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Deriving the London's equation for superconductor

  1. Dec 16, 2012 #1
    The equation can be obtained from the fact that the "canonical momentum of the ground grstate of superconductor is zero", but where does this fact follow from.
    P.S. Jackson gives a vague reference to Kittel, which I couldn't find in his Introduction_to_solid_state/Quantum_theory_of_Solids.
     
  2. jcsd
  3. Dec 17, 2012 #2
  4. Dec 17, 2012 #3

    DrDu

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    There are plenty of explanations around. Maybe the most general is due to broken symmetry arguments as expounded by Weinberg:
    http://ptp.ipap.jp/link?PTPS/86/43/ [Broken]
     
    Last edited by a moderator: May 6, 2017
  5. Dec 17, 2012 #4
    Thank you for the gem, DrDu - it is beautiful.
     
  6. Dec 18, 2012 #5
    Actually, it is beautiful, but I don't understand much. Can you provide an explanation without calling for field theory?
     
  7. Dec 18, 2012 #6

    DrDu

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    In Kittel Quantum theory of solids he discusses how a superconductor reacts to an applied transversal field of long wavelength. He finds that the wavefunction remains unchanged to lowest order. Hence the expectation of the momentum <p>=0 also does not change with A. This has already been called "rigidity of the wavefunction" by the Londons.
     
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