Ground State of the Simple Harmonic Oscillator in p-space

[qubyte]

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?

 

[qubyte]

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.