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Class Notes for Quantum Field Theory: Section IIS-Matrix, Wick’s Theorem, Feynman Rules, Specific Sample Calculations

Primary Reference: Mandl & Shaw, “Quantum FieldTheory”

See also: Peskin and Schroeder, “Quantum FieldTheory”

Jack GunionU.C. Davis

230A – Part II

J. Gunion

The S-matrix

• Ideally, we would like to solve the coupled differential equations that resultfrom introducing interactions among the free fields, as for instance E&Minteractions found by the minimal substitution rules.

• An exact solution has not been found, but we have made much progressusing a perturbative approach. So far, we have been fortunate that naturehas always allowed such a perturbative approach to be fruitfully comparedto experiment.

QED, in particular, has passed extremely precise tests because we cancompute very precisely due to the small size of α ∼ 1/137.

• The perturbative approach is most conveniently derived in the interactionpicture in which most of the time evolution of the states (namely, thatassociated with the free particle part of the Hamiltonian) is removedleaving behind only the (small) time evolution from the perturbativelytreated interaction(s).

J. Gunion 230A – Part II 1

So far, we have been using the Heisenberg picture in which state vectors(in the Foch space sense) are constant in time while the field operatorscontain all the time dependence.

• The solution to the full problem can be formulated using the Dysonexpansion which is ideally suited for obtaining perturbation results systematically.

The Interaction Picture

• Consider QED as the example theory.

• We divide up the full L into

L = L0 + LI (1)

with

L0 =: ψ(x)(i∂µγµ −m)ψ(x)−1

2(∂µAν)(∂

µAν) : (2)

(using the Lorentz condition formulation for the Aµ fields) and

LI =: eψ(x)Aµγµψ(x) : . (3)


• A full (non-perturbative) solution to the theory would arise if we couldcompute

U ≡ U(t, t0) = e−iH(t−t0) (4)

where U defines the time evolution of the Schroedinger picture states,

|A, t〉S = U |A, t0〉S (5)

from which one has

id

dt|A, t〉S = H|A, t〉S . (6)

U also defines the relation between the Schroedinger and Heisenbergpictures:

|A, t〉H = U†|A, t〉S = |A, t0〉S (7)

andOH(t) = U†OSU , (8)

where the Schroedinger picture is defined by OS being independent of time.Of course, O = H is time-independent and is the same in both pictures.

At t = t0, the states and operators in the two pictures are the same.However, the Schroedinger state changes with time whereas the Heisenbergpicture state is constant in time: one could write |A, t〉H = |A〉H, but MSdoesn’t.


• Of course, expectation values are the same in the two pictures since U isunitary; i.e.

S〈B, t|OS|A, t〉S =H 〈B, t|OH(t)|A, t〉H . (9)

• From Eq. (8),

id

dtOH(t) = [OH(t),H] . (10)

• Now write H = H0 +HI, where the I subscript is for interaction and notfor interaction picture!

• The I.P. is defined relative to the S.P. by using

U0 ≡ U0(t, t0) = e−iH0(t−t0) (11)

to define the I.P. state

|A, t〉I = U†0 |A, t〉S (12)


and I.P. operatorsOI(t) = U†0OSU0 . (13)

Here the subscript I on |A, t〉I and the superscript I on OI(t) refer to theinteraction picture (I.P.) and not to the interaction part of the Hamiltonian.

Note that HI0 = HS

0 ≡ H0 .

• Differentiating Eq. (13) gives

id

dtOI(t) = [OI(t),H0] (14)

and we also find

id

dt|A, t〉I = i

d

dt

[e+iH0(t−t0)|A, t〉S

]= −H0e

+iH0(t−t0)|A, t〉S + e+iH0(t−t0)id

dt|A, t〉S

= −e+iH0(t−t0)HS0 |A, t〉S + e+iH0(t−t0)HS|A, t〉S

= e+iH0(t−t0)HSI |A, t〉S


= e+iH0(t−t0)HSI e−iH0(t−t0)|A, t〉I

= HII |A, t〉I , (15)

where: 1) we used the fact that HS ≡ H in this notation; 2) we used thefact that HS

0 = H0 (see below Eq. (13)); 3) we used Eq. (6); and 4) wewere very careful to preserve the placing of the H operator next to |A, t〉S.

• Note that we can rewrite the last few lines above as

HII = e+iH0(t−t0)HS

I e−iH0(t−t0)

= e+iH0(t−t0)e−iH(t−t0)HHI (t)eiH(t−t0)e−iH0(t−t0)

= U ′†HHI U

′ , (16)

where U ′ is a unitary operator that takes us from the H.P. to the I.P. forany operator (not just the Hamiltonian).


The S-matrix expansion

• This expansion is based upon the I.P. reviewed above, in which:

– the operators satisfy the Heisenberg-like equations of motion but involvingthe free Hamiltonian H0 only, not the complete H, as algebraically statedin Eq. (14):

id

dtOI = [OI(t),H0] ; (17)

– if the interaction LI does not involve field derivatives (as for all casesof interest to us), the fields canonically conjugate to the interactingfields and those conjugate to the free fields are identical, e.g. in QEDπα = ∂L

∂ψα= ∂L0

∂ψα, where we are in the H.P. when writing such equations

and thinking about conjugate fields.This implies that since the I.P. and the H.P. are related by a unitarytransformation, then in the I.P. the interacting fields satisfy the samecommutation or anticommutation relations as the free fields. Using scalarfield notation, this is shown algebraically as follows:

[φI(~x, t), πI(~y, t)] = [U ′†φH(~x, t)U ′, U ′

†πH(~y, t)U ′]


= U ′†[φH(~x, t), πH(~y, t)]U ′

= U ′†iδ3(~x− ~y)U ′

= iδ3(~x− ~y) . (18)

This in turn implies that we will 2nd quantize the I.P. fields in a mannerthat is essentially identical to what we did in the free field case.

– Note: there is a subtlety since the asymptotic states in the I.P. can havea different mass and different normalization than the asymptotic statesin the free-particle case.This difference arises because the asymptotic states must contain notjust the free particle behavior, but the full collection of virtual processesassociated with the presence of interactions (think bare line + line withall possible insertions, loops, ....).This is a subtlety that we will eventually address, but not in detail untilwe come to renormalization theory.

• At this point, MS simplifies the notation a bit and writes in the I.P.

id

dt|Φ(t)〉 = HI(t)|Φ(t)〉 (19)


whereHI(t) ≡ eiH0(t−t0)HS

I e−iH0(t−t0) = HI

I (t) ; (20)

that is, we will be dropping the I superscript reminding us that everythingis in the I.P.

• It will also be important to note that if HSI is a product of fields in the S.P.

(i.e product of operators in the S.P.), then HI will be the same product offields/operators where all the fields/operators are in the I.P. (This is shownby inserting a whole bunch of 1’s in between the operators in the form1 = e−iH0(t−t0)eiH0(t−t0).)

• If we start with a state at some initial time, |Φ(ti)〉 = |i〉, the solution ofEq. (19) gives the state |Φ(t)〉 at any other time t.

Since HI(t) is hermitian, this time development is given by a unitarytransformation. Accordingly, it preserves the normalization of the state:

〈Φ(t)|Φ(t)〉 = const. (21)

• This formalism we are going to develop will not be appropriate for boundstates. It will apply to the situation where we consider an initial state in


which there are a certain number of widely separated particles which thenconverge upon a small interaction region, interact via the interactions, andthen at large times a (usually different) set of widely separated particlesemerge from the interaction region.

Eq. (19) determines the state |Φ(t = ∞)〉 into which |Φ(t = −∞)〉evolves long after the scattering is over and all particles are far apart again.

• The S-matrix is defined by

|Φ(t =∞)〉 = S|Φ(t = −∞)〉 = S|i〉 . (22)

Note that Eq. (21), conservation of normalization, implies

〈Φ(∞)|Φ(∞)〉 = 〈Φ(−∞)|S†S|Φ(−∞)〉 = 〈Φ(−∞)|Φ(−∞)〉 (23)

which requires that S†S = 1, i.e. S is a unitary operator.

• S|i〉 can consist of a large selection of possible final states |f〉.The probability that after the collision (i.e. at t = ∞) the system is instate |f〉 is given by

|〈f |Φ(∞)〉|2 , (24)


assuming unity normalizations for |Φ(−∞)〉 = |i〉 (and hence also for|Φ(∞)〉 given conservation of normalization as noted above). Thecorresponding probability amplitude is

〈f |Φ(∞)〉 = 〈f |S|i〉 ≡ Sfi . (25)

in terms of which|Φ(∞)〉 =

∑f

|f〉Sfi . (26)

The unitarity of the S-matrix can be written in this basis as∑f

|Sfi|2 = 1 , (27)

as follows from

1 = 〈i|i〉 = 〈i|S†S|i〉 = 〈i|S†|f〉〈f |S|i〉 = S∗fiSfi . (28)

This equation expresses the conservation of probability. It is more generalthan the corresponding conservation of particles in NRQM since nowparticles can be created and/or destroyed.


+(−i)2

∫ t

−∞dt1

∫ t1

−∞dt2HI(t1)HI(t2)|i〉+ . . . (31)

yielding (as t→∞)

S =

∞∑n=0

(−i)n∫ ∞−∞

dt1

∫ t1

−∞dt2 . . .

∫ tn−1

−∞dtnHI(t1)HI(t2) . . .HI(tn) .

(32)

• Now comes a very crucial trick that partly motivates the use of timeordering. The above expression can be rewritten as

S =

∞∑n=0

(−i)n

n!

∫ ∞−∞

dt1

∫ ∞−∞

dt2 . . .

∫ ∞−∞

dtnTHI(t1)HI(t2) . . .HI(tn) .

(33)Here, the time ordering operation has been generalized to include anarbitrary list of operators. It is necessary to insert the T instruction, sincein the original form, Eq. (32), the HI operators (which do not commutewith one another when evaluated at different times) were very definitively


ordered so that HI’s with earlier times were always written to the right ofHI’s with later times.

Let me show how one arrives at this using the 2nd order term as an example.We write (dropping the subscript I for the moment):

∫ t

−∞dt1

∫ t1

−∞dt2H(t1)H(t2)

=

∫ t

−∞dt2

∫ t

t2

dt1H(t1)H(t2) interchange of integration order

=

∫ t

−∞dt1

∫ t

t1

dt2H(t2)H(t1) relabel t1 ↔ t2

=

∫ t

−∞dt1

∫ t

t1

dt2TH(t1)H(t2) no change because of T instruction .

(34)


Of course, it is also trivially true that

∫ t

−∞dt1

∫ t1

−∞dt2H(t1)H(t2) =

∫ t

−∞dt1

∫ t1

−∞dt2TH(t1)H(t2) (35)

since the H’s are already in the correct time ordering.

Thus, we can write

∫ t

−∞dt1

∫ t1

−∞dt2H(t1)H(t2)

=1

2

∫ t

−∞dt1

∫ t1

−∞dt2TH(t1)H(t2)

+1

2

∫ t

−∞dt1

∫ t

t1

dt2TH(t1)H(t2)

=1

2

∫ t

−∞dt1

∫ t

−∞dt2TH(t1)H(t2) . (36)


Given that we want to include − signs in the definition of T when fermionicfields are involved, the absence of any extra signs in the 2nd version of theabove requires that HI contains an even number of fermion factors (as inQED) and that HI be written in terms of a local interaction (all fields atsame space-time x point). With this latter point in mind we can write

S =

∞∑n=0

(−i)n

n!

∫ ∞−∞

dt1

∫ ∞−∞

dt2 . . .

∫ ∞−∞

dtnTHI(t1)HI(t2) . . .HI(tn)

=

∞∑n=0

(−i)n

n!

∫d4x1

∫d4x2 . . .

∫d4xnTHI(x1)HI(x2) . . .HI(xn) .

(37)

where the∫d4x integrals are over all space and all times.

• MS now embarks on a discussion of why it is justified to use non-interactingstates to specify |i〉 and |f〉 instead of those including the photon and loopclouds. This discussion is simply incorrect. In fact, in the interacting theory,|i〉 and |f〉 are not described using a free-particle state Foch state pictureand basis. They will be defined using creation and annihilation operators


that create single particle states in the sense that the states can exist ontheir own even in the presence of interactions. Such states will have a massthat is different from the bare mass appearing in H0. This difference arisesfrom the fact that as an isolated particle moves along in the interactingtheory, it is continually emitting and reabsorbing virtual quanta in a mannerdictated by Feynman diagrams. This cloud of virtual processes alters thephysical particle mass.

It is only these states which include these virtual particle clouds that can beused to construct a proper basis for the theory after including interactions.So let us imagine that we have creation and annihilation operators for theseisolated single particle states including their separated virtual clouds. Thestates |i〉 and |f〉 will be defined by using these operators. Effectively, wewill mimic this by using I.P. basis operators to form |i〉 and |f〉, but withthe mass appearing in the ω~p contained in the φI(x) exponential factorsbeing the full mass obtained after including the virtual particle clouds.

In addition, the creation operator, used to quantize φI(x) in the usual way,will lead to a strange normalization of the states |i〉 and |f〉 when they aredefined using this same creation operator. The state created will have asingle particle component present with some reduced normalization Z < 1.Very roughly this means that |~p〉 = a†(~p)|Ω〉, where |Ω〉 is now the vacuum


in the presence of interactions, has normalization 〈~p|~p〉 = Z < 1. This factorof Z arises because the creation operator a† appearing in the field φI(x)(which we are assuming has conventional normalization, as required for thestandard canonical commutation 2nd quantization) also ’knows’ that theparticle it is creating has connections to multi-particle states as well as to asingle particle state, and it is only after summing over all such possibilitiesthat full unity normalization holds.

The√Z factor arising for each single particle state used to construct |i〉

or |f〉 must be carefully treated. It used to be that Peskin and Schroedertried to give a naive discussion of this in their earlier chapters. I see thatin the latest edition of their book they have given up trying to do this, justas I will and as the Mandl-Shaw book implicitly does.

Thus, we will simply drop the√Z factors that should really be present in

the scattering formulas. Perturbatively, this can actually be justified if oneis only interested in computing so-called tree-level amplitudes in which noclosed loops of virtual particles appear. The reason is that in perturbationtheory Z ∼ 1 +O(α) (e.g. in QED) while the tree-level amplitudes definethe lowest order in perturbation theory at which a process can take place.Thus, including the correction to Z ∼ 1 is a higher order correction.

In contrast, we must use the physical mass (which includes loops ....) of


a free-particle-like state in defining our I.P. states and fields. It is not thesame as the mass of the free-particle theory.

Wicks Theorem

• We must now develop a computational technique for writing down anexpression for S given a certain form for HI(x).

• To get a non-zero contribution to Sfi we must have an appropriate numberof creation and/or annihilation operators to match those appearing in thedefinitions of the states |i〉 and |f〉.

Creation operators in the product of HI’s for particles that do not appearin |i〉 or |f〉 must be compensated by annihilation operators; the result isemission and later absorption of a virtual particle that only appears in anintermediate state (i.e. one that mediates between |i〉 and |f〉, but is notpart of either).

• The only way to make this language clear is to give a specific example.

MS chooses e−(p) + γ(k) → e−(p′) + γ(k′) at lowest possible order inperturbation theory.


You thus start with an initial state

|i〉 = c†r(~p)a†s(~k)|0〉 (38)

and end with a final state

|f〉 = c†r′(~p′)a†s′(

~k′)|0〉 . (39)

• So, now the trick is to go to the form of S given in Eq. (33) and to usean n such that there are enough HI’s in between 〈f | and |i〉 as to get anon-zero result. If you insert the minimal number, i.e. use the smallest n,then this defines the “tree-level” amplitude.

Since 〈f | has a different photon in it than does |i〉, you will need twoA fields, one to annihilate the initial photon and one to create the finalphoton. This means you will need two HI’s.

• Since the minimal substitution interaction is HI = −LI = −e : ψA/ ψ :,this also implies that that you will have an extra ψ and ψ in addition tothe ψ and ψ needed to annihilate the electron present in the initial stateand create the one present in the final state.


These extra ψ and ψ will have inside them creation and annihilationoperators that will basically annihilate one another.

• For now, let us not worry about these two extra guys and focus only on thefields necessary to get from the initial state to the final state.

The only non-zero contribution to

〈f |(−e : ψA/ ψ :x1

) (−e : ψA/ ψ :x2

)|i〉 (40)

will come from the structure (notice I am not yet specifying whether thefields below are at x1 or x2 – we will come to the appropriate mixtures):

〈f |ψ−A−ψ+A+|i〉 = 〈0|as′(~k′)cr′(~p ′)ψ−A−ψ+A+c†r(~p)a†s(

~k)|0〉 (41)

where the A+ has an a in it to “kill” the a†s(~k) used to define |i〉 and ψ+

has the c in it to “kill” the c†r(~p) that is also part of defining |i〉. Similarly,

the c† in ψ−

will “kill” the cr′(~p′) and the a† in A− will “kill” the as′(~k

′).

By “kill” I mean the operation (to give one of the 4 kills)

ai(~l)a†s(~k)|0〉 =

([ai(~l), a

†s(~k)] + a†s(

~k)ai(~l))|0〉 = δisδ~l,~k|0〉+ 0 . (42)


Here i,~l are the dummy summation variables used in defining

Aµ+(x2) =∑i,~l

1√2V |~l|

εµi (~l)ai(~l)e−il·x2 . (43)

• Note: The ordering which makes the killing simple, as given above, is thenormal ordering of the required operators.

Thus, in general, we will want to expand the S-matrix as a sum of normalproducts. The method for doing so is due to Dyson and Wick.

• First, we repeat the general definition of a normal product. LetQ,R, . . . ,Wbe operators of the type ψ±, A±, etc. Each contains a creation orannihilation operator. Then,

: QR . . .W := (−1)P × (reordering) (44)

where the reordering is such that all absorption operators (i.e. +

components) are to the right of all creation operators (i.e. − components).


The exponent P is the number of interchanges of fermion operators requiredfor this reordering.

We also require that normal ordering be a distributive operation: : RS . . .+VW . . . :=: RS . . . : + : VW . . . :.

• Now, we note that in all cases HI(x) =: A(x)B(x) . . . : is already normalordered. (Remember if A = A+ +A− then use distributive rule to do thenormal ordering.)

What appears in S is THI(x1) . . .HI(xn). We must learn how to dealwith this structure.

• First, note that (with A = A(x1) and B = B(x2))

AB− : AB :=

[A+, B−]+ , for two fermion fields[A+, B−] , otherwise

(45)

Proof, e.g. for two fermions fields:

AB− : AB : =[A

+B

++A

+B−

+A−B

++A−B−]− [A+

B+ − B−A+

+A−B

++A−B−]

= A+B−

+ B−A

+=[A

+, B−]

+. (46)


• Now, the above commutators or anticommutators are c-numbers and sotheir vacuum expectation values are the same as the numbers themselves.

So consider, for example, (remembering A+|0〉 = 0 since A+ contains theannihilation operator)

〈0|[A+, B−]+|0〉 = 〈0|[A+B− +B−A+]|0〉= 〈0|A+B−|0〉= 〈0|(A+ +A−)(B+ +B−)|0〉= 〈0|AB|0〉 (47)

where the next to last line follows since 〈0|A− = 0 and B+|0〉 = 0. Thecommutator case follows a similar sequence of steps:

〈0|[A+, B−]|0〉 = 〈0|[A+B− −B−A+]|0〉= 〈0|A+B−|0〉= 〈0|(A+ +A−)(B+ +B−)|0〉= 〈0|AB|0〉 (48)


As a result, in both cases we have

AB =: AB : +〈0|AB|0〉. (49)

• Further, we note that : AB := ± : BA :, with the − sign applying in thecase of two fermion fields, and the plus sign otherwise.

Just to make sure you understand this statement, lets do a fermion case.We must remember that two creation operators anticommute as do twoannihilation operators.

: AB : = : (A+ +A−)(B+ +B−) :

= A+B+ +A−B+ +A−B− −B−A+

= −B+A+ +A−B+ −B−A− −B−A+

= − : (B+ +B−)(A+ +A−) :

= − : BA : , (50)

where we used − : B+A− := −(−)A−B+ = A−B+.


• Using the above, it follows that if x01 > x0

2 we have

TA(x1)B(x2) = A(x1)B(x2)

= : A(x1)B(x2) : +〈0|A(x1)B(x2)|0〉= : A(x1)B(x2) : +〈0|TA(x1)B(x2)|0〉 (51)

whereas if x01 < x0

2, then

TA(x1)B(x2) = ±B(x2)A(x1)

= ± [: B(x2)A(x1) : +〈0|B(x2)A(x1)|0〉]= ± [± : A(x1)B(x2) : ±〈0|TA(x1)B(x2)|0〉]= : A(x1)B(x2) : +〈0|TA(x1)B(x2)|0〉 , (52)

i.e. exactly the same result since the ± sign changes cancel one another.In short, we have

TA(x1)B(x2) =: A(x1)B(x2) : +A(x1)B(x2) (53)


where we have introduced the shorthand notation:

A(x1)B(x2) ≡ 〈0|TA(x1)B(x2)|0〉 (54)

which is simply the Feynman propagator that we have discussed.

This Feynman propagator is, of course, only non-zero when A and B havecompensating creation and annihilation operators. Sometimes the abovestructure is also called the “contraction” of the two fields.

• The Feynman propagators we have so far considered are:

φ(x1)φ(x2) = i∆F (x1 − x2)

φ(x1)φ†(x2) = φ†(x2)φ(x1) = i∆F (x1 − x2)

ψα(x1)ψβ(x2) = −ψβ(x2)ψα(x1) = iSFαβ(x1 − x2)

Aµ(x1)Aν(x2) = iDµν

F (x1 − x2) . (55)


• We now need to generalize this beyond just 2 operators in the followingway. We define a generalized normal product containing many operators,some of which have been contracted, as

: ABCDEF . . . JKLM . . . :

= (−1)PAKBCEL . . . : DF . . . JM . . . : . (56)

Here, P is the number of interchanges of neighboring fermion operatorsrequired to change the order ABC . . . to AKB . . .; for example,

: ψα(x1)ψβ(x2)Aµ(x3)ψγ(x4)ψδ(x5) :

= (−1)ψβ(x2)ψδ(x5) : ψα(x1)Aµ(x3)ψγ(x4) : (57)

• With this definition, Wick has proven the following generalization of

TAB =: AB : +AB (assuming that none of the times are the same inthe operators):

TABCD . . .WXY Z


= : ABCD . . .WXY Z :

+ : ABC . . . Y Z : +all other single contraction cases

+ : ABCDEF . . . Y Z : +all other double contraction cases

+ . . .

+all maximal contraction cases . (58)

In the maximal contraction cases, if there are matching fields then no fieldwill be left over.

The proof of this theorem is by induction and can be found in the old bookby Bjorken and Drell on Relativistic Field Theory (2nd volume). Anotherform of the inductive proof appears at the end of Peskin and Schroeder,section 4.3. We will not go through the proof here, but will give someillustration below of how it works in a specific case of interest.

• Finally, we must figure out what the correct procedure is for the case ofinterest where we are dealing with

THI(x1) . . .HI(xn) = T: AB . . . :x1 . . . : AB . . . :xn . (59)


Let us consider a simplified version of this sort of structure. For x01 > x0

2

we have, assuming bosonic fields for simplicity,

TA(x1) : B(x2)C(x2) := (A+ +A−)(B+C+ +B−C− +B−C+ + C−B+)

= A−(B+C+ +B−C− +B−C+ + C−B+)

+(B+C+ +B−C− +B−C+ + C−B+)A+

+[A+, B−]C− +B−[A+, C−] + [A+, B−]C+ + [A+, C−]B+

= : ABC : +[A+, B−]C + [A+, C−]B

= : ABC : +〈0|TAB|0〉C + 〈0|TAC|0〉B

= : ABC : +ABC +ACB (60)

where we of course wrote : BC : in normal ordered form to begin with, andthen moved the A+ part of the A field from left to right until we also gotcomplete normal ordering. In so doing, we had to cross the B− and C−

fields with which A+ has non-zero contractions (assuming the same kindof field). But, it was never necessary to commute a B component past aC component, since they were already in normal ordering. In the above,


we also used the fact that [A+, B−] = 〈0|TAB|0〉 by virtue of the factthat (by our initial assumption) A had later time than B. A very similarproof would have given the same general result had the time ordering beenreversed.

The net result in all cases is that Eq. (58) applies with the proviso that noequal time contractions are to be included, which in the present contextmeans that fields inside the same HI are not to be contracted.

• You should note that this result is critically dependent upon using : HI :(i.e. normal ordering the interaction Hamiltonian from the beginning) ratherthan HI. In the latter case, you would include equal time contractions inapplying Wicks theorem. Using normal ordering for HI, in a certain sense,obscures what is really happening.

Peskin and Schroeder discuss the situation without normal ordering HI. Inthis PS approach (also quite common in other texts), all the contractions offields in the same HI with one another are kept. Such contractions, as wealready know, are associated with infinities. We have discussed how suchinfinities can be thrown away in the linear free particle context (i.e. whenconsidering just one H0 in the free particle case). It is quite another matterto show that they don’t matter when considering HI multiply repeated


inside the S matrix.

In fact, such a proof can be carried out and is contained in Sec. 4.4 ofPeskin-Schroeder. In order to understand this proof, you must follow thedevelopment of the concept of the true vacuum state. The true vacuumstate contains all sorts of infinities associated with not normal ordering HI.But, when computing an S matrix element, the infinities arising in the Smatrix calculation associated with not normal ordering HI are canceled bythe infinities associated with simply defining the true interacting vacuum|Ω〉 relative to the non-interacting vacuum |0〉.

Pictorially, the vacuum should be visualized as a bare vacuum plus all kindsof Feynman graphs containing various sorts of closed bubbles, the latterarising only when we include contractions of fields within the same HI.Meanwhile any given S matrix calculation will have a “connected” partassociated with the process at hand, multiplied by this same collection ofclosed bubble graphs. But the closed bubble factor is simply absorbed intothe definition of the “true” vacuum and thus does not affect the final resultfor the physical calculation.

Well, I am sure that all this discussion is somewhat mysterious; it is only in230C that it will really be clarified. For now, we must be satisfied with thefact that at tree-level all the naive manipulations are ok.


Feynman Diagrams and QED rules

Calculation of the Matrix element

• Let us return to the the process

e−(p) + γ(k)→ e−(p′) + γ(k′) (61)

process. We wish to evaluate

〈f |S|i〉 = 〈0|as′(~k′)cr′(~p ′)Sc†r(~p)a†s(~k)|0〉 (62)

with the part of S that we keep being the minimal version capable ofyielding the process. As described earlier this is what MS calls S(2), i.e. the2nd order component:

S(2) = −e2

2

∫d4x1

∫d4x2T: ψ1A/ 1ψ1 :: ψ2A/ 2ψ2 : (63)


and, in fact, only certain components of this S(2) that emerge in the Wickexpansion survive: namely, those with two contracted fermion fields withthe others left over to “kill” the creation and annihilation operators definingthe initial state and the hermitian conjugate of the final state:

S(2) 3 −e2

2

∫d4x1

∫d4x2 : ψ1A/ 1ψ1ψ2A/ 2ψ2 :

−e2

2

∫d4x1

∫d4x2 : ψ2A/ 2ψ2ψ1A/ 1ψ1 : (64)

From the symmetry under 1 ↔ 2 above, it should be obvious that thesetwo terms are equal and that we need only evaluate one of these two termswith a factor of −e2 — let us take the 1st term to evaluate.

• In this 1st term we must employ the particular fields components ψ−1 3 c†

and ψ+2 3 c. We also need one A+ 3 a and one A− 3 a†, leaving:

S(2) 3 −e2

∫d4x1

∫d4x2 : ψ

−1 A/

−1 ψ1ψ2A/

+2 ψ

+2 :

−e2

∫d4x1

∫d4x2 : ψ

−1 A/

+1 ψ1ψ2A/

−2 ψ

+2 : (65)


• Now we need a few intermediate results related to the “killing” process.Our first example is:

ψ+(x)c†r(~p)|0〉 =∑~q,t

1√2V E~q

ct(~q)ut(~q)e−iq·xc†r(~p)|0〉

=1√

2V E~pur(~p)e−ip·x|0〉 (66)

after using [ct(~q), c†r(~p)]+ = δrtδ~q,~p.

Similarly, we find

〈0|cr′(~p ′)ψ−

(x) = 〈0|cr′(~p ′)∑~q,t

1√2V E~q

c†t(~q)ut(~q)e+iq·x

=1√

2V E~p ′ur′(~p

′)e+ip′·x〈0| . (67)


Finally, we have

Aµ+(x)a†s(~k)|0〉 =

∑t,~q

1√2V |~q|

εµt (~q)at(~q)e−iq·xa†s(

~k)|0〉

=1√

2V |~k|εµs (~k)e−ik·x|0〉 , (68)

and

〈0|as′(~k′)Aµ−(x) = 〈0|as′(~k′)∑t,~q

1√2V |~q|

εµt (~q)a†t(~q)e+iq·x

=1√

2V |~k′|εµs′(

~k′)eik′·x〈0| . (69)

So let us use these results to first evaluate the contribution of the first termin the part of S(2) that appears in Eq. (65). Using 〈0|0〉 = 1, we obtain,dropping temporarily the 1√

2V ...factors:


〈0|as′(~k′)cr′(~p

′)[−e2

∫d

4x1

∫d

4x2 : ψ

−1 A/

−1 ψ1ψ2A/

+2 ψ

+2 :]c†r(~p)a

†s(~k)|0〉

= 〈0|as′(~k′)cr′(~p

′)[−e2

∫d

4x1

∫d

4x2ψ

−1 A/

−1 ψ1ψ2A/

+2 ψ

+2

]c†r(~p)a

†s(~k)|0〉

= 〈0|as′(~k′)cr′(~p

′)[−e2

∫d

4x1

∫d

4x2ψ

−1 γµA

µ−1 ψ1ψ2γνA

ν +2 ψ

+2

]c†r(~p)a

†s(~k)|0〉

= ur′(~p′)γµε

µ

s′(~k′)[−e2

∫d

4x1

∫d

4x2ψ1ψ2e

ix1·(k′+p′)−ix2·(k+p)

]γνε

νs (~k)ur(~p)

= ur′(~p′)ε/ s′(

~k′)[−e2

∫d

4x1

∫d

4x2ψ1ψ2e

ix1·(k′+p′)−ix2·(k+p)

]ε/ s(~k)ur(~p)

= ur′(~p′)ε/ s′(

~k′)[−e2

∫d

4x1d

4x2

i

(2π)4

∫d

4qq/ +m

q2 −m2e−iq·(x1−x2)

eix1·(k

′+p′)−ix2·(k+p)]ε/ s(~k)ur(~p)

= ur′(~p′)ε/ s′(

~k′)[−e2 i

(2π)4

∫d

4qq/ +m

q2 −m2(2π)

4δ

4(k′

+ p′ − q)(2π)

4δ

4(q − p− k)

]ε/ s(~k)ur(~p)

= −e2(2π)

4δ

4(p′

+ k′ − p− k)ur′(~p

′)ε/ s′(

~k′)

[iq/ +m

q2 −m2

]q=p+k

ε/ s(~k)ur(~p)

×1√

2V E~p√

2V E~p ′√

2V |~k|√

2V |~k′|(70)


where we have extended the d4x1 and d4x2 integrals over all space andtime, but have temporarily kept the volume normalization for the externalfactors (writing them explicitly in the last line of the equation after havingdropped them earlier). This latter will be convenient when it comes timeto compute a cross section.

Of course, we could have kept finite volume in all the non-external stuffas well and arrived at an exactly analogous answer. The integral overcontinuous ~q values traces all the way back to QFT-I where we wrote forthe scalar field case

[φ+(x), φ−(y)] =1

2V

∑~k~k′

1√ω~kω~k′

[a(~k), a†(~k′)]e−ik·x+ik′·y

V→∞→1

2(2π)3

∫d3~k

ω~ke−ik·(x−y)

≡ i∆+(x− y) . (71)

We could have simply written i∆+(x − y) = 12V

∑~k

1ω~ke−ik·(x−y). When

we reexpressed as a contour integral, the dk0 would still have been a

continuous integral, but the d3~k(2π)3 would have been replaced by 1

V

∑~k . And


in the algebra on the preceding page, the∫d3x1

∫d3x2 would yield∫

d3x1

∫d

3x2e

ix1·(k′+p′)−ix2·(k+p)−iq·(x1−x2)

= V2δ~k ′+~p ′,~q δ~q,~p+~k , (72)

(the dx01dx

02 would still be over all time and you would have Dirac delta

functions from these integrals still). The final stages of the calculationwould have then been:

ur′(~p′)ε/ s′(

~k′)[−e2

∫d

4x1

∫d

4x2ψ1ψ2e

ix1·(k′+p′)−ix2·(k+p)

]ε/ s(~k)ur(~p)

= ur′(~p′)ε/ s′(

~k′)[−e2

∫d

4x1d

4x2

i

(2π)V

∫dq

0∑~q

q/ +m

q2 −m2e−iq·(x1−x2)

eix1·(k

′+p′)−ix2·(k+p)]ε/ s(~k)ur(~p)

= ur′(~p′)ε/ s′(

~k′)[−e2 i

(2π)V

∫dq

0∑~q

q/ +m

q2 −m2(2π)δ([k

′+ p′ − q]

0)(2π)δ([q − p− k]

0)]

×V

2δ~k ′+~p ′,~q δ~q,~p+~k

ε/ s(~k)ur(~p)

= −e2(2π)V δ([p

′+ k′ − p− k]

0)δ~p ′+~k ′,~p+~k

ur′(~p′)ε/ s′(

~k′)

[iq/ +m

q2 −m2

]q=p+k

ε/ s(~k)ur(~p)

×1√

2V E~p√

2V E~p ′√

2V |~k|√

2V |~k′|(73)


This shows that the infinite volume limit corresponds to

V δ~a,~b→ (2π)3δ(~a−~b) . (74)

In any case, putting aside the normalization factors and exact versionof three momentum conservation delta function, the structure we haveobtained makes a lot of sense. We observe an over all coupling strengthof e2, overall 4-momentum conservation δ functions, and a very specificalgebraic structure that can be associated with a momentum-space Feynmandiagram as shown in the first diagram of the figure.

!

"#$&%('*))+,

-

.

/0 1

/20

3 4561

7

8 5 19

/#5$&%'*))+:,

.

-

/(0

/(0

Figure 1: The two Feynman diagrams contributing to e−γ → e−γ.


The 2nd term of Eq. (65) gives a similar result. We just have to be a littlemore careful with the Dirac structure. We have

〈0|as′(~k′)cr′(~p

′)[−e2

∫d

4x1

∫d

4x2 : ψ

−1 A/

+1 ψ1ψ2A/

−2 ψ

+2 :]c†r(~p)a

†s(~k)|0〉

= 〈0|as′(~k′)cr′(~p

′)[−e2

∫d

4x1

∫d

4x2 : ψ

−1 γνA

ν +1 ψ1ψ2γµA

µ−2 ψ

+2 :]c†r(~p)a

†s(~k)|0〉

= 〈0|as′(~k′)cr′(~p

′)[−e2

∫d

4x1

∫d

4x2A

µ−2 ψ

−1 γνψ1ψ2γµψ

+2 A

ν +1

]c†r(~p)a

†s(~k)|0〉

= εs′µ

(~k′)ur′(~p

′)γν[−e2

∫d

4x1

∫d

4x2ψ1ψ2e

ix2·k′+ix1·p

′−ix1·k−ix2·p]γµur(~p)εs

ν(~k)

= ur′(~p′)ε/ s(~k)

[−e2

∫d

4x1d

4x2

i

(2π)4

∫d

4qq/ +m

q2 −m2e−iq·(x1−x2)

eix2·k

′+ix1·p′−ix1·k−ix2·p

]ε/ s′(

~k′)ur(~p)

= ur′(~p′)ε/ s(~k)

[−e2 i

(2π)4

∫d

4qq/ +m

q2 −m2(2π)

4δ

4(p′ − k− q)(2π)

4δ

4(q + k

′ − p)]

ε/ s′(~k′)ur(~p)

= −e2(2π)

4δ

4(p′

+ k′ − p− k)ur′(~p

′)ε/ s(~k)

[iq/ +m

q2 −m2

]q=p−k′

ε/ s′(~k′)ur(~p)

×1√

2V E~p√

2V E~p ′√

2V |~k|√

2V |~k′|(75)


• So, the rules that you would use to get the two contributions to 〈f |S|i〉are the following.

1. Draw the diagrams involving two external photons, a final e− and aninitial e− (the latter connected by a continuously uni-directional e−-fermion flow line) and only e−→ e−γ “vertices” as specified by the formof HI and having no loops.You should quickly understand that you must have exactly two verticesand that there are only the two topologically distinct (including the arrowon the electron line) diagrams that we have drawn.

2. Now write down an algebraic expression for each of the two diagramsusing the following rules. For diagram 1 we do the following.

(a) For the outgoing e−(p′, r′) write on the far left of the Dirac structurea ur′(~p

′).(b) With the outgoing external photon associate a polarization vector

εµs′(~k′) (in a complex polarization basis this would actually be ε∗).

(c) Attach the outgoing photon εµs′(~k′) to a ieγµ written just to the right

of the ur′(~p′).

(d) Then write down the momentum space virtual e− propagator

iq/ +m

q2 −m2 + iε=

i

q/ −m+ iε(76)


with the momentum q carried by the virtual e− given by momentumconservation, i.e. q = p+ k for the first diagram.

(e) Then comes the ieγν to which we attach the incoming photon ενs(~k).

(f) Finally write down to the far right of the Dirac structure the spinorur(~p) associated with the incoming e−.

3. For the 2nd diagram, we follow a very similar procedure except for thefollowing.

(a) We first attach the incoming ενs(~k) to a ieγν just to the right of the

ur′(~p′).

(b) We employ the e− propagator using the appropriate momentumconservation value of q = p− k′.

(c) We then attach εµs′(~k′) for the outgoing photon to a ieγµ written to

the right of the Feynman propagator.(d) We finish the Dirac structure with the ur(~p).

4. These rules define the amplitudes (denoted Ma and Mb in MS).

Ma = −e2ur′(~p′)ε/ s′(~k

′)

[iq/ +m

q2 −m2

]q=p+k

ε/ s(~k)ur(~p) (77)

Mb = −e2ur′(~p′)ε/ s(~k)

[iq/ +m

q2 −m2

]q=p−k′

ε/ s′(~k′)ur(~p) (78)


The full amplitude is the sum M =Ma +Mb, and is to be multipliedby (2π)4δ4(p′ + k′ − p − k), the overall momentum conservation deltafunction, and the product of the 1/

√2V E normalization factors.

• One of the key ingredients above is the ieγµ “vertex”. Algebraically, youcan think of this factor as coming from:

1. −i for each power of HI appearing in the required level of S-matrixexpansion;

2. −eγµ as appearing in each HI in association with constructing the A/appearing therein.

This vertex is associated with each interaction of a photon and a fermionwhether these are real or virtual.

In the above case, the initial or final e− is real, the photons are real, butthe intermediate e− is virtual.

Regardless, one writes down (assuming incoming or outgoing electrons, andnot positrons)

ur(p)ieγµus(q)εµt (k) (79)

for each vertex, even when one of the particles is virtual.


In addition, each internal virtual propagator is then associated with thei/(q2 −m2) factor. We will shortly understand why I have left out theq/ +m part.

In the above process, this procedure works as follows, using the example ofdiagram 1.

1. We have a εµs′(k′)ur′(~p

′)ieγµut(q) for the upper vertex. (I am free towrite the ε, which is just a number, anywhere I want.)

2. We have a ut(q)ieγνur(~p)ενs(~k) for the lower vertex.

3. We have the i/(q2 − m2) factor with q = p + k by momentumconservation.

4. We sum over possible intermediate spin states t for the virtual electron.5. We use

∑t ut(q)ut(q) = q/ +m which is true even if we allow the q to

have an off-shell energy.6. The result,

∑t

εµs′(~k′)ur′(~p

′)ieγµut(q)i

q2 −m2ut(q)ieγνur(~p)ενs(

~k)

= εµs′(~k′)ur′(~p

′)ieγµi

[q/ +m

q2 −m2

]q=p+k=p′+k′

ieγνur(~p)ενs(~k)


= ur′(~p′)ieε/ s′(~k

′)i

[q/ +m

q2 −m2

]q=p+k=p′+k′

ieε/ s(~k)ur(~p) , (80)

is precisely what we had before.

• There are a few more rules that we can only get by considering a few otherprocesses. For the rules for external positrons, we need the appropriate“killing” operations. For an incoming positron contained in |i〉 we have:

ψ+

(x)d†r(~p)|0〉 =∑~q,t

1√2V E~q

dt(~q)vt(~q)e−iq·xd†r(~p)|0〉

=1√

2V E~pvr(~p)e−ip·x|0〉 (81)

after using [dt(~q), d†r(~p)]+ = δrtδ~q,~p.

Similarly, for an outgoing positron in |f〉, we find

〈0|dr′(~p ′)ψ−(x) = 〈0|dr′(~p ′)∑~q,t

1√2V E~q

d†t(~q)vt(~q)e+iq·x


=1√

2V E~p ′vr′(~p

′)e+ip′·x〈0| . (82)

Note how these differ from the corresponding electron results. Where wehave a u(~p) for an incoming electron we find v(~p) for an incoming positron,and where we had u(~p) for an outgoing electron, we find v(~p) for anoutgoing positron. As a result, the v and u spinors both appear at the endof the line directed in the direction of electron (i.e. fermion vs. antifermion)flow. For instance, an incoming positron has the arrow of fermion flowdirected in the outgoing direction, and one writes v.

• As an example of how to use the above results, let us consider e+(p)γ(k)→e+(p′)γ(k′) scattering. There would be the same two types of contributions,but we would have to focus on different contributions to the S matrix. Onecontribution would be:

〈0|as′(~k′)dr′(~p

′)[−e2

∫d

4x1

∫d

4x2 : ψ

+1 α

[A/−1 ψ1ψ2A/

+2

]αβψ−2 β

:]d†r(~p)a

†s(~k)|0〉

= 〈0|as′(~k′)dr′(~p

′)[+e

2∫d

4x1

∫d

4x2ψ−2 β

[A/−1 ψ1ψ2A/

+2

]αβψ

+1 α

]d†r(~p)a

†s(~k)|0〉

= 〈0|as′(~k′)dr′(~p

′)[+e

2∫d

4x1

∫d

4x2ψ−2 β

[γµA

µ−1 ψ1ψ2γνA

ν +2

]αβψ

+1 α

]d†r(~p)a

†s(~k)|0〉


= vr′(~p′)β

[γµε

µ

s′(~k′)[+e

2∫d

4x1

∫d

4x2ψ1ψ2e

ix2·p′+ix1·k

′−ix1·p−ix2·k]γνε

νs (~k)

]αβvr(~p)α

= vr(~p)ε/ s′(~k′)[+e

2∫d

4x1

∫d

4x2ψ1ψ2e

ix2·p′+ix1·k

′−ix1·p−ix2·k]ε/ s(~k)vr′(~p

′)

= vr(~p)ε/ s′(~k′)[+e

2∫d

4x1d

4x2

i

(2π)4

∫d

4qq/ +m

q2 −m2e−iq·(x1−x2)

eix2·p

′+ix1·k′−ix1·p−ix2·k

]ε/ s(~k)vr′(~p

′)

= vr(~p)ε/ s′(~k′)[+e

2 i

(2π)4

∫d

4qq/ +m

q2 −m2(2π)

4δ

4(k′ − p− q)(2π)

4δ

4(q + p

′ − k)]

ε/ s(~k)vr′(~p′)

= +e2(2π)

4δ

4(p′

+ k′ − p− k)vr(~p)ε/ s′(

~k′)

[iq/ +m

q2 −m2

]q=k′−p

ε/ s(~k)vr′(~p′)

×1√

2V E~p√

2V E~p ′√

2V |~k|√

2V |~k′|(83)

Note how, in going from the 1st line to the 2nd line above, I moved the

ψ+

1 α all the way to the right, past the [. . .]αβ matrix and past ψ−2 β. Thisis allowed since I retain proper Dirac matrix contractions so long as I keeptrack of the α index sums later, as I do. A similar statement applies tomoving the ψ−2 β over to the left.

The final expression above can be mapped to the second of the Feynman


diagrams shown in the figure below.

"!

#$%'&)(+**,-

.

/

021 34

01 5

6 5 "!

7$%'&)(8**,-

/

.

091 3

0:1 3;

Figure 2: The two Feynman diagrams contributing to e+γ → e+γ.

This figure will be confusing at first. The arrows show the direction offermion flow. But we must remember that the incoming (antifermion)e+ has momentum p entering the diagram from the bottom right, andthe outgoing e+ has momentum p′ exiting the diagram. I.e. anti-fermion momenta and fermion flow are oppositely directed. Momentumconservation for the virtual propagator is correctly given (for the directionof fermion flow) by k′ − p for the 2nd diagram. That is, we always writethe Feynman propagator in the canonical form where q is given by themomentum carried in the fermionic direction, rather than the antifermionicdirection.

I will not work out the details of the 2nd contribution, which gives you thefirst (left-hand) diagram of the above figure. Using our Feynman rules we


will get

+e2(2π)

4δ

4(p′

+ k′ − p− k)vr(~p)ε/ s(~k)

[iq/ +m

q2 −m2

]q=−k−p

ε/ s′(~k′)vr′(~p

′)

×1√

2V E~p√

2V E~p ′√

2V |~k|√

2V |~k′|. (84)

A homework problem will be to verify this result in the same kind of detailas given for the right-hand diagram.

• One important note on the above. We were careful to keep the signresulting from strict application of Wick’s theorem, which ended up givingus a +e2 overall sign for this positron case as opposed to the −e2 overallsign for the electron case.

Such an overall sign is not actually observable, since we will eventuallysquare these amplitudes.

• However, one should be very careful with signs in cases where differentdiagrams correspond to the exchange of identical fermions in the final state

An example of this, which is also useful for getting the correct result for avirtual photon propagator, is provided by e−e−→ e−e− scattering.


For this process, we need two ψ fields and two ψ fields in order “kill”the creation operators defining the incoming and outgoing states. More

precisely, we will need 2 ψ+’s and 2 ψ−

’s. This means we are again dealingwith S(2) the 2nd order term in the S-matrix expansion, which I repeatbelow. Note that since each of the two HI’s required contains an A field,we will have to contract these two A fields against one another in order togenerate a non-zero contribution to the scattering process of interest. Therelevant stuff is thus:

S(2) = −e2

2

∫d4x1

∫d4x2T: ψ1A/ 1ψ1 : : ψ2A/ 2ψ2 :

3 −e2

2

∫d4x1

∫d4x2 : ψ

−1 α

(γµ)αβψ+1 βAµ1A

ν2ψ−2 δ

(γν)δρψ+2 ρ

:

= −e2

2

∫d4x1

∫d4x2ψ

−2 δψ−1 α

(γµ)αβAµ1A

ν2(γν)δρψ

+1 βψ+

2 ρ, (85)

where two changes of sign occurred when we passed two fermionic operatorspast one another in order to get to the explicitly normal ordered form given.

We must now adopt a convention for our |i〉 and |f〉 states. We use p, rand k, s for the incoming guys and p′, r′ and k′, s′ for the outgoing guys


so that we define

|i〉 = c†r(~p)c†s(~k)|0〉 , |f〉 = c†r′(~p

′)c†s′(~k′)|0〉 ,

⇒ 〈f | = 〈0|cs′(~k′)cr′(~p ′) , (86)

where the precise order of the c† operators must be carefully adhered tobecause of the minus signs from various anticommutators.

Now, let us consider the structure

ψ+2 ρc†r(~p)c†s(

~k)|0〉

=∑t,~l

1√2V E~l

ct(~l)ut(~l)ρe−il·x2c†r(~p)c†s(

~k)|0〉 (87)

Contained within this sequence is the structure

ct(~l)c†r(~p)c†s(

~k)|0〉 =[δtrδ~l~pc

†s(~k)− c†r(~p)ct(~l)c

†s(~k)]|0〉

=[δtrδ~l~pc

†s(~k)− c†r(~p)δtsδ~l~k

]|0〉 . (88)


If we now bring into play the

ψ+1 β

=∑u,~q

1√2V E~q

cu(~q)uu(~q)βe−iq·x1 (89)

we will need the structure

cu(~q)ct(~l)c†r(~p)c†s(

~k)|0〉 = cu(~q)[δtrδ~l~pc

†s(~k)− c†r(~p)δtsδ~l~k

]|0〉

=[δtrδ~l~pδusδ~q~k − δurδ~q~pδtsδ~l~k

]|0〉 . (90)

Obviously, the two terms differ by the interchange of t,~l ↔ u, ~q or,equivalently, r, ~p ↔ s,~k and the minus sign which is keeping track of thefermionic statistics.

Using these δ functions to eliminate the∑u,~q and

∑t,~l yields

1√2V E~p

√2V E~k

[e−ik·x1−ip·x2us(~k)βur(~p)ρ

− e−ip·x1−ik·x2ur(~p)βus(~k)ρ

](91)


These clearly differ by the interchange r, ~p ↔ s,~k and the fermionicstatistics sign.

We would now have to carry out an exactly equivalent game for the lefthand part of the expression in Eq. (85). Without giving the details, I hopeit is relatively straightforward to understand the result. One simply has toswitch everything to primes and do the barring (which includes a complexconjugation of the exponential factors and u → u) and change the Diracindices appropriately (β → α and ρ→ δ):

〈0|cs′(~k′)cr′(~p ′)ψ−2 δψ−1 α

=1√

2V E~p ′√

2V E~k′

[e+ik′·x1+ip′·x2us′(~k

′)αur′(~p′)δ

− e+ip′·x1+ik′·x2ur′(~p′)αus′(

~k′)δ

]. (92)

Note the antisymmetry under r′, ~p ′ ↔ s′, ~k ′ corresponding to Fermistatistics. This and the initial-state antisymmetry under r, ~p ↔ s,~kguarantee that we get zero if the initial or final electrons are in exactly thesame physical state.


We now combine the results of Eqs. (91) and (92) with the remainder ofthe stuff in Eq. (85) to obtain the result

〈f |S(2)|i〉 =−e2

2

1√2V E~p ′

√2V E~k′

√2V E~p

√2V E~k

∫d

4x1

∫d

4x2 ×

[e

+ik′·x1+ip′·x2us′(~k′)αur′(~p

′)δ − e

+ip′·x1+ik′·x2ur′(~p′)αus′(

~k′)δ

]

(γµ)αβ

[i

(2π)4

∫d

4qe−iq·(x1−x2) −gµν

q2 + iε

](γν)δρ[

e−ik·x1−ip·x2us(~k)βur(~p)ρ − e

−ip·x1−ik·x2ur(~p)βus(~k)ρ

](93)

where we inserted the Fourier representation of Aµ1Aν2 = iDµν

F (x1 − x2).

Of the four terms in the above expression, two are equal to correspondingones of the remaining two. This will cancel the 1

2appearing in e2/2. This

always happens in QED. We saw another example of this in our earliercalculations. There is equivalence under interchange of the x1 and x2vertices (i.e under H1 ↔H2). Let us focus on two inequivalent terms. Webegin with (dropping some external factors for the moment) taking the 1st


term in each of the large brackets of the above expression:

∫d

4x1

∫d

4x2

[e

+ik′·x1+ip′·x2us′(~k′)αur′(~p

′)δ

](γµ)αβ

[i

(2π)4

∫d


q2 + iε

](γν)δρ


]

= (2π)4δ

4(p + k− p′ − k′)us′(~k

′)γµus(~k)

[−igµνq2 + iε

]q=k′−k=p−p′

ur′(~p′)γνur(~p) (94)

where we did the d4x1 and d4x2 integrals to get (2π)4δ4(k′ − q − k)and (2π)4δ4(p′ + q − p) respectively and used one of these to do the d4qintegral. Now, we take the 2nd term of the 1st large bracket and the 1stterm of the 2nd large bracket. We have

∫d

4x1

∫d

4x2

[−e+ip′·x1+ik′·x2ur′(~p

′)αus′(

~k′)δ

](γµ)αβ

[i

(2π)4

∫d


q2 + iε

](γν)δρ


]

= −(2π)4δ

4(p + k− p′ − k′)ur′(~p

′)γµus(~k)

[−igµνq2 + iε

]q=p′−k=p−k′

us′(~k′)γνur(~p)(95)


where we did the d4x1 and d4x2 integrals to get (2π)4δ4(p′ − q − k)and (2π)4δ4(k′ + q − p) respectively and used one of these to do thed4q integral. A quick inspection of Eq. (94) compared to Eq. (95) showsthat the relation corresponds to interchanging the r′, ~p ′ final state e− withthe s′, ~k′ final state e− and introducing a relative minus sign between thetwo contributions. The remaining two contributions obtained by using the2nd term of the 2nd large bracket in Eq. (93) simply duplicate the twocontributions that we have already obtained. Thus, after extracting thestandard (2π)4δ4(p′ + k′ − p − k) overall momentum conservation factorwith its (2π)4 factor and the four 1/

√2V E factors, we end up with the

two invariant amplitudes:

Ma = −e2us′(~k′)γµus(~k)

[−igµνq2 + iε

]q=k′−k=p−p′

ur′(~p′)γνur(~p)

Mb = +e2ur′(~p′)γµus(~k)

[−igµνq2 + iε

]q=p′−k=p−k′

us′(~k′)γνur(~p)(96)

the factor of 12

having been canceled by the above-noted factor of 2duplication. The relative − sign becomes part of our Feynman rules:

– Whenever there are two contributing diagrams that are related to one


another by interchanging fermions of the same type, one writes the sametype of expression for each but introduces a relative minus sign.

We note that while the relative sign betweenMa andMb is determined, thesigns of both could be switched without any physical impact in computingprobabilities (related to |Ma+Mb|2). The overall sign is pure convention,related to how we defined |i〉 and |f〉. Had we reordered the c† operatorsin the definition of one of these two states, both Ma and Mb would havechanged sign.

The Feynman diagrams associated with Ma and Mb are shown in thefigure below.

!#"%$'&( ) )*#+-,

./'01#

.2'03

40.

5

67 !#"%$8&( ) 9 *#+-,

.:403

Figure 3: The two Feynman diagrams contributing to e−e−→ e−e−.

• So, by combining the results for the various processes we have computed


we obtain an almost complete set of Feynman rules for QED.

1. Draw all possible tree-level (lowest possible order) diagrams that can giverise to the process of interest making use of just the fermion → fermion+ photon “vertex”.In so doing, keep track of the fermion line direction. An incomingfermion line must be traced continuously to an outgoing fermion line (or,equivalently, to an incoming antifermion line).

2. For each vertex, write ieγµ, where a different Lorentz index µ must beused for each vertex and this µ must be connected either to the µ on avirtual photon propagator or to the µ of an external photon polarizationvector.

3. For an outgoing photon γ, p, s, write ε∗s(~p), where I here allow for acomplex polarization basis.

4. For an incoming photon γ, p, s write εs(~p).5. For an incoming e−, p, r write ur(~p).6. For an outgoing e−, p, r write ur(~p).7. For an incoming e+, p, r write vr(~p).8. For an outgoing e+, p, r write vr(~p).9. For the positron cases, note that the momentum direction will be opposite

the fermion arrow direction.


10. For an internal (virtual) electron propagator write

iq/ +m

q2 −m2 + iε(97)

where q is the momentum in the direction of the fermion arrow and thevalue of q is obtained by momentum conservation at the vertices.

11. For an internal photon propagator, write

i−gµν

q2 + iε(98)

where q can be chosen in any direction and is given by momentumconservation at the vertices.

12. A fully contracted Dirac structure should be constructed for each fermionline, beginning with a u or v to the far left of the expression and workingback against the fermion arrow flow direction until terminating on a u orv spinor.Along the way, one will encounter first a vertex and then, possibly, aninternal fermion line, and then, possibly, another vertex, and so forthuntil ending with a vertex and then the final u or v.


13. If there are diagrams that differ by simply interchanging:– two initial e−’s;– or two initial e+’s;– or two final state e−’s;– or two final state e+’s;– or an initial e+ with a final e−;– or a final e+ with an initial e−;then a relative minus sign should be introduced between the correspondingFeynman amplitudes.

14. For each closed loop, there will be an integration over an unconstrainedmomentum, call it qi. One should perform the integral 1

(2π)4

∫d4qi for

each such unconstrained loop momentum.If the closed loop is a continuous fermion line, an explicit minus signshould be introduced.

• A note on quickly checking the sign of the vertex. We began with L =ψiD/ ψ with Dµ = ∂µ + iqAµ to obtain LI = −qψA/ ψ. The conventionfor QED is that q = −e so we end up with LI = eψA/ ψ = −HI. Theevolution operator appearing in the S matrix of the interaction picture isbasically e−iHIt which implies that S contains (−iHI)

n powers. The lowest


order is just −iHI = +ieψA/ ψ. If you remove the external ψ,ψ,Aµ fieldsthe corresponding vertex in Dirac space is just +ieγµ.

• It is the last of the Feynman rules above that we have yet to derive usingan explicit example.

We choose to use the virtual e− loop correction to the photon propagatoras our basic example. The basic Feynman diagram that corresponds toour calculation is below. There, I use the label p for the free momentumassociated with the loop.

! "!

Figure 4: The e− loop correction to the photon propagator.

The computation begins with |i〉 = a†r(k)|0〉 and ends with |f〉 = a†r(k)|0〉and contains two vertices, so that means we are looking at S(2). We then


have

〈0|ar(~k)

[−e2

2

∫d

4x1

∫d

4x2T

: ψ1γµA

µ1ψ1 :: ψ2γνA

ν2ψ2 :

]a†r(~k)|0〉

(99)

To reduce this expression, we must employ Wick’s theorem. In this case,ψ1 must contract with ψ2 and vice versa. (Recall: no contractions betweenfields in same HI.) In getting to the contraction configuration given below,I must do an odd number of fermion field interchanges — hence the verycrucial extra − sign! Keeping in mind that we must have one A+ and oneA− in order to kill the a and a† operators, the above reduces to (using the1st line to simply display all the Dirac indices of the previous form above,and the 2nd line to display the particular contraction term in Wicks theoremthat will survive as well as the two different choices as to which A is A+

and which is A− – we need one of each and there are two possibilities)

〈0|ar(~k)

[−e2

2

∫d

4x1

∫d

4x2T

: ψ1α(γµ)αβA

µ1ψ1β :: ψ2ρ(γν)ρσA

ν2ψ2σ :

]a†r(~k)|0〉

= 〈0|ar(~k)

[+e2

2

∫d

4x1

∫d

4x2 : ψ2σψ1α(γµ)αβA

µ+1 ψ1βψ2 ρ(γν)ρσA

ν−2 :

]a†r(~k)|0〉

+〈0|ar(~k)

[+e2

2

∫d

4x1

∫d

4x2 : ψ2σψ1α(γµ)αβA

µ−1 ψ1βψ2 ρ(γν)ρσA

ν +2 :

]a†r(~k)|0〉


We thus need only evaluate∫d

4x1

∫d

4x2ε

νr(~k)e

ik·x2

[+e


]εµr (~k)e

−ik·x1

=

∫d

4x1

∫d

4x2ε

νr(~k)e

ik·x2

+e2[

i

(2π)4

∫d

4qe−iq·(x2−x1)

(q/ +m

q2 −m2

)σα

](γµ)αβ

[i

(2π)4

∫d

4pe−ip·(x1−x2)

(p/ +m

p2 −m2

)βρ

](γν)ρσ

εµr (~k)e−ik·x1

= ενr(~k)

+e2[

i

(2π)4

∫d

4q

(q/ +m

q2 −m2

)σα

](γµ)αβ

[i

(2π)4

∫d

4p

(p/ +m

p2 −m2

)βρ

](2π)

4δ

4(k− q + p)(2π)

4δ

4(q − p− k)(γν)ρσ

εµr (~k)

= (2π)4δ

4(k− k)(+e

2)

∫d4p

(2π)4

i( q/ +m

q2 −m2

)q=p+k

ε/ r(~k)i

(p/ +m

p2 −m2

)ε/ r(~k)

σσ

(102)

After including the 1/

√2V |~k| factors we can summarize our final result


as

〈γ(r,~k)|S(2)|γ(r,~k)〉 = (2π)4δ4(k − k)1√

2V |~k|√

2V |~k|M ,(103)

with

M = +e2

∫d4p

(2π)4Tr

[i

(q/ +m

q2 −m2

)q=p+k

ε/ r(~k)i

(p/ +m

p2 −m2

)ε/ r(~k)

](104)

Note that this result has the opposite sign compared to what you wouldwrite down just naively using ieγµ and ieγν for the two vertices and i q/ +m

q2−m2

and i p/ +mp2−m2 for the internal fermion propagators.

Note also how the Dirac structure is written by starting to the left withthe p+ k propagator and then working “backward” to the p propagator asshown in the earlier figure.

In this final form, we have also used the Trace notation. The Dirac indexstructure was such that one ended up with a continuous connection of


Dirac indices to one another as we moved along the fermion line, with theresult corresponding to the trace of the complicated Dirac matrix productindicated.


Tree-level QED Processes

Computing a cross section

• We will consider colliding two particles in the initial state |i〉 with momentapi = (Ei, ~pi), i = 1, 2. (Sorry about the confusion of notation; I am simplyfollowing MS here.) The final state |f〉 is assumed to contain N particleswith p′f = (E′f , ~p

′f), f = 1, . . . N .

• The S-matrix, which always has an overall δ4 momentum conservationfactor, can be written as

Sfi = δfi + (2π)4δ4(∑f

p′f −∑i

pi)∏i

(1

2V Ei

)1/2∏f

(1

2V E′f

)1/2

M

(105)The δ4 function in the above is obtained in the limit of an infinite timeinterval T →∞ and an infinite volume, V →∞. For finite T and V , we


would have obtained the same expression with

(2π)4δ4(∑f

p′f −∑i

pi) =lim

T →∞

V →∞δTV (

∑f

p′f −∑i

pi)

≡lim

T →∞

V →∞

∫ +T/2

−T/2

dt

∫V

d3~x exp

ix · (∑f

p′f −∑i

pi)

(106)

replaced by δTV (∑f p′f −

∑i pi). In deriving the cross section, it is helpful

to take T and V finite temporarily.

• For finite T and V , we can define the transition probability per unit of timeas

w = |Sfi|2/T (107)

which will contain the factor[δTV (

∑f p′f −

∑i pi)

]2. For one of these

factors, we revert back to the continuum result δTV (∑f p′f −

∑i pi) →

(2π)4δ4(∑f p′f −

∑i pi). For the other factor, we can take (

∑f p′f −∑

i pi) = 0 and use δTV (0) = TV , keeping the finite volume and timeinterval.


• We then get

w = V (2π)4δ4(∑f

p′f −∑i

pi)∏i

(1

2V Ei

)∏f

(1

2V E′f

)|M|2 . (108)

This is the result for transition from state i to a single final state f . It isof course vanishingly small in the continuum limit of V → ∞ as the finalstate possibilities become a continuum or possibilities.

In the V → ∞ limit, it only makes sense to compute the transitionprobability to a group of states centered on some central state; one takesstates in the interval (~p ′f , ~p

′f + d~p ′f), f = 1, . . . , N . The number states in

such an interval is given by

∏f

V d3~p ′f

(2π)3. (109)

The factors of V , above, cancel those in the denominator in the final statepart of the expression for w, leaving one V in the numerator from the δTVand two V ’s in the denominator from the initial state killing operations.


• The differential cross section, dσ, is defined to be the transition rate intothis group of final states for one scattering center divided by the flux ofincident particles incident on the volume containing this one scatteringcenter.

With our choice of normalization for the states, the volume V which weare considering contains just one scattering center. To show this (again),just compute

〈~p|~p〉 = 〈0|a(~p)a†(~p)|0〉 = δ~p~p〈0|0〉 = 1 , (110)

where we have used the finite volume result for the commutator,

[a(~p), a†(~p ′)] = δ~p~p ′ . (111)

And, the incident flux is vrel/V where vrel is the relative velocity of thecolliding particles. To see this, picture a box of volume V traveling withvelocity vrel towards a fixed (for convenience) box of volume V . Orientthe traveling box so that one face is parallel to the front face of the fixedbox. Start at t = 0 with the front face of the traveling box touching thefront face of the fixed box. If the facing faces of the two boxes have area


A, then the fraction of the traveling volume V that passes the front faceof the fixed box per unit time is vrelA/V . Since there is only one particlein the volume V , this is also the number of particles that pass through thefront face of the fixed box (the number is fractional). The flux of particlesis the number of particles per unit time passing through per unit area, i.e.it is this number divided by A, giving flux = vrel/V .

Thus, we obtain

dσ = w1

[vrelV

]

∏f

V d3~p ′f

(2π)3

= (2π)4δ4(∑f

p′f −∑i

pi)1

4E1E2vrel

∏f

d3~p ′f

(2π)32E′f|M|2 .(112)

Note that this differs from the result in MS (with the extra 2m factorsfor fermions) by virtue of the different normalization for fermion fields andoperators. Using my normalizations (which are the same as essentially allmodern treatments), there is no distinction in the basic normalizations forbosons and fermions.

Note that the δ4 function above implies that not all of the ~p ′f are


independent. In any given situation, we will have to integrate out acertain number of the ~p ′f to eat up the δ4 function.

• The above result holds in any Lorentz fame in which the colliding particlesmove collinearly.

In such a frame, and assuming the particles are moving in opposite directionto one another (it is also ok if one is a rest), the relative velocity is givenby

vrel =

∣∣∣∣|~p1|E1

+|~p2|E2

∣∣∣∣=

1

E1E2

|E2|~p1|+ E1|~p2|| . (113)

Let us compare this expression with (remember opposite directions for thecolliding particles are assumed)

(p1 · p2)2 −m2

1m22 = (E1E2 + |~p1||~p2|)2 −m2

1m22

= E21E

22 + |~p1|2|~p2|2 + 2E1E2|~p1||~p2| −m2

1m22

= (|~p1|2 +m21)(|~p2|2 +m

22) + |~p1|2|~p2|2 + 2E1E2|~p1||~p2| −m2

1m22


= 2|~p1|2|~p2|2 +m21|~p2|2 +m

22|~p1|2 + 2E1E2|~p1||~p2|

= E21|~p2|2 + E

22|~p1|2 + 2E1E2|~p1||~p2|

= (E2|~p1| + E1|~p2|)2(114)

In short, we have

E1E2vrel =[(p1 · p2)

2 −m21m

22

]1/2. (115)

This last result, substituted in the general form of dσ, implies that dσ is a

Lorentz invariant. (Recall that d3~p(2π)32E~p

is a Lorentz invariant.)

• Some useful special cases are the com system, with |~p1| = |~p2| resulting in

E1E2vrel = |E2|~p1|+ E1|~p2|| = |~pcom|Etot , (116)

(where Etot ≡ E1 + E2) and the laboratory frame in which ~p2 = 0 and

E1E2vrel = |E2|~p1|+ E1|~p2|| = |~p1|m2 , (117)

• Since the center of mass system is so common and since we often consider2→ 2 processes, it is useful to give some results specific to this case.


We use Eq. (116) and, after eliminating ~p ′2, the result that |~p ′1| = |~p ′2| ≡|~p ′com| to obtain the form

dσ =1

16|~pcom|Etot(2π)2E′1E′2

δ4(p′1 + p

′2 − p1 − p2)d

3~p′1d

3~p′2|M|

2

=1

64π2|~pcom|EtotE′1E′2

δ(E′1 + E

′2 − Etot)|~p ′1|

2d|~p ′1|dΩ

′1|M|

2

=1


[∂(E′1 + E′2)

∂|~p ′com|

]−1

|~p ′com|2dΩ′1|M|

2

=1


∂(√

(m′1)2 + |~p ′com|2 +√

(m′2)2 + |~p ′com|2)

∂|~p ′com|

−1


2

=1


[|~p ′com|E′1

+|~p ′com|E′2

]−1


2

=1

64π2Etot(E′1 + E′2)

|~p ′com||~pcom|

dΩ′1|M|

2

=1

64π2E2tot

|~p ′com||~pcom|

dΩ′1|M|

2, (118)


where in the last step we used E′1 + E′2 = Etot. Thus, our final result is(dσ

dΩ′1

)com

=1

64π2E2tot

|~p ′com||~pcom|

|M|2 . (119)

• The above and other results for differential cross sections apply irregardlessof whether or not we are dealing with identical particles.

However, if there are two or more identical particles in the final state, wemust not duplicate integration regions in obtaining the total cross section.For example, in the 2 → 2 process just computed, if the final particleswere 2 e−’s, then the case where (θ′1, φ

′1) = (α, β) is not physically

distinguishable from the case where (θ′1, φ′1) = (π − α, π + β). In this

case, we should only integrate over 0 ≤ θ′1 ≤12π, or, equivalently, we could

integrate over all θ′1 and then divide the final result by 2!:

σtotcom =

1

2

∫4π

dΩ′1

(dσ

dΩ′1

)com

. (120)

In the more general case of n identical particles in the final state, we canobtain the correct result for σtot by integrating over all of phase space andthen dividing by n!.


In a very rough sense, the 1/2! is partly compensating for the fact that fortwo identical particles there will be two Feynman diagram amplitudes thatdiffer only by interchanging the two final particles (as in e−e− → e−e−),which, again very roughly, means that the amplitude-squared would be 4times as large as compared to the case where the final particles are notidentical (e.g. e−µ−→ e−µ−).

(As we shall discuss, the e− and µ− are not identical particles. They carrya lepton identifier quantum number that distinguishes them. See MS formore details.)

• As an example of a particularly useful cross section, let us consider theprocess e+(p)e−(k)→ µ+(p′)µ−(k′).

We consider how to deal with two different types of leptons. As notedabove muons and electrons are experimentally distinguishable — e.g. theyhave different mass and have a different lepton number.

What this means in our theoretical framework is that there are creation andannihilation operators for electrons that are distinct from those for muons.


Each lepton has its own free-particle Lagrangian so that

L0 =∑

l=e,µ,τ

ψl(x)(i∂/ −ml)ψl(x) . (121)

Making the minimal substitution (with q = −e for each) gives

HI(x) = −LI(x) = −e∑l

: ψl(x)A/ (x)ψl(x) : (122)

Note that in the above there is no term like

− e : ψµ(x)A/ (x)ψe(x) : (123)

Such a term would constitute what is called a flavor-changing neutralcurrent (FCNC) interaction. Electromagnetism does not have such FCNCinteractions when constructed using the minimal substitution rule fromfree-particle Lagrangians for the individual leptons. FCNC interactions forthe leptons are predicted in the context of the weak interactions now thatwe know that neutrinos have mass and that the mass eigenstates are clearlynot the same as the Lagrangian eigenstates.


• So, now let’s get back to the cross section computation making the absolutelepton number conservation assumption.

The initial and final states are:

|i〉 = c†e(~k)d†e(~p)|0〉 , |f〉 = c†µ(~k ′)d†µ(~p ′)|0〉 . (124)

The relevant Feynman diagram is

Figure 5: The one Feynman diagram contributing to e+e− → µ+µ−. Thearrows show the directions for the momenta. For the µ+ and e+, thefermion-flow direction is opposite the indicated momentum direction.


Can you figure out from the Feynman ’rules’ (generalized to include µvertices as well as e vertices) what to write down without going throughthe derivation below?

Now, since we need a e+ and e− in the initial state we will need a ψ+e and

a ψ+

e for the killing operations. For the final state µ+ and µ− we will need

a ψ−µ and a ψ−µ for the killing operations. This means we must go to S(2)

which now includes

S(2) 3 −

e2

2

∫d

4x1

∫d

4x2

[T: ψµ(x1)A/ (x1)ψµ(x1) :: ψe(x2)A/ (x2)ψe(x2) :

+Tµ↔ e, x1 ↔ x2]

(125)

To see this, we must remember that S(2) contains

THI(x1)HI(x2) = T[HeI(x1) +HµI (x1)][HeI(x2) +HµI (x2)] (126)

and that the terms of interest then arise as the two cross terms containingthe product of HeI with HµI . Of course, the x1 ↔ x2 2nd term above givesexactly the same contribution as the 1st term and so we will work with theequivalent form

S(2) 3 −e2

∫d

4x1

∫d

4x2T: ψµ(x1)A/ (x1)ψµ(x1) :: ψe(x2)A/ (x2)ψe(x2) : (127)


For the process being considered we keep the already delineated parts ofS(2), namely:

S(2) 3 −e2

∫d

4x1

∫d

4x2T: ψ

−µ (x1)A/ (x1)ψ

−µ (x1) :: ψ

+e (x2)A/ (x2)ψ

+e (x2) :

→ −e2∫d

4x1

∫d

4x2ψ

−µ (x1)γ

ρψ−µ (x1)Aρ(x1)Aσ(x2)ψ

+e (x2)γ

σψ

+e (x2) (128)

We now insert this between our initial and final states and use the standardkilling operations. Thus, we compute (you will have noted that I am notattempting to keep spin indices on the creation and annihilation operatorsdefining |i〉 and |f〉)

〈f |S(2)|i〉 = 〈0|dµ(~p′)cµ(~k

′)

[−e2

∫d

4x1

∫d

4x2ψ

−µ (x1)γ

ρψ−µ (x1)Aρ(x1)Aσ(x2)

ψ+e (x2)γ

σψ

+e (x2)

]c†e(~k)d†e(~p)|0〉

= −e2uµ(~k

′)γρvµ(~p

′)ve(~p)γ

σue(~k)×∫

d4x1

∫d

4x2e

ix1·(k′+p′)−ix2·(k+p)

[i

(2π)4

∫d

4q−gρσq2 + iε

e−iq·(x1−x2)

]= −e2

uµ(~k′)γρvµ(~p

′)ve(~p)γ

σue(~k)

1

(2π)4

∫d

4q ×

(2π)4δ

4(k′

+ p′ − q)(2π)

4δ

4(q − p− k)

[i−gρσq2 + iε

]q=p+k


= (2π)4δ

4(p′

+ k′ − p− k)M , with

M = uµ(λ′, ~k′)(ieγ

ρ)vµ(λ

′, ~p′)ve(λ, ~p)(ieγ

σ)ue(λ,~k)

[i−gρσq2 + iε

]q=p+k

(129)

which is what you would get by “naive” application of the Feynmanrules, generalized to include muons as well as electrons. In the very lastexpression, we have included the spin/helicities of the various particles thathad been dropped until now.

• So, now let’s insert this into our 2→ 2 cross section expression:(dσ

dΩ′1

)com

=1

64π2E2tot

|~p ′com||~pcom|

|M|2 (130)

using the approximation of neglecting me and mµ, so that |~p ′|com =|~p|com = Etot/2, to obtain (dropping the ′1 notation on Ω, as is conventional)

(dσ

dΩ

)com

=1

64π2E2tot

∣∣∣uµ(λ′, ~k ′)(ieγρ)vµ(λ′, ~p ′)

ve(λ, ~p)(ieγρ)ue(λ,~k)

[ −iq2 + iε

]q=p+k

∣∣∣2 (131)


The above result is for fixed spins for all the initial and final particles.More typically, we might wish to average over initial spins (correspondingto unpolarized e+ and e− incoming beams) and sum over final spins(corresponding to not measuring the final spins). In this case, we can makesome simplifications. To show how this works, we consider for the momentthe electron subcomponent of this process:

∑λ,λ

[ve(λ, ~p)(ieγρ)ue(λ,~k)][ve(λ, ~p)(ieγρ′)ue(λ,~k)]∗ (132)

where, since ρ is a dummy index eventually to be summed over, we mustuse ρ′ for the other half of the absolute square. Now,

[ve(λ, ~p)(ieγρ′)ue(λ,~k)]∗ = ue(λ,~k)†(ieγρ′)†(ve(λ, ~p)†γ0)†

= ue(λ,~k)†γ0γ0(−ieγ†ρ′)(γ0)†ve(λ, ~p)

= ue(λ,~k)(−ieγρ′)ve(λ, ~p) (133)

so that we have (exposing Dirac indices in the 3rd line before proceeding)


∑λ,λ

[ve(λ, ~p)(ieγρ)ue(λ,~k)][ve(λ, ~p)(ieγρ′)ue(λ,~k)]∗

=∑λ,λ

[ve(λ, ~p)(ieγρ)ue(λ,~k)][ue(λ,~k)(−ieγρ′)ve(λ, ~p)]

=∑λ,λ

e2ve(λ, ~p)γ(γρ)γδue(λ,~k)δue(λ,~k)ε(γρ′)εβve(λ, ~p)β

=∑λ,λ

e2ve(λ, ~p)βve(λ, ~p)γ(γρ)γδue(λ,~k)δue(λ,~k)ε(γρ′)εβ

= e2(p/ −me)βγ(γρ)γδ(k/ +me)δε(γρ′)εβ

= e2Tr [(p/ −me)γρ(k/ +me)γρ′] . (134)

Similarly, we find (note the matching of the ρ and ρ′ Lorentz indices ismaintained within M and M∗)∑

λ′,λ′

[uµ(λ′, ~k ′)(ieγρ)vµ(λ′, ~p ′)][uµ(λ′, ~k ′)(ieγρ

′)vµ(λ

′, ~p ′)]∗

= e2Tr [(k/ ′ +mµ)γρ(p/ ′ −mµ)γρ′] . (135)


• Thus, we now need to figure out how to evaluate this kind of trace of Diracmatrices. There are a series of rules that we will slowly add to that oneneeds to derive.

1. First, just by the definition of the trace, we have

Tr[Γ1Γ2] = Tr[Γ2Γ1] (136)

for any two of our 16 matrices.2. Next, we can show that the trace of an odd number of γ matrices is

zero:Tr[γαγβ . . . γν] = 0 (137)

if there are an odd number of matrices in [. . .]. This is shown as follows:

Tr[γαγβ . . . γν]

= Tr[γ25γαγβ . . . γν] (since γ2

5 = 1)

= −Tr[γ5γαγβ . . . γνγ5] (since γ5 anticommutes with all the γα, . . .)

= −Tr[γ5γ5γαγβ . . . γν] (by the cyclic property of the trace)

= −Tr[γαγβ . . . γν] (since γ25 = 1) (138)


so that the only possibility is that the original Tr was equal to 0.3. Next, we consider

Tr[γαγβ] =1

2Tr[γαγβ + γβγα] =

1

2Tr[[γα, γβ]+]

=1

2Tr[2gαβ14×4] = 4gαβ (139)

4. A particularly important one is:

Tr[γαγβγεγδ]

= Tr[(2gαβ − γβγα)γεγδ]

= Tr[2gαβγεγδ]− Tr[γβ(2gαε − γεγα)γδ]

= Tr[2gαβγεγδ]− Tr[γβ2gαεγδ] + Tr[γβγε(2gαδ − γδγα)]

= Tr[2gαβγεγδ]− Tr[γβ2gαεγδ] + Tr[γβγε2gαδ]− Tr[γβγεγδγα]

= 8[gαβgεδ − gβδgαε + gβεgαδ]− Tr[γαγβγεγδ] . (140)

Since the last term above duplicates the starting Tr, we can move it to


the lhs of the equation and divide both sides by a factor of 2 to obtain

Tr[γαγβγεγδ] = 4[gαβgεδ − gβδgαε + gβεgαδ] (141)

Note that if we contract some momenta with some of the free γ indiceswe will get things like

Tr[p/ γβk/ γδ] = pαkεTr[γαγβγεγδ]

= 4pαkε[gαβgεδ − gβδgαε + gβεgαδ]

= 4[pβkδ − gβδp · k + pδkβ] (142)

Well, that’s all we need for the moment. The Appendix of MS has more.

• We now return to the initial spin average, final spin sum version of ourcross section:

(dσ

dΩ

)com

=1

64π2E2tot

14

∑λ,λ,λ′,λ′

∣∣∣∣∣uµ(λ′, ~k′)(ieγ

ρ)vµ(λ

′, ~p′)ve(λ, ~p)(ieγρ)ue(λ,~k)

[ −iq2 + iε

]q=p+k

∣∣∣∣∣2

(143)


We use our conversion of the spin sums to traces in the form

e2Tr [(p/ −me)γρ(k/ +me)γρ′]e

2Tr [(k/

′+mµ)γ

ρ(p/′ −mµ)γ

ρ′]

= 16e4[pρkρ′ − gρρ′(p · k +m

2e) + pρ′kρ][k

′ρp′ρ′ − gρρ

′(k′ · p′ +m

2µ) + k

′ρ′p′ρ

]

= 32e4[p · p′k · k′ + p · k′p′ · k]

(144)

where we used the above theorems, including Tr[odd number] = 0,then neglected m2

e and m2µ and performed the remaining Lorentz index

contractions.

We insert this to obtain:(dσ

dΩ

)com

=1

8π2E2tot

e4[p · p′k · k′ + p · k′p′ · k]

∣∣∣∣ 1

(p+ k)2 + iε

∣∣∣∣2(145)

• We now do a little bit of kinematics, writing (neglecting masses so thatE = E′ = Etot/2 for all particles — for inclusion of masses in all this, see


MS)

p = (E, 0, 0, E) , k = (E, 0, 0,−E)

p′ = (E,E sin θ, 0, E cos θ) , k′ = (E,−E sin θ, 0,−E cos θ)

(146)

yielding

p · p′ = k · k′ = E2(1− cos θ)

p · k′ = k · p′ = E2(1 + cos θ)

(p+ k)2 = 2p · k = 4E2 (147)

so that we obtain(dσ

dΩ

)com

=1

8π2E2tot

e4E4[(1− cos θ)2 + (1 + cos θ)2]

[1

4E2

]2

=1

8π2E2tot

(4πα)2 2

16[1 + cos2 θ]


=α2

4E2tot

[1 + cos2 θ] , and

σtot =4π

3

α2

E2tot

=πα2

3E2.

(148)

It is common convention to write

s ≡ (p+ k)2 = (p′ + k′)2 , (149)

where s = E2tot = 4E2, the latter holding when masses are neglected. Also,

we frequently write

t ≡ (p′ − p)2 = (k′ − k)2 , u ≡ (k′ − p)2 = (k − p′)2 . (150)

In the massless limit, t = −2p′ · p = −2k′ · k and u = −2k′ · p =−2k · p′. Further, since t = −2E2(1− cos θ) we have dt = 2E2d cos θ =(E2/π)dΩ = 1

4(s/π)dΩ. With these substitutions, our differential cross


section can be expressed in the form

dσ

dt=

4π

s

1

8π2s(4πα)2

[t2

4+u2

4

]1

s2

=2πα2

s2

[(t

s

)2

+

(u

s

)2]. (151)

This is a Lorentz invariant form that actually holds in any frame, since it isexpressed entirely in terms of Lorentz invariants.

A note on dimensions: dσ/dt has ’energy’ dimensions of 1/E4 in theh = c = 1 units. In the massless limit, we see that all these dimensions areprovided by the energy scale of the process, as encoded in s. This is alwaysthe case when the underlying theory has a dimensionless coupling constantsuch as e, which in turn is always the case if we generate interactions usingthe minimal substitution rule. In minimal substitution, ∂µ → ∂µ + iqAµ.Since ∂µ and Aµ have the same dimension, q must be dimensionless. Tocheck the dimensions of Aµ remember that

∫d4xL must be a dimensionless

action. With L = −14(∂µAν − ∂νAµ)(∂µAν − ∂νAµ), it is clear that Aµ

must have the same dimensions as ∂µ in order to cancel the dimensions ofd4x.


Photon polarization sums for a cross section

• WriteM = εαr1

(~k1)εβr2

(~k2) . . .Mαβ...(~k1, ~k2, . . .) . (152)

The εµ(~k) are gauge dependent objects. In particular, if we write (we keepto a basis in which our initial choices — you will see why I use the word“initial” in a moment — for the polarization ε’s are real)

Aµ+(x) =∑~k

∑r

1√2V ω(~k)

εµr (~k)e−ik·xar(~k)

Aµ−(x) =∑~k

∑r

1√2V ω(~k)

εµr (~k)e+ik·xa†r(~k) (153)

and perform the gauge transformation Aµ(x) → Aµ(x) + ∂µf(x) with

f(x) = f(k)(ar(~k)e−ik·x − a†(~k)eik·x

), then the corresponding change

in εµ(~k) isεµr → εµr (~k)− ikµf(k) . (154)


(Note that this choice of gauge transformation keeps theAµ field hermitian).In QED, we can apply this for each of the external ε’s, each with their own~ki. Thus, if M is GI, we must have

0 = kα1 εβr2

(~k2) . . .Mαβ...(~k1, ~k2, . . .)

= εαr1(~k1)k

β2 . . .Mαβ...(~k1, ~k2, . . .)

= . . . . (155)

MS extrapolates this to say that

kα1Mαβ...(~k1, ~k2, . . .) = kβ2Mαβ...(~k1, ~k2, . . .) = . . . = 0 . (156)

This is true only in the case of Abelian gauge theory (which QED is). Ina non-abelian gauge theory, this final form is not true and only a singlephoton can be “removed” at a time — all the other polarizations must bekept in place for the identity to hold.

• Another point of view that gives this same result is current conservation.

Let us focus on just one of the photons and write M = εµ(~k)Mµ(k).We know that the photon is seeing an interaction of the form L =


∫d4xejµ(x)Aµ(x), where jµ = ψγµψ. Therefore,

Mµ(k) =

∫d4xeik·x〈f |jµ(x)|i〉 . (157)

Then

kµMµ(k) =

∫d4x

[(−i∂µ)eik·x

]〈f |jµ(x)|i〉

= −∫d4xeik·x[−i∂µ〈f |jµ(x)|i〉]

= 0 (158)

by virtue of current conservation, ∂µ〈f |jµ(x)|i〉 = 0. (The current is notnecessarily conserved until placed in the context of definite on-shell statesfor the other particles involved in the process, as encapsulated in the 〈i| and|f〉 notation.) We know that such current conservation applies for the freefields. In the interacting case, this current conservation is a consequence ofgauge invariance. The identity kµMµ(k) = 0 is called the Ward Identity.

• With this result, we can now simply perform polarization sums for the cross


section.

Writing Mr(k) = εµr (~k)Mµ(~k), the cross section will be proportional to

X =∑r=1,2

∣∣∣Mr(~k)∣∣∣2 =Mµ(~k)M∗ν(~k)

∑r=1,2

εµr (~k)ενr(~k) . (159)

For simplicity, we go to a Lorentz frame (M is a Lorentz invariant sowe can choose any frame we like) where kµ = (k, 0, 0, k) has only adirection 3 vector component. Then, in the Lorentz gauge we know thatε1(~k) = (0, 1, 0, 0) and ε2(~k) = (0, 0, 1, 0), i.e. ε1

1 = 1 and ε22 = 1, all

others zero. So,X = |M1|2 + |M2|2 , (160)

where the subscripts are the Lorentz indices. Now, for the above k, theWard identity reduces to

k(M0 +M3) = 0 , ⇒ M0 = −M3 . (161)

Then we can write equally well

X = |M1|2 + |M2|2 + |M3|2 − |M0|2 = −gµνMµM∗ν = −MνM∗ν .(162)


So, effectively, we have made the replacement∑r=1,2

εµr (~k)ενr(~k)→ −gµν . (163)

This is not an actual equality. In fact (for k2 = 0)

∑r=1,2

εµr (~k)ενr(~k)→ −gµν−

1

(k · n)2[kνkµ−(k ·n)(kµnν+kνnµ)] , (164)

where nµ = (1, 0, 0, 0), but the other terms don’t contribute because ofthe Ward identity.

• Let us now use this technology for the case of Compton scattering. Thiswas discussed beginning with Eq. (61) in some detail with the result thatwe developed the two contributing amplitudes

Ma = −e2ur′(~p′)ε/ s′(~k

′)

[iq/ +m

q2 −m2

]q=p+k

ε/ s(~k)ur(~p) (165)


Mb = −e2ur′(~p′)ε/ s(~k)

[iq/ +m

q2 −m2

]q=p−k′

ε/ s′(~k′)ur(~p) (166)

corresponding to the Feynman diagrams

!

"#$&%('*))+,

-

.

/0 1

/20

3 4561

7

8 5 19

/#5$&%'*))+:,

.

-

/(0

/(0

Figure 6: The two Feynman diagrams contributing to e−γ → e−γ.

We need14

∑pol, s,s′

∑spin, r,r′

|Mrsr′s′|2 = 14

∑spin

MαβM∗αβ (167)

where we have written (dropping the r and r′ indices for the moment)

Mss′ = εαs (~k)εβs′(~k ′)Mαβ (168)


and used the following procedure:∑s,s′

|Mss′|2 =∑s,s′

[εαs (~k)εβs′(

~k ′)Mαβ

] [εα′

s (~k)εβ′

s′ (~k ′)M∗α′β′

]

=

(∑s

εαs (~k)εα′

s (~k)

)(∑s′

εβs′(~k ′)εβ

′

s′ (~k ′)

)MαβM∗α′β′

= (−gαα′)(−gββ

′)MαβM∗α′β′

= MαβM∗αβ . (169)

In the above, we have

Mαβ =Maαβ +Mb

αβ (170)

with

Maαβ = −ie2ur′(~p

′)γβ

[p/ + k/ +m

(p+ k)2 −m2

]γαur(~p)

Mbαβ = −ie2ur′(~p

′)γα

[p/ − k/ ′ +m

(p− k′)2 −m2

]γβur(~p) (171)


Using p2 = p′2

= m2 and k2 = k′2

= 0, the denominators above reduce to2p · k and −2p · k′, respectively, and we find

14

∑pol, s,s′

∑spin, r,r′

|M|2 = 14

∑spin

MαβM∗αβ

=e4

16

(Xaa

(pk)2+

Xbb

(pk′)2−Xab +Xba

(pk)(pk′)

)(172)

where

Xaa =∑spin

[ur′(~p′)γβ

(f/ 1 +m)γαur(~p)][ur′(~p

′)γβ(f/ 1 +m)γαur(~p)]

∗

=∑spin

[ur′(~p′)γβ

(f/ 1 +m)γαur(~p)][ur(~p)γα(f/ 1 +m)γβur′(~p

′)]

= Tr

γβ(f/ 1 +m)γα

(∑r

ur(~p)ur(~p)

)γα(f/ 1 +m)γβ

∑r′ur′(~p

′)ur′(~p

′)

= Tr

[γβ

(f/ 1 +m)γα

(p/ +m)γα(f/ 1 +m)γβ(p/′

+m)]

(173)

Xbb =∑spin

[ur′(~p′)γα

(f/ 2 +m)γβur(~p)][ur′(~p

′)γα(f/ 2 +m)γβur(~p)]

∗

= Tr[γα

(f/ 2 +m)γβ

(p/ +m)γβ(f/ 2 +m)γα(p/′

+m)]

(174)


Xab =∑spin

[ur′(~p′)γβ

(f/ 1 +m)γαur(~p)][ur′(~p

′)γα(f/ 2 +m)γβur(~p)]

∗

= Tr[γβ

(f/ 1 +m)γα

(p/ +m)γβ(f/ 2 +m)γα(p/′

+m)]

(175)

Xba =∑spin

[ur′(~p′)γα

(f/ 2 +m)γβur(~p)][ur′(~p

′)γβ(f/ 1 +m)γαur(~p)]

∗

= Tr[γα

(f/ 2 +m)γβ

(p/ +m)γα(f/ 1 +m)γβ(p/′

+m)]

(176)

where f1 = p+ k and f2 = p− k′.

Under the substitutions k ↔ −k′, α ↔ β we find f1 ↔ f2 and henceXaa ↔ Xbb and Xba ↔ Xab. Thus, we need only compute Xaa and Xab

and then use the substitution rules to get Xbb and Xba.

If we look closely at Xab as compared to Xba, we see that they arerelated by trace reversal. In the Appendix of MS, we see that the traceof a product of γ matrices is equal to the trace of the reverse product ofthe γ matrices by virtue of the fact that there is a matrix C such thatCγµC−1 = −[γµ]T . Assuming an even number of γ matrices, the result isthat inserting CC−1 = 1 throughout the trace converts the original traceto a trace such that the transpose of each γ matrix appears inside the traceand then we employ Tr[ATBTCT . . . ZT ] = Tr[Z . . . CBA].


• To simplify these traces, we need some more trace identities.

1. First, we have

γαγα = gαβγ

αγβ

=1

2gαβ[γ

α, γβ

]+ =1

2gαβ2g

αβ= 4 . (177)

2. Next, we have

γαγνγα = (2g

αν − γνγα)γα = (2− 4)γν

= −2γν

(178)

3. Thirdly, we have

γαγνγµγα

= (2gνα − γ

νγα)γ

µγα

= 2γµγν − γν(−2γ

µ)

= 2[γµ, γν

]+

= 4gµν

(179)

4. Finally, we need

γαγνγµγσγα

= (2gνα − γ

νγα)γ

µγσγα

= 2γµγσγν − γνγαγµγσγα

= 2γµγσγν − 4g

µσγν

= 2γµγσγν − 2(γ

µγσ

+ γσγµ

)γν

= −2γσγµγν. (180)


• Using these identities we may now evaluate our X’s.

Xaa = Tr[γβ

(f/ 1 +m)γα

(p/ +m)γα(f/ 1 +m)γβ(p/′

+m)]

= Tr[(f/ 1 +m)γ

α(p/ +m)γα(f/ 1 +m)γβ(p/

′+m)γ

β]

= Tr[(f/ 1 +m)(−2p/ + 4m)(f/ 1 +m)(−2p/

′+ 4m)

]= 4Tr

[(f/ 1 +m)(p/ − 2m)(f/ 1 +m)(p/

′ − 2m)]

= 4Tr[f/ 1p/ f/ 1p/

′] +m

2[−4Tr[f/ 1p/ ]− 4Tr[f/ 1p/

′] + Tr[p/ p/

′] + 4Tr[f/ 1f/ 1]

]+ 4m

4Tr[1]

= 16

[2(f1p)(f1p

′)− (pp

′)(f1f1)] +m

2[−4(f1p)− 4(f1p

′)

+4(f1f1) + (pp′)] + 4m

4

= 32[(pk)(pk′) +m

2(pk) +m

4] , (181)

where in the last step we used

(f1f1) = (p+ k)2 = m2 + 2(pk)

(f1p) = (p+ k) · p = m2 + (pk)

(f1p′) = (p′ + k′) · p′ = m2 + (k′p′) = m2 + (kp)

(pp′) = p · (p+ k − k′) = m2 + (pk)− (pk′) (182)


as follow from p2 = p′2

= m2, (pk) = (p′k′) and (pk′) = (p′k).

From the substitution rule, we immediately obtain

Xbb = 32[(pk)(pk′)−m2(pk′) +m4] . (183)

For Xab we find

Xab = Tr[γβ

(f/ 1 +m)γα

(p/ +m)γβ(f/ 2 +m)γα(p/′

+m)]

= Tr[γβ

(f/ 1 +m)(−2f/ 2γβp/ + 4mf2β + 4mpβ −m22γβ)(p/

′+m)

]= Tr

[−8(f1f2)p/ + 4mf/ 2p/ + 4mf/ 2(f/ 1 +m) + 4mp/ (f/ 1 +m)

−2m2(−2f/ 1 + 4m)

(p/′

+m)

]

= 4

[−8(f1f2)(pp

′) + 4m

2(f2p′) + 4m

2(pp′) + 4m

2(f1p′)

+

4m

2(f2p)

+4m2(f2f1) + 4m

2(f1p)− 8m

4]

1st from p/ ′, 2nd from m

= 4

[−8m

2(pp′) + 4m

2[(pp′)− (pk)] + 4m

2(pp′) + 4m

2[m

2+ (pk)] + 4m

2[m

2 − (pk′)]


+4m2m

2+ 4m

2[m

2+ (pk)]− 8m

4]

= 16[2m4

+m2(pk)−m2

(pk′)] , (184)

where we had to use

(f1f2) = (p + k) · (p− k′) = (p + k) · p− (p′

+ k′) · k′ = m

2+ (pk)− (p

′k′) = m

2

(f2p′) = (p− k′) · p′ = (pp

′)− (k

′p′) = (pp

′)− (pk)

(f1p′) = (p + k) · p′ = (p

′+ k′) · p′ = m

2+ (k

′p′) = m

2+ (kp)

(f2p) = (p− k′) · p = m2 − (pk

′)

(f1p) = (p + k) · p = m2

+ (pk) .

Using the substitution rule (k ↔ −k′) to get Xba, we verify thatXba = Xab as expected.

• Using these results in

14

∑pol, s,s′

∑spin, r,r′

|M|2 = 14

∑spin

MαβMαβ

=e4

16

(Xaa

(pk)2+

Xbb

(pk′)2−Xab +Xba

(pk)(pk′)

)(185)


we obtain

14

∑pol, s,s′

∑spin, r,r′

|M|2 = 2e4

(p · kp · k′

+p · k′

p · k

)+ 2m

2(

1

p · k−

1

p · k′

)+m

4(

1

p · k−

1

p · k′

)2

(186)

This takes a very simple form in the laboratory system defined by p =(m, 0, 0, 0) for which we can write k = (ω,~k), k′ = (ω′, ~k ′) and p′ =(E′, ~p ′). For these definitions, we have

14

∑pol, s,s′

∑spin, r,r′

|M|2 = 2e4

(ω

ω′+ω′

ω

)+ 2m

(1

ω−

1

ω′

)+m

2(

1

ω−

1

ω′

)2. (187)

This can be further simplified by using the Compton scattering relation

1

ω−

1

ω′=

1

m(cos θ − 1) , (188)

where θ is the angle between ~k and ~k′, which gives

14

∑pol, s,s′

∑spin, r,r′

|M|2 = 2e4

ω

ω′+ω′

ω− sin2 θ

. (189)


To derive Eq. (188) we note that since p = p′ + k′ − k we also have

p · k = (p′ + k′ − k) · k = p′ · k + k′ · k = p · k′ + k′ · k , (190)

where we used (p− k′)2 = m2− 2p · k′ = (p′− k)2 = m2− 2p′ · k, whichin turn implies that p · k′ = p′ · k, for the 2nd equality.

If we take ~k along the z axis and ~k′ at an angle θ with respect to the zaxis, then ~k′ · ~k = ωω′ cos θ and the above equation reduces to

ωm = ω′m+ ωω′(1− cos θ) (191)

which, in turn, gives Eq. (188). For the cross section, we return to theexpression

dσ = (2π)4δ4(∑

p′f −∑

pi)1

4E1E2vrel

∏f

d3~p′f

(2π)32E′f

|M|2 (192)

with E1E2vrel = [(p1 ·p2)2−m2

1m22]

1/2. In our case we can identify particle1 as a photon with m1 = 0 and particle 2 as the target proton. Then we


get

dσ = (2π)4δ4(p′ + k′ − p− k)1

4ωm

(d3~p′

(2π)32E′

)(d3~k′

(2π)32ω′

)|M|2

=1

64π2ωω′mE′δ(ω′ + E′ −m− ω)ω′

2dω′dΩ |M|2

=ω′

64π2ωmE′

[∂(ω′ + E′)

∂ω′

]−1

dΩ |M|2

(193)

where we have used the notation dΩ for the ~k′ solid angle. To computethe required derivative, we note that

E′ = [m2 + (~k − ~k′)2]1/2 = [m2 + ω2 + ω′2 − 2ωω′ cos θ]1/2 (194)

so that∂E′

∂ω′=ω′ − ω cos θ

E′(195)


implying that

∂(E′ + ω′)

∂ω′=

ω′ − ω cos θ

E′+ 1

=ω′ + E′ − ω cos θ

E′

=ω +m− ω cos θ

E′

=mω

E′ω′, (196)

where for the next-to-last step we used energy conservation, ω′ + E′ =ω + E, and for the last step we used Eq. (188). Altogether, we get

dσ

dΩ=

ω′

64π2ωmE′E′ω′

mω2e4

ω

ω′+ω′

ω− sin2 θ

=

e4

32π2m2

(ω′

ω

)2ω

ω′+ω′

ω− sin2 θ


=α2

2m2

(ω′

ω

)2ω

ω′+ω′

ω− sin2 θ

. (197)

In the non-relativistic limit, ω m, which means ω ∼ ω′ (see Eq. (188)),and we get

dσ

dΩ→

α2

2m2(1 + cos2 θ) , (198)

which is the Thomson cross section. Obviously, the full relativistic resultof Eq. (197) has been tested against experimental data in a highly detailedway and excellent agreement has been found (after including radiative, i.e.higher order, corrections).

• Finally, we wish to return to the gauge invariance issue. You will recall thatwe had the two amplitudes:

Ma = −e2ur′(~p′)ε/ s′(~k

′)

[iq/ +m

q2 −m2

]q=p+k

ε/ s(~k)ur(~p) (199)

Mb = −e2ur′(~p′)ε/ s(~k)

[iq/ +m

q2 −m2

]q=p′−k

ε/ s′(~k′)ur(~p) , (200)


where for q in Mb we write q = p′ − k instead of the equivalent (bymomentum conservation) form q = p− k′ employed earlier.

The gauge invariance claim is that if we replace ε(~k′) by k′ or ε(~k) by k,then M = Ma +Mb → 0. Let’s check this for the case of ε(~k) → k.We get, dropping spin indices for convenience, and writing in the explicit qvalues for each amplitude

Ma→ −e2u(~p ′)ε/ (~k′)

[ip/ + k/ +m

(p+ k)2 −m2

]k/ u(~p) (201)

Mb→ −e2u(~p ′)k/

[ip/ ′ − k/ +m

(p′ − k)2 −m2

]ε/ (~k′)u(~p) (202)

In Ma, we write

p/ + k/ +m

(p+ k)2 −m2k/ u(~p) =

p/ + k/ +m

(p+ k)2 −m2(k/ + p/ − p/ )u(~p)

=p/ + k/ +m

(p+ k)2 −m2(k/ + p/ −m)u(~p)


=(p+ k)2 −m2

(p+ k)2 −m2u(~p)

= u(~p) , (203)

so thatMa→ −ie2u(~p ′)ε/ (~k′)u(~p) . (204)

In Mb, we write

u(~p ′)k/p/ ′ − k/ +m

(p′ − k)2 −m2= u(~p ′)(k/ − p/ ′ + p/ ′)

p/ ′ − k/ +m

(p′ − k)2 −m2

= u(~p ′)(k/ − p/ ′ +m)p/ ′ − k/ +m

(p′ − k)2 −m2

= u(~p ′)−(p′ − k)2 +m2

(p′ − k)2 −m2

= −u(~p ′) , (205)

so thatMb→ +ie2u(~p ′)ε/ (~k′)u(~p) . (206)


Obviously, these two results (after the substitution) forMa andMb cancelto give zero, the requirement of gauge invariance.


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