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Can someone provide me with a pedagogical introduction into the role of criticality in BCS theory?

The QCD condensate is due to strong coupling. The BCS condensation involves only weak coupling - nevertheless we get a condensate. As far as I know, this can only happen if our model involves criticality. How exactly does the formation of the condensate work, and which parameters play a crucial role?

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    $\begingroup$ I'm not sure exactly what you are asking. But one reason that Cooper pairing is possible with arbitrarily weak attractive interactions is that limiting dynamics to the Fermi surface effectively reduces the dimensionality of the problem. You may enjoy this blog post: thiscondensedlife.wordpress.com/2015/10/27/… $\endgroup$
    – Rococo
    Nov 27 '15 at 23:19
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    $\begingroup$ Thanks a lot for this great link, that illuminates a very interesting and important aspect of BCS theory! $\endgroup$
    – Thomas
    Nov 30 '15 at 14:39
  • $\begingroup$ @Rococo It would be great if you could define what you mean by criticality. $\endgroup$
    – FraSchelle
    Dec 5 '15 at 18:48
  • $\begingroup$ @MengCheng It would be great if you could define what you mean by criticality. $\endgroup$
    – FraSchelle
    Dec 5 '15 at 18:48
  • $\begingroup$ @LCF It would be great if you could define what you mean by criticality. $\endgroup$
    – FraSchelle
    Dec 5 '15 at 18:49
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BCS theory deals with superconductivity in a metal, or basically a finite density of non-interacting fermions. There is a Fermi surface with tons of gapless particle-hole excitations, so you can say it is critical. As long as the Fermi surface has certain symmetry (time-reversal or inversion), the pairing instability is infinitesimal, meaning that the condensate forms for arbitrarily small attractive interaction.

To be more concrete, a simplified model which nevertheless captures the essence of the BCS theory is the following:

$H=\sum_k \xi_k c_{k}^\dagger c_k+ g\sum_{k,k'}c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger c_{-k'\downarrow}c_{k'\uparrow}$

Here $\xi_k$ is the single-particle spectrum of the fermions. The transition temperature is given by $T_c\sim e^{-\frac{1}{g\nu}}$, where $\nu$ is the density of states at Fermi surface.

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    $\begingroup$ In what sense does a Fermi surface with "tons of gapless particle-hole excitations" imply criticality? $\endgroup$
    – Rococo
    Nov 27 '15 at 23:19
  • $\begingroup$ OK, I'm not exactly sure what you mean by criticality. It's a gapless system, but not "critical" in the same way as Dirac fermions since it has "more" gapless excitations and there is a length scale, the Fermi wavelength determined by the fermion density. $\endgroup$
    – Meng Cheng
    Nov 27 '15 at 23:33
  • $\begingroup$ You're not sure what I mean by criticality? I don't understand. You are the one who used that terminology in your answer. What did you mean by it? $\endgroup$
    – Rococo
    Nov 28 '15 at 4:23
  • $\begingroup$ It's a Fermi surface, so whatever you mean by criticality, whether it is the same as my understanding or not, just compare it with what happens with a Fermi surface. $\endgroup$
    – Meng Cheng
    Nov 28 '15 at 4:37
  • $\begingroup$ Thanks for your answer and comments. However, I agree to Rococo's first question and would like to ask Meng Cheng what physical requirements a system needs to fulfill in order to obtain criticality? $\endgroup$
    – Thomas
    Nov 30 '15 at 14:41

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