130_notes.dvi

i ̄h

dcn

dt

=

∑

~k,α

∑

j

Hnje−iωtcjeiωnjt

In the equations forcn, weexplicitly account for the fact that an intermediate state can
make a transition back to the initial state. Transitions through another intermediate state
would be higher order and thus should be neglected. Note that the matrix elements for the transitions
to and from the initial state are closely related. We alsoinclude the effect that the initial state
can become depletedas intermediate states are populated by usingcn(instead of 1) in the
equation forcj. Note also that all the photon states will make nonzero contributions to the sum.

Our task is tosolve these coupled equations. Previously, we did this by integration, but needed
the assumption that the amplitude to be in the initial state was 1.

Since we are attempting tocalculate an energy shift, let us make that assumptionand plug
it into the equations to verify the solution.

cn=e

−i∆Ent

h ̄

∆Enwill be a complex number, thereal part of which represents an energy shift, and the
imaginary part of which represents the lifetime (and energywidth)of the state.

i ̄h

dcj

dt

=

∑

~k,α

Hjneiωtcne−iωnjt

cn = e

−i∆ ̄hEnt

cj(t) =

1

i ̄h

∑

~k,α

∫t

0

dt′Hjneiωt

′

e

−i∆Ent′

̄h e−iωnjt

′

cj(t) =

1

i ̄h

∑

~k,α

∫t

0

dt′Hjnei(−ωnj−∆ωn+ω)t

′

cj(t) =

∑

~k,α

Hjn

[

ei(−ωnj−∆ωn+ω)t

′

̄h(ωnj+ ∆ωn−ω)

]t

0

cj(t) =

∑

~k,α

Hjn

ei(−ωnj−∆ωn+ω)t− 1

̄h(ωnj+ ∆ωn−ω)

Substitutethis back into the differential equation forcnto verify the solution andto find out
what∆Enis. Note that the double sum over photons reduces to a single sum because we must
absorb the same type of photon that was emitted. (We have not explicitly carried along the photon
state for economy.)

i ̄h

dcn

dt

=

∑

~k,α

∑

j

Hnje−iωtcjeiωnjt

cj(t) =

∑

~k,α

Hjn

ei(−ωnj−∆ωn+ω)t− 1

̄h(ωnj+ ∆ωn−ω)