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}

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    $\begingroup$ How did you get that? What is $\zeta_k$? As a warm up, do you manage to get the correct BCS ground state energy? $\endgroup$ – Adam Dec 24 '20 at 15:23
  • $\begingroup$ 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. $\endgroup$ – Audrique Dec 24 '20 at 20:29
  • $\begingroup$ I edited the hamiltonian. I didn't notice there was an epsilon. $\endgroup$ – Audrique Dec 25 '20 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 $-v^*_k+u^*_k\hat{c}_{k\uparrow}^\dagger\hat{c}_{-k\downarrow}^\dagger$.

  • $\begingroup$ 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? $\endgroup$ – Audrique Dec 25 '20 at 14:27
  • $\begingroup$ You were right. I meant $\left<0|\hat{c}_{k''\uparrow}\hat{c}_{-k'\downarrow}\hat{c}_{-k'\downarrow}^\dagger\hat{c}_{-k'\downarrow}\hat{c}_{-k'\downarrow}^\dagger\hat{c}_{k''\uparrow}^\dagger|0\right>$. $\endgroup$ – C Tong Dec 25 '20 at 14:56
  • $\begingroup$ Thanks, now it makes sense to me! $\endgroup$ – Audrique Dec 25 '20 at 15:28
  • $\begingroup$ Maybe one last question, in the definition of the excitated state, is the product over k only for the positive k's (because otherwise pairs would be counted double) but in the sommations in the hamiltonian also over the negative k's? $\endgroup$ – Audrique Dec 25 '20 at 15:34
  • $\begingroup$ 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. $\endgroup$ – C Tong Dec 25 '20 at 16:12

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