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I am having troubles to understand the concept of the Berry curvature in a hexagonal lattice.

What I know is: The Berry curvature $\Omega_n (\vec{k})$ for the $n$-th band reads

\begin{equation} \Omega_n(\vec{k}) = \mathrm{i} \nabla_{\vec{k}} \times \langle u_{\vec{k}}^n \,| \, \nabla_{\vec{k}}\, | \, u_{\vec{k}}^n \rangle \text{,} \end{equation}

where $u_{\vec{k}}^n$ denotes the function coming from the Bloch-wavefunctions $\psi_{\vec{k}}^n (\vec{r}) = \mathrm{e}^{\mathrm{i} \, \vec{k} \cdot \vec{r}} u_{\vec{k}}^n(\vec{r})$.

In this arxive article it states, that for a honeycomb lattice in the tight-binding limit ...

(in which the second quantized Hamiltonian $\hat{H}_{\text{tb}}$ assumes the form

\begin{equation} \begin{aligned} H_{\text{tb}} &= \sum_{\vec{k}} \left( a_1(\vec{k})\text{,} \quad a_2(\vec{k}) \right) \begin{pmatrix} \Delta & f(\vec{k}) \\ f(\vec{k})^{*} & -\Delta \end{pmatrix} \left( \begin{array}{c} a_1(\vec{k})\\ a_2(\vec{k})\\ \end{array}\right) \\ &=: \sum_{\vec{k}} \left( a_1(\vec{k})\text{,} \quad a_2(\vec{k}) \right) H_{\text{tb}}(\vec{k}) \left( \begin{array}{c} a_1(\vec{k})\\ a_2(\vec{k})\\ \end{array}\right) \text{,} \end{aligned} \end{equation} where $a_{1,2}^{(\dagger)}(\vec{k})$ correspond to annihilation (creation) operators at the first, second atom of the unit cell and $\Delta \rightarrow 0$)

... the Berry curvature is given by the above formula, but $u_{\vec{k}}^n$ correspond (for $n = 1,2)$ to the two eigenstates of the matrix $H_{\text{tb}}(\vec{k})$.

The conclusion I am having issues with is identifying the eigenstates of the Hamiltonian with the functions $u_{\vec{k}}^n$. (For example: Where is the $\vec{r}$-dependance in the eigenstates of the Hamiltonian? There is none. (?))

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  • $\begingroup$ It's not clear what your question is exactly. Which part are you stuck on? $\endgroup$ – d_b Jun 17 '19 at 6:53
  • $\begingroup$ Thank you for your input. I edited the question. If there is still some unclarity, let me know. $\endgroup$ – Antihero Jun 17 '19 at 23:41
  • $\begingroup$ I don't know if this is an answer but the Bloch states $|u_{n,\mathbf{k}}\rangle$ are distinct from the wavefunctions $u_{n,\mathbf{k}}(\mathbf{r}) = \langle \mathbf{r}| u_{n,\mathbf{k}}\rangle$. $\endgroup$ – d_b Jun 21 '19 at 21:49

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