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How to determine the orientation of the massive Dirac Hamiltonian?

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In the calculation of the Chern number within a 2D lattice model, let's take the Haldane model for example, the Chern number$=\pm1$ has 2 contributions coming from 2 Dirac points described by $$h_1(\mathbf{q})=q_y\sigma_x-q_x\sigma_y-\sigma_z$$ and $$h_2(\mathbf{q})=-q_y\sigma_x-q_x\sigma_y+\sigma_z$$.

Both of the above 2 Hamiltonians contribute the same 1/2(or -1/2) Chern number with the same orientation (i.e., the sign of Chern number).

My question is: How to judge whether two massive Dirac Hamiltonians(e.g. $h_1$ and $h_2$) have the same or opposite orientations simply from the form of the Hamiltonian?

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy
asked Feb 19, 2014 in Theoretical Physics by Kai Li (975 points) [ no revision ]

1 Answer

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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
answered Feb 24, 2014 by Everett You (660 points) [ no revision ]
@ Everett You Thanks a lot. And is the parameter $m$ (mass term) always assumed to be positive?

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy
@K-boy Yes, you need to define which mass is called positive. If you flip the sign of the mass, the Chern number will also flip the sign. The chirality of your Dirac Hamiltonian is only defined if the fermion is gapped, then the sign of the mass is a mater of how you choose to gap the system.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Everett You

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