QM: Difference between these Initial Wavefunctions

[WWCY]

Homework Statement

I've been asked as a part of some school project to find the Fourier transform, and time evolution of the following initial wavefunctions:

1. $\Psi(x,0) = Ae^{\frac{-x^2}{2\sigma ^2}}$
2. $\Psi(x,0) = Be^{\frac{-x^2}{2\sigma ^2}}e^{\frac{ipx}{\hbar}}$

What physical difference does the $e^{ipx}$ term make?

To find the time evolution of 1 and 2, do I follow the following steps?
1. Normalize them
2. Find their Fourier transform
3. Plug it into their inverse Fourier transform $\int \frac{\tilde{\psi}}{\sqrt{2\pi}} e^{i(kx - \frac{\hbar k^2}{2m} t)}$

I was told to take $\hbar = 1$ and therefore $p = k$

Assistance is greatly appreciated

Homework Equations

The Attempt at a Solution

 

[hilbert2]

The factor $e^{ipx}$ gives the wavepacket a nonzero expectation value of momentum. So, it's a wavepacket that is "shot" to one direction with a certain velocity, unlike the real-valued Gaussian wavepacket that spreads to both directions equally fast.

 

[WWCY]

The factor $e^{ipx}$ gives the wavepacket a nonzero expectation value of momentum. So, it's a wavepacket that is "shot" to one direction with a certain velocity, unlike the real-valued Gaussian wavepacket that spreads to both directions equally fast.

Thanks for the response!

So is it right to think of this as a Gaussian wavepacket with some initial "average" momentum?

Edit:

I'd also like to ask about the $p$ terms in (2) and the Fourier transform $\frac{1}{\sqrt{2\pi}} \int \Psi(x,0) e^{-ipx}dp$. Are they the same thing? If not, what do they mean?