A brief view of Quantum Electrodynamic - LLRpolypaganini/ch3.pdf · Chapter 3 A brief view of Quantum Electrodynamic Few references: I.J.R Aitchison & A.J.G. Hey,“Gauge theories

Chapter 3

A brief view of QuantumElectrodynamic

Few references:I.J.R Aitchison & A.J.G. Hey,“Gauge theories in particle physics”, vol1, Chapters 6,7,8, IoP2003.M.E. Peskin & D.V. Schroeder, “An introduction to Quantum Field Theory”, Chapter 4,5,Westview Press Inc 1995

With the first two chapters, we saw that the quantization of the solutions ofthe wave equations leads to quantum fields, a well adapted framework to treatstates made of many particles that can be created or annihilated because ofinteractions. This chapter will briefly try to explain how this can be adaptedto the interaction of electrons with photons. I want to warn the reader: thiscourse is not a quantum field course. The goal here, is to give the conceptswithout complete demonstrations and prepare the reader to be able to do simplecalculation of scattering processes at the lowest order.

3.1 The action and Lagrangians

3.1.1 The least action principle

Classical case: In the hamiltonian formulation of classical mechanics, the equations of motionare deduced by minimizing a quantity called the action S. Let us recall how it works using asimple example. Consider a non-relativistic particle moving in 1 dimension x with a kineticenergy T = 1

2

mx2 and with a potential energy V (that can be mgx for instance). The quantity:

L = T � V

is called a Lagrangian with the dimension of the energy and depends obviously on x and x. Themomentum and energy are given by:

p = @L@x

E = px� L

63

64 A brief view of Quantum Electrodynamic

The equation of motion between the time t1

and t2

is deduced by minimizing the action Sdefined by:

S =

Z t2

t1

L(x, x)dt

The meaning of minimization is that among all trajectories x(t) that can be imagined startingfrom x

1

at t1

and finishing at x2

at t2

, the one which will have the lowest value of the actionwill be the one adopted by the particle. Let us call x0(t) the trajectory we want to find. Anytrajectory can be written x(t) = x0(t) + �x(t). The action becomes:

S =

Z t2

t1

L(x0 + �x, x0 +d

dt�x)dt =

Z t2

t1

✓L(x0, x0) +

@L

@x�x +

@L

@x

d

dt�x

◆dt

Integrating by part:

Z t2

t1

@L

@x

d

dt�x dt =

@L

@x�x

�t2

t1

�Z t

2

t1

d

dt

✓@L

@x

◆�x dt = �

Z t2

t1

d

dt

✓@L

@x

◆�x dt

where �x(t1

) = �x(t2

) = 0 has been used (all trajectories start from the same point and finishat the same point). Hence the variation of the action is just:

�S =

Z t2

t1

✓@L

@x� d

dt

@L

@x

◆�x dt

and thus:

�S = 0) @L

@x� d

dt

@L

@x= 0

which is the Euler-Lagrange equation. Applying it to our simple example:

@L@x = �@V

@xddt@L@x = mx

�mx = �@V

@x= F

So, we find the usual Newton’s law.

Simple quantum case: Let us call qr (r = 1, n) the generalized coordinates (was x before,with n = 1). In quantum mechanics, we have the additional constraint:

[qr, pr0 ] = i�rr0

(with ~ = 1), qr and pr0 being operators. The Lagrangian L(q1

, . . . , qn, q1

, . . . , qn) is now anoperator and we still have:

pr = @L@qr

H = (Pn

r=1

pr qr)� L

where H the hamiltonian, gives the total energy of the system. Since the hamiltonian is her-metian (the energy is real), the Lagrangian must be hermetian too. Now, the Euler-Lagrangeequation is simply:

@L

@qr� d

dt

@L

@qr= 0

P.Paganini Ecole Polytechnique Physique des particules

Free electron Lagrangian 65

Quantum field case: in the continuous case ie n ! 1, the n qr operators become a field�(~x, t) depending on spacetime coordinates, and the summation must become an integral. There-fore, instead of considering the Lagrangian L itself, we consider the Lagrangian density L:

L =

ZL d3~x (3.1)

and hence, the action becomes:

S =

ZL d4x (3.2)

As before L depends on � and � but since � is a continious function of spacetime, L can alsodepend on @�/@x, @�/@y, @�/@z so that L is now L(�, @µ�).The momentum p becomes a “momentum field” ⇡(xµ):

⇡(xµ) =@L

@�(xµ)(3.3)

and the hamiltonian density field is:

H = ⇡(xµ)�(xµ)� L (3.4)

The quantization generalizes what we had in the previous case1:

[�(xµ),⇡(xµ)] = i�(3)(~x� ~y) (3.5)

This is this procedure which yields the quantized field of the previous chapter, the commutatorbetween the creation and annihilation operators being a simple consequence of the previousrelation. Finally, the Euler-Lagrange equation becomes:

@L@�� @µ

@L@(@µ�)

= 0 (3.6)

3.1.2 Free electron Lagrangian

When the wave equation is known (in our case the Dirac equation), it is not too di�cult to findthe Lagrangian. The free electron Lagrangian, namely the Dirac Lagrangian density, is:

LD = (i/@ �m) (3.7)

Let us check that it gives the Dirac’s equation. The variables that have to be considered arethe 4 components of the Dirac’s field ↵, its adjoint field ↵ (since is complex, we can vary and independently) and the derivatives @µ ↵ and @µ ↵ with ↵ = {1, 4}. Making explicit theLagrangian in terms of components:

L =X↵,�

↵i�µ↵�@µ � �m�↵� ↵ �

where �↵� is the Kronecker symbol. Applying the Euler-Lagrange equation for � = ↵:

@L@ ¯ ↵

=P

� i�µ↵�@µ � �m�↵� �

@L@(@µ ¯ ↵)

= 0

)X�

i�µ↵�@µ � �m ↵ = 0

1one has to use anticommutator in case of fermions as in the previous chapter.

P.Paganini Ecole Polytechnique Physique des particules avancee


Since this equation is valid for the 4 components ↵, we do recover the Dirac’s equation: i�µ@µ �m = 0. Now, applying the Euler-Lagrange equation for � :

@L@ �

=P

↵�m�↵� ↵@L

@(@µ �)

=P

↵ ↵i�µ↵�

)�m � �

X↵

@µ ↵i�µ↵� = 0

where putting together the 4 equations (1 per component), we find i@µ �µ+m = 0, the adjointequation 2.25.

3.1.3 Free photon Lagrangian

As seen in the previous chapter, the free photon obeys the wave equation (2.72) namely ⇤Aµ �@µ(@⌫A⌫) = 0 (before exploiting the gauge invariance). The corresponding Lagrangian is then:

L� = �1

4Fµ⌫F

µ⌫ (3.8)

where Fµ⌫ is the usual electromagnetic tensor. Let us check this by applying the Euler-Lagrangeequation. Rewriting first the Lagrangian:

L� = �1

4(@µA⌫ � @⌫Aµ)Fµ⌫ = �1

4(@µA⌫F

µ⌫ + @⌫AµF ⌫µ) = �1

2@µA⌫F

µ⌫

= �1

2@µA⌫(@

µA⌫ � @⌫Aµ)

= �1

2gµ↵g⌫�@µA⌫(@↵A� � @�A↵)

where in the first line we used successively Fµ⌫ = �F ⌫µ and µ$ ⌫. Thus:

@L�@(@µA⌫)

= �1

2gµ↵g⌫�(@↵A� � @�A↵)� 1

2gµ↵g⌫�@µA⌫

⇢@(@↵A�)

@(@µA⌫)� @(@�A↵)

@(@µA⌫)

�

= �1

2(@µA⌫ � @⌫Aµ)� 1

2@↵A�

n�µ↵�

⌫� � �

µ��

⌫↵

o

= �1

2(@µA⌫ � @⌫Aµ)� 1

2{@µA⌫ � @⌫Aµ}

= �(@µA⌫ � @⌫Aµ)

the � denoting the Kronecker symbol (�µ↵ = 1 if µ = ↵ and 0 otherwise.).The Euler-Lagrange

equation is then:

@L�@A⌫

� @µ@L�

@(@µA⌫)= 0) 0 + @µ@

µA⌫ � @µ(@⌫Aµ) = ⇤A⌫ � @µ(@⌫Aµ) = 0

which is the wave equation of a free photon.

3.1.4 Gauge invariance consequences

In 1917, Emmy Noether published a theorem that states that every continuous symmetry yieldsa conservation law, and conversely every conservation law is a sign of an underlying symmetry.One of the famous consequences of her theorem is the fact that the momentum is conserved fora system which is invariant under translations in space. Similarly, for the conservation of theangular momentum and the invariance under rotations.


Gauge invariance consequences 67

Global phase invariance: Now, consider the Dirac’s Lagrangian 3.7 and the transformationof the field:

! 0 = e�iq� (3.9)

where q and � are simple real numbers. The Lagrangian becomes:

LD ! L0 = e+iq� (i/@ �m)e�iq� = LD

Thus, the Lagrangian is invariant under this transformation. Such transformation has all theproperties of a symmetry namely 1) the product of 2 transformations is also a transformation, 2)there exists an identity transformation, 3) there exists an inverse transformation and 4) there isassociativity: if R

1,2,3 are 3 transformations, R1

(R2

R3

) = (R1

R2

)R3

. The set of transformations3.9 forms a group called U(1) and the properties just mentioned are actually the ones of a group.Moreover, we see that the order of 2 transformations doesn’t matter, or in other words, the U(1)group elements commute. Such group is called Abelian. According to Noether’s theorem, theremust be a conservation law. Consider an U(1) infinitesimal transformation of parameter ��:

� = �iq��

and imposing the Lagrangian to remain unchanged �L�� = 0

�L = @L@ � + @L

@(@µ )

�(@µ ) + � @L@ ¯

+ �(@µ ) @L@(@µ ¯ )

= �@L@ iq�� @L

@(@µ )

iq��@µ + iq�� @L@ ¯

+ iq��@µ @L

@(@µ ¯ )

@L�� = �iq

nh@L@ � @µ

⇣@L

@(@µ )

⌘ + @µ

⇣@L

@(@µ )

⌘i�h @L@ ¯ � @µ

⇣@L

@(@µ ¯ )

⌘+ @µ

⇣ @L@(@µ ¯ )

⌘io

The first two terms in the two brakets cancel because of the Euler-Lagrange equation appliedto and , so that:

@L�� = @µ

n�iq

⇣@L

@(@µ )

� @L@(@µ ¯ )

⌘o= 0

and hence the current jµ = �iq⇣

@L@(@µ )

� @L@(@µ ¯ )

⌘is conserved. Using the Dirac’s Lagrangian

3.7, @L@(@µ )

= i �µ and @L@(@µ ¯ )

= 0, thus we finally have

jµ = q �µ (3.10)

which is the charge-current we introduced in equation 2.28! The continuity equation @µjµ thenmeans that the charge Q =

Rd3x j0 must be conserved.

Local phase invariance: Now, let us assume that the phase � in 3.9 is a real functiondepending on the spacetime coordinates:

! 0 = e�iq�(x) (3.11)

How is the Dirac’s Lagrangian transformed?

LD ! L0 = e+iq�(x) (i/@ �m)e�iq�(x) = e+iq�(x) i�µ

⇥(�iq@µ�(x))e�iq�(x) + e�iq�(x)@µ

⇤�m

= LD + q �µ @µ�(x)= LD + jµ@µ�(x)



If we wish to have a Lagrangian invariant under this local transformation, we need to add anextra term Lint to the Dirac Lagrangian:

Lint = �jµAµ = �q �µ Aµ (3.12)

which will compensate the term jµ@µ�(x). Lint couples the vector Aµ to the charge-current ofthe Dirac’s field jµ. Let us see how the modified Lagrangian now transforms:

LD + Lint ! L0 = LD + jµ@µ�(x)� j0µA0µ

But j0µ = q 0�µ 0 = q �µ = jµ, thus:

LD + Lint ! L0 = LD + jµ�@µ�(x)�A0

µ

�If Aµ transforms as:

Aµ ! A0µ = Aµ + @µ�(x) (3.13)

the new Lagrangian LD + Lint would be invariant. We have already encountered the abovetransformation: it is the same as the one allowed by the gauge invariance of the electromagneticfield 2.73!

3.1.5 QED Lagrangian

Let us recap: promoting the free Dirac Lagrangian to be invariant under a local phase trans-formation U(1), requires an additional term that couples a vector field Aµ, consistent with thephoton field (having the same transformations), to the electron charge-current. Furthermore, ifwe use the correspondence principle 2.46 as in the case of the Dirac’s equation in presence of anelectromagnetic field:

@µ ! Dµ = @µ + iqAµ (3.14)

the Dirac Lagrangian becomes:

LD ! (i�µDµ �m) ! LD � q �µ Aµ

! LD + Lint

(3.15)

where Lint is the same as in 3.12! Dµ is called the covariant derivative. Conclusion: the localgauge invariance requirement, ie the local phase invariance, included in the free electrons theory,has generated a new field, the photon, that couples to the electron, through the Lagrangian Lint.This is an example of the “gauge principle” which elevates a global symmetry (U(1) here), intoa local one. The “gauge principle” requires to replace the normal derivative by the covariantderivative.

We can now specify the full QED Lagrangian:

L = LD + L� + Lint = (i/@ �m) � 1

4Fµ⌫F

µ⌫ � q �µ Aµ (3.16)

It is the sum of the free electron Lagrangian (ie the Dirac Lagrangian), the free photon La-grangian, and the interaction Lagrangian between electrons2 and photons. The recipe to obtainthe QED Lagrangian is to start from the free Lagrangians and replace the derivative by the co-variant derivative. Doing so, we obtain a QED Lagrangian that is locally U(1) gauge invariant3:the field transformation (3.11) being absorbed by the Aµ transformation (3.13).

2For electrons q = �e has to be used.3The term Fµ⌫ being obviously invariant under transformation 3.13.


Perturbation with interacting fields: the S-matrix 69

3.2 Perturbation with interacting fields: the S-matrix

We saw in the previous section that electron field and photon field interact. We then expectprocesses where for example, the electron will be scattered by a photon. Hence, we need to beable to calculate measurable quantities as the cross-section using time-dependent perturbationtheory as exposed in section 1.6. This section aims at making a connection between the fieldsand the matrix element appearing in formulas such as 1.62.

3.2.1 Schrodinger, Heisenberg and Interaction representations

In the well known Schrodinger reprensentation, the time evolution of a system is given by:

i@

@t s(t) = H s(t) (3.17)

where we use the subscript s to clearly remind that we work in the Schrodinger representation.If the hamiltonian H doesn’t depend on time, s can be computed for any time t provided thestate is known at another time let us say t = 0 via4:

s(t) = US(t, 0) s(0) = e�itH s(0) (3.18)

where US is unitary (U †S = U�1

S ) since H is hermitian. Thus, the state vectors carry the entiretime dependance. The field can be expanded in terms of creation and annihilation operators att = 0. However, the problem with this representation, is this particular time: it is not manifestlyLorentz invariant.

Now, note that the matrix element of an operator A between state s and 0s can be estimated

at the time t with:

h 0s|A| si =

Z 0

s⇤(t) A s(t) =

Z 0

s⇤(0) US(t, 0)†A US(t, 0) s(0)

Now, this expression can be seen di↵erently, with an operator AH depending on time:

AH(t) = US(t, 0)†A US(t, 0)

and states H = s(0), 0H = 0

s(0) without time dependance, such as:

h 0s|A| si = h 0

H |AH | Hi

This new representation is the Heisenberg one (thus the subscript H) where operators dependon time and states do not. The time-dependency is entirely transferred to operators via theHeisenberg equation:

id

dtAH(t) = [AH(t), H]

With this representation, the evolution of the creation and annihilation operators appearingin the field expansion, is given by the previous equality, which requires the usage of the fullhamiltonian. When this hamiltonian is the one of free particles, we know how to do it. For morecomplicated cases, the full hamiltonian is not necessarily known. Note that the Hamiltonian H

4The exponential means e�itH =P1

k=0

(�itH)k/k!



is identical in the 2 representations since it commutes with itself. Also, note that H can bewritten as:

H = US(t, 0)† s(t) = eitH s(t)

Finally, there is the Interaction representation which is a hybrid of the two previous repre-sentations. It is well adapted for the perturbative calculations. In this representation, both thestates and the operators have a time dependency. Consider an hamiltonian:

H = H0

+ Hint

where H0

describes a free system and Hint the perturbation source of interactions. In theInteraction picture, the state I is defined as:

I(t) = eitH0 s(t) (3.19)

Notice the H0

instead of H as in the Heisenberg representation. Since (t) obeys to 3.17, it iseasy to conclude that I(t) obeys to:

i @@t s(t) = H s(t) ) i @@t(e�itH

0 I(t)) = (H0

+ Hint)e�itH0 I(t)

) i ddt I(t) = eitH

0 Hint e�itH0 I(t) = HI int(t) I(t)

(3.20)

with:HI int(t) = eitH

0 Hint e�itH0

Similarly as for the Hamiltonian, any operator in the Interaction picture can be expressed fromthe corresponding operator in the Schrodinger picture as:

AI(t) = eitH0A e�itH

0 (3.21)

which means that its time evolution satisfies the Heisenberg equation:

id

dtAI(t) = [AI(t), H0

] (3.22)

with H replaced by H0

. The field operators satisfy to this equation which is valid even in thepresence of an interaction. In other words, the field can still be expressed as a linear combinationof the free field solution.

3.2.2 Dyson expansion

We wish to find the evolution operator U which allows to calculate the state at any time tknowing the state at a given time t

0

:

I(t) = U(t, t0

) I(t0) (3.23)

Thus:d

dt I(t) =

dU(t, t0

)

dt I(t0)

but I also satisfies:

id

dt I(t) = HI int(t) I(t) (3.24)


Dyson expansion 71

so that:

idU(t, t

0

)

dt I(t0) = HI int(t) I(t) = HI int(t) U(t, t

0

) I(t0)

and hence:

id

dtU(t, t

0

) = HI int(t) U(t, t0

) (3.25)

From now on, we are going to use reduced notation by forgetting the subscript I referring tothe Interaction representation. Equation 3.25 can be solved by an iterative procedure:

U(t, t0

) = U(t0

, t0

)� i

Z t

t0

dt1

Hint(t1) U(t1

, t0

)

But U(t0

, t0

) is by definition 1 (see 3.23), so that:

U(t, t0

) = 1� iR tt0

dt1

Hint(t1) U(t1

, t0

)

= 1� iR tt0

dt1

Hint(t1)h1� i

R t1

t0

dt2

Hint(t2) U(t2

, t0

)i

= 1� iR tt0

dt1

Hint(t1) + (�i)2R tt0

dt1

R t1

t0

dt2

Hint(t1) Hint(t2) U(t2

, t0

)

Pursuing the same approach at higher order, we finally get:

U(t, t0

) =1X

n=0

(�i)n

Z t

t0

dt1

Z t1

t0

dt2

· · ·Z tn�1

t0

dtn Hint(t1) Hint(t2) · · · Hint(tn)

Now, we can re-arrange the integrals to have the same range of integration [t0

, t]. Consider theintegrals of the second order term:Z t

t0

dt1

Z t1

t0

dt2

Hint(t1) Hint(t2) =

Z t

t0

dt1

Z t

t0

dt2

✓(t1

� t2

) Hint(t1) Hint(t2)

where ✓ is the usual step function, so the implicit condition t1

> t2

is imposed. Now, splittingin 2 and changing the order of the integrals in the second term, it comes:R t

t0

dt1

R t1

t0

dt2

Hint(t1) Hint(t2)

= 1

2

R tt0

dt1

R tt0

dt2

✓(t1

� t2

) Hint(t1) Hint(t2) + 1

2

R tt0

dt2

R tt0

dt1

✓(t1

� t2

) Hint(t1) Hint(t2)

= 1

2

R tt0

dt1

R tt0

dt2

✓(t1

� t2

) Hint(t1) Hint(t2) + 1

2

R tt0

dt1

R tt0

dt2

✓(t2

� t1

) Hint(t2) Hint(t1)

= 1

2

R tt0

dt1

R tt0

dt2

✓(t1

� t2

) Hint(t1) Hint(t2) + ✓(t2

� t1

) Hint(t2) Hint(t1)(3.26)

where in the last line, the integration labels were interchanged. The trick now is to introducethe T -product (or time-ordered product) of operators defined as:

T (Hint(t1) Hint(t2)) = Hint(t1) Hint(t2) if t1

> t2

Hint(t2) Hint(t1) if t2

> t1

= Hint(t1) Hint(t2)✓(t1 � t2

) + Hint(t2) Hint(t1)✓(t2 � t1

)(3.27)

which for the product of n operators A reads:

T (A(t1

) A(t2

) · · · A(tn)) = A(ti1

) A(ti2

) · · · A(tin) with ti1

> ti2

> · · · > tin (3.28)

Hence, 3.26 becomes:Z t

t0

dt1

Z t1

t0

dt2

Hint(t1) Hint(t2) =1

2

Z t

t0

dt1

Z t

t0

dt2

T (Hint(t1) Hint(t2))



For the higher order terms, one has to take into account the number of permutations in theT -product, and the final formula is:

U(t, t0

) =1X

n=0

(�i)n

n!

Z t

t0

dt1

Z t

t0

dt2

· · ·Z t

t0

T (Hint(t1) Hint(t2) · · · Hint(tn)) (3.29)

which can be simply written in the symbolic form:

U(t, t0

) = T e�i

R tt0

d⌧Hint(⌧) (3.30)

where again, Hint is understood here in the interaction representation. Now let us consider thefollowing scattering process: long before the interaction, say at t! �1, Hint is negligible andaccording to 3.24, is a steady state in the Interaction representation. That’s one of the maininterest of this representation where the kets evolve with time only when there is an interaction.Let us denote this initial state |ii = | (�1)i. Experimentally, |ii is typically the 2 particlesthat are supposed to collide in an accelerator. As time goes by, the particles described by |iimay scatter and Hint is not negligible anymore. The time evolution of is then given by:

(t) = U(t,�1) (�1)

Long after the interaction, for the same reason, the system potentially constituted of manyparticles resulting of the collision, is in another steady state | (+1)i. The amplitude of theprobability for finding a given state |fi is then:

Sfi = hf | (+1)i = hf |U(+1,�1) (�1)i = hf |U(+1,�1)|ii

The matrix:S = U(+1,�1) = T e�i

R+1�1 d⌧Hint(⌧) (3.31)

is called the scattering matrix (S-matrix). Its element Sfi gives the amplitude probability forhaving a transition i! f . According to 3.29 the S-matrix expression is thus:

S =1X

n=0

S[n] (3.32)

where the nth order term is:

S[n] =(�i)n

n!

Z· · ·

Zdt

1

· · · dtn T (Hint(t1) · · · Hint(tn)) (3.33)

3.2.3 Connection with the general formula of transition rate

We consider an initial state |ii made of several particles with definite momenta and similarlyfor the final state |fi. |ii and |fi are the states corresponding to the asymptotic conditions att = �1 and t = +1. The probability of a transition from |ii to |fi is then given by:

Pi!f =| hf |S|ii |2hi|ii hf |fi (3.34)

If there is no reaction at all, we must have S = l1 (the identity operator), and i = f and so we dohave P = 1. Therefore, we expect S to be written as S = l1+ · · · where the dots are an expression


Connection with the general formula of transition rate 73

of the operators for a non trivial interaction. We do not need to have an explicit formulation ofthese operators (which could be horribly complicated), only the matrix element hf |S|ii for theinitial and final states envisaged is required. How can we write it? It is convenient to factor outa delta function implementing the 4-momentum conservation5 to define the Lorentz invariantscattering amplitude, Mfi:

hf |S|ii = hf |ii+ (2⇡)4�(4)(Pf � Pi) iMfi

where Pf and Pi are respectively the final and initial total momenta. The factor (2⇡)4 isconveniently factorized so that iMfi will be the result of a Feynman diagram calculation. Sincewe are interested in non trivial interactions, we have:

hf |S|ii = (2⇡)4�(4)(Pf � Pi) iMfi , i 6= f (3.35)

Now let us make explicit the terms of (3.34). Using the commutation rules for bosons 2.14 (butwe would find the same result with fermions), we notice that:

hp0|pi = h0|ap0a†p|0i = h0|a†

pap0 |0i+ (2⇡)32Ep �(3)(~p0 � ~p) = (2⇡)32Ep �

(3)(~p0 � ~p)

where a† is the creation operator of the corresponding particles (boson or fermion). This resultis known as the relativistic normalization of momenta states. However, that means that if weconsider for simplicity an initial state with just one particle with momentum pi, we have:

hi|ii = (2⇡)32Epi�(3)(0)

which is an infinite quantity! If instead of plane waves (with definite momenta) we had consideredlocalized wave packets for the asymptotic initial and final states, the result would have beenfinite. The problem appears here because we implicitly use plane waves in an infinite volume(the whole 3-D space). Indeed, according to formula 1.69, we have:

�(3)(p) =1

(2⇡)3

Zeip.xd3x = lim

L!1�L(p)

with�L(p) = 1

(2⇡)

3

R L/2

�L/2

dx eipxxR L/2

�L/2

dy eipyyR L/2

�L/2

dz eipzz

= 1

(2⇡)

3

L sin(pxL/2)

pxL/2

L sin(pyL/2)

pyL/2

L sin(pzL/2)

pzL/2

= V(2⇡)

3

sin(pxL/2)

pxL/2

sin(pyL/2)

pyL/2

sin(pzL/2)

pzL/2

where V = L3 is the volume of 3-space (a box with sides of length L). We conclude that�L(0) ⌘ V/(2⇡)3 (since limx!0

sin xx = 1) is equivalent to �(3)(0) in the limit V !1 and hence

hi|ii is equivalent to 2EpiV. We can then do all our calculations with a finite volume V, and forphysics calculations, the volume should cancel out so that the infinite volume limit can be takensafely. It is straightforward to generalize to n particles in the initial state:

hi|ii =nY

k=1

2EkV = VnnY

k=1

2Ek

5All interactions are assumed to conserve the 4-momenta.



and similarly for n0 particles in the final states:

hf |fi = Vn0n0Y

k=1

2E0k

The last term in 3.34 to be determined is | hf |S|ii |2. According to 3.35, it is:

| hf |S|ii |2 = (2⇡)4�(4)(PF � PI)(2⇡)4�(4)(PF � PI)|Mfi|2

Here, we have to evaluate the square of the delta-function �(4)(PF � PI). Being applied twice,the first delta-function imposes PF = PI into the second and hence �(4)(PF � PI)�(4)(PF �PI) = �(4)(PF � PI)�(4)(0). The question is thus to estimate �(4)(0) = �(3)(0)�(0) = V

(2⇡)

3

�(0).

Proceeding as for our estimation of �(3)(0), we have:

�(E) =1

(2⇡)

ZeiE.tdt =

1

(2⇡)lim

T!1

Z T/2

�T/2

dt eiEt =1

(2⇡)lim

T!1T

sin(ET/2)

ET/2

and hence we conclude �(0) ⌘ T(2⇡)

leading to:

| hf |S|ii |2 = (2⇡)4�(4)(PF � PI)VT |Mfi|2

Putting all the pieces together, we finally establish:

Pi!f = V1�n�n0T (2⇡)4�(4)(PF � PI)|Mfi|2

nYk=1

1

2Ek

n0Yk=1

1

2E0k

or more appropriately in terms of transition rate (i.e. probability per unit of time):

�i!f = V1�n�n0(2⇡)4�(4)(PF � PI)|Mfi|2

nYk=1

1

2Ek

n0Yk=1

1

2E0k

However, we forgot something: the final states have a continuous energy spectrum. The particlesin the final state have not an infinitely accurate momenta/energies but should rather be describedby a set of final states belonging to a given phase space element (i.e. belonging to px ±�px, py ±�py, pz ± �pz). Let us consider an infinitesimal phase space element so that we can assume that|Mfi| remains constant for these dN states. The infinitesimal transition probability per unittime is then the sum of the probabilities of each individual states:

d�i!f = V1�n�n0(2⇡)4�(4)(PF � PI)|Mfi|2

nY

k=1

1

2Ek

n0Yk=1

1

2E0k

!dN

As well known from quantum mechanics, in case of a single particle in a box of sides L, thecomponents of its momentum are quantized as: px,y,z = (2⇡/L) nx,y,z, and therefore the numberof states in d3~p = dpxdpydpz is dN = dnxdnydnz = Vd3~p/(2⇡)3. Therefore for n0 particles in

the final state, dN becomes dN =Qn0

k=1

Vd3 ~p0k/(2⇡)3 and thus:

d�i!f = V1�n (2⇡)4�(4)(PF � PI)|Mfi|2nY

k=1

1

2Ek

n0Yk=1

d3 ~p0k

(2⇡)32E0k

(3.36)


Feynman rules and diagrams 75

which is exactly the formula 1.62 we used in the first chapter to establish the formulas of thedecay rate and cross-section where, as it should, the volume V cancels out.

We have seen in equation (3.35) that the matrix element Mfi is related to hf |S|ii definedat a order [n] by (3.33). Hence:

(2⇡)4�(4)(Pf � Pi) iM[n]

fi =(�i)n

n!

Zdt

1

· · · dtn T (hf |Hint(t1) · · · Hint(tn)|ii)

In order to make explicit the Lorentz invariance, it is better to use the hamiltonian density H(H =

Rd3xH(x)). The final formula for the matrix elements then becomes:

(2⇡)4�(4)(Pf � Pi) iM[n]

fi =(�i)n

n!

Zdx

1

· · · dxn T (hf |Hint(x1

) · · · Hint(xn)|ii) (3.37)

where, once again, Hint is the interaction hamiltonian given in the Interaction representation(meaning that the product of fields constituting the hamiltonian are themselves given in theinteraction representation).

3.3 Feynman rules and diagrams

Let us consider the Lagrangian of QED 3.16. We wish to find the expression of Hint knowingthat according to 3.4 and 3.3:

H =@L

@�(xµ)�(xµ)� L (3.38)

Fields derivative in 3.16 only enters in the two free Lagrangians LD and L� . Hence, Hint reducesto:

Hint = �Lint = q �µ Aµ (3.39)

3.3.1 Electron-photon vertex

Consider the first order where q has to be understood as the charge of the particle, not the oneof the antiparticle6:

S[1] = �i

Zd4x Hint(x) = �iq

Zd4x �µ Aµ

Now, inject the 3 fields development for (2.67), (2.68), and Aµ (2.76):

S[1] = �iqR

d4xR d3~p

(2⇡)

3

2Ep

d3~p0

(2⇡)

3

2Ep0d3~q

(2⇡)

3

2Eq

Pr=1,2

Pr0

=1,2

P3

�=0hc†~prurpe+ipx + d~prvrpe�ipx

i�µ

hc~p0r0ur0p0e�ip0x + d†

~p0r0vr0p0e+ip0xi

h✏µq�↵~q�e

�iq.x + ✏⇤µq�↵†~q�e

+iq.xi

6For electron-positron, q = �|e|.



Integrating over x (using 1.69) and rearranging, we finally obtain:

S[1] = (2⇡)4R d3~p

(2⇡)

3

2Ep

d3~p0

(2⇡)

3

2Ep0d3~q

(2⇡)

3

2Eq

Pr=1,2

Pr0

=1,2

P3

�=0

{c†~prc~p0r0↵~q� (urp (�iq�µ) ur0p0)✏µq� �(4)(p� p0 � q)+

c†~prd

†~p0r0↵~q� (urp (�iq�µ) vr0p0)✏µq� �(4)(p + p0 � q)+

d~prc~p0r0↵~q� (vrp (�iq�µ) ur0p0)✏µq� �(4)(p + p0 + q)+

d~prd†~p0r0↵~q� (vrp (�iq�µ) vr0p0)✏µq� �(4)(p0 � p� q)+

c†~prc~p0r0↵†

~q� (urp (�iq�µ) ur0p0)✏⇤µq� �(4)(p� p0 + q)+

c†~prd

†~p0r0↵

†~q� (urp (�iq�µ) vr0p0)✏⇤µq� �

(4)(p + p0 + q)+

d~prc~p0r0↵†~q� (vrp (�iq�µ) ur0p0)✏⇤µq� �

(4)(q � p� p0)+

d~prd†~p0r0↵

†~q� (vrp (�iq�µ) vr0p0)✏⇤µq� �

(4)(p0 � p + q) }

(3.40)

Thus, S[1] is constituted of 8 terms7 having a similar form: product of 3 operators, a charge-current (product of an adjoint-spinor and a spinor sandwiching a �µ matrix), a photon polarisa-tion vector and a delta function. Looking at the operators, one notices that the total charge isalways zero: for example the first line creates an electron, annihilates an electron and annihilatesa photon, while the second creates an electron and a positron and annihilates a photon.

Let us detailed the first line as an example. Clearly, it gives a contribution only if it is sand-wiched between an initial state containing a photon and an electron |ei, �i (that are annihilated)and a final state containing an electron |ef i (that is created):

hef | c†~prc~p0r0↵~q� |ei, �i = h0| c~pf rf c

†~prc~p0r0↵~q�c

†~piri

↵†~q��

|0i= h0| c~pf rf c

†~prc~p0r0c†

~piri↵~q�↵

†~q��

|0i= h0|

⇣�c†

~prc~pf rf + (2⇡)32Ef�(3)(~pf � ~p)�rf r

⌘⇣�c†

~piric~p0r0 + (2⇡)32Ei�(3)(~p0

i � ~p0)�rir0

⌘⇣↵†~q��

↵~q� � g�� (2⇡)32E��(3)(~q � ~q�)⌘

|0i= �(2⇡)32Ef�(3)(~pf � ~p)�rf r(2⇡)32Ei�(3)(~p0

i � ~p0)�rir0

g�� (2⇡)32E��(3)(~q � ~q�) h0|0i

Inserting this result into the S[1] expression and performing the integrations and summations,we get a term8

(2⇡)4�(4)(pf � pi � q�)�urfpf (�iq�µ) uripi

�✏µq�� (3.41)

Comparing with 3.37 we conclude in that case:

iM =�urfpf (�iq�µ) uripi

�✏µq�� (3.42)

The creation of the final electron is thus associated to an adjoint spinor u while the annihilationof the initial electron is associated a spinor u. Similarly, the annihilation of the photon isassociated to a vector polarization. This yields our first Feynman diagram shown on figure 3.1.The arrow of time on the horizontal axis is going from the left to the right while the spacialdimension is vertical. The photon is symbolized by a wavy line, and the electron (actually anyfermion) by a solid line. The intersection of the lines is called a vertex. The direction of the

73 fields with 2 operators give 23 = 8 combinations.8Real photons have only transverse polarisation, ie �� = 1 or 2. Thus g�� = �1.


Electron-photon vertex 77

Figure 3.1: A basic vertex in QED.

arrow on the electron line has a meaning: for electrons in the initial state, it goes toward thevertex while for electrons in the final state it points away from the vertex. In other words, thearrow of a fermion is always oriented from the past to the future.

The development we made to the first order of perturbation leads to diagrams with only 1vertex. The number of vertices represents the order of the perturbation development. The 7other terms of S[1] development yield similar kind of diagrams. They are all shown on the figure3.2. Positrons (or any antifermions) are represented by the same line as electrons. What di↵ers

Figure 3.2: The vertices of QED at first order of perturbation.

is only the direction of the arrow. For example, the second diagram on the first row, depicts aphoton that annihilates in 1 electron (line going up) and 1 positron. Note that for the outgoingpositron, the arrow “goes back in time”. Similarly, in the fourth diagram of the first row, theincoming particle is a positron, the arrow going back in time. On the photon, there is no arrowsince it is its own antiparticle. According to 3.40, following the same procedure that yields 3.41,we see that antifermions in the initial state are associated to adjoint antispinor v and antispinorv in the final state. For photons, a polarisation vector ✏µ is associated to an incoming photonsand ✏⇤µ for an outgoing one.

Looking back to formula 3.40 and 3.42 we see that all terms contain a �iq�µ contributionsandwiched by 2 (anti-)spinors. This factor is logically associated to the vertex itself in thediagram. The charge q = �|e| is the coupling between electrons or positrons and photons.Development at nth order would produce n vertices, leading to a factor |e|n in the amplitude.The convergence of the perturbation series depends on |e|, which finally represents the “strength”of the perturbation. In natural units |e| '

p4⇡/137 ' 0.3.

Finally, There is a last term that we haven’t exploited yet in 3.41: the delta function. Itclearly imposes the energy-momentum conservation at the vertex. Well, it appears that none



of these 8 terms are physically possible with free particles! An obvious one: the third onthe first row of figure 3.2. The energy conservation imposed by �(4)(p + p0 + q) means thatEe� + E� + Ee+ = 0 ) Ee� = E� = Ee+ = 0 which is impossible (there is at least the restenergy since electrons are massive). Even, more conventional diagrams as the first one, areimpossible. Both energy conservation and momentum conservation can’t be satisfied at thesame time (because the photon mass is 0). The only possibility would be to envisage that oneof these particles does not have its normal mass value (or more appropriately that it does notrespect the relativistic relation E2 = p2 + m2). Such a particle is called a virtual particle and isallowed by the uncertainty inherent in quantum mechanics:

�E � ~/T

The relation can be interpreted as follow: the greater the energy (or mass) is shifted from itsnominal value, the shorter the particle will live. In other words, virtual particles (thus notsatisfying E2 = m2 +p2 with m the usual mass), must live a short time: they have to be quicklyreabsorbed (annihilated) by another process9. Conclusion: another vertex is needed and onehas to go to the second order of the perturbation development.

3.3.2 Photon and electron propagators

Consider the second order of perturbation:

S[2] =(�i)2

2q2

Zd4x d4y T ( (x)�µ (x)Aµ(x) (y)�⌫ (y)A⌫(y)) (3.43)

If we develop all quantities as in the previous section we have 82 = 64 terms, and thus 64diagrams. Two such diagrams are shown in figure 3.3. In the first diagram, we see that a

Figure 3.3: Example of QED diagrams at the second order.

(virtual) photon is created at the spacetime point x (assuming tx < ty) and annihilated aty. Similarly, the diagram on the right-hand side shows a virtual electron being exchanged.We can de-couple the initial and final states and consider that a virtual particle must be ableto be created from the vacuum and annihilated. Looking at the formula 3.43, we see that,for the photon, the product Aµ(x)A⌫(y) has to be involved. Indeed, developed in terms of

creation-annihilation operators, the combination ↵~q�(x)↵†~q0�0(y) appears, which will give a non-

zero contribution when sandwiched with the vacuum |0i. Similarly, for the fermions, the only

product giving non-zero term involved c~pr(x)c†~p0r0(y) or d~pr(x)d†

~p0r0(y) which originates only from

9In fact, all particles can be considered as virtual: for instance a photon emitted by a star is detected on earthby a photomultiplier which annihilates the photon. Its “degree of virtuality” is however ridiculously tiny to beable to live during million years!


Summary of QED Feynman rules 79

(x) (y). Conclusion: the two quantities in 3.43 that can describe the propagation of a virtualparticle from a point x to a point y (ie its creation followed by its annihilation) are:

h0| T (Aµ(x)A⌫(y)) |0i , h0| T ( (x) (y)) |0i (3.44)

They are called propagators (respectively of the photon and a fermion). The calculation (notso complicated) can be found in any quantum field theory book (see for instance [8, p. 62]).We simply give here the results in the momentum space10 (i.e. after having integrated on allpossible position for x and y):

Photon propagator:�igµ⌫

p2 + i✏(3.45)

where p is the 4-momentum of the (virtual) photon and ✏ is an infinitesimal real positive numberto avoid singularities when p ! 0. (In practice, ✏ is dropped). This formula corresponds to aparticular choice of gauge called Feynman gauge. For the electron, we have:

spin 1/2 propagator: i/p + m

p2 �m2 + i✏(3.46)

where again, p is the 4-momentum of the (virtual) fermion and m its mass (ie the nominal massused in E2 = |~p|2 + m2). Remember that /p = �µpµ is a matrix. Hence, diagrams, as the first offigure 3.43 correspond mathematically to an amplitude:

iM =�vre+pe+

(�iq�µ) ure�pe�

� �igµ⌫

p2 + i✏

⇣ur0

e�p0e�

(�iq�⌫) vr0e+

p0e+

⌘(3.47)

where the first charge current jµ = vre+pe+(q�µ) ure�pe� corresponds to the first perturbation

term between the electron, positron and the virtual photon, �igµ⌫/(p2 + i✏) corresponds to thepropagation of the virtual photon, and the second charge-current j⌫ = ur0

e�p0e�

(q�⌫) vr0e+

p0e+

to the second perturbation term between the electron, positron and the virtual photon, whichannihilates the virtual photon.

Before concluding this section, let us examine the left diagram of figure 3.4. At first sight, itseems hard to interpret in term of a particle being created and then absorbed since both verticesare at the same time (time goes from left to right here). But, consider the 2 other diagrams in3.4. When t

1

< t2

, in the centre diagram, a virtual positron (= an electron moving backward intime on the diagram) is first created and then absorbed whereas when t

2

> t1

, in the right handside diagram, a virtual electron is first created and then absorbed. Since, in the definition of thepropagators 3.44, there is a T-product of field operators, the Feynman diagram on the left-handside, includes the 2 possibilities depicted by the centre and right pictures! Both time-orderingsare always included in each propagator line! It makes perfectly sense, since there is no way todistinguish experimentally the two modalities.

3.3.3 Summary of QED Feynman rules

In this section, we wish to summarise the recipe to calculate the amplitude of a process.

1. Draw, at a given order, all diagrams describing the transition between an initial state (inpractice, 1 or 2 particles) and a final state (can contain many particles). A QED vertexmust always connect 2 (anti-)fermions and 1 photon.For each vertex:

10Generally, the positions of particles in a reaction are not known but their momenta are.



1

e!"

e!" e!" e!"

e!"e!" γ"

γ"γ"

γ"γ"

γ"

1"

2"

e+" e!"

Figure 3.4: Left: e�� ! e�� Feynman diagram. Centre and right: the corresponding 2 time-ordereddiagrams.

• Energy-momentum is conserved.

• Charge of the incoming particle(s) is always equal to the charge of the outgoingparticle(s). QED can be seen as a flow of charges (that’s why the charge-current isinvolved).

2. Calculate iM of each diagram by using the factors given in table 3.1. Incoming and out-going particles are connected to vertices exchanging virtual particles. The correspondingfactors in 3.1 are respectively the ones of the “External lines”, “Vertex” and “Propagators”.

3. Combine the amplitudes of all diagrams to get the total amplitude of the process.

• If 2 diagrams di↵er only by the exchange of 2 external identical fermions (2 outgoing,or 2 incoming, or 1 incoming fermion and 1 outgoing anti-fermion, or 1 incominganti-fermion and 1 outgoing fermion), subtract the amplitudes11.

• add all other amplitudes.

For completeness, we also give the factors for massive spin 0 boson interacting with photons.The procedure is similar to the fermion case: from the wave equation, the free Lagrangianis deduced, allowing the quantization of fields. The propagator can then be calculated. Theinteraction term is then obtained, by replacing in the Lagrangian, the derivatives by covariantderivatives 3.14. The vertex factors and external lines factors are determined accordingly.

3.4 QED and helicity/chirality

We have seen that QED interactions involve the charge current. So terms as qu�µu, qv�µv,qu�µv, qv�µu are expected. Now, we saw in section 2.3.4.3, that any spinor can be decomposedin left-handed and right-handed components of chirality through the projectors 2.61. Hence, thecurrent can be decomposed:

qu�µu = q(uL + uR)�µ(uL + uR)= quL�µuL + quR�µuR + quL�µuR + quR�µuL

11This can be understood as a consequence of the fact that when 2 identical fermions are interchanged, thewave function gets a minus sign. It is clearly related to the spin-statistic theorem.


QED and helicity/chirality 81

External lines

Spin 0 Boson incoming 1Boson outgoing 1anti Boson incoming 1anti Boson outgoing 1

Spin 1

2

Fermion incoming u spinorFermion outgoing u spinoranti Fermion incoming v spinoranti Fermion outgoing v spinor

Spin 1 Photon incoming ✏µPhoton outgoing ✏⇤µ

Propagators

Spin 0 Boson ip2�m2

+i✏

Spin 1

2

Fermion i /p+m

p2�m2

+i✏

Spin 1 Photon �igµ⌫p2

+i✏

Vertex

Spin 0 �iq(pµ + p0µ)

Spin 1

2

�iq�µ

q = �|e| for e±

Table 3.1: Feynman rules in QED.

But:

uL�µuR = u† 1

2

(1� �5

†)�0�µ 1

2

(1 + �5)u= 1

4

u†(1� �5)�0�µ(1 + �5)u= 1

4

u†�0(1 + �5)(1� �5)�µu= 0



And similarly:uL�µuR = uR�µuL = vL�µvR = vR�µvL = 0uL�µvL = uR�µvR = vL�µuL = vR�µuR = 0

(remember that vL = PRv). Now, at high energy in the ultra-relativistic regime where E � m,the helicity states and chirality states are the same (see section 2.3.4.3).We can then conclude that:

In scattering processes (t-channel): only the combinations:

uL�µuL + uR�

µuR or vL�µvL + vR�

µvR

give non-zero contribution. The helicity of the particle after the scattering is the same as theone before. Helicity is conserved in the ultra-relativistic limit by QED (and chirality is alwaysconserved).

In annihilation/pair creation processes (s-channel): only the combinations:

uL�µvR + uR�

µvL or vL�µuR + vR�

µuL

give non-zero contribution. The helicity of the particle is the opposite of the one of the anti-particle. In our jargon, we’re still talking about helicity conservation in the sense that onlyspecific helicity configurations are possible. Hence, in the centre of mass frame of the pair, thespins of the particle and antiparticle must be aligned, leading to Jz = ±1. The following figuresummarize the situation.

Figure 3.5: The conservation of Helicity in QED.

3.5 Simple examples of graph calculation

3.5.1 Spin and polarisation summations: traces theorems

Consider the simplest QED graph e�µ� ! e�µ� shown on figure 3.6.


Spin and polarisation summations: traces theorems 83

µ�

e�

µ�

e�

Figure 3.6: Lowest order diagram e�µ� ! e�µ�

Let us denote, p and k the 4-momenta of respectively e� and µ� in the initial state. And samenotation with a prime (’) for the final state. The photon has a momentum q = p� p0 = k0 � k.Following Feynman rules 3.1, the amplitude is:

iM = (uk0,r0(i|e|�µ)uk,r)�igµ⌫

q2

(up0,s0(i|e|�⌫)up,s) = i|e|2q2

(uk0,r0�µuk,r)(up0,s0�µup,s)

The cross section formula is proportional to the probability of the process and thus to |M|2.Generally, the initial state is not polarized and the spin of the final state is not measured. Hence,the measured cross-section is an average over the spins of the initial state and a sum over thespins of the final state12. Let us denote:

|M|2 =1

2

Xs

1

2

Xr

Xs0

Xr0

|M|2 =|e|4q4

Lµ⌫(µ�)Lµ⌫(e�) (3.48)

with:Lµ⌫(µ�) = 1

2

Pr,r0

⇥uk0,r0�µuk,r

⇤⇤ ⇥uk0,r0�⌫uk,r

⇤Lµ⌫(e�) = 1

2

Ps,s0

⇥up0,s0�µup,s

⇤⇤ ⇥up0,s0�⌫up,s

⇤ (3.49)

Now, noting that:

(uk0,r0�µuk,r)⇤ = u†

k,r�µ†u†

k0,r0 = u†k,r�

0�µ�0(u†k0,r0�

0)† = uk,r�µuk0,r0

Lµ⌫ can be rewritten:

Lµ⌫(µ�) = 1

2

Pr,r0

⇥uk,r�µuk0,r0

⇤ ⇥uk0,r0�⌫uk,r

⇤= 1

2

Pr uk,r�µ

�Pr0 uk0,r0 uk0,r0

��⌫uk,r

= 1

2

Pr uk,r�µ

�/k0 + m0� �⌫uk,r

where we used the completeness relation 2.66. Now, we write explicitly the matrix element ofLµ⌫ with label ↵,�:

Lµ⌫ = 1

2

P↵,�

Pr u(↵)

k,r

⇥�µ

�/k0 + m0� �⌫⇤

(↵�)

u(�)

k,r

= 1

2

P↵,�

hPr u(�)

k,r u(↵)

k,r

i ⇥�µ

�/k0 + m0� �⌫⇤

(↵�)

= 1

2

P↵,� [/k + m](�↵)

⇥�µ

�/k0 + m0� �⌫⇤

(↵�)

= 1

2

Tr�(/k + m)�µ

�/k0 + m0� �⌫�

12One may wonder why we don’t sum over the amplitudes of the di↵erent spin states and take the modulus-squared of the sum. The key here is that a particle with spin-up or spin-down defines 2 di↵erent spin states thatare in principle distinguishable (even if spin is not measured).



Tr denoting the trace13. We obtained these useful equalities:

Lµ⌫uu = 1

2

Pr,r0

⇥uk0,r0�µuk,r

⇤⇤ ⇥uk0,r0�⌫uk,r

⇤= 1

2

Tr�(/k + m)�µ

�/k0 + m0� �⌫�

Lµ⌫vv = 1

2

Pr,r0

⇥vk0,r0�µvk,r

⇤⇤ ⇥vk0,r0�⌫vk,r

⇤= 1

2

Tr�(/k �m)�µ

�/k0 �m0� �⌫� (3.50)

where the result for anti-spinor was obtained similarly. Following the same approach, one canalso show that:

Lµ⌫vu = 1

2

Pr,r0

⇥vk0,r0�µuk,r

⇤⇤ ⇥vk0,r0�⌫uk,r

⇤= 1

2

Tr�(/k + m)�µ

�/k0 �m0� �⌫�

Lµ⌫uv = 1

2

Pr,r0

⇥uk0,r0�µvk,r

⇤⇤ ⇥uk0,r0�⌫vk,r

⇤= 1

2

Tr�(/k �m)�µ

�/k0 + m0� �⌫� (3.51)

Now recall that:

Tr(↵A + �B) = ↵Tr(A) + �Tr(B) , Tr(AB) = Tr(BA)

so that when m = m0 as in our case:

Lµ⌫uu(m = m0) =

1

2

�k⇢k

0⌘Tr(�⇢�µ�⌘�⌫) + mk⇢Tr(�⇢�µ�⌫) + mk0

⌘Tr(�µ�⌘�⌫) + m2Tr(�µ�⌫)�

(3.52)Therefore, we need to evaluate traces of products of � matrices. Since, the �’s satisfies theCli↵ord algebra 2.17 ({�µ, �⌫} = �µ�⌫ + �⌫�µ = 2gµ⌫ l1), we conclude the following rules:

Tr(�µ�⌫) = gµ⌫Tr(l1) = 4gµ⌫ (3.53)

In passing, we thus have:Tr(/p/k) = 4p.k (3.54)

Now recall �5 properties 2.55 and 2.58 (ie (�5)2 = 1, �5�µ = ��µ�5)

Tr(�µ1 · · · �µn) = Tr(�µ

1 · · · �µn�5�5)= Tr(�5�µ

1 · · · �µn�5)= (�1)nTr(�µ

1 · · · �µn�5�5)= (�1)nTr(�µ

1 · · · �µn)

Hence:Tr(odd nb of �) = 0 (3.55)

Now, for 4 matrices:

Tr(�⇢�µ�⌘�⌫) = Tr(�µ�⌘�⌫�⇢)= �Tr(�µ�⌘�⇢�⌫) + 2g⇢⌫Tr(�µ�⌘)= �Tr(�µ�⌘�⇢�⌫) + 8g⇢⌫gµ⌘

= Tr(�µ�⇢�⌘�⌫)� 8g⌘⇢gµ⌫ + 8g⇢⌫gµ⌘

= �Tr(�⇢�µ�⌘�⌫) + 8g⇢µg⌘⌫ � 8g⌘⇢gµ⌫ + 8g⇢⌫gµ⌘

Thus:Tr(�⇢�µ�⌘�⌫) = 4(g⇢µg⌘⌫ � g⇢⌘gµ⌫ + g⇢⌫gµ⌘) (3.56)

13If A and B are 2 matrices, the product C = AB has elements Cij =P

k AikBkj and thus Tr(C) =P

i Cii =Pi,k AikBki.


Electron-muon scattering: e�µ� ! e�µ� 85

In passing, we thus have:

Tr( /p1

/p2

/p3

/p4

) = 4(p1

.p2

p3

.p4

� p1

.p3

p2

.p4

+ p1

.p4

p2

.p3

) (3.57)

Hence, coming back to 3.50

Lµ⌫uu(m = m0) =

1

2(4k⇢k

0⌘(g

⇢µg⌘⌫ � g⇢⌘gµ⌫ + g⇢⌫gµ⌘) + 4m2gµ⌫)

so that:Lµ⌫

uu(m = m0) = Lµ⌫vv (m = m0) = 2(kµk0⌫ + k⌫k0µ + (m2 � k.k0)gµ⌫) (3.58)

For completeness, we give other useful formulas:

�µ�µ = 4�µ�⌫�µ = �2�⌫

�µ�⌫�⇢�µ = 4g⌫⇢

�µ�⌫�⇢�⌘�µ = �2�⌘�⇢�⌫

(3.59)

Tr(�5�µ�⌫) = 0Tr(�5�µ�⌫�⇢�⌘) = �4i✏µ⌫⇢⌘

(3.60)

with ✏µ⌫⇢⌘ = 1 for µ, ⌫, ⇢, ⌘ an even permutation of 1, 2, 3, 4, �1 for odd permutation, 0 otherwise.One can deduce the following properties of ✏µ⌫⇢⌘:

✏µ⌫⇢⌘ = �✏⌫µ⇢⌘ = ✏⌫⇢µ⌘ = �✏⌫⇢⌘µ✏µ⌫⇢⌘ = �✏⌫⇢⌘µ = ✏⇢⌘µ⌫ = �✏⌘µ⌫⇢✏µ⌫⇢⌘✏µ⌫↵� = �2(�⇢↵�

⌘� � �

⇢��⌘↵)

(3.61)

Another equality which is worth mentioning is that for any 4-vector p, we have /p2 = pµp⌫�µ�⌫ =pµp⌫(2gµ⌫ � �⌫�µ) = 2p2 � /p2 and thus:

/p2 = p2 (3.62)

Polarization summation: for photons, there is a similar summation as for fermions (iePr urur = /k + m). The result is [8, p. 159]:

4X�=1

✏⇤µ�✏⌫ � = �gµ⌫ (3.63)

3.5.2 Electron-muon scattering: e�µ� ! e�µ�

We can now complete the calculation of the amplitude of e�µ� ! e�µ�. Coming back to 3.48and inserting 3.58, we have:

|M|2 =|e|4q4

Lµ⌫(µ�)Lµ⌫(e�)

Lµ⌫(µ�)Lµ⌫(e�) = 2(kµk0⌫ + k⌫k0µ + (m2

µ � k.k0)gµ⌫)2(pµp0⌫ + p⌫p0

µ + (m2

e � pp0)gµ⌫)= 4

�2k.p k0.p0 + 2k.p0 k0.p + 2(m2

e � p.p0)k.k0 + 2(m2

µ � k.k0)p.p0+(m2

e � p.p0)(m2

µ � k.k0)4�

= 8�k.p k0.p0 + k.p0 k0.p�m2

ek.k0 �m2

µp.p0 + 2m2

em2

µ

�(3.64)



Let us consider the ultra-relativistic limit neglecting terms with me and mµ. We can evaluatethe matrix element in the centre of mass frame and using the Mandelstam variables:

s = (k + p)2 = (k0 + p0)2 ' 2k.p ' 2k.0p0

t = (p� p0)2 = (k0 � k)2 = q2

u = (p� k0)2 = (p0 � k) ' �2p.k0 ' �2p0.k

so that:

|M|2 =2|e|4t2

(s2 + u2) (3.65)

3.5.3 Electron-positron annihilation: e�e+ ! µ+µ�

The graph for e�(p) e+(k)! µ+(k0) µ�(p0) is shown in figure 3.7. Of course, we can follow the

e+

e�

µ�

µ+

Figure 3.7: Lowest order diagram e�e+ ! µ+µ�

same kind of calculation as in the previous example, but we can do better using the crossingsymmetry (and neglecting the masses). Comparing graph 3.7 and 3.6 we see that the t-channelof e�µ� ! e�µ� is just the s-channel of e�e+ ! µ+µ� as illustrated in the figure 3.8.

(2) µ�

(1) e�

(4) µ�

(3) e�

(3) e+

(1) e�

(4) µ�

(2) µ+

(3) e+

(1) e�

(4) µ�

(2) µ+

Figure 3.8: Left: the reaction e�µ� ! e�µ� considered as the s-channel i.e. s = (p1

+ p2

)2, t =(p

1

� p3

)2 u = (p1

� p4

)2. Centre: in order to obtain the t-channel of the left diagram, one has to takethe opposite of p

2

and p3

(thus s$ t) and interpret the result as anti-particles. Right: same diagram asthe centre but redraw in a more conventional way (after “horizontal stretching”).

Hence, we just have to replace t$ s in the previous matrix element! That’s a nice example ofcrossing. We get:

|M|2e�e�!µ+µ� =2|e|4s2

(t2 + u2) (3.66)

Let us go further by computing the di↵erential cross-section. Using 1.86, we have:

d�

d⌦⇤ =1

64⇡2s2|e|4 (t2 + u2)

s2

|~pf ||~pi|


Compton scattering: e�� ! e�� 87

Inserting the fine structure constant14:

↵ =|e|24⇡

(3.67)

and developing the Mandelstam variables in the center of mass frame where |~pi| = |~pf | = p⇤

(because the masses are neglected):

s = (k + p)2 = 4p⇤2

t = (p� k0)2 = �2p⇤2(1� cos ✓)u = (p� p0)2 = �2p⇤2(1 + cos ✓)

where ✓ is the angle (in the center of mass frame) between in incident e� and the outgoing µ+.we have:

d�d⌦

⇤ = ↵2

2s(4p⇤4

(1�cos✓)2+4p⇤4

(1+cos✓)2

16p⇤4

= ↵2

4s (1 + cos2 ✓)

Integrating over the � angle, we conclude:

d�

d(cos ✓)=↵2⇡

2s(1 + cos2 ✓)

Experimentally, the angle which is used is generally the one between the 2 particles e�, µ�

(which is equal to the one between e+, µ+). Thus ✓ ! ✓ � ⇡ leaving however unchanged theformula. Is this prediction in agreement with the experimental data? Look at figure 3.9 on thetop which shows the s d�

d cos ✓ distribution for the reaction e+e� ! µ+µ�. The data were collectedat an e�e+ collider at

ps = 29 GeV (PEP collider at SLAC in the 1980’). The plot on the

top is for e+e� ! µ+µ� and the dashed line describes QED predictions but at higher order(but largely dominated by the cos2 ✓ behaviour). Comparing to the data point, the agreementis good but the solid line gives a better fit. It corresponds to QED and additional electroweakcorrections where e+e� ! µ+µ� can occur via Z boson exchange. We will see this in the nextchapters.

To conclude this calculations, we can determine the total cross-section by integration overcos ✓:

� =

Z1

�1

d(cos ✓)↵2⇡

2s(1 + cos2 ✓) =

4⇡↵2

3s(3.68)

which is in agreement with the data at the % level.

3.5.4 Compton scattering: e�� ! e��

We are going to calculate the cross-section of the Compton scattering:

e�(p) �(k)! e�(p0) �(k0)

There are 2 diagrams, shown in figure 3.10, contributing to this process at the lowest order:s-channel (left-hand side of 3.10) and u-channel15 (right-hand side). Since photons are bosons,

14It’s a dimensionless measure of the strength of the electromagnetic interaction15It may not be obvious that the 2 contributions are s and u-channels. Let us assign to the particles in

e�(p) �(k) ! e�(p0) �(k0) the labels (1) + (2) ! (3) + (4). Then the u-channel is obtained by swapping (2) and(4) (namely the 2 photons) in the left-hand side diagram of figure 3.10. It gives the diagram on the right after“stretching” in the vertical direction.



VOLUME 517 NUMBER 21 PHYSICAL REVIEW LETTERS 21 NOVEMBER 198$

60

50

40

30— (a)

50

40

30— (b)

8Vl

ob 00

b

I.05—

1.00

0.95(c)

0.90-0.75 0.750

cosaFIG. 1, The differential cross sections for (a) Reac-

tion (2) and (b) Beaction (3), normalized to the QEDprediction. (c) The ratio of the differential cross sec-tion for Reaction (1) to the g(~3) QED prediction. Theexperimental cross section in (c) is normalized to theelectroweak cross section taken from a simultaneousfit to all three reactions. In (a)—(c) the dashed linecorresponds to the O(0. ) QED cross section and thesolid line to the fitted cross section.

es due to the limited azimuthal coverage of theLA system, event acceptances at cos0 =0.0 are77%%u~, 80%%ug, and 60%%uo for Reactions (1), (2), and (3),respectively, and fall to 73%, 77%, and 52/0 atIcosa' =0.6.The acceptance- corrected, background- sub-

tracted, differential cross sections are shownin Figs. 1(a)-1(c) and are compared to the O(n')QED prediction. Third-order QED processes cancreate angular asymmetries and normalizationchanges which are comparable to the weak-inter-action effects. Asymmetries in the angular dis-tribution arise from the interference between dia-grams of opposite charge-conjugation parity. 'The QED prediction is derived from a MonteCarlo calculation" that includes contributionsfrom initial- and final-state radiation, vertexcorrections, and leptonic and hadronic vacuumpolarization. Within the experimental acceptancethe QED forward-backward asymmetry is calcu-lated to be A„=1.0% in Reaction (2) and A,o=0.5/o in Reaction (3).Systematic errors in the relative normalization

of Reactions (2) and (3) to Reaction (1) arise from

three major sources: (1) errors in the back-ground estimates, 0.9% for e'e —g'y, and 1.7'%for e "e —~'7, (2) errors in detector simula-tion, 0.6/o in each process, and (3) uncertainty inthe decay modes of the tau lepton. The error(1.8/o) due to uncertainties in tau branching frac-tions is determined by folding the sensitivity toindividual decay modes with the errors in themeasurements of the branching fractions. ' Wehave neglected higher-order corrections to theO(n') QED cross section. These contributions(= 1~1%) are estimated from a leading-log calcu-lation" and are included in the systematic errorestimate. Uncertainties in the hadronic vacuum-polarization corrections introduce an additionalerror in the relative normalizations of roughly0.5/o. Combining the errors from all sourceslisted in quadrature, we find a 1.6%%uo normaliza-tion error on the ratio c»/o'„and a 2.8/o erroron the ratio cr«/c„.Reactions (1)-(3) are fitted by the O(a') QED

cross section plus the weak contributions. ' Frommaximum likelihood fits to the corrected cos0distributions, setting g„'= 0, me find g, 'g, &

=0.32 + 0.07+ 0.02 and g,'g, = 0.19+ 0.09+ 0.02,where the first error is statistical and the secondsystematic. ' The fits yield acceptance-, back-ground-, and QED-corrected forward-backwardelectroweak asymmetries, extrapolated to thefull cosG interval, of A„"" =—7.1/o+ 1.7/o andA, """=—4.2/o+ 2.0% where the standard modelpredicts" A " =—5.7%%uo. The fitted cross sec-tions are superimposed on the data in Figs. 1(a)and 1(b). The p, -~ universality of the axial-vectorcurrent is tested, and we find g, '/g, "=0.6+ 0.3.If we assume e -p -7 universality, then g,' =0.27+ 0.06+ 0.02 from Reactions (2) and (3). From asimultaneous fit to all three reactions which takesinto account the angular distributions and relativecross sections, we find g, ' =0.23 + 0.05 + 0.02 andg„' =0.03+ 0.03+ 0.03 where systematic errorshave been explicitly included in the fit, and thefitted cross section is superimposed on the datain Fig. 1(c). These results are in good agree-ment with the standard model and with other ex-periments. '" The probability that QED alonewould have led to the results presented here is6x10 '. The weak couplings above have beencomputed in the limit ~,—~. Alternatively, thevalues quoted above may be considered to meas-ure the product g'[M, '/(M, ' —s)]. Also g, ' andg„' only correspond to the couplings of the lowest-order &' exchange diagram. Inclusion of themass dependence and radiative corrections" to

1943

Figure 3.9: s d�d cos ✓ vs cos ✓ for e+e� ! µ+µ� in (a), e+e� ! ⌧+⌧� in (b). In (c) the ratio of d�

d cos ✓with the expected QED contribution for the reaction e+e� ! e+e�. Dashed line: QED contribution [10]

e�

�

�

e�

e�

�

�

e�

Figure 3.10: The 2 lowest order compton diagrams

the total amplitude is the sum of the two. Following then Feynman rules 3.1, the amplitude ofthe first diagram is:

iM1

= up0(�iq�µ)✏⇤µ k0 i/p + /k + m

(p + k)2 �m2

(�iq�⌫)✏⌫ kup

There is no ambiguity in the position of the di↵erent terms in the equation: since the amplitudeis a simple complex number, the adjoint spinor has to be on the left, and hence the second vertex



terms are on the left. Similarly, the second amplitude is:

iM2

= up0(�iq�µ)✏µ k i/p� /k0 + m

(p� k0)2 �m2

(�iq�⌫)✏⇤⌫ k0up

We can simplify a bit these expressions by using the fact that incoming/outgoing particles areon shell so that we can use: p2 = p02 = m2 and k2 = k02 = 0. Hence, the amplitudes becomesusing q = �|e|:

M1

= �|e|2✏⇤µ k0✏⌫ k up0�µ /p+

/k+m

2p.k �⌫up

M2

= |e|2✏⇤µ k0✏⌫ k up0�⌫ /p�/k0+m

2p.k0 �µup

where we have changed in M2

: µ$ ⌫. Now, noting that /p�⌫ = �⇢p⇢�⌫ = p⇢(��⌫�⇢ + 2g⇢⌫) =��⌫/p + 2p⌫ and similarly /p�µ = ��µ/p + 2pµ, it comes for the total amplitude:

M = �|e|2✏⇤µ k0✏⌫ k up0

⇣�µ ��⌫/p+2p⌫+

/k�⌫+�⌫m

2p.k � �⌫ ��µ/p+2pµ�/k�µ+�µm

2p.k0

⌘up

= �|e|2✏⇤µ k0✏⌫ k up0

⇣�µ 2p⌫+

/k�⌫

2p.k � �⌫ 2pµ�/k�µ

2p.k0

⌘up

= �|e|2✏⇤µ k0✏⌫ k up0

⇣�µ 2p⌫+

/k�⌫

s�m2

+ �⌫ 2pµ�/k�µ

u�m2

⌘up

where the identity (/p � m)up = 0 has been used in the second line for on-shell fermions ands = (p + k)2 = m2 + 2p.k ) p.k = (s�m2)/2 and p.k0 = �(u�m2)/2 has been used in the lastline.We wish to calculate the un-polarized cross-section. Thus we have to average over the 2 trans-verse polarizations of the incoming real photon, and the 2 spins of the electron. And we haveto sum over the final polarization states. Hence:

|M|2 =1

2

X�=1,2

1

2

Xs

X�0

=1,2

Xs0

|M|2 = |M1

|2 + |M2

|2 + M1

M⇤2

+�M

1

M⇤2

�⇤

with:

|M1

|2 = |e|44(s�m2

)

2

Ph✏⇤µ k0✏⌫ k up0�µ(2p⌫ + /k�⌫)up

i h✏⇤⇢ k0✏⌘ k up0�⇢(2p⌘ + /k�⌘)up

i⇤

|M2

|2 = |e|44(u�m2

)

2

Ph✏⇤µ k0✏⌫ k up0�⌫(2pµ + /k�µ)up

i h✏⇤⇢ k0✏⌘ k up0�⌘(2p⇢ + /k�⇢)up

i⇤

M1

M⇤2

= |e|44(s�m2

)

2

(u�m2

)

2

Ph✏⇤µ k0✏⌫ k up0�µ(2p⌫ + /k�⌫)up

i h✏⇤⇢ k0✏⌘ k up0�⌘(2p⇢ + /k�⇢)up

i⇤

We are just going to detail the first calculation. No need to calculate |M2

|2 since the secondgraph is just the u-channel of the first: replacing s$ u will give the answer. For the interferenceterm, there is no other solution than doing the calculation (which is very similar to the firstterm).

|M1

|2 =|e|4

4(s�m2)2

X�

✏⌫ k✏⇤⌘ k

X�0

✏⇤µ k0✏⇢ k0Xs,s0

⇥up0�µ(2p⌫ + /k�⌫)up

⇤ ⇥up0�⇢(2p⌘ + /k�⌘)up

⇤⇤

According to 3.63,P

� ✏⌫ k✏⇤⌘ k = �g⌫⌘ andP

�0 ✏⇤µ k0✏⇢ k0 = �gµ⇢ and:

⇥up0�⇢(2p⌘ + /k�⌘)up

⇤⇤= u†

p(2p⌘ + �⌘†k↵�↵†)�⇢†�0

†up0 = u†

p(2p⌘ + �0�⌘�0k↵�0�↵�0)�0�⇢�0�0up0

= u†p(2p⌘ + �0�⌘/k�0)�0�⇢up0 = up(2p⌘ + �⌘/k)�⇢up0



Hence:

|M1

|2 = |e|44(s�m2

)

2

g⌫⌘gµ⇢P

s,s0⇥up0�µ(2p⌫ + /k�⌫)up

⇤ ⇥up(2p⌘ + �⌘/k)�⇢up0

⇤= |e|4

4(s�m2

)

2

Ps,s0

⇥up0�µ(2p⌫ + /k�⌫)up

⇤ ⇥up(2p⌫ + �⌫/k)�µup0

⇤= |e|4

4(s�m2

)

2

Tr�(/p0 + m)�µ(2p⌫ + /k�⌫)(/p + m)(2p⌫ + �⌫/k)�µ

�= |e|4

4(s�m2

)

2

{ Tr�/p0�µ(2p⌫ + /k�⌫)/p(2p⌫ + �⌫/k)�µ

�+m2Tr (�µ(2p⌫ + /k�⌫)(2p⌫ + �⌫/k)�µ)

(using 3.55)

Let us consider the first trace:

Tr�/p0�µ(2p⌫ + /k�⌫)/p(2p⌫ + �⌫/k)�µ

�= 4p2Tr

�/p0�µ/p�µ

�+ 2Tr

�/p0�µ/k/p2�µ

�+ 2Tr

�/p0�µ/p2/k�µ

�+Tr

�/p0�µ/k�⌫/p�⌫/k�µ

�= 4m2

�Tr

�/p0�µ/p�µ

�+ Tr

�/p0�µ/k�µ

��+ Tr

�/p0�µ/k�⌫/p�⌫/k�µ

�= 4m2

��2Tr

�/p0�⌫

�(p⌫ + k⌫)

�� 2p⇢Tr

�/p0�µ/k�⇢/k�µ

�= �8m2 (4p0⌫(p⌫ + k⌫)) + 4p⇢Tr

�/p0/k�⇢/k

�= �32m2(p0.p + p0.k) + 4Tr

�/p0/k/p/k

�= �32m2(p0.p + p0.k) + 16(p0.k p.k � p0.p k2 + p0.k k.p)= 32(�m2p0.p + p0.k(p.k �m2))

where we have used successively the properties /p2 = p2 = m2, (3.59), (3.53), (3.57) and k2 = 0.Using a similar approach, one finds for the second trace:

Tr (�µ(2p⌫ + /k�⌫)(2p⌫ + �⌫/k)�µ) = 4p2Tr(�µ�µ) + 2Tr(�µ/p/k�µ) + 2Tr(�µ/k/p�µ) + Tr(�µ/k�⌫�⌫/k�µ)

= 16m2Tr( l1) + 8p.kTr( l1) + 8k.pTr( l1) + 4Tr(�µ/k2�µ)= 64m2 + 64p.k

and thus, injecting the results of the two traces:

|M1

|2 =4|e|4

(s�m2)2��2m2p0.p + 2p0.k p.k � 2m2p0.k + 4m4 + 4m2p.k

�

Now expressing the expression as function of the 2 Mandelstam variables s and u (rememberingthat s + t + u = 2m2):

s = (p + k)2 = m2 + 2p.k ) p.k = s�m2

2

t = (p0 � p)2 = 2m2 � 2p0.p) p0.p = m2 � t2

= s+u2

u = (k0 � p)2 = (k � p0)2 = m2 � 2p0.k ) p0.k = m2�u2

It finally comes:

|M1

|2 = 4|e|4

2m4

(s�m2)2+

m2

s�m2

� 1

2

u�m2

s�m2

�

The expression for |M2

|2 is obtained by swapping s and u:

|M2

|2 = 4|e|4

2m4

(u�m2)2+

m2

u�m2

� 1

2

s�m2

u�m2

�

and for the interference (after a similar boring calculation):

M1

M⇤2

= 2|e|4m2

1

u�m2

+1

s�m2

+4m2

(s�m2)(u�m2)

�



so that finally |M|2 = |M1

|2 + |M2

|2 + M1

M⇤2

+ (M1

M⇤2

)⇤ reads:

|M|2 = �2|e|4"

u�m2

s�m2

+s�m2

u�m2

� 4

m2

s�m2

+m2

u�m2

+

✓m2

s�m2

+m2

u�m2

◆2

!#

Let us denote p = (m,~0), k = (!,~k), p0 = (E0, ~p0), k0 = (!0,~k0) and ✓ the angle between ~k and~k0. We notice that:

u = (p� k0)2 = m2 � 2m!0 ) u�m2 = �2m!0

s = (p + k)2 ) s�m2 = 2m!m2 = p02 = (p + k � k0)2 = m2 � 2k.k0 + 2p.k � 2p.k0 = m2 � 2!!0(1� cos✓) + 2m! � 2m!0

) 1

!0 � 1

! = 1

m(1� cos ✓)

and hence:

|M|2 = 2|e|4h!0

! + !!0 + 4

�m2! �

m2!0 + ( m

2! �m2!0 )2

�i= 2|e|4

h!0

! + !!0 + 2

��(1� cos ✓) + 1

2

(1� cos ✓)2�i

= 2|e|4h!0

! + !!0 � sin2 ✓

i

We can then express the cross-section using formula 1.76:

d� = 1

4

p(p.k)

2�0

(2⇡)4�(4)(p0 + k0 � p� k)|M|2 d3~p0

(2⇡)

3

2E0d3 ~k0

(2⇡)

3

2!0

= 1

8m!(2⇡)

2

�(4)(p0 + k0 � p� k)|e|4h!0

! + !!0 � sin2 ✓

id3~p0

E0d3 ~k0!0

Rd3~p0

= 1

8m!(2⇡)

2

�(E0 + !0 �m� !)|e|4h!0

! + !!0 � sin2 ✓

i!02d!0d⌦

E0!0 R

d�

= 1

8m!2⇡ �(E0 + !0 �m� !)|e|4

h!0

! + !!0 � sin2 ✓

i!0d!0d(cos ✓)

E0

But:

E0 =p

m2 + |~p0|2 =q

m2 + (~k � ~k0)2 =p

m2 + !2 + !02 � 2!!0 cos ✓

And thus:

d�d(cos ✓) = 1

16m!⇡ |e|4h!0

! + !!0 � sin2 ✓

i!0d!0

E0

�(p

m2 + !2 + !02 � 2!!0 cos ✓ + !0 �m� !)

= 1

16m!⇡ |e|4h!0

! + !!0 � sin2 ✓

i!0

E01

2!0�2! cos ✓2E0 +1

= 1

16m!⇡ |e|4h!0

! + !!0 � sin2 ✓

i!0

!0�! cos ✓+E0 E0 = m + ! � !0

= 1

16m!⇡ |e|4h!0

! + !!0 � sin2 ✓

i!0

m+!(1�cos ✓) 1� cos ✓ = m( 1

!0 � 1

! )

= 1

16m2!⇡|e|4

h!0

! + !!0 � sin2 ✓

i!02

!

where we used the delta propertyR

+1�1 �(g(x)) dx =

Pi

1

| @g@x(xi)| to integrate over !0. Inserting

e2 = 4⇡↵, we obtain the formula:

d�

d(cos ✓)=⇡↵2

m2

!0

!+!

!0 � sin2 ✓

� ✓!0

!

◆2



which is known as the Klein-Nishina formula. At low energy, when ! ! 0 (and thus !!0 =

1 + !m(1� cos ✓)! 1)

d�

d(cos ✓)' ⇡↵2

m2

⇥2� sin2 ✓

⇤=⇡↵2

m2

⇥1 + cos2 ✓

⇤) � =

Z1

�1

⇡↵2

m2

⇥1 + cos2 ✓

⇤=

8⇡↵2

3m2

In this expression, we recognize the classical radius of the electron16 re = ↵/m =p

3�/(8⇡) '3 10�13 cm (with � ' 6 10�25 cm2).

3.6 Few words about the renormalization

Let us consider again the e�µ� scattering 3.6:

e� (p1

) µ�(p2

)! e� (p01

) µ�(p02

)

for which the amplitude is:

iM1

= up01

i|e|�µup1

�igµ⌫

q2

up02

i|e|�⌫up2

= 4⇡↵ up01

�µup1

igµ⌫

q2

up02

�⌫up2

with ↵ = |e|2/4⇡ is the usual electromagnetic fine structure constant. Suppose you want tomeasure ↵ with this process. Let us imagine, we have a beam of e� and µ� and we detect thescattered e� and µ�. By playing with the beam energy and the scattering angle, we can countthe number of recorded events and so measure the cross-section at a given momentum transferq2. From the measurement, we would then get the value of ↵meas = ↵.

Now, there are higher order corrections to this process. One is shown on figure 3.11. The

µ�

e�

µ�

e�

Figure 3.11: A virtual photon fluctuating in e�e+ pair.

virtual photon splits into a pair of virtual electron-positron (the fermion loop). If the photonhas a momentum q, then one member of the pair carries a momentum p and the other q � p,so that the energy-momentum is conserved. however, there is no constraint on the value of p!Therefore, all possible values have to be taken into account. The amplitude of this diagram canbe calculated but it requires additional rules with respect to the one given in section 3.1, namelya factor �1 for the closed fermion loop17, and the trace of the associated photon vertex must beused. Justification for the trace can be found in [8, p. 120], but a naive “explanation” is that,

16The classical radius of the electron is the size the electron that is determined assuming that only the electro-static potential energy contributes to its mass. Namely, in natural unit: E = m = |e|2/(4⇡re) ) re = ↵/m.

17The electron and positron in the loop can be interchanged, thus the �1.


Few words about the renormalization 93

basically we have a � ! � process which cannot depend on spinor indices. The amplitude isthen given by:

iM2

= up01

i|e|�µup1

n�igµµ0q2

R d4p(2⇡)

4

(�1)Trhi|e|�µ0

i /p+m

p2�mi|e|�⌫0

i /p�/q+m

(p�q)2�m

i�ig⌫0⌫

q2

o⇥

up02

i|e|�⌫up2

= up01

i|e|�µup1

n�1

q4

R d4p(2⇡)

4

(�1)(i|e|)2Trh�µi /p+m

p2�m�⌫i

/p�/q+m

(p�q)2�m

io⇥

up02

i|e|�⌫up2

Adding the 2 amplitudes, we finally have:

iM = up01

i|e|�µup1

⇢�igµ⌫

q2

+�i⇧

[2]

µ⌫(q2

)

q4

�up0

2

i|e|�⌫up2

M = up01

|e|�µup1

⇢gµ⌫q2

+⇧

[2]

µ⌫(q2

)

q4

�up0

2

|e|�⌫up2

where:

i⇧[2]

µ⌫(q2) =

Zd4p

(2⇡)4(�1)(i|e|)2Tr

�µi

/p + m

p2 �m�⌫i

/p� /q + m

(p� q)2 �m

�

Hence, we see that adding this contribution can be interpreted as a modification of the initialphoton propagator as:

�igµ⌫

q2

! �igµ⌫

q2

+�i⇧[2]

µ⌫(q2)

q4

We can consider this new photon propagator in the graph calculation, but there is a problemsince the integral in ⇧µ⌫ diverges! More precisely, one can show that ⇧µ⌫ can be reduced to theonly contributing term:

⇧[2]

µ⌫(q2) = �q2gµ⌫

↵

3⇡

✓Z 1

m2

dp2

p2

�� f(q2)

◆+ · · · + O(↵)

with f(q2) a finite contribution function given by:

f(q2) = 6

Z1

0

dz z(1� z) log

✓1� q2

m2

z(1� z)

◆(3.69)

ClearlyR1m2

dp2

p2

diverges as a logarithm. Let us introduce a cut-o↵ ⇤2, namely a maximum valuefor the upper bound of the integral. Is it shocking to truncate the integral this way? Well, anytheory must have its domain of validity. We can see ⇤ as a parametrization of our ignorance,

above which new physics must appear. Then, with this prescription, ⇧[2]

µ⌫ becomes:

⇧[2]

µ⌫(q2) = �q2gµ⌫

↵

3⇡

✓log

✓⇤2

m2

◆� f(q2)

◆+ · · · + O(↵)

and hence, the amplitude becomes (rewritten in terms of ↵):

M = up01

�µup1

gµ⌫

q2

4⇡↵

⇢1� ↵

3⇡log

✓⇤2

m2

◆+

↵

3⇡f(q2)

�up0

2

�⌫up2

Now in order to avoid confusion, we are going to rename the ↵ above as ↵0

:

M = up01

�µup1

gµ⌫

q2

4⇡↵0

⇢1� ↵

0

3⇡log

✓⇤2

m2

◆+↵

0

3⇡f(q2)

�up0

2

�⌫up2



The label 0 refers to the fact that this is the ↵ that is used in the case of 0 loop as at thebeginning of this section. ↵

0

is called the “bare” constant. The equation now is how can weinterpret the measurement of the cross section we have made? The actual scattering amplitudeM is not supposed to depend on an arbitrary parameter ⇤. If ⇤ is changed, we don’t wantthat our predicted cross-section (that we’re going to compare to the measured one) changes. Inother words, we would have to change ↵

0

itself in order to compensate the change of ⇤ ! ↵0

,the constant of the QED lagrangian is not a physical parameter! The key of the renormalizationidea is to express what we measure with physical (measured) parameters. Here, with our secondorder development, if our experiment is made at q2 = µ2, we would identify:

↵meas = ↵0

✓1� ↵

0

3⇡log

✓⇤2

m2

◆+↵

0

3⇡f(µ2)

◆= ↵

0

� ↵2

0

3⇡

✓log

✓⇤2

m2

◆� f(µ2)

◆+ O(↵2

0

) (3.70)

Our measurement was performed at q2 = µ2. To clearly point this out, I now use the name↵meas(µ2). Let’s revert the previous equation:

↵0

= ↵meas(µ2) +↵2

0

3⇡

⇣log

⇣⇤

2

m2

⌘� f(µ2)

⌘+ O(↵2

0

)

= ↵meas(µ2) + ↵2

meas(µ2

)

3⇡

⇣log

⇣⇤

2

m2

⌘� f(µ2)

⌘+ O(↵2

meas(µ2))

The second equality is equivalent to the first one, to the second order used here. But what ifinstead of µ2, we would have chosen a arbitrary value q2? We would have:

↵meas(q2) = ↵0

� ↵2

0

3⇡

⇣log

⇣⇤

2

m2

⌘� f(q2)

⌘+ O(↵2

0

)

= ↵meas(µ2) + ↵2

meas(µ2

)

3⇡

⇣log

⇣⇤

2

m2

⌘� f(µ2)

⌘�↵2

meas(µ2

)

3⇡

⇣log

⇣⇤

2

m2

⌘� f(q2)

⌘+ O(↵2

meas(µ2))

= ↵meas(µ2) + ↵2

meas(µ2

)

3⇡

�f(q2)� f(µ2)

�+ O(↵2

meas(µ2))

The good news now, is that the arbitrary ⇤ doesn’t appear anymore! We can express any processat any energy if we have measured ↵meas at a given reference point. This is the miracle of therenormalization. But it has a price to pay: what we called a coupling constant is not a constant!It depends on the energy scale. We manage to eliminate ⇤ at the second order of perturbation.The renormalization works if we can eliminate it at any order. In QED, it turns out to be thecase.

Coming back to f(q2), when the momentum transfer is large with respect to m (which is

the case in high energy physics experiments), by developing f(q2) at order O( m2

�q2

), it is not

too di�cult to show that: f(q2) ! � log⇣

m2

�q2

⌘� 5/3 = log

⇣�q2

e5/3m2

⌘. Note that q2 < 0 with

scattering processes. We can define the positive quantity Q2 = �q2. In this approximation,↵meas(Q2) that we now simply denote by ↵(Q2) becomes18:

↵(Q2) = ↵(µ2)

1 +

↵(µ2)

3⇡log

✓Q2

µ2

◆�+ O(↵2)

Now, this result is valid at order ↵2. We can go at higher order by adding more loops. Thedependency of ↵ with Q2 then becomes:

↵(Q2) =↵(µ2)

1� ↵(µ2

)

3⇡ log⇣

Q2

µ2

⌘ (3.71)

18µ2 is now understood as a positive quantity


A major test of QED: g-2 95

Let us interpret this formula valid only in the regime Q2 � m2: it gives the value of thefine structure constant at a given energy, if it has been measured at another energy scale. Itsvariation is only logarithmic with the scale. In Coulomb scattering, at Q2 ' 0, ↵ = 1/137 whileat the Z0 pole, it is about 1/128. The variation of the charge (or ↵) as function of the scale is areal physical e↵ect. It is due to the vacuum polarization. Imagine you want to test the charge ofan electron by approaching a probe-charge. In the vicinity of the electron, virtual charged pairscan be created for a short amount of time �t ⇠ ~/mc2. They can spread apart at a maximumdistance c�t, creating a dipole. The virtual positrons tend to be closer to the electron thanthe virtual electrons. If the probe-charge doesn’t approach enough (ie d � c�t), it will “see”mainly the surrounding clouds of positron, yielding to a measurement underestimating the realcharge of the electron. In other words, at small Q2 (ie high distance), the charge seems smaller.This phenomena is usually called a “screening”. On the contrary, at high Q2, the probe chargepenetrates the virtual cloud and can see the whole electron charge.19

In summary, the renormalization is a procedure that allows to absorb all infinite quantitiesby a redefinition of a finite number of parameters. In QED, they are the charge of the electron(or equivalently the fine structure constant), the mass of the electron (yes, the mass!) and thenormalization of the fields (thus the name renormalization).

3.7 A major test of QED: g-2

There are many tests that validate QED, both at high energy (in accelerator) and low energy.One of the most impressing test is probably the prediction of the gyromagnetic ratio g whichenters in the magnetic moment of the electron.

3.7.1 Prediction with Dirac’s equation

Non relativistic prediction without spin: Let us recall first, the classical formula forthe magnetic moment of charged particle moving in a circular orbit. Classically, the magneticmoment is defined as:

µ = current⇥ area

For a circular orbit or radius r, the area is ⇡r2. And the current is the charge per unit time,namely the charge times the frequency of rotation (the velocity divided by the circumference ofthe orbit). Hence:

µ = qv

2⇡r⇥ ⇡r2 = q

v

2r

Expressing as function of the angular momentum ~L = ~r ⇥ ~p = ~r ⇥m~v, we have:

~µ = gq

2m~L g = 1

g = 1 in the classical case is called the Lande g-factor or gyromagnetic ratio. Electrons placedin a magnetic field ~B would then have an additional potential energy due to the field:

EB = �~µ. ~B = �gq

2m~L. ~B

19Well, it’s not true: virtual pairs of higher mass particles can be created. There is always a virtual cloud seenby the probe.



which can be confirmed by considering the non-relativistic hamiltonian appearing in the Schrodingerequation is

H =p2

2m

In the presence of a magnetic field, we make as usual the substitution p! p� q ~A, and then:

H =p2

2m� q

m~A.p +

q2A2

2m' p2

2m� q

m~A.p

where the term in q2 is neglected. The additional energy is then given by:

EB = � q

m~A.p

But, ~B = ~r⇥ ~A and ~A can be chosen satisfying the gauge ~r. ~A = 0. Thus, ~A = 1

2

B⇥~r and EB

becomes:

EB = � q

2m( ~B ⇥ ~r).p = � q

2m(~r ⇥ p). ~B = � q

2m~L. ~B = �~µ. ~B

where we have applied a scalar triple product property. We finally find the expected expressionfor ~µ. Since electrons have an intrinsic angular momentum, the spin ~S, we would expect anadditional energy:

EB = �gq

2m~S. ~B with g = 1

However, at the time of Dirac, the measurements for g excluded the classical value 1, and wascompatible with 2.

Non relativistic prediction with spin: we can follow the same procedure, starting fromDirac’s equation instead of Schrodinger one. Hence, starting from the momentum space equation2.30:

(/p�m)u = (�µpµ �m)u = 0pµ!pµ�qAµ��! (�µpµ � q�µAµ �m)u = 0

(�0E � q�0�� ~�.~p + q~�. ~A�m)u = 0

In terms of 2-components, this equation reads:

(E � q��m) �~�.~p + q~�. ~A

~�.~p� q~�. ~A �(E � q�+ m)

!✓ua

ub

◆= 0

(~�.~p� q~�. ~A)ub = (E � q��m)ua

(~�.~p� q~�. ~A)ua = (E � q�+ m)ub

Now, to be able to easily identify the magnetic momentum as in the case of the Schrodingerequation, we wish to find a relation satisfied by the spinors in the non-relativistic limit of theDirac’s equation. In this limit, E ' m, but E�m = T and T ⌧ m where T is the kinetic energy.For usual electromagnetic field, the field energy is negligible with respect to mc2. Therefore,m� q�. and thus:

(~�.~p� q~�. ~A)ub = (T � q�)ua

(~�.~p� q~�. ~A)ua ' 2m ub


Higher order corrections 97

So that by multiplying the first equation by 2m and inserting the second:

(~�.~p� q~�. ~A)(~�.~p� q~�. ~A)ua = 2m(T � q�)uah(~�.~p)2 + q2(~�. ~A)2 � q

⇣(~�.~p)(~�. ~A) + (~�. ~A)(~�.~p)

⌘iua = 2m(T � q�)ua

(1)h|~p|2 + q2| ~A|2 � q

⇣~p. ~A + i~�.(~p⇥ ~A) + ~A.~p + i~�.( ~A⇥ ~p)

⌘iua = 2m(T � q�)uah

(~p� q ~A)2 � iq~�.⇣~p⇥ ~A + ~A⇥ ~p

⌘iua = 2m(T � q�)ua

where in (1), we have used the equality (~�.~a)(~�.~b) = ~a.~b+ i~�.(~a⇥~b). Going back to the languageof wave functions (and thus replacing ua by = uae�ip.x and ~p by �i~r), we have:

(�i~r�q ~A)

2

2m � q2m~�.

⇣~r⇥ ( ~A ) + ~A⇥ ~r

⌘= (T � q�)

(�i~r�q ~A)

2

2m � q2m~�.

⇣~r⇥ ~A

⌘ = (T � q�) h

(�i~r�q ~A)

2

2m � q2m~�. ~B + q�

i = T

The quantity in the bracket:

HE'm =(~p� q ~A)2

2m� q

2m~�. ~B + q�

corresponds to the non-relativistic hamiltonian of the Dirac’s equation. Recalling that the spinoperator for 2-components spinors is ~S = 1

2

~�, we identify:

~µ =q

2m~� = 2

q

2m~S

meaning that:g = 2

Hence, the non-relativistic limit of the Dirac’s equation does imply a gyromagnetic ratio of 2.It was one of the successes of Dirac’s prediction.

3.7.2 Higher order corrections

Consider the electron charge current involved in the basic vertex of the left diagram of figure3.12. It can be decomposed in 2 parts (as known as Gordon decomposition):

jµ = �e f�µ i

= � e2

[uf�µui + uf�µui] e�i(pi�pf ).x = ue�ip.x

= � e2m [ufpf⌫�⌫�µui + uf�µ�⌫pi⌫ui] e�i(pi�pf ).x uf (/pf

�m) = 0 , (/pi�m)ui = 0

= � e2m uf [pf⌫ (2gµ⌫ � �µ�⌫) + �µ�⌫pi⌫ ] ui e�i(pi�pf ).x �µ�⌫ + �⌫�µ = 2gµ⌫

= � e2m uf

h2pµ

f + �µ�⌫(pi⌫ � pf⌫)iui e�i(pi�pf ).x

Now, using the matrix:

�µ⌫ =i

2(�µ�⌫ � �⌫�µ)) �µ�⌫ = gµ⌫ � i�µ⌫

so that we get:

jµ = � e2m uf

h2pµ

f + gµ⌫(pi⌫ � pf⌫)� i�µ⌫(pi⌫ � pf⌫)iui e�i(pi�pf ).x

=h� e

2m uf (pµf + pµ

i )ui + iufe

2m�µ⌫(pi⌫ � pf⌫)ui

ie�i(pi�pf ).x

(3.72)



Referring to the Feynman rules of table 3.1, we see that the first term in the bracket � e2m uf (pµ

f +

pµi )ui is analogous to the interaction of a charged spin 0 particle. We then expect the second

term:

jµs = iuf

e

2m�µ⌫(pi⌫ � pf⌫)ui e�i(pi�pf ).x

to be related to the interaction involving the spin of the electron. To check this assumption,consider, to the first order of perturbation, the amplitude given by the interaction of such currentwith an electromagnetic field Aµ:

S[1] = �i

Zd4x jµ

s Aµ =

Zd4x uf

e

2m�µ⌫(pi⌫ � pf⌫)ui e�i(pi�pf ).xAµ

Let us consider a static field (coulomb scattering) for simplification. Then Aµ does not dependon time. We can perform the integration over the time:

S[1] = 2⇡ �(Ei � Ef )

Zd3x uf

e

2m�µ⌫(pi⌫ � pf⌫)ui e�i(pi�pf ).xAµ

Because of the Dirac-delta function, the time-like component of the 4-momenta in the integralhas to be equal, and hence cannot contribute to the amplitude. Only the spatial-like componentsmatter, and after re-arrangement, it reads:

S[1] = 2⇡i �(Ei � Ef )

Zd3x uf

e

2m�µjui @j

⇣e�i(pi�pf ).x

⌘Aµ

In the expression above, only the exponential and Aµ depend on space coordinates (the spinorsdepend only on 4-momenta). Hence, after integration by part (the field Aµ vanishing at infinity):

S[1] = �2⇡i �(Ei � Ef )

Zd3x uf

e

2m�µjui e�i(pi�pf ).x @jAµ

Figure 3.12: Left: simplest QED vertex. Right: a correction to the vertex.

As in the previous section, in order to see clearly the spin contribution, we are going to considerthe low energy limit, where the mass dominates over the 3-momentum. Then, only the uppercomponent of the spinors matters since in the lower component, there is the term ~�.~p/(E + m)(see formula 2.40). In other words, spinors simplify to:

u =

✓ua

~�.~pE+mua

◆!

✓ua

0

◆


Higher order corrections 99

In these conditions, the term uf�µjuiAµ = uf�0juiA0

+ uf�kjuiAk ⇡ uf�kjuiAk. Indeed,according to the Dirac’s representation of the � matrices:

�0j = i2

✓l1 0000 �l1

◆✓00 �j

��j 00

◆�✓

00 �j

��j 00

◆✓l1 0000 �l1

◆�= i

✓00 �j

�j 00

◆

and hence uf�0jui ⇡ 0. Thus, only uf�kjui gives a significant contribution to the amplitude:

S[1] = �2⇡i �(Ei � Ef )

Zd3x uf

e

2m�kjui e�i(pi�pf ).x @jAk

Let us express �kj :

�kj = i2

✓00 �k

��k 00

◆✓00 �j

��j 00

◆�✓

00 �j

��j 00

◆✓00 �k

��k 00

◆�

= i2

✓�(�k�j � �j�k) 00

00 �(�k�j � �j�k)

◆

= i2

✓�2i

Pl ✏kjl�l 0000 �2i

Pl ✏kjl�l

◆

=P

l ✏kjl

✓�l 0000 �l

◆

where ✏kjl is the usual antisymmetric tensor (✏kjl = 1 for cyclic permutation of 123, = �1 foranticyclic permutation, 0 otherwise). Hence:

uf�kjui =

Xl

✏kjl(u†af

, 0)

✓l1 0000 � l1

◆✓�l 0000 �l

◆✓uai

0

◆=X

l

✏kjl u†af�luai

Now, noticing for example that:

uf�1jui@j = uf�11@1

ui + uf�12@2

ui + uf�13@3

ui

=P

l ✏11l u†af�

l@1

uai +P

l ✏12l u†af�

l@2

uai +P

l ✏13l u†af�

l@3

uai

= u†af

�Pl ✏11l�l@

1

+P

l ✏12l�l@2

+P

l ✏13l�l@3

�uai

= u†af

��3@

2

� �2@3

�uai

= �u†af

h~� ⇥ ~@

i1

uai

and similarly for the other components. Thus, we conclude:

uf�kjui @jAk = �u†

af

h~� ⇥ ~@

ikAk uai = �u†

af

h~� ⇥ ~@

i. ~A uai = �u†

af

h~@ ⇥ ~A

i.~� uai

Hence, since ~B = ~@ ⇥ ~A, we finally have:

S[1] = �2⇡i �(Ei � Ef )

Zd3x u†

af

⇣� e

2m

⌘~B.~� uai e�i(pi�pf ).x

where we identify the magnetic moment of the electron ~µ = � e2m~�. Conclusion, as announced

at the beginning of this section, the current:

jµs = iuf

e

2m�µ⌫q⌫ui e�i(pi�pf ).x



with q = pi � pf , yields naturally the gyromagnetic ratio.

Now, consider the second diagram on the right hand side of figure 3.12. The computa-tion of the correction leads to a modification of the current20 in the low momentum transferapproximation [9, p. 160] (dropping the exponential):

jµ = uf

n� e

2m(pµf + pµ

i ) + i e2m�

µ⌫q⌫o

ui

! uf

n�e�µ

h1 + ↵

3⇡q2

m2

⇣log( m

m�� 3

8

⌘i+⇥↵2⇡ i e

2m�µ⌫q⌫

⇤oui

where m� is a cut-o↵ to avoid an infinite term due to the so called infra-red divergence. We canimmediately deduce that adding the 2 diagrams will yield a modification of the gyromagneticratio:

~µ = � e

2m~� ! � e

2m

⇣1 +

↵

2⇡

⌘~� ) ae =

g � 2

2=

↵

2⇡

where ae, the deviation with respect to the value g = 2, is called the anomalous magneticmoment. Numerically, we obtain ae ' 0.001 16, to be compared to the most precise experimentalvalue obtained so far [11]:

aexpe = 1159 652 180.73 (0.28)⇥ 10�12 (3.73)

the digits in parentheses denoting measurement uncertainty in the last two digits at one standarddeviation. Pretty close! Such experimental accuracy pushes very far the theory and requires acalculation at least at the 8th order (ie ↵4). The 10th order, which represents 12672 vertex-typeFeynman diagrams (!!), is already partially evaluated [12]. Hadronic (vacuum polarization),electroweak e↵ects and small QED contributions from virtual muon and tau-lepton loops con-tributions have also to be taken into account. Fortunately, the diagrams are now evaluatednumerically and a recent computed value of ae is [12]:

athe = 1159 652 181.13 (0.11)(0.37)(0.02)(0.77)⇥ 10�12

where the first, second, third, and fourth uncertainties come from the calculated eighth-orderQED term, the crude tenth-order estimate, the hadronic and electroweak contributions, and thefine structure constant, respectively. Both theory and experiment are in very good agreementsince:

aexpe � ath

e = �0.40 (0.88)⇥ 10�12

The largest source of uncertainty of athe is now the experimental value used for the fine structure

constant (obtained from Cesium or Rubidium atom experiments), and not anymore an uncer-tainty coming from the calculation itself! Therefore, it makes sense now to obtain ↵ from thetheory and the measured value of ae instead of the contrary.

The accuracy of such computation of the g-factor is probably one of the most impressivetriumphs of the theory of quantum electrodynamics.

3.7.3 Measurement of the g-factor

We have already seen that an electron in a magnetic field gets extra contribution to its energywith a term �~µ. ~B = �g q

2m~S. ~B. A field on the Z-axis will yield �g �e

2m~msBz = h⌫sms with

20after renormalization.


Measurement of the g-factor 101

ms = ±1

2

and the spin frequency ⌫s is related to the cyclotron frequency ⌫c with ⌫s = g2

⌫c and

⌫c = eBz2⇡m . The principle of the measurement is thus to measure both ⌫c and ⌫s or more precisely

⌫c and ⌫a = ⌫s � ⌫c, the “anomaly” frequency since:

g

2=⌫s

⌫c= 1 +

⌫s � ⌫c

⌫c= 1 +

⌫a

⌫c

The experimental set-up [11] is based on a Penning-trap which suspends a single electronthanks to a strong magnetic field (5.36 T) along the z-axis and an electrostatic quadrupolepotential as shown on figure 3.13. The e↵ect of the electric field is to confine the electronin a potential well in which it makes small vertical oscillation. The Penning trap is used toartificially bind the electron in an orbital state, as if it was an electron of an atom. Actually, theelectrostatic potential shifts the cyclotron frequency from ⌫c to ⌫c and thus ⌫a to ⌫a = ⌫s � ⌫c.The lowest energy level including the leading relativistic correction can be approximated by [11]:

En,ms = h⌫sms +

✓n +

1

2

◆h⌫c �

1

2h�(n +

1

2+ ms)

2

where the quantity � is of the order � ⇡ 10�9⌫c. According to this formula giving the energylevels, the transitions |n, msi = |1, 1

2

i ! |0, 1

2

i ! |1,�1

2

i ! |0,�1

2

i:

E1, 1

2

� E0, 1

2

= h(⌫c � 3

2

�)

E0, 1

2

� E1,� 1

2

= h⌫a

E1,� 1

2

� E0,� 1

2

= h(⌫c � 1

2

�)

allow to measure ⌫c, � and ⌫a and hence g/2. It turns out that ⌫c ⇡ 150 GHz while ⌫a ⇡ 174MHz. Since ⌫c ' ⌫c is proportional to B, a high magnetic field is necessary to increase thespacing between the cyclotron energy levels. The cavity of the Penning-trap is maintained at a

Figure 3.13: Schema of a Penning trap [13]. The constant electric field (blue) is generated by aquadrupole (a and b). The superposed constant and homogeneous magnetic field (red) is generated by asurrounding cylinder magnet (c). A particle, indicated in red (here positive) is stored in between caps ofthe same polarity. The particle is trapped inside a vacuum chamber.

very low temperature (100 mK) to avoid transitions from the ground state to other states due



to blackbody photons. The electron is initially prepared in the state |0, 1

2

i. The spin-up state isobtained by playing with the electrostatic potential (which induced an additional magnetic field).The higher cyclotron state are artificially excited by injecting microwaves into the trap cavity.Then, the measurement detects the anomaly transition |0, 1

2

i ! |1,�1

2

i followed by the decayto the ground state |1,�1

2

i ! |0,�1

2

i. Four measurements are performed at slightly di↵erentcyclotron frequencies (by varying the B field) in order to estimate the frequency shift due to thecavity trap (the shape of the trap cavity modifies the density of states of the radiation modes offree space). They are presented on figure 3.14, from which the measured value of equation 3.73is deduced.

Figure 3.14: Four measurements of g/2 without (open) and with (filled) cavity-shift corrections. Thelight gray uncertainty band shows the average of the corrected data. From [11].

3.8 Exercises

Exercise 3.1 Draw the diagrams at the minimal order for the following reactions and writedown the corresponding probability amplitudes.

1. e+µ� ! e+µ�. Use the 4-momenta: e+(k) µ�(p)! e+(k0) µ�(p0).

2. e+µ+ ! e+µ+. Use the 4-momenta: e+(k) µ+(p)! e+(k0) µ+(p0).

3. e+e� ! µ+µ�. Use the 4-momenta: e+(k) e�(p)! µ+(k0) µ�(p0).

What about e+µ� ! e�µ+?

Exercise 3.2 Draw all diagrams at the first order (in ↵e) for the reactions below. If there areseveral diagrams per reaction, precise whether you have to add or subtract the diagrams.

1. e�e� ! e�e� (Moller scattering)

2. e+e� ! e+e� (Bhabba scattering)


Exercises 103

3. e�� ! e�� (Compton scattering)

4. e�e+ ! �� (pair annihilation)

Exercise 3.3 Electron-positron annihilation into a muon pair, e�e+ ! µ�µ+. We consideronly the ultra-relativistic regime and use the notation for the 4-momenta: e�(p)e+(k)! µ�(p0)µ+(k0).

1. List the helicity combinations that contribute to the process.

2. For each of them, compute the probability amplitude M. (Use the formulas 2.59 for thehelicity states spinors).

3. In order to determine the average probability < |Mtot|2 > of the process, do you have toadd first the di↵erent amplitudes and square the result or do you have to add the individualsquared amplitudes? Justify your answer.

4. Using the Mandelstam variables, show that < |Mtot|2 > previously determined can be

written 2e4

s2

(t2 + u2).

5. Redo the calculation of |Mtot|2 by using the spin summations and trace theorems. Checkthe consistency of your result.


Documents

A brief view of Quantum Electrodynamic - LLRpolypaganini/ch3.pdf · Chapter 3 A brief view of Quantum Electrodynamic Few references: I.J.R Aitchison & A.J.G. Hey,“Gauge theories