Expectation value of momentum of wavefunction

[Reshma]

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]

0 momentum average seems reasonable to me because the wavefunction you're dealing with is that of the infinite well of length 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...

 

[quantumdude]

quasar has given the correct physical answer. Now just for completeness here is the mathematical answer. The expectation value of the momentum is zero because the integrand is odd with respect to the range of integration (the two sine functions are odd, and so is the momentum operator). Functions that exhibit odd symmetry on an interval always integrate to zero on that interval. In fact you could have validly done this integral by inspection, stating that as the reason.

 

[Reshma]

Thanks quasar and Tom Mattson!