0
$\begingroup$

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 $(1.1)$?

$\endgroup$
2
  • $\begingroup$ Can you elaborate on why you think the minus sign should not be there? $\endgroup$ Jul 28, 2021 at 10:20
  • $\begingroup$ 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? $\endgroup$
    – Hermi
    Jul 28, 2021 at 11:09

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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.. $\endgroup$
    – Hermi
    Jul 28, 2021 at 11:46
  • $\begingroup$ Unitarity is not just a unitary determinant, but $|S|=1$ and $S^\dagger=S^{-1}$. $\endgroup$ Jul 28, 2021 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.