130_notes.dvi

For~p= 0 and~k= 0, a delta function requires thatp~′′= 0. It turns out thatu(r

′′)†

0 γ^4 γnu

(r)
0 = 0, so
that the cross section is zero in this limit.

γi =

(

0 −iσi

iσi 0

)

γ 4 =

(

1 0

0 − 1

)

γ 4 γi =

(

0 −iσi

−iσi 0

)

This matrix only connectsr= 1,2 spinors tor= 3,4 spinors because of its off diagonal nature. So,
the calculation yields zero for a cross section in contradiction to theother two calculations. In fact,
since the photon momentum is not quite zero, there is a small contribution, but far too small.

The above calculation misses someimportant terms due to the “negative energy” sea. There
are additional terms if we consider the possibility that the photon can elevate a “negative energy”
electron to have positive energy.

In one term, the initial state photon is absorbed by a “negative energy” electron, then the initial
state electron fills the hole in the “negative energy” sea emitting thefinal state photon. In the other
term, even further from the mass shell, a “negative energy” electron emits the final state photon
and moves to a positive energy state, then the initial state electron absorbs the initial photon and
fills the hole left behind in the sea. These terms are larger because theγ 4 γimatrix connects positive
energy and “negative energy” states.

c(2)p~′,r′;k~′ǫˆ′(t) =

ie^2 c^2

2 V

√

ω′ω

∑

p~′′r′′=3, 4

[

〈p~′′r′′|iγ 4 γnǫ′ne−ik~′·~x|~pr〉〈~p′r′|iγ 4 γnǫnei~k·~x|p~′′r′′〉

E′−E′′− ̄hω

+

〈p~′′r′′|iγ 4 γnǫnei~k·~x|~pr〉〈p~′r′|iγ 4 γnǫ′ne−ik~′·~x|p~′′r′′〉

E′′−E′+ ̄hω′

]∫t

0

dt 2 ei(E

′−E+ ̄hω′− ̄hω)t 2 / ̄h