Momentum matrix elements in a crystal

Question

I am trying to follow along a derivation (E. I. Blount, Solid State Phys. 13, 305 (1962)) in which he derives the matrix elements of the true momentum $p_{n,n'}(k,k')$ (not the crystal momentum). He arrives at the following expression:

$p_{n,n'}(k,k') = \delta(k-k')(\hbar k\delta_{n,n'}-i\hbar\int u_n^*\frac{\partial u_{n'}}{\partial x}\,d\tau)$,

where $n$ labels the band index, $k$ is the Bloch vector, $u_n$ is the Bloch function with the periodicity of the lattice, and the integral is over a unit cell. Can anyone help me to derive this result?

Crucially, the problem boils down to evaluating

$\int dx \, e^{-ikx}u_{nk}^*(x)e^{ik'x}u_{n'k'}(x)$,

which, according to Blount ought to equal $\delta_{n,n'}\delta(k,k')$. This makes sense if our wave functions

$\psi_{n,k}(x) = e^{ikx}u_{nk}(x)$

(i.e. Bloch waves) are to be normalized, but I just can't seem to figure out how to 'pluck' those two delta functions out from the integration.

Answer 1

I will use atomic units throughout. The Bloch states are formed by the product of a plane wave times a periodic part $|u_{n\mathbf{k}}\rangle$, \begin{equation} \langle \mathbf{r} |\psi_{n\mathbf{k}} \rangle = \frac{1}{\sqrt{V}} e^{i \mathbf{k}\cdot \mathbf{r}} \langle \mathbf{r} | u_{n\mathbf{k}} \rangle \end{equation} where $V$ is the volume of the solid. The normalization of the Bloch states is \begin{equation} \begin{split} \langle \psi_{n'\mathbf{k}'} | \psi_{n\mathbf{k}} \rangle &= \frac{1}{V}\int d\mathbf{r} e^{-i \mathbf{k}'\cdot\mathbf{r}} e^{i \mathbf{k}\cdot\mathbf{r}} u_{n\mathbf{k}'}^*(\mathbf{r}) u_{n\mathbf{k}}(\mathbf{r}) \\ &=\frac{1}{V}\sum_\mathbf{R} e^{i (\mathbf{k}'-\mathbf{k})\cdot \mathbf{R}} \int_{V_\text{UC}} u_{n\mathbf{k}'}^*(\mathbf{r}) u_{n\mathbf{k}}(\mathbf{r}) \\ &=\delta_{\mathbf{k}\mathbf{k}'} \delta_{nn'} \end{split} \end{equation} In step two we expressed the integral over all the volume $V$ as a sum over the unit cells of volume $V_{\text{UC}}$, with $\mathbf{R}$ a lattice vector. In the last step, we used the lattice sum rule, \begin{equation} \frac{1}{N}\sum_\mathbf{R} e^{i (\mathbf{k}'-\mathbf{k})\cdot \mathbf{R}} = \delta_{\mathbf{k}\mathbf{k}'} \end{equation} where $N$ is the number of lattice cells, and the normalization of the periodic part of the Bloch wavefunction, \begin{equation} \frac{1}{V_\text{UC}}\int_{V_\text{UC}} u_{n'\mathbf{k}}^*(\mathbf{r}) u_{n\mathbf{k}}(\mathbf{r}) = \delta_{nn'} \end{equation}

With this, the derivation of the momentum operator follows, \begin{equation} \begin{split} \langle \psi_{n \mathbf{k}} | \mathbf{p} | \psi_{n' \mathbf{k}'} \rangle &= \frac{1}{V}\int d\mathbf{r} e^{-i\mathbf{k}\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) (-i) \nabla_\mathbf{r} \left(e^{i \mathbf{k}'\cdot \mathbf{r}} u_{n'\mathbf{k}'}(\mathbf{r})\right) \\ &=\frac{1}{V}\int d\mathbf{r} e^{-i\mathbf{k}\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) (-i) \left[ i\mathbf{k}' e^{i \mathbf{k}'\cdot \mathbf{r}} u_{n'\mathbf{k}'}(\mathbf{r}) + e^{i \mathbf{k}'\cdot \mathbf{r}}\nabla_\mathbf{r} u_{n'\mathbf{k}'}(\mathbf{r})\right] \\ &= \frac{1}{V} \int d\mathbf{r} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) \left[ \mathbf{k}' u_{n'\mathbf{k}'}(\mathbf{r}) -i \nabla_\mathbf{r} u_{n'\mathbf{k}'}(\mathbf{r})\right] \\ & = \frac{1}{V} \sum_\mathbf{R} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{R}} \left[ \mathbf{k}' \int_{V_\text{UC}} u_{n\mathbf{k}}^*(\mathbf{r}) u_{n'\mathbf{k}'}(\mathbf{r}) - i\int_{V_\text{UC}} u_{n\mathbf{k}}^* (\mathbf{r}) \nabla_\mathbf{r} u_{n'\mathbf{k}'}(\mathbf{r})\right] \\ &= \delta_{\mathbf{k}\mathbf{k}'} \left[ \mathbf{k} \delta_{nn'} - i\int_{V_\text{UC}} u_{n\mathbf{k}}^* (\mathbf{r}) \nabla_\mathbf{r} u_{n'\mathbf{k}}(\mathbf{r})\right] \end{split} \end{equation} as shown by Blount.

Answer 2

I agree with aljg up to the third line from the bottom of his derivation. I would argue that from there, the derivation should go as follows. \begin{align} \langle \psi_{n \mathbf{k}} | \frac{\mathbf{p}}{\hbar} | \psi_{n' \mathbf{k}'} \rangle &= \frac{1}{V} \int d\mathbf{r} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) \left[ \mathbf{k}' u_{n'\mathbf{k}'}(\mathbf{r}) -i \nabla_\mathbf{r} u_{n'\mathbf{k}'}(\mathbf{r})\right] \\ & = \frac{1}{V}\sum_\mathbf{R}e^{i(\mathbf{k'-k)\cdot R}}\int_{V_{UC}}d\mathbf{r} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) \left[ \mathbf{k}' u_{n'\mathbf{k}'}(\mathbf{r}) -i \nabla_\mathbf{r} u_{n'\mathbf{k}'}(\mathbf{r})\right] \\ & = \frac{N}{V}\delta_{\mathbf{kk'}}\int_{V_{UC}}d\mathbf{r} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) \left[ \mathbf{k}' u_{n'\mathbf{k}'}(\mathbf{r}) -i \nabla_\mathbf{r} u_{n'\mathbf{k}'}(\mathbf{r})\right] \\ & = \frac{N}{V}\delta_{\mathbf{kk'}}\int_{V_{UC}}d\mathbf{r} u_{n\mathbf{k}}^*(\mathbf{r}) \left[ \mathbf{k} u_{n'\mathbf{k}}(\mathbf{r}) -i \nabla_\mathbf{r} u_{n'\mathbf{k}}(\mathbf{r})\right] \\ & = \frac{N}{V}\delta_{\mathbf{kk'}}\left[\mathbf{k}V_{UC}\delta_{nn'}-i\int_{V_{UC}}d\mathbf{r}u_{n\mathbf{k}}^*(\mathbf{r})\nabla_\mathbf{r}u_{n'\mathbf{k}}(\mathbf{r}) \right] \\ & =\delta_{\mathbf{kk'}}\left[\mathbf{k}\delta_{nn'}-\frac{i}{V_{UC}}\int_{V_{UC}}d\mathbf{r}u_{n\mathbf{k}}^*(\mathbf{r})\nabla_\mathbf{r}u_{n'\mathbf{k}}(\mathbf{r}) \right]. \end{align}

However, it seems I have picked up an unwanted factor of $\frac{1}{V_{UC}}$ in the second term, which messes up the dimensions.