- #1
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I'm trying to figure out the second order extension of the "trick" used on page 92 (https://www.damtp.cam.ac.uk/user/tong/aqm/solid3.pdf) for the calculation of the effective mass matrix ##m^{\star}_{ij} = \hbar^2 (\partial^2 E/ \partial k_i \partial k_j)^{-1}## on page 94. I think for this one would need to consider the following perturbation:\begin{align*}
\delta H &= \frac{\partial H_{\mathbf{k}}}{\partial \mathbf{k}} \cdot \mathbf{q} + \frac{1}{2}\frac{\partial^2 H_{\mathbf{k}}}{\partial k_i \partial k_j} q_i q_j \\
&= \frac{\hbar^2}{m}\mathbf{q} \cdot (-i\nabla + \mathbf{k}) + \frac{\hbar^2}{m} \delta_{ij} q_i q_j
\end{align*}Then I can equate the second order perturbation expansion to the Taylor expansion of the exact result ##E(\mathbf{k} + \mathbf{q})##,\begin{align*}
\frac{\partial E}{\partial \mathbf{k}} \cdot \mathbf{q} + \frac{1}{2}\frac{\partial^2 E}{\partial k_i \partial k_j} q_i q_j &= \langle \psi_{n,\mathbf{k}}| \delta H | \psi_{n,\mathbf{k}} \rangle + \sum_{n \neq n'} \frac{|\langle \psi_{n,\mathbf{k}} | \delta H | \psi_{n', \mathbf{k}} \rangle|^2}{E_{n}(\mathbf{k}) - E_{n'}(\mathbf{k})} \\
\frac{1}{2}\frac{\partial^2 E}{\partial k_i \partial k_j} q_i q_j &= \frac{\hbar^2}{m} \delta_{ij} q_i q_j + \frac{\hbar^2}{m^2} \sum_{n \neq n'} \frac{|\langle \psi_{n,\mathbf{k}} | \mathbf{q} \cdot -i \hbar \nabla | \psi_{n', \mathbf{k}} \rangle|^2}{E_{n}(\mathbf{k}) - E_{n'}(\mathbf{k})}
\end{align*}where I canceled the first order terms from both sides, and also dropped the fourth order terms in ##q_i## from the second term. Then\begin{align*}
\hbar^2 \left(\frac{\partial^2 E}{\partial k_i \partial k_j} \right)^{-1} &= \frac{m}{2}\left[ \delta_{ij} - \frac{1}{m} \sum_{n\neq n'} \frac{\langle \psi_{n,\mathbf{k}} | p_i | \psi_{n',\mathbf{k}} \rangle \langle \psi_{n',\mathbf{k}} |p_j | \psi_{n,\mathbf{k}} \rangle}{E_n(\mathbf{k}) - E_{n'}(\mathbf{k})} \right]
\end{align*}What's wrong?
\delta H &= \frac{\partial H_{\mathbf{k}}}{\partial \mathbf{k}} \cdot \mathbf{q} + \frac{1}{2}\frac{\partial^2 H_{\mathbf{k}}}{\partial k_i \partial k_j} q_i q_j \\
&= \frac{\hbar^2}{m}\mathbf{q} \cdot (-i\nabla + \mathbf{k}) + \frac{\hbar^2}{m} \delta_{ij} q_i q_j
\end{align*}Then I can equate the second order perturbation expansion to the Taylor expansion of the exact result ##E(\mathbf{k} + \mathbf{q})##,\begin{align*}
\frac{\partial E}{\partial \mathbf{k}} \cdot \mathbf{q} + \frac{1}{2}\frac{\partial^2 E}{\partial k_i \partial k_j} q_i q_j &= \langle \psi_{n,\mathbf{k}}| \delta H | \psi_{n,\mathbf{k}} \rangle + \sum_{n \neq n'} \frac{|\langle \psi_{n,\mathbf{k}} | \delta H | \psi_{n', \mathbf{k}} \rangle|^2}{E_{n}(\mathbf{k}) - E_{n'}(\mathbf{k})} \\
\frac{1}{2}\frac{\partial^2 E}{\partial k_i \partial k_j} q_i q_j &= \frac{\hbar^2}{m} \delta_{ij} q_i q_j + \frac{\hbar^2}{m^2} \sum_{n \neq n'} \frac{|\langle \psi_{n,\mathbf{k}} | \mathbf{q} \cdot -i \hbar \nabla | \psi_{n', \mathbf{k}} \rangle|^2}{E_{n}(\mathbf{k}) - E_{n'}(\mathbf{k})}
\end{align*}where I canceled the first order terms from both sides, and also dropped the fourth order terms in ##q_i## from the second term. Then\begin{align*}
\hbar^2 \left(\frac{\partial^2 E}{\partial k_i \partial k_j} \right)^{-1} &= \frac{m}{2}\left[ \delta_{ij} - \frac{1}{m} \sum_{n\neq n'} \frac{\langle \psi_{n,\mathbf{k}} | p_i | \psi_{n',\mathbf{k}} \rangle \langle \psi_{n',\mathbf{k}} |p_j | \psi_{n,\mathbf{k}} \rangle}{E_n(\mathbf{k}) - E_{n'}(\mathbf{k})} \right]
\end{align*}What's wrong?
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. I’m still trying to see how ##p_i## comes about. Then all one needs is the missing factor of ##m^2## and the sign error…..
