# A simple conjecture on the Chern number of a 2-level Hamiltonian $H(\mathbf{k})$?

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For example, let's consider a quadratic fermionic Hamiltonian on a 2D lattice with translation symmetry, and assume that the Fourier transformed Hamiltonian is described by a $2\times2$ Hermitian matrix $H(\mathbf{k})=a(\mathbf{k})\sigma_x+b(\mathbf{k})\sigma_y+c(\mathbf{k})\sigma_z$ and has a finite energy gap, then the Chern number $N$ can be determined.

If $H(-\mathbf{k})=H(\mathbf{k})$ holds for all $\mathbf{k}\in BZ$, then the Chern number $N$ is always an even number, am I right? This seems to be true from the geometrical interpretation of Chern number as a winding number covering a unit sphere, but I have not yet found a rigorous mathematical proof.

Remark: The necessary condition finite energy gap ($\Leftrightarrow$ The map $(a(\mathbf{k}),b(\mathbf{k}),c(\mathbf{k}))/\sqrt{a(\mathbf{k})^2+b(\mathbf{k})^2+c(\mathbf{k})^2}$ from BZ(2D torus) to the unit sphere is well defined) is to ensure that the Chern number/winding number is well defined.

This post imported from StackExchange Physics at 2014-03-09 08:40 (UCT), posted by SE-user K-boy

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I just found a relative rigorous argument supporting my conjecture:

The Chern number $N=\frac{1}{2\pi}\int _{BZ}b(\mathbf{k})$, where $b(\mathbf{k})$ is the Berry curvature. Since $H(-\mathbf{k})=H(\mathbf{k})$, it's easy to show that $b(-\mathbf{k})=b(\mathbf{k})$, accordingly, we can divide the $BZ$ into two halfs called $BZ_1$ and $BZ_2$, therefore, $N=\frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})+\frac{1}{2\pi}\int _{BZ_2}b(\mathbf{k})=2\times \frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})$. Now there are two ways to show that $\frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})$ is an integer,

(1)Due to the relation $b(-\mathbf{k})=b(\mathbf{k})$ and periodic structure of the $BZ$, the half Brillouin zone $BZ_1$ is topologically equivalent to a sphere, and the 'flux' through a sphere(closed surface) $\frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})$ must be quantized as an integer;

(2)Since $H(-\mathbf{k})=H(\mathbf{k})$, the eigenfunction $\psi(\mathbf{k})$ is also even(i.e. $\psi(-\mathbf{k})=\psi(\mathbf{k})$), then the Berry connection $\mathbf{a(k)} \propto \left \langle \psi(\mathbf{k})\mid \bigtriangledown_{\mathbf{k}} \psi(\mathbf{k})\right \rangle$ is odd, thus, it's easy to show that $\int _{BZ_1}b(\mathbf{k})=\oint _ {\partial BZ_1}\mathbf{a(k)}\cdot d\mathbf{k}=0$(where $\partial BZ_1$ is the boundary of $BZ_1$), however, to be consistent, the number '0'(Berry phase) here should be understood as $2\pi\times integer$.

Remark: The key point in our argument (1) is that the points $\mathbf{k}$ and $-\mathbf{k}$ are equivalent, and hence the half $BZ$ is topologically equivalent to a sphere which is a closed surface.

This post imported from StackExchange Physics at 2014-03-09 08:40 (UCT), posted by SE-user K-boy
answered Nov 13, 2013 by (980 points)

See the "supplemental material " in arXiv:1604.04781

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