# Momentum matrix elements in a crystal

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.

• The wave functions $\psi_{n, k}$ are, by definition, eigenfuntions of a (periodic) Hamiltonian, and so are orthogonal Commented May 12, 2019 at 10:48
• Yes, its clear to me that the $\psi_{n,k}$ should be orthogonal, but it is unclear to me how to show orthogonality when you write the $\psi_{n,k}$ as Bloch waves. $\int dx e^{i(k-k’)x} = \delta(k-k’)$ but why explicitly is $\int dx e^{i(k-k’)x}u_{nk}^*u_{n’k’} = \delta(k-k’)\delta_{n,n’}$? Commented May 12, 2019 at 15:43

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

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

• In the second line of your evaluation of $\left\langle\psi_{n \mathbf{k}}|\mathbf{p}| \psi_{n^{\prime} \mathbf{k}^{\prime}}\right\rangle,$ should't the first term become negative since you are multiplying $(-i)\times(-i)?$ Commented Feb 20, 2020 at 14:28
• Now it should be correct, sorry for the confusion.
– aljg
Commented Feb 20, 2020 at 19:37
• Hey again. I'm sorry, I'm still a little bit confused. How do you go from the 4th line to the 5th line in your evaluation of $\langle\psi_{nk}|\mathbf{p}|\psi_{n'k'}\rangle$? In particular, how do you go from the volume integral $\int e^{i(k'-k)\cdot r}u_{nk}^*\nabla_r u_{n'k'}$ to the sum over unit cell integrals $\sum_R e^{i(k'-k)\cdot R}\int_{UC}u_{nk}^*\nabla_r u_{n'k'}$? Commented Mar 31, 2020 at 22:34
• Ah. I think I see. Since both $e^{ik\cdot r}\nabla_ru_{nk}(r)$ and $e^{ik\cdot r}u_{nk}(r)$ satisfy Bloch's theorem, we may replace the integral over the crystal with a sum of integrals over unit cells. I was worried that the argument wouldn't hold because of the presence of the gradient. Commented Apr 3, 2020 at 0:45

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.