**Required math: calculus **

**Required physics: **Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.22.

While analyzing the free particle, we saw that we could construct a normalizable combination of stationary states by writing

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

The integral 1 cannot usually be done in closed form, but one case where it can is if the initial wave function is a Gaussian, of the form

To find , we normalize:

The integral comes out to from which we get

Finding requires finding via equation 2. So we get (using Maple to do the integral):

We can now use 1 again using Maple to do the integral:

where Maple was used for the integral.

Calculating can be done using Maple, but it seems to require a bit of help. First we write out the complex conjugate:

Then we calculate using the Maple command *simplify(evalc()) assuming positive* and we get

with

At , , so which is correct. The wave packet at is therefore a Gaussian centred at . As increases, gets smaller but the centre of the Gaussian does not move from so the packet spreads out. A couple of plots show this behaviour. We’ve set in both plots. In the red plot so and in the green plot is larger, at a value such that .

We can get the mean values of position and momentum by integration, although it takes a bit of work. By symmetry, . To get the other two average values, we use integration with Maple.

This shows that the wave function spreads out with time. At , but it then increases quadratically with .

Calculating starts with:

This can be evaluated with the Maple command *simplify(evalc(int(-h^2*simplify(evalc(psixtconj*(diff(psixt, x$2)))), x = -infinity .. infinity))) assuming positive *where *psixtconj* and *psixt* are the Maple expressions for and respectively. The result is:

The uncertainty principle thus becomes

The system has the least uncertainty at . Uncertainty increases with time as the wave packet spreads out.

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PeroInstead of calculating using $\Psi$, I found using $\phi(k)^2$ as the pdf for k, then got from that. This avoided any diffilcult integrals.

growescienceI guess it’s a bit easier, although you need to express in momentum space, which is

so finding still involves taking the second derivative of , although it’s true this is a bit easier than taking the second derivative of .