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The Aharonov-Bohm effect is discussed for the case of particles moving along a closed loop through a region with zero magnetic field, however I was wondering whether it still holds for arbitrary fields where particles move through regions with arbitrary non-zero magnetic fields. I suspect this still applies.

The reason I think this is that:

  1. We know from the Aharonov-Bohm effect that the phase only depends upon the flux enclosed by the loop, so a magnetic field outside the loop would not change anything. Let’s turn on a magnetic field there.

  2. The proof of the Aharonov-Bohm effect also does not specify the width of the solenoid—it only specifies that the particles do not move through it—so let’s turn on a magnetic field inside the loop too by making the solenoid infinitesimally thinner than the loop.

So we have a non-zero magnetic field everywhere except on the path of the particle. The proof of the Aharonov-Bohm effect as given in the original paper and the proof using Berry phases relies on the fact that, along the path of the particle, as $\mathbf{B} = \nabla \times \mathbf{A} = 0$, one can construct local patches to express $\mathbf{A} = \nabla \chi$ for some $\chi$ for each patch. This tells us that the vector potential in these regions is a pure gauge and we can solve the Schrodinger equation by simply viewing the presence of this vector potential as a gauge transformation from the $\mathbf{A}=0$ case, something like $\psi= e^{i e \chi}\psi_0$.

For the case of an arbitrary magnetic field, I would not be able to say that the magnetic field is zero on the path of the particle and therefore $\nabla \times \mathbf{A} \neq 0$ so I cannot use the pure gauge argument.

I believe a simple calculation can demonstrate this via path integrals but I imagine I’d now have to insert some potential term into the Lagrangian which models an experimentalist physically restricting the particle to move along a particular path (e.g. for the double slit experiment) otherwise the particles would just undergo cyclotron motion now as they are in a magnetic field.

My question is: does the Aharonov-Bohm effect still apply for arbitrary magnetic fields? And if so, how do I show it?

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  • $\begingroup$ You keep referring to "the loop"; do you intend this to be the path of the particle, or something else? $\endgroup$ – probably_someone May 25 '20 at 5:06
  • $\begingroup$ Yes it’s the path of the particles $\endgroup$ – Matt0410 May 25 '20 at 9:12
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The Lagrangian of a charged particle in a general electromagnetic field is given by:

$$L=\frac{1}{2}m\dot{\vec{x}}^2+\frac{q}{c}\dot{\vec{x}}\cdot\vec{A}+qV$$

You have no electric fields in your setup, so $V=0$. Let $S_0$ be the action of a free particle; then the action in your setup is:

$$S=\int L\;dt=S_0+\frac{q}{c}\int\dot{\vec{x}}\cdot\vec{A}\;dt=S_0+\frac{q}{c}\int\vec{A}\cdot\vec{dl}$$

The phase acquired along a given path is given by $e^{iS/\hbar}$, so you can see that, regardless of whether the magnetic field is nonzero along the particle's path or not, the path integral of the vector potential will give you the phase contribution for that path.

So as long as you have the vector potential for your particular setup, you should be able to determine the acquired phase for any path. The issue at this point is that, unlike the original Aharonov-Bohm setup, for your setup the enclosed magnetic flux is different for every pair of paths, which means that the phase difference is path-dependent in a way that may be nontrivial.

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  • $\begingroup$ Is it possible to show it using wavefunctions and the schrodinger equation? The original proof of the effect solves it by treating the effect of the solenoid as a gauge transformation. We cannot do this now $\endgroup$ – Matt0410 May 25 '20 at 10:12

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