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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?

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    $\begingroup$ Why do you say that "it is clear that this is not a linear combination of the stationary states that he found previously in the chapter"? It is not at all clear to me! Have you looked at Mehler's formula? (see Wikipedia for this) $\endgroup$ – mike stone Sep 28 at 12:12
  • $\begingroup$ @mikestone I should have said it seems not a linear combination of the stationary states, thanks for this note!, I'll edit it $\endgroup$ – Physor Sep 28 at 12:15
  • $\begingroup$ @mikestone should I remove the question `? $\endgroup$ – Physor Sep 28 at 12:18
  • $\begingroup$ No! I just posted an "answer" of a sort. $\endgroup$ – mike stone Sep 28 at 12:29
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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.

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