I Expectation value of the momentum for an electron in a box

[Haorong Wu]

TL;DR Summary

Why is the expectation value of the momentum 0 for a electron confined in a box?

In studying the Aharonov-Bohm effect, a model of an electron confined in a box is used, for example, on page 353 of Modern Quantum Mechanics by Sakurai et al. The box makes one turn along a closed loop surrounding a magnetic flux line.

In the derivation, there will be an integration involving the expectation value of momentum as \begin{equation}
\int d^3 x \psi^*_n(\mathbf r-\mathbf R)\nabla_{\mathbf R}\psi_n(\mathbf r-\mathbf R),
\end{equation}
where $\mathbf R$ is the vector connecting the origin and a reference point in the box, $\mathbf r$ is the position vector of the electron, and $\psi_n$ is the $n$th wave function of the electron in the box. This term is equivalent to $$-\int d^3 x \psi^*_n(\mathbf r-\mathbf R)\nabla_{\mathbf r}\psi_n(\mathbf r-\mathbf R)\propto \left < \mathbf p\right >.$$

The textbook just reads: The (second) term under the integral vanishes for the electron in the box. I do not see how this comes. Some other materials say that the eigenstate $\psi_n$ is stationary, so its expectation value of the position is constant and the expectation value of the momentum vanishes.

This is still not clear to me. The box is moved around the magnetic flux line. Does not the position of the electron change in this process? If the expectation value of the momentum is zero, how can the electron in the box move around the magnetic flux?

 

[PeterDonis]

the expectation value of momentum

As I understand it, it's the expectation value of the momentum of the electron relative to the center of mass of the box that is claimed to be zero. That is separate from the momentum of the box itself.

 

[Haorong Wu]

@PeterDonis , from the image in the book, it appears that $\mathbf r$ is not the relative position vector.

I may solve the problem. Denote $\mathbf x=\mathbf r-\mathbf R$. Then, $\nabla_{\mathbf R}\psi_n(\mathbf r-\mathbf R)=(\frac \partial {\partial R^1}\psi_n(\mathbf r-\mathbf R),\cdots, \frac \partial {\partial R^N}\psi_n(\mathbf r-\mathbf R))$, with $\frac \partial {\partial R^i}\psi_n(\mathbf r-\mathbf R)=\frac {\partial \psi_n(\mathbf x)} {\partial x^j}\frac {\partial x^j}{\partial R^i}=-\delta^j_i \frac {\partial \psi_n(\mathbf x)} {\partial x^j}=-\frac {\partial \psi_n(\mathbf x)} {\partial x^i}$. Of course, the average of $\frac {\partial} {\partial x^i}$ is zero.

I hope this is right. I think the statement in Sakurai's book is somewhat misleading. It reads that $\nabla_{\mathbf R}$ is simply a gradient operator in the space and direction of $\mathbf R$. This causes me to think this is somehow related to directional derivatives $\nabla_{\mathbf v}=\mathbf v \cdot \nabla$.

 

[PeterDonis]

it appears that $\mathbf r$ is not the relative position vector.

I didn't say it was. I said that the expectation value is for the relative position. The expectation value integral has $\mathbf{r} - \mathbf{R}$ as the argument of the wave function, not just $\mathbf{r}$.