Time Dependence of a Gaussian Wave Packet

Assume we start with our Gaussian (minimum uncertainty) wavepacket
at
.
We are not interested in careful normalization here so we will drop constants.

We write explicitly that depends on . For our free particle, this just means that the energy depends on the momentum. To cover the general case, lets expand around the center of the wave packet in k-space.

We anticipate the outcome a bit and name the coefficients.

We still need to do the integral as before. Make the substitution giving . Factor out the constant exponential that has no dependence.

We now **compare this integral to the one we did earlier** (so we can avoid the work of completing the square again).
Dropping the constants, we had

Our new integral is the same with the **substitutions**
,
,
and
.
We can then write down the answer

Jim Branson 2013-04-22