130_notes.dvi

Relativistic corrections become quite important for highZatoms in which the typical velocity of
electrons in the most inner shells is of orderZαc.

36.16Thomson Scattering

The cross section forThomson scattering illustrates the need for “negative energy” states
in our calculations. Recall that we got the correct cross section from the non-relativistic calcula-
tion and that Thomson also got the correct result from classical E&M.

In the Dirac theory, we have only one term in theinteraction Hamiltonian,

Hint=ieγ 4 γkAk

Because it is linear inAit can create a photon or annihilate a photon. Photon scattering is therefore
second order (and proportional toe^2 ). Thequantized photon fieldis

Aμ(x) =

1

√

V

∑

kα

√

̄hc^2

2 ω

ǫ(μα)

(

ak,α(0)eikρxρ+a†k,α(0)e−ikρxρ

)

.

The initial and final states aredefinite momentum states, as are the intermediate electron states.
We shallfirst do the calculation assuming no electrons from the “negative energy” sea
participate, other than to exclude transitions to those “negative energy” states. The initial and

final states are therefore the positive energy plane wave statesψ
(r)
~p forr= 1,2. The intermediate
states must also be positive energy states since the “negative energy” states are all filled.

The computation of the scattering cross section follows the same steps made in the development
of the Krammers-Heisenberg formula for photon scattering. There is noA^2 term so we are just
computing thetwo second order terms.

c(2)p~′,r′;k~′,ǫˆ′(t) =

−e^2

̄h^2

∑

I

1

V

̄hc^2

2

√

ω′ω

∫t

0

dt 2 〈p~′r′;k~′ˆǫ(α

′)

|iγ 4 γn(ǫ(nα)ak,αei(

~k·~x−ωt 2 )

+ǫ(α

′)

n a

†

k′,α′e

i(−~k·~x+ω′t 2 ))|I〉ei(E′−E′′)

×

∫t^2

0

dt 1 ei(E

′′−E)t 1 / ̄h

〈I|iγ 4 γn(ǫ(nα)ak,αei(

~k·~x−ωt 1 )

+ǫ(α

′)

n a

†

k′,α′e

i(−~k·~x+ω′t 1 ))|~pr;~kˆǫ(α)〉

c(2)p~′,r′;k~′ǫˆ′(t) =

e^2 ̄hc^2

2 V

√

ω′ω

∑

p~′′r′′=1, 2

[

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

E′′−E− ̄hω

+

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

E′′−E+ ̄hω′

]

i

̄h

∫t

0

dt 2 ei(E

′−E+ ̄hω′− ̄hω)t 2 / ̄h

As in the earlier calculation, the photon states have been eliminated from the equation since they
give a factor of 1 with the initial state photon being annihilated and the final state photon being
created in each term.

Now lets take alook at one of the matrix elements. Assume theinitial state electron is at
rest and that the photon momentum is small.

〈p~′′r′′|iγ 4 γnǫnei

~k·~x

|~pr〉