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.

  • $\begingroup$ How does your final equation follow from your first equation? $\endgroup$ – d_b Jun 17 '19 at 6:40
  • $\begingroup$ You identify $\langle \vec{r} \, | \, \Psi(\vec{R}) \rangle \equiv \psi_{\vec{k}}^n(\vec{r})$. Then the plane-wave-factors cancel out. $\endgroup$ – Antihero Jun 17 '19 at 23:54
  • $\begingroup$ Do you have a reference? Is it possible that the choice $|\Psi(\mathbf{R})\rangle = |u_n(\mathbf{k})\rangle$ is being made? $\endgroup$ – d_b Jun 21 '19 at 22:45
  • $\begingroup$ Thank you for your answer. I also considered this identification. :-) My first reference is this arxiv article. arxiv.org/pdf/1509.02295.pdf; Eq. (2.20) is the definition of the Berry connection as above. Looking at eqs (2.39) and (2.40) supports your choice of identification. However... What is bothering me with this interpretation is that the function $| u_n(\vec{k}) \rangle$ is not a physical state in a Hilbert space, is it? Only the full Bloch-wavefunction (with plane-wave-factor) should be a physical state. (?) $\endgroup$ – Antihero Jun 22 '19 at 7:50
  • $\begingroup$ I don't see why the latter statement should be true. $\exp\left(-i\mathbf{k}\cdot\mathbf{r}\right)$ is a unitary operator. Applying it to a physical state should give back a physical state. $\endgroup$ – d_b Jun 22 '19 at 19:52

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