The conventional superconductors can be explained using the BCS theory. Usually, the BCS theory is introduced as follows:

We would like to consider the Hamiltonian that describes the system of electrons. This system is specific in a sense that electrons with opposite momenta and spins $(\mathbf{k},\uparrow)$, $(\mathbf{-k},\downarrow)$ have some kind of attractive interaction (no matter how weak it is): $$ H_{eff} = \sum_{\mathbf{k}\sigma} \xi_{\mathbf{k}} c^\dagger_{\mathbf k, \sigma} c_{\mathbf k, \sigma} + \frac{1}{N} \sum_{\mathbf k \mathbf k'} V_{\mathbf k \mathbf k'} c^\dagger_{\mathbf k \uparrow} c^\dagger_{-\mathbf k \downarrow} c_{-\mathbf k' \downarrow} c_{\mathbf k' \uparrow} $$ This unusual behavior can be explained due to the presence of electron-phonon interaction. This effective Hamiltonian is very complicated, but it can be simplified using the mean field theory. Then it can be diagonalized using the Bogoliubov transformations: $$ H = \sum_{\mathbf k, \sigma} E_{\mathbf k} \;\gamma^\dagger_{\mathbf k,\sigma} \gamma_{\mathbf k,\sigma} + const. $$ where $\gamma_{\mathbf k,\sigma}$ is the annihilation operator of newly obtained Bogoliubons (new particles) and $E_{\mathbf k} = \sqrt{\xi_{\mathbf k}^2 + |\Delta_{\mathbf k}|^2}$ is the corresponding dispersion relation. We can see that this new dispersion relation has a gap $|\Delta_{\mathbf k}|$ (I could write the analytical expression for this quantity, but it doesn't matter at this point).

Everything so far is clear to me. But how do we see that the system which is described with this formalism describes the superconductor? How to show that the resistivity is indeed vanishing?

Also, I've often seen that at finite temperature, the current has a contribution that correspond to 'carriers of superconductor' (Cooper pairs) and normal charge carriers (electrons). Even if these Cooper pairs have no resistivity (which I also do not understand, this is asked in first part of the question), why don't we get any dissipation due to the presence of normal electrons that are not bound in Cooper pairs?

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    $\begingroup$ Does this answer your question? What is the link between the BCS ground state and superconductivity? In a nutshell, your Hamiltonian is written in the center-of-mass reference frame Cooper pairs. $\endgroup$ Dec 3, 2021 at 14:18
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    $\begingroup$ Well, no. The answers that you are referring to are only talking about the first part of my question. They do not say anything about normal and supperconducting terms in the current. One of the things I am interested in is: how is it possible to have a state with no dissipation if there is normal current present. $\endgroup$
    – RedGiant
    Dec 3, 2021 at 15:51
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    $\begingroup$ no dissipation means no excitations above the gap. It is like a scattering state: there's a flux, but the energy doesn't change. $\endgroup$ Dec 3, 2021 at 16:47
  • $\begingroup$ Very interesting question. Related: physics.stackexchange.com/q/651336/226902 (this is the "experimental" complement to this "theoretical" question). More about the nature of BCS state: physics.stackexchange.com/q/345321/226902 $\endgroup$
    – Quillo
    Jun 2 at 18:01
  • $\begingroup$ The argument in the book of Mahan "Many-Particle Physics" (Plenum Press, 1981, page 793) goes like this: A current flow corresponds to a shift of the center of the Fermi sphere away from the origin of momentum space. Normal conductor: the electrons can relax "independently" via scattering (the center of the Fermi sphere will relax to the origin because of individual scattering processes). Superconductor: all Cooper pairs have to relax at once as they all share a common state (pair condensation): they are correlated to the displaced centre of the Fermi sphere. $\endgroup$
    – Quillo
    Jun 2 at 18:35


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