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Superconductors and Fermi surfaces

  1. May 7, 2010 #1

    The dispersion of Bogolyubov quasiparticles in a d-wave superconductor is


    E(\mathbf k) = \pm \sqrt{\varepsilon (\mathbf k)^2+\Delta (\mathbf k)^2},

    where ε(k) is the normal-state dispersion and ∆(k) is the gap dispersion. My question is: The Fermi surface (FS) of the normal state is just ε(k). Is this also the FS of the superconductor?
  2. jcsd
  3. May 8, 2010 #2
    Superconducting states (i.e. BCS) are not Fermi balls, and so don't have Fermi surfaces.
  4. May 8, 2010 #3
    Thanks for replying. I don't quite understand; at low temperatures, the electrons at the FS become unstable, and join in Cooper pairs to reduce energy. Then how can the BCS state not have a FS?

    Again, thanks. I really appreciate it.
  5. May 8, 2010 #4

    Physics Monkey

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    Hi Niles,

    The Fermi surface in an interacting system may be defined in terms of a discontinuity in the electron occupation number [tex] n_k [/tex] in momentum space. For free electrons this discontinuity is one while for interacting electrons in a Fermi liquid this discontinuity is less than one. In an s-wave superconductor there is no longer this discontinuity because the qausiparticle energies no longer touch zero i.e. [tex] E_k = \epsilon_k - \mu [/tex] is replaced by [tex] E_k = \pm \sqrt{(\epsilon_k - \mu)^2 + \Delta^2} [/tex]. The original Fermi surface is smeared on the energy scale of [tex] \Delta [/tex]. However, sometimes people refer to the surface of minimum gap in momentum space as the "Fermi surface," and this surface tracks the original Fermi surface in many cases, but they are strictly speaking different concepts.

    Hope this helps.
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