BCS Theory: Bogoliubov Transformation for Fermions

Question

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 $(1.1)$?

Answer 1

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.