130_notes.dvi

perturbation carefully.

V(~r,t) = 2V(~r) cos(ωt)→ 2 Vcos(ωt) =V

(

eiωt+e−iωt

)

We have introduced the factor of 2 for later convenience. With that factor, we haveV times a
positive exponential plus a negative exponential. As before,V depends on position but we don’t
bother to write that for most of our calculations.

Putting this perturbation into the expression forcn(t), we get

cn(t) =

1

i ̄h

∫t

0

eiωnit

′

Vni(t′)dt′

=

1

i ̄h

Vni

∫t

0

dt′eiωnit

′(

eiωt

′

+e−iωt

′)

=

1

i ̄h

Vni

∫t

0

dt′

(

ei(ωni+ω)t

′

+ei(ωni−ω)t

′)

Note that the terms in the time integral willaverage to zero unless one of the exponents is
nearly zero. If one of the exponents is zero, the amplitude to be in the stateφnwill increase with
time. To make an exponent zero we must have one of two conditions satisfied.

ω = −ωni

ω = −

En−Ei

̄h

̄hω = Ei−En

Ei = En+ ̄hω

This is energy conservation for the emission of a quantum of energy ̄hω.

ω = ωni

ω =

En−Ei

̄h

̄hω = En−Ei

Ei = En− ̄hω

This is energy conservation for the absorption of a quantum of energy ̄hω. We can see the possibility
of absorption of radiation or of stimulated emission.

Fort→ ∞, the time integral of the exponential gives (some kind of)delta function of energy
conservation. We will expend some effort to determine exactly what delta functionit is.

Lets take thecase of radiation of an energy quantum ̄hω. If the initial and final states have
energies such that this transition goes, the absorption term is completely negligible. (We can just
use one of the exponentials at a time to make our formulas simpler.)

The amplitude to be in stateφnas a function of time is

cn(t) =

1

i ̄h

Vni

∫t

0

dt′ei(ωni+ω)t

′