1 Bhabha scattering (1936)

Consider the scattering of an electron and a positron. We have already computed the annihilation of an electron/positron in finding the cross-section for producing muon pairs. It is also possible for the final particles to be an electron-positron pair. In addition, the initial particles may exchage a photon without annihilating. Since the particles are distinguish- able, we have only the t-channel diagram for this, without the crossed diagram. Let the in and out electrons have momenta p, p0, respectively and the initial and final positrons have momenta k,k0. Let the angle between the initial and final electrons be θ. The general calculation is lengthy; here we restrict our attention to the ultrarelativistic case, where we can treat the m2 electron and positron as massless. This means we neglect terms of order E2 . For GeV or TeV colliders, this term is of order 10−8 or 10−14, so we incur no great error. We may drop m whenever it occurs additively.

1.1 The scattering matrix We have two Feynman diagrams, so the matrix element is the sum

s t iM fi = iM fi + iM fi where s and t refer to the photon channel. The first, s-channel, diagram describes the annihilation-creation. The matrix element is the same as we use for the muon problem, except the particle masses are all equal, and here we want to call p,k the incoming momenta and p0,k0 the outgoing momenta. Therefore, instead of   s 0
α −i qα qβ  β  0
iM = v¯ p (−ieγ )u(p) η − (1 − ξ) u¯(k) −ieγ v k fi q2 αβ q2 we write   s α −i qα qβ 0
 β  0
iM = v¯(k)(−ieγ )u(p) η − (1 − ξ) u¯ p −ieγ v k fi q2 αβ q2 that is, we just need to interchange p0 ↔ k in our previous result, and we have q = p + k = p0 + k0. As before, the gauge terms vanish immediately. We replace q2 = s, leaving

2 s ie α 0
β 0
iM = η v¯(k)γ u(p)u¯ p γ v k fi s αβ The second Feynman diagram has the matrix element   t 0
α −i qα qβ  β  0
iM = u¯ p (−ieγ )u(p) η − (1 − ξ) v¯(k) −ieγ v k fi q2 αβ q2

where p, p0 are the incoming and outgoing electron, respectively; k,k0 are the incoming and outgoing proton (or any other fermion of charge Q, except the electron which would also require the u channel), respectively; q = p− p0 = k0 −k is the t-channel 4-momentum; ξ allows a choice of gauge. The gauge term does not immediately drop in this case. We have:   −i qα q   i u¯p0
(−ieγα )u(p) −(1 − ξ) β v¯(k) −ieγβ vk0
= (1 − ξ)e2u¯p0
(iq/)u(p)v¯(k)(iq/)vk0
q2 q2 t2 i = (1 − ξ)e2u¯p0
ip/ − ip/0
u(p)v¯(k)ik/0 − ik/
vk0
t2 i = (1 − ξ)e2u¯p0
2mu(p)v¯(k)2mvk0
t2 4im2e2 = (1 − ξ)e2u¯p0
u(p)v¯(k)vk0
t2 where we use the Dirac and conjugate Dirac equations in the middle step. The matrix element becomes

2 2 2 t ie 0
α 0
β 4im e 2 0
0
iM = η u¯ p γ u(p)u¯ k γ u(k) − (1 − ξ)e u¯ p u(p)v¯(k)v k fi t αβ t2

1 where we set q2 = t. We should get the same result in any gauge, so we could just set ξ = 1, but in any case the term of of order m2 and we drop it. The resulting matrix element is s t iM fi = iM fi + iM fi ie2 ie2 = η v¯(k)γα u(p)u¯p0
γβ vk0
+ η u¯p0
γ µ u(p)v¯(k)γν vk0
s αβ t µν Squaring the matrix element leads to four terms, 2 s 2 s t∗ t s∗ t 2 iM fi = iM fi + M fiM fi + M fiM fi + iM fi We work them out one at a time.

1.2 Matrix squared: s channel 2 s We already have iM fi from the muon case. Averaging over initial spins and summing over final spins, we found 4 1 s 2 8e 0
0
0 0
0
0
2
0 0
2 0 0
2 4
∑ M fi = 2 p · p k · k − k · p (k · p) + p · k p · k − m (k · p) + (k · p) k · p + m k · p + 2m (k · p) + 2m 4 all spins s 8e4 = p · p0
k · k0
+ p0 · k
p · k0
+ m2 k0 · p0
+ m2 (k · p) + 2m4
s2 where some simplification occurs because of the equal masses.

1.3 Matrix squared: t channel For the final term, we have  2  2  1 t 2 1 ie 0
µ ν 0
ie 0
ρ σ 0
∑ M fi = ∑ ηµν u¯ p γ u(p)v¯(k)γ v k − ηρσ v¯ k γ v(k)u¯(p)γ u p 4 all spins 4 all spins t t 4 e 0
µ ν 0
0
ρ σ 0
= ∑ 2 ηµν ηρσ u¯ p γ u(p)v¯(k)γ v k v¯ k γ v(k)u¯(p)γ u p all spins 4t 4 e 0
µ ν 0
0
ρ σ 0
= ∑ 2 ηµν ηρσ u¯a p γabub (p)v¯c (k)γcdvd k v¯e k γe f v f (k)u¯g (p)γghuh p all spins 4t 4 14 e µ σ 0
0
ν 0
0
ρ = (−1) ∑ 2 ηµν ηρσ γabub (p)u¯g (p)γghuh p u¯a p γcdvd k v¯e k γe f v f (k)v¯c (k) all spins 4t e4 = η η γ µ (p/ + m) γσ p/0 + m
γν k/0 − m
γρ (k/ − m) 4t2 µν ρσ ab bg gh ha cd de e f f c e4 = η η tr γ µ (p/ + m)γσ p/0 + m

tr γν k/0 − m
γρ (k/ − m)
4t2 µν ρσ where the factor (−1)14 keeps track of the number of times we commute fermion fields. The traces give tr γ µ (p/ + m)γσ p/0 + m

= tr γ µ p/γσ p/0 + mγ µ γσ p/0 + mγ µ p/γσ + m2γ µ γσ
= tr γ µ p/γσ p/0
+ m2tr (γ µ γσ ) 0  µ λ σ τ  2 µσ = pλ pτtr γ γ γ γ + 4m η

0  µλ στ µσ λτ µτ λσ  2 µσ = 4pλ pτ η η − η η + η η + 4m η

0  µλ στ µσ λτ µτ λσ  2 µσ = 4pλ pτ η η − η η + η η + 4m η = 4pµ p0σ − p · p0
η µσ + p0µ pσ + m2η µσ
tr γν k/0 − m
γρ (k/ − m)
= 4k0ν kρ − k · k0
ηνρ + kν k0ρ + m2ηνρ

2 Combining these results,

4 1 t 2 4e µ 0σ 0
µσ 0µ σ 2 µσ
0ν ρ 0
νρ ν 0ρ 2 νρ
∑ M fi = 2 ηµν ηρσ p p − p · p η + p p + m η k k − k · k η + k k + m η 4 all spins t 4e4 = p p0 k0ν kρ − k · k0
ηνρ + kν k0ρ + m2ηνρ
t2 ν ρ 4e4 − p · p0
η k0ν kρ − k · k0
ηνρ + kν k0ρ + m2ηνρ
t2 νρ 4e4 + p0 p k0ν kρ − k · k0
ηνρ + kν k0ρ + m2ηνρ
t2 ν ρ 4m2e4 + η k0ν kρ − k · k0
ηνρ + kν k0ρ + m2ηνρ
t2 νρ Performing the remaining contractions,

4 1 t 2 4e 0
0
0
0
0 0
2 0

∑ M fi = 2 p · k p · k − k · k p · p + (p · k) p · k + m p · p 4 all spins t 4e4 − p · p0
k · k0
− 4k · k0
+ k · k0
+ 4m2
t2 4e4 + p0 · k0
(p · k) − k · k0
p · p0
+ p0 · k
p · k0
+ m2 p · p0

t2 4m2e4 + k · k0
− 4k · k0
+ k · k0
+ 4m2
t2 8e4 = p · k0
p0 · k
− k · k0
p · p0
+ (p · k)p0 · k0
+ m2 p · p0
+ m2 − p · p0

−k · k0
+ 2m2

t2 8e4 = p · k0
p0 · k
+ (p · k)p0 · k0
− m2 p · p0
− m2 k · k0
+ 2m4
t2

1.4 Matrix squared: first cross term Now consider the first of the cross terms. We need to make sure we’ve assigned the same momenta to the four particles. But for the s channel we’ve set p0 as the initial positron momentum instead of the final.

 2  2  1 s t∗ 1 ie α 0
β 0
ie 0
ν µ 0
∑ M fiM fi = ∑ ηαβ v¯(k)γ u(p)u¯ p γ v k − ηµν v¯ k γ v(k)u¯(p)γ u p 4 all spins 4 all spins s t 4 e α 0
β 0
0
ν µ 0
= ηαβ ηµν ∑ v¯a (k)γabub (p)u¯c p γcdvd k v¯e k γe f v f (k)u¯g (p)γghuh p 4st all spins 4 15 e α β 0
0
ν µ 0
0
= (−1) ηαβ ηµν ∑ γabub (p)u¯g (p)γcdvd k v¯e k γe f v f (k)v¯a (k)γghuh p u¯c p 4st all spins e4 = − η η k/ γα p/ γ µ p/0 γβ k/0 γν 4st αβ µν f a ab bg gh hc cd de e f e4   = − η η tr k/γα p/γ µ p/0γβ k/0γν 4st αβ µν where we have dropped all mass terms. Notice that the fermion exchanges have introduced an overall sign. α β
1  α β
Now evaluate the trace. First, we eliminate the free γ-matrices, using tr ηαβ γ γ = 2tr ηαβ γ ,γ = 4. To do this, we must bring the two relevant gamma matrices next to each other. Also, notice that

α α β α α γ p/ = pβ γ γ (2p − p/γ )  αβ β α  = pβ 2η − γ γ = 2pα − p/γα

3 and for any of p,k, p0,k0,

α β p/p/ = pα pβ γ γ 1 n o = p p γα ,γβ 2 α β 1 = p p 2ηαβ 2 α β = p2 = 0

The trace is

4 1 s t∗ e  α µ 0 β 0 ν  ∑ M fiM fi = − ηαβ ηµνtr k/γ p/γ p/ γ k/ γ 4 all spins 4st

Using conservation of momentum, we replace k0 = p + k − p0,

4 1 s t∗ e  α µ 0 β 0
ν  ∑ M fiM fi = − ηαβ ηµνtr k/γ p/γ p/ γ p/ + k/ − p/ γ 4 all spins 4st 4 e
= − T + T − T 0 4st p k p where

 α µ 0 β ν  Tp = ηαβ ηµνtr k/γ p/γ p/ γ p/γ

 α µ 0 β ν  Tk = ηαβ ηµνtr k/γ p/γ p/ γ k/γ

 α µ 0 β 0 ν  Tp0 = ηαβ ηµνtr k/γ p/γ p/ γ p/ γ

 0 ν α µ 0 β  = ηαβ ηµνtr p/ γ k/γ p/γ p/ γ

0 where we used the cyclic property on Tp0 to check that it is related to Tk by the cyclic substitution, k → p → p → k, and renaming the dummy Lorentz indices. We therefore only need to compute the first two. For Tp,

 α µ 0 β ν  Tp = ηαβ ηµνtr k/γ p/γ p/ γ p/γ

 α µ µ 0  β β  ν  = ηαβ ηµνtr k/γ (2p − γ p/) p/ 2p − p/γ γ

0
 α 0 β  = 4tr k/p/p/ p/ − 2ηαβ tr k/γ p/ p/γ p/

µ 0 ν
 α µ 0 β ν  −2ηµνtr k/p/γ p/p/ γ + ηαβ ηµνtr k/γ γ p/p/ p/γ γ

0
0

 α 0  β β   = 4tr k/p/ 2 p · p − p/p/ − 2ηαβ tr k/γ p/ 2p − γ p/ p/

µ µ 0 ν
 α µ 0
0
β ν  −2ηµνtr k/p/(2p − p/γ ) p/ γ + ηαβ ηµνtr k/γ γ 2 p · p − p/ p/ p/γ γ

Simplifying, and using p2 = m2 = 0,

0
2 0
0
 α 0 β 2 Tp = 8 p · p tr (k/p/) − 4tr k/p p/ − 4tr k/p/p/ p/ + 2ηαβ tr k/γ p/ γ p

0
2 µ 0 ν
0
 α µ β ν   α µ 0 β ν  −4tr k/p/p/ p/ + 2ηµνtr k/p γ p/ γ + ηαβ ηµν 2 p · p tr k/γ γ p/γ γ − ηαβ ηµνtr k/γ γ p/ p/p/γ γ = 32p · p0
(k · p) − 4tr k/p/p/0 p/
− 4tr k/p/p/0 p/

4 0
 α µ β ν  +2 p · p ηαβ ηµνtr k/γ γ p/γ γ = 32p · p0
(k · p) − 8tr k/2p · p0
− p/0 p/
p/
0
 α µ β ν  +2 p · p ηαβ ηµνtr k/γ γ p/γ γ

Look at

 α µ β ν   α α µ  βν ν β  ηαβ ηµνtr k/γ γ p/γ γ = ηαβ ηµνtr (2k − γ k/)γ p/ 2η − γ γ µ ν = 4tr (k/p/) − 2ηµνtr (γ p/γ k/) α µ µ ν −2ηαµtr (γ k/γ p/) + 4ηµνtr (k/γ p/γ ) µ ν = 4tr (k/p/) − (2 + 2 − 4)ηµνtr (γ p/γ k/) = 16(k · p)

Therefore,

0
0
0
0
Tp = 32 p · p (k · p) − 16 p · p tr (k/p/) + 8tr k/p/ p/p/ + 32 p · p (k · p) = 32p · p0
(k · p) − 64p · p0
(p · k) + 32p · p0
(k · p) = 0

Now compute Tk, which is easier because the repeated ks are closer:

 α µ 0 β ν  Tk = ηαβ ηµνtr k/γ p/γ p/ γ k/γ

 α µ 0 β ν ν  = ηαβ ηµνtr k/γ p/γ p/ γ (2k − γ k/)

 α 0 β   α µ 0 β ν  = 2ηαβ tr k/γ p/k/p/ γ − ηαβ ηµνtr k/γ p/γ p/ γ γ k/

 α α 0 β  = 2ηαβ tr (2k − γ k/) p/k/p/ γ

0
 α 0 β  = 4tr p/k/p/ k/ − 2ηαβ tr γ k/p/k/p/ γ = 4tr p/k/p/0k/
− 8tr k/p/k/p/0
= −4tr p/k/2p0 · k
− k/p/0

= −8p0 · k
tr (p/k/) + 4tr p/k/k/p/0
= −32p0 · k
(p · k)

and cycling k → p0 → p → k we have

0
0
Tp0 = −32 p · p k · p

The full form of this cross term contribution to the squared matrix element it therefore,

4 1 s t∗ e
∑ M fiM fi = − Tp + Tk − Tp0 4 all spins 4st e4 = − 0 − 32p0 · k
(p · k) + 32p · p0
k · p0

4st 8e4 = − p · p0
k · p0
− p0 · k
(p · k)
st

1.5 Matrix squared: second cross term The second cross term, averaged and summed over spins is

5  2  2  1 t s∗ 1 ie 0
µ ν 0
ie 0
β 0
α ∑ M fiM fi = ∑ ηµν u¯ p γ u(p)v¯(k)γ v k − ηαβ v¯ k γ u p u¯(p)γ v(k) 4 all spins 4 all spins t s 4 e 0
µ ν 0
0
β 0
α = ηαβ ηµν ∑ u¯a p γabub (p)v¯c (k)γcdvd k v¯e k γe f u f p u¯g (p)γghvh (k) 4st all spins 4 15 e µ ν 0
0
β 0
0
α = (−1) ηαβ ηµν ∑ γabub (p)u¯g (p)γcdvd k v¯e k γe f u f p u¯a p γghvh (k)v¯c (k) 4st all spins e4 = − η η γ µ u (p)u¯ (p)γα v (k)v¯ (k)γν v k0
v¯ k0
γβ u p0
u¯ p0
4st αβ µν ab b g gh h c cd d e e f f a e4   = − η η tr γ µ p/γα k/γν k/0γβ p/0 4st αβ µν We need     tr γ µ p/γα k/γν k/0γβ p/0 = tr p/γα k/γν k/0γβ p/0γ µ Compare this to the first cross term, where we calculated e4   8e4 η η tr k/γα p/γ µ p/0γβ k/0γν = p · p0
k · p0
− p0 · k
(p · k)
4st αβ µν st These differ only by the exchange p ↔ k, p0 ↔ k0. Therefore,

4 1 t s∗ 8e 0
0
0

∑ M fiM fi = − k · k p · k − k · p (k · p) 4 all spins st

1.6 Final matrix squared The final matrix squared, averaged/summed over spins, is

2 s 2 s t∗ t s∗ t 2 iM fi = iM fi + M fiM fi + M fiM fi + iM fi 8e4 = p · p0
k · k0
+ p0 · k
p · k0
+ m2 k0 · p0
+ m2 (k · p) + 2m4
s2 8e4 − p · p0
k · p0
− p0 · k
(p · k)
st 8e4 − k · k0
p · k0
− k0 · p
(k · p)
st 8e4 + p · k0
p0 · k
+ (p · k)p0 · k0
− m2 p · p0
− m2 k · k0
+ 2m4
t2 Dropping the mass terms and combining,

4 4 2 8e 0
0
0
0

8e 0
0
0 0

iM fi = p · p k · k + p · k p · k + p · k p · k + (p · k) p · k s2 t2 8e4 − p · p0
k · p0
− p0 · k
(p · k) − k0 · p
(k · p) + k · k0
p · k0

st

1.7 Relativistic kinematics Consider the collision of an electron on a positron in the CM frame. Then the momenta are p = (E,p) k = (E,−p) p0 = E,p0
k0 = E,−p0

6 Then, dropping masses, and setting p2 = E2 − m2 = E2, all quantities may be expressed in terms of E and θ:

p · k = E2 + p2 = 2E2 − m2 = 2E2 p0 · k = E2 + p · p0 = E2 + p2 cosθ = E2 (1 + cosθ) k0 · k = E2 − p · p0 = E2 (1 − cosθ) p · p0 = E2 (1 − cosθ) p · k0 = E2 (1 + cosθ) p0 · k0 = E2 + p0 · p0 = 2E2 t = p − p0
2 = 2m2 − 2p · p0 = −2E2 (1 − cosθ) s = (p + k)2 = 4E2

Substituting into the squared matrix element,

4 4 1 2 8e 0
0
0
0

8e 0
0
0 0

iM fi = p · p k · k + p · k p · k + p · k p · k + (p · k) p · k 4 ∑ s2 t2 8e4 − p · p0
k · p0
− p0 · k
(p · k) − k0 · p
(k · p) + k · k0
p · k0

st 8e4 8e4 = E2 (1 − cosθ)E2 (1 − cosθ) + E2 (1 + cosθ)E2 (1 + cosθ)
+ E2 (1 + cosθ)E2 (1 + cosθ) + 2E22E2
4E24E2 2E2 (1 − cosθ)2E2 (1 − cosθ) 8e4 − E2 (1 − cosθ)E2 (1 + cosθ) − E2 (1 + cosθ)2E2 − E2 (1 + cosθ)2E2 + E2 (1 − cosθ)E2 (1 + cosθ)
4E2t Then the differential cross section is

dσ 1 2 = M ji dΩ 64π2s   2 4 2 4 + (1 + cosθ) e 2
=  1 + cos θ +  64π24E2 (1 − cosθ)2 e4  1  + 21 − cos2 θ
− 4(1 + cosθ)
64π24E2 1 − cosθ   2  4 2 4 + (1 + cosθ) 2
e 2
2 1 − cos θ 4(1 + cosθ) =  1 + cos θ + + −  64π24E2 (1 − cosθ)2 1 − cosθ 1 − cosθ ! e4 1 1 + cos4 θ θ 2cos2 θ = 1 + cos2 θ
+ 2 + 2cos2 − 2 64π22E2 2 4 θ 2 2 θ sin 2 sin 2 ! e4 1 1 + cos4 θ 2cos4 θ = 1 + cos2 θ
+ 2 − 2 64π22E2 2 4 θ 2 θ sin 2 sin 2 This is the ultrarelativistic limit of th Bhabha cross section (1936).

7 1.8 Traces of gamma matrices Now compute the traces using the fundamental relation γα ,γβ = 2ηαβ , and the cyclic property of the trace, tr (A...BC) = tr (CA...B). First, we can show that the trace of the product of any odd number of γ-matrices van- 2 a ishes by using γ5 = 1 and {γ5,γ } = 0,      α β  α β tr γ ...γ  = tr 1γ ...γ | {z } 2n+1  α β  = tr γ5γ5γ ...γ

 α β  = −tr γ5γ γ5 ...γ

2n+1  α β  = (−1) tr γ5γ ...γ γ5

2n+1  α β  = (−1) tr γ5γ5γ ...γ   = −tr γα ...γβ = 0

For even products, we will need traces of products of 2,4,6 and 8 gamma matrices.     tr γα γβ = tr −γβ γα + 2ηαβ 1   = −tr γβ γα + 2ηαβ tr (1)   = −tr γα γβ + 8ηαβ   tr γα γβ = 4ηαβ

and      tr γα γβ γ µ γν = tr −γβ γα + 2ηαβ 1 γ µ γν   = −tr γβ γα γ µ γν + 2ηαβ tr (γ µ γν )   = −tr γβ (−γ µ γα + 2η µα )γν + 2ηαβ tr (γ µ γν )     = tr γβ γ µ γα γν − 2η µαtr γβ γν + 2ηαβ tr (γ µ γν )     = tr γβ γ µ (−γν γα ) + 2ηνα − 2η µαtr γβ γν + 2ηαβ tr (γ µ γν )       = −tr γβ γ µ γν γα + 2ηναtr γβ γ µ − 2η µαtr γβ γν + 2ηαβ tr (γ µ γν )       2tr γα γβ γ µ γν = 2ηναtr γβ γ µ − 2η µαtr γβ γν + 2ηαβ tr (γ µ γν )   tr γα γβ γ µ γν = 4ηνα ηβ µ − 4η µα ηβν + 4ηαβ η µν

For six, we use the simple pattern to more quickly find      tr γα γβ γ µ γν γρ γσ = tr −γβ γα + 2ηαβ 1 γ µ γν γρ γσ   = tr −γβ (2ηαµ − γ µ γα )γν γρ γσ + 2ηαβ γ µ γν γρ γσ   = tr −γβ γ µ γν γρ γσ γα + 2ηασ γβ γ µ γν γρ − 2ηαρ γβ γ µ γν γσ + 2ηαν γβ γ µ γρ γσ − 2ηαµ γβ γν γρ γσ + 2ηαβ γ µ γν γρ γσ

8 and from here we can use the result for the trace of four,     tr γα γβ γ µ γν γρ γσ = tr ηασ γβ γ µ γν γρ − ηαρ γβ γ µ γν γσ + ηαν γβ γ µ γρ γσ − ηαµ γβ γν γρ γσ + ηαβ γ µ γν γρ γσ     = 4ηασ ηβ µ ηνρ − ηβν η µρ + ηρβ η µν − 4ηαρ ηβ µ ηνσ − ηβν η µσ + ησβ η µν     +4ηαν ηβ µ ηρσ − ηβρ η µσ + ηβσ η µρ − 4ηαµ ηβν ηρσ − ηβρ ηνσ + ηβσ ηνρ

+4ηαβ (η µν ηρσ − η µρ ηνσ + η µσ ηνρ )

or, perhaps more mnemonically,     tr γα γβ γ µ γν γρ γσ = 4ηαβ (η µν ηρσ − η µρ ηνσ + η µσ ηνρ ) − 4ηαµ ηβν ηρσ − ηβρ ηνσ + ηβσ ηνρ     +4ηαν ηβ µ ηρσ − ηβρ η µσ + ηβσ η µρ − 4ηαρ ηβ µ ηνσ − ηβν η µσ + ησβ η µν   +4ηασ ηβ µ ηνρ − ηβν η µρ + ηρβ η µν

From this, if we don’t run out of Greek letters, we can immediately write the result for eight gammas:     tr γα γβ γ µ γν γρ γσ γλ γτ = 4ηαβ (η µν ηρσ − η µρ ηνσ + η µσ ηνρ ) − 4ηαµ ηβν ηρσ − ηβρ ηνσ + ηβσ ηνρ     +4ηαν ηβ µ ηρσ − ηβρ η µσ + ηβσ η µρ − 4ηαρ ηβ µ ηνσ − ηβν η µσ + ησβ η µν   +4ηασ ηβ µ ηνρ − ηβν η µρ + ηρβ η µν

9