I have recently been reading about the interpretation of the Aharonov-Bohm effect via Feynman's path integral (see viXra:1403.0950). I do not know whether I am missing something, but I do not understand why when evaluating the action they have to take into account the potential even if the electron does not "feel" the field. As far as I understand Feynman's theory, the action is a classical entity which relies on local information (in this case the path the electron follows). Why then does one have to take into account the potential?

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    $\begingroup$ Of course the path integral is not local in the sense you describe, even though the action is local. You're literally supposed to be integrating over all possible paths (with "weight" $e^{iS}$). Besides, the notion of "the path the electron follows" is non-existent (in general)---that's one of the main points of quantum theory! $\endgroup$
    – Danu
    Jul 11, 2016 at 13:45
  • $\begingroup$ That's my point. To take two paths and assign them the classical action with the respective potential parts is an over simplification then, isn't it? Instead, one should take all the possible paths (the group of paths for which the electron does not feel the field and the group of paths for which the electron does feel the field) and add them up, so I was wondering whether there is a justification to ignore the contributions of the other paths (which do not feel the field) or it is just an over simplification. $\endgroup$
    – J.J.
    Jul 11, 2016 at 14:20
  • $\begingroup$ @Juan It's not a simplification so much as a very specific situation. Neglecting the contributions that pass over regions with a nonzero $B$ amounts to saying you've trapped the electron in a potential well such that it CAN'T pass over the region where $B$ is nonzero. This allows you to isolate the effect of the vector potential from the effect of the actual $B$ field. $\endgroup$ Jul 11, 2016 at 15:19
  • $\begingroup$ @Jahan Claes But according to Feynman's path integral theory, one should consider all possible paths joining the points x to x' not only the ones which are trapped in a potential well. $\endgroup$
    – J.J.
    Jul 11, 2016 at 15:58
  • $\begingroup$ @Juan Yes, but if Feynmann's path integral is equivalent to normal QM (and it is!) those paths can't contribute to the propagator, since in normal QM the propagator doesn't have any amplitude outside the potential well. $\endgroup$ Jul 11, 2016 at 16:22

1 Answer 1


The key here is that, just like the infinite walls of the potential well, where the wave function is identically zero outside the well, the wave function in the Aharonov-Bohm effect vanishes in the region where $\mathbf{B} \neq 0$. Further, this impenetrable zone is excluded for all paths. No path that passes through the forbidden zone contributes to the path integral. In the case of a cylinder with an enclosed field, then, all paths pass over or under the cylinder, none through.

Sakurai tackles this problem in Modern Quantum Mechanics, first starting on p. 136 by showing the shift in the energy eigenvalues in the bound-state problem (concentric cylinders), where he states that

The wave function is required to vanish on the inner and outer walls.

referring to the walls of the concentric cylinders.

He then goes to solve the unbounded problem via Feynman's path integral approach, starting on p. 138. On the following page, he concludes that

[T]he interference effect discussed here is purely quantum mechanical...the Lorentz force is identically zero in all regions where the particle wave function is finite. Yet there is a striking interference pattern that depends on the presence or absence of a magnetic field inside the impenetrable cylinder. This point has led some people to conclude that in quantum mechanics it is $\mathbf{A}$ rather than $\mathbf{B}$ that is fundamental.

Indeed, the reason this effect appears is that even though the magnetic field $\mathbf{B}$ vanishes in all regions where the particle may be found, the vector potential $\mathbf{A}$ is nonzero. In this sense, the particle does "feel" the magnetic field (specifically the vector potential), inasmuch as its canonical momentum is affected (through the additional term $\frac{e}{c}\frac{d\mathbf{x}}{dt}\cdot \mathbf{A}$ in the Lagrangian), its wave function experiences a phase shift, and accordingly the probabilities associated with finding it in various regions is affected.


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