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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 $F = 0$) 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 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 $\mathbb{R}^2 - \{0\}$ 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
Accordingly we have the differential 1-forms
hence
Here the expression on the right extends smoothly also to the $x$-axis and this extension we call
From the way this is constructed it is clear that $\theta$ is a closed differential form
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
then:
the field strength vanishes $F = \mathbf{d}A = 0$;
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.
L. Mangiarotti, Gennadi Sardanashvily, section 6.6 of Connections in Classical and Quantum Field Theory, World Scientific, 2000
Mikio Nakahara, Section 10.5.3 of: Geometry, Topology and Physics, IOP 2003 (doi:10.1201/9781315275826, pdf)
See also
Last revised on October 19, 2020 at 05:53:57. See the history of this page for a list of all contributions to it.