nLab Aharonov-Bohm effect




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The Aharonov-Bohm effect is a configuration of the electromagnetic field which has vanishing electric/magnetic field strength (vanishing Faraday tensor F=0F = 0) and but is nevertheless non-trivial, in that the vector potential AA 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 doesn’t 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.


Let 2{0}\mathbb{R}^2 - \{0\} be the plane with the origin removed, and consider the space ( 2{0})×(\mathbb{R}^2 - \{0\}) \times \mathbb{R} (thought of as 3d Cartesian space with the z-axis removed) and spacetime ( 2{0})× 2(\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 zz-coordinate and the time-coordinate, so for notational simplicity we may suppress these and consider just 2{0}\mathbb{R}^2 - \{0\}.

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

x=rcos(ϕ),y=rsin(ϕ). x = r cos(\phi)\,,\;\;\; y = r sin(\phi) \,.

Accordingly we have the differential 1-forms

dx=cos(ϕ)drrsin(ϕ)dϕ \mathbf{d}x = cos(\phi)\mathbf{d}r - r sin(\phi) \mathbf{d}\phi
dy=sin(ϕ)dr+rcos(ϕ)dϕ \mathbf{d}y = sin(\phi)\mathbf{d}r + r cos(\phi) \mathbf{d}\phi


dϕ =1rcos(ϕ)dy1rsin(ϕ)dx =1r 2xdy1r 2ydx. \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 xx-axis and this extension we call

θ1r 2xdy1r 2ydxΩ 1( 2{0}). \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 θ\theta is a closed differential form

dθ=0. \mathbf{d}\theta = 0 \,.

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

Aθ A \coloneqq \theta


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

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

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


See also

Last revised on June 4, 2022 at 10:00:21. See the history of this page for a list of all contributions to it.