Static linear response: theory: Difference between revisions

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Let’s consider three types of static perturbations

atomic displacements ${\textstyle u_m}$ with $m=I\alpha$ with $I=\{1..N_\text{atoms}\}$ and $\alpha=\{1..3\}$
homogeneous strains ${\textstyle \eta_j}$ with ${\textstyle j=\{1..6\}}$
static electric field ${\textstyle \mathcal{E}_\alpha}$ with ${\textstyle \alpha=\{1..3\}}$

By performing a Taylor expansion of the total energy in terms of these perturbations we obtain^[1]

{\displaystyle \begin{aligned} E(u,\mathcal{E},\eta) = &E_0 + \\ &\frac{\partial E}{\partial u_m} u_m + \frac{\partial E}{\partial \mathcal{E}_\alpha} \mathcal{E}_\alpha+ \frac{\partial E}{\partial \eta_j} \eta_j + \\ &\frac{1}{2} \frac{\partial^2 E} {\partial u_m \partial u_n } u_m u_n + \frac{1}{2} \frac{\partial^2 E} {\partial \mathcal{E}_\alpha \partial \mathcal{E}_\beta } \mathcal{E}_\alpha \mathcal{E}_\beta + \frac{1}{2} \frac{\partial^2 E} {\partial \eta_j \partial \eta_k} \eta_j \eta_k + \\ &\frac{\partial^2 E}{\partial u_m \partial \mathcal{E}_\alpha} u_m \mathcal{E}_\alpha + \frac{\partial^2 E}{\partial u_m \partial \eta_j} u_m \eta_j + \frac{\partial^2 E}{\partial \mathcal{E}_\alpha \partial \eta_j} \mathcal{E}_\alpha \eta_j + \text{terms of higher order} \end{aligned} }

The derivatives of the energy with respect to an electric field are the polarization, with respect to atomic displacements are the forces, with respect to changes in the lattice vectors are the stress tensor.

{\displaystyle P_\alpha = -\frac{\partial E}{\partial \mathcal{E}_\alpha} \qquad \text{polarization} }

{\displaystyle F_m = -\Omega_0\frac{\partial E}{\partial u_m} \qquad \text{forces} }

{\displaystyle \sigma_j = \frac{\partial E}{\partial \eta_j} \qquad \text{stresses} }

This leads to the following ‘clamped-ion’ or ‘frozen-ion’ definitions:

{\displaystyle \overline{\chi}_{\alpha\beta} = - \frac{\partial^2 E}{\partial \mathcal{E}_\alpha \partial \mathcal{E}_\beta} |_{u,\eta} \qquad \text{dielectric susceptibility} }

{\displaystyle \overline{C}_{jk} = \frac{\partial^2 E}{\partial \eta_j \partial \eta_k} |_{u,\mathcal{E}} \qquad \text{elastic tensor} }

{\displaystyle \Phi_{mn}=\Omega_0 \frac{\partial^2 E}{\partial u_m \partial u_n} |_{\mathcal{E},\eta} \qquad \text{force-constants} }

{\displaystyle \overline{e}_{\alpha k} = \frac{\partial^2 E}{\partial \mathcal{E}_\alpha \partial \eta_k} |_{u} \qquad \text{piezoelectric tensor} }

{\displaystyle Z_{m\alpha}=-\Omega_0 \frac{\partial^2 E}{\partial u_m \partial \mathcal{E}_\alpha} |_{\eta} \qquad \text{Born effective charges} }

{\displaystyle \Xi_{mj}=-\Omega_0 \frac{\partial^2 E}{\partial u_m \partial \eta_j} |_{\mathcal{E}} \qquad \text{force response internal strain tensor} }

References

↑ X. Wu, D. Vanderbilt, and D. R. Hamann, Phys. Rev. B 72, 035105 (2005).

[wu:prb:2005-1] X. Wu, D. Vanderbilt, and D. R. Hamann, Phys. Rev. B 72, 035105 (2005).

[1]

@@ Line 1: / Line 1: @@
 Let’s consider three types of static perturbations
-# atomic displacements <math display="inline">u_m</math>
+# atomic displacements <math display="inline">u_m</math> with <math>m=I\alpha</math> with <math>I=\{1..N_\text{atoms}\}</math> and <math>\alpha=\{1..3\}</math>
 # homogeneous strains <math display="inline">\eta_j</math> with <math display="inline">j=\{1..6\}</math>
 # static electric field <math display="inline">\mathcal{E}_\alpha</math> with <math display="inline">\alpha=\{1..3\}</math>
-By performing a Taylor expansion of the total energy in terms of these perturbations we obtain
+By performing a Taylor expansion of the total energy in terms of these perturbations we obtain{{cite|wu:prb:2005}}
 :<math>
 \begin{aligned}
@@ Line 83: / Line 83: @@
 \qquad \text{force response internal strain tensor}
 </math>
+== References ==