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 $\mathbf{A}$?

  • $\begingroup$ The answer provided below is of course correct. However, you might find it interesting to look at e.g. Section 2.6 in Modern Quantum Mechanics by Sakurai. The phase shift can be understood within the Feynman path integral formalism, which I personally find more intuitive. If you are already comfortable with path integrals, the argument is also much simpler. $\endgroup$ – Joshua Barr Mar 25 '13 at 3:49

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.

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    $\begingroup$ thanks som uch for your answer. Why the phase factor of the interference term of the superposition wave function can be written in: $ e^{\mathbf{i}q\left(\chi_{\uparrow}\left(b\right)-\chi_{\downarrow}\left(b\right)\right)/\hbar}$ Why does it depend on the chosen gauge $\chi$. $\endgroup$ – Timothy Mar 24 '13 at 20:53
  • $\begingroup$ In practise it does not depend on the gauge choice, since it is related to the flux only. I just used this specific choice to show that the effect exists even if there is no magnetic field. This is the hypothesis of Aharonov and Bohm: the phase shift is only sensible to the flux piercing the interference trajectory. A good exercise for you could be to do the same calculation without imposing a gauge :-) Have fun. PS: You should put some dollars symbols $ to displays Latex mathematics formulas. $\endgroup$ – FraSchelle Mar 24 '13 at 21:18

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