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I've been studying introductory superconductor theory from Solid State Physics textbook (Kittel / Ashcroft), and I found some conflicting statement from each textbook. In Kittel 8th edition, it is said in p.264 that "Some or all of the electrons thermally excited in the normal state are ordered in the snperconducting state." On the other hand, Ashcroft p. 741 says ground state of BCS model is constructed with pairing all electrons, given by $\Psi(r_i,s_i)=\phi(r_1s_1,r_2s_2)...\phi(r_{N-1}s_{N-1},r_Ns_N)$. I also found this representation in original work from Schrieffer (Theory of Superconductivity p.42), as a coordinate representation of N-particle projection of BCS ground state.

There are many arguments mentioning that the cooper pairs are only formed near Fermi surface $E_F$, and inner electrons are frozen without experiencing effective phonon-mediated interaction. I didn't fully accept this statement, because of the fact that carrier density of supercurrent $n_s$ is known to approach full electron density $n$ at zero temperature. Since typical penetration depth of magnetic field is near 100Å, the formula from London equation $\lambda=\sqrt{mc^2/4\pi n_se^2}$ suggests $n_s\sim10^{22}cm^{-3}$, which is order of normal conduction electron density. So it seems that experiment confirms most of conduction electrons act as superconducting electron. Furthermore, in zero temperature the system will be at BCS ground state, so there must be no elementary excitation (Bogolyubov), and the supercurrent must be either from one-electron or cooper pair. Due to dimension of $n_s$, it is unlikely to attribute supercurrent to tiny amount of electrons near Fermi surface ($|\epsilon-\epsilon_F|<\Delta\sim10^{-4}eV)$.

In addition, according to Barden-Pines interaction describing overscreening of dressed phonon, $$V_{k,k'}=|g_{qj}|^2\frac{2\hbar w_{qj}}{(\epsilon_k-\epsilon_{k'})^2+(\hbar w_{qj})^2}$$ two electrons with similar energy can be attracted by overscreening, and I can't find what part of this formula describes the limitation that both electron must be near Fermi surface. So I wonder that it may be possible to pair electrons deep inside Fermi surface, supporting that all electrons can be paired to cooper pair. (I heard exact interaction involves exchange of virtual phonon and change of state, so I'm quite not sure of this point...)

What point am I missing in this paragraph?

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I think mathematically you do a Bogoliubov transformation at each k-point. This would mean that all electrons could be potentially paired (qualitative description).

However in practice, for many k-points the transformation could be trivial (u=1 and v=0). This would mean no pairing for such k-points. As a consequence the pairing would involve only a limited portion of the k-space, i.e. a partial electron density.

The k-space size would depend on the strength of the interaction and other factors. You need to get exact values to know the answer (quantitative description).

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  • $\begingroup$ Thank you for your answer. What I'd understood is that: the Bogoliubov transformation becomes trivial in some k-points due to simplified model of $V_{kk'}$, assigning zero value for $|\epsilon-\epsilon_F|>\hbar w_{cutoff}$ so that $\delta_{k}=-\sum_{k'}V_{kk'}u_{k'}v_{k'}$ is zero far from Fermi surface. So, the detailed model for V_{kk'} can change this conclusion; large portion of $u_k$ and $v_k$ can be nonzero, and this will lead to nontrivial Bogoliubov transformation and more cooper pair coupling inside the Fermi surface. $\endgroup$ Jul 4, 2022 at 7:16
  • $\begingroup$ Am I having this right? In addition, does detailed field-theoretical computation including full electron-phonon interaction give accurate value of $n_s$, coinciding with penetration depth measurements in some degree? (I tried to consult chapter 6 of Schrieffer, but soon found lack of background information to understand such a qualitative argument...) $\endgroup$ Jul 4, 2022 at 7:26
  • $\begingroup$ You are getting my answer right. I'm not able to answer to the last question. $\endgroup$ Jul 4, 2022 at 20:44
  • $\begingroup$ Thank you for your help! $\endgroup$ Jul 5, 2022 at 1:15
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We may look at the problem by another way. The pairs can occur below the Fermi surface (FS), but the pairs on FS are stable (permanent), whereas below FS the pairs can be created and annihilated, what destroys a supercurrent. Indeed, if the pairing occurs under FS, then there are electrons with energy higher than the energy of electrons at the pairing level. These higher energy electrons can fall down to the pairing level below FS and, thus, form new pairs, replacing the “old” ones. If the pairs emerge on FS, then there are no electrons above FS, no electrons can fall down to FS, so new pairs don’t arise and the “old” ones become permanent, keeping the supercurrent and zero resistivity. Thus only pairs on FS are permanent and superconducting.

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