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I have been reading about the quantum hall effect during which i had to read about the AB effect used in the Laughlin gauge argument. In many sources, it is directly assumed that the flux quantum in the AB effect is $\frac{\hbar c}{e}$. How do I prove this? How do I prove this is the smallest change in magnetic flux that can occur? Why is it called the flux quantum if $\Phi$ is not quantized? How can I prove that it is the smallest flux change that would leave the physics same?

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Not sure if this will help: flux quantization arises in superconducting loops more because of the electrons than the electromagnetic field. The property of being a superconductor sets the boundary conditions which link the electrons and the electromagnetic field, and it's the fact that electrons can only have discrete (orbital) angular momentum around the loop that lead to discrete values for the flux. In turn, the discrete values of the angular momentum come from the purely geometrical fact that the electron wavefunction must have an integer number of cycles (wiggles) around the loop. – Jess Riedel Apr 22 '13 at 14:20

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What exactly you mean by "flux quantum in AB"? You may consider any flux you want and then notice that exactly this value corresponds to periodicity in change of transmission coefficient (or any other value you want to measure) as a function of external field. Thus, effect magnetic flux quantum appears naturally in AB effect, but not "assumed", whatever.

For the reasoning behind "magnetic flux quantum" itself, in addition to a quantization of the magnetic flux from an electron which angular momenta is quantized; less trivial problem where this value appears is the following. If you consider periodic 2d structure with "element" area S and switch on the magnetic field, then you may choose gauge consistent with periodicity of the structure iff the flux through the area S is exactly integer number of $\frac{\hbar c}e$.

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In your last sentence, the gauge that you chose is for the vector potential right? Or is it a phase change in the wavefunction? – user7757 Apr 24 '13 at 8:30
@ramanujan_dirac By gauge I mean the choice of the vector potential. – Misha Apr 24 '13 at 9:25
Its ok. I got all my doubts resolved. Ryders QFT has an excellent section on QFT. I was trying to figure it out with no knowledge of the AB effect at all. – user7757 Apr 24 '13 at 13:31
Thanks for your answer, though. +1. – user7757 Apr 24 '13 at 13:31

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