**Required math: calculus **

**Required physics: **Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.43.

We’ve looked at the stationary Gaussian wave packet for the free particle. The initial wave function in that case was

We can turn this into a travelling Gaussian wave by adding a factor to the wave function:

where is a real constant.

Since we have added only a complex exponential, the normalization condition is the same as for the stationary case:

To find we follow the same procedure as in the stationary case. So we get

Given the initial wave function, we can find via Plancherel’s theorem:

So we can now find the general solution:

where Maple was used for the integral.

Calculating can be done using Maple, with the result:

with

The results above reduce to the stationary wave packet when .

At , , so which is correct. The wave packet at is therefore a Gaussian centred at . As increases, gets smaller but in this case, the peak of the Gaussian moves according to . The speed of the peak is .

By direct integration we find, . Calculating the other means requires a bit of effort but we can use Maple to do most of it. The results are:

All these results reduce to those for the stationary wave packet from problem 2.22 when .

The uncertainty principle thus becomes

which is the same result as in the stationary wave packet. Thus although the packet here travels with a constant speed, it spreads out at the same rate as the stationary packet.

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