I have a question about the mechanism of Cooper instability. I know the naive picture with two interacting electrons and noninteracting fermi sea wchich leads to the bound state formation. However, if you take all electrons into acount and introduce these effective attractive interaction coming from phonon exchange the Cooper pair will also form.

My question is the following: do the Cooper pairs form only in the vicinity of the Fermi level (magic $2\hbar\omega_{D}$ interval) or is the whole Fermi sea composed of Cooper pairs like the BCS ground state? Which situtation is the correct one?

Is the BCS ground state just an approximation?

  • $\begingroup$ Please check the corrections I've done to your question, and tell me if they are relevant. $\endgroup$
    – FraSchelle
    Jul 15, 2014 at 18:08

1 Answer 1


The original Cooper instability as treated by Cooper in his seminal paper is indeed an approximation for only two electrons on top of the Fermi sea. Nevertheless, you can make the notion precise for the whole gas using mean-field approximation and renormalisation procedure.

The Cooper pairs form only in the vicinity of the Fermi level. Then this region is depopulated and the BCS condensate is gapped (picturesquely said). In the BCS mechanism, when the attractive interaction is driven by the phonon field, it seems consistent to use the Debye energy $\hbar \omega_{D}$ as a cut-off for your theory, since this is the characteristic energy of the phonons.

The remaining of the Fermi sea, as for the transport in normal metals, do not participate in the superconducting mechanism, except they define the Fermi energy of course. This role is important, the electrons deep inside the Fermi sea are not properly speaking superconducting.

The BCS ground state is variational exact for a point-contact interaction between the electrons in the range of energy given by (twice) the Debye energy, as it is discussed in the original BCS paper. See also Bogoliubov's treatment of superconductivity (far more complicated though).

I do not detail more the answer, since you can find more precise answers in any textbook on superconductivity: Tinkham is a really popular one. The variational exactness of the BCS Ansatz is treated in great details in deGennes. Both books called superconductivity something...

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    $\begingroup$ Which quantity calculated from.bcs wavefunction can show that the cooper pairs mainly formed near fermi level, is it the expectation value of $b_k^\dagger b_k$, where $b_k$ is the pair creation operator $\endgroup$ Mar 22, 2015 at 16:24
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    $\begingroup$ @buzhidao Nice question indeed. It's hidden in the calculation for the gap-parameter. You can calculate the gap exactly, and show the isotopic effect (variation of the gap with the atomic mass) as due to the restriction of the energy in the Debye range around the Fermi level. The Fermi level appears from the beginning in the calculation, in the form of the chemical potential. Fermionic problems are all limited to the proximity with the Fermi level in condensed matter problems (at least I do not know any problem involving electrons deep inside the Fermi sea). $\endgroup$
    – FraSchelle
    Mar 23, 2015 at 9:20
  • $\begingroup$ @buzhidao I detailed the gap-calculation in a long post there : physics.stackexchange.com/a/65444/16689 You can verify what I'm talking about the Debye frequency as a cut-off for the theory in the weak-interaction approximation. $\endgroup$
    – FraSchelle
    Mar 23, 2015 at 9:23

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