David Strubbe edited Sternheimer2.tex  over 9 years ago

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We can solve for linear response to various different perturbations. The most straight-forward case is the response of a finite system to an electric field \(\mathcal{E}_{i,\omega}\) with frequency \(\omega\) in the direction \(i\) , \(i\),  where the perturbation operator is \(\delta \hat v = \hat r_i\)~\cite{Andrade_2007}. In this case the polarizability can be calculated as %  \begin{equation}  \alpha_{ij}\left( \omega \right) = - \sum_n^{\rm occ} \left[ \langle \varphi_n \vert r_i \vert \frac{\partial \varphi_n}{\partial \mathcal{E}_{j, \omega}} \rangle + \langle \frac{\partial \varphi_n}{\partial \mathcal{E}_{j, -\omega}} \vert r_i \vert \varphi_n \rangle \right] \label{eq:sternheimer_polarizability}\ . 

\end{align}  We can also make the above expression Eq. \ref{eq:Born} for Born charges from the vibrational perturbation usable in a periodic system with the replacement $\vec{r} \varphi \rightarrow -i \partial \varphi/\partial k$.  Unfortunately the $\vec{k} \cdot \vec{p}$ perturbation is not usable to calculate the polarization \cite{Resta_2007}, and a sum over strings of k-points on a finer grid is required. We have implemented the special case of a $\Gamma$-point calculation for a large supercell, where the single-point Berry phase can be used \cite{Yaschenko1998}. For cell sizes $L_i$ in each direction, the dipole moment is given by derived from the determinant of a matrix whose basis is the occupied KS orbitals:  \begin{align}  \mu_i = - \frac{e L_i}/{2*\pi} L_i}/{2 \pi}  \mathcal{I} {\rm ln} ln}\  {\rm det} det}\  \left< \varphi_n \left| exp(- 2*\pi 2 \pi  i x_i/L_i) \right| \varphi_m \right> \end{align}  % magnetic is non-self-consistent if wfns are real.