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The computation of the band structure of Graphene basically leads to the diagonalization of the following Hamiltonian:

$$ H = -t \left( \begin{array}{cc} 0 & \epsilon(\vec{k}) \\ \epsilon^*(\vec{k})&0 \end{array} \right), $$

where $\epsilon(\vec{k})= \sum_{i=1}^3e^{i\vec{k}\cdot\vec{\delta}_i}.$ The vectors

$$ \delta_1 =\frac{a}{2}(1,\sqrt{3})\qquad\delta_2 = \frac{a}{2}(1,-\sqrt{3})\qquad\delta_3 = -a(1,0) $$

are the vectors connecting one point of the honeycomb lattice to the three nearest neighbours at a distance $a$ from that point.

Using $\xi = \pm$ as a label for the valence ($\xi = -1$) and the conduction ($\xi = +1$) band, the eigenvalues of this Hamiltonian are given by $\lambda_\xi = \xi t |\epsilon|$ and the eigenvectors are

$$ \lvert\Psi_\xi\rangle=\frac{1}{\sqrt{2}}\left( \begin{array}{c} 1 \\ -\xi e^{-i\phi_\epsilon} \end{array} \right), $$

where $\phi_\epsilon = \arg\epsilon$.

Now to my question: Several months ago I attended a presentation about second quantization in condensed matter physics and the speaker mentioned that besides helicity (which is conserved for small $|\vec{k}|$) the Chern number is a conserved quantity for all $\vec{k}$. I tried to check that using the above eigenvectors and the Berry potential, which gave me

$$ \vec{A}(\vec{k}) = i\left( \begin{array}{c} \langle\Psi_+\rvert \frac{\partial}{\partial k_x}\lvert\Psi_+\rangle + \langle\Psi_-\rvert \frac{\partial}{\partial k_x}\lvert\Psi_-\rangle \\ \langle\Psi_+\rvert \frac{\partial}{\partial k_y}\lvert\Psi_+\rangle + \langle\Psi_-\rvert \frac{\partial}{\partial k_y}\lvert\Psi_-\rangle \end{array} \right) = \vec{\nabla}\phi_\epsilon. $$

The Berry curvature is then given by

$$ \Omega = \frac{\partial A_y}{\partial k_x}-\frac{\partial A_x}{\partial k_y} = 0. $$

Since the Chern number is the integral over the Berry curvature, it simply vanishes. I don't think this was not meant by "the Chern number is a conserved quantity in Graphene". So, what am I doing wrong? And in case I am not wrong, what could make the Chern number be non-zero in Graphene? I'm open to any suggestions, literature recommendations and calculations.

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    $\begingroup$ You are not doing wrong --- simple hopping model does not give you a nontrivial Chern number. A nonzero Chern# would require modifications in the model. For example, in the Haldane model, the hopping is carrying a phase which is equivalent to a flux inserted in a plaquette that would eventually gives you a nonzero Chern#, as long as your symmetry setting is proper. $\endgroup$ – Kite.Y Jun 8 '18 at 15:02
  • $\begingroup$ @Kite.Y Thank you for the explanation. The speaker mentioned that the conserved quantity (which according to him is the Chern number) for low energies (small $|\vec{k}|$) reduces to helicity which is conserved in the low-energy description. So, before I start computing the Chern number using the model you mentioned, can you tell whether there is some model in which the non-zero Chern number really reduces to helicity for small $|\vec{k}|$? $\endgroup$ – MeMeansMe Jun 13 '18 at 7:39
  • $\begingroup$ Firstly, by small $|\vec{k}|$ I assume you are talking about momentums around band minimum. And YES: actually the Chern number in Chern insulator is always related to the wrapping number of $k$-dependent parameters in Hamiltonian. The model I give you is the most well-know one with non-trivial Chern-number. In materials, you can design some spin-orbit coupling. $\endgroup$ – Kite.Y Jun 13 '18 at 14:35

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