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
    Commented Jun 17, 2019 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
    Commented Jun 17, 2019 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
    Commented Jun 21, 2019 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
    Commented Jun 22, 2019 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
    Commented Jun 22, 2019 at 19:52

1 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 $H_{\mathbf{k}} = e^{-i \mathbf{k \cdot r}}He^{i \mathbf{k \cdot r}}$.


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