Feynman Calculus: Examples from QED

Feynman rules The Feynman rules tell us how to go from a diagram to

the corresponding matrix element (or amplitude) whichis necessary to calculate σσσσ and ΓΓΓΓ.

There are 3 kinds of ingredients: external lines, internallines and propagators. To set up the Feynmandiagrams we need the free particle wave functions.Also everything is in momentum space. The rules areAlso everything is in momentum space. The rules aregiven in Table 6.2 of Halzen and Martin, or in Chapter 7of Griffith (((( ))))

(((( ))))

(((( )))) xki

xpi

xpi

ekAphoton

eu

eu

⋅⋅⋅⋅−−−−

⋅⋅⋅⋅−−−−

⋅⋅⋅⋅−−−−

≈≈≈≈

≈≈≈≈

≈≈≈≈

h

h

h

/*

/2

/1

)(

fermions2

1

µµµµµµµµ εεεεψψψψψψψψ

Feynman rules In the previous equation εεεε is the polarization

vector. There are two independent polarizationdirections. A photon has spin 1, but because m=0its polarization can have only two spinprojections, called transverse because

00 ====∂∂∂∂====⋅⋅⋅⋅ µµµµεεεε Agaugethefromk

We do not really care about εεεε since we are notgoing to deal with external photons.

00 ====∂∂∂∂====⋅⋅⋅⋅ µµµµµµµµεεεε Agaugethefromk

Summary of the Feynman rules(1) Momenta. Label incoming and outgoing momenta p1,

p2…, pn and spin s1, s2,…, sn. Label internal 4-momenta q1, q2,…, qn. Put an arrow on each line to keep track of the “positive” direction (in time). In general an out-going anti-particle is equivalent to an incoming particle. Therefore external lines indicate e- or e+. The direction of internal fermion lines are assigned according to the “flow”. The arrows on external photon lines points forward, on internal lines is arbitrary.

p p'

Summary of the Feynman rules2) External lines

Electrons

Positronsoutgoinguincomingu

outgoingvincomingv

Photons incomingµεoutgoing*µε

In writing the factors follow the arrows backward

3) Vertex Factor: Each vertex contributes

4) Propagator: Each internal line contributes a propagator

Quantum Electro Dynamics

cegig ee

h

πππππαπαπαπαγγγγ µµµµ 44with ========

222

)(positronselectrons

cmq

mcqi

−−−−++++

⇒⇒⇒⇒++++ µµµµµµµµγγγγ

2photonsq

igµνµνµνµν−−−−⇒⇒⇒⇒

Feynman Rules

Note that for the photon propagator:

where q is the photon four momenta. The propagator carries Lorentz indices because the photon is a spin 1 particle. The

2q

igµν−

Lorentz indices because the photon is a spin 1 particle. The4-momenta q is given by 4-momentum conservation. This photon is virtual or “off mass shell” since

02 ≠q

Feynman Rules(4) Conservation of energy momenta. Energy and momentum must be conserved at each vertex. We impose energy-momentum conservation by assigning a δ-function to each vertex

where the incoming particles are taken with a (+) sign and

)()2( 314 qpp +−δπ

where the incoming particles are taken with a (+) sign andoutgoing particles with a (-) sign.(5) Integration over internal momenta. For each internal line we have an integral

Therefore we have to integrate over all internal momenta.

44)2(

1dq

π

Feynman Rules

(6) Cancel the final δ-function. After we follow step 1-5, we get a final result which will include a δ-function of the type

This δ-function enforces overall energy-momentum

).......()2( 214

nppp −−−−++++δδδδππππ

This δ-function enforces overall energy-momentumconservation. Cancel this factor then what remains is proportional to –iM(7) Antisymmetrization: Include a (-) sign between the diagram that differ only in the interchange of two incoming (or outgoing) electrons (or positron) or of an incoming electron with an outgoing positron (or viceversa).Note that you can find the Feynman’s rules in chapter 7

Summary of the Feynman rules

e+e-→µ→µ→µ→µ+µµµµ-

We are now ready to calculate the cross section for theprocess

−−−−++++−−−−++++ →→→→ µµµµµµµµee

p4,s4

t

e+ µ+

e- q

p3,s3

p4,s4

e+e-→µ→µ→µ→µ+µµµµ-

Apply the Feynman rules:

( ) ( )( ) qdppqqpppuigpv

q

igpvigpu

se

s

se

s

434

421

412

2434

)()()()(

)()(2

12

43

−−−+

−∫

δδγ

γπ

µ

µνµ

e+ µ+

e- q

p3,s3

p4,s4

Integration over dq and the δ-function of the verticesyields:

( )e 342112

Muon piece Electron- piece

4321 pppp ++++====++++

( )[ ] ( )[ ])()()()()( 12432

21

21243 pupvpvpu

pp

gM sssse

µµ γγ

+−=

e+e-→µ→µ→µ→µ+µµµµ-

Notice that all indices Lorentz and Dirac are contractedso that M is a pure complex number

In principle we can have also the Z0 exchange( remember electro-weak unification).

Cross section for e+e-→µ→µ→µ→µ+µµµµ-

Now that we have determined the matrix element we cancalculate the cross section for

because it is an easy two body final state which corresponds to only 1 diagram in QED. Since the cross

−−−−++++−−−−++++ →→→→ µµµµµµµµee

corresponds to only 1 diagram in QED. Since the cross section is given by:

Therefore we can divide our calculation in 3 steps:Kinematics calculations of |M|2 from Feynman diagrams phase space integration and cross section

spacephase2 ××××≈≈≈≈ Mdσσσσ

Cross section for e+e-→µ→µ→µ→µ+µµµµ-

We want to determine:

with ECM>>m µ so that we will neglect e and µ masses.We also want ECM<m Z so we can neglect Z exchange diagram. The only contributing diagram is via photon

)()()()( 4321 pppepe ++++−−−−++++−−−− →→→→++++ µµµµµµµµ

diagram. The only contributing diagram is via photonexchange.

We will work in the e+ e- center of mass frame, but tokeep things as general as possible we will use as muchas we can Lorentz invariant quantities.

Kinematics The kinematics constraints offers us useful information.

Consider the final state: there are 6 degrees of freedomalias the 3-momenta of the µ- and the µ+ (p3 and p4). Butconservation of 4-momentum gives us 4-constraints:

therefore there are only two independent variables in the 4321 pppp ++++====++++

final state. To get the total cross section we will only have to do two

integrals, the δ4 functions take care of the rest. Let

where we have neglected the electron and muon masses.

4321

243

221

2

22

)()(

pppps

ppppsEs cm

⋅⋅⋅⋅====⋅⋅⋅⋅====++++====++++========

Kinematics The quantity s is a Lorentz invariant quantity. Let us now

consider the e+ e- center of mass frame. The e+ and e-

collide head-on with equal and opposite momenta and theµ + and µ - go off also back to back with equal and oppositemomenta but in some other directions.

If we neglect the masses, all particles have the sameenergyenergy

e- (p1)z

e+ (p2)

µ- (p3)

µ+ (p4)

θ

22cmEs ====

Kinematics Taking the e- direction to be in the z direction , setting c=1 and

neglecting the masses:

+

−

−=

=

2,0,0,

2

2,0,0,

2

2

1

cmcm

cmcm

EEE

eEE

p

eEE

p

We have taken the plane of the momenta to define φ =0. Ecm is considered to be fixed, so that the two degrees of

freedom are the angles θ and φ. There is azimuthal symmetryand we do not expect a dependence on φ⇒easier integration.

+

−

−−=

=

µθθ

µθθ

cos2

,sin2

,0,2

cos2

,sin2

,0,2

4

3

cmcmcm

cmcmcm

EEEp

EEEp

Recall the Feynman diagram

Calculation of |M|2

243

221 )()( pppps ++++====++++====

e+ µ+

e- q

p4,s4

The first step to determine |M|2, is to calculate M*

( )[ ] ( )[ ])()()()()( 12432

21

2

1243 pupvpvpupp

gM ssss

µµ γγ

+−=

p3,s3

Calculation of |M|2 Recall the Feynman diagram [ (AB)+=B+A+ ]

Let us use

( )[ ] ( )[ ]=+

−= ++++++ )()()()()( 21342

21

2*

2134 pvpupupvpp

gM

sss sν

ν γγ

µµ γγγγγψψ == ++ 000 and

Since

γγγγγψψ == and

I====00γγγγγγγγ

[ ] [ ])()()()()( 2

0001300042

21

2*

2134 pvpupupvpp

gM

sss s γγγγγγγγ νν

++

+−=

[ ] [ ])()()()()( 21342

21

2*

2134 pvpupupvpp

gM

sss sν

ν γγ+

−=

Calculation of |M|2 If everything is unpolarized we

average over the initial helicities

and we sum over the final helicities

∑∑

212

1

2

1

ss

∑∑

This is done when we do not use polarized initial statesand we do not measure the polarization in the final state( i.e we accept all polarizations). Note that these areincoherent sums and that different helicity states do notinterfere. The sum over the helicity states will allow us touse the properties of the Dirac equation ( see 7.5 and 7.7)

∑∑

43 ss

Calculation of |M|2 The final result

[ ] [ ][ ] [ ])()()()(

)()()()(

)(4

1

4

1

4334,

221

2

,,,

*2

4334

43

4321

pupvpvpu

pvpupupv

pp

gMMM

ssss

ss

ssss

µν

µν

γγ

γγ

∑

∑

∑

×

×+

==

We define two tensors

[ ] [ ])()()()( 1221,

1221

12

pupvpvpuss ss

ss

µν γγ∑

[ ] [ ][ ] [ ])()()()(

)()()()(

1221,

4334,

1221

12

4334

43

pupvpvpuB

pvpupupvA

ss

ssss

ssss

ss

µννµ

µννµ

γγ

γγ

∑

∑

=

=

Calculation of |M|2

Note that Aνµ depends only on the muon momenta while Bµν

depends on the electron momenta. This means that Aνµ isrelated to the final states (the µ +µ - vertex) and Bµν isrelated to the initial states (the e+e- vertex). Therefore we can calculate Aνµ and Bµν separately and then contract them to get the average value of |M|2. (See Griffith’s 7.7)

Then we have:

where the indices a, b, c and d refer to the Dirac indices. Remember that the Dirac indices goes from 1 to 4 and that repeated indices are summed (Einstein convention).

( ) ( ) )()()()( 4334 4334

43

pvpupupvAdcdab

ss

sc

sb

sas µννµ γγ∑∑=

Calculation of |M|2

We can group the elements of Aνµ as:

Since the u and v spinors satisfy the Dirac equation inmomentum space we have:

( ) ( ) 00 =+=− vmcpumcp µµ

µµ γγ

[ ]( ) [ ]( )cd

sabd

s

pupupvpvAc

sb

sass µννµ γγ )()()()( 3344 33

3

44

4

∑∑=

v and u are orthogonal

v and u are normalized

( ) ( )( ) ( ) 00

00

=+=−

=+=−

mcpvmcpu

vmcpumcp

µµ

µµ

µµ

γγ

γγ

( ) ( ) ( ) ( ) 00 2121 == vvuu

mcvvmcuu 22 −==

Calculation of |M|2 v and u are complete

If we neglect the mass m:

mcpvvmcpuu ss

s

ss

s

−=+= ∑∑==

µµ

µµ γγ

2,12,1

ppvvppuu ss

s

ss

s/==/== ∑∑

==µ

µµ

µ γγ2,12,1

Therefore

Since Aνµ is a trace , we can write is as:

ss == 2,12,1

( ) ( ) ( ) ( ) ( )µνµννµ γγγγ 3434 ppTrppAcdbcabda //−=//=

( )[ ] [ ]νµνµµννβαµνµαββµαν

βα

µβναβα

νµ γγγγ

34433434

34

44 ppgppppggggggpp

TrppA

+⋅−−=+−−=

=−=

Calculation of |M|2 The Aνµ is symmetric in µ and ν

Bµν is also a trace

µννµ AA =

( ) [ ]νµνµµνµν

νµ γγ 21212121 4 ppgppppppTrB +⋅−−=//−=

In fact Aνµ and Bµν are similar because the interactionvertices are similar.

Putting everything together :

νµνµ BA

pp

gM

421

22

)(4

1

+=

Calculation of |M|2 Note:

[ ][ ]

)11()11(

16

23141324

212121

343434

pppppppp

gpppppp

gppppppBA

⋅⋅++⋅⋅+

=⋅−+

⋅−+=νµνµµν

µννµµννµ

νµ

Recall that

( ) )(216

)11()11(16

23141324

2143

21434321

pppppppp

ggpppp

pppppppp

⋅⋅+⋅⋅×=

=

⋅⋅

+⋅⋅+−⋅⋅+−µν

νµ

4=µννµ gg

Finally since (p1+p2)2=s

Note that this is a real number and it is a Lorentz invariant asit is always true for |M|2. Note:

Calculation of |M|2

22 22)()( pppppppps ⋅=⋅≈+=+=

[ ]))(())((8

324142312

42

pppppppps

gM ⋅⋅+⋅⋅=

Then

23412

232

41

24312

242

31

43212

432

21

22)()(

22)()(

22)()(

ppppppppu

ppppppppt

pppppppps

⋅−=⋅−≈−=−=

⋅−=⋅−≈−=−=

⋅=⋅≈+=+=

[ ]222

42 2

uts

gM +=

|M|2 in CM system Let:

be the energy of any of the particles ( remember that the energies of all 4 particles are the same if we neglect the masses). Then

1622

24 s

EEs

E cm ===

313131

ppandppSince

ppEEpprrrr

rr

−=−=⋅−=⋅

Therefore :42

231

4321

cos ppEpp

ppandppSincerrrr

rrrr

⋅==⋅

−=−=

θ

( )( )( )( ) 24

3241

244231

)cos1(

)cos1(

θθ

+=⋅⋅

−=⋅⋅

Epppp

Epppp

[ ])cos1( 242θ+= gM

Phase Space Integration Now we can use 6.47 that yields the general expression for

dσ/dΩ for a 2 by 2 scattering process in the CM system.

For e+e-→µ+µ- scattering we have:

1,1)(

1

64

1 2

221

2===

+=

ΩScforM

EEp

p

d

d

i

fhr

r

πσ

For e e →µ µ scattering we have:

[ ])cos1( 242θ+= gM

221

221 )()(

0

ppsEE

mmifpp efi

+==+

=== µrr

Phase Space Integration Therefore.

If

πα

2eg=

( )s

gee

d

d e

CM

θπ

µµσ 2

2

4 cos1

64)(

+=→Ω

−+−+

Therefore

πα

4=

( )θαµµσ 22

cos14

)( +=→Ω

−+−+

see

d

d

CM

Phase Space Integration To get the total cross section we have to integrate in dΩ,

since

∫ = πφ 2d ( )3

2coscos

1

1

2 =∫−

θd

13

4)(

2

===→ −+−+ cs

ee hαπµµσ

In other books you will find

3 s

( )

SIGHL qqq

s

cee

0

2

14

3)(

επ

απµµσ

==

=→ −+−+ h

( )s

cee

2

3

4)(

hαπµµσ =→ −+−+

Comments on total cross section

So far as the total cross sectionis concerned the only Lorentzinvariant variable in this

We could have got the same result with a simple “order ofmagnitude” argument. See T. D. Lee : “Particle Physics andintroduction to field theory” Chapter 8.

e+ µ+

p4,s4

problem is the 4-momentum qcarried by the virtual γ. In theCM system

22 qEs CM −== Therefore the total cross section must be of the form

( )µασ mmsf e ,,2=

e- q

p3,s3

Comments on total cross section When ECM>> me, mµ we may set me=mµ =0. Since α is

dimensionless and

Therefore in natural units h=c=1

[ ] [ ]22 −== LsLσ

2ασ ≈ As you can see we get the correct estimation albeit without

the factor of 4π/3 Suppose ECM=35 GeV

s

ασ ≈

( ) ( ) pbGeV

b

GeV70

1

398

35

1

137

1

3

422

2

≅

≈ −

µπσ