# nLab Aharonov-Bohm effect

## Surveys, textbooks and lecture notes

#### Differential cohomology

differential cohomology

# Contents

## Idea

The Aharonov-Bohm effect is a configuration of the electromagnetic field which has vanishing electric/magnetic field strength (vanishing Faraday tensor $F = 0$) and but is nevertheless non-trivial, in that the vector potential $A$ is non-trivial. Since the vector potential affects the quantum mechanical phase on the wavefunction of electrons moving in an electromagnetic field, in such a configuration classical physics sees no effect, but the phase of quantum particles, which may be observed as a interference pattern on some screen, does.

More technically, a configuration of the electromagnetic field is generally given by a circle-principal connection and an Aharonov-Bohm configuration is one coming from a flat connection, whose curvature/field strength hence vanishes, but which is itself globally non-trivial. This is only possible on spaces (spacetimes) which have a non-trivial fundamental group, hence for instance it does never happen on Minkowski spacetime.

In practice one imagines an idealized electric current-carrying solenoid in Euclidean space. Away from the solenoid itself the magnetic field produced by it gives such a configuration.

## Details

Let $\mathbb{R}^2 - \{0\}$ be the plane with the origin removed, and consider the space $(\mathbb{R}^2 - \{0\}) \times \mathbb{R}$ (thought of as 3d Cartesian space with the z-axis removed) and spacetime $(\mathbb{R}^2 - \{0\}) \times \mathbb{R}^2$ (thought of as the previous configuration statically moving in time).

For the following argument only the topological structure of the space matters, and nothing needs to explicitly depend on the $z$-coordinate and the time-coordinate, so for notational simplicity we may suppress these and consider just $\mathbb{R}^2 - \{0\}$.

On this space minus the x-axis consider the polar coordinates $(\phi,r)$ with

$x = r cos(\phi)\,,\;\;\; y = r sin(\phi) \,.$

Accordingly we have the differential 1-forms

$\mathbf{d}x = cos(\phi)\mathbf{d}r - r sin(\phi) \mathbf{d}\phi$
$\mathbf{d}y = sin(\phi)\mathbf{d}r + r cos(\phi) \mathbf{d}\phi$

hence

\begin{aligned} \mathbf{d}\phi & = \frac{1}{r}cos(\phi)\mathbf{d}y - \frac{1}{r}sin(\phi) \mathbf{d}x \\ & = \frac{1}{r^2} x \mathbf{d}y - \frac{1}{r^2} y \mathbf{d}x \end{aligned} \,.

Here the expression on the right extends smoothly also to the $x$-axis and this extension we call

$\theta \coloneqq \frac{1}{r^2} x \mathbf{d}y - \frac{1}{r^2} y \mathbf{d}x \;\; \in \Omega^1(\mathbb{R}^2 - \{0\}) \,.$

From the way this is constructed it is clear that $\theta$ is a closed differential form

$\mathbf{d}\theta = 0 \,.$

However, on $\mathbb{R}^2 - \{0\}$ this is not an exact form. In other words, if one regards $\theta$ as the vector potential being the configuration of an electromagnetic field

$A \coloneqq \theta$

then:

1. the field strength vanishes $F = \mathbf{d}A = 0$;

2. but there is no gauge transformation relating $A$ to the trivial field configuration.

This is possible because $\mathbb{R}^2 - \{0\}$ is not simply connected and hence the Poincaré lemma does not apply.

## References

Revised on May 15, 2014 05:26:26 by Urs Schreiber (145.116.148.168)