What does it mean to break a Cooper pair?

Question

I am studying BCS Theory for the first time, and I did encounter the Bogoliubov-Vitalin transformation for the BCS hamiltonian, that gives $$ \hat{\mathcal{H}} = - \sum_\mathbf{k} \sqrt{\xi_\mathbf{k}^2 + \Delta^2} + \sum_{k} \sqrt{\xi_\mathbf{k}^2 + \Delta^2} \left[ \hat{\gamma}_{\mathbf{k}\uparrow}^\dagger \hat{\gamma}_{\mathbf{k}\uparrow} + \hat{\gamma}_{\mathbf{k}\downarrow}^\dagger \hat{\gamma}_{\mathbf{k}\downarrow} \right] $$ where $\xi_\mathbf{k}\equiv\epsilon_\mathbf{k}-\epsilon_F$, and the $\gamma$ operators are: $$ \begin{cases} \hat{\gamma}_{\mathbf{k}\uparrow} = u_\mathbf{k} \hat{c}_{\mathbf{k}\uparrow} - v_\mathbf{k} \hat{c}_{-\mathbf{k}\downarrow}^\dagger \\ \hat{\gamma}_{\mathbf{k}\uparrow}^\dagger = u_\mathbf{k} \hat{c}_{\mathbf{k}\uparrow}^\dagger - v_\mathbf{k} \hat{c}_{-\mathbf{k}\downarrow} \end{cases} \qquad\qquad \begin{cases} \hat{\gamma}_{\mathbf{k}\downarrow} = u_\mathbf{k} \hat{c}_{\mathbf{k}\downarrow} + v_\mathbf{k} \hat{c}_{-\mathbf{k}\uparrow}^\dagger \\ \hat{\gamma}_{\mathbf{k}\downarrow}^\dagger = u_\mathbf{k} \hat{c}_{\mathbf{k}\downarrow}^\dagger + v_\mathbf{k} \hat{c}_{-\mathbf{k}\uparrow} \end{cases} $$ To derive this hamiltonian, following BCS Theory, the parameter $\Delta_\mathbf{k}$ was supposed to be a constant $\Delta$ inside a shell of extension $\hbar\omega_D$ (Debye energy) both inside and outside the Fermi sphere (in momentum space). Also, I used $u_\mathbf{k},v_\mathbf{k} \in \mathbb{R}$. By variational approach to the same problem, it is found $$ \Delta = 2\hbar\omega_D e^{-2/\rho_0 V_0} $$ with: $\rho_0$ the single particle DoS at the Fermi surface, $V_0$ the magnitude of the potential inside the interaction shell. This is precisely the binding energy of one Cooper pair (worked out starting from a "rigid" filled Fermi sphere and two electrons added outside).

The system was mapped onto a system of composite fermions and the $\hat{\gamma}$ operators create and destroy said fermions. It seems that the ground state of the system is the one with no $\gamma$ particles at all, and that should be the BCS ground state.

Context: Often textbooks plot the function $\sqrt{\xi^2 + \Delta^2}$, and argue that the system is gapped by an amount $\Delta$ and that explains the rigidity of the BCS ground state. That is the $\gamma$ band, so its states may be reached by applying the $\hat{\gamma}$ operator, which however does not conserve the number of particles, creating an electron-hole superposition.

Now: I expect that pumping in the system an energy $\Delta$ or greater, I can break a Cooper pair (i wouldn't know how to do it). But it seems like the first excited state does not contain the two "escaped" electrons or something similar, but instead one $\gamma$ fermion. In the sense, if I count the electrons in the state one is missing. I understand that the system is grand-canonical so particle number is not conserved, but physically I have trouble accepting this fact.

M. Thinkam (Introduction to Superconductivity, 2 Ed., pp.68-69) shows that the action of the $\hat{\gamma}$ operators on the BCS ground state is: $$ \hat{\gamma}_{\mathbf{k}\uparrow}^\dagger \bigotimes_\mathbf{q} \left[ u_\mathbf{q} + v_\mathbf{q} \hat{c}_{\mathbf{q}\uparrow}^\dagger \hat{c}_{-\mathbf{q}\downarrow}^\dagger \right] |\Omega\rangle = \hat{c}_{\mathbf{k}\uparrow}^\dagger \bigotimes_{\mathbf{q}\neq\mathbf{k}} \left[ u_\mathbf{q} + v_\mathbf{q} \hat{c}_{\mathbf{q}\uparrow}^\dagger \hat{c}_{-\mathbf{q}\downarrow}^\dagger \right] |\Omega\rangle $$ so, to add one $\gamma$ particle means to have one electron in $\mathbf{k}\uparrow$, one hole in $-\mathbf{k}\downarrow$, and a BCS-like state everywhere else. Evidently doing this action we are modifying the number of particles of the states.

Question: If I take one Cooper pair and break it, and I want to conserve the number of electrons (in the sense that my final states has all the other Cooper pairs and two more particles floating around somehow but not as a Cooper pair), is it required more energy than to "adding one $\gamma$ particle"? And physically, if yes, why is the energy required higher than the pair binding energy?

tl;dr: "Oh, jeez, I lost one electron..."

Answer 1

$\Delta$ is the single-electron gap. I think the binding energy of a two-electron Cooper pair is $2\Delta$. That solves your problem.

Even though U(1) is spontaneously broken, (charge mod 2) is still conserved. And the $\gamma$ excitation does have a well-defined charge of (1 mod 2). Unless you specifically inject one unit of charge, no amount of pair breaking can give you a single $\gamma$.

(The grand canonical ensemble only permits fluctuation of particle number, not total electric charge. Granted, you can allow charge fluctuation with a "charge-dependent chemical potential", but that is for all intent and purposes an external electric potential.)

By the way, before you do Bogoliubov transform, the off-diagonal term in the Hamiltonian looks like pairs of electrons are constantly entering and leaving the condensate with equal probability. A pair-breaking perturbation as seen in the literature usually refers to something that breaks the balance.