In the following link https://docs.abinit.org/theory/wavefunctions/#symmetry-properties, there's Eqn. (8) that describes for degenerate states how wavefunctions of k-points connected by symmetry $\mathcal R$ are connected (transformed from each other):

$$ \begin{equation} \Psi_{n\bf k} \left( \mathcal R^{-1} \left( \bf r-t \right) \right) = \sum_{\alpha \in \mathcal C_{n\bf k}} D_{\alpha n}\left( \mathcal R \right) \Psi_{\alpha\mathcal R \bf k} \left( \bf r \right) \end{equation} $$

My question is, how does one calculate or obtain the matrix elements $D_{\alpha n}$?


1 Answer 1


Let's just take your equation seriously. We have

$$ \Psi_{n\bf k} \left( \mathcal R^{-1} \left( \bf r-t \right) \right) = \sum_{\alpha \in \mathcal C_{n\bf k}} D_{\alpha n}\left( \mathcal R \right) \Psi_{\alpha\mathcal R \bf k} \left( \bf r \right) $$

This tells us that $\Psi_{n\bf k} \left( \mathcal R^{-1} \left( \bf r-t \right) \right)$ can be written as a linear combination of $\Psi_{\alpha\mathcal R \bf k} \left( \bf r \right)$. If we want to know the coefficients of the linear combination, we should just multiply both sides of your equation by $\Psi^*_{\beta\mathcal R \bf k} \left( \bf r \right)$ and integrate with respect to $\mathbf{r}$.

\begin{align*} \int d\mathbf{r}\ \Psi^*_{\beta \mathcal{R}\mathbf{k}}(\mathbf r)\Psi_{n\bf k} \left( \mathcal R^{-1} \left( \bf r-t \right) \right) &= \sum_{\alpha \in \mathcal C_{n\bf k}} D_{\alpha n}\left( \mathcal R \right) \int d\mathbf{r}\ \Psi^*_{\beta\mathcal R \bf k} \left( \bf r \right)\Psi_{\alpha\mathcal R \bf k} \left( \bf r \right)\\ &= \sum_{\alpha \in \mathcal C_{n\bf k}} D_{\alpha n}\left( \mathcal R \right) \delta_{\alpha\beta} \qquad\text{(because of orthogonality)}\\ &= D_{\beta n}\left( \mathcal R \right) \qquad\text{(using the $\delta$ to do the sum)} \end{align*}

So we have our formula for $D_{\beta n}$, as desired!

  • $\begingroup$ This is a "numerical" solution, but unfortunately this is not what I'm looking for. I'm more looking for a "general" and analytical solution that makes connection to the considered symmetry $\mathcal R$. In other words, how does $D$ look like as a function of $\mathcal R$? $\endgroup$
    – rnels12
    Commented Apr 9, 2019 at 8:53
  • 1
    $\begingroup$ @rnels12 You might want to look up "representation theory" in that case. I'll note that $D$ is not uniquely determined by $R$, or even $R$ and the number of degenerate bands, because you can always change your basis at each $k$ point and change the form of $D$. So if you want the actual matrix of $D$, there IS no shorter recipe than (1) find the $\Psi_n$ (2) Do the integral. $\endgroup$ Commented Apr 9, 2019 at 19:43

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