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

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}_{k\uparrow}^\dagger \hat{c}_{-k\downarrow}^\dagger)|0\rangle \end{align} The hamiltonian is the reduced BCS hamiltonian: \begin{align} \hat{H} &= \sum_{k, \: \sigma}\zeta_{k} \hat{c}_{k\sigma}^\dagger\hat{c}_{k\sigma} + \frac{1}{\Omega}\sum_{k,k'}V_{k'-k} \hat{c}_{k'\uparrow}^\dagger \hat{c}_{-k'\downarrow}^\dagger \hat{c}_{-k\downarrow} \hat{c}_{k\uparrow} \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''}2\zeta_{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}

• How did you get that? What is $\zeta_k$? As a warm up, do you manage to get the correct BCS ground state energy?
Dec 24, 2020 at 15:23
• It is the kinetic energy minus the chemical potential. And yes I got the correct BCS ground state energy. When I calculate this expectation value I split the sommation up in many case. For example for the kinetic energy term the terms where the operators in the hamiltonian are the same as k', k'' or a k not equal to any of those (and thus inside the product part of the excited state). The same logic I used for the interaction part. Thanks for the comment. Dec 24, 2020 at 20:29
• I edited the hamiltonian. I didn't notice there was an epsilon. Dec 25, 2020 at 11:51

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 $$\left< \Psi_{ex}|\epsilon_k\hat{c}_{-k'\downarrow}^\dagger\hat{c}_{-k'\downarrow}|\Psi_{ex} \right>$$, you will get a $$1$$ instead of $$v_{-k'}$$ when you move $$\hat{c}_{-k'\downarrow}$$ toward $$\left.|0\right>$$, 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 $$-v^*_k+u^*_k\hat{c}_{k\uparrow}^\dagger\hat{c}_{-k\downarrow}^\dagger$$.
• Is this not equal to $\left< 0| \hat{c}_{k''\uparrow} \hat{c}_{-k'\downarrow} \hat{c}_{k'\downarrow}^\dagger\hat{c}_{k'\downarrow} \hat{c}_{-k'\downarrow}^\dagger \hat{c}_{k''\uparrow}^\dagger|0 \right>$ which gives zero right? Dec 25, 2020 at 14:27
• You were right. I meant $\left<0|\hat{c}_{k''\uparrow}\hat{c}_{-k'\downarrow}\hat{c}_{-k'\downarrow}^\dagger\hat{c}_{-k'\downarrow}\hat{c}_{-k'\downarrow}^\dagger\hat{c}_{k''\uparrow}^\dagger|0\right>$. Dec 25, 2020 at 14:56
• It would actually need something like $(u'+v' \hat{c}_{-k'\downarrow}^\dagger \hat{c}_{k''\uparrow}^\dagger)\left.|\Psi_{BCS}\right>$ to produce the two summation terms. Dec 25, 2020 at 16:12