Phy5646/Simpe Example of Time Dep Pert

This example taken from "Quantum Physics" 3rd ed., Stephen Gasiorowicz, p. 237-238.

Problem: A particle of charge $q$ in a one-dimensional harmonic oscillator of characteristic frequency $\omega$ is placed in an electric field that is turned on and off so that the potential energy is

$\lambda V(t)=qExe^{-t^{2}/\tau ^{2}}$

If the particle is initially in the ground state, what is the probability that after time $t$, such that $t>>\tau$, the particle is found in the first excited state of the harmonic oscillator? What is the probability that it is found in the second excited state?

Solution: The harmonic oscillator eigenstates: $|n\rangle$, with eigen-energies $h\omega (n+1/2)$. Then

${\begin{aligned}\langle n|\psi (t)\rangle &=\langle n|0\rangle +{\frac {1}{i\hbar }}\int _{-\infty }^{t}dt'\langle n|V(t')|0\rangle \\&\end{aligned}}$

cn{t) = %r \ dt' <?"■*'(n|jc|0)c" in J Vf Since (»T,we may take the upper and lower limits in the time integral in (15-10) to be ±oo. We therefore get c«(°°) = ^-<«W0> I dt' e-'^'e-'"'72 in J — <x = ~<w|jt|0> J dt ■' e-""'e-2h2 = \^T^{n\x\0)e-w«  in This means that q2Ey h2 Pn{^ = ^~r-\(n\x\0)\2e^Wl2 To calculate («|jc|0) we use (6-35), to get x= l^—{A+A+) and recall that A|0> = 0 and A*|0) = |1>. Thus |<nbc|0>|2 = -^- S„. and therefore 2ma) 2 mtuo We also see that Pn = 0 for n = 2, 3, Note the following 1. As t —> oo Pt —> 0. When the electric field is turned on very slowly, then the transition probability goes to zero; that is, the system adjusts adiabatically to the presence of the field, without being "jolted" into making a transition. 2. The potential, through its jc-dependence, will allow only certain transitions to take place. In other words, we see that there is a selection rule that operates here. If the potential were proportional to x2, for example, then transitions to n = 2 would be allowed.