# Aharonov-Bohm effect as a geometric phase-Adiabatic transfer not needed?

In his 1984 paper, Michael Berry proved that the Aharonov-Bohm effect is the same as a geometric phase. He did this by transferring a box containing charged particles around a solenoid. However, he mentions between equations 33 and 34 that the box need not be transferred adiabatically.

Why is this not needed?

To clarify the last point: Let $\Psi_0(\mathbf{r},t)$ be the freely propagating Schrödinger wavefunction in the absence of any electromagnetic fields. When we switch on the EM field, the Hamiltonian changes, in fact you now have $$H = \frac{(\mathbf{p} - q \mathbf{A})^2} {2m} + q \phi,$$ where $\mathbf{A}$ and $\phi$ are EM vector and scalar potential. A solution to the Schrödinger equation with EM field is then given by $\Psi(\mathbf{r},t) = \exp\left(\frac{i}{\hbar} \zeta(\mathbf{r},t) \right)\Psi_0(\mathbf{r},t),$ where $$\zeta(\mathbf{r},t) = \int_\Gamma A^\mu(x') \text{d} x'_\mu .$$ Here I have used the four-vector notation $A = ( \phi / c, \mathbf{A})$, x = $( ct, \mathbf{r})$, and $\Gamma$ denotes a path from some $x_0$ to $x$. This transformation is the origin of the Aharonov-Bohm phase. At no point any adiabatic assumption had to be forced on the solution. The interpretation however is similar. You can view the four-vector potential as a connection in a fibre bundle just like the Berry connection.