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Microscopic theory of superconductivity

  1. May 6, 2009 #1
    hello everybody
    i want informational about microscopic theory of superconductivity but without sophisticated quantum mechanics
    can you help me ?
  2. jcsd
  3. May 7, 2009 #2
    Re: Superconductivity

    the electrons form 'cooper pairs'. somehow that results in virtually zero resistance
  4. May 7, 2009 #3
    Re: Superconductivity

    Not virtually zero, but actually zero resistance.

    The Cooper pair is a quasiparticle. Unlike the two electrons inside it, the Cooper pair is a Boson, not a Fermion. Therefore the Cooper pair is not limited by the Pauli exclusion principle and all the Cooper pair are in the ground state (lowest quantum numbers).

    In the case of a free electron (Fermion) conductivity requires electrons to be in excited states because they are above the Fermi energy so scattering of the excited electrons drops them back into the ground state. For the Boson there is no such scattering process.
  5. May 7, 2009 #4

    Vanadium 50

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    Re: Superconductivity

    No. The microscopic theory requires "sophisticated quantum mechanics". This is fundamentally a quantum mechanical process.
  6. May 7, 2009 #5
    Re: Superconductivity

    do the cooper pairs form a degenerate gas?
  7. May 7, 2009 #6
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