Phase shift in electromagnetic potential In Aharonov-Bohm effect, how to derive that the wave function of a electric charge $q$ acquires a phase shift $\phi=\frac{q}{\hbar}\int \mathbf{A} \cdot d\mathbf{x}$ after travelling in the non-zero magnetic vector potential $\mathbf{A}$?
 A: The Aharonov-Bohm effect arrises in a hypothetical situation when
the magnetic field $B=\nabla\times A=0$ whereas the vector potential
$A$ may not be null. This situation in particular arrise when we
choose the gauge in such a way that $A'=0\Leftrightarrow A=-\nabla\chi$,
which indeed verifies $B=-\nabla\times\nabla\chi=0$. Aharonov and
Bohm gave the particular example of a two-slit interference setup,
when a magnetic flux is enclosed inside the interfering paths.
This can be done using a infinitely long solenoid, which do produce
a magnetic flux, but without magnetic field outside the solenoid.
Now, one can integrate the gauge choice along a given path such that
$$
A=-\nabla\chi\Rightarrow\int_{a}^{b}A\cdot dl=-\int_{a}^{b}\nabla\chi\cdot dl=\chi\left(a\right)-\chi\left(b\right)
$$
for a path from $a$ to $b$. Then, the phase shift along a path is
given by the circulation of the vector potential along the same path
in the chosen gauge.
Now, as conventional for a two-slit experiment, one can separate the
total wave function $\Psi$ as a superposition of the wave function
$\Psi_{\uparrow}$ passing on the upper slit and the wave function
$\Psi_{\downarrow}$ passing throw the lower slit.
Suppose we choose to separate the $\Psi$ wave function at the point
$a$ and recollect it at the point $b$. Then, one has
reads 
$$
\left\vert \Psi\left(b\right)\right\vert ^{2}=\left\vert \Psi_{\uparrow}\left(b\right)\right\vert ^{2}+\left\vert \Psi_{\downarrow}\left(b\right)\right\vert ^{2}+2\text{Re}\left[\Psi_{\uparrow}\left(b\right)\Psi_{\downarrow}\left(b\right)^{*}e^{\mathbf{i}q\left(\chi_{\uparrow}\left(b\right)-\chi_{\downarrow}\left(b\right)\right)/\hslash}\right]
$$
 and, using the above definition for the phase drop along a path
$$
\chi_{\uparrow}\left(b\right)-\chi_{\downarrow}\left(b\right)=\int_{a\rightarrow b;\downarrow}A\cdot dl-\int_{a\rightarrow b;\uparrow}A\cdot dl
$$
 where the two integral paths are from $a$ to $b$. The first integral
follows this path using the bottom slit, whereas the second integral
follows the upper slit. Thus, one has, for the probability to find the particle after its passage though the system 
$$
\int_{a\rightarrow b;\downarrow}A\cdot dl-\int_{a\rightarrow b;\uparrow}A\cdot dl=\left(\int_{a\rightarrow b;\downarrow}+\int_{b\rightarrow a;\uparrow}\right)A\cdot dl=\int_{a\circlearrowleft a}A\cdot dl
$$
 and thus corresponds to the integral along the closed contour made
by the two interfering paths.
Using that 
$$
\oint A\cdot dl=\iint B\cdot dS=\Phi
$$
 with $\Phi$ the flux enclosed in between the two interfering paths,
one finally obtain that the total probability amplitude $\left\vert \Psi\left(b\right)\right\vert ^{2}$
is 
$$
\left\vert \Psi\left(b\right)\right\vert ^{2}=\left\vert \Psi_{\uparrow}\left(b\right)\right\vert ^{2}+\left\vert \Psi_{\downarrow}\left(b\right)\right\vert ^{2}+2 {\Re}\left[\Psi_{\uparrow}\Psi_{\downarrow}^{*}\right]\cos\frac{2\pi\Phi}{\Phi_{0}}+2 {\Im}                                                                                                         \left[\Psi_{\uparrow}\Psi_{\downarrow}^{*}\right]\sin\frac{2\pi\Phi}{\Phi_{0}}
$$
 with $\Phi_{0}=2\pi\hslash/q$ is called the flux quantum. 
In a superconductor, the basic charge is $2e$, thus the flux quantum
becomes $\Phi_{0}=\pi\hslash/e$ which is a fundamental constant of
quantum circuitry.
