130_notes.dvi

=

Vni

i ̄h

[

ei(ωni+ω)t

′

i(ωni+ω)

]t′=t

t′=0

=

Vni

i ̄h

[

ei(ωni+ω)t− 1

i(ωni+ω)

]

=

Vni

i ̄h

ei(ωni+ω)t/^2

[

ei(ωni+ω)t/^2 −e−i(ωni+ω)t/^2

i(ωni+ω)

]

=

Vni

i ̄h

ei(ωni+ω)t/^2

2 sin ((ωni+ω)t/2)

i(ωni+ω)

Pn(t) =

Vni^2

̄h^2

[

4 sin^2 ((ωni+ω)t/2)

(ωni+ω)^2

]

In the last line above we have squared the amplitude to get the probability to be in the final state.
The last formula is appropriate to use, as is, for short times. For long times (compared toωni^1 +ω
which can be a VERY short time), the term in square brackets looks like some kind of delta function.

We will show (See section 28.4.1), that the quantity in square brackets in the last equation is
2 πt δ(ωni+ω). The probability to be in stateφnthen is

Pn(t) =

Vni^2

̄h^2

2 πt δ(ωni+ω) =

2 πVni^2

̄h^2

δ(ωni+ω)t=

2 πVni^2

̄h

δ(En−Ei+ ̄hω)t

The probability to be in the final stateφnincreases linearly with time. There is a delta function
expressing energy conservation. The frequency of the harmonicperturbation must be set so that
̄hωis the energy difference between initial and final states. This is trueboth for the (stimulated)
emission of a quantum of energy and for the absorption of a quantum.

Since the probability to be in the final state increases linearly with time, it is reasonable to describe
this in terms of atransition rate. The transition rate is then given by

Γi→n≡

dPn

dt

=

2 πVni^2

̄h

δ(En−Ei+ ̄hω)

We would get a similar result for increasingE(absorbing energy) from the other exponential.

Γi→n=

2 πVni^2

̄h

δ(En−Ei− ̄hω)

It does not make a lot of sense to use this equation with a delta function to calculate the transition
rate from a discrete state to a discrete state. If we tune the frequency just right we get infinity
otherwise we get zero. This formula is what we need if either the initialor final state is a continuum
state. If there is a free particle in the initial state or the final state, we have a continuum state. So,
the absorption or emission of a particle, satisfies this condition.

The above results are very close to a transition rate formula knownasFermi’s Golden Rule.
Imagine that instead of one final stateφnthere are acontinuum of final states. The total rate
to that continuum would be obtained by integrating over final stateenergy, an integral done simply
with the delta function. We then have