2
$\begingroup$

In F. Schwabl, Quantum Mechanics p.148 it is explained that if we have a particle in an electromagnetic field given by potentials $\varphi$ and $\mathbf{A}$ with wave function $\psi$, then a gauge transformation $$\mathbf{A}'=\mathbf{A} + \nabla\chi\ \ \ \ \ \ \varphi'=\varphi - \frac{1}{c}\frac{\partial\chi}{\partial t}$$ induces a change on the wave function given by $$\psi' = e^{-\frac{ie}{\hbar c}\chi}\psi$$ Then it explains how this doesn't affect the physical meaning of $\varphi$ since it is only a change of phase and $|\varphi'|^2$ remains unchanged.

A couple of pages later, the Aharonov-Bohm effect is presented. We have the classical setting

Aharonov-Bohm experiment

The first thing Schwabl does is to compute (with the help of a gauge transformation) the wave function when one slit is closed. We get $$\psi = e^{\frac{ie}{\hbar c}\Lambda}\psi_0$$ where $\Lambda(\mathbf{x}) = \int_{\mathrm{origin}}^\mathbf{x}\mathbf{A}\cdot ds$ and $\psi_0$ is the wave function of the free particle. Then he sums the two wave functions (one for each slit) to get the interference on the screen.

My problem is: How comes that a change of phase, which we said doesn't change the physical meaning of $\psi$ when speaking of gauge transformations, now affects the physical meaning of $\psi$ (namely producing an interference pattern)? Is this due to the fact that $\mathbb{R}^3$ with a straight line removed is not simply connected, and thus the integral defining $\Lambda$ is not path independent when both slits are open? Or is the cause something else entirely?

$\endgroup$
  • 1
    $\begingroup$ In the first case it's a global phase (which is physically irrelevant), in the final case it's a relative phase between two wave functions (which is not irrelevant). $\endgroup$ – Malabarba Aug 4 '13 at 11:10
  • $\begingroup$ @BruceConnor Yes, I see that, but if we were to compute the wave function with the two slits open, wouldn't we obtain the same kind of phase shift on the global wave function and thus get (physically) the same thing as the motion of a free particle through the two slits? $\endgroup$ – Daniel Robert-Nicoud Aug 4 '13 at 11:12
  • $\begingroup$ No. When you have two slits open, the phase acquired is different depending on the slit crossed. So the net result is not a global phase, it's a relative phase given by the difference between the two phases. $\endgroup$ – Malabarba Aug 4 '13 at 12:13
  • 1
    $\begingroup$ @BruceConnor Ok. And what is the cause of the fact that the phase is not global? Is it due to the fact that if you open both slits then the space you're working on is no more simply connected? $\endgroup$ – Daniel Robert-Nicoud Aug 4 '13 at 12:29
  • $\begingroup$ You're looking at it wrong. The only reason the phase is global with a single slit is because the wave function is composed of only a single state (in the relevant basis). In this situation ANY phase is a global phase. When you open two slits the wave function is a superposition of two states (in that same basis), the two possible paths. Each of these states acquires a phase according to the path it represents. It's the same thing that happened with the single slit, except now you have a superposition of two states instead of just one, and they each acquire their different phases. $\endgroup$ – Malabarba Aug 4 '13 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.