the BCS ground state is:

$$|\psi_G\rangle= \prod_k(u_k+v_kc_{k,\uparrow}^+ c_{-k,\downarrow}^+)|\phi_0\rangle $$

now, i always believed that the vacuum state $|\phi_0\rangle$ was a state with no particles, whatever their nature. (i.e. no single electrons and also no Cooper pairs).
the physical meaning of $|\psi_G\rangle$ so was clear: the probability amplitudes $u_k , v_k$ described the probability, for any k-value, of creating a couple of electrons or leaving the state empty. This couples of electrons ARE NOT necessarily Cooper pairs, they are Cooper pairs only when the value of k is such that the energy associated to the electron lays in a range $\hbar \omega_D$ from the fermi energy. The BCS ground state so is a mixture of Cooper condensate and not paired electrons. Now this seems in contrast with what i read here:https://doi.org/10.1119/1.5093291, quoting:
"*Now, let us consider a superconductor, where a new
ground state $|\Omega_S\rangle$ has developed. The pair condensate can be
thought of as the superconducting vacuum (since paired electrons have minimal energy), which we denote as the field
$\Delta(x)$. The electrons in the superconductor may interact pairwise with the vacuum, either vanishing and creating a pair or
appearing and breaking a pair, and as a result electron number is no longer conserved and states of different electron
numbers may now have a nonzero overlap.
In particular,*
$$\langle \Omega_S| \Psi_{\downarrow} \Psi_{\uparrow} |\Omega_S\rangle = \langle \Omega_S|\Omega_S -2e^- + pair\rangle \neq 0$$

*The ground state may now begin with N electrons, interact
with the condensate and exchange any number of pairs and
electrons, and then be found in a state with M electrons.*

Now this seems in contrast with my previous picture: if we take $|\Omega_S\rangle =|\psi_G\rangle$ the electron annihilation operators $\Psi_{\downarrow} \Psi_{\uparrow}$ acting on this state will annihilate a couple of electrons, but this will not imply a creation of a Cooper pair. This made me think that maybe the vacuum state $|\phi_0\rangle$ is not a state completely empty, but is the state of the Cooper condensate, i.e. the state where all the electrons form Cooper pairs. the meaning of the BCS ground state thus is that we have probabilities $u_k , v_k$ of not destroying or destroying, respectively, a Cooper pair. Now i'm wondering, what is the correct interpretation of the vacuum state?

This post imported from StackExchange Physics at 2023-06-10 21:25 (UTC), posted by SE-user Valerio Actis Dato Casale