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Xavier Andrade edited Sternheimer2.tex
over 9 years ago
Commit id: 41da50474f01fbb42345001389ec5b0c8dd06730
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\begin{multline}
m^{-1}_{ijn\vec{k}} = \frac{\partial^2 \epsilon_{n\vec{k}}}{\partial k_i \partial k_j}
= \delta_{ij} + \left< \varphi_{n\vec{k}} \left| \frac{\partial \hat{H}}{\partial
\vec{k}} k_i} \right| \frac{\partial \varphi_{n\vec{k}}}{\partial
\vec{k}} k_j} \right> + \mathrm{cc.}
\\
+ \left< \varphi_{n\vec{k}} \left| \left[
\hat{\vec{r}}, \hat{r}_i, \left[
\hat{\vec{r}}, \hat{r}_j, \hat{v}_{\alpha} \right] \right] \right| \varphi_{n\vec{k}} \right>\ .
\end{multline}
The $\vec{k} \cdot \vec{p}$ wavefunctions can be used to compute response to electric fields in periodic systems.
In finite systems, a homogeneous electric field can be represented simply via the position operator $\vec{r}$.
However, this operator is not well defined in a periodic system and cannot be used. According to the quantum theory of polarization,
the solution is to replace $\vec{r} \varphi$ with $-i d/dk
\varphi$. \varphi$~\cite{Resta_2007}. % cite Baroni, also Gonze papers
The response to Using this
perturbation can be used to compute expression in eq.~(\ref{eq:sternheimer_polarizability}) give obtain a formula for the polarization of the
polarizability using from eq.~\ref{eq:sternheimer_polarizability} crystal,
\begin{align}
\alpha_{ij} \left( \omega \right) = i
\sum_k f_k \sum_n f_n \left[ \left. \left< \frac{\partial
\varphi_k}{\partial \varphi_n\vec{k}}{\partial k_i} \right| \frac{\partial
\varphi_k}{\partial \varphi_v\vec{k}}{\partial \mathcal{E}_{j, \omega}} \right> + \left. \left< \frac{\partial
\varphi_k}{\partial \varphi_n\vec{k}}{\partial \mathcal{E}_{j, -\omega}} \right| \frac{\partial
\varphi_k}{\partial \varphi_n\vec{k}}{\partial k_j} \right>
\right] \right]\ .
\end{align}
The polarizability is most usefully represented in a periodic system via the dielectric constant
\begin{align}