I do not manage to understand a simplification in a quantum mechanics demonstration that computes the momentum matrix element in a semi conductor band structure. In particular how to pass from first to second line :

\begin{eqnarray} \left\langle \phi_{\lambda'k'}| \mathbf p | \phi_{\lambda k} \right\rangle &=& \int_V \phi_{\lambda'k'}^*(\mathbf r) \mathbf p \phi_{\lambda k}(\mathbf r) d^3r \\ &=&\frac{1}{V}\int_V e^{-i(\mathbf k'- \mathbf k)\cdot \mathbf r}\left\langle\lambda'|\mathbf r\right\rangle (\hbar\mathbf k+\mathbf p )\left\langle \mathbf r|\lambda \right\rangle d^3r \end{eqnarray}

Where $\phi_{\lambda k}(\mathbf r)$ is the wave function of an electron with wave vector $\mathbf k$ in band $\lambda$. We know from the Bloch theorem that to first order approximation :

\begin{equation} \phi_{\lambda k}(\mathbf r)\approx e^{i\mathbf k\cdot \mathbf r} \left\langle\mathbf r | \lambda\right\rangle \end{equation}

Where $\left\langle\mathbf r | \lambda\right\rangle$ is the state of the bottom electron in band $\lambda$ with wave vector $\mathbf k = 0$.

  • $\begingroup$ Did you try to just substitute the $\phi$-s and use ${\bf p} = -i\hbar \nabla$? $\endgroup$ – udrv Feb 21 '17 at 19:26
  • $\begingroup$ Indeed that's all there is to do in fact. I Forgot to differentiate the exponential. Thanks ! $\endgroup$ – Ronan Tarik Drevon Feb 21 '17 at 21:06

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