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