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}