I find the BCS ground state confusing and I have some questions about its interpretation.
I have looked at this answer to this question; below I intersperse my (potentially incorrect) understanding and questions.
Tinkham (2nd ed., Eq. 3.14) writes the ground state as
$$ |\Psi_G\rangle =\prod_{{\bf k}={\bf k}_1}^{\,{\bf k}_M} \bigl(u_{\bf k}+v_{\bf k}\,c^{\dagger}_{{\bf k}\uparrow}\, c^{\dagger}_{-{\bf k}\downarrow}\bigr)\,|\phi_0\rangle, $$
with $|u_{\bf k}|^2+|v_{\bf k}|^2=1$; the probability that the pair $({\bf k}\uparrow,-{\bf k}\downarrow)$ is occupied is $|v_{\bf k}|^2$.
Normal metal, (T=0).
Even without a pairing interaction one may write the ground state in the same form, taking
$u_{\bf k}=0,\,v_{\bf k}=1$ for $|{\bf k}|<|{\bf k}_F|$ and vice-versa. That merely states every momentum below the Fermi surface is occupied; it does not yet imply Cooper pairing. Explicitly, with $v_{\bf k}=1$ for $|{\bf k}|<|{\bf k}_F|$, $$|\Psi_G\rangle =\prod_{{\bf k}<{\bf k}_F} c^{\dagger}_{{\bf k}\uparrow}c^{\dagger}_{-{\bf k}\downarrow}\,|\phi_0\rangle =\prod_{{\bf k}<{\bf k}_F,\;\sigma} c^{\dagger}_{{\bf k}\sigma}\,|\phi_0\rangle. $$ For $T>0$, and for a normal metal, I believe this form is insufficient, because single-particle excitations ${\bf k}\uparrow$ must be allowed (?)
Adding the BCS pairing interaction
Tinkham writes the BCS pairing Hamiltonian $$ \mathcal H =\sum_{{\bf k}\sigma}\epsilon_{\bf k}\,n_{{\bf k}\sigma} +\sum_{{\bf k}{\bf l}}V_{{\bf k}{\bf l}}\, c^{\dagger}_{{\bf k}\uparrow}\,c^{\dagger}_{-{\bf k}\downarrow}\, c_{-{\bf l}\downarrow}\,c_{{\bf l}\uparrow}. $$
The most generic Hamiltonian that we could have instead considered is $$ \underbrace{\sum_{{\bf i}{\bf j}} t_{{\bf i}{\bf j}}\, c^{\dagger}_{\bf i}c_{\bf j}}_{\text{1-body}} \;+\; \underbrace{\frac14\sum_{{\bf i}{\bf j}{\bf k}{\bf l}} V^{(2)}_{{\bf i}{\bf j},{\bf k}{\bf l}}\, c^{\dagger}_{\bf i}c^{\dagger}_{\bf j} c_{\bf l}c_{\bf k}}_{\text{2-body}} \;+\; \underbrace{\frac1{36}\sum_{{\bf i}{\bf j}{\bf k}{\bf l}mn} V^{(3)}_{{\bf i}{\bf j}{\bf k},{\bf l}mn}\, c^{\dagger}_{\bf i}c^{\dagger}_{\bf j}c^{\dagger}_{\bf k} c_{n}c_{m}c_{\bf l}}_{\text{3-body}} \;+\;\cdots $$
The BCS model keeps only the two-body term and "projects" it onto the $({\bf k}\uparrow,-{\bf k}\downarrow)$ channel.
Gap equation and model interaction
$$ \Delta_{\bf k} = -\sum_{\bf l}V_{{\bf k}{\bf l}}\,u_{\bf l}v_{\bf l}. $$
$$ V_{{\bf k}{\bf l}}= \begin{cases} -V,& -\hbar\omega_D<\xi_{\bf k},\xi_{\bf l}<\hbar\omega_D,\\[4pt] 0,&\text{otherwise}, \end{cases} $$
and likewise
$$ \Delta_{\bf k}= \begin{cases} \Delta,& -\hbar\omega_D<\xi_{\bf k}<\hbar\omega_D,\\[4pt] 0,&\text{otherwise}. \end{cases} $$
This yields the usual coherence factors
$$ u_{\bf k}^2=\tfrac12\!\left(1+\dfrac{\xi_{\bf k}} {\sqrt{\xi_{\bf k}^2+\Delta^2}}\right), \quad v_{\bf k}^2=\tfrac12\!\left(1-\dfrac{\xi_{\bf k}} {\sqrt{\xi_{\bf k}^2+\Delta^2}}\right). $$
Question
In the product state, not every $({\bf k}\uparrow,-{\bf k}\downarrow)$ pair is a Cooper pair.
Yet the $u_{\bf k},v_{\bf k}$ derived above apply to all pairs, Cooper or not.
Near some momentum where ${\bf k}_i<k_F<{\bf k}_j$$$ \prod\!\cdots\! \bigl(u_{{\bf k}_i}+v_{{\bf k}_i}\, c^{\dagger}_{{\bf k}_i\uparrow} c^{\dagger}_{-{\bf k}_i\downarrow}\bigr) \bigl(u_{{\bf k}_j}+v_{{\bf k}_j}\, c^{\dagger}_{{\bf k}_j\uparrow} c^{\dagger}_{-{\bf k}_j\downarrow}\bigr) \!\cdots $$
the first factor mixes an unoccupied and an occupied pair state but is not a Cooper pair,
whereas the second mixes states within the Debye shell and is a Cooper pair.
How should one interpret these “mixed” non-Cooper factors?For ${\bf k} \ll -\Delta$ we have that $v_{\bf k}^2 \approx 1,$ but crucially it is still true that $v_{\bf k}^2 \neq 1!$ The "smearing" of the momentum states applies not just to those ${\bf k}$ near the Fermi surface.
Edit: I am now realizing that I made a silly error. Because $\Delta =0$ for $|\xi_{\bf k}| > \hbar \omega_D$ this means that $v_k = 1$ exactly, not as an approximation, when $\xi_{\bf k} < -\hbar \omega_D,$ and $v_k=0$ when $\xi_{\bf k} > \hbar \omega_D.$ This deflates some of my question, but I am still curious. How should I interpret the states inside the Debye window, $|\xi_{\bf k}| < \hbar\omega_D$? Outside this window, it seems like the electrons behave exactly like a normal metal at $T=0$ (since $\Delta_{\bf k}=0$, the occupations are strictly 1 or 0). But inside the window, the ground state is a coherent superposition with fractional occupations, leading to the "smearing" of the Fermi surface.
This seems to create a three-part picture of the electron sea:
- Inert, fully occupied pairs for $\xi_{\bf k} < -\hbar\omega_D$.
- Inert, empty states for $\xi_{\bf k} > \hbar\omega_D$.
- A "superconducting soup" of correlated pairs in the shell $|\xi_{\bf k}| < \hbar\omega_D$.
Is this three-part picture the correct physical interpretation of the BCS ground state (within this model)? And within that middle shell, how should I think about the pairs? Are all of them "Cooper pairs" by virtue of being in that shell, or does that term only meaningfully apply to the states closest to the Fermi energy (e.g.,~$|\xi_{\bf k}| \sim \Delta$) where the superposition between empty and occupied states is strongest? Does it even make sense to speak of individual Cooper pairs or do you just have to think of the whole condensate as being one effective state?
