Lecture Notes Particle Physics - Physikalisches Institutsschaetz/teaching/PP15/... · In the ﬁnal step we have used the high energy physics unit for cross sections, the barn, which

Particle Physics

Lecture Notes

Priv.-Doz. Dr. S. Schätzel

Master’s Degree Course MKEP1

Ruprecht-Karls-Universität Heidelberg

2015

Version as of 4 May 2015 (Chapter 1)

ii

Contents

1 Theoretical basis and QED 1

1.1 Lagrangian field theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Natural units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.2 Relativistic notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.3 Lagrange equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.4 Continuity equation and conserved charge . . . . . . . . . . . . . . . . . . . 4

1.1.5 Particle creation and annihilation . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.1 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.2 Solutions for particles at rest . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2.3 General solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.4 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.2.5 Spinor normalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.2.6 Transformation properties of spinors . . . . . . . . . . . . . . . . . . . . . . 14

1.3 Quantum electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.3.1 Lagrange density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.3.2 Electromagnetic current density and conserved charge . . . . . . . . . . . 17

1.3.3 Local phase transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.3.4 The photon field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.3.5 Principle of gauge theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.3.6 Electron-photon vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.3.7 Antiparticles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.3.8 The unphysical photon vertex . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.3.9 Virtual photons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.3.10 Photon propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.3.11 Units of electric charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.4 Electromagnetic scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.4.1 Golden rule for scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.4.2 2 → 2 scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.4.3 Cross section for eµ→ eµ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

1.4.4 Mott and Rutherford scattering . . . . . . . . . . . . . . . . . . . . . . . . . . 36

1.4.5 Mandelstam variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

1.4.6 Crossing symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1.4.7 e+e− →µ+µ− . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

1.4.8 Bhabha scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

1.4.9 Møller scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

1.4.10 Dependence on scattering angle . . . . . . . . . . . . . . . . . . . . . . . . . 44

1.4.11 Helicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

iii

CONTENTS CONTENTS

1.4.12 Chirality (Handedness) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

1.4.13 Chirality in QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

1.4.14 Parity conservation in QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

1.4.15 e+e− → qq . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

1.4.16 Adding amplitudes vs. adding cross sections . . . . . . . . . . . . . . . . . . 52

1.4.17 Rhad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

1.5 Higher order corrections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

1.5.1 Loop diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

1.5.2 Renormalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

1.5.3 Resolving small structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

1.5.4 Vacuum polarisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

1.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

iv

Chapter 1

Theoretical basis and QED

1.1 Lagrangian field theory

1.1.1 Natural units

This section follows chapter 6.1 of [1].

Formulas in particle physics often contain the speed of light c and Planck’s constant ħ≡ h2π :1

E 2 = (mc2)2 + (pc)2 Sz =1

2ħ (1.1)

For clarity we drop these factors from the formulas:

E 2 = m2 +p2 Sz =1

2(1.2)

so that momentum and mass are measured in units of energy (for which we use the electron-

volt, eV) and spin is measured in units of Planck’s constant. These units are the natural units for

particle physics because they make the physics content of formulas stand out much clearer. For

example: the energy of a particle is given by its mass and its momentum, added in quadrature.

The natural units of other quantities, like cross sections and lifetimes, are not straightforward

and not always easy to understand. To convert from natural to ordinary units, we insert into the

formula factors of c and ħ. The following conversion factor is very convenient and worthwhile

to remember:

ħc = 197 MeV fm (1.3)

1In this text, three-vectors are denoted by a bold-face letter, for example: p.

1

1.1. LAGRANGIAN FIELD THEORY CHAPTER 1. THEORETICAL BASIS AND QED

Example 1.1 (Thomson scattering cross section)

Thomson scattering is the low-energy limit (Eγ ≪ me ) of Compton scattering (γ+e → γ+e) and

the cross section is in natural units given by

σ= 8π

3

α2

m2e

(1.4)

With me = 0.511 MeV, the cross section has the dimension of a squared inverse energy. To get

the cross section in ordinary units, i.e., with dimension of an area, we multiply by (ħc)2:

σ= 8π

3

α2

m2e

(ħc)2 = 8π

3

α2

m2e

(197 MeV fm)2 (1.5)

= 6.65 ·10−25 cm2 = 0.665 b (1.6)

In the final step we have used the high energy physics unit for cross sections, the barn, which

is defined as

1 b ≡ 10−24 cm2 = 10−28 m2 (1.7)

Example 1.2 (Positronium lifetime)

The lifetime of the e+e− ground state (1 1S0) is τ= 2

α2

1

mein natural units. To get the lifetime in

seconds we multiply by ħc and divide by c:

τ= 2

α2

1

me197 MeV fm

1

3 ·108 ms

= 1.24 ·10−10 s (1.8)

1.1.2 Relativistic notation


The usual 4-vectors are used in this text:

• space-time coordinates: x = xµ = (t , x, y, z) = (t ,x)

• energy and momentum: p = (E ,p)

The metric tensor

gµν = diag(1,−1,−1,−1) =

1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1

= gµν (1.9)

relates covariant and contravariant vectors:

aµ =3∑

ν=0

gµνaν ≡ gµνaν (1.10)

aµ = gµνaν (1.11)

in which the Einstein summation convention was used which implies summation over indices

that appear twice.

2

CHAPTER 1. THEORETICAL BASIS AND QED 1.1. LAGRANGIAN FIELD THEORY

Example 1.3

a0 = g 0νaν = g 00a0 + g 01

︸︷︷︸

0

a1 + g 02

︸︷︷︸

0

a2 + g 03

︸︷︷︸

0

a3 = a0 (1.12)

ai =−ai (i = 1,2,3) (1.13)

For the summation convention Greek indices run from 0 to 3, Roman indices from 1 to 3 (spatial

indices).

Lorentz transformations are rotations in Minkowski space-time represented by the rotation

matrix Λ

x ′µ =Λµν xν (1.14)

Λ is a real 4×4 matrix and satisfies the condition

ΛµνΛ

τµ = δ τ

ν =

1 ν= τ

0 ν 6= τ(1.15)

Scalar products are defined as

x · y ≡ xµ yµ = x0 y0 +xi yi = x0 y0 −xi yi = x0 y0 −x ·y (1.16)

Example 1.4

p1 ·p2 = E1E2 −p1 ·p2 = E1E2 −|p1||p2|cosθ12 (1.17)

Scalar products are invariant under Lorentz transformations:

x ′ · y ′ = x ′µy ′µ =Λ

µνxν

Λτµ yτ = xν

ΛµνΛ

τµ

︸︷︷︸

δ τν

yτ = xνyν = x · y (1.18)

4-vector derivatives are denoted by

∂µφ≡ ∂

∂xµφ (1.19)

with, for example, the zero-component

∂0 ≡∂

∂x0= ∂

∂t= ∂t (1.20)

1.1.3 Lagrange equations


For a set of fields φr (r = 1,2, . . . , N ) the Lagrange density L (φr ,∂µφr ) is a function of the fields

and their derivatives. The action S in a space-time region Ω is given by the integral

S =∫

Ω

d 4x L (1.21)

3


Equations of motion are obtained from the variational principle which postulates that varia-

tions of the fields φr (x) → φ′r (x) = φr (x)+δφr (x) leave S unchanged if δφr vanishes on the

surface that encloses Ω (the boundary ∂Ω):

δφr (x ∈ ∂Ω) = 0 (1.22)

The variation of S is given by the integral of the total derivative of L :

0 = δS =∫

Ω

d 4x δL (φr ,∂µφr ) (1.23)

=∫

Ω

d 4x

[∑

r

∂L

∂φrδφr

]

+

∑

r

∂L

∂(∂µφr )δ(∂µφr )︸︷︷︸

∂µδφr ∗

(1.24)

=∗∗∫

Ω

d 4x∑

r

[∂L

∂φr−∂µ

(∂L

∂(∂µφr )

)]

δφr (1.25)

+∑

r

∂µ

(∂L

∂(∂µφr )δφr

)

︸︷︷︸

aµr (x)

(1.26)

Annotation 1.1

(*) δφ=φ′−φ⇒ δ(∂µφ) = ∂µφ′−∂µφ= ∂µ(φ′−φ) = ∂µδφ

(**) here N zeros have been added: 0 = ∂µ

(∂L

∂(∂µφr )

)

δφr −∂µ

(∂L

∂(∂µφr )

)

δφr

The divergence ∂µaµr is equivalent to the surface integral of ar over ∂Ω (4-dimensional diver-

gence theorem). However, the 4-vector ar vanishes on the surface of Ω because of (1.22), so the

terms in (1.26) vanish. From (1.25) it then follows that, for δS to vanish for arbitrary δφr , the

following N Euler-Lagrange equations need to hold:

∂L

∂φr= ∂µ

(∂L

∂(∂µφr )

)

r = 1,2, . . . , N (1.27)

These are the equations of motions for the fields φr . Adding to the Lagrange density a term

which does not depend on a field or its derivative will leave the equations of motion unchanged.

1.1.4 Continuity equation and conserved charge


In this section we examine the role of symmetry transformations. These transformations act

on the fields φr and change them by the amount δφr :

φr →φ′r =φr +δφr

If this replacement leaves the Lagrange density invariant (or adds only a constant term) then

the equations of motion are unchanged and the system possesses a symmetry.

4


An invariant L implies a conserved quantity, as we can see by looking at the total derivative

0 = δL = ∂L

∂φr︸︷︷︸

∗

δφr +∂L

∂(∂µφr )δ(∂µφr ) (1.28)

= ∂µ

(∂L

∂(∂µφr )

)

δφr +∂L

∂(∂µφr )∂µ(δφr ) (1.29)

= ∂µ

(∂L

∂(∂µφr )δφr

)

≡ ∂µ f µ (1.30)

Annotation 1.2

(*) according to the Euler-Lagrange equations (1.27) this term is equal to ∂µ

(∂L

∂(∂µφr )

)

The current density f µ

f µ =(

∂L

∂(∂µφr )δφr

)

(1.31)

fulfils the continuity equation

∂µ f µ = 0 (1.32)

which is equivalent to

∂t f 0 =−∇ · f =−div f (1.33)

Upon integration over a space-volume V we obtain

∂t F 0 ≡ ∂t

∫

Vd 3x f 0 =−

∫

Vd 3x div f =−

∮

∂VdA · f (1.34)

so that the temporal change of the zero-component of f inside volume V is given by the flux

outward through the surface A of that volume.

The current density f (often referred to simply as current) is related to the fields φr . If we are

looking at a configuration where all fields are restricted to a certain region in space and let V

encompass all of that space then the flux through the surface of V will be zero. In this case the

space-integral of the zero-component of f in that region is constant in time:

∂t F 0 = 0 (1.35)

A symmetry of the Lagrange density hence corresponds to a conserved quantity which is usu-

ally identified with a charge. This is an example of Noether’s Theorem.

1.1.5 Particle creation and annihilation


We will use the example of an U (1) symmetry to illustrate how particles are created and annihi-

lated in Quantum Field Theory (QFT). We will use this concept later when discussing Feynman

diagrams in which particles are created and destroyed at each vertex.

From the definition of the current density (1.31), the zero-component is given by

f 0 = ∂L

∂(∂0φr )δφr =

∂L

∂φr

δφr ≡∑

r

πrδφr (1.36)

5


with

πr =∂L

∂φr

(conjugate momentum) (1.37)

As an example we consider the global U (1) phase transformation which acts on a complex

scalar field φ. The field φ has two degrees of freedom, the real scalar fields φa and φb :

φ= 1p

2

(

φa + iφb

)

(1.38)

Instead of using φa and φb , we work with φ and its Hermitian conjugate φ† and treat them as

two independent fields:2

φ† = 1p

2

(

φa − iφb

)

(1.39)

This is possible because φ and φ† are linear combinations of φa and φb and working with the

former pair instead of the latter corresponds to choosing a different set of coordinates.

Our set of fields φr is given by φ,φ† (r = 1,2) and the corresponding conjugate momenta are

π,π†. The fields are transformed by U (1) as follows:

φ → φ′ = e iεφ= (1+ iε)φ (1.40)

φ† → φ†′ = e−iεφ† = (1− iε)φ† (1.41)

from which we find the changes

δφ = φ′−φ= iεφ (1.42)

δφ† = −iεφ† (1.43)

so that

f 0 =∑

r

πr δφr =π δφ+π† δφ† = iε[

π φ−π† φ†]

(1.44)

and

F 0 =∫

d 3x f 0 = iε

∫

d 3x[

π φ−π† φ†]

(1.45)

We assume that the U (1) transformation leaves L invariant. This implies that F 0 (and every

product of it) is constant in time. We define a new quantity

Q ≡−q

εF 0 (1.46)

which is also constant in time. The quantity q is a real number and ε is the same as in (1.40)

and (1.41).

For the question of how Q should be interpreted, we change from a classical picture to QFT. In

QFT, the fields φr and the conjugate momenta πr become operators which (in general) do not

commute. The following commutation relation holds [1]:

[

φr (x, t ), πs(x′, t )]

= iδr s δ(x−x′) (1.47)

In this formula, r and s take the values 1 and 2. Note that φr and πs are taken at the same time

t .

2The Hermitian conjugate is defined as the transposed complex conjugate. Here it simplifies to the complex conju-

gate because φ is a scalar: φ† = (φ∗)T =φ∗.

6


We let the operator Q act on the field φ and take Q and φ at the same time t (we hence drop the

time from the following formula to make it more readable):

Q φ(x) = −i q

∫

d 3x ′[

π(x′) φ(x′)−π†(x′) φ†(x′)]

φ(x) (1.48)

We will now move φ(x) from the right to the left of the integral. The fields φr commute, and

also φ and π† commute because of the δr s in (1.47) (φ corresponds to r = 1 and π† to s = 2) but

π(x′) φ(x) =φ(x) π(x′)− iδ(x−x′) (1.49)

This results in

Q φ(x) = −i q φ(x)

∫

d 3x ′[

π(x′) φ(x′)−π†(x′) φ†(x′)− iδ(x−x′)]

(1.50)

= −i q φ(x)

∫

d 3x ′[

π(x′) φ(x′)−π†(x′) φ†(x′)]

−q φ(x)

∫

d 3x ′δ(x−x′) (1.51)

= φ(x) Q −qφ(x) (1.52)

which is a commutation relation (at equal times) between Q and φ:

[

Q, φ]

=−q φ (1.53)

We assume that |Q ′⟩ is an eigenstate of Q with eigenvalue Q ′:

Q |Q ′⟩ = Q ′︸︷︷︸

eigenvalue

|Q ′⟩︸︷︷︸

eigenstate

(1.54)

and let Q act on φ |Q ′⟩:

Q(

φ |Q ′⟩)

= (φQ + [Q,φ]︸︷︷︸

−qφ

) |Q ′⟩ =φ(

Q −q)

|Q ′⟩ (1.55)

= φ(

Q ′−q)

|Q ′⟩ =(

Q ′−q)

φ |Q ′⟩ (1.56)

So φ |Q ′⟩ is an eigenstate with eigenvalue Q ′−q .

We interpret these results as follows:

• Q is the charge operator

• φ annihilates a particle with charge q

• similarly: φ† creates a particle with charge q

This example shows one of the basic features of QFT: the fields are operators which create and

annihilate particles.3 This concept is crucial for understanding the calculations behind Feyn-

man diagrams which we will discuss later on.

Example 1.5

• |Q ′⟩: state with 2 electrons

• Q ′: charge of the state = 2q

• φ |Q ′⟩: state with charge Q ′−q = q , corresponding to only one electron

• The operator φ has destroyed one electron.

3A more in-depth treatment of QFT is beyond the scope of this lecture and the interested reader is referred to the

dedicated lectures held at this university or textbooks like [1].

7

1.2. DIRAC EQUATION CHAPTER 1. THEORETICAL BASIS AND QED

As a summary: the charge Q is conserved if L is invariant under the U (1) transformation (1.40),

(1.41). This charge can be the ordinary electric charge (U (1)EM) or hypercharge (U (1)Y ). Simi-

larly, conservation of momentum, energy, and angular momentum follows from the invariance

of L under continuous transformations as follows:

transformation L invariance implies conservation of

U (1) charge

translation in space momentum

translation in time energy

rotation in space angular momentum (and spin)

Questions to Section 1.1

1. Which factor is used to convert HEP (High Energy Physics) units to ordinary units?

2. What is k in the following equations (aµ is a 4-vector)?

• a0 = k a0

• ai = k ai

3. Complete the sentence: Invariance of the Lagrangian L under symmetry transfor-

mations leads to a continuity equation for . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . of which

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . is conserved in time (Noether’s theorem).

4. Which quantity is conserved if L is invariant under U (1) transformations?

5. Which role do fields play in Quantum Field Theory?

1.2 Dirac equation


1.2.1 Derivation

The Dirac equation is the equation of motion for spin ½ fermions. It is consistent with

p2 = pµ ·pµ = E 2 −p2 = m2 (1.57)

but first order in time.

When applying the usual quantum mechanical substitutions

p → −i∇ (1.58)

E → i∂t = i∂

∂t(1.59)

pµ → i∂µ (1.60)

in (1.57) one arrives at

(

− ∂2

∂t 2+∇2

)

φ= m2φ (Klein-Gordon) (1.61)

8

CHAPTER 1. THEORETICAL BASIS AND QED 1.2. DIRAC EQUATION

which is the Klein-Gordon equation for spin 0 particles which we will not discuss here. Dirac

noted that (1.57) can be turned into an equation first order in time by writing it as follows:

0 = pµpµ−m2 =(

γµpµ+m) (

γνpν−m)

(1.62)

The quantities γµ are four 4 × 4 matrices. For the equation to be true, (at least) one of the

two expressions in parentheses needs to be zero and the conventional choice is the latter one,

which when applying the replacement (1.60), leads to the Dirac equation:

(

iγµ∂µ−m)

ψ= 0 (Dirac) (1.63)

Note that summation over µ is implied here and that the mass m is multiplied by the 4×4 unit

matrix. For better readability we do not show the unit matrix in equations. The quantity ψ is a

spinor:

ψ=

ψ1

ψ2

ψ3

ψ4

(1.64)

It is not a 4-vector because it does not transform likeΛx under Lorentz transformations, cf. (1.14).

The γ matrices (also referred to as Dirac matrices) obey the following rules:

γµ = gµνγν (1.65)

(

γ0)2 = γ0γ0 = 1 (1.66)

(

γi)2

= −1 (1.67)

γµ,γν ≡ γµγν+γνγµ = 0 (ν 6=µ) (1.68)

Note that 1 denotes the 4×4 unit matrix. The expression γµ,γν is called anti-commutator of

γµ and γν. From (1.65) follow

γ0 = γ0 (1.69)

γi = −γi (1.70)

A fifth gamma matrix is defined as

γ5 ≡ iγ0γ1γ2γ3 (1.71)

with

γµ,γ5 = γµγ5 +γ5γµ = 0 (1.72)

We use the following representation:

γ0 =(

1 0

0 −1

)

, γi =(

0 σi

−σi 0

)

, γ5

(0 1

1 0

)

(1.73)

(where 1 denotes the 2×2 unit matrix) with the Pauli matrices

σ1 =σx =(

0 1

1 0

)

, σ2 =σy =(

0 −i

i 0

)

, σ3 =σz =(

1 0

0 −1

)

(1.74)

9


1.2.2 Solutions for particles at rest

For a particle at rest (p = 0) the replacement (1.60) implies

∂xψ= 0 = ∂yψ= ∂zψ (1.75)

and the Dirac equation simplifies to

(

iγ0∂0 −m)

ψ= 0 (1.76)

We define ψA and ψB as the upper two and lower two components of ψ:

ψA ≡(ψ1

ψ2

)

, ψB ≡(ψ3

ψ4

)

(1.77)

so that (1.76) reads

(1 0

0 −1

)(∂tψA

∂tψB

)

=−i m

(ψA

ψB

)

(1.78)

which separates to the following equations

∂tψA = −i mψA (1.79)

∂tψB = i mψB (1.80)

with solutions

ψA = ψA(0)e−i mt (1.81)

ψB = ψB (0)e i mt (1.82)

The time-dependence of a quantum state follows from the Schroedinger equation

Hψ= i∂tψ (1.83)

and is given by

ψ=ψ0e−i Et (1.84)

In our case E = m and ψA behaves as expected. However, ψB seems to correspond to a par-

ticle with negative energy E = −m (we consider masses to be positive). The solution for ψB is

reinterpreted to describe an antiparticle with positive energy (cf. Section 1.3.7).

10


The solutions for the four spinor components are (up to a multiplicative factor):

ψ1 = e−i mt

1

0

0

0

electron, spin up (1.85)

ψ2 = e−i mt

0

1

0

0

electron, spin down (1.86)

ψ3 = e i mt

0

0

0

1

positron, spin down (1.87)

ψ4 = e i mt

0

0

1

0

positron, spin up (1.88)

These are the only independent solutions. There is, for example, no solution

ψ= e i mt

1

0

0

0

(1.89)

because the left hand side of (1.76) does not vanish for such a ψ.

1.2.3 General solutions

For particles with non-vanishing momentum we make the following Ansatz of a plane wave:

ψ= a e−i k·x u(k) (1.90)

in which a is a normalisation factor, x is the ordinary space-time 4-vector, and u is a spinor that

depends on the wave 4-vector k.

The derivative of ψ is given by

∂µψ=−i kµψ (1.91)

so that it follows from the Dirac equation (1.63) that

(

iγµ∂µ−m)

ψ = 0(

γµkµ−m)

a e−i k·x u(k) = 0 | ·e i k·x and a 6= 0 (1.92)(

γµkµ−m)

u(k) = 0 (1.93)

11


With the following relation (derivation left as an exercise to the reader):

γµkµ =(

k0 −k ·σk ·σ −k0

)

(1.94)

in which

k ·σ = kxσx +kyσy +kzσz =(

kz kx − i ky

kx + i ky −kz

)

(1.95)

(1.93) can be written in components:

0 =(

γµkµ−m)

u(k) =(

k0 −m −k ·σk ·σ −k0 −m

)(uA

uB

)

(1.96)

where we used uA for the two upper components of u and uB for the two lower components.

From (1.96) we find

0 = (k0 −m)uA −k ·σuB (1.97)

0 = k ·σuA − (k0 +m)uB (1.98)

corresponding to

uA = k ·σk0 −m

uB (1.99)

uB = k ·σk0 +m

uA (1.100)

Using (1.100) in (1.99) yields

uA = (k ·σ)2

(k0)2 −m2uA (1.101)

From (1.95) it follows that

(k ·σ)2 = k2 (1.102)

so that (1.101) leads to

k2 = k20 −m2 (1.103)

from which we find the following relation to the 4-momentum p:

k2 = m2 = p2 (1.104)

The relative sign between k and p determines the time-dependence of the spinor ψ (cf. (1.90)):

k =

+p : time dependence e−i Et (particles)

−p : time dependence e i Et (antiparticles)(1.105)

There exist four independent solutions:

uA(1) =(

1

0

)

≡χ1 and (cf. (1.100)) uB (1) = 1

E +m

(pz

px + i py

)

(k = p) (1.106)

uA(2) =(

0

1

)

≡χ2 and (cf. (1.100)) uB (2) = 1

E +m

(px − i py

−pz

)

(k = p) (1.107)

uB (3) =χ1 and (cf. (1.99)) uA(3) = 1

E +m

(pz

px + i py

)

(k =−p) (1.108)

uB (4) =χ2 and (cf. (1.99)) uA(4) = 1

E +m

(px − i py

−pz

)

(k =−p) (1.109)

12


Solutions 1 (1.106) and 2 (1.107) are the particle solutions with k = p. Solutions 3 (1.108) and 4

(1.109) are the antiparticle solutions with k =−p.

Written as spinors, the solutions are

u1 = N

1

0

pz

E+m

px+i py

E+m

, u2 = N

0

1

px−i py

E+m−pz

E+m

, u3 = N

pz

E+m

px+i py

E+m

1

0

≡−v2, u4 = N

px−i py

E+m−pz

E+m

0

1

≡ v1 (1.110)

with a normalisation factor N .

Particles are described by

ψ1 = ae−i p·x u1, ψ2 = ae−i p·x u2 (1.111)

Antiparticles are described by

ψ3 = ae i p·x v1, ψ4 = ae i p·x v2 (1.112)

The quantity a is a factor to keep units consistent. For p → 0, the solutions approach the ones

for the particles-at-rest case ((1.85)–(1.88)).

1.2.4 Spin

The spin operator S is given by

S = 1

2Σ with Σ=

(σ 0

0 σ

)

(1.113)

The z-component is

Sz =1

2Σz =

(σz 0

0 σz

)

= 1

2diag(1,−1,1,−1) = 1

2

1 0 0 0

0 −1 0 0

0 0 1 0

0 0 0 −1

(1.114)

The spinors u1, u2, v1, v2 are in general no eigenstates of Sz . As an example, consider u1:

Sz u1 = Sz N

1

0

pz

E+m

px+i py

E+m

= 1

2N

1

0

pz

E+m

− px+i py

E+m

(1.115)

The sign of the lowest component is flipped and so u1 is no eigenstate.

The spinors can be made eigenstates by choosing the z-axis along the direction of motion of

the particle. Then px = 0 = py and the spinors become

u1 = N

1

0

pz

E+m

0

, u2 = N

0

1

0

−pz

E+m

, v1 = N

0

−pz

E+m

0

1

, v2 =−N

pz

E+m

0

1

0

(1.116)

13


and

Sz u1 = 1

2u1 spin up (1.117)

Sz u2 = −1

2u2 spin down (1.118)

Sz v1 = −1

2v1 spin down (1.119)

Sz v2 = 1

2v2 spin up (1.120)

Plane-wave solutions to the Dirac equation are eigenstates of Sz only if p = (0,0, pz ). One can

construct eigenstates of the helicity operator Σ ·p/|p| (cf. Section 1.4.11) from the plane-wave

solutions in the general case (i.e., the momentum does not have to be along z).

1.2.5 Spinor normalisation

By convention [2, 3], spinors are normalised so that

u†u = 2E (1.121)

We will calculate N for u1 in (1.116), i.e., with px = 0 = py :

u†1 = N (1 0

pz

E +m0) (1.122)

pz = |p| =√

E 2 −m2 =√

(E +m)(E −m) (1.123)

2E = u†u = N 2 (1+p2

z

(E +m)2) (1.124)

p2z

(E +m)2= (E +m)(E −m)

(E +m)2= E −m

E +m(1.125)

N 2 = 2E

1+ E−mE+m

= 2EE+m+E−m

E+m

= E +m (1.126)

N =p

E +m (1.127)

As an example, u1 from (1.116) is given by

u1 =

pE +m

0p

E −m

0

for p =

0

0

pz

(1.128)

1.2.6 Transformation properties of spinors


It can be shown that the spinor ψ viewed from a system moving with speed β along the x-axis

is given by

ψ′ = Sψ with S =

a+ 0 0 a−

0 a+ a− 0

0 a− a+ 0

a− 0 0 a+

(1.129)

14


in which

a± =±√

1

2(γ±1) with γ= 1

√

1−β2(1.130)

From this we find that ψ†ψ is not Lorentz-invariant:

ψ′†ψ′ = (Sψ)†(Sψ) =ψ† S†S︸︷︷︸

S2

ψ 6=ψ†ψ (1.131)

because

S2 6= 1 (1.132)

as is evident from looking at the 11-component:

(S2)11 = a2++a2

− = 1

2(γ+1)+ 1

2(γ−1) = γ 6= 1 (1.133)

We define an adjoint spinor:

ψ ≡ ψ†γ0 = (ψ∗1 ψ∗

2 ψ∗3 ψ∗

4 )

1

1

−1

−1

(1.134)

= (ψ∗1 ψ∗

2 −ψ∗3 −ψ∗

4 ) (1.135)

which is a “row spinor.”

The quantity ψψ is a scalar. It is Lorentz-invariant as shown in the following:

ψ′ψ′ =ψ′†γ0ψ′ =ψ† S†γ0S︸︷︷︸

γ0

ψ=ψψ (1.136)

The parity transformation can be shown to correspond to multiplication by γ0:

Pψ= γ0ψ (1.137)

We find that ψψ has even parity:

P (ψψ) = γ0ψγ0ψ= (γ0ψ)†γ0γ0

︸︷︷︸

1

ψ=ψ† γ0†

︸︷︷︸

γ0

ψ=ψψ (1.138)

The quantity ψγ5ψ is a Lorentz-invariant scalar but it has odd parity:

P (ψγ5ψ) =−ψγ5ψ (1.139)

It is therefore called a pseudo-scalar.

One finds that ψγµψ Lorentz-transforms like a vector and has odd parity, whereas ψγµγ5ψ

transforms like an axial vector (pseudo-vector). The transformation properties of these bilinear

covariants are listed in the following table.

15

1.3. QUANTUM ELECTRODYNAMICS CHAPTER 1. THEORETICAL BASIS AND QED

bilinear covariant transformation property

ψψ scalar

ψγ5ψ pseudoscalar

ψγµψ vector (V)

ψγµγ5ψ axial vector (A)

Later we will discuss the V-A structure of the weak interaction which is called like that because

of the occurrence of terms like ψ(γµ−γµγ5)ψ=ψγµ(1−γ5)ψ which is the difference between

a vector (V) and an axial vector (A).


1. Which particles are described by the Dirac equation?

2. What is the difference between the four solutions to the Dirac equation?

3. How do the following bilinear covariants transform under a combined Lorentz and parity

transformation?

• ψγµψ

• ψγµγ5ψ

1.3 Quantum electrodynamics

In this section we are going to build the Lagrangian of Quantum Electrodynamics (QED). Our

basis is the Dirac equation which describes the propagation of free electrons and positrons.

As first step we construct a Lagrangian L from which the Dirac equation follows as one of

the Euler-Lagrange equations. This is the free Lagrangian. Interactions between electrons and

positrons are mediated by the photon and we will introduce it into L in a second step. In the

construction of L we will encounter a symmetry which corresponds to the conservation of

electric charge.

1.3.1 Lagrange density


We will show that the following Lagrange density leads to the Dirac equation:

L =ψ[iγµ∂µ−m]ψ (1.140)

The Euler-Lagrange equations (1.27) are written in terms of the fields φr , of which there are two

here: φ1 =ψ and φ2 =ψ=ψ†γ0.

For the Euler-Lagrange equation for ψ we first assemble the pieces:

∂L

∂ψ= −ψm (1.141)

∂L

∂(∂µψ)= ψiγµ (1.142)

∂µ

(∂L

∂(∂µψ)

)

= (∂µψ)iγµ (1.143)

16

CHAPTER 1. THEORETICAL BASIS AND QED 1.3. QUANTUM ELECTRODYNAMICS

with which the Euler-Lagrange equation for ψ reads

−ψm = (∂µψ)iγµ ⇔ i (∂µψ)γµ+mψ= 0 (1.144)

For the Euler-Lagrange equation for ψ we find:

∂L

∂ψ= (iγµ∂µ−m)ψ (1.145)

∂L

∂(∂µψ)= 0 (1.146)

∂µ

(∂L

∂(∂µψ)

)

= 0 (1.147)

so that

(iγµ∂µ−m)ψ= 0

which is the Dirac equation (1.63).

The density L is invariant under the U (1) transformation

ψ → ψ′ = e−iαqψ= (1− iαq)ψ (1.148)

ψ† → ψ′† = e iαqψ† | ·γ0 (1.149)

ψ → ψ′ = e iαqψ (1.150)

The quantity q is a real number and will be identified with the charge of the particle.

Proof of invariance:

L′ = ψ′ [iγµ∂µ−m] ψ′ = e iαq ψ [iγµ∂µ−m] e−iαq ψ (1.151)

= e iαq e−iαq ψ [iγµ∂µ−m] ψ=ψ [iγµ∂µ−m] ψ=L (1.152)

1.3.2 Electromagnetic current density and conserved charge

As discussed in Section 1.1.4, the invariance of L leads to a current density f µ which fulfils the

continuity equation (1.32). The current density is given by (1.31):

f µ = ∂L

∂(∂µφr )δφr

= ∂L

∂(∂µψ)︸︷︷︸

(1.142)

δψ+ ∂L

∂(∂µψ)︸︷︷︸

=0 (1.146)

δψ (1.153)

= ψiγµ δψ=αqψγµψ (1.154)

in which we have used (1.148):

δψ=ψ′−ψ=−iαq ψ (1.155)

We define a new current density

jµ = f µ/α= qψγµψ (1.156)

which fulfils

∂µ jµ = 0 (1.157)

17


The space-integral of the zero-component of the current density is constant in time (con-

served) and is given by

Q =∫

d 3x j 0 = q

∫

d 3x ψγ0

︸︷︷︸

ψ†γ0γ0=ψ†

ψ (1.158)

= q

∫

d 3x ψ†ψ (1.159)

Classically (i.e., when not working with operators), ψ† =ψ∗ andψ∗ψ= |ψ(x)|2 is the probability

density of finding the particle at x, so that∫

d 3x |ψ|2 = 1 and Q = q , the charge of the particle.

In QFT, the fields are operators and Q is the charge operator. For QED the charge is the electric

charge.

1.3.3 Local phase transformation

The transformation (1.148)–(1.150) is called a global phase transformation because the param-

eter α is independent of the space-time coordinate x: ∂µα= 0 (q is a real number that also does

not depend on x).

We are now going to investigate whether L is also invariant under a local phase transformation

in which α depends on x:

ψ→ψ′ = e−iα(x)qψ (1.160)

We relabel L as L0 to indicate that this is the Lagrangian for a free field ψ, i.e., this field does

not interact with another particle (the Dirac equation is the equation of motion of a freely prop-

agating particle). The transformed L0 is given by

L′

0 = ψ′ [iγµ∂µ−m] ψ′ = e iα(x)q ψ [iγµ∂µ−m] e−iα(x)q ψ (1.161)

= e iα(x)q ψ iγµ∂µ e−iα(x)q ψ−e iα(x)qψme−iα(x)qψ (1.162)

The second term simplifies because e−iα(x)q is just a number which can be moved to the left

side of ψ:

e iα(x)qψme−iα(x)qψ= e iα(x)q e−iα(x)qψmψ=ψmψ (1.163)

In the first term, both e−iα(x)q and ψ depend on x so when applying the derivative, we get from

the product rule

∂µ e−iα(x)q ψ=−i q(∂µα)e−iα(x)q ψ+e−iα(x)q∂µψ (1.164)

Therefore

L′

0 = q(∂µα)ψγµψ+ iψγµ∂µψ−ψmψ (1.165)

= ψ [iγµ∂µ−m] ψ︸︷︷︸

L0

+q(∂µα)ψγµψ (1.166)

We see that L0 is not invariant. The term that destroys the invariance is q(∂µα)ψγµψ.

A different Lagrangian can be constructed which is invariant. To do this, we replace the ordi-

nary derivative ∂µ with the covariant derivative

Dµ = ∂µ+ i q Aµ(x) (1.167)

18


in which Aµ is a new field which we postulate to transform as follows:

A′µ = Aµ+∂µα (1.168)

The new Lagrangian is

L =ψ [iγµDµ−m] ψ=L0 +LI (1.169)

in which LI is the interaction Lagrangian

LI =−qψγµψAµ (1.170)

named in this way because it introduces a coupling of ψ to Aµ (see below). LI transforms to

L′

I = −qψ′γµψ′A′µ (1.171)

= −qe iαqψγµe−iαqψ(Aµ+∂µα) (1.172)

= −qψγµψAµ︸︷︷︸

LI

−qψγµψ(∂µα) (1.173)

So also LI is not invariant. But the term that destroys the invariance is the negative of the term

that destroyed the invariance of L0 in (1.166). The full Lagrangian which is the sum of L0 and

LI is therefore invariant:

L′ =L

′0 +L

′I =L0 +LI =L (1.174)

We constructed a Lagrangian L that is invariant under local U(1) transformation of the matter

fields ψ,ψ by introducing a new gauge field Aµ with certain transformation properties. The

gauge field couples to the matter fields.

For completeness: the gauge field Aµ can also propagate freely and the corresponding La-

grangian is

Lγ =−1

4Fµν Fµν (1.175)

with the field tensor

Fµν = ∂νAµ−∂µAν (1.176)

(The field tensor is not related to the space-integrated current density (1.34), they are just re-

ferred to by the same symbol.)

The full Lagrangian of QED is given by

L =L0 +LI +Lγ (1.177)

1.3.4 The photon field

In classical electrodynamics, potentials φ and A are introduced such that the magnetic and

electric fields are given by

B = rot A =∇×A = curl A (1.178)

E =−grad φ−∂t A =−∇φ−∂t A (1.179)

The potentials are not unique because the same fields B and E are obtained for

φ′ =φ+∂tα (1.180)

A′ = A−∇α (1.181)

19


We define the 4-potential

Aµ =(φ

A

)

, Aµ =(φ

−A

)

(1.182)

and the components (µ = 0,1,2,3) of this 4-potential transform like (1.180) and (1.181), and

these transformations can be written in 4-vector notation as in (1.168).

The 4-potential (1.182) has the exact transformation properties that are required for the gauge

field Aµ in Section 1.3.3. We therefore identify the gauge field with the 4-potential and refer to

it as the photon field because of its relation to the electromagnetic fields. The label gauge field

stems from the fact that (1.180) and (1.181) are called gauge transformations. A theory that is

based on the introduction of a gauge field is called a gauge theory.

1.3.5 Principle of gauge theories

Due to the success of QED in describing all electromagnetic (EM) phenomena to extremely

high precision, the theories of the weak and strong interactions were modelled after it as gauge

theories. Gauge theories are built according to the following recipe.

The starting point is a free-field Lagrangian L0 which is invariant under global phase transfor-

mation of the matter fields (ψ,ψ in our example). Global means that the phase is changed by

the same amount at every space-time point. The invariance of L0 implies a conserved quantity

or several conserved quantities (depending on the transformation).

At this point we demand invariance under local phase transformation, i.e., L should not change

even if the transformation of the field phases varies from space-time point to space-time point.

This is achieved through the introduction of one or more gauge fields that couple to the matter

fields (and may couple among themselves and gauge fields of other transformations). These

gauge fields provide the interaction between the matter fields.

Example 1.6 (QED)

• L0 follows from Dirac equation

• global phase transformation: U (1), e−iαq , ∂µα= 0

• conserved quantity: electric charge

• local phase transformation: U (1), e−iα(x)q , ∂µα 6= 0

• gauge field: Aµ (photon)

• interactions of matter particles with photon: LI

1.3.6 Electron-photon vertex

The electromagnetic current density jµ is defined by (1.156) so that the QED interaction La-

grangian (1.170) is given by the negative of the product of the current density and the photon

field:

LI =− jµ Aµ =− j · A (1.183)

The Feynman diagram of the electron-photon vertex is shown in Figure 1.1. The current density

jµ =ψ qγµ ψ is in this case called the electron current (it is understood that it really is a density).

The spinor ψ represents an electron and q is the electric charge of the electron (q = −e). This

charge is conserved in the electron-photon interaction: the charge in the initial state is −e and

20


e e

γ

jμ

(Aμ)time

Figure 1.1: Electron-photon vertex. The time axis points from left to right.

the charge in the final state is also −e. In the theory, the reason for charge conservation is that

L is invariant under the U (1) transformation.

Feynman diagrams are graphical representations of interaction amplitudes. The diagram in

Figure 1.1 is not a physical amplitude as we will see below but we will use it to illustrate the

building principle of these diagrams (Feynman rules). The spinor ψ in the current is an op-

erator which destroys an electron at the photon vertex. qγµ is the vertex factor and q is the

coupling strength between the electron and the photon. The spinor ψ creates the outgoing

electron.

1.3.7 Antiparticles

The time dependence of a quantum state is given by e−i Et (cf. (1.84)) with positive energy and

time (E > 0, t > 0). In Feynman diagrams we symbolise such a state as follows (in all Feynman

diagrams in this text the time evolves from left to right):

E

t

e−i Et

This is a particle with positive energy going forward in time.

A state that behaves like e−i (−E)(−t ) is a particle with negative energy that is going backward in

time and it is symbolised by

-Ee−i (−E)(−t )

The two are equivalent because e−i (−E)(−t ) = e−i Et and therefore

E = -Ee−i Et e−i (−E)(−t )

QED interactions depend on the current (1.156). For an electron with 4-momentum pµ = (E ,p)

the current jµ can be shown, e.g., by using explicit solutions to the Dirac equation, to be pro-

portional to the product of the electron charge and the momentum:

jµe− ∝ qe− pµ = qe+ (−pµ) (1.184)

which corresponds to the charge of a positron multiplied by the 4-momentum −pµ = (−E ,−p).

We now look again at the diagram of the photon vertex and write the energy and momentum

of the particle next to its line. A positron going backward in time is symbolised by

21


e+ e+ (E, )p

e i E(−t )

This corresponds to a state behaving like e i E(−t ). It is equivalent to a state behaving like e i (−E)t

so that the previous diagram is equivalent to the following in which we turned the arrows on

the positron line:

e+ e+ (-E,- )p

e i (−E)t

jµ

e+ ∝ qe+

(−E

−p

)

This corresponds to an electron with positive energy and momentum because we change the

charge and the sign of the 4-momentum vector simultaneously so that the current is unchanged:

e- e- (E, )p

jµe− ∝ qe−

(E

p

)

The three previous diagrams are therefore equivalent:

e+ e+ (E, )p e+ e+ (-E,- )pe- e- (E, )p

= =

Comparing the left-most diagram and the right-most one, we see that an antiparticle going

backward in time corresponds to the corresponding particle going forward in time (with the

same energy and momentum).

In the same way, the following two diagrams are equivalent:

e+ e+ (E, )p e- e- (E, )p

=

which means that an antiparticle going forward in time corresponds to the corresponding par-

ticle going backward in time (with the same energy and momentum).

In the following, we will assume all particles and antiparticles to have positive energy and mo-

mentum (and therefore not write energy and momentum next to the line in the Feynman dia-

gram).

In this text we draw Feynman diagrams according to the following convention, which is used

for example also in [2]:

22


• The fermion lines in Feynman diagrams correspond to particles (not to antiparticles).

• As the fermion lines are always particles, we simply write “e” instead of “e−”.

The positron-photon vertex is hence symbolised by

e e

1.3.8 The unphysical photon vertex

The process f → f γ (where f stands for any electrically charged fermion) is kinematically for-

bidden.

f f

In the rest frame of the initial fermion, the total initial state energy E is given by the mass of the

fermion: E = m f because the momentum is zero (rest frame, p f = 0).

The total final state energy E ′ is given by

E ′ = E ′f

︸︷︷︸

>m f

+ Eγ︸︷︷︸

>0

> m f = E (1.185)

The energy of the photon is larger than zero and the final state fermion has non-zero momen-

tum because it has to balance the photon momentum (recoil). Therefore, E ′ > E in violation of

energy conservation.

Similarly forbidden is the process f f → γ:

f

f

We choose the f f centre-of-mass system to analyse the process (and use a script ‘*’ to denote

variables in the centre-of-mass frame). The total momentum of the initial state is zero:∑

p∗i=

0. But a photon always as non-zero momentum: |p∗γ| = E∗

γ > 0. So momentum conservation is

violated in this process.

A second vertex has to be introduced in these diagrams to model physical processes.

1.3.9 Virtual photons

We consider electron-muon scattering, e−µ− → e−µ−, through photon exchange:

23


e e

μ μ

γ

The arrow next to the photon line represents the photon 4-momentum which we choose to

point away from the electron-photon vertex (this is an arbitrary choice and the results do not

depend on it). For the 4-momenta at the vertices we have from energy and momentum con-

servation (final state quantities are primed):

pe = pγ+p ′e (1.186)

pγ+pµ = p ′µ (1.187)

which when combined yield

pe +pµ = p ′e +p ′

µ (1.188)

The mass of a photon is zero, therefore one expects

p2γ = E 2

γ−p2γ = m2

γ = 0 (1.189)

Instead we find

p2γ = (pe −p ′

e )2 = p2e

︸︷︷︸

m2e

−2pe ·p ′e + p ′2

e︸︷︷︸

m2e

(1.190)

= 2m2e −2

[

Ee E ′e −pe ·p′

e

]

(1.191)

The scalar product of a 4-vector is Lorentz-invariant. We can evaluate it in any coordinate

system. For the following analysis we choose the centre-of-mass system (CMS) of the initial

state e−µ−. In this system, the incoming and outgoing electron energies and momenta are

equal because the collision is elastic (see exercise 3 on sheet 1):

E ′∗e = E∗

e (1.192)

|p′∗e | = |p∗

e | (1.193)

Therefore

p2γ = p∗2

γ = 2m2e −2

[

E∗e E ′∗

e −p∗e ·p′∗

e

]

(1.194)

= 2m2e −2

E∗2e

︸︷︷︸

m2e+p∗2

e

−p∗2e cosθ∗ee ′

(1.195)

=−2p∗2e

[

1−cosθ∗ee ′]

(1.196)

in which θ∗ee ′ is the angle in the CMS between the directions of the incoming electron and the

outgoing electron:

e μ

e

μ

θ*

24


From (1.196) we see that p2γ depends on the scattering angle θ∗

ee ′ :

θ∗ee ′ cosθ∗

ee ′ p2γ

0 1 0 no scattering (e and µ do not interact)

180 −1 −4p∗2e back-scattering

0 < θ∗ee ′ ≤ 180 −4p∗2

e ≤ p2γ < 0 scattering

For our theory to describe scattering we need p2γ < 0. The exchanged photon is not a real

photon (for which the mass vanishes). Instead it is a virtual particle. Virtual particles have

a squared 4-momentum that is different from the square of the rest mass and are therefore also

said to be off their mass shell. QED describes electromagnetic interactions very well and the

concept of virtual particles proves to be successful.

The quantity p2γ specifies how far the photon is off its mass shell. It therefore quantifies how

virtual the photon is. Because p2γ is negative but values of p2

γ near zero mean low virtuality, one

introduces a new variable Q2 as the negative of p2γ:

Q2 ≡−p2γ (1.197)

Large values of Q2 correspond to high virtuality and one therefore calls Q2 the virtuality of the

photon.

1.3.10 Photon propagator

The photon in eµ scattering “travels” from one vertex to the other:

e e

μ μ

γ

1

2

When calculating the probability amplitude of a given Feynman diagram, the term that stands

for such a propagation of an exchange particle between two vertices is called propagator term.

In general,

propagator ∝ 1

q2 −m2(1.198)

in which q is the 4-momentum of the exchanged particle and m is the real or on-shell mass of

the exchange particle. The usage of q to denote the momentum is customary but it unfortu-

nately clashes with our previous usage of q for the charge. However, it should be clear from the

context whether a charge is meant or a momentum.

For the photon, the mass m vanishes and

photon propagator ∝ 1

q2(1.199)

For other exchange particles like the W and Z bosons, the mass is non-zero and has to be

considered in (1.198).

Using the nomenclature of Section 1.3.9, q = pγ and Q2 =−q2.

25


Feynman rules are recipes to calculate transition amplitudes from diagrams. These rules fol-

low from the expansion of the scattering matrix and details can be found in books on QFT, for

example in [1]. We will use some Feynman rules here to illustrate the concept and to develop

an understanding of how cross sections are related to the coupling strength between particles.

The transition amplitude M in eµ scattering is given by

−iM = current at vertex 1 × propagator × current at vertex 2 (1.200)

= ue

(

i geγν)

ue

−i gνβ

q2uµ

(

i geγβ)

uµ (1.201)

in which gµν is the metric tensor (1.9) and ge is the electric coupling. The summation over β

leads to

M = ue

(

i geγν)

ue1

q2uµ

(

i geγν)

uµ (1.202)

= −g 2

e

q2ueγ

νue uµγνuµ (1.203)

The spinor ue destroys the incoming electron at vertex 1 and the adjoint spinor ue creates the

outgoing electron. The spinors uµ and uµ play similar roles for the muons (here the subscript

‘µ’ specifies the particle associated with the spinor, it is not a Greek index of a 4-vector). A

short-hand notation often encountered is to write e for the electron spinor, etc., so that (1.203)

reads

M =−g 2

e

q2eγνe µγνµ (1.204)

The coupling ge is related to the electromagnetic constant α as follows:

ge =q

e

p4πα (1.205)

in whichqe

is the charge of the particle (not antiparticle!) at the vertex in units of the positron

charge. Examples are

e−, µ−, τ−: ge =−p

4πα

u-type quark: ge = 23

p4πα

d-type quark: ge =−13

p4πα

The cross section is proportional to the absolute square of the amplitude (Golden Rule, see next

section), so that the dependence of the cross section on α is

σ∝|M |2 ∝α2

1.3.11 Units of electric charge

Different textbooks use different units of electric charge. Here is an overview which shows how

these units are related.

system SI Gauss Heaviside-Lorentz

unit of charge Coulomb electrostatic units (e.s.u.) e.s.u.

(“statcoulomb”)

α (= 1137

)e2

SI

4πε0e2

G

e2HL

4π

positron charge eSI =p

4παε0 eG =pα≈ 0.085 eHL =

p4πα≈ 0.3

26

CHAPTER 1. THEORETICAL BASIS AND QED 1.4. ELECTROMAGNETIC SCATTERING

For the electric coupling ge , the charge of the particle and the positron charge have to be taken

in the same system. For example, [1, 3] use the Heaviside-Lorentz system of charge and ge is

given by

ge = qHL

eHL

p4πα= qHL (used in [1, 3]) (1.206)

In [2], the Gauss system is used and

ge = qG

eG

p4πα= qG

p4π (used in [2]) (1.207)


1. What is the signature conserved quantity of QED? Why is it conserved?

2. Why is the photon called a gauge boson?

3. What is the principle of Gauge Theories?

4. What determines the strength of the QED coupling?

5. What is a virtual particle?

6. What is the photon propagator?

1.4 Electromagnetic scattering

1.4.1 Golden rule for scattering

This section follows chapter 6.2.2 of [2].

The cross section σ for the scattering of two particles with given 4-momenta p1 and p2 which

produces many particles in the final state

1+2 → 3+4+ . . .+n

is given by

σ= S

4√

(p1 ·p2)2 −m21m2

2

∫

|M |2 (2π)4 δ4(p1 +p2 −p3 −p4 − . . .−pn)×

×n∏

j=3

2π δ(p2j −m2

j ) θ(p0j )

d 4p j

(2π)4(1.208)

The integral is over the outgoing particle momenta. The first delta function ensures energy and

momentum conservation between the initial and final state. The second delta function ensures

that the outgoing particles are real, i.e., on their mass-shell. The theta function leads to positive

outgoing particles energies. The dynamics of the scattering are contained in |M |2.

S is a statistical factor which accounts for identical particles in the final state. For each group

gi of identical final state particles, S contains a factor 1gi !

(permutation factor).

27

1.4. ELECTROMAGNETIC SCATTERING CHAPTER 1. THEORETICAL BASIS AND QED

Example 1.7

Consider

1+2 → 3+4

in which 3 and 4 are identical particles. Integration over d 4p3 and d 4p4 yields the cross section

for the same process twice (double counting). This can be seen for example by looking at the

scattering angles: we are integrating over all angles 0 ≤ θ ≤ π for both particle 3 and 4 but we

should use only 0 ≤ θ3 < π2

for 3 and π2≤ θ4 ≤π for 4 because the final state is physically identical

under 3 ↔ 4. Therefore we need to use S = 12

.

We bring (1.208) into a form more suitable for the calculations of the following sections. We

rewrite the delta function which ensures real outgoing particles:

δ(p2 −m2) = δ(E 2 −p2 −m2) = δ(E 2 − (p2 +m2)) (1.209)

We exploit the following property of the delta function:

δ(x2 −a2) = 1

2a[δ(x −a)+δ(x +a)] for a > 0 (1.210)

and use

x = E = p0 (1.211)

a =√

p2 +m2 > 0 (1.212)

so that (1.209) becomes

δ(p2 −m2) = 1

2√

p2 +m2

δ(E −

√

p2 +m2)+δ(E +√

p2 +m2)︸︷︷︸

∗

(1.213)

(1.214)

The term (*) does not contribute to the integral in (1.208) because θ(p0) ensures that E > 0. We

can therefore replace the delta function δ(p2j−m2

j) in the integral with

1

2√

p2j+m2

j

δ(E j −√

p2j+m2

j)

so that

σ= S

4√

(p1 ·p2)2 −m21m2

2

∫

|M |2 (2π)4 δ4(p1 +p2 −p3 −p4 − . . .−pn)×

×n∏

j=3

2π1

2√

p2j+m2

j

δ(E j −√

p2j+m2

j) θ(p0

j )d 4p j

(2π)4(1.215)

We carry out the integrations over p0j= E j and obtain

σ= S

4√

(p1 ·p2)2 −m21m2

2

∫

|M |2 (2π)4 δ4(p1 +p2 −p3 −p4 − . . .−pn)×

×n∏

j=3

1

2√

p2j+m2

j

d 3p j

(2π)3(1.216)

with positive outgoing particle energies E j =√

p2j+m2

jin M and the delta function.

28


1.4.2 2 → 2 scattering

This section follows p. 209 and following in [2].

As a special case of the general Golden Rule for scattering introduced in Section 1.4.1, we con-

sider 2 → 2 scattering:

1+2 → 3+4

As a first step, we calculate√

(p1 ·p2)2 −m21m2

2 which is a Lorentz-invariant scalar that we can

evaluate in any coordinate system. In the centre-of-mass system,

p∗2 =−p∗

1 (1.217)

and

p1 ·p2 =(

E∗1

p∗

)T

·(

E∗2

−p∗

)

= E∗1 E∗

2 +p∗2 (1.218)

m2i = E∗2

i −p∗2 (1.219)

m21m2

2 = E∗21 E∗2

2 −E∗21 p∗2 −p∗2E∗2

2 +p∗4 (1.220)

(p1 ·p2)2 = E∗21 E∗2

2 +2E∗1 E∗

2 p∗2 +p∗4 (1.221)

(p1 ·p2)2 −m21m2

2 = E∗21 p∗2 +p∗2E∗2

2 +2E∗1 E∗

2 p∗2 (1.222)

= p∗2(

E∗21 +E∗2

2 +2E∗1 E∗

2

)

(1.223)

= p∗2(

E∗1 +E∗

2

)2(1.224)

With this result, (1.216) becomes

σ= S

64π2(E∗1 +E∗

2 )|p∗|

∫

|M |2 δ4(p∗1 +p∗

2 −p∗3 −p∗

4 )d 3p∗

3√

p∗23 +m2

3

d 3p∗4

√

p∗24 +m2

4

(1.225)

The 4-dimensional delta function separates into an energy part and a momentum part:

δ4(p∗1 +p∗

2 −p∗3 −p∗

4 ) = δ(E∗1 +E∗

2 −E∗3 −E∗

4 ) δ3(0−p∗3 −p∗

4 )︸︷︷︸

δ3(p∗3+p∗

4 )

(1.226)

and the d 3p∗4 integration in (1.225) leads to

p∗4 =−p∗

3 (1.227)

We therefore get

σ= S

(8π)2(E∗1 +E∗

2 )|p∗|

∫

|M |2δ(E∗

1 +E∗2 −

√

p∗23 +m2

3 −√

p∗23 +m2

4)√

p∗23 +m2

3

√

p∗23 +m2

4

d 3p∗3 (1.228)

To solve the integral we introduce spherical coordinates

p3 = |p∗3 |

sinθ∗ cosφ∗

sinθ∗ sinφ∗

cosθ∗

(1.229)

r ≡ |p∗3 | (1.230)

dΩ∗ = dφ∗ d(cosθ∗) (1.231)

d(cosθ∗)

dθ∗=−sinθ∗ (1.232)

29


so that

d 3p∗3 = r 2 dr dΩ

∗ (1.233)

The differential cross section dσdΩ∗ is defined as

σ=∫

dΩ∗ dσ

dΩ∗ (1.234)

and is given by

dσ

dΩ∗ = S

(8π)2(E∗1 +E∗

2 )|p∗|

∫

|M |2δ(E∗

1 +E∗2 −

√

r 2 +m23 −

√

r 2 +m24)

√

r 2 +m23

√

r 2 +m24

r 2 dr (1.235)

We change from r to the variable u:

u ≡√

r 2 +m23 +

√

r 2 +m24 (1.236)

du

dr= r

√

r 2 +m23

+ r√

r 2 +m24

= r u√

r 2 +m23

√

r 2 +m24

(1.237)

so that the integral in (1.235) is given by∫

|M |2 δ(E∗1 +E∗

2 −u)r

udu (1.238)

Upon integration, the delta-function sends u to the centre-of-mass energy of the collision:

u = E∗1 +E∗

2 = ECM (1.239)

From (1.236) it follows then after some algebra that

r = 1

2ECM

√

E 4CM

+m43 +m4

4 −2m23m2

4 −2E 2CM

(m23 +m2

4) = |p∗3 | ≡ |p∗

f | (1.240)

which is by definition (1.230) the final state momentum |p∗3 | = |p∗

4 | that is consistent with en-

ergy and momentum conservation and which we label |p∗f|.

In summary, the cross section for the 2 → 2 process is given by

dσ

dΩ∗ = S

(8π)2 E 2CM

|p∗f|

|p∗i||M |2 (1.241)

which in the case of elastic scattering (|p∗f| = |p∗

i|) simplifies to

dσ

dΩ∗ = S

(8π)2 E 2CM

|M |2 (for elastic scattering) (1.242)

If there are no identical particles in the final state, the permutation factor S = 1.

Energy-momentum conservation implies that the only free parameters are the two angles θ∗

and φ∗ which specify the flight direction of particle 3, cf. (1.229). In general, |M |2 depends on

these angles and then so does the cross section dσdΩ∗ = d 2σ

dφ∗d(cosθ∗).

As a consistency check we analyse the number of variables and constraints, starting from the

four 4-momenta p1, p2, p3, p4 (16 unknowns). From energy-momentum conservation, p1 +p2 = p3 + p4, we have four constraints. We know the particle masses mi , which give another

four constraints. The experimental setup determines the three-momenta p1 and p2 (six con-

straints). In total we get 14 constraints, leaving two unknown variables.

30


1.4.3 Cross section for eµ→ eµ

This section follows in part chapters 7.5–7.7 of [2].

We are going to calculate the cross section for electron-muon scattering in QED using Feynman

rules. The goal of this section is to demonstrate how cross sections are calculated using first

principles. It is not expected that such a calculation can be performed in the exam.

The process under study is

e−(p1, s1)+µ−(p2, s2) → e−(p3, s3)+µ−(p4, s4) (1.243)

in which the pi denote the 4-momenta and the si the spin configurations of the particles.

The Feynman diagram looks as follows:

e e

μ μ

γ

1

2p2

p1 p3

p4

q

The arrows next to the fermion and photon lines denote the directions of the particle momenta.

The QED Feynman rules to calculate −iM are as follows. Follow the arrows on the fermion

lines and write the following terms from right to left:

• external lines:

– incoming particle ( ): u

– incoming antiparticle ( ): v

– outgoing particle ( ): u

– outgoing antiparticle ( ): v

• vertex factor: −i geγµ with ge = q

e

p4πα and

qe

the electric charge of the particle in units

of the positron charge

• photon propagator: − i gµν

q2 (q is here the 4-momentum of the photon)

• energy-momentum conservation:

– for each vertex write (2π)4δ4(k1+k2+k3) in which the ki are the 4-momenta coming

into the vertex

– integrate over internal momenta:∫ d 4q

(2π)4

– drop the overall δ-function (2π)4δ4(p1 +p2 −p3 −p4)

Following this recipe we get

−iM =∫

u(s3)(p3)(

−i geγµ)

u(s1)(p1)−i gµν

q2u(s4)(p4)

(

−i geγν)

u(s2)(p2)× (1.244)

× (2π)4 δ4(p1 −p3 −q)︸︷︷︸

∗

(2π)4δ4(p2 +q −p4)︸︷︷︸

∗∗

d 4q

(2π)4(1.245)

31


The δ-function * sends q to p1 −p3 and ** becomes (2π)4δ4(p1 +p2 −p3 −p4) which we drop

according to the last Feynman rule. We therefore obtain

−iM =i g 2

e

(p1 −p3)2

[

u(s3)(p3)γµu(s1)(p1)][

u(s4)(p4)γµu(s2)(p2)]

(1.246)

Each of the brackets encloses a 1×1 number and M is a number which can be calculated when

the momenta pi and spin configurations si are specified.

Spin averaging

A situation frequently encountered in high energy physics is that particle beams are unpo-

larised and the detector does not distinguish between spin states. In that case, the measured

cross section corresponds to a combination of different spin configurations. An unpolarised

beam means that the probability of having the incoming electron spin in the up state is 50%

and the probability of having it in the down state is also 50%. The same is true for the incoming

muon spin. To obtain the unpolarised cross section one therefore has to average over the four

initial state spin configurations.

For the final state, the fact that the detector does not distinguish between the spin states up and

down for the outgoing particles means that what is measured are all possible combinations of

spin final states, i.e., the sum of the processes that lead to (up,up), (down,up), (up,down), and

(down,down).

We are dealing with probabilities here, so the cross sections need to be added. The only part of

the cross section that depends on the spin is |M |2 (cf. Golden Rule), and we average over the

initial spin configurations (s1, s2) and sum over the final spin configurations (s3, s4) of (1.246):

|M |2 = 1

4

∑

s1

∑

s2︸︷︷︸

averaging

∑

s3

∑

s4︸︷︷︸

summing

|M (s1, s2, s3, s4)|2 (1.247)

We now calculate |M |2 and use a simplified notation in which u(i ) stands for u(si )(pi ).

|M |2 =MM∗ =

g 4e

(p1 −p3)4

[

u(3)γµu(1)][

u(4)γµu(2)][

u(3)γνu(1)]∗ [

u(4)γνu(2)]∗

(1.248)

=g 4

e

(p1 −p3)4

[

u(3)γµu(1)][

u(3)γνu(1)]∗ [

u(4)γµu(2)][

u(4)γνu(2)]∗

(1.249)

where we could re-order the bracketed terms because each of them corresponds to a number.

Casimir’s trick

We encounter here twice the generic form

G =[

u(a)Γ1u(b)][

u(a)Γ2u(b)]∗

(1.250)

=[

u(a)Γ1u(b)][

u(a)Γ2u(b)]†

(1.251)

32


in which Γ stands for a γ matrix. We examine the second bracket:

[

u(a)Γ2u(b)]† =

[

u†(a)γ0Γ2u(b)

]†(1.252)

=[

γ0Γ2u(b)

]†u(a) (1.253)

= u†(b)Γ†2 γ0†

︸︷︷︸

γ0

u(a) (1.254)

= u†(b)γ0γ0

︸︷︷︸

1

Γ†2γ

0u(a) (1.255)

= u(b)γ0Γ

†2γ

0

︸︷︷︸

≡Γ2

u(a) (1.256)

(1.257)

With this definition of Γ2, the generic form reads

G = [u(a)Γ1u(b)] [u(b)Γ2u(a)] (1.258)

Summing over the spin orientations of particle b,

∑

sb

G = u(a)Γ1

(∑

sb

u(b)u(b)

︸︷︷︸

∗

)

Γ2u(a) (1.259)

The expression * corresponds to

2∑

sb=1

u(sb )(pb)u(sb )(pb) = γµpb,µ+mb (completeness relation) (1.260)

This relation is proven in exercise 4 on sheet 2.

We introduce the short-hand slash notation

γµpµ ≡ /p (1.261)

so that∑

sb

G = u(a) Γ1( /pb +mb)Γ2︸︷︷︸

≡Q

u(a) (1.262)

with a 4×4 matrix Q.

We now sum over the spin configurations of particle a:∑

sa ,sb

G =∑

sa

u(a) Q u(a) (1.263)

We write it in components so we can re-order the terms:

∑

sa ,sb

G =2∑

sa=1

4∑

µ,ν=1

u(sa )µ (pa) Qµν u

(sa )ν (pa) (1.264)

=∑

µ,νQµν

∑

sa

u(sa )ν (pa)

︸︷︷︸

4×1

u(sa )µ (pa)

︸︷︷︸

1×4

(1.265)

=∑

µ,νQµν

[

∑

sa

u(sa )(pa) u(sa )(pa)

]

νµ

(1.266)

33


which, because of (1.260), corresponds to

∑

sa ,sb

G =∑

µ,νQµν

[

/pa +ma

]

νµ (1.267)

=∑

µ

[

Q ( /pa +ma)]

µµ = Tr(

Q ( /pa +ma))

(1.268)

in which Tr(

Q ( /pa +ma))

stands for the trace of the 4× 4 matrix Q ( /pa +ma). Inserting the

definitions of G (1.250) and Q (1.262) we have just proven the following relation which is known

as Casimir’s trick:

∑

sa ,sb

[

u(a)Γ1u(b)][

u(a)Γ2u(b)]∗ = Tr

(

Γ1( /pb +mb)Γ2( /pa +ma))

(1.269)

Calculation of the spin-averaged cross section

To apply Casimir’s trick to our case, we use

Γ1 = γµ (1.270)

Γ2 = γν (1.271)

Γ2 = γ0γν†γ0 = γν (exercise) (1.272)

and with (1.249), (1.247) becomes

|M |2 = 1

4

∑

s1,s2,s3,s4

g 4e

(p1 −p3)4

[

u(3)γµu(1)][

u(3)γνu(1)]∗ [

u(4)γµu(2)][

u(4)γνu(2)]∗

(1.273)

=g 4

e

4(p1 −p3)4Tr

(

γµ( /p1 +m1)γν( /p3 +m3))

Tr(

γµ( /p2 +m2)γν( /p4 +m4))

(1.274)

This corresponds to equation (7.126) in [2] and for the following we adopt the notation used

there, which is to denote the electron mass with m and the muon mass with M :

m1 = m3 = me ≡ m (1.275)

m2 = m4 = mµ ≡ M (1.276)

In calculating the traces in (1.274), we employ trace theorems, which are, for example given

in [2] on pages 252 and 253. Those are 16 rules for taking traces of combinations of γ matrices,

of which we will quote only those that are needed for our purpose.

First we will calculate the electron trace

Tr(

γµ( /p1 +m)γν( /p3 +m))

= Tr(

(γµ /p1γν+mγµγν)( /p3 +m)

)

(1.277)

= Tr(

γµ /p1γν

/p3 +mγµγν /p3 +mγµ /p1γν+m2γµγν

)

(1.278)

= Tr(

γµ /p1γν

/p3

)

+m Tr(

γµγν /p3

)

︸︷︷︸

0

+m Tr(

γµ /p1γν)

︸︷︷︸

0

+m2 Tr(

γµγν)

(1.279)

where we used the known relations from matrix algebra: Tr(A+B) = Tr(A)+Tr(B) and Tr(αA) =αTr(A) (for a scalarα). The second and third trace in (1.279) vanish on account of rule 10 (using

the numbering in [2]) which states that the trace of a product of an odd number of γ matrices

vanishes. Remember that there is a γ matrix “hidden” in the slash notation: /p = γλpλ.

34


The fourth trace in (1.279) is solved using

Tr(

γµγν)

= 4gµν (rule 12) (1.280)

For the first trace we resolve the slash notation:

Tr(

γµ /p1γν

/p3

)

= Tr(

γµγλp1,λγνγσp3,σ

)

(1.281)

= p1,λp3,σTr(

γµγλγνγσ)

(1.282)

and use

Tr(

γµγλγνγσ)

= 4[

gµλgνσ− gµνgλσ+ gµσgλν]

(rule 13) (1.283)

so that

Tr(

γµ /p1γν

/p3

)

= 4p1,λp3,σ

[

gµλgνσ− gµνgλσ+ gµσgλν]

(1.284)

= 4(

pµ1 pν

3 − gµν pσ1 p3,σ

︸︷︷︸

p1·p3

+pµ3 pν

1

)

(1.285)

The muon trace results from the electron trace by replacing m with M , lowering the Greek

indices, and replacing p1 → p2 and p3 → p4, so that (1.274) becomes

4(p1 −p3)4

g 4e

1

16|M |2 = (p

µ1 pν

3 − gµνp1 ·p3 +pµ3 pν

1 +m2gµν)×

× (p2,µp4,ν− gµνp2 ·p4 +p4,µp2,ν+M 2gµν) (1.286)

= (p1 ·p2) (p3 ·p4)− (p1 ·p3) (p2 ·p4)+ (p1 ·p4) (p2 ·p3)+M 2 p1 ·p3

− (p2 ·p4) (p1 ·p3)+ gµνgµν︸︷︷︸

4

(p1 ·p3) (p2 ·p4)− (p1 ·p3) (p2 ·p4)−4M 2 p1 ·p3

+ (p3 ·p2) (p1 ·p4)− (p3 ·p1) (p2 ·p4)+ (p3 ·p4) (p1 ·p2)+M 2 p1 ·p3

+m2 p2 ·p4 −4m2 p2 ·p4 +m2 p4 ·p2 +4m2M 2 (1.287)

= 2(p1 ·p2) (p3 ·p4)+2(p1 ·p4) (p2 ·p3)+4m2M 2 −2M 2p1 ·p3 −2m2p2 ·p4

(1.288)

which we bring into the final form

|M |2 =8g 4

e

(p1 −p3)4

[

(p1 ·p2) (p3 ·p4)+ (p1 ·p4) (p2 ·p3)+2m2M 2 −M 2p1 ·p3 −m2p2 ·p4

]

(1.289)

This equation corresponds to equation (7.129) in [2].

With the spin-averaged squared amplitude (1.289), the cross section for unpolarised e−µ− →e−µ− scattering is therefore given by (cf. (1.242))

dσ

dΩ∗ = 1

(8π)2 (p1 +p2)2|M |2 (1.290)

35


1.4.4 Mott and Rutherford scattering

This section follows example 7.7 on pages 254 and 255 of [2].

We will use the result we have derived in Section 1.4.3 to derive the cross section for electrons

scattering off the Coulomb potential of a spin ½ particle. This is known as Rutherford scattering

if the electron is non-relativistic and as Mott scattering in the general case.

Scattering off the Coulomb potential here means that we take the target particle to be infinitely

heavy. We will obtain the Coulomb potential cross section by using the electron-muon cross

section (1.290) and let the muon mass go to infinity (M → ∞). Due to its infinite mass, the

target particle does not recoil. In the rest-frame of the target particle, the initial state momenta

are given by

pe = p1 = (E1,p1) pµ = p2 = (M ,0) (1.291)

and the final state momenta by

p ′e = p3 = (E3,p3) p ′

µ = p4 = (M ,0) (1.292)

and the following sketch shows the scattering process and the definition of the scattering angle

θ:

eμ

e

θp1

p3

The square of the centre-of-mass energy is given by

(p1 +p2)2 = m2 +M 2 +2 p1 ·p2︸︷︷︸

E1M

≈ M 2 (1.293)

where we used that M ≫ E1 and the electron mass m is neglected against M . We are deal-

ing with elastic scattering and therefore (cf. exercise 3 on sheet 1) the incoming and outgoing

electron energy and momentum are equal in the centre-of-mass system:

E∗1 = E∗

3 (1.294)

|p∗1 | = |p∗

3 | (1.295)

The speed of the centre-of-mass frame in the target particle rest-frame is given by

βCM = |p1|E1 +M

→ 0 (1.296)

which implies that the two frames are identical in our limit and we can identify the energies

and momenta in the CM frame with the ones in the rest-frame:

p∗1 = p1, p∗

2 = p2, p∗3 = p3, p∗

4 = p4 (1.297)

and because of (1.294) and (1.295)

E1 = E∗1 = E∗

3 = E3 ≡ E (1.298)

|p1| = |p∗1 | = |p∗

3 | = |p3| ≡ |p| (1.299)

36


where we have defined E and |p| as the energy and momentum of the electron, respectively.

For the terms in (1.289) we find with the scattering angle θ =∡(p1,p3):

p1 ·p2 = E M = p3 ·p4 = p1 ·p4 = p2 ·p3 (1.300)

p1 ·p3 = E 2 −p1 ·p3 = E 2 −p2 cosθ = m2 +p2 −p2 cosθ (1.301)

= m2 +p2 (1−cosθ)︸︷︷︸

2sin2 θ2

(1.302)

p2 ·p4 = M 2 (1.303)

(p1 −p3)2 = 2m2 −2p1 ·p3 = 2[m2 −E 2︸︷︷︸

−p2

+p2 cosθ] = 2p2(cosθ−1) (1.304)

=−4p2 sin2 θ

2(1.305)

so that (1.289) becomes

|M |2 =8g 4

e

16(p2 sin2 θ2

)2

[

(E M)2 + (E M)2 +2m2M 2 −M 2(m2 +2p2 sin2 θ

2)−m2M 2

]

(1.306)

=g 4

e

2(p2 sin2 θ2

)2

[

2(E M)2 −2M 2p2 sin2 θ

2

]

(1.307)

=g 4

e M 2

(p2 sin2 θ2

)2

E 2 − p2

︸︷︷︸

E 2v2

sin2 θ

2

(1.308)

in which v is the electron velocity:

E = γm (1.309)

|p| = γm v = E v (1.310)

Using

ge =q

e

p4πα=−

p4πα (1.311)

we obtain

|M |2 = (4πα)2 M 2

E 2v4 sin4 θ2

(

1− v2 sin2 θ

2

)

(1.312)

and for the cross section (cf. (1.290)):

dσ

dΩ= α2

4E 2v4 sin4 θ2

(

1− v2 sin2 θ

2

)

(Mott scattering) (1.313)

This is the Mott scattering cross section and our formula corresponds to formula (8.93b) in [1].

If the electron is non-relativistic,

v2 =( v

c

)2=β2 ≪ 1 (1.314)

|p| = γm v = γmβ≪ m (1.315)

E =√

m2 +p2 ≈ m (1.316)

37


then we obtain the Rutherford cross section as an approximation of the Mott cross section:

dσ

dΩ= α2

4m2v4 sin4 θ2

(Rutherford scattering) (1.317)

This formula corresponds to formula (8.95) in [1]. It describes the scattering of a non-relativistic

electron off the proton to good approximation.

1.4.5 Mandelstam variables

For 2 → 2 scattering

p1

p3

p2

p4

the following Mandelstam variables are defined:

s ≡ (p1 +p2)2, t ≡ (p1 −p3)2, u ≡ (p1 −p4)2 (1.318)

which are related by

s + t +u =4∑

i=1

m2i (1.319)

Hence, if the masses of the particles are known, only two of the variables s, t , and u are inde-

pendent.

When neglecting the electron and muon masses, the |M |2 for electron-muon scattering (1.289)

can be written using the Mandelstam variables as

|Meµ|2 = 2g 4e

s2 +u2

t 2(for me = 0 = mµ) (1.320)

To prove this relation, we start with (1.289) and set the masses to zero so that

|M |2 =8g 4

e

(p1 −p3)4

[

(p1 ·p2) (p3 ·p4)+ (p1 ·p4) (p2 ·p3)]

(1.321)

Comparing with (1.320) we have to show

1

4(s2 +u2) = (p1 ·p2) (p3 ·p4)+ (p1 ·p4) (p2 ·p3) (1.322)

With vanishing masses, the following relations hold:

s = (p1 +p2)2 = p21

︸︷︷︸

0

+ p22

︸︷︷︸

0

+2p1 ·p2 ⇒ s2 = 4(p1 ·p2)2 (1.323)

u = (p1 −p4)2 =−2p1 ·p4 ⇒ u2 = 4(p1 ·p4)2 (1.324)

(1.325)

38


so that

1

4(s2 +u2) = (p1 ·p2)2 + (p1 ·p4)2 (1.326)

From this we find that (1.322) is proven if

p1 ·p2 = p3 ·p4 (1.327)

p1 ·p4 = p2 ·p3 (1.328)

Those equations follow from energy-momentum conservation

p1 +p2 = p3 +p4 (1.329)

for vanishing masses:

2p1 ·p2 = (p1 +p2)2 = (p3 +p4)2 = 2p3 ·p4 (for mi = 0) (1.330)

−2p1 ·p4 = (p1 −p4)2 = (p3 −p2)2 =−2p3 ·p2 (for mi = 0) (1.331)

and we have hence proven (1.320). Expressing the spin-averaged squared amplitude in terms

of Mandelstam variables is useful in the context of crossing as will become apparent in the next

section.

1.4.6 Crossing symmetry

We have seen in Section 1.3.7 that

E = -E

Starting from the diagram for e−µ− → e−µ−,

e e

μ μp2

p1 p3

p4

we can flip (“cross”) the lines of the outgoing electron and the incoming muon. The diagram

we obtain is

e

e μ

μ

p2

p1

p3

p4

which can be redrawn as

e

e

μ

μ

p2p1

p3 p4

39


and corresponds to the amplitude for the process e+e− → µ+µ−. Note that the momentum ar-

rows for the crossed fermion lines still point in the same direction as the arrows on the fermion

lines. This means that the 4-momentum of the incoming positron is −p3 and the 4-momentum

of the outgoing antimuon is −p2.

In the definition of the Mandelstam variables (cf. Section 1.4.5), the momenta p1 and p2 denote

the incoming momenta while p3 and p4 are the outgoing momenta. We now apply the same

nomenclature to the crossed diagram but use primed variables:

e

e

μ

μ

p3p1

p2 p4'

' '

'

The primed momenta correspond to the original momenta as follows:

p ′1 = p1, p ′

2 =−p3, p ′3 =−p2, p ′

4 = p4 (1.332)

which results in the primed Mandelstam variables

s′ = (p ′1 +p ′

2)2 = (p1 −p3)2 = t (1.333)

t ′ = (p ′1 −p ′

3)2 = (p1 +p2)2 = s (1.334)

u′ = (p ′1 −p ′

4)2 = (p1 −p4)2 = u (1.335)

The amplitudes of the original diagram and the crossed diagram are equal (crossing symmetry)

provided we account for the change of sign of the two momenta in (1.332). This implies that

the cross section for e+e− → µ+µ− in the massless limit is obtained from (1.320) simply by

interchanging s and t .

This result is summarised in the following table.

e−µ− → e−µ− e+e− →µ+µ−

e e

μ μp2

p1 p3

p4

qe

e

μ

μ

p3p1

p2 p4q

t-channel exchange s-channel exchange

t = (p1 −p3)2 = q2 s = (p1 +p2)2 = q2

in massless limit: in massless limit:

|M |2 = 2g 4e

s2+u2

t 2 |M |2 = 2g 4e

t 2+u2

s2

Depending on the variable to which the photon momentum corresponds, the process is said to

proceed via t-channel or s-channel photon exchange. As a mnemonic device, the denominator

is formed by the photon momentum q to the fourth power:

|M |2 ∝ x2 + y2

q4(1.336)

40


with x and y being the two Mandelstam variables not corresponding to q2.

1.4.7 e+e− →µ+µ−

In this section we are going to analyse the cross section for muon pair production in e+e− an-

nihilation

e

e

μ

μ

p3p1

p2 p4q

By employing crossing we found from (1.320) that the spin-averaged square of the amplitude

is, in the massless limit, given by

|M |2 = 2g 4e

t 2 +u2

s2(for me = 0 = mµ) (1.337)

and the differential cross section is obtained by using this |M |2 in (1.290). To calculate the to-

tal cross section we relate (1.337) to an angle over which we then integrate. As we are dealing

with an annihilation process in which the two initial state particles are absent in the final state,

we cannot define a scattering angle in the sense that the direction of an incoming particle is

altered. Instead, the angle describes the direction under which one of the two final state parti-

cles emerges relative to some arbitrarily chosen direction. In the CMS, the only other direction

of the process is the collision axis of the initial state particles and the angle is the one between

the incoming and outgoing particle axes. We call that angle θ∗ in the following.

It is proven in exercise 1 of sheet 3 that in the massless limit

t 2 +u2

s2= 1+cos2θ∗

2(1.338)

t =−s1−cosθ∗

2(1.339)

u =−s1+cosθ∗

2(1.340)

and hence the cross section is given by

dσ

dΩ∗ = 1

(8π)2

1

s2g 4

e

1+cos2θ∗

2(1.341)

= α2

4s(1+cos2θ∗) (1.342)

Integrating over the angle we find for the total cross section of unpolarised e+e− → µ+µ− scat-

tering:

σ=∫

dΩ∗ dσ

dΩ∗ = α2

4s

∫2π

0dφ∗

∫1

−1d(cosθ∗) (1+cos2θ∗) (1.343)

= α2

4s2π

8

3(1.344)

= 4π

3

α2

s(1.345)

41


This cross section formula describes the rate of muon pair production in e+e− scattering very

well at centre-of-mass energies below the Z boson mass. This is shown for example in Figure

6.6 of [3].

The α2 dependence of the cross section is easily understood: it follows from the fact that there

are two vertices and each vertex factor carries onepα, so M ∝α and |M |2 ∝α2.

Concerning the dependence on s = (p1 +p2)2, we have from (1.290)

dσ

dΩ∗ = 1

(8π)2 s|M |2

This dσdΩ∗ ∝ 1

s|M |2 behaviour follows from the Golden Rule for elastic 2 → 2 scattering, as

shown in the calculations leading to (1.242). Comparing with (1.345), we see that the ampli-

tude does not depend on s. This is puzzling at first, because the photon is exchanged in the

s-channel and the propagator is therefore 1s

: from the Feynman rules (cf. Section 1.4.3) we get

M ∝ (v2γµu1)

1

s(u4γµv3) (1.346)

The answer lies in the fact that spinors are normalised according to (1.121). In the CMS,

u1u†1 = 2E∗ =

ps = v2v†

2 = u4u†4 = v3v†

3 (1.347)

so that

|M |2 =MM∗ ∝ 1

s2(v2γ

µu1) (u4γµv3) (v2γνu1)† (u4γνv3)† (1.348)

= 1

s2(v2γ

µu1)(v2γνu1)†

︸︷︷︸

∝s

(u4γµv3)(u4γνv3)†

︸︷︷︸

∝s

(1.349)

∝ 1

s2s2 = independent of s (1.350)

The spinor normalisation therefore cancels the s-dependence of the propagator.

1.4.8 Bhabha scattering

The process e+e− → e+e− (Bhabha scattering) can proceed via two diagrams:

e e

e e

e e

e e

s-channel t-channel

The amplitudes for the s-channel and t-channel have to be combined. Diagrams which differ

only in the interchange of two incoming or outgoing fermions or of an incoming fermion and

an outgoing anti-fermion or of an incoming anti-fermion and an outgoing fermion receive a

relative minus sign (p. 245 of [2]). As described on p. 248 of [2], the t-channel diagram is

obtained from the s-channel diagram by the interchange of the incoming positron and the

outgoing electron:

42


Therefore the amplitudes carry a relative minus sign:

M =Ms −Mt (1.351)

|M |2 = |Ms |2 +|Mt |2 + interference term (1.352)

In the massless limit (electron mass and muon mass neglected), the result for unpolarised

Bhabha scattering is

|M |2 = 2g 4e

(u2 + t 2

s2+ s2 +u2

t 2+ 2u2

t s

)

(1.353)

in which the last term results from the interference.

1.4.9 Møller scattering

The process e−e− → e−e− is called Møller scattering and proceeds via two diagrams. The first

is the familiar t-channel:

e e

e ep2

p1 p3

p4

q

but one also has to cover the case in which the p3-electron comes from the vertex with the

p2-electron:

e e

e ep2

p1

p3

p4q

This corresponds to the first diagram with p3 ↔ p4, so that q2 = (p1 − p4)2 = u, and is called

u-channel exchange. The difference between the two diagrams lies in the interchange of the

two outgoing electrons, so the diagrams need to be subtracted to get the total amplitude. The

result for unpolarised Møller scattering in the massless limit is

|M |2 = 2g 4e

(s2 + t 2

u2+ s2 +u2

t 2+ 2s2

tu

)

(1.354)

43


1.4.10 Dependence on scattering angle

We investigate the scattering angle dependence of the unpolarised cross sections of Bhabha,

Møller, and muon pair-production in e+e− annihilation. The general formula for elastic scat-

tering is given by (1.242) in which we use the spin-averaged squared amplitude to account for

the fact that we are considering unpolarised scattering:

dσ

dΩ∗ = S

(8π)2 s|M |2 (1.355)

The permutation factor S is ½ in the case of Møller scattering and 1 for Bhabha scattering and

in e+e− →µ+µ−.

Muon pair production

The differential cross section for e+e− →µ+µ− is given by (1.342).

Bhabha scattering

For Bhabha scattering we get from (1.353) and (1.205):

dσ

dΩ∗ = α2

2s

(u2 + t 2

s2+ s2 +u2

t 2+ 2u2

t s

)

(1.356)

For the scattering angle dependence, we use (1.338) and the relations (1.339) and (1.340) from

which it follows that

2u2

t s=

2(

−s 1+cosθ∗

2

)2

−s 1−cosθ∗

2s

=−2cos4 θ∗

2

sin2 θ∗

2

(1.357)

s2 +u2

t 2=

1+cos4 θ∗

2

sin4 θ∗

2

(1.358)

and we get for Bhabha scattering:

dσ

dΩ∗ = α2

2s

(1+cos2θ∗

2︸︷︷︸

s-channel

+1+cos4 θ∗

2

sin4 θ∗

2︸︷︷︸

t-channel

−2cos4 θ∗

2

sin2 θ∗

2︸︷︷︸

interference

)

(1.359)

The cross section dσ/d(cosθ∗) is shown in Figure 1.2 forp

s = 30GeV with the contributions

of the three terms shown separately. The t-channel diagram introduces a divergence at θ∗ = 0

(forward scattering peak). This term is dominant (except at θ∗ = π where the t- and s-channel

contributions are equal) so that most electrons are scattered only by small angles.

In the limit of small angles for which cosθ∗ ≈ 1 ≈ cos θ∗

2, it follows from (1.359) that

2s

α2

dσ

dΩ∗ ≈(

1+ 2

sin4 θ∗

2

− 2

sin2 θ∗

2

)

(1.360)

≈ 2

sin4 θ∗

2

(1.361)

dσ

dΩ∗ ≈ α2

s

1

sin4 θ∗

2

(1.362)

44


0.5 1 1.5 2 2.51

10

210

310

410

510

* (pb)θ/dcosσd = 30 GeVs

* (rad)θ

(Bhabha)-e+ e→ -e+e

sum

t-channel

s-channel

|interference|

Figure 1.2: Differential cross sections dσ/d(cosθ∗) as a function of the scattering angle θ∗ in

the centre-of-mass system atp

s = 30GeV for Bhabha scattering (e+e− → e+e−). The individual

contributions from the s-channel diagram, the t-channel diagram, and the interference (mul-

tiplied by -1) are also shown. The range of the horizontal axis is chosen to avoid the forward

divergence at θ∗ = 0.

This is the small scattering angle limit of relativistic e+e− → e+e− scattering (masses neglected).

The Rutherford formula (1.317) has the same angle dependence but it was derived in the limit

of an infinitely heavy target particle hit by a non-relativistic electron.

Møller scattering

From (1.354) and (1.205) we get with S = ½:

dσ

dΩ∗ = α2

4s

(s2 +u2

t 2+ 2s2

tu+ s2 + t 2

u2

)

(1.363)

which has the same forward peak as Bhabha scattering from the t-channel diagram. For the

other terms,

2s2

tu= 8

1−cos2θ∗(1.364)

s2 + t 2

u2=

1+ sin4 θ∗

2

cos4 θ∗

2

(1.365)

The u-channel term diverges at θ∗ = π (backward peak) and the interference term diverges

(but only quadratically) for both θ∗ = 0 and θ∗ =π. The differential cross section dσ/d(cosθ∗)

is shown in Figure 1.3 forp

s = 30GeV.

A comparison between the three types of scattering cross sections is shown in Figure 1.4. The

measurements of these processes are well described by QED, see for example results of the

CELLO Collaboration at the PETRA collider [4] for Bhabha scattering (the differential cross sec-

tion is reproduced in Figure 8.4 of [1]).

45


0.5 1 1.5 2 2.5

210

310

410

510


* (rad)θ

(Moller)-e- e→ -e-e

sum

u-channel t-channel

interference



s = 30GeV for Møller scattering (e−e− → e−e−). The individual

contributions from the s-channel diagram, the u-channel diagram, and the interference are

also shown. The range of the horizontal axis is chosen to avoid the forward and backward

divergences at θ∗ = 0 and θ∗ =π.

1.4.11 Helicity

This section and the next follow section 9.7.1 of [2].

Helicity is the spin orientation along the direction of motion of the particle. The helicity oper-

ator is given by

p

|p|·Σ (1.366)

in which p is the momentum operator and Σ is two times the spin operator defined in (1.113).

The solutions u1, u2, u3, u4 to the Dirac equation given in (1.110) are not helicity eigenstates

but helicity eigenstates can be constructed from linear combinations of them. For example,

u(±) = A(±)

(y (±)

±|p|E+m

y (±)

)

(1.367)

in which

A(±) =√

E +m

2|p|(|p|±pz )(1.368)

y (±) =(

pz ±|p|px ± i py

)

(1.369)

are solutions of the Dirac equation and eigenstates of the helicity operator.

46


0.5 1 1.5 2 2.5

210

310

410

510


* (rad)θ

-e- e→-e-e

-e+ e→-e+e

-µ+µ →-e+e



s = 30GeV for Møller scattering (e−e− → e−e−), Bhabha scatter-

ing (e+e− → e+e−) and muon pair-production in e+e− annihilation. The range of the horizontal

axis is chosen to avoid the forward and backward divergences at θ∗ = 0 and θ∗ =π.

Example 1.8

For p =

0

0

p

the helicity operator is

p

|p|·Σ=Σz =

(σ3 0

0 σ3

)

(1.370)

The eigenvalues are −1 if the spin points in the direction opposite to the direction of motion

and +1 if the spin points in the direction of motion.

Helicity is not handedness! Helicity is determined by the alignment of spin and momentum.

Its eigenvalues are +1 or −1. Handedness is discussed in the next section.

1.4.12 Chirality (Handedness)

The word chirality comes from the Greek word for hand (χǫιρ) and literally means handedness.

Handedness is a particle property that determines if and how a particle takes part in weak in-

teractions (discussed below).

The left-handed component of a particle spinor u is defined to be

PL u ≡ 1−γ5

2u = uL (1.371)

and the right-handed component is defined to be

PR u ≡ 1+γ5

2u = uR (1.372)

47


PL and PR and are projection operators. They satisfy PL +PR = 1 and therefore u = uL +uR .

It can be shown that for particles (u1,u2):

γ5u1,2 =( p·σ

E+m0

0p·σ

E−m

)

u1,2 (1.373)

For antiparticles,

vL = PR v, vR = PL v (1.374)

Relation between chirality and helicity for massless particles

For massless particles, E = |p| and from (1.373) it follows that

γ5u =(p·σ

|p| 0

0p·σ|p|

)

u =(

p

|p|·Σ

)

u (1.375)

so that γ5 corresponds to the helicity operator.

For the case that u is a helicity eigenstate (eigenvalue +1 or −1), we find

helicity of u

+1 −1

γ5u = u −u

(1−γ5)u = 0 2u

PLu = uL = 0 u

(1+γ5)u = 2u 0

PR u = uR = u 0

In that sense, a massless helicity −1 state is left-handed and a massless helicity +1 state is right-

handed.

If u− denotes the helicity −1 eigenstate and u+ the helicity +1 eigenstate, then a general mass-

less state is given by a linear combination of the helicity eigenstates:

u =αu−+βu+ (1.376)

and

PLu =αu− (1.377)

PR u =βu+ (1.378)

so that PL projects out the helicity −1 component and PR the helicity +1 component.

Relation between chirality and helicity for massive particles

In general handedness and helicity are different things.

If the particle mass can be neglected against the total particle energy, then a correspondence

between handedness and helicity can be found as discussed above.

48


1.4.13 Chirality in QED

QED interactions are described by the electromagnetic current jµ (1.156). We will see in the

following that the chirality does not change at the photon vertex.

jµ = quγµu = quL +uRγµ(uL +uR ) (1.379)

uL +uR = (uL +uR )†γ0 = u†Lγ0 +u†

Rγ0 = uL +uR (1.380)

jµ/q = (uL +uR )γµ(uL +uR ) = uLγµuL +uRγ

µuR +uLγµuR

︸︷︷︸

0

+uRγµuL

︸︷︷︸

0

(1.381)

The mixed terms vanish because they contain PLPR u or PR PLu which are zero (the left-handed

component of a right-handed state is zero and vice versa). The proof goes as follows:

uL = (PLu)†γ0 = u†P †Lγ0 = u†

(1−γ5†

2

)

γ0 (1.382)

Since γ5 is real and anti-diagonal (cf. 1.73),

(γ5)† = ((γ5)T )∗ = γ5 (1.383)

so that

uL = u† 1−γ5

2γ0 = u† γ

0 −γ5γ0

2(1.384)

= u†γ0 1+γ5

2= uPR (1.385)

where (1.72) was used. The next step is to move the projection operator PR past the γµ matrix

in uLγµuR :

PRγµ = 1+γ5

2γµ = γµ+γ5γµ

2= γµ−γµγ5

2= γµPL (1.386)

So

uLγµuR = uPRγ

µuR = uγµPLPR u = 0 (1.387)

because

PLPR ∝ (1−γ5)(1+γ5) = 1− (γ5)2 = 1−1 = 0 (1.388)

Because of (1.381), the handedness of the fermion does not change at the photon vertex:

fL fL fR fR

For annihilation processes, the amplitudes of the following diagrams vanish:

fL

fL

fR

fR

49


For example in the left diagram, using the Feynman rules of Section 1.4.3, the term for the left

vertex is proportional to

vLγµuL (1.389)

From (1.374) it follows that

vL = (PR v)†γ0 = v†P †Rγ0 = v†PRγ

0 = v†γ0PL = vPL (1.390)

so that

vLγµuL = vPLγ

µuL = vγµPR PLu = 0 (1.391)

At the photon vertex, the fermion and antifermion therefore need to have opposite handedness

which can be interpreted as the following crossing relations (cf. [3], p. 127):

fR fR fR

fL

crossing−−−−−−→

fL fL fL

fR

crossing−−−−−−→

For example, in e+e− →µ+µ− scattering only the following configurations are possible:

e+L e−R →µ+Lµ

−R (1.392)

e+L e−R →µ+Rµ

−L (1.393)

e+R e−L →µ+Lµ

−R (1.394)

e+R e−L →µ+Rµ

−L (1.395)

1.4.14 Parity conservation in QED

Parity is conserved in QED because the photon couples with equal strength to left-handed and

right-handed particles:

fL fLgefR fRge

The photon sees only the electric charge but not the handedness.

As an example, consider the processes (1.392) and (1.395). The two processes are related by

the parity transformation which becomes obvious in the massless limit in which a left-handed

(right-handed) lepton has negative (positive) helicity:

50


e- e+

μ+

μ-

⇒ ⇒⇒

⇒e+ e-

μ-

μ+

⇒ ⇒⇒

⇒

e+R e−L →µ+Rµ

−L P−→ e+L e−R →µ+

Lµ−R

In this diagram the arrows denote the momenta and the double arrows indicate the spin orien-

tation.

The parity operator multiplies a spinor by γ0 (cf. 1.137). Calculation shows that the parity op-

erator transforms the amplitude of the process e+R e−L →µ+Rµ

−L into the amplitude of the process

e+L e−R →µ+Lµ

−R . The amplitudes are equal because the photon coupling to left-handed and right-

handed particles is identical.

1.4.15 e+e− → qq

For the total e+e− →µ+µ− cross section we found in (1.345):

σ= 4π

3

α2

s

in which the dependence on α results from the two photon vertices

e

e

μ

μ

ge(μ-)

ge(e-)

with

ge (µ−) = ge (e−) = Qe−

e

p4πα=−

p4πα (1.396)

M ∝ g 2e (e−) ∝α (1.397)

σ∝|M |2 ∝α2 (1.398)

For the production of quark pairs in e+e− annihilation, e+e− → qq , the vertex factor at the

photon-quark vertex carries the charge of the quark:

e

e

q

q

ge(q)

ge(e-)

51


with

ge (q) =qq

e

p4πα (1.399)

M ∝ ge (e−)ge (q) ∝αqq

e(1.400)

σ∝|M |2 ∝α2

(qq

e

)2

(1.401)

The cross section is therefore sensitive to the fractional quark charges. For example,

σ(e+e− → uu) = 4π

3

α2

s

(2

3

)2

3︸︷︷︸

colour

(1.402)

The colour factor 3 occurs because the qq pair can have three colour configurations (red-

antired, blue-antiblue, green-antigreen). We did not specify the final state colour configura-

tion and therefore all configurations have to be considered by adding the cross sections for

these different final states.

1.4.16 Adding amplitudes vs. adding cross sections

Amplitudes are added (or subtracted if there is a relative minus sign between them) if several

diagrams lead to the same final particle state.

Cross sections are added if the final state is different and one has actually different processes.

One adds cross sections usually because the measurement to which the calculation is com-

pared is not sensitive to the difference in the final state. Then the measurement is really the

sum of different processes and the theoretical prediction has to accommodate that (like in the

colour.

Example 1.9 (e+e− → e+e−)

The two diagrams (s-channels and t-channel) lead to the same final state and therefore the

amplitudes need to be added (subtracted in this case).

Example 1.10 (e+e− → uu, e+e− → dd , e+e− → ss)

The final state particle content is different. Measurements cannot distinguish between the final

states so the cross sections need to be added.

1.4.17 Rhad

In an experiment, the measured final state particles are hadrons, not quarks, because of colour

confinement. Often one does not distinguish between different hadron flavour content and

just measures the production of hadrons inclusively. Then

σ(e+e− → hadrons) = 4π

3

α2

s3∑

q

(qq

e

)2

(1.403)

and the sum is over all kinematically allowed flavours. It is customary to normalise the hadron

production cross section to the cross section for muon pair-production and call the ratio Rhad,

Rhad ≡ σ(e+e− → hadrons)

σ(e+e− →µ+µ−)= 3

∑

q

(qq

e

)2

(1.404)

52

CHAPTER 1. THEORETICAL BASIS AND QED 1.5. HIGHER ORDER CORRECTIONS

The prediction is in good agreement with measurements of Rhad as shown for example in Fig.

8.4 of [2].


1. Which of the following diagrams represents t-channel photon exchange and which one

s-channel exchange?

a) b)

t

2. What is the total cross section for e+e− →µ+µ− scattering? Sketch it as a function ofp

s.

3. In which direction are most particles scattered in t-channel photon exchange processes?

What is the functional form of the dependence on the scattering angle (in the CMS)?

4. What is helicity?

5. What is chirality (handedness)?

6. What is the relation between chirality and helicity?

7. What is the reason for the scattering angle dependence in e+e− scattering?

8. How do we know experimentally that quarks have fractional charges?

1.5 Higher order corrections

1.5.1 Loop diagrams

The QED scattering matrix is expanded in terms of απ and each Feynman diagram is a term

in this expansion. The expansion series converges early because α is small ( 1137

with small

corrections at high energies, see next section) and usually only a few diagrams are necessary

to obtain a prediction with uncertainties comparable to the measurement precision of high

energy physics experiments.

As an example, consider photon exchange in electron-muon scattering. The lowest order dia-

gram (tree level) is

e e

μ μwith an amplitude proportional to α because each vertex adds a factor

pα to the amplitude.

Higher order Feynman diagrams have more internal lines and vertices. The photon can split

53

1.5. HIGHER ORDER CORRECTIONS CHAPTER 1. THEORETICAL BASIS AND QED

into a fermion-antifermion pair which subsequently recombines (vacuum polarisation):

e

μ

e

μThe loop diagram and the tree level diagram lead to the same final particle state, and conse-

quently the amplitudes have to be added and then squared to obtain the cross section. The

amplitude of the loop diagram is ∝ α2 and the cross section therefore receives only a small

correction from the loop diagram.

1.5.2 Renormalisation

This section follows section 7.9 of [2].

The 4-momentum k of the particle (mass m) in the loop of the previous section is not fixed and

this leads to terms like∫∞

m

d |k||k|

= ln |k|∣∣∣

∞

m→∞ (1.405)

The solution is to introduce a cutoff M ,

∫M

m

d |k||k|

= lnM

m(1.406)

and let M go to infinity at the end of the calculation.

The introduction of the cutoff has two consequences. First, the amplitude can be written as a

function of a renormalised coupling

gR = ge

√

1− g 2e

12π2ln

M 2

m2(1.407)

in which ge corresponds to the bare charge. The renormalised coupling gR corresponds to the

physical charge which can be determined experimentally.

The second consequence is the appearance of a correction term f (Q2/m2) in the amplitude

where Q2 is the photon virtuality −q2 (q is the photon 4-momentum):

M ∝ g 2R

(

1+g 2

R

12π2f

(Q2

m2

))

(1.408)

This correction too is absorbed into gR :

gR (Q2) = gR (0)

√

1+ gR (0)2

12π2f

(Q2

m2

)

(1.409)

which translates to

α(Q2) =α(0)

[

1+ α(0)

3πf

(Q2

m2

)]

(1.410)

54


The function f is given in eq. (7.184) on page 264 of [2] but we will not make further use of it

here.

If contributions from multi-loop corrections are taken into account, the dependence of α on

Q2 becomes

α(Q2) = α(0)

1− α(0)3π ln

Q2

m2

forQ2

m2≫ 1 (1.411)

with α(0) = 1137

. This expression for the so-called running of α holds for one particle (e.g.,

an electron) in the loops but corrections also result from other particles. One has to take all

corrections into account and then the running is given by eq. (6.21) on page 184 of [5]):

α(Q2) = α(µ2)

1− α(µ2)π ln

Q2

µ2

forQ2

µ2≫ 1 (1.412)

Since we now have several particles with different masses, the mass in (1.411) is replaced by a

scale µ.

The running of α can be determined experimentally, for example, by measuring the cross sec-

tion (1.345) for e+e− → µ+µ− which is proportional to α2. The photon virtuality Q2 is in this

process given by s, the square of the centre-of-mass energy. By changing s through adjust-

ments of the positron and electron beam energies, α can be determined as a function of Q2.

One finds that the value of α at Q2 = m2Z (where mZ = 91.2GeV is the mass of the Z boson) is

≈ 1129

[6]. This is a 6% increase from 1137

which is a small change compared to the running of

the strong coupling αs .

A physical interpretation for the running of the coupling will be given in the next sections.

1.5.3 Resolving small structures

We will examine the statement “High energies are required to resolve small structures” and start

from the Heisenberg uncertainty relation

∆x ∆px ≥ ħ2

(1.413)

The quantities ∆x and ∆px are inherent quantum mechanical fluctuations, not measurement

uncertainties.

Consider a proton at rest and an electron that is hitting a quark inside the proton:

1 fm

electron

pe quark

proton (at rest) px

x

55

1.5. HIGHER ORDER CORRECTIONS CHAPTER 1. THEORETICAL BASIS AND QED

The quark momentum in the x direction after the collision is px . The goal is to measure the

spatial position x < 12

fm by which the quark is displaced from the centre of the proton. To

measure a value different from zero, by the uncertainty principle (1.413), ∆x has to be smaller

than x (otherwise the measured values would be smeared out about zero). Also the outgoing

quark momentum px is measured and must be larger than ∆px to obtain a value significantly

different from zero.

Therefore

px >∆px ≥ ħ2∆x

> ħ2x

> ħ1fm

= ħc

1fm= 197MeVfm

1fm= 197MeV (1.414)

The momentum px is transverse to the electron momentum pe , the magnitude of which has

to be larger than px : |pe | > px . Electron energies larger than 200MeV are therefore required

to resolve structures inside the proton. The current upper limit on the quark radius is 10−18 m

and requires electron energies above 200GeV. Resolving small structure indeed requires high

energy.

Role of the photon, Q2

The electron does not couple directly to the quark though, the interaction occurs through the

exchange of a photon:

γ

p

e

quark

e

The photon 4-momentum is given by q ,

q2 = E 2γ−p2

γ (1.415)

−q2 =−E 2γ+p2

γ (1.416)

At high Q2 =−q2, p2γ ≫ E 2

γ and |q | is approximately equal to the photon momentum:

|q| ≈ |pγ| (1.417)

The photon momentum has to be larger than the quark momentum px and from (1.414) it

follows that a highly virtual photon (large Q2) resolves small structures.

1.5.4 Vacuum polarisation

The physics interpretation of the Q2 dependence of the electromagnetic coupling α is vac-

uum polarisation which leads to a shielding of the electric charge. In the loop diagram in Sec-

tion 1.5.1, a fermion-antifermion pair is produced by the exchanged photon. The fermion pair

forms an electric dipole with the positively charged side pointing to the electron:

56


+

-

e-

At higher orders, corresponding to several loops, several dipoles are formed:

+

-

e-+

+

+

-

-

-

At high Q2, the photon resolves the bare electron charge. At low Q2, the photon “sees” the

charge in a larger area and part of the electron charge is shielded by the dipoles:

+

-

e-+

+

+

-

-

-

area seenby photon

large Q2

+

-

e-+

+

+

-

-

-

small Q2

The electron charge seen by the photon at low Q2 is smaller than the bare charge.


1. How does the electromagnetic coupling depend on the scale Q2? How can this behaviour

be interpreted?

57

1.6. SUMMARY CHAPTER 1. THEORETICAL BASIS AND QED

1.6 Summary

Some of the important points discussed in this chapter are

• L invariant under global U(1) symmetry transformations (e−iαq ) of the matter fields

EM current density jµ

EM= qψγµψ

• Space integral of j 0EM

(=electric charge) is conserved (constant in time)

• L invariance under local U(1) symmetry transformations (e−iα(x)q ) of the matter fields

requires photon field Aµ that transforms like A′µ = Aµ+∂µα (=gauge transformations of

electrodynamical potentials)

• interaction Lagrangian Lint =− jEM · A

• strength of coupling between photon and matter fields is given by electric charge

• photon propagator 1q2 (q is here the 4-momentum of the photon)

• cross sections and decay rates are calculated using the Golden Rule and Feynman rules

– have to take spin of particles into account; average/sum over spin configurations

for unpolarised scattering

– crossing relates t- and s-channel photon exchange

• conservation of angular momentum leads to dependence of cross sections on scattering

angle (cf. exercise 1 on sheet 4)

• t-channel produces forward scattering peak

• QED preserves parity because photon couples with equal strength to left- and right-

handed particles

• Rhad proves fractional quark charges and colour factor 3

• vacuum polarisation leads to running of fine structure constant

58

Bibliography

[1] F. Mandl and G. Shaw, Quantum Field Theory. Wiley, 2nd ed., 2010.

[2] D. Griffiths, Introduction to Elementary Particles. Wiley-VCH, 2nd ed., 2008.

[3] F. Halzen and A. Martin, Quarks and Leptons. Wiley, 1984.

[4] CELLO Collaboration, H. Behrend et al., Measurement of e+e− → e+e− and e+e− → γγ at

Energies Up to 36.7 GeV, Phys. Lett. B103 (1981) 148.

[5] D. Perkins, Introduction to High Energy Physics. Cambridge University Press, 4th ed., 2000.

[6] S. Mele, Measurements of the running of the electromagnetic coupling at LEP,

arXiv:hep-ex/0610037 [hep-ex].

59

http://dx.doi.org/10.1016/0370-2693(81)90690-0

http://arxiv.org/abs/hep-ex/0610037

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