# Expectation value interaction term BCS ground state

Let's define the pair-creation and pair-annihilation operators: $$b_{\bf{k}} = c_{-\bf{k}\downarrow}c_{\bf{k}\uparrow},\\ b_{\bf{k}}^{\dagger} = c_{\bf{k}\uparrow}^{\dagger}c_{-\bf{k}\downarrow}^{\dagger}.$$

The BCS ground state reads: $$|\Psi_{\mathrm{BCS}}\rangle = \prod_{\bf{k}}(u_k+v_k b_{\bf{k}}^{\dagger})|0\rangle,$$ where $$u_k$$ and $$v_k$$ satisfy $$|u_k|^2+|v_k|^2=1$$, and $$|0\rangle$$ denotes the vaccum.

The most basic interacting term in the pairing Hamiltonian reads $$V = \sum_{\bf{k},\bf{k}'}V_{\bf{k},\bf{k}'}b_{\bf{k}}^{\dagger}b_{\bf{k}'}.$$

It is a standard result (see for instance Tinkham Introduction to superconductivity, p. 53, or the orginal BCS paper from 1957) that $$\langle V \rangle = \sum_{\bf{k},\bf{k}'}V_{\bf{k},\bf{k}'} u_k v_k^{*}u_{k'}^{*}v_{k'},$$ where $$\langle\dots\rangle$$ denotes the average over the BCS ground state, but I am unable to show this result. Below I detail my attempt.

$$\langle V \rangle = \sum_{\bf{k},\bf{k}'}V_{\bf{k},\bf{k}'} \langle b_{\bf{k}}^{\dagger}b_{\bf{k}'} \rangle = \sum_{\bf{k} \neq \bf{k}'}V_{\bf{k},\bf{k}'} \langle b_{\bf{k}}^{\dagger}b_{\bf{k}'} \rangle + \sum_{\bf{k}}V_{\bf{k},\bf{k}} \langle b_{\bf{k}}^{\dagger}b_{\bf{k}} \rangle$$

The first term yields the following: \begin{align} \langle b_{\bf{k}}^{\dagger}b_{\bf{k}'} \rangle &= \langle (u_k^* + v_k^* b_{\bf{k}})(u_{k'}^* + v_{k'}^* b_{\bf{k}'})b_{\bf{k}}^{\dagger}b_{\bf{k}'}(u_k + v_k b^{\dagger}_{\bf{k}})(u_{k'} + v_{k'} b^{\dagger}_{\bf{k}'})\prod_{\bf{q}\neq \bf{k},\bf{k}'}(u_q^*+v_q^* b_{\bf{q}})(u_q+v_q b^{\dagger}_{\bf{q}})\rangle\\ &= u_{k'}^*v_k^*u_k v_{k'}\prod_{\bf{q}\neq \bf{k},\bf{k}'}(|u_q|^2 + |v_q|^2) = u_{k'}^*v_k^*u_k v_{k'}, \end{align} which resembles the result in the literature. However, for the second term, \begin{align} \langle b_{\bf{k}}^{\dagger}b_{\bf{k}} \rangle &= \langle (u_k^* + v_k^* b_{\bf{k}})b_{\bf{k}}^{\dagger}b_{\bf{k}}(u_k + v_k b^{\dagger}_{\bf{k}})\prod_{\bf{q}\neq \bf{k}}(u_q^*+v_q^* b_{\bf{q}})(u_q+v_q b^{\dagger}_{\bf{q}})\rangle\\ &= |v_k|^2\prod_{\bf{q}\neq \bf{k},\bf{k}'}(|u_q|^2 + |v_q|^2) = |v_k|^2. \end{align}

Putting all together I find \begin{align} \langle V \rangle &= \sum_{\bf{k} \neq \bf{k}'}V_{\bf{k},\bf{k}'} u_{k'}^*v_k^*u_k v_{k'} + \sum_{\bf{k}}V_{\bf{k},\bf{k}} |v_k|^2 = \sum_{\bf{k} \bf{k}'}V_{\bf{k},\bf{k}'} u_{k'}^*v_k^*u_k v_{k'} - \sum_{\bf{k}}V_{\bf{k},\bf{k}}|u_k|^2 |v_k|^2 + \sum_{\bf{k}}V_{\bf{k},\bf{k}} |v_k|^2 =\\ &= \sum_{\bf{k} \bf{k}'}V_{\bf{k},\bf{k}'} u_{k'}^*v_k^*u_k v_{k'} + \sum_{\bf{k}}V_{\bf{k},\bf{k}} |v_k|^4. \end{align}

Is there a reason why the last term should vanish ? What went wrong in my calculation ?