Why does gaussian wavepacket have a minimum uncertainty product? Is it because of the way variance is defined?
The Gaussian distribution is the only mathematical function that has the same form in x-space and k-space. That should point you in the right direction. I recommend http://farside.ph.utexas.edu/teaching/qmech/lectures/node24.html if you want more information.
You can derive this: In the derivation of the uncertainty relation you use two inequalities: Cauchy Schwarz and Im(z)
The gaussian distribution is the max entropy solution on the open interval (-infinity , infinity) whether it is expressed in position or momentum space. Since it is a maximum in both domains, it has to be an equality. Note I have never seen this expressed explicitly, but it is another way to think about it, which the above arguments are more standard way to talk about it QM.
The situation is a little bit more complicated if the Gaussian has non-zero covriance. Such states saturate the Robertson-Schrödinger inequality. The latter is, anyway, better to use in all theoretical questions since it is symplectically covariant.
This may be helpful ...
Consider a Gaussian "packet" in the time (sec) domain with a variance of sig^2.
Compute its Fourier transform.
In the frequency (rad/sec) domain,
this "packet" is also Gaussian, but now it has a variance of 1/sig^2.
If we take the variance to be a measure of "uncertainty" of a pulse or packet, then the product of the uncertainties in both domains is unity!
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