QM - Position/Momentum representation problem

[knowlewj01]

Homework Statement

Write down the time independant Schrodinger eqn in the momentum representation for a particle with mass m when the potential is given by $V(x) = \frac{1}{2} \gamma x^2$
Given that a possible solution is given by $\Phi(p) = e^{\frac{-Bp^2}{2}}$

determine B and the corresponding energy eigenvalue.

Homework Equations

position / momentum

$x \rightarrow i\hbar \frac{d}{dp}$
$-i\hbar \frac{d}{dx} \rightarrow p$

The Attempt at a Solution

in position representation the full time independant schrodinger eqn is:

$- \frac{\hbar^2}{2m}\frac{d^2}{dx^2} \Psi(x) + \frac{1}{2}\gamma x^2 \Psi(x) = E\Psi(x)$

becomes

$\frac{p^2}{2m}\Phi(p) - \frac{\hbar^2 \gamma}{2}\frac{d^2}{dp^2}\Phi(p) = E\Phi(p)$
in the momentum representation, where Phi is the FT of Psi.

After plugging in the Trial Solution I get:

$E = \frac{p^2}{2m} - \frac{\hbar^2 \gamma}{2}\left( B^2 p^2 - B\right)$

Not sure what to do after this bit, I tried to normalise the wavefunction $A = \left(\frac{B}{\pi}\right)^{\frac{1}{4}}$

But I don't think that helps.

Any ideas on where i could get a second equation to find B and E?

Thanks

 

[vela]

The energy E needs to be a constant, so it can't depend on p. For what value of B will the dependence of E on p vanish?

 

[knowlewj01]

ok, the only way i can see of doing this is, first rearrange to get:

$p^2 \left(\frac{1}{2m} - \frac{B^2 \hbar^2 \gamma}{2}\right) + \frac{B\hbar^2 \gamma}{2} = E$

choose B such that:

$B^2 \frac{\hbar^2 \gamma}{2} = \frac{1}{2m}$

which gives:

$B = \frac{1}{\hbar}\sqrt{\frac{1}{\gamma m}}$

So the energy eigenvalue would be:

$\frac{\hbar}{2}\sqrt{\frac{\gamma}{m}}$

 

[vela]

Looks good.

 

[knowlewj01]

thanks very much