130_notes.dvi

As long as the E field is weak, the initial state will not be significantly depleted and the assumption
we have made concerning that is valid. We do see that the transition probability oscillates with the
time during which the E field is applied. We would get a (much) larger transition probability if we
applied an oscillating E field tuned to have the right frequency to drivethe transition.

Clearly the probability to make a transition to the second excited state is zero in first order. If we
really want to compute this, we can use our first order result forc 1 and calculate the transition
probability to then= 2 state from that. This is a second order calculation. Its not too bad to do
since there is only one intermediate state.

28.4 Derivations and Computations

28.4.1 The Delta Function of Energy Conservation

For harmonic perturbations, we have derived a probability to be in the final stateφnproportional
to the following.

Pn∝

[

4 sin^2 ((ωni+ω)t/2)

(ωni+ω)^2

]

For simplicity of analysis lets consider the characteristics of the function

g(∆≡ωni+ω) =

[

4 sin^2 ((ωni+ω)t/2)

(ωni+ω)^2 t^2

]

≡

4 sin^2 (∆t/2)

∆^2 t^2

for values oft >>∆^1. (Note that we have divided our function to be investigated byt^2. For ∆ = 0,
g(∆) = 1 while for all other values for ∆,g(∆) approaches zero for larget. This is clearly some
form of a delta function.

To find out exactly what delta function it is, we need to integrate over ∆.

∫∞

−∞

d∆f(∆)g(∆) = f(∆ = 0)

∫∞

−∞

d∆g(∆)

= f(∆ = 0)

∫∞

−∞

d∆

4 sin^2 (∆t/2)

∆^2 t^2

= f(∆ = 0)

∫∞

−∞

d∆

4 sin^2 (y)

4 y^2

= f(∆ = 0)

2

t

∫∞

−∞

dy

sin^2 (y)

y^2

= f(∆ = 0)

2

t

∫∞

−∞

dy

sin^2 (y)

y^2

We have made the substitution thaty=∆ 2 t. The definite integral overyjust givesπ(consult your