4
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

enter image description here

$\endgroup$
  • $\begingroup$ :- the answer is nice. But you already assumed that the $\psi(A\neq0)$ is of a particular form. Can we do better than this ? $\endgroup$ – user38579 Feb 22 '14 at 14:48
  • $\begingroup$ @user38579 I am not aware of a "better" procedure. $\endgroup$ – Hunter Feb 22 '14 at 14:58
2
$\begingroup$

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 $\phi=\int A \cdot\mathrm{d}x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.