# Aharonov-Bohm effect and its topological connection

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?

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.

• Comments are not for extended discussion; this conversation has been moved to chat. – rob Jun 2 '17 at 17:35

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.)

• But can AB effect be explained using only topological/geometrical arguments? In particular, if I do physics in which I really have only a manifold with a hole and there really is no "space within the hole" in which I can put a magnetic field, what geometric parameters of my manifold would decide if I would have an AB effect or not? Thanks! – Dvij Mankad Nov 14 '18 at 15:50
• @DvijMankad Good question. It's whether the electron's wavefunction is defined on a trivial or nontrivial $U(1)$ fiber bundle. – tparker Nov 16 '18 at 5:01

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.

• A tree will not make universe non simply connected unless it is an infinitely long one. The real point behind AB effect is that the topology of space may allow for a field ($\vec A$) which is irrotational but with non zero circulation. If there is no field, there is no effect. – Diracology May 31 '17 at 2:58
• What has the length of the tree got to do with it? When we do experiments with solenoids, which surely are not infinitely long, the interference pattern on the screen in the double slit experiment, while putting the (not infinitely long) solenoid between the slits and the screen will still show a phase displacement of the entire interference pattern. And what is the effect on the topology of space which may allow for a circulation free $\vec A$ field with zero $\vec B$ field? – descheleschilder May 31 '17 at 21:05
• Certainly, an $\vec A$ field comes into play, just as in making a continuous local (independent of time and place) phase transformation, of which the AB effect is just a little piece. – descheleschilder May 31 '17 at 21:05
• The length of the tree has a lot to do! A space is said to be simply connected if any closed curve can be continuously contracted to a point. Therefore a tree will never make space simply connected. In the A-B effect, the solenoid has to be long enough so that space (the region of the experiment) is non simply connected. – Diracology May 31 '17 at 22:05
• Regarding the effect of topology, please check this question and its answers: Why is this vector field curl-free? and also this answer – Diracology May 31 '17 at 22:07

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.

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:

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.