On the Bogoliubov transformation in the BCS I have a question regarding the diagonalization of the BCS-Hamiltonian using the Bogoliubov-DeGennes-transformation. I hope someone can help me, so I start with the following Hamiltonian, it is related to the kinetic term in the equation and follows Tinkham's book.
$$
\sum_{\sigma k} \xi_k c_{\sigma k}^\dagger c_{\sigma k} - \dots
$$
we go on to define the quasiparticles
$$
 c_{k\uparrow} = u^*_{k}\gamma_{k0} + v_{k}\gamma^\dagger_{k1},
$$
and
$$
  c^\dagger_{-k\downarrow} = -v^*_{k}\gamma_{k0} + u_{k}\gamma^\dagger_{k1}
$$
where $\gamma_{k0}$ participates in destroying an electron with $k\uparrow$ or creating one in $-k\downarrow$. 
Now, if I do the spin summation I will get one term for up and one for down. The down-term is
$$
\sum_k \xi_k c^\dagger_{k\downarrow} c_{k\downarrow} 
$$
In the subscript there is no minus-sign, and so it does not "fit" with any of the defined quasiparticles. I could of course use the latter one, and let $k \rightarrow -k$ but then I should be left with terms proportional to $v_{-k}$ and so on, however in the final solution there are no such terms. What am I missing?
The final expression (the kinetic part) is
$$
\sum_k \xi_k [(|u_k|^2 - |v_k|^2)(\gamma^\dagger_{k0}\gamma_{k0} + \gamma_{k1}^\dagger \gamma_{k1}) + 2|v_k|^2 + 2u_k^*v_k^* \gamma_{k1}\gamma_{k0} + 2u_k v_k \gamma_{k0}^\dagger \gamma_{k1}^\dagger].
$$
 A: I think you are missing the meaning of a Bogoliubov-DeGennes transformation. This transformation is useful to write a Hamiltonian in its diagonal pertubations coordinates. The pertubations must have some conserved properties, for example:
Supose that $c_{\textbf{k} \uparrow}^\dagger$ creates a pertubation in some state (in your case it creates a electron with energy $\epsilon_\textbf{k}$ in a vaccum state, but pay attention to the general discussion). The Bogoliubov-DeGennes idea is to write that pertubation as a linear combination of other two:
$$c_{\textbf{k} \uparrow}^\dagger=\alpha_\textbf{k} \ d_{\textbf{k} \uparrow}^\dagger+\beta_\textbf{k} \ d_{-\textbf{k} \downarrow}$$
Look what the "other two" have in common: the $c_{\textbf{k} \uparrow}^\dagger$ pertubation creates a linear combination of a $d$ particle with spin $\uparrow$ and momentum $\textbf{k}$ and a $d$ particle with spin $\downarrow$ and momentum $-\textbf{k}$, so the net spin flux remains the same (the spin $\downarrow$ goes in $-\textbf{k}$ direction, equivalent to the spin $\uparrow$ going in $\textbf{k}$ direction). Assuming the $d$ particles have the same commutation properties than the $c$, you find that this transformation must be unittary. In a general form:
$$
\quad
\begin{pmatrix} 
c_{\textbf{k} \uparrow}^\dagger \\
c_{-\textbf{k} \downarrow} 
\end{pmatrix}
=
\begin{pmatrix} 
\alpha_\textbf{k} & \beta_\textbf{k} \\
\gamma_\textbf{k} & \upsilon_\textbf{k} 
\end{pmatrix}
\begin{pmatrix} 
d_{\textbf{k} \uparrow}^\dagger \\
d_{-\textbf{k} \downarrow} 
\end{pmatrix}
=
P \begin{pmatrix} 
d_{\textbf{k} \uparrow}^\dagger \\
d_{-\textbf{k} \downarrow} 
\end{pmatrix},
$$
where $P$ matrix must be unitary. The constants must also satisfy other condition you fix on then (usually, equations that diagonalize the Hamiltonian).
A: Consider $c^+_{\boldsymbol{k}\uparrow}=(c_{\boldsymbol{k}\uparrow})^+=(u^*_\boldsymbol{k}γ_\boldsymbol{k0}+v_\boldsymbol{k}γ^+_\boldsymbol{k1})^+=u_{\boldsymbol{k}}γ^{+}_\boldsymbol{k0}+v^{*}_\boldsymbol{k}γ_\boldsymbol{k1}$, it yields the right result.
