Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  How to determine the orientation of the massive Dirac Hamiltonian?

+ 3 like - 0 dislike
1817 views

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 (980 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

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 (785 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

How did you know Chern number is 1/2 or -1/2??

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysics$\varnothing$verflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...