I'm trying to understand how in Hartree-Fock-Bogoliubov (HFB) or BCS theory we can write a product of creation/annihilation operators as single-particle densities under the guise of "Wick's theorem".
In this set of slides, on slide no. 20 (and I've seen this in many many, many papers) the author makes the claim $c_a^\dagger c_b = \rho_{ba} + :c_a^\dagger c_b:$, invoking Wick's theorem, where $\rho_{ba} \equiv \left.\langle\Phi\right| c_a^\dagger c_b\left|\Phi\rangle\right.$. Essentially, the Wick contraction of $c_a^\dagger c_b$ is equal to $\rho_{ba}$. I'd like to put this on firm footing. My two questions are:
- How do we know the contraction $\langle c_a^\dagger c_b \rangle = \rho_{ba}$?
- Given #1, the normal ordered product $:c_a^\dagger c_b:$ must be 0 in the state $\left|\Phi\rangle\right.$ by definition. Are we free to choose "normal-ordering" with respect to any state we like?
That is, can we say $c_a^\dagger c_b$ is normal-ordered with respect to the vacuum $\left|-\rangle\right.$, so $\langle c_a^\dagger c_b \rangle$ should be 0 there but isn't normal-ordered with respect to the state $\left|\Phi\rangle\right.$ where $c^\dagger_a c_b \ne :c^\dagger_a c_b:$ (whatever $:c^\dagger_a c_b:$ actually is)? Is there a way to prove that some combination $c_a^\dagger c_b$, $c_a c_b$, etc. always has a vanishing normal-ordered product (I think I can show this with Boguliubov quasiparticles in this case)?
If we have the freedom of #2, then #1 is trivial without having to calculate anything, since $$ \rho_{ba} = \left.\langle\Phi\right| c_a^\dagger c_b\left|\Phi\rangle\right. = \left.\langle\Phi\right| \langle c_a^\dagger c_b \rangle + :c_a^\dagger c_b:\left|\Phi\rangle\right. = \langle c_a^\dagger c_b \rangle. $$
From there we can do the more general decomposition $$\left.\langle\Phi\right| c_a^\dagger c_b^\dagger c_d c_c \left|\Phi\rangle\right. = \rho_{ca} \rho_{db} - \rho_{da}\rho_{cb} + \kappa_{ba}^* \kappa_{cd},$$
where $\kappa_{ba} = \left.\langle\Phi\right| c_a c_b \left|\Phi\rangle\right.$, using Wick's theorem.
