# Calculation of a Berry phase in the Aharonov-Bohm effect

In his seminal paper, where he introduced the concept of geometric phase, Berry investigates, among other things, a quantum system in a box encircling the infinitely thin solenoid carrying flux $\Phi$. I am having problems correctly interpreting his work.

The setup: Consider a particle of charge $q$ and mass $m$ which is confined to a box $\mathcal{B}$. We can solve the Schrodinger equation for a particle in a box $$\left(\frac{\vec{p}'^2}{2m} + V(\vec{r}')\right) \psi_0(\vec{r}') = E \psi_0 (\vec{r}').$$ Here, $\vec{r}'$ is the position of the particle with the respect to the center of the box and $\vec{p}'$ its conjugate momentum. The potential $V(\vec{r}')$ can be arbitrary (but well-defined) and has to confine the particle so that its wave-function vanishes both at the boundary of the box $\psi(\partial \mathcal{B}) = 0$ and outside of it.

The plot thickens: Now introduce an infinitely long but infinitesimally thin solenoid that runs along the $z$ axis and produces the vector potential $$\vec{A} = \frac{\Phi}{2\pi r} \hat{\varphi},$$ written in cylindrical coordinates centered at some point on the solenoid. The Schrodinger equation now changes to $$\left(\frac{(\vec{p} - q \vec{A}(\vec{r}))^2}{2m} + V(\vec{r} - \vec{R})\right) \psi(\vec{r}) = E \psi(\vec{r}).$$ In these coordinates, we have $\vec{r}' = \vec{r} - \vec{R}$ and $\vec{p}' = \vec{p}$, with $\vec{R}$ being the position of center of the box as measured from the new origin. If the box $\mathcal{B}$ does not intersect the solenoid, the particle does not feel any magnetic field and the two wave-functions are related by the gauge transformation $$\psi (\vec{r}) = \exp\left(\frac{iq}{\hbar} \int_{\vec{r}_0}^\vec{r} \vec{A}(\vec{s}) \cdot d\vec{s}\right) \psi_0(\vec{r} - \vec{R}),$$ where $\vec{r}_0$ is an arbitrary reference point.

The problem: In his paper, Berry takes the arbitrary reference point to be the center of the box $\vec{r}_0 = \vec{R}$. Then he moves the box around the solenoid and shows there is the non-vanishing Berry phase $$\gamma = i \oint \langle \psi \vert \vec{\nabla}_\vec{R} \vert \psi \rangle \cdot d \vec{R} = \frac{q \Phi}{\hbar},$$ which is exactly equal to the Aharonov-Bohm phase. Now, since the choice of the point $\vec{r}_0$ is (or should be) arbitrary, I would expect to get the same Berry phase when I move the box around the solenoid, but keep the reference point $\vec{r}_0$ fixed in space. Unfortunately, in this way I lose the dependence on $\vec{R}$ in the gauge transformation and the $\vec{\nabla}_\vec{R}$ has nothing to act upon and gives zero for the Berry phase! What's even more interesting, I can do the calculation in which I don't move the box around the solenoid, instead I move the reference point $\vec{r}_0$ around the solenoid and get the same Berry phase as before $$\gamma = i \oint \langle \psi \vert \vec{\nabla}_{\vec{r}_0} \vert \psi \rangle \cdot d \vec{r}_0 = \frac{q \Phi}{\hbar}.$$

My question: I don't know how to interpret these results. It seems that Berry fine-tuned his arbitrary point, $\vec{r}_0 = \vec{R}$ so that he gets the Aharonov-Bohm phase as he moves the box around the solenoid. If one does not make that choice, one does not obtain the correct Aharonov-Bohm phase by moving the box, but by moving the arbitrary point $\vec{r}_0$. On the other hand, the arbitrary point does not even enter the Hamiltonian and so should not be even considered to be the parameter of the system which can be adiabatically changed etc. Is there a sensible way out of this mess?

Update: I have just found a potentially very relevant comment in Griffiths' QM textbook (first edition). In Chapter 10, he discusses the motion of the box around the solenoid and writes the following comment after equation (10.94), in a footnote:

It is convenient to set the reference point $\mathcal{O}$ at the center of the box, for this guarantees that we recover the original phase convention for $\psi$ when we complete the journey around the solenoid. If you use a fix point in space, you'll have to readjust the phase "by hand", at the far end; this leads to exactly the same answer, but it's a crude way to do it.

What does he mean by readjusting the phase "by hand" and how will it lead to the same answer for the Berry phase?

• Does the choice of reference point generate a family of phases that are related in some simple way? Then that is probably what he means? – Emil Jan 30 '18 at 7:28