A simple method to judge the chirality (or in your words "orientation") of the Hamiltonian is to evaluate the following quantity
$$f=\frac{i}{2}\mathrm{Tr}\frac{\partial h}{\partial{q_x}}\frac{\partial h}{\partial{q_y}}\frac{\partial h}{\partial{m}}.$$
The sign of this quantity $f$ gives the chirality of the Hamiltonian.
Example: Given the two Hamiltonians $h_1=q_y\sigma_x-q_x\sigma_y-m\sigma_z$ and $h_2=-q_y\sigma_x-q_x\sigma_y+m\sigma_z$, we can evaluate
$$f_1=\frac{i}{2}\mathrm{Tr}(-\sigma_y)\sigma_x(-\sigma_z)=1,$$
$$f_2=\frac{i}{2}\mathrm{Tr}(-\sigma_y)(-\sigma_x)\sigma_z=1.$$
Because $f_1$ and $f_2$ are of the same sign, so $h_1$ and $h_2$ are of the same chirality.
The reason that this trick works is that it basically estimates the Berry curvature, which is defined as
$$F=\frac{i}{2}\mathrm{Tr}\,G^{-1}\mathrm{d}G\wedge G^{-1}\mathrm{d}G\wedge G^{-1}\mathrm{d}G,$$
where $G=(i\omega - h)^{-1}$ is the single particle Green's function. The Chern number is then simply an integral of the Berry curvature, i.e. $C=\frac{1}{2\pi}\int F$. Since the Berry curvature mostly concentrated around the origin of the momentum-frequency space, one just need to estimate the Berry curvature at that point to determine the sign of the Chern number. While in the formula for $F$, $$G^{-1}\mathrm{d}G = (i\omega-h)d(i\omega-h)^{-1}\sim G dh,$$ which gives the $\mathrm{d} h$ terms. And because the Hamiltonian is gaped, so in the zero momentum and frequency limit, the Green's function is a constant $G\propto \partial_m h$ that is proportional to the mass term. Putting all these pieces together, and to the leading order of momentum and frequency, we find the Berry curvature can be roughly estimated from $F\sim f$. So the quantity $f$ is of the same sign as the Chern number, and can be used to determine the chirality of the Hamiltonian. This estimate is exact around the Dirac point, which is just the case of the examples you provided.
This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Everett You