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The Aharanov–Bohm Effect

  1. Nov 23, 2014 #1
    I was reading this paper http://venables.asu.edu/quant/Gas/gassupp16.pdf on the Aharanov-Bohm effect and became confused regarding the formula (16A-7) which reads
    [tex] e \hbar \vec \nabla \Lambda (\vec r) + e \vec A = 0. [/tex]

    Now i understand that (16A-7) can only be true if ## \vec B = 0 ##. This is solved to obtain the gauge transformation (16A-5)
    [tex] \psi (\vec r) = e^{-i (e / \hbar) \int _{P} ^{r} \vec {dl} \cdot \vec A} \psi _{0} (\vec r). [/tex]

    Now my source of confusion comes from equation (16A-12) which is by my calculation
    [tex] \oint \vec {dl} \cdot \vec A = \iint \vec {dS} \cdot \vec \nabla \times \vec A = - \hbar \iint \vec {dS} \cdot \vec \nabla \times \vec \nabla \cdot \Lambda (\vec r) = 0. [/tex]

    However in the paper this happens:
    [tex] \oint \vec {dl} \cdot \vec A = \iint \vec {dS} \cdot \vec \nabla \times \vec A = \iint \vec {dS} \cdot \vec B = \Phi. [/tex]
    I honestly cannot figure out the justification for using a result in which ## \vec B = 0 ## is required to calculate a result in which clearly ## \vec B \neq 0 ##. I mean, we have fixed ## \vec A ## to be a gradient of our gauge function ## \Lambda (\vec r) ##, so obviously ## \vec \nabla \times \vec A = 0 ## at all points. Having a non-zero ## \vec B ## should kill our (16A-7) and also (16A-5) destroying all results obtained with this for non-zero magnetic fields. What am i missing here?

    Thanks.
     
  2. jcsd
  3. Nov 23, 2014 #2

    vanhees71

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    That's an example for the importance to emphasize the condition that the region in space must be simply connected for the Poincare lemma to hold:

    From [itex]\vec{\nabla} \times \vec{A}=0[/itex] the path independence of line integrals over [itex]\vec{A}[/itex] only follows, if only paths in a simply connected region are considered.

    The typical discussion of the Aharanov-Bohm effect is exactly about the case, where this is not fulfilled. There is a region in space, e.g., a solenoid, where you have a (strong) magnetic field while outside is none (or a very weak one). Then in a naive particle picture, when the particles move in the field-free region, it is surprising that there are quantum mechanical interference effects due to the phase shift between particles running around the solenoid in one direction compared to those running around in the other. It's derived in your manuscript. The phase shift is
    [tex]\Delta \varphi \propto \int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{A}=\int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}=\Phi,[/tex]
    i.e., the phase shift between the two paths is proportional to the magnetic flux through the surface [itex]F[/itex] with the two paths as a boundary (the phase shift is of course the integral along one path minus the integral along the other, so that you have a closed path taken both together), which is a gauge invariant quantity although the phase factor is initially expressed in terms of the gauge-dependent vector potential.

    Of course, a gauge transformation must not have any measurable effect, and as this consideration shows, that's also not the case with the Aharonov-Bohm effect. It rather demonstrates the non-locality effects of quantum theory: The naive idea of particle trajectories in the field-free region makes only sense in classical mechanics but is misleading in the quantum realm.
     
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