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?

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

Let $\psi_0$ denote the wave function that satisfies the free Schrodinger equation: $$i \frac{\partial \psi_0}{\partial t} = -\frac{1}{2m}\mathbf{\nabla}^2 \psi_0 + V \psi_0 \tag{1}$$ Furthermore, let $\psi$ be the wave function that obeys the Schrodinger equation for a non-vanishing vector potential $\mathbf{A}$: $$i \frac{\partial \psi}{\partial t} = -\frac{1}{2m}(\mathbf{\nabla}-i\mathbf{A})^2 \psi+ V \psi \tag{2}$$ Let us now write: $$\psi=\exp \left( i \int_{\gamma} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right)\psi_0$$ where $\gamma$ is a path from some arbitrary point $\mathbf{x}_0$ to some other point $\mathbf{x}_1$. We can then write: $$\left( \mathbf{\nabla} -i \mathbf{A} \right)^2 \psi = \exp \left( i \int_{\gamma} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right) \mathbf{\nabla}^2 \psi_0$$ 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{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} We can use Stoke's theorem on the first term inside the brackets, because $\gamma_1-\gamma_2$ is a closed path: $$\int_{\gamma_1} \mathbf{A} \cdot \mathrm{d} \mathbf{l} - \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} = \int \mathbf{B} \cdot \mathrm{d}\mathbf{S} = F$$ 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: $$\psi_{\scriptscriptstyle Q} = \exp \left( i \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right) \left( \exp \left( i F \right)\psi_{0}(\mathbf{x},\gamma_1) + \psi_{0}(\mathbf{x},\gamma_2) \right)$$ 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.

• :- the answer is nice. But you already assumed that the $\psi(A\neq0)$ is of a particular form. Can we do better than this ? Feb 22, 2014 at 14:48
• @user38579 I am not aware of a "better" procedure. Feb 22, 2014 at 14:58

To simplify the problem, we may neglect the potential energy term $V(r)$, as it is simply irrelevant to our derivation. So we write the Hamiltonian as $$H=\frac{1}{2}(-i\partial_x-A)^2.$$ The ground state is given by minimization of the energy. As the Hamiltonian is a square of $(-i\partial_x-A)$, so it is minimized when $(-i\partial_x-A)=0$. Which means on the ground state, we roughly have $$(-i\partial_x-A)\psi=0.$$ If we only care about the phase configuration of the wave function, we may write $\psi\sim e^{i\phi}$, and substitute into the above equation, $$(\partial_x\phi -A)e^{i\phi}=0,$$ which means $\partial_x\phi=A$, and its solution is $\phi=\int A \cdot\mathrm{d}x$.

• Hi Everett. Sorry for the bump on this old answer. Aren't you worried that \psi defined this way is not single-valued unless the flux is an integer? I found a similar discussion by Berry in this paper iopscience.iop.org/article/10.1088/0143-0807/1/3/008 Jul 27, 2022 at 2:50
• Hi Ryan, nice comment. You are right, $\psi$ is not single-valued, which means the wavefront is not globally defined in the presence of non-integer flux. I think a better way to formulate the AB effect is to use the path integral, which does not rely on a globally defined wavefront. Jul 28, 2022 at 6:36
• I've gone quite far down this rabbit hole since yesterday! I find it to be a fascinating problem. The original Aharonov-Bohm paper even talks about this, and in fact they gave a single-valued solution. The most beautiful derivation (from Berry) I found of this solution actually does follow your trick of singular gauge transformations, but in a resummation of the usual angular expansion is.muni.cz/el/sci/jaro2015/F8592/um/Berry.pdf . It seems to match the path integral picture of summing over the ways that the path can encircle the flux, but I don't know how to relate them directly. Jul 28, 2022 at 6:47
• @RyanThorngren I see. Thanks for pointing out Berry's nice paper. What I got from the paper is that the two least-whirling paths dominate the path integral as they accumulate the most stationary actions, so it is fair to approximate the phase structure by the singular gauge transformation. But if we rotate around the flux, those subleading paths will become leading, so as to fix the non-single-value problem. I feel that the non-integer flux leads to a quantum anomaly in SO(2) rotation, such that angular momentum is allowed to take fractional values (symmetry fractionalization?) Jul 28, 2022 at 8:48
• Yes! As you know there's an angular momentum pump in the simplified problem of a particle on a circle with a flux inside. I think a similar thing is going on here. It's an anomalous thing because it wants to be at the edge of a 1d Thouless pump. I think you can actually see something like that here the way that the dislocation gets "pumped" along the forward scattering direction as flux is varied from 0 to 2pi. Jul 28, 2022 at 17:32