# Bogoliubov transformation BCS Hamiltonian

I am reading on the BCS theory and the bogoliubov transformation to diagonilize the BCS Hamiltonian. And there is one step that I really can't seem to get.

So the Hamiltonian looks like this:

$$$$H=\sum_{\mathbf{k} \sigma} \xi_{\mathbf{k}} c_{\mathbf{k} \sigma}^{\dagger} c_{\mathbf{k} \sigma}-\sum_{\mathbf{k}}\left(\Delta_{\mathbf{k}} c_{\mathbf{k} \uparrow}^{\dagger} c_{-\mathbf{k} \downarrow}^{\dagger}+\Delta_{\mathbf{k}}^{*} c_{-\mathbf{k} \downarrow} c_{\mathbf{k} \uparrow}\right)+\sum_{\mathbf{k}} \Delta_{\mathbf{k}}\left\langle c_{\mathbf{k} \uparrow}^{\dagger} c_{-\mathbf{k} \downarrow}^{\dagger}\right\rangle$$$$

And then we define the operators \begin{aligned} c_{\mathbf{k} \uparrow} &=u_{\mathbf{k}}^{*} \gamma_{\mathbf{k} \uparrow}+v_{\mathbf{k}} \gamma_{-\mathbf{k} \downarrow}^{\dagger} \\ c_{-\mathbf{k} \downarrow}^{\dagger} &=u_{\mathbf{k}} \gamma_{-\mathbf{k} \downarrow}^{\dagger}-v_{\mathbf{k}}^{*} \gamma_{\mathbf{k} \uparrow} \end{aligned}

and they have the normalization condition: $$$$\left|u_{\mathbf{k}}\right|^{2}+\left|v_{\mathbf{k}}\right|^{2}=1$$$$

So that the first term becomes: \begin{aligned} \sum_{\mathbf{k} \sigma} \xi_{\mathbf{k}} c_{\mathbf{k} \sigma}^{\dagger} c_{\mathbf{k} \sigma} &=\sum_{\mathbf{k}} \xi_{\mathbf{k}}\left[c_{\mathbf{k} \uparrow}^{\dagger} c_{\mathbf{k} \uparrow}+c_{-\mathbf{k} \downarrow}^{\dagger} c_{-\mathbf{k} \downarrow}\right] \\ &=\sum_{\mathbf{k}} \xi_{\mathbf{k}}\left[\left(\left|u_{\mathbf{k}}\right|^{2}-\left|v_{\mathbf{k}}\right|^{2}\right)\left(\gamma_{\mathbf{k} \uparrow}^{\dagger} \gamma_{\mathbf{k} \uparrow}+\gamma_{-\mathbf{k} \downarrow}^{\dagger} \gamma_{-\mathbf{k} \downarrow}\right)+2\left|v_{\mathbf{k}}\right|^{2}+2 u_{\mathbf{k}} v_{\mathbf{k}} \gamma_{\mathbf{k} \uparrow}^{\dagger} \gamma_{-\mathbf{k} \downarrow}^{\dagger}+2 u_{\mathbf{k}}^{*} v_{\mathbf{k}}^{*} \gamma_{-\mathbf{k} \downarrow} \gamma_{\mathbf{k} \uparrow}\right] \end{aligned}

It is the last step I don't understand, when I put the operators into the hamiltonian I get something like this:

\begin{aligned} c_{\mathbf{k} \uparrow}^{\dagger} c_{\mathbf{k} \uparrow} &=\left(u_{\mathbf{k}} \gamma_{\mathbf{k} \uparrow}^{\dagger}+v_{-\mathbf{k}}^{*} \gamma_{\mathbf{k} \downarrow}\right) \left(u_{\mathbf{k}}^{*} \gamma_{\mathbf{k} \uparrow}+v_{\mathbf{k}}\gamma_{\mathbf{-k} \downarrow}^{\dagger}\right)\\ &=|u_{\mathbf{k}}|^2\gamma_{\mathbf{k} \uparrow}^{\dagger}\gamma_{\mathbf{k} \uparrow} + u_{\mathbf{k}} v_{\mathbf{k}}\gamma_{-\mathbf{k} \uparrow}^{\dagger} \gamma_{-\mathbf{k} \downarrow}^{\dagger} + u_{\mathbf{k}}^{*} v_{\mathbf{k}}^{*}\gamma_{-\mathbf{k} \downarrow}\gamma_{\mathbf{k} \uparrow} +|v_{\mathbf{k}}|^2\gamma_{-\mathbf{k} \downarrow}\gamma_{-\mathbf{k} \downarrow}^{\dagger}\\ c_{\mathbf{k} \downarrow}^{\dagger} c_{\mathbf{k} \downarrow} &=\left(-v_{\mathbf{k}}^{*} \gamma_{\mathbf{-k} \uparrow}+u_{\mathbf{k}} \gamma_{\mathbf{k} \downarrow}^{\dagger}\right) \left(-v_{\mathbf{k}} \gamma_{\mathbf{-k} \uparrow}^{\dagger}+u_{\mathbf{k}}^{*}\gamma_{\mathbf{k} \downarrow}\right)\\ &=|v_{\mathbf{k}}|^2\gamma_{-\mathbf{k} \uparrow} \gamma_{\mathbf{-k} \uparrow}^{\dagger} - u_{\mathbf{k}}^{*} v_{\mathbf{k}}^{*} \gamma_{\mathbf{-k} \uparrow} \gamma_{\mathbf{k} \downarrow} - u_{\mathbf{k}} v_{\mathbf{k}}\gamma_{\mathbf{k} \downarrow}^{\dagger}\gamma_{-\mathbf{k} \uparrow}^{\dagger} +|u_{\mathbf{k}}|^2 \gamma_{\mathbf{k} \downarrow}^{\dagger}\gamma_{\mathbf{k} \downarrow} \end{aligned}

I really don't see how adding these terms together forms the equation that is above. Especially the $$2\left|v_{\mathbf{k}}\right|^{2}$$ term confuses me how it is supposed to be there, with no operator terms, my guess is that the operators cancel out somehow but I don't see it. Maybe I did something wrong above or am missing something crucial. I could really use a push in the right direction.

We want the Bogoliubov transformation to be canonical, so it must preserve the canonical anticommutation relations. That is, you need $$\delta_{\mathbf{k},\mathbf{k}'}\delta_{\sigma\sigma'}=\{\gamma_{\mathbf{k}\sigma}, \gamma_{\mathbf{k}'\sigma'}^\dagger\}$$. In particular this implies $$\gamma_{\mathbf{k}\sigma}\gamma_{\mathbf{k}\sigma}^\dagger = 1 - \gamma_{\mathbf{k}\sigma}^\dagger \gamma_{\mathbf{k}\sigma}$$. The constant gives you the $$2|v_\mathbf{k}|^2$$ term.