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Aharonov Bohm Effect (gauge invariance)

  1. Nov 1, 2009 #1
    I was reading an article about the Aharonov - Bohm effect and gauge invariance ( J. Phys. A: Math. Gen. 16 (1983) 2173-2177 ) and there is something I really don't get it.

    The facts are:

    The problem is the familiar Aharonov-Bohm one, in which we have a cylinder and inside the cylinder [tex]\rho < R[/tex] there is a magnetic field [tex]\vec{B} = B \hat{e}_z[/tex].

    The writers wanted to use a different gauge (than the usual written in books) which is, in cylindrical coordinates [tex](\rho, \phi, z)[/tex],
    [tex]A_{\rho} = - \rho B_z \phi[/tex]

    At this gauge the vector potential [tex]\vec{A}[/tex] vanishes when [tex]\vec{B}[/tex] does, i.e. when [tex]\rho > R[/tex].
    Furthermore the vector potential is a multivalued function.

    The writer in order to "fix" this problem cuts the space and considers the space as a union of two regions:
    [tex]0 < \phi < 2 \pi[/tex] and [tex]-\pi < \phi < \pi[/tex]

    He defines in these regions two different potentials
    [tex]A_{\rho}^1 = - \rho B_z \phi ,\ \ \ 0 < \phi < 2\pi[/tex], and
    [tex]A_{\rho}^2 = - \rho B_z \phi' ,\ \ \ -\pi < \phi' < \pi[/tex].

    Ok here is my question...
    How can I calculate the correct Flux for a curve [tex]C[/tex] for [tex]\rho > R[/tex] ?

    i.e. [tex]\Phi = \int \vev{A} \cdot dl[/tex] which must be [tex]\Phi = \pi R^2 B_z[/tex].
     
  2. jcsd
  3. Nov 1, 2009 #2
    Don't tell me that with everything given you cannot calculate an integral. Isn't dl=Rdφ?
     
  4. Nov 1, 2009 #3
    Well that's the point Bob. [tex]dl=Rd\phi[/tex], while [tex]A[/tex] only has a radial component in both cases of the vector potential. So if you take C to be a circle, [tex]A\cdot dl =0[/tex] and the flux would always be zero. That would be my first attempt as well, so I can see where the confusion arises.

    But the problem with this approach is that you integrate from 0 to 2pi, assuming that the endpoints correspond to the same point. But they don't. There is a "branch cut" corresponding to the line [tex]\phi = 0[/tex], and you implicitly cross it if you just integrate over the circle. To account for this, you need to go around the branch cut (which starts at the origin).

    So the contour you use is as follows: it starts out at [tex](\phi=0+\epsilon,\rho=R)[/tex], where [tex]\epsilon[/tex] is a small number that is taken to zero in the end. Then you integrate over almost a full circle with radius R, from [tex]0+\epsilon[/tex] to [tex]\phi = 2\pi - \epsilon[/tex] -- since A has no radial component this integral is zero.

    At [tex]\phi =2\pi - \epsilon[/tex] you integrate over [tex]\rho[/tex], from R to 0. You make a small circle around the origin, and you integrate over [tex]\rho[/tex] again, from 0 to R (with [tex]\phi=\epsilon[/tex]). The small circle around the origin is zero again, since you integrate over [tex]\phi[/tex].

    A picture of the contour looks like this:
    180px-Keyhole_contour.svg.png

    The two integrals that remain give:

    [tex]\int_0^R (-B_z \rho \epsilon)d\rho + \int_R^0(-B_z\rho (2\pi-\epsilon))d\rho
    = -B_z \epsilon\int_0^R\rho d\rho + B_z(2\pi-\epsilon) \int_0^R \rho d\rho
    = R^2 B_z (\pi-\epsilon) [/tex]

    Taking [tex]\epsilon[/tex] to zero corresponds to integrating over the full circle, and it gives you the answer that you want. I've used [tex]A_\rho^1[/tex] in this procedure ofcourse, but you can use the other one as well -- it just corresponds to a different path.
     
  5. Nov 1, 2009 #4
    Thanks, xepma, your explanation is very instructive!
     
  6. Nov 2, 2009 #5
    But what happens in the gauge [tex] A = -1/2 \vec{r} \times \vec{B} [/tex], with [tex] \vec{B} = B_z \hat{e}_z [/tex]? Here you still have the problem that the circuitation outside the cylinder vanishes, but the function defining A is no longer multivalued. Of course, there is a discontinuity, but this happens also in the gauge used in the quoted article.
     
  7. Nov 2, 2009 #6
    In that case the vector potential does have a component in the [tex]\phi[/tex] direction, so integrating over a circle does give a contribution.

    You can use the exact same contour is I used earlier (although you don't need to). This time the radial integration is automatically zero, the inner circle tends to zero if you shrink it and the outer circle gives you the answer.
     
  8. Nov 2, 2009 #7
    Thank you very much xepma !!!!

    Your solution solves my problem. I didn't know about these techniques with contour integrals.

    But I would also like to make a comment about the answer,
    The correct answer is only [tex]R^2 B_z (\pi - \epsilon )[/tex] when [tex]\epsilon \to 0 [/tex], and is not valid when [tex]\epsilon = 0 [/tex] because I have discontinuity on [tex] A_{\rho}[/tex].

    Can you make a comment about this ?? I know that you can solve this problem, [tex]\epsilon = 0 [/tex], with thinking this line ([tex]\epsilon = 0 [/tex]) as a Dirac String.

    Thank you again!!!
     
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