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the solution is to replace $\vec{x} \varphi$ with $-i d/dk \varphi$. % cite Baroni, also Gonze papers  The response to this perturbation can be used to compute the polarizability:  \begin{align}  \alpha_{ij} \left( \omega \right) = - \sum_k f_k \left[ \left< \varphi_k \left| r_i \right| \frac{\partial \varphi_k}{\partial \mathcal{E}_{j, \omega}} \right> + \left< \frac{\partial \varphi_k}{\partial \mathcal{E}_{j, -\omega}} \left| \r_i r_i  \right| \varphi_k \right> \right] \\ \alpha_{ij} \left( \omega \right) = i \sum_k f_k \left[ \left. \left< \frac{\partial \varphi_k}{\partial k_i} \right| \frac{\partial \varphi_k}{\partial \mathcal{E}_{j, \omega}} \right> + \left. \left< \frac{\partial \varphi_k}{\partial \mathcal{E}_{j, -\omega}} \right| \frac{\partial \varphi_k}{\partial k_j} \right> \right]  \end{align}  The polarizability is most usefully represented in a periodic system via the dielectric constant