significant changes tothe decay rate. The decay b → dγ has recently been observed by the BaBar and Belleexperiments, but c→ u transitions have not yet been seen.

12.4 The Cabibbo-Kobayashi-Maskawa (CKM) Matrix

The 3×3 CKM matrix is an extension of the 2×2 Cabibbo matrix to include the heavyquarks:

d

s

b

=

Vud Vus Vub

Vcd Vcs Vcb

Vtd Vts Vtb

dsb

(12.12)

where the mass eigenstates of the Q = −1/3 quarks are d, s, b and the weak eigenstateswhich couple to the W are d

, s

, b

. The matrix is unitary, and its elements satisfy:

i

V2ij

= 1

j

V2ij

= 1 (12.13)

i

VijVik = 0

j

VijVkj = 0 (12.14)

By considering the above unitarity constraints it can be deduced that the CKM matrixcan be written in terms of just four parameters. Starting from the definition of theCabibbo angle, the four CKM parameters can be writen as three angles si = sin θi,ci = cos θi, and a complex phase δ:

c1 s1c3 s1s3

−s1c3 c1c2c3 − s2s3eiδ

c1c2s3 + s2c3eiδ

s1s2 −c1s2c3 − c2s3eiδ −c1s2s3 + c2c3e

iδ

(12.15)

It is more common to see the CKM matrix written as an expansion in powers of theCabibbo angle λ = sin θC . This is known as the Wolfenstein parametrization. It satisfies

85

unitarity to O(λ3):

1− λ

2/2 λ Aλ

3(ρ− iη)−λ 1− λ

2/2 Aλ

2

Aλ3(1− ρ− iη) Aλ

2 1

(12.16)

The relative sizes of the CKM elements in powers of λ vary from 1 on the diagonal, toλ

3 in the off-diagonal corners. Note that the complex phase η is associated with thesmallest elements of the matrix Vub and Vtd.

Figure 12.1: Constraints on the ρ and η parameters of the CKM matrix.Plot comes from the CKM fitter group at http://ckmfitter.in2p3.fr. More de-tails on the measurements can be found at the Heavy Flavour Averaging Grouphttp://www.slac.stanford.edu/xorg/hfag/.

The measured values of the CKM elements are:

• |Vcs| = 0.97(12) from D → Kν.

• |Vcd| = 0.224(12) from D → πν and neutrino production of charm νd→ c.

• |Vcb| = 0.042(1) from b→ cν. This determines the parameter A.

• |Vub| = 0.0044(4) from b→ uν.

• |Vtd| ≈ 0.008 and |Vts| ≈ 0.04 from ∆md and ∆ms in Bd and Bs mixing(see Lecture 13).

• |Vtb| ≈ 1 from top decays at the Tevatron.

The constraints on the ρ and η parameters of the CKM matrix are shown in Figure 12.We will discuss K and the angles α, β and γ of the “unitarity triangle” in Lecture 13.

From a full fit of all the constraints on the CKM matrix the Wolfenstein parametersare:

λ = 0.225(1) A = 0.81(2) ρ = 0.14(3) η = 0.34(2)

These are fundamental parameters of the Standard Model.

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13 Symmetries in Particle Physics

Symmetries play in important role in particle physics. The mathematical descriptionof symmetries uses group theory, examples of which are SU(2) and SU(3):

A serious student of elementary particle physics should plan eventually tostudy this subject in far greater detail. (Griffiths P.115)

There is a relation between symmetries and conservation laws which is known asNoether’s theorem. Examples of this in classical physics are:

• invariance under change of time → conservation of energy

• invariance under translation in space → conservation of momentum

• invariance under rotation → conservation of angular momentum

In particle physics there are many examples of symmetries and their associated con-servation laws. There are also cases where a symmetry is broken, and the mechanismhas to be understood. The breaking of electroweak symmetry and the associated Higgsfield will be discussed in lectures 15 and 16.

13.1 Gauge Symmetries

The Lagrangian is L = T − V , where T and V are the kinetic and potential energies ofa system. It can be used to obtain the equations of motion. The Dirac equation followsfrom a Lagrangian of the form:

L = iψγµδµψ −mψψ (13.1)

It can be seen that this Lagrangian is invariant under a phase transformation:

ψ → eiα

ψ ψ → e−iα

ψ (13.2)

This is an example of a gauge invariance.

13.1.1 U(1) Symmetry of QED

The Lagrangian for QED is written:

L = ψ(iγµδµ−m)ψ + eψγµA

µψ −

1

4FµνF

µν (13.3)

where Aµ represents the photon, and the second term can be thought of as JµA

µ whereJµ = eψγ

µψ is an electromagnetic current. Fµν is the electromagnetic field tensor:

Fµν = δµAν − δνAµ (13.4)

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The gauge transformation is:

ψ → eiα

ψ Aµ → Aµ +1

eδµα (13.5)

This has the property that it conserves the current, and hence conserves electric charge.A mass term mAµA

µ is forbidden in the Lagrangian by gauge invariance. This explainswhy the photon must be massless.

The gauge invariance of QED is described mathematically by a U(1) group.

13.1.2 SU(3) Symmetry of QCD

This gauge symmetry was already introduced in Lecture 8. A rotation in colour spaceis written as:

U = e−iαa·λa

(13.6)

where αa are the equivalent of “angles”, and λa are the generators of SU(3). QCD

amplitudes can be shown to be invariant under this non-Abelian gauge transformation.

The transformations of the quark and gluon states are:

q → (1 + iαaλa)q G

a

µ→ G

a

µ−

1

gs

∂µαa − fabcαbGc

µ(13.7)

The Lagrangian for QCD is written:

L = q(iγµδµ−m)q + gsqγµλ

aG

µ

aq −

1

4G

a

µνG

µν

a(13.8)

where the gluon states Ga

µreplace the photon, and gs replaces e. The gluon field energy

contains a term for the self-interactions of the gluons:

Ga

µν= δµG

a

ν− δνG

a

µ− gsfabcG

b

µG

c

ν(13.9)

The absence of a mass term mGa

µG

µ

amakes the gluon massless.

13.2 Flavour Symmetries

In Lecture 10 we met isospin symmetry, which is a flavour symmetry of the stronginteractions between u and d quarks. It is described by SU(2), and can be extended toSU(3) with the addition of the s quark. The SU(3) symmetry is partially broken by thes quark mass. In principle this could be extended further to an SU(6) symmetry be-tween all the quark flavours, but at this point the level of symmetry breaking becomesrather large. The interesting question, to which we do not yet have an answer, is whatcauses the breaking of quark flavour symmetry.

Similar questions can be asked about lepton flavour symmetry. There are precise testsof lepton flavour conservation, and of the universality of lepton couplings in electro-magnetic and weak interactions. However, flavour symmetry is broken by the lepton

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masses for reasons that are not understood.

There is overall conservation of baryon number (B) and lepton number (L). However,to generate the matter-antimatter asymmetry of the universe B has be violated. Inmany models for this B−L is then conserved.

13.3 Discrete Symmetries

13.3.1 Parity

The parity operation P performs a spatial inversion through the origin:

Pψ(r) = ψ(−r) (13.10)

This is NOT a mirror reflection through an axis, e.g. ψ(x) → ψ(−x).Many books get this wrong!

Applying parity twice restores the original state, P2 = 1. From this the parity of

a wavefunction ψ(r) has to be either even, P = +1, or odd, P = −1. For exampleψ(x) = cos kx is even, and ψ(x) = sin kx is odd.

The hydrogen atom wavefunctions are a product of a radial function f(r) and thespherical harmonics Y

m

L(θ,φ), where L and m are the orbital angular momentum of

the state and its projection along an axis. In spherical polar coordinates the parityoperation changes r, θ, φ −→ r, π − θ,π + φ. From the properties of Y

m

Lthe wave-

functions have parity P = (−1)L. It is observed that single photon transitions betweenatomic states obey the selection rule ∆L = ±1. From this it can be deduced that theintrinsic parity of the photon is:

(−1)L = (−1)L±1× Pγ Pγ = −1 (13.11)

The parity of the photon can also be obtained from the gauge symmetry of QEDdiscussed in the previous section.

13.3.2 Intrinsic Parity of Fermions

Applying a spatial inversion to the Dirac equation gives

iγ0 δ

δt− iγ · −m

ψ( −r, t) = 0 (13.12)

This is not the same as the Dirac equation because there is a change of sign of the firstderivative in the spatial coordinates. If we multiply from the left by γ

0 and use therelations (γ0)2 = 1 and γ

0γ

i + γiγ

0 = 0 we get back a valid Dirac equation:

iγ0 δ

δt+ iγ · −m

γ

0ψ( −r, t) = 0 (13.13)

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We identify the parity operator with γ0:

ψ(r, t) = Pψ(−r, t) = γ0ψ(−r, t) (13.14)

Applying this to the Dirac spinors:

Pu1 = Pu2 = +1 Pv1 = Pv2 = −1 (13.15)

The intrinsic parity of fermions is P = +1 (even)The intrinsic parity of antifermions is P = −1 (odd)

Parity is a multiplicative quantum number, so the parity of a many particle systemis equal to the product of the intrinsic parities of the particles, and the parity of thespatial wavefunction which is (−1)L. As an example, positronium is an e

+e− atom

with:P (e+

e−) = Pe−Pe+(−1)L = (−1)L+1 (13.16)

where L is the relative orbital angular momentum between the e+ and e

−.

13.3.3 Charge Conjugation

Charge Conjugation is a discrete symmetry that reverses the sign of the charge andmagnetic moment of a particle. Like the parity operator it satifies C

2 = 1, and haspossible eigenvalues C = ±1. Electromagnetism is C invariant, since Maxwell’s equa-tions apply equally to + and − charges. However the electromagnetic fields change signunder C, which means the photon has:

Cγ = −1 (13.17)

For fermions charge conjugation changes a particle into an antiparticle, so fermionsthemselves are not eigenstates of C, but combinations of fermions are. Positroniumhas:

C(e+e−) = (−1)L+S (13.18)

where S is the sum of the spins which can be either 0 or 1. Positronium states witheven L + S decay to two photons, and those with odd L + S decay to three photons.This shows that electromagnetic interactions are invariant under charge conjugationand parity, and conserve C and P .

We can also determine the P and C states of mesons. The lowest pseudoscalar mesonshave J

PC = 0−+ and the vector mesons have JPC = 1−−.

13.3.4 Time Reversal

Time reversal Tψ(t) = ψ(−t), is another discrete symmetry operator with T2 = 1, and

possible eigenvalues T = ±1. The solutions of the Dirac equation describe antifermionstates as equivalent to fermion states with the time and space coordinates reversed.

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13.3.5 Summary of Discrete Symmetry Transformations

• A polar vector such as momentum, p, transforms under parity P = −1.

• An axial vector such as angular momentum, L = r × p transforms as P L = L.This implies that parity does not affect the spin of a particle.

• Charge conjugation reverses the charge, but does not change the direction of thespin vector or the momentum of a particle.

• Time reversal changes the sign of both the spin and momentum.

Quantity Notation P C TPosition r -1 +1 +1Momentum (Vector) p -1 +1 -1Spin (Axial Vector) σ = r × p +1 +1 -1Helicity σ.p -1 +1 +1Electric Field E -1 -1 +1Magnetic Field B +1 -1 -1Magnetic Dipole Moment σ. B +1 -1 +1Electric Dipole Moment σ. E -1 -1 -1Transverse Polarization σ.(p1 × p2) +1 +1 -1

13.4 Parity Violation in Weak Interactions

In contrast to electromagnetic interactions it is found that weak interactions maxi-mally violate both parity and charge conjugation symmetries. The original evidencefor parity violation came from the study of the β decay of polarized 60Co, where it wasobserved that the electron was emitted preferentially in the direction opposite to thespin of the nucleus. The distribution of the decay electrons can be described by:

dN

dΩ= 1−

σ.p

E(13.19)

The parity operator reverses the direction of the electron but not the spin of the nucleus,so the σ.p term is parity-violating. A similar parity violation is observed in muon decay.A µ

+ emits an e+ preferentially along the direction of the µ

+ spin, whereas a µ− emits

an e− preferentially in the direction opposite to the µ

− spin. The difference betweenµ

+ and µ− shows that charge conjugation is also violated. However a comparison of

the two decay distributions shows that the combined operation CP is conserved.

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14 CP and CP Violation

14.1 CPT Theorem

The CPT theorem requires that all interactions that are described by localized Lorentzinvariant gauge theories must be invariant under the combined operation of C, P andT in any order. The proof of the CPT theorem is based on very general field theoreticassumptions. It can be thought of as a statement about the invariance of Feynmandiagrams under particle/antiparticle interchange, and interchange of the initial andfinal states.

The CPT theorem predicts that particles and antiparticles must have the same massand lifetime, but opposite electric charge and magnetic moment. Experimental tests ofthe CPT theorem have shown very precise agreement.

The CPT theorem also means that the transformation properties of gauge theoriesunder the discrete symmetries C, P and T are related to each other:

CP ↔ T CT ↔ P PT ↔ C (14.1)

The first of these establishes that time reversal invariance is equivalent to CP invariance.

In the Big Bang model of the universe there is an arrow of time so T may notbe a valid symmetry of the universe. It is believed that matter and antimatter wereoriginally created in equal amounts, but we observe that we live in a matter dominateduniverse, with a baryon density compared to photons of Nb/Nγ = 10−9, and no evidencefor primordial antibaryons. In 1966 Sakharov postulated three conditions that arenecessary for our matter-dominated universe to exist:

• An epoch with no thermal equilibrium

• Baryon number violation

• CP Violation (or equivalently T violation)

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15 Measuring CP Violation

15.1 Neutral Meson Mixing

The couplings Vtd and Vts have been inferred from second order weak processes in whichtop quarks appear inside loops with W boson. These involve mixing diagrams whichdescribe the transitions between neutral mesons K

0 ↔ K0, B0 ↔ B0 and Bs ↔ Bs.

Figure 15.1: Second order weak diagram for neutral meson mixing.

15.1.1 Mixing of Neutral Kaons

A state that is initially K0 or K

0 will evolve as a function of time due to the mixingdiagram:

ψ(t) = a(t)|K0> +b(t)|K0

> idψ

dt= Hψ(t) (15.1)

where H is the effective Hamiltonian which can be written in terms of 2× 2 mass anddecay matrices M and Γ:

H = M−i

2Γ (15.2)

The diagonal elements of these matrices are associated with flavour-conserving transi-tions, while the off-diagonal elements are associated with the mixing transitions K

0 ↔

K0. The matrix H has two eigenvectors corresponding to the mass and weak decay

eigenstates KL and KS. The real parts of the eigenvectors are the masses, mL and mS,and the imaginary parts are the decay widths. The eigenstates can be expressed as alinear superposition of K

0 and K0:

|KS >= p|K0

> +q|K0

> |KL >= p|K0

> −q|K0

> (15.3)

where |q|2 + |p|2 = 1 and:q

p=

2M∗12 − i/2Γ∗

12

∆mK − i/2∆ΓK

(15.4)

The differences in the masses and decay widths of the weak eigenstates are:

∆mK = mL −mS = (3.52± 0.01)× 10−12MeV = 0.529× 1010s−1 (15.5)

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∆ΓK =1

τL

−1

τS

= 1.1× 1010s−1 (15.6)

The mass difference ∆mK is very small compared to the neutral Kaon mass!

Figure 15.2: Time evolution of an initial K0 state

The time evolution of a state that is initially pure K0 is given by:

|ψK0(t)|2 =1

4

e−ΓLt + e

−ΓSt + 2e−(ΓL+ΓS)

2 t cos ∆mKt

(15.7)

|ψK0(t)|2 =1

4

e−ΓLt + e

−ΓSt− 2e−

(ΓL+ΓS)2 t cos ∆mKt

(15.8)

Neutral meson mixing leads to flavour oscillations, with a frequency given by themass difference between the weak eigenstates.

15.1.2 Mixing of B mesons

There are two systems of neutral B mesons, the Bd states B0 and B

0, and the equivalentBs states where an s quark replaces the d quark. They are expected to mix in a similarway to the K

0 states, but in this case the mixing diagram is dominated by the topquark, and the off-diagonal elements of the mixing matrix are given by

M12 ∝ (VtbV∗td

)2 q

p=

V∗tbVtd

VtbV∗td

(15.9)

Oscillations of Bd mesons have been observed with:

∆md = 0.508(4)ps−1τBd

= 1.53(1)ps (15.10)

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Bs oscillations were observed at the Tevatron in 2006 using an amplitude scan to Fourieranalyse their Bs decays.

∆ms = 17.8(1)ps−1τBd

= 1.47(6)ps (15.11)

Note the much larger oscillation frequency which makes the direct observation of theoscillations difficult, although it should be possible at the LHC.

From the ratio of the two oscillation frequencies it is possible to determine:

|Vtd

Vts

| = 0.206(1) (15.12)

The main uncertainty in this ratio is now coming from the theoretical calculation of thehadronic properties of B mesons, the decay constants fB and the “bag” constants BB.It should be noted that most of the uncertainties cancel in the ratio, and the individualdeterminations of |Vtd| and |Vts| have theoretical errors which are ×10 larger.

15.2 CP Violation in Neutral Kaon Decays

The CP eigenstates of neutral Kaons are:

K1 =1√

2[K0 + K

0] K2 =1√

2[K0

− K0] (15.13)

CP |K1 = +1 CP |K2 = −1 (15.14)

If CP is violated the weak decay eigenstates are not the same as the CP eigenstates:

KL =1

√1 + 2

[K1 + K2] KS =1

√1 + 2

[K1 − K2] (15.15)

where is a complex number.

The notation KL and KS refers to the long and short lifetimes:

τL = 5.2× 10−8s τS = 0.9× 10−10

s (15.16)

95

15.2.1 CP Violation in K → 2π Decays

There are neutral Kaon decays to two and three pion final states with:

CP |π+π− = CP |π

0π

0 = +1 (15.17)

CP |π+π−π

0 = CP |π

0π

0π

0 = −1 (15.18)

If CP is conserved it is expected that K1 → 2π and K2 → 3π, with K1 = KS andK2 = KL.

In 1964 the decay KL → π+π− was observed which violates CP .

The ratios of decays are written as:

η+− =KL → π

+π−

KS → π+π−= +

(15.19)

η00 =KL → π

0π

0

KS → π0π0= − 2 (15.20)

where the parameter represents a “direct” CP violation between the ∆I = 1/2 and

∆I = 3/2 amplitudes in K → 2π decays.

The η have measured magnitudes and phases:

|η+−| = 2.286(14)× 10−3φ+− = 43.4(7) (15.21)

|η00| = 2.276(14)× 10−3φ00 = 43.6(8) (15.22)

From a comparison of the charged and neutral pion decays:

Re(/) = 1.67(26)× 10−3 (15.23)

15.2.2 CP and T Violation in Semileptonic Decays

In semileptonic decays the charge of the lepton is given by the charge of the W boson.Thus a K

0 decay by an s → u transition gives an +, and a K

) decay by an s → u

transition gives an −. This is known as the ∆Q = ∆S rule. The charge of the lepton

gives a flavour tag to the neutral Kaon decay.

If there is no CP violation, the KL is an equal superposition of K0 and K

0, so itshould decay equally to

+ and − with no charge asymmetry. If we add in the small

amount of CP violation , then a charge asymmetry is predicted:

δSL =Γ(KL → π

−+ν)− Γ(KL → π

+−ν)

Γ(KL → π−+ν) + Γ(KL → π+−ν)(15.24)

δSL =(1 + )2 − (1− )2

(1 + )2 + (1− )2= 2Re() (15.25)

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The experimental measurement of this asymmetry is:

δSL = 3.27(12)× 10−3 (15.26)

There is another elegant measurement that can be made with semileptonic decays thatexplicitly demonstrates time-reversal violation. We start with a pure K

0 or K0 state,

and let it oscillate and then decay semileptonically. The T violation is observable as arate asymmetry:

Γ(K0→ K

0→ π

+−ν) = Γ(K0

→ K0→ π

−+ν) (15.27)

The amount of T violation corresponds to the amount of CP violation, so CPT sym-metry is preserved. A direct test of CPT violation in semileptonic decays would be:

Γ(K0→ π

+−ν) = Γ(K0

→ π−+ν) (15.28)

The bounds on a CPT violating parameter δCPT in neutral Kaon decays are actuallyonly one order of magnitude below :

δCPT = (2.9± 2.7)× 10−4 (15.29)

15.3 General Formalism for CP violation

There is an excellent review of CP violation in the Particle Data Group compilation athttp://pdg.lbl.gov/2009/reviews.

In a more general notation the weak eigenstates are labelled ML and MH (for lightand heavy mass), and are not assumed to be the same as the CP eigenstates:

ML = pM0 + qM

0MH = pM

0− qM

0 (15.30)

with the normalisation |p|2 + |q|2 = 1.

We use the following notation for the mass and decay width differences:

∆m = MH −ML ∆Γ = ΓH − ΓL (15.31)

Γ =ΓH + ΓL

2x =

∆m

Γy =

∆Γ

Γ(15.32)

The amplitudes for the decays of the flavour eigenstates M0 and M

0 to a final state f

or f , are written as A and A. If the final state is a CP eigenstate f = f , but A and A

are not necessarily equal.

The time dependent decay rates of the flavour eigenstates to a CP eigenstate M0 → f

and M0 → f , are given in the most general form by:

dΓ

dt= e

−Γt [α cosh(∆Γt) + β cos(∆mt) + 2Re[γ] sinh(∆Γt)− 2Im[γ] sin(∆mt)] (15.33)

97

α = |A|2 + |

q

pA|

2β = |A|

2− |

q

pA|

2γ =

q

pA∗A (15.34)

dΓ

dt= e

−Γtα cosh(∆Γt)− β cos(∆mt) + 2Re[γ] sinh(∆Γt) + 2Im[γ] sin(∆mt)

(15.35)

α = |p

qA|

2 + |A|2

β = |p

qA|

2− |A|

2γ =

p

qAA

∗ (15.36)

Note the changes in sign of the second and fourth terms in the decay rates.

The sin and cos terms give the mixing oscillations with frequency ∆m. The ampli-tudes of these oscillations depend on γ, and include a possible CP violation throughmixing.

15.3.1 Types of CP violation

There are three types of CP violation that can be observed:

• CP violation in the mixing amplitude, due to the mass eigenstates being differentfrom the CP eigenstates, |q/p| = 1.In the neutral Kaon system this is represented by the semileptonic charge asym-metry δSL.

• CP violation in the amplitudes A and A for decays to a particular final state,|A/A| = 1 and phase differences between them. This is commonly known asdirect CP violation. It does not require mixing, and can be found in bothcharged and neutral meson decays.In the decays KL,S → 2π it is represented by

.

• CP violation in the interference between mixing and decay amplitudes, whichrequires an overall weak phase Im[λ] = 0, where λ = qA/pA.In the decays KL,S → 2π it is represented by .

15.4 CP Violation in B Meson Decays

All three types are CP violation are expected to occur in the decays of neutral B

mesons. Due to the dominance of the t quark contribution inside mixing and penguindiagrams, and the presence of the suppressed CKM couplings Vub and Vtd, measurementsof CP violation in B decays provide important additional information compared to theneutral Kaon system.

15.4.1 Bd → J/ψKS and sin 2β

CP violation through interference between mixing and decay amplitudes was first ob-served in the decay Bd → J/ψKS in 2001 by the BaBar and Belle experiments. For

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this decay:

λ =qAf

pAf

=

V∗tbVtd

VtbV∗td

V∗cs

Vcb

VcsV∗cb

V∗cs

Vcd

VcsV∗cd

(15.37)

where the three set of CKM factors respectively account for Bd mixing, the B → J/ψKS

decay amplitude and final state K0 mixing.

The measurement is done by producing a B0B

0 pair in e+e− → Υ(4S) collisions. One

of the B mesons decays into a flavour-specific final state providing a tag of the flavourof the other B at the time of the first B decay. Then the CP asymmetry is measuredas a function of the time difference, ∆t, between the B and B decays. The amplitudeof the asymmetry as a function of the mixing oscillations sin ∆md∆t is proportional toIm(λ). The measured CP violation is large, Im(λ) = sin 2β = 0.67(2), with the angle

figure=psiks.eps,width=0.8

Figure 15.3: The BaBar measurement of the CP asymmetry: a) and b) in B0 → J/ψKS,

c) and d) in B0 → J/ψKL.

β = 21 being the complex phase of Vtd in the Standard Model.

15.4.2 Bd → ππ, Bd → ρρ and the angle α

A similar measurement of time-dependent CP asymmetries can be made with the rarehadronic final states B

0 → π+π− and B

0 → ρ+ρ−. In this case the decay amplitude is

proportional to Vub, and the angle α is measured which is the complex phase betweenVtd and Vub in the Standard Model.

There is a complication due to an additional contribution to the amplitude from ab → d penguin diagram. This is rather important for B → ππ, and its effect has to bedetermined from an isospin analysis including the modes B

+ → π+π

0 and B0 → π

0π

0.

The BaBar experiment obtains α = (96 ± 6) from a fit to B → ρρ decays where thepenguin diagram has a much smaller effect.

15.4.3 Direct CP violation in B → Kπ

In 2004 the BaBar and Belle experiments made the first observation of a direct CP

violation in the decay amplitudes for B0 → Kπ decays:

ACP =Γ(B0 → K

−π

+)− Γ(B0 → K+π−)

Γ(B0 → K−π+)− Γ(B0 → K+π−)(15.38)

The latest world average for this is ACP = −0.10 ± 0.01. However, significant directCP asymmetries have not been observed in the decays B± → K±

π0 and B

± → KSπ±,

for reasons that are not well understood.

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15.5 CP violation in D mesons

CP violation in D mesons was observed for the first time in 2011 by the LHCb collab-oration at CERN.

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16 Neutrino Oscillations

This topic is well covered by Chapter 11 of Griffiths, and there is also a good web siteat http://neutrinooscillation.org/.

In the Standard Model neutrinos are described as massless neutral fermions which comein three different flavours, νe, νµ and ντ . It is necessary to invoke separate conservationof each lepton flavour number Le, Lµ and Lτ , to avoid flavour-changing weak couplingsof leptons to W and Z bosons.

If lepton flavours are not separately conserved, and neutrinos have finite masses, thenit might be possible for the different neutrino flavours to mix in a similar way to theneutral mesons in the last lecture. This was first suggested by Pontecorvo in 1957. Theweak interaction eigenstates remain the flavour eigenstates, νe, νµ and ντ , but the masseigenstates ν1, ν2 and ν3 are linear superpositions of the weak eigenstates.

Note that neutrino mixing requires a new lepton-flavour violating interaction which is

not present in the Standard Model. We do not yet know what this interaction is!

16.0.1 Neutrino States

A massless neutrino is purely left-handed, and a massless antineutrino is purely right-handed. The P operator changes the direction of p but not σ, so it reverses the helicitystate. The C operator changes a neutrino into an antineutrino. Each of these operatorsby itself changes a physical state into a forbidden state, again showing that P andC must be maximally violated in weak interactions with neutrinos. The combinedoperator CP changes a left-handed neutrino into a right-handed antineutrino which isallowed.

16.1 Description of Oscillations

16.1.1 Two Neutrino Flavours

Neutrinos are produced in weak decays, so they start off as weak eigenstates. How-ever, they propagate through space-time as plane waves corresponding to their masseigenstates:

ν1(t) = ν1(0)e−iE1t

ν2(t) = ν2(0)e−iE2t (16.1)

We have the equivalent of the Cabibbo angle to describe the mixing of two neutrinostates:

νe

νµ

=

cos θ12 sin θ12

− sin θ12 cos θ12

ν1

ν2

(16.2)

where θ12 is a new parameter not present in the Standard Model.

If the neutrino masses m1 and m2 are different, the energies E1 and E2 are different.

101

Assuming highly relativistic neutrinos with m E, p ≈ E:

Ei = p +mi

2p2∆E =

∆m212

E(16.3)

If we start off with a pure νe beam, the amplitude for νe at a later time t is:

νe(t) = νe(0)1− sin θ12 cos θ12(−e

−iE1t + e−iE2t)

(16.4)

and the probability of observing an oscillation to νµ is:

P (νe → νµ) = |νµ(t)|2 = 1− |νe(t)|2 = sin2 2θ12 sin2 (E2 − E1)

2t (16.5)

For experimental convenience this is usually expressed as:

P (νe → νµ) = sin2 2θ12 sin2

1.27∆m

212L

E

(16.6)

where the numerical factor 1.27 applies if we express ∆m2 in eV2, the distance from

the source L in metres, and the neutrino energy E in MeV.

To observe these oscillations experimentally a “near” detector measures the initial νe

flux, and a “far” detector measures either disappearance of νe, or appearance of νµ. Thechoice of the “baseline”, L, has to be matched to the oscillation frequency 1.27∆m

212/E,

and the amplitude of the oscillations is related to the mixing angle by sin2 2θ12.Note that the maximum possible mixing is for θ12 = 45.

16.1.2 The PMNS Mixing Matrix

For the full case of three neutrinos we have the equivalent of the CKM matrix whichis known as the PMNS (Pontecorvo, Maki, Nakagawa, Sakata) matrix. It is usuallywritten out as the product of three matrices representing the three different types oftwo neutrino mixings:

νe

νµ

ντ

= VPMNS

ν1

ν2

ν3

(16.7)

VPMNS =

1 0 00 c23 s23

0 −s23 c23

c13 0 s13e

iδ

0 1 0−s13e

−iδ 0 c13

c12 s12 0−s12 c12 0

0 0 1

(16.8)

It is parameterised by three angles, where sij = sin θij, cij = cos θij, and one complexphase δ. The observations of neutrino oscillations, described in the next sections, canbe accounted for by small mass differences ∆m12 and ∆m23, and large mixing anglesθ12 and θ23. At present there is only an upper limit on the third angle θ13. As in theCKM case, the phase δ can give rise to CP violation in neutrino oscillations, but onlyif θ13 = 0.

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16.2 Solar Neutrinos

16.2.1 The Standard Solar Model

The sun creates energy by fusion of light nuclei. During this process a large flux of lowenergy electron neutrinos are released from β

+ decays of the fusion products. Mostof the flux comes from the p-p fusion process, in which neutrinos are emitted up to amaximum energy of 400 keV. There is a small component of higher energy neutrinos,up to a maximum of 15 MeV, associated with 8B.

A large amount of work, mostly by Bahcall, has gone into calculating the flux of solarneutrinos using a Standard Solar Model (SSM).

16.2.2 The Davis Experiment

From 1970-1995 Ray Davis looked for solar neutrinos using a large tank containing100,000 gallons of cleaning fluid placed in a mine in South Dakota. The neutrinos from8B and 7Be are detected by the interaction:

νe +37Cl →37Ar + e− (16.9)

Only 0.5 Argon atoms are produced per day(!). The whole cleaning tank is analysedradiochemically every few months to count these atoms. The observed rate is 2.56±0.23SNU, where 1 SNU(solar neutrino unit) is 10−36 captures per atom per second. Thepredicted rate from the SSM is 7.7± 1.2 SNU.

This is the famous solar neutrino deficit factor of 0.33±0.06. There was a long discussionabout whether the radiochemical extraction of the Argon atoms was reliable, and an

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equally long discussion about whether the predictions of the Standard Solar Model werereliable. Now it is accepted that the deficit is real, and is attributable to νe → νµ (orντ ) oscillations.Note that due to the low neutrino energy it is impossible to detect νµ or ντ by charged

current interactions.

16.2.3 Recent Solar Neutrino Experiments

The Kamiokande experiment used 50,000 tons of water as a Cherenkov detector for the8B neutrinos. The electron scattering process νe + e

− → νe + e− gives a recoil electron

that produces light which is detected by 11,000 photomultiplier tubes. This methodallows the direction of the neutrinos to be determined, proving that they come fromthe sun. Like the Davis experiment, Kamiokande measures a solar neutrino deficit of0.45± 0.02.

Two experiments, Gallex and SAGE, used large quantities of metallic Gallium to mea-sure the flux of the lower energy p-p neutrinos. In this case νe+71Ga→71Ge+e

−, andthe Germanium atoms are again counted by radiochemical means. Both experimentsmeasure 71 ± 5 SNU, compared to the SSM prediction of 129 SNU. The importanceof these measurements is that the rate of p-p neutrinos is determined precisely by thethermal output of the sun, so the SSM prediction is rather reliable. At this point mostpeople believed that the solar neutrino deficit was due to about half of the electronneutrinos oscillating into another neutrino flavour.

The SNO experiment proved this between 2000 and 2006 using 1,000 tons of heavywater (D2O) to detect neutrinos in three different ways:

• Scattering on electrons νe + e− → νe + e

−

• Charged current scattering on deuterium νe + d→ p + p + e−

• Neutral current scattering on deuterium ν + d→ n + p + ν

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As indicated by the lack of a subscript, the last process does not distinguish betweenνe, νµ and ντ . The difference between the neutral and charged current scattering ondeuterium shows that the νµ (or ντ ) flux is exactly what is required to account for thesolar neutrino deficit.

16.2.4 The MSW Effect

In 1978 Wolfenstein noted that the effect of flavour-specific neutrino interactions mustbe taken into account when considering neutrino propagation in the presence of matter.Since matter contains electrons but not muons, electron neutrinos experience a potentialenergy due to interactions, Ue ∝ GF Ne, where Ne is the electron density of the matter.This potential has an equivalent effect to a mass difference, i.e. it changes the energywith which the electron neutrinos propagate. This leads to matter-induced electronneutrino oscillations, with an effective mixing angle in matter θm, which differs from θ

in vacuum:

sin2 2θm =sin2 2θ

(cos 2θ − a)2 + sin2 2θa ∝ GF EνNe/∆m

2 (16.10)

In the sun, the electron density Ne varies with radius, and there can be a radius wherea = cos 2θ and sin2 2θm = 1 leads to resonance-enhanced oscillations of electronneutrinos. This is known as the MSW effect.

Combining all solar neutrino results, and including the MSW effect, the parametersof the solar neutrino oscillations have been determined to be:

∆m212 = (7.6± 0.2)× 10−5eV2 sin2 2θ12 = 0.87± 0.03 (16.11)

There is an alternative solution with vacuum oscillations and no MSW effect. This hasa much smaller ∆m

212, but it is ruled out by reactor experiments.

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16.3 Atmospheric Neutrinos

Neutrinos are produced in the upper atmosphere by the interactions of cosmic rays. Theinitial strong interaction of protons with nuclei produces charged (and neutral) pions.The charged pions decay via π

+ → µ+νµ, µ

+ → e+νeνµ, and the charge conjugate π

−

decays. This gives ratios of two muon (anti)neutrinos to one electron (anti)neutrino.Note that atmospheric neutrinos have much higher energies than solar neutrinos, in theGeV range.

SuperKamiokande detected atmospheric neutrinos via the charged current interactionsνe +p→ p+ e

−, νµ +p→ p+µ−. The muon and electron can be identified and used to

tag the flavour of the incoming neutrino. What is observed is a deficit of upward goingmuons, produced by muon neutrinos coming from the atmosphere on the other side ofthe earth:

Note that there is no up-down asymmetry for electrons from the electron neutrinos, andthat the expected µ : e ratio of a factor of two is observed for the downward neutrinos.

This observation is interpreted as the oscillation of muon neutrinos into unobserved tauneutrinos over the earth’s diameter, with parameters:

∆m223 = (2.4± 0.1)× 10−3eV2 sin2 2θ23 = 1.00± 0.05 (16.12)

Note that this mass difference squared is 30 times larger than the solar neutrino massdifference, and that the mixing is consistent with being maximal.

16.4 Accelerator Neutrino Experiments

A typical accelerator neutrino beam is either νµ or νµ, produced from the decays of π±

and K± mesons. There is ≈ 1% contamination of νe from semileptonic decays. The

beam energies are in the range 100 MeV to 10 GeV, and the corresponding baselines

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range from 1 to 1000km.

Accelerator beams have been used to confirm the oscillations of νµ → ντ . The K2Kexperiment fired a 1 GeV beam across Japan from KEK to Kamiokande (L=250km).They measured the disappearance of νµ, and obtained results consistent with the at-mospheric neutrinos. More recently the MINOS experiment fired a 10 GeV beam fromFermilab to Soudan (L=735km), to obtain the world’s most accurate values for ∆m

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and sin2 2θ23.

Both MINOS and a Japanese experiment, T2K, are now looking for νµ → νe appearanceto try and measure the small mixing angle θ13, for which the current limit is sin2 2θ13 <

0.15.

16.5 Open Questions on Neutrinos

In the past decade we have confirmed the existence of neutrino oscillations and explainedthe solar neutrino deficit, but there remain several open questions:

• We do not know the absolute neutrino mass scale. It could be m ≈ ∆m, or themasses could be degenerate m ∆m.

• We do not know the mass hierarchy, because we determine the magnitudes butnot the signs of the mass differences. It could be normal m1, m2 < m3, or invertedm3 < m2, m1.

• We have measured two mass differences and two mixing angles. The third massdifference must be ∆m13 ≈ 10−3eV2, but the third mixing angle only has an upperlimit sin2 2θ13 < 0.15.

• If θ13 is large enough, it may eventually be possible to measure the CP-violatingphase δ. This will require an accelerator known as a “neutrino factory” which iscurrently being designed.

• Why is the PMNS matrix very close to tri-bimaximal mixing:sin2

θ23 = 1/2, sin2θ12 = 1/3, θ13 = 0?

16.6 Neutrinos in Astrophysics*

Finally some comments on the role of neutrinos in astrophysics:

• The Big Bang model predicts a large relic density of very low energy neutrinos,similar to the microwave background of photons. However, the mass of neutrinosis too small to account for dark matter.

• In 1987 a few electron neutrinos with E= 10−40 MeV were observed coming fromthe Supernova SN1987A in the Large Magellanic Cloud (L=175k light years).

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The energies and spread of arrival times constrain the neutrino mass, and mayeventually provide information on the initial stages of a supernova explosion. Wejust have to wait for the next one...

• Detectors such as AMANDA at the South Pole, and ANTARES in the Mediter-ranean detect very high energy neutrinos from outer space. The advantage ofneutrinos is that they are unaffected on their path from a point source to theearth.

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