Aharonov-Bohm effect and its topological connection

Question

The topological explanation of Bohm-Aharonov effect assumes that the presence of a solenoid makes the configuration space non-simply connected.

1. Now assume that the magnetic field inside the solenoid is switched off by turning off the current through it. However, the solenoid is still present but with $B=0$.

Does this situation correspond to a hole in space?

If yes, then the topological explanation, "presence of hole leads to a shift of the fringes" doesn't work. Because in this situation we don't observe a shift. If no, then the topological explanation works fine for me.

So should I conclude that there is no hole unless I turn on the magnetic field in the solenoid?

Answer 1

No, there is no contradiction. The Aharonov-Bohm effect specifies that there can be a shift in the interference fringes whenever there is a superposition of waves with different winding numbers about the hole, and it gives that phase shift in the interference pattern as $$ \oint \mathbf A\cdot \mathrm d\mathbf l. $$ The presence of the hole-in-space at the solenoid enables the existence of a vector potential $\mathbf A(\mathbf r)$ with nonzero circulation, while keeping a zero magnetic field $\mathbf B(\mathbf r) = \nabla \times \mathbf A$ everywhere in space. However, it does not require that this circulation be nonzero - and, in fact, it can be arbitrary, and among the possible arbitrary values of the circulation is the real number $\oint \mathbf A\cdot \mathrm d\mathbf l=0$.

So, to be crystal-clear on this: the presence of a doubly-connected region of space is perfectly consistent with a null value of the magnetic flux at the hole-in-space. Making a hole in space does not add a shift to the interference pattern, it's what you do later (i.e. put a magnetic flux through that hole) that does.

Answer 2

The nontrivial topology of the spatial region that the electron wavefunction can access is an important part of one possible explanation for the effect, but it's not enough to produce the interference fringe shifts by itself - the magnitude of the magnetic flux is another crucial ingredient. The electrons inside a donut don't pick up phase shifts just because they circle the donut hole. (It's also possible to explain the A-B effect without any direct reference to the topology of the configuration space.)

Answer 3

I think the assumption that the configuration space is still not simply connected is wrong. The solenoids become objects just as any other object. Even the solenoid (with a current) itself doesn't "puncture space", even when it's infinite long. Space is only non-connected if we remove some infinite string of space which leaves literally a hole in space. And we don't say that the configuration space around us, including trees, dogs people, you name it, is not simply connected. The fact that there is an $\vec A$ field without a $\vec B$ field is the same as a literal gauge (phase) transformation on for example an all single-electron wave functions participating in a double slit experiment.I think shutting down the current through the solenoid is accompanied by the emission of a photon.

A single solenoid produces a phase shift in the interference pattern when placing it between a double slit and the screen where the electrons kick in. The phase of every single wave function of the electrons is changed by the same amount. Now imagine placing a lot of those solenoids between a lot of double slits and the screens. You can vary the strength of the current and let is vary in time and place, in which case you make a concrete (though discrete) realization of a local gauge (phase) transformation on the electron field. Like a continuous gauge transformation (though on the Lagrangian, which is the same as a concrete gauge on the particles which belong to the Lagrangian), I think this is accompanied by the emergence of an $\vec A$ field, though the discreteness of the transformation makes me doubt a little.

Answer 4

Yes, it is in contradiction. Topological explanation fails. If the experimentalist has a means to know what magnetic field inside the solenoid is, the scattered electrons have this means too ;-)

As waves, the electrons have access to (because they are sourced with) any point of space. If one does not use the boundary conditions, but solves the complete set of equations (the boundary conditions are simplified and approximate solutions), then one can see that the electrons have access to everything.

Answer 5

Quote from Wikipedia

The Aharonov–Bohm effect, sometimes called the Ehrenberg–Siday–Aharonov–Bohm effect, is a quantum mechanical phenomenon in which an electrically charged particle is affected by an electromagnetic potential (V, A), despite being confined to a region in which both the magnetic field B and electric field E are zero.

Having a setup where the magnetic field is shielded perfectly I ask in PSE about the electric field of such a shielded current:

1. What is the electric field outside a cylindrical solenoid?

2. If outside a cylindrical solenoid exist an electrical field what does that mean to the Aharonov-Bohm Effect?

The answer to the first question was enlightening:

For what it's worth, it is stated in http://arxiv.org/abs/1407.4826 and references therein in the context of the Aharonov-Bohm effect that even a constant-current solenoid has outside electric fields: "always there is an electric field outside stationary resistive conductor carrying constant current. In such ohmic conductor there are quasistatic surface charges that generate not only the electric field inside the wire driving the current, but also a static electric field outside it...These fields are well-known in electrical engineering." Sorry, I have not checked that, but it sounds plausible. EDIT (07/25/2014) Seems there is a confirmation here: http://www.astrophysik.uni-kiel.de/~hhaertel/PUB/voltage_IRL.pdf , see, especially, Fig.4 therein.