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BCS theory

  1. Oct 6, 2009 #1
    In the theory of superconductivity BCS theory is given eigen - problem

    [tex]-\frac{\hbar^2}{2m}(\Delta_{\vec{r}_1}+\Delta_{\vec{r}_2})\psi(\vec{r}_1-\vec{r}_1)=(E+2\frac{\hbar^2k^2_F}{2m})\psi(\vec{r}_1-\vec{r}_1)[/tex]

    Why [tex]E+2\frac{\hbar^2k^2_F}{2m}[/tex]?

    Maybe because is Fermi sphere is centered in origin?
     
  2. jcsd
  3. Oct 8, 2009 #2

    olgranpappy

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    This is just a convenient redefinition of the zero of energy. There are two electrons involved in the pair so it is convenient to measure energy with respect to *twice* the Fermi energy. Where the Fermi energy is supposed to be given in terms of the Fermi momentum [itex]k_F[/itex] by
    [tex]
    E_F=\frac{k_F^2}{2m}
    [/tex]
     
  4. Nov 6, 2009 #3
    Is there a clear boundary between low temperature and high temperature superconductors? Is it maybe 40K? I ask because BCS theory is theory for low temperature superconductors. And from this theory we know that the energy of bond of Cooper pair is

    [tex]E=-2\hbar\omega_De^{-\frac{1}{WN(0)}}[/tex]

    where [tex]WN(0)<<1[/tex]
     
  5. Nov 6, 2009 #4
    Yes, above 30K anything is considered high Tc superconductivity, because that's the predicted upper limit of BCS theory.

    Müller and Bednorz received 1986 Nobel prize immediately after discovering a ceramic structure with Tc=35K.
     
  6. Nov 6, 2009 #5

    olgranpappy

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    In low temperature superconductors it is thought that phonons make up the "glue" for the Cooper pairs (hence the \omega_D in your formula, etc). In high temperature superconductors the "glue" is thought to be something else...
     
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