quantum mechanics - How to derive the Aharonov-Bohm effect result?

In the derivations of the Aharonov-Bohm phase, it is directly mentioned that due to the introduction of the vector potential $A$, an extra phase is introduced into the wavefunction for case $A\neq0$ i.e.

$$\psi(A\neq0) = \exp(\iota\varphi)\psi(A=0),$$

where

$$\varphi = \frac{q}{\hbar} \int_P \mathbf{A} \cdot d\mathbf{x}.$$

How to derive it from the following Schordinger equation

$$\left[\frac{1}{2m}(\frac{\hbar}{i}\triangledown-eA)^{2}+V(r)\right]\psi=\epsilon\psi.$$

I tried taking the terms containing $A$ on the right and treating the equation as an inhomogeneous equation but it just becomes tedious. What is the straightforward simple way?

Answer

First, I will set $e=1$ for simplicity.

Let $\psi_0$ denote the wave function that satisfies the free Schrodinger equation:

$$\begin{equation} i \frac{\partial \psi_0}{\partial t} = -\frac{1}{2m}\mathbf{\nabla}^2 \psi_0 + V \psi_0 \tag{1} \end{equation}$$ Furthermore, let $\psi$ be the wave function that obeys the Schrodinger equation for a non-vanishing vector potential $\mathbf{A}$: $$\begin{equation} i \frac{\partial \psi}{\partial t} = -\frac{1}{2m}(\mathbf{\nabla}-i\mathbf{A})^2 \psi+ V \psi \tag{2} \end{equation}$$ Let us now write: $$\begin{equation} \psi=\exp \left( i \int_{\gamma} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right)\psi_0 \end{equation}$$ where $\gamma$ is a path from some arbitrary point $\mathbf{x}_0$ to some other point $\mathbf{x}_1$. We can then write: $$\begin{equation} \left( \mathbf{\nabla} -i \mathbf{A} \right)^2 \psi = \exp \left( i \int_{\gamma} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right) \mathbf{\nabla}^2 \psi_0 \end{equation}$$ Substituting this expression into equation $(2)$ gives equation $(1)$. This implies that the wave function of an electrically charged particle travelling through space where $\mathbf{A} \neq 0$ will gain an additional phase.

We know that the wave function at the point $Q$ (see the figure below) is a result of quantum superposition, i.e. we can write:

$$\begin{equation} \begin{aligned} \begin{split} \psi_{\scriptscriptstyle Q} & = \psi(\mathbf{x},\gamma_1) + \psi(\mathbf{x},\gamma_2) \\& = \exp \left( i \int_{\gamma_1} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right)\psi_{0}(\mathbf{x},\gamma_1) + \exp \left( i \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right)\psi_{0}(\mathbf{x},\gamma_2) \\& = \exp \left( i \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right) \left( \exp \left( i \int_{\gamma_1} \mathbf{A} \cdot \mathrm{d} \mathbf{l} - i \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right)\psi_{0}(\mathbf{x},\gamma_1) + \psi_{0}(\mathbf{x},\gamma_2) \right) \end{split} \end{aligned} \end{equation}$$ We can use Stoke's theorem on the first term inside the brackets, because $\gamma_1-\gamma_2$ is a closed path: $$\begin{equation} \int_{\gamma_1} \mathbf{A} \cdot \mathrm{d} \mathbf{l} - \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} = \int \mathbf{B} \cdot \mathrm{d}\mathbf{S} = F \end{equation}$$ where $F$ is the total magnetic flux due to the solenoid through a surface defined by the closed boundary $\gamma_2-\gamma_1$. The wave function at $Q$ can now be written as: $$\begin{equation} \psi_{\scriptscriptstyle Q} = \exp \left( i \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right) \left( \exp \left( i F \right)\psi_{0}(\mathbf{x},\gamma_1) + \psi_{0}(\mathbf{x},\gamma_2) \right) \end{equation}$$ This shows that the relative phase difference, and thus the interference pattern, is dependent on the magnetic flux due to the solenoid. This is the Aharonov-Bohm effect.