# QM 2D Gaussian wave packet translation

I've been reading a lot but cannot find an example of 2D Gaussian wave packet moving in a particular direction. I've done some of the math myself, in a 1D case, and then kind of guessed the generalization to 2D. I'd be thrilled if someone could confirm or correct the expression. This is supposed to model a free particle moving in the x axis with a fairly definite momentum:

$\Psi (x,y,t) = \frac{1}{\sqrt{\alpha + i\beta t}} \exp \left[-\dfrac{(x-vt)^2+y^2}{4(\alpha + i\beta t)} + i(kx-\omega t) \right]$

The wave function is not normalized and the constants $\alpha$ and $\beta$ are not relevant in my opinion, I could add how they relate with mass and $\hbar$ if someone thinks they're necessary.

Also I would like to add to the question what's the relation between $k$, $\omega$ and $v$ and what does each one represent. My guess is that $v$ is the group velocity, and you'd get the average momentum from the fourier transform of that wave function, but I'm not sure what's $k$ then. Is there any need at all to add the phasor $e^{i(kx-wt)}$, does it change anything?? The packet is going to move anyway and the amplitude is not going to change unless you add other wave functions.

Thank you very much!

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Could you show the work you've done in the one-dimensional case? It might help us answer your question more accurately. –  Sanath Devalapurkar Jan 25 '14 at 3:52
Saying "myself" was rather optimistic, we did it in class. I believe the reasoning was the following: assume Gaussian distribution of momentum, take taylor expansion of w(k) up to second power, then take fourier transform. Then w, v and beta came from that expansion. I'm guessing k was the mean momentum in the initial gaussian momentum distribution then. –  freejuices Jan 25 '14 at 4:04
You could use multivariable calculus to take a $2$-dimensional Fourier transform of the two-dimensional analogs of the calculations you did in class. –  Sanath Devalapurkar Jan 25 '14 at 22:13