Valid Solutions for Harmonic Oscillator Potentials

[Mohandas]

\psi(x,0) = N exp[-\alpha(x-a)^2] is a solution to the time-independent SE at time t = 0 for the potential

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

where N is a constant and \alpha = m\omega/(2\hbar).

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 \ V(x) = (1/2)*m\omega^2(x-a)^2 ?

 

[kdv]

\psi(x,0) = N exp[-\alpha(x-a)^2] is a solution to the time-independent SE at time t = 0 for the potential

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

where N is a constant and \alpha = m\omega/(2\hbar).

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 \ V(x) = (1/2)*m\omega^2(x-a)^2 ?

The obvious thing to do is to simply plug this wavefunction in the time-ind Schroedinger equation and show that the equation is satisfied only if a=0

 

[jtbell]

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

Yes, that's what "show that function f is a solution to the differential equation D" means.

(in which case, what am i looking for?) ?

After you plug the function and its derivatives into the differential equation, if the function is in fact a solution, you can cancel everything out (usually after some algebraic manipulation) and end up with 0 = 0.

 

[faen]

\psi(x,0) = N exp[-\alpha(x-a)^2] is a solution to the time-independent SE at time t = 0 for the potential

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

where N is a constant and \alpha = m\omega/(2\hbar).

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 \ V(x) = (1/2)*m\omega^2(x-a)^2 ?

Thats right, put it in the time independent schrödinger equation and do the gay maths.. Do you happen to do your home exam in the university in oslo? Me too, i got exactly the same task ^^