# Berry Curvature in a hexagonal lattice

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

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

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{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} 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. (?))

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