Pictures Paths Particles Processes

November 7, 2013 187
〈
|A2 | 2〉 = 1 4 Tr ( (/q + m) γ β (/q − /k 1 + m) γ α (/p + m) γ α (/q − /k 1 + m) γ β
)
〈A 1 A ∗ 2〉 = 〈A 2 A ∗ 1〉
= 16m 4 − 8(pq)m 2 + 8(pk 1 )(qk 1 ) + 8(pk 1 )m 2 − 16(qk 1 )m 2 ,
= 1 4 Tr ( (/q + m) γ α (/p + /k 1 + m) γ β (/p + m) γ α (/q − /k 1 + m) γ β
)
= 8(pq)(pk 1 ) − 8(pq)(qk 1 ) + 16(pq)m 2 − 8(pq) 2
−4(pk 1 )m 2 + 4(qk 1 )m 2 . (7.45)
We can most easily evaluate this in the photon-electron centre-of-mass frame 11 .
In this frame, we have
p 0 = q 0 = s + m2
2 √ s
, |⃗p| = |⃗q| = | ⃗ k 1 | = | ⃗ k 2 | = K
2 √ s , (7.46)
where K = s − m 2 : and the angle between ⃗q and ⃗ k 1 is denoted by θ. Putting
everyhting together, we find
〈
|M|
2 〉 = 16π 2 α
(8 2 m2
K + 8 m4
K 2 + 2
m 4
− 8
(qk 1 )K +
The phase space integration element is given by
dV (p + k 1 ; q, k 2 ) = 1 1
(2π) 2 8
m4
(qk 1 ) 2 − 4 m2
K
(qk 1 ) + 4(qk 1)
K
K
s
(qk 1 )
)
. (7.47)
dΩ , (7.48)
where Ω is the solid angle of the emitted electron. The flux factor is
1
2λ(s, m 2 , 0) 1/2 = 1
2K . (7.49)
The only nontrivial quantity in the computation is
(
)
0
(qk 1 ) = k 1 q 0 − |⃗q| cos θ = K (
)
(s + m 2 ) − K cos θ
4s
and we can find the angular averages
∫
1
4π
∫
1
dΩ 1
4π (qk 1 )
∫
1 1
dΩ
4π (qk 1 ) 2 =
dΩ (qk 1 ) = K(s + m2 )
,
4s
= 2s (
K 2 log 1 + K )
m 2
,
, (7.50)
4s
m 2 K 2 . (7.51)
11 In the actual experiment, the photon will of course be impingeing on the stationary
electron ; but since the cross section is invariant we may choose any frame we want.