[tex]\psi(x,0) = N exp[-\alpha(x-a)^2][/tex] is a solution to the time-independent SE at time(adsbygoogle = window.adsbygoogle || []).push({}); t = 0for the potential

[tex]\ V(x) = (1/2)*m\omega^2x^2[/tex]

whereNis a constant and [tex]\alpha = m\omega/(2\hbar)[/tex].

I'm asked to show that the solution is valid only if a = 0.

I'm a little at loss as to what strategy i should use to show this.

My initial thought was to somehow exploit the fact that the potential can be expanded in a power series about the origin, and that the solution (for a yet unknown reason) should also be expanded about the origin (and not about 'a'), but it kinda stopped right there..

Of course it might be as easy as just to differentiate (for a=0 and a=constant) the solution and plug it into SE and evaluate the result (in which case, what am i looking for?) ?

Btw, would the solution be valid for any 'a' if the potential was [tex]\ V(x) = (1/2)*m\omega^2(x-a)^2[/tex] ?

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# Homework Help: Harmonic Oscillator (QM)

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