I am trying to solve a problem where a system (a quantum harmonic oscillator) is suddenly perturbed by a large field of strength $E$. I want to calculate the transition probability for it to go from the ground state of the original Hamiltonian ($\frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$) to the $n$th state of the new Hamilitonian ($\frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$ – $qEx$).
If this were a small perturbation, then I would simply use first-order perturbation theory to calculate the transition probability.
However, in my case, the perturbation is not small. Therefore, first order approximations are not valid, and I would have to use the more general form given below:
My question is: Where does the perturbation effects (namely, $-qEx$ in the new Hamiltonian) come into play if I were to use Eq (10.73)? This seems like a far too general form, but I know that it is correct. Any suggestions on how I should approach this?
Best Answer
So the assumption that is being applied is that the underlying wavefunction $$|\Psi\rangle = \sum_n \alpha_n e^{-iE_nt/\hbar} |\psi_n\rangle,$$ is not going to change in the time it takes the Hamiltonian to change. (Let's assume $t=0$ when the Hamiltonian changes for simplicity.)
Therefore we should find a new basis $|\phi_n\rangle$ for the new Hamiltonian $\hat H$ and then re-expand $|\Psi\rangle$ in terms of the new basis. Since it is an orthonormal basis, $$\hat 1 = \sum_m |\phi_m\rangle\langle \phi_m|,$$and therefore we have $$|\Psi\rangle = \hat 1|\Psi\rangle = \sum_{mn}\alpha_n\langle\phi_m|\psi_n\rangle~|\phi_m\rangle.$$Your expression comes when $\alpha_n = \{1 \text{ if } n=k\text{ else } 0\}$, in which case we can simplify that sum to only $$|\psi_k\rangle = \sum_m \langle \phi_m|\psi_k\rangle ~|\phi_m\rangle.$$We see that the probability of being in the state $|\phi_\ell\rangle$ is $\big|\langle \phi_\ell|\psi_k\rangle\big|^2,$ as promised.
Observing as @jim says that the two harmonic oscillator solutions are shifted versions of each other in position space, this integral (before being squared) is $$\langle \phi_m|\psi_0\rangle = \int_{-\infty}^\infty dx~\psi_m^*(x - \lambda)~\psi_0(x),$$ and you can hypothetically just plug in those functions and evaluate that integral.
There may also be a more theoretical way to your solution by looking at the eigenstates of the usual bosonic annihilator (sometimes called a "ladder operator") $\hat a = \sqrt{m\omega\over2\hbar}\left(\hat x + \frac{i}{m\omega} \hat p\right)$ of the harmonic oscillator. These eigenstates look like $\hat a|\alpha = \alpha |\alpha\rangle$ and generally look like copies of the vacuum state shifted in their phase space to the position $x = \sqrt{2\hbar\over m\omega}~\Re \alpha$, $p = \sqrt{2\hbar m\omega} ~\Im \alpha,$ if I'm remembering all of the constants correctly.
Since $a |n\rangle = \sqrt{n} |n-1\rangle$ we can obtain a series expansion of $|\alpha\rangle = \sum_n c_n|\alpha\rangle$ where the equation $\hat a |\alpha\rangle = \alpha |\alpha\rangle$ becomes the expression $$c_n = \frac{\alpha}{\sqrt n} c_{n-1}.$$Solving we have $c_n = Z \alpha^n/\sqrt{n!}$ for some normalization constant $Z$ which must force $Z^2 \sum_n |\alpha|^{2n} / n! = 1$ and therefore $Z=\exp\big(-\frac12 |\alpha|^2\big).$
If all of that is correct then your transition probability will be something of the form $$\big|\langle \phi_k|\psi_0\rangle\big|^2 = \frac1{k!}~\exp\left(-\frac{m\omega \lambda^2}{2\hbar}\right)~\left(\frac{m\omega \lambda^2}{2\hbar}\right)^k.$$