Berry connection in a solid

Question

I am having troubles to understand an equation-sign for the Berry connection in a solid.

The general formula reads

\begin{equation} \vec{A}(\vec{R}) = \mathrm{i} \langle \Psi(\vec{R}) \, | \nabla_{\vec{R}} \, | \, \Psi(\vec{R}) \rangle \text{.} \end{equation}

Now assuming that \begin{equation} H_0 \psi_{\vec{k}}^n (\vec{r}) = E_n(\vec{k}) \psi_{\vec{k}}^n (\vec{r}) \text{,} \end{equation}

where $u_{\vec{k}}^n$ denotes the function coming from the Bloch-wavefunctions $\psi_{\vec{k}}^n (\vec{r}) = \mathrm{e}^{\mathrm{i} \vec{k} \cdot \vec{r}} u_{\vec{k}}^n(\vec{r})$, it seems (for $\vec{R} \equiv \vec{k}$) to be too obvious to explain why ...

\begin{equation} \vec{A^n}(\vec{k}) = \mathrm{i} \cdot \left( \mathrm{i} \cdot \langle u_{\vec{k}}^n \, | \vec{r} \, | \, u_{\vec{k}}^n \rangle + \langle u_{\vec{k}}^n \, | \nabla_{\vec{k}} \, | \, u_{\vec{k}}^n \rangle \right) = \mathrm{i} \cdot \langle u_{\vec{k}}^n \, | \nabla_{\vec{k}} \, | \, u_{\vec{k}}^n \rangle \end{equation}

... the first term vanishes.

I would be grateful if someone could help me out.

Answer 1

$\mathbf{k}$ is not a parameter of the Hamiltonian $H_0$ for the eigensystem of your second equation. For the general definition of Berry connection, the Hamiltonian $H(\mathbf{R})$ depends on the parameter $\mathbf{R}$. So for Bloch states, one should use $u_{n\mathbf{k}}$ for the eigensystem $H_{\mathbf{k}}u_{n\mathbf{k}}=E_{n\mathbf{k}}u_{n\mathbf{k}}$, where $H_{\mathbf{k}} = e^{-i \mathbf{k \cdot r}}He^{i \mathbf{k \cdot r}}$.