BCS theory: where did all the bosons go?

Question

After perusing BCS theory, I am perplexed by a seeming inconsistency. On the one hand, at the motivational level, it introduces bound, hence bosonic, Cooper pairs, which justify condensation bypassing Pauli exclusion. Then it treats bogolons as fermionic electron-hole superpositions, obeying a Fermi-Dirac distribution (see eq. (60) in 1, (3.48) in [2]). In the actual BCS derivation of there is no trace of Bose-Einstein statistics, which should describe the Cooper pairs, since they are bosons. According to Wikipedia "multiple Cooper pairs [being bosons] are allowed to be in the same quantum state, which is responsible for the phenomena of superconductivity". If they are bosons, how can they be described by Fermi statistics, which relies on Pauli exclusion?

1 Rafael M. Fernandes, Lecture Notes: BCS theory of superconductivity

[2] Tinkham M. Introduction to Superconductivity (2ed., MGH, 1996)

PS1 The following tentative summary is directly inspired by Artem's answer below, which I would accept if it were not declaredly a draft version (indeed some of the final remarks need clarification).

BCS theory does not deal with Cooper pairs. Still, it describes how pairs of electrons of opposite spin and momentum acquire a non-zero gap when they are subjected to an attractive potential below a certain temperature. The fact that, once the gap arises, they give rise to bosonic Cooper pairs is independent of BCS theory. I was confused by statements such as this in 1 referring to the effective Hamiltonian: "The second term describes the destruction of a Cooper pair (two electrons with opposite momenta and spin) and the subsequent creation of another Cooper pair". In my current understanding, those are not Cooper pairs, but two electrons with opposite momenta and spin, which become Cooper pairs only once the gap arises, i.e. below the critical temperature. In a nutshell, BCS is about how the gap arises, not about what happens after. Ain't so?

PS2 Summary

My understanding is now the following. As T reaches and passes Tc from above, a pole appears in the vertex $\Gamma$ on the real line and moves to the upper half-plane, meaning that the system becomes unstable (see 15.4 and 15.7 in [3]). This points to electron pairs with opposite momentum and spin appearing spontaneously in the system. You may call them Cooper pairs, however, BCS does not tell us that those pairs are bosons. Since there is nothing in BCS that changes the statistics from Fermi to Bose, that's something you have to put in by hand, as hinted at in Artem's remark about introducing anomalous averages and as indeed is done in 1 just after (32) through the ansatz that the mean value $⟨c^\dagger_{k\uparrow} c^\dagger_{-k\downarrow}���$ is not zero. It then turns out that above Tc there are no Cooper pairs, i.e, the ansatz does not hold. I now accept Artem's answer. After pondering the issue I may come up with a new question.

[3] R.D. Mattuck, A Guide to Feynman Diagrams in the Many-Body Problem, 2nd Edition.

Answer 1

It is the draft version of the answer. It will be upgraded (if needed)

First of all, when you perform Bogoliubov transformation, you just choose a correct vacuum of theory. This transformation does not change statistics: one start from fermionic operators and introduce new fermionic operators. In case of SC, attractive interaction between fermions modifies vacuum of theory and we should find a correct ladder operator. For me, eq. (60) from 1 simply desribes thermal averages of fermionic operators and for me introducing of new fictive particle, ''bogolon'' is unnecessary.

To understand how bosons appear, it is convenient to the following thing. One start from the 4-fermion attractive interaction and take into account that the vacuum of theory modifies. With this fact, one should introduce anomalous averages, $$\langle c^{\dagger}c^{\dagger}\rangle,\quad \langle c c\rangle,$$ where I omit spin indices. This averages obey bosonic statistics.

The key point of BCS is attractive interaction, so Cooper pairs "implcitly" exists in BCS, but to see we should use mean-field theory. For me, it seems that the author is not so accurate in description of eq. (31). The interaction term describes (as I understand and I am sure that it is correct) 4-fermion scattering process by interaction $V_{kk'}$. From this hamiltonian, we can see the Cooper instability (as it was shown earlier). I do not know about your background, but I try to sketch the idea for 2-particle process. To see the instability, one should sum all the 2-particle (2PI) proceses and find the vertex function $\Gamma$. As you can know, an appearance of a quasiparticle in a theory can be prooved by checking existence of pole of 1PI (1-particle) Green function. For the vertex function, the idea is the same: if $\Gamma$ has a pole, it means that there is a two particle bound state in theory.

In 1, eq. (32) means exactly this fact. We take interaction only in the sense that it modifies ground state of theory (mean-field approximation) and writing down expression for $\Gamma$ (it is just 4-operator average). In this expression the anomalous averages appear and they correspond to Cooper pairs. Hope that it is clear. As reference, you can check the Ch. 7 of Altland & Simons book.

Also, if you are familar with Green function, you can check this question and this

Answer 2

The BCS theory is about paired fermions. So no bosons anywhere, and Pauli exclusion principle is always respected by the theory.

However, at very low densities, i.e. when all occupations are much smaller than one, and thus Fermi exclusion principle can be safely ignored, it can be shown that the operator which creates a cooper can be approximated as a bosonic operator. Moreover it can be shown that BCS --> BEC in the low (vanishing) density limit. But this is just an approximation which breaks down by increasing the cooper pairs density.