Ground State of the Simple Harmonic Oscillator in p-space

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Homework Statement

A particle is in the ground state of a simple harmonic oscillator, potential → V(x)=$\frac{1}{2}$mω$^{2}$x$^{2}$
Imagine that you are in the ground state |0⟩ of the 1DSHO, and you operate on it with the momentum operator p, in terms of the a and a† operators. What is the eigenstate? Then, how would you represent this operation in terms of a probability density in momentum space?

Homework Equations

p=i$\sqrt{\frac{\hbar m ω}{2}}$(a†-a)
Fourier Transform, x→p-space

The Attempt at a Solution

p|0⟩=i$\sqrt{\frac{\hbar m ω}{2}}$|1⟩
So now the particle is in the first excited state of the SHO

But I don't understand the next part of the problem. How can I represent this terms of probability density in p-space?

I can perform the Fourier Transform and find that,

$\Phi$(p,t)=$\sqrt{2}$($\frac{\pi \hbar}{m ω}$)$^{\frac{1}{4}}$ e$^{\frac{-p^{2}}{2 \hbar m ω} - \frac{i t ω}{2}}$

And the probability density is,

|$\Phi$(p,t)|$^{2}$=2$\sqrt{\frac{\pi \hbar}{m ω}}$ e$^{\frac{-p^{2}}{\hbar m ω}}$

I don't believe I understand what this part of the question is asking. Any suggestions/ideas would be greatly appreciated.

 

[vela]

It doesn't make sense to me either. Is that the problem statement exactly as it was given to you?

 

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Yes. Word for word.

Maybe it is supposed to be something like; "Find the expectation value of p...and then do the same thing in p-space."
But that seems too simple and we've already covered that. The p operator in p-space is just p.