Time evolution of the Gaussian packet

Question

I am trying to get the time evolution for the following initial condition: $$ \Psi(x,0) = \left(\frac{1}{2\pi \sigma^2} \right)^{\frac{1}{4}} e^{- \left(\frac{ x-x_{0}}{2 \sigma}\right)^{2}} e^{i\frac{ p_{0} x}{\hbar}}$$ I have got the Fourier transform of this function, $$\Phi(p) = \left(\frac{2 \sigma^{2}}{\pi \hbar^{2}} \right)^{\frac{1}{4}} e^{ - \frac{\sigma^{2} (p-p_{0})^{2}}{\hbar^{2}}}e^{i \sigma x_{0}},$$ and my question is:

To get the time evolution wave equation, should I transform using $$\Psi(x,t) = \frac{1}{\sqrt {2\pi\hbar}} \int_{-\infty}^{\infty} \Phi(p) e^{i\frac{px}{\hbar}} e^{-i \frac{p^{2}t}{2m\hbar}} \,dp$$

Or should I use $$\Psi(x,t) = \frac{1}{\sqrt {2\pi\hbar}} \int_{-\infty}^{\infty} \Phi(p) e^{i\frac{p(x-x_{0})}{\hbar}} e^{-i \frac{p^{2}t}{2m\hbar}}\,dp~?$$ I ask that because at $t=0$, $\langle x\rangle = x_{0}$. However, when I get the time evolution using the first equation I get that $\langle x\rangle = \frac{p_{0}}{m} t$, which means $\langle x\rangle=0$ at $t=0$.

My professor said that we should get $\langle x\rangle = x_{0} + \frac{p_{0}}{m} t$.

Could someone help me?

Answer 1

Or should I use...

No! You should not use flakey wrong expressions simply because they might get you to the right known answer by cancelling errors.

You should do the FT correctly, since you know it's wrong, as, for t =0, its inverse FT you start with $$\Psi(x,0) = \frac{1}{\sqrt {2\pi\hbar}} \int_{-\infty}^{\infty} \Phi(p) ~ e^{i\frac{px}{\hbar}} \,dp$$ fails to recover your original expression!

Your $\Phi(p)$ is clearly wrong; you spend too much time chasing after superfluous factors and symbols and muff the central point involved. You may first set $\hbar=1$ and $\sigma= 1/\sqrt 2$ to deal with balanced FT Gaussians, if you are not familiar with non-dimensionalization (which you should be), and reinstate those in your final expressions after you appreciate the point involved.

Ignoring overall normalizations which do not matter, you should be able to see that $$\Psi\propto e^{ip_0x_0} ~ e^{ip_0(x-x_0)} ~ e^{-(x-x_0)^2/2}~~~\leadsto \\ \Phi(p)\propto e^{ix_0(p-p_0)} ~ e^{-(p-p_0)^2/2} , $$ which will get you back to $\Psi$, this time centered on $x_0$.

You may then plug in the free kinetic energy propagator, as you started, to obtain the standard expressions, whose x amounts to your $x-x_0$.