A particle with mass m is in a one-dimensional infinite square-well potential of width a, so V(x)=0 for 0 <= x <= a, and there are infinite potential barriers at x=0 and x=a. Recall that the normalized solutions to the Schrodinger equation are
psi_n(x) = sqrt(2/a)sin[(n pi x)/a]
E_n = (hbar^2 (pi^2 n^2)/(2m a^2)
where n = 1,2,3,...
The particle is initially in the ground state. A delta-function perturbation
H_1 = K(delta(x-a/2))
(where K is a constant) is turned on at time t=-t_1, and turned off at t=t_1. A measurement is made at some later time t_2, where t_2 > t_1.
a) What is the probability that the particle will be found to be in the excited state n=3?
b) There are some excited states n in which the particle will never be found, no matter what values are chosen for t_1 and t_2. Which excited states are these?
Denote the eigenstates corresponding to the eigenvalue E_n as |n>. The wavefunction if the mass is in state
|n> is then
psi_n(x) = <x|n> = sqrt(2/a) sin(n pi x/a)
We can always express the state of the mass in the form:
|psi(t)> = sum over n from n = 1 to infinity of c_n(t) Exp(-i E_n t/hbar) |n>
The amplitude c_n to first order in perturbation theory is given by:
c_n(t) = 1/(i hbar) Integral from -t_1 to t of <n|V(t)|1> exp(i omega t) dt
where V(t) is the perturbation Kdelta(x-a/2)F(t) and omega = (E_n - E_1)/hbar = hbar pi^2(n^2 -1)/(2ma^2)
and F(t) = 1 if -t_1 < t< t_1
The c_n(t) are constant when t > t_1, as the perturbation is switched off and the system's evolution proceeds from that ...
We explain how, using first order time dependent perturbation theory, one can compute the transition probabilities. We give a rigorous proof that only transitions to odd n states are possible to all orders in perturbation theory.