Probability wave function is still in ground state after imparting momentum

[xoxomae]

Homework Statement

An interaction occurs so that an instantaneous force acts on a particle imparting a momentum $p_{0} = \hbar k_{0}$ to the ground state SHO wave function. Find the probability that the system is still in its ground state.

Homework Equations

$\psi _{0} =\left( \frac{mw}{\hbar\pi} \right )^\frac{1}{4} e^{-mwx^{2}/2\hbar}$

The Attempt at a Solution

[/B]
$\Psi(x)=\psi_{0}e^{ik_{0}x}$
This wave function gives a <p> =$\hbar k_{0}$
Im confused whether this is the correct Fourier transform to do.
$c(k)=\frac{1}{2\pi}^{0.5}\int_{-\infty}^{\infty}e^{-ikx}\psi_{0}e^{ik_{0}x}dx$
And then solving for when the wavenumber of the ground state using E0=0.5* hbar * w.
Therefore
$c(k)^2$ = Probability

Is this correct?

 

[Simon Bridge]

You want to know the probability of a particular result of a measurement of energy.

 

[xoxomae]

Ah right. So would this be the probability of the system still being in the ground state?
$\left | \int_{-\infty}^{\infty} \psi_{0} \psi_{0} e^{ik_{o}x}dx\right |^2=P$

 

[Simon Bridge]

Well done - left out a star, but $\psi_0$ is real so...

 

[xoxomae]

Thank you. Do you by any chance know any way to solve that integral by hand? I solved it on mathematica and it gave me $P=e^\frac{-{k_{0}}^{2}\hbar}{4mw}$ but I have no idea how to solve it by hand. Thanks again.

 

[Simon Bridge]

Hint: integrate by parts. (That's normal for quantum.)
You may be able to shortcut using $\int_\infty \psi_0^2\; dx = 1$