A system is in an eigenstate | psi_i > with energy E_i. The perturbation
V(t) = H'exp(-((alpha)^2)(t^2))
is turned on at t_i = -infinity and left on until t_f = infinity. Here H' is independent of time, and alpha is a constant. Show that at t_f = infinity, the probability that the system has evolved into the eigenstate | psi_f > with energy E_f is
P(i -> f) = (pi)/((hbar^2)(alpha^2)|< psi_f | H' |psi_i >|^2 exp[-((E_f - E_i)^2)/(2(hbar^2)(alpha^2)].© BrainMass Inc. brainmass.com March 21, 2019, 8:25 pm ad1c9bdddf
If we denote the time dependent wave function of the system by |psi(t)> and the eigenstates with energy E_r by |r>, then we can write:
|psi> = sum over r of c_r (t) exp(-E_r t/hbar) |r>
The coefficients c_r(t) would not depend on time if there were no time dependent perturbation. If the system is initially in the state |i>, then that means that initially c_i = 1 and all other coefficients are zero. If we compute the coefficients in perturbation theory, we find ...
We consider a time dependent perturbation of the form
V(t) = H'exp[-((alpha)^2)(t^2)]
and compute to first order in perturbation theory the transition probability for an eigenstate with energy E_1 at
t = -infinity to evolve to an eigenstate with energy E_2 at t = infinity.