Which is the energy of one Cooper pair? Is it below some energy bands (1s,2s) since a Cooper pair is a boson ?Which is the ground state of a boson (electron -phonon-electron)?
Superconductivity – Understanding the Energy of Cooper Pairs
superconductivity
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It is indeed a counterintuitive fact. Let us go slowly through the argument.
You start with a degenerate gaz of fermions. They are piled-up in the momentum space, up to the so-called Fermi level. It means that the last electrons entering the game have the Fermi energy, which is pretty huge ($\sim 10\;000$ Kelvin). Also, these electrons are the only accessible ones, since the ones deep inside the Fermi sea are frozen up by the other ones.
On top of this stable situation (this is called a metal after all), you allow coupling of two-electrons through phonon exchange. This happens locally in space, since this is the simpler hypothesis. This is the usual naive picture: an electron creates a vibration of the lattice (i.e. the electron releases a phonon) which perturbs the next electron flowing (i.e. the next electron catches up the phonon), but this picture is confusing and you should not bother yourself with I believe.
Let us come back to our two-electrons interaction mediated locally by a phonon. It turns out that the interaction is slightly attractive for some parameters. You can verify that this small attraction leads to a huge effect : two electrons which were previously at the Fermi level collapse (so to say) to the zero-energy state (picturesquely said, they lost roughly $10\;000$ Kelvin in the process, but this also is a naive picture). This is called the Cooper instability of the Fermi sea. In fact, the electrons do not collapse, they bound to each other, like in a covalent bound if you wish. And so the kinetic energy of the bound state is zero, although the two electrons forming the bound state were previously at the Fermi energy... so how to reconcile these two ideas ? Well, you say that the former electrons were in states $k_{F}$ and $-k_{F}$, the Fermi momentum, such that the bound state has momentum $k_{F}+(-k_{F})=0$ is at rest ! So you indeed couple two electrons with opposite momenta into a bound state with no momentum.
Since the interaction is local in space, the bounding can only happens for singlet pairing: the former independent electrons were in opposite spin states, thanks to Pauli's principle.
Now, it is customary to call this state a Cooper pair, and to note it as $\left|k_{F},\uparrow;-k_{F},\downarrow\right\rangle $.
So far so good for two electrons and one phonon, what about the condensate ? You can verify that the condensate of paired fermions minimised the energy of the system of interacting electrons. Bardeen, Cooper and Schrieffer first demonstrated it, and so is the BCS explanation for superconductivity.
So to conclude: yes the two electrons forming the Cooper pair have opposite momenta (and spin, but I think your question was mostly about the momentum) but the resulting momentum of the bound state is still zero. The trick is that a Cooper pair is a novel object, which has not so much to do with the properties of the two electrons it's made with, and that's why the condensate is at rest of course ! It's even better to not see a Cooper pair as the pairing of 2 electrons at all, since it is a new state. Say differently, the Cooper instability absorbs two particles (picturesquely called electrons) at momentum $\pm k_{F}$ and with opposite spins and create a new particle (picturesquely coined Cooper pair) with zero momentum and zero spin. Then spin and momentum are conserved (hourra !). To conserve the charge, the new particle (the Cooper pair) must have twice the charge of the former ones (the electrons).
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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The binding energy of a Cooper pair is the energy of the gap that exists in the superconducting state. Since the size of the gap is temperature dependent, so is the binding energy. An approximate formula for the temperature dependence of the energy gap is: $$\Delta(T)=3.2kT_c(1-\frac{T}{T_c})^{0.5}$$ where $T_c$ is the superconducting transition temperature.
Since Cooper pairs involve electrons in the conduction band, that is where the gap is formed. The lowering of energy of a Cooper pair does not push their energy below that of core levels of the ions that constitute the lattice. Typical values for the energy gap at T=0K are in units of $10^{-4}$eV. For example, aluminum is 3.4 x $10^{-4}$eV. This is far smaller than the binding energy of core electrons in the ions.
Be careful too! A Cooper pair is not really a Boson. I know that the total spin of a Cooper pair is 0, but that is not all that goes into defining what type of particle it is. There is also how is responds under exchange of particles. Eisberg and Resnick make this clear in their book Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles:
With regard to what an Energy Gap actually is, the following figure may help: