# 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 ? – user38579 Feb 22 '14 at 14:48
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 $\phi=\int A \cdot\mathrm{d}x$.