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Every explanation of the Aharonov–Bohm effect that I have seen seems to justify the phase that shows up due to different paths that the particles (electrons) take to reach some point in space.

How does this make any sense in (standard = Kopenhagen) Quantum Mechanics where there are no trajectories?

In the double slit experiment (standard = Kopenhagen) Quantum Mechanics does not allow any trajectories so why is it with the Aharonov–Bohm effect legit to reason based on trajectories?

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The Bohm-Aharonov effect (well, really the double-slit experiment) does not concern one single path, but rather a sum over all paths, each being weighted by a phase (of unit modulus). That is what gives the interference pattern. The presence of a magnetic vector potential with non-zero circulation about a particular region - that region being the solenoid's interior in the case of the AB-effect - changes the respective phases of all those paths, and this shows up in the interference pattern.

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  • $\begingroup$ But in Standard QM there are no paths. Therefore the sum should be zero. $\endgroup$
    – Thomas Elliot
    Commented Nov 30, 2016 at 19:05
  • $\begingroup$ @ThomasElliot en.wikipedia.org/wiki/Path_integral_formulation $\endgroup$
    – Arturo don Juan
    Commented Nov 30, 2016 at 19:49
  • $\begingroup$ @ThomasElliot There are paths in the same way that there are positions, or any other observable with a continuous spectrum, in quantum mechanics. $\endgroup$
    – Arturo don Juan
    Commented Nov 30, 2016 at 19:51
  • $\begingroup$ Nitpick: The magnetic vector potential in the Aharonov-Bohm effect is rotationless in the area of interest - there is no magnetic field outside the solenoid. The crux is that the area of interest is not simply-connected so the rotationlessness does not force the field to be a gradient or to have zero integral along loops. $\endgroup$
    – ACuriousMind ♦
    Commented Dec 1, 2016 at 14:11
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    $\begingroup$ @ACuriousMind Oh my gosh yeah, that was clumsy of me. I wouldn't call that being "nitpicky" at all. I corrected it in a way that preserves my diction. $\endgroup$
    – Arturo don Juan
    Commented Dec 2, 2016 at 17:22

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