# Expectation value of momentum of wavefunction

I have a wavefunction given by:
$$\psi = \sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}$$
With boundary conditions 0<x<L.

When I compute the expectation value for the momentum like this:

$$\overline{p_x} = \int_0^L \sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L} \left(-i\hbar \frac{\partial}{\partial x}\right)\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L} dx$$

On evaluation I get:
$$\frac{2n\pi \hbar}{iL^2}\left[\frac{L}{2n\pi}\sin^2 \frac{n\pi x}{L}\right]_0^L = 0$$
Why is the expectation value of the momentum Zero?

quasar987
Homework Helper
Gold Member
0 momentum average seems reasonable to me because the wavefunction you're dealing with is that of the infinite well of lenght L. And this setting being perfectly symetrical, it seems reasonable to believe that once out of two, we'll observe the momentum to be positive (particle voyaging in the positive x direction), and once out of two negative (particle voyaging in the negative x direction). Just as once out of two, you'll observe the particle in the left side of the box, and the other hald of the time in the positive section of the box.

The expected value of the SQUARE of the momentum is not zero though! As expected...

Tom Mattson
Staff Emeritus