# 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 $$(1.1)$$?

• Can you elaborate on why you think the minus sign should not be there? Jul 28, 2021 at 10:20
• Sure! I assumed $v_k$ to be real, so $v_k^{\star} = v_k$. The adjoint of $d_{-k, \downarrow}^{\dagger}$ is just $d_{-k, \downarrow}$. Now we just have to change the sign of the $k$ and flip over the spin, but why is there a minus sign? Jul 28, 2021 at 11:09

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.
• Okay, but if I did not write a minus sign, then for the determinant, I would simply get the condition $\vert S\vert = u_k^2 - v_k^2 = 1$, so I am not sure this explains why we have to put there a minus sign.. Jul 28, 2021 at 11:46
• Unitarity is not just a unitary determinant, but $|S|=1$ and $S^\dagger=S^{-1}$. Jul 28, 2021 at 11:47