A problem with the BCS energy expectation value of an excited state

Question

I want to calculate the energy expectation value of the following state.
begin{align}
         |Psi_{ex}rangle = hat{c}_{-k'downarrow}^dagger hat{c}_{k''uparrow}^dagger prod_{k neq k', k''}(u_{k} + v_{k}hat{c}_{kuparrow}^dagger hat{c}_{-kdownarrow}^dagger)|0rangle
        end{align}
The hamiltonian is the reduced BCS hamiltonian:
begin{align}
         hat{H} &= sum_{k, : sigma}zeta_{k} hat{c}_{ksigma}^daggerhat{c}_{ksigma} + frac{1}{Omega}sum_{k,k'}V_{k'-k} hat{c}_{k'uparrow}^dagger hat{c}_{-k'downarrow}^dagger hat{c}_{-kdownarrow} hat{c}_{kuparrow}
        end{align}
I tried this before but I got a wrong result (which I know is wrong when I compare it with the original BCS paper):
begin{align}
 E_{ex} =  sum_{k neq k', k''}2zeta_{k}|v_{k}|^2 + frac{1}{Omega}sum_{k,l neq k', k''}V_{l-k}u_{k}^*v_{k}v_{l}^*u_{l}
       end{align}

C Tong · Answer

It seems to be almost right, except some missing terms. In the kinetic energy part, you should have the additional $zeta_{-k'}+zeta_{k''}.$ This is because when evaluating $left< Psi_{ex}|epsilon_khat{c}_{-k'downarrow}^daggerhat{c}_{-k'downarrow}|Psi_{ex} right>$, you will get a $1$ instead of $v_{-k'}$ when you move $hat{c}_{-k'downarrow}$ toward $left.|0right>$, which is multiplied to $1$ instead of $v_{k'}^*$ when you move $hat{c}_{k'downarrow}^dagger$ toward $left<0|right.$.

The interaction term for this particular excitation is actually what you suspected $0$. but the actual excitation is really what is described by a combination of $hat{c}$ on top of the BCS ground state which can be described by the Bogoliubov procedure $hat{b}$, which turns the $k$ related term into $-v^*_k+u^*_khat{c}_{kuparrow}^daggerhat{c}_{-kdownarrow}^dagger$.

A problem with the BCS energy expectation value of an excited state

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