130_notes.dvi

an electron is bound than they are if it is free. We will thereforecompute the amplitude from
the second diagram.

Hint = −

e

mc

A~·~p

A~ = √^1

V

∑

~k,α

√

̄hc^2

2 ω

ˆǫ(α)

(

a~k,αei(

~k·~x−ωt)

+a~†k,αe−i(

~k·~x−ωt))

This contains a term causing absorption of a photon and another term causing emission. We separate
the terms for absorption and emission and pull out the time dependence.

Hint =

∑

~k,α

(

Hk,α~abse−iωt+H~k,αemiteiωt

)

Habs = −

√

̄he^2

2 m^2 ωV

a~k,αei

~k·~x

~p·ǫˆ(α)

Hemit = −

√

̄he^2

2 m^2 ωV

a†~k,αe−i

~k·~x

~p·ˆǫ(α)

The initial and final state is the same|n〉, and second order perturbation theory will involve asum
over intermediate atomic states,|j〉and photon states. We will use the matrix elements of
the interaction Hamiltonian between those states.

Hjn = 〈j|H~k,αemit|n〉

Hnj = 〈n|H~k,αabs|j〉

Hnj = Hjn∗

We have dropped the subscript onHjnspecifying the photon emitted or absorbed leaving a reminder
in the sum. Recall from earlier calculations that the creation and annihilation operators just give a
factor of 1 when a photon is emitted or absorbed.

From time dependent perturbation theory, the rate of change ofthe amplitude to be in a state is
given by

i ̄h

∂cj(t)

∂t

=

∑

k

Hjk(t)ck(t)eiωjkt

In this case, we want to use the equations for the the state we arestudying,ψn, and all intermediate
states,ψjplus a photon. Transitions can be made by emitting a photon fromψnto an intermdiate
state and transitions can be made back to the stateψnfrom any intermediate state. We neglect
transitions from one intermediate state to another as they are higher order. (The diagram is emit a
photon fromψnthen reabsorb it.)

Thedifferential equations for the amplitudesare then.

i ̄h

dcj

dt

=

∑

~k,α

Hjneiωtcne−iωnjt