Solutions of the Harmonic Oscillator are $not$ always a Combination of Separable Solutions?

Question

Are there solutions of the Schrödinger equation that are not a linear combination of separable solutions and how do we find them?

In Griffiths, Quantum, Prob. 2.49, there is a solution of the (time-dependent) Schrödinger equation, which reads $$ \Psi(x,t)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left[-\frac{m\omega}{2\hbar}\left(x^2+\frac{a^2}{2}(1+e^{-2i\omega t})+\frac{i\hbar t}{m}-2axe^{-i\omega t} \right)\right]. $$ It seems that this is not a linear combination of the stationary states that he found previously in the chapter.

If it is the caes, does that mean that solving the time-dependent Schrödinger equation by separation of variables does not yield the general solution as the author claimed? if so, how do we find the other solutions?

Answer 1

Sometimes the expansions are not obvious. For example The harmonic oscillator time-dependent Schr"odinger equation $$ i\partial_t \psi = -\frac 12 \partial^2_x \psi +\frac 12 \omega^2 x^2 \psi $$ has a ``breathing'' solution $$ \psi(x,t)= \left(\frac{\omega}{\pi}\right)^{1/4}\frac 1{\sqrt{e^{i \omega t} +R e^{-i\omega t}}}\exp\left\{ - \frac \omega 2 \left(\frac{1-R\,e^{-2i\omega t}}{1+R\,e^{-2i\omega t}}\right)x^2\right\}, $$ where the parameter $|R|<1$.

Mehler's formula gives expansion in terms of the states as
$$\psi(x,t) {=}\pi^{1/4}\sum_{n=0}^\infty e^{-i(n+1/2) \omega t} \varphi_n(0)(i\sqrt R)^n \frac{\varphi_n(\sqrt{\omega} x)}{(\omega)^{1/4}}. $$ Here $$ \varphi_n(x)\equiv \frac{1}{\sqrt{2^n n! \sqrt{\pi}}} H_n(x) e^{-x^2/2} $$ is the normalized $\omega=1$ harmonic oscillator wavefunction. Now $\varphi_n(0)$ vanishes if $n$ is odd, and $$ \pi^{1/4}\varphi_{2n}(0)= \frac{1}{\sqrt{4^n (2n)! } } \frac{(2n)!}{n!}(-1)^{n}. $$ so one has found as set of quite "non obvious" expansion coefficients.