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BCS Hamiltonian

  1. Apr 29, 2010 #1

    In superconductors, the fermions are interacting. In order to diagonalize our Hamiltonian (which contains the product of four fermion operators), we use Wick's theorem to approximate the product of four fermion operators by the product of two fermion operators.

    Now, a Hamiltonian consisting of the product between two fermion operators describes a non-interacting system. Isn't that a contradiction?
  2. jcsd
  3. Apr 29, 2010 #2


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    Yes, that's a generalized Hartree Fock procedure. The quasi-particles only interact via a mean field which has to be determined self-consistently.
  4. May 2, 2010 #3
    That's exactly the point. We are transforming from a basis spanned by [tex]{ a^\dag_{k\sigma}, a_{k'\sigma}}[/tex] to a new basis [tex]{\alpha^\dag_{k\sigma},\alpha_{k'\sigma}}[/tex] related to the old one by a linear transformation (Boguliubov transformation). Then requiring that the coefficient of the quartic term in the resulting expression should vanish, leads us to the gap equation. In the vacuum defined by the new basis, the vacuum for the old basis, is now an excited state containing non-interacting "quasiparticles". Of course, this procedure only works and the quasiparticles can be said to be "non-interacting" only if higher-order terms can be neglected.

    Classically, the whole process is simply a canonical transformation on the phase space to new variables in terms of which an interacting Hamiltonian can be diagonalized and written as a sum of harmonic oscillators.
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