How is the geometry of the magnetic vector potential determined in the original Aharonov-Bohm experiment?

Question

I've tried but I can't find anything about the geometry of the gauge field, which is mentioned in an article in Scientific American 1981, by Bernstein and Phillips.

They say, without explaining it, that the phase shift in the electron beams is modeled by parallel transport around a cone. The cone extends outside the solenoid, and inside it the geometry is a hemisphere. So that gives a cone truncated at a line of latitude on a hemisphere.

This field and its geometry are not detectable in any real sense, unless electrons interact with such a field, so I guess a takeaway is that this gauge field exists in the sense it interacts with quantum phases.

Answer 1

The truncated cone from Bernstein & Phillips is a very good visualization of the following facts that were found by Aharonov & Bohm:

• The vector potential $\mathbf{A}$ is non zero everywhere but has non zero curl ("curvature" on the fiber bundle) only inside the solenoid. That's where the magnetic field $\mathbf{B}=\nabla\times\mathbf{A}$ is non zero.

• The truncated cone with a cap as hemisphere is a two-dimensional surface that (similarly to the fiber bundle) has non zero curvature on the hemisphere and zero curvature on the cone part. That's the basic idea of their visualization.

• The rest is better explained in their own words below. A 2D capped cone is not the fiber bundle one really needs to handle this rigorously. This is done for example in ch. 10 of M. Nakahara, Geometry, Topology and Physics.