Time evolution of a particle in momentum space

[Foracle]

Homework Statement

At time t=0, the wave function of a particle is
$\Psi(x,0)=(\frac{\alpha}{\pi})^{\frac{1}{4}}e^{ik_{0}x-ax^{2}/2}$
$\alpha$ and $k_{0}$ are real constants

What is the wavefunction at time t in momentum space, $\tilde{\Psi}(k,t)$?

Relevant Equations

$\tilde{\Psi}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty }^{\infty}dxe^{ikx}\Psi(x)$

$U(t,t_{0})=e^{-\frac{i}{\hbar}\hat{H}t} = e^{-\frac{i}{\hbar}\frac{\hat{p^2}t}{2m}}$

Since it asks for the time evolution of the wavefunction in the momentum space, I write : $\tilde{\Psi}(k,t) = < p|U(t,t_{0})|\Psi> = < U^\dagger(t,t_{0})p|\Psi>$

Since $U(t,t_{0})^\dagger = e^{\frac{i}{\hbar}\frac{\hat{p^2}t}{2m}}$, the above equation becomes
$\tilde{\Psi}(k,t) = e^{\frac{-i}{\hbar}\frac{p^2t}{2m}} < p|\Psi> = e^{\frac{-i}{\hbar}\frac{p^2t}{2m}} \frac{1}{\sqrt{2\pi}}\int_{-\infty }^{\infty}dxe^{ikx}\Psi(x)$

Evaluate this with $\Psi(x)=(\frac{\alpha}{\pi})^{\frac{1}{4}}e^{ik_{0}x-ax^{2}/2}$, I end up getting :
$\tilde{\Psi}(k,t) = \frac{1}{\sqrt{a}}(\frac{\alpha}{\pi})^{1/4}e^{-\frac{(k-k_{0})^2}{2a}}e^{\frac{-i}{\hbar}\frac{p^2t}{2m}}$

Since $p=\hbar k$,
$\tilde{\Psi}(k,t) = \frac{1}{\sqrt{a}}(\frac{\alpha}{\pi})^{1/4}e^{-\frac{(k-k_{0})^2}{2a}}e^{\frac{-i}{\hbar}\frac{\hbar^2k^2t}{2m}}$

Is this the right way to solve this problem?

 

[TSny]

Your work looks good to me except for some sign issues.

Relevant Equations:: $\tilde{\Psi}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty }^{\infty}dxe^{ikx}\Psi(x)$

Check to see if the factor of $e^{ikx}$ in the integrand should actually be $e^{-ikx}$. With $e^{-ikx}$, then I think your result for $\tilde{\Psi}(k, t)$ is correct. But, if you use your way of writing $\tilde{\Psi}(k)$, then I believe you would get a result for $\tilde{\Psi}(k, t)$ with a factor of $\large e^{-\frac{(k+k_0)^2}{2a}}$ instead of $\large e^{-\frac{(k-k_0)^2}{2a}}$.

(But maybe I'm the one who's getting the signs wrong. )