BCS Theory: Bogoliubov Transformation for Fermions I would like to follow up on this question. The Bogoliubov transformation is written as follows (I assume that $u_{\mathbf{k}} = u_k$ and $v_{\mathbf{k}} = v_k$ as well as $u_k, v_k\in\mathbb{R}$):
\begin{equation}\begin{aligned}
c_{k, \uparrow} &=u_{k} d_{k, \uparrow}+v_{k} d_{-k,  \downarrow}^{\dagger} \qquad (1.1)
                                \\
c_{-k, \downarrow}^{\dagger} &=u_{k} d_{-k \downarrow}^{\dagger} - v_{k} d_{k,  \uparrow} \qquad \ (1.2)
\end{aligned}\end{equation}
Question: Why do we get in the transition from Eq. $(1.1)$ to $(1.2)$ a minus sign in front of the second term in Eq. $(1.2)$? After all, shouldn't we able to derive $(1.2)$ from $(1.1)$?
 A: The transformation should be unitary. In other words, if we write
$$
\begin{pmatrix}
c_{k,\uparrow}\\
c_{-k,\downarrow}
\end{pmatrix}
=
\begin{pmatrix}
u_k & v_k\\
-v_k & u_k
\end{pmatrix}
\begin{pmatrix}
d_{k,\uparrow}\\
d_{-k,\downarrow}
\end{pmatrix}=
\mathcal{S}
\begin{pmatrix}
d_{k,\uparrow}\\
d_{-k,\downarrow}
\end{pmatrix},
$$
the determinant of the transformation matrix should be $1$, and $\mathcal{S}^\dagger=\mathcal{S}^{-1}$, i.e.:
$$
|\mathcal{S}|=u_k^2+v_k^2=1,\\
\mathcal{S}^{-1}=\frac{1}{u_k^2+v_k^2}
\begin{pmatrix}
u_k & -v_k\\
v_k & u_k
\end{pmatrix}
=\mathcal{S}^T
$$
with the parametrization as is, it can be viewed as a simple rotation with
$$
u_k=\cos\phi_k,v_k=\sin\phi_k
$$
Note also that the transformation is that of a pair of states with different quantum numbers: $k,\uparrow$ and $-k,\downarrow$ - they have different spin and momentum, i.e.$c_{-k,\downarrow}^\dagger$ is not the same as $c_{k,\uparrow}^\dagger= u_kd_{k,\uparrow}^\dagger + v_kd_{-k,\downarrow}$, which the OP seems to suggest.
