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Superconductor question

  1. Sep 25, 2005 #1
    in deriving quantization of flux in superconductor ring, the momentum of cooper pair p:
    [tex]p=\hbar\nabla\theta=e^*(\Lambda J_s + A)[/tex]
    then integrate around the ring,
    [tex]\hbar\oint\nabla\theta dl=e^*\oint(\Lambda J_s + A)dl[/tex]
    using stoke's theorem and integrate sufficiently deep in the ring where current density is very small, the RHS becomes
    and the left hand side,becomes hn where n is integer.
    So the quantized flux is
    e star is the effective cooper pair charge which is -2e.

    i got confused here,why the left hand side integral becomes hn?

    The argument used by the book (Van Duzer, superconductivity page 116) is that because theta is unique or differ by a multiple of 2 Pi at each point, so the integral
    [tex]\oint\nabla\theta dl=2\pi n[/tex]
    why?? where does n come from? theta is a scalar function of r.
    please help.

    Last edited: Sep 25, 2005
  2. jcsd
  3. Sep 25, 2005 #2


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    [itex]\theta[/itex] is the phase of the order parameter [itex] \psi = |\psi|e^{i \theta} [/itex]. Since [itex]\psi[/itex] must be single valued at some specific angular position [itex]\phi + 2k \pi[/itex] on the ring, we need, [itex]\psi(\phi ) = \psi(\phi + 2m \pi ) [/itex], or we need [itex]e^{i \theta ( \phi )} = e^{i \theta ( \phi + 2m \pi)}[/itex], for all m. For the phase factor to remain unchanged over integral number of traversals of the loop, the phase [itex]\theta[/itex] must itself change by only an integer multiple of [itex]2 \pi [/itex] (since [itex]e^{2in \pi} = 1 [/itex]).

    [tex]\theta ( \phi + 2m \pi ) = \theta ( \phi) + 2n \pi [/tex].

    In other words
    [tex]\hbar\oint\nabla\theta dl = \hbar \Delta\theta = \hbar (\theta ( \phi + 2 \pi ) - \theta ( \phi)) = 2 \pi n \hbar = nh[/tex]
    Last edited: Sep 25, 2005
  4. Sep 26, 2005 #3
    Good answer, Gokul.

    In your opinion, from an experimental point of view, what would be a simple method to effect a change in the phase of the wavefunction of the supercurrent while remaining below T(c)?

    Creator :biggrin:
    Last edited: Sep 26, 2005
  5. Sep 26, 2005 #4
    change the radius of the ring.
  6. Sep 27, 2005 #5


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    Have a magnetic flux through the loop. That, after all, is how a SQUID works.

  7. Sep 28, 2005 #6
    True, in theory; but by what method are you going to change the length of a brittle superconducting wire while it is in the superconducting state?
  8. Sep 28, 2005 #7
    Of course; externally applying B thru the loop. I should be more specific. I guess I am referring to a non-electromagnetic method of altering the quantum phase.

    Last edited: Sep 28, 2005
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