Given a massive Dirac Hamiltonian of $H = k_x\sigma_x + k_y\sigma_y + m\sigma_z$

The Chern number can be found as $Q = \frac{1}{2\pi}\int d^2k (\nabla_k\times A(k))$

Where $A(k)$ is the Berry's connection and is given as $A(k) =-i<\psi(k)|\nabla_k|\psi(k)>$

Is there any worked example of this type of question? I'm trying to calculate the Chern number here but is having trouble calculating $A(k)$. Also, what are the boundaries of the integral supposed to be?

EDIT: I have the eigenvalue of $E_\pm = \sqrt{k_x^2+k_y^2+m^2}$

Eigenvector of $|\psi(k)> = (\matrix{-m-E_\pm-k_x+ik_y \\ m - E_\pm-k_x-ik_y})$

I'm not sure how to calculate $A(k)$. I split the $\nabla_k$ into $\nabla_{k_x}$ and $\nabla_{k_y}$. Hence I have $A(k_x) = -i<\psi(k)|\nabla_{k_x}|\psi(k)>$

and $A(k_y) = -i<\psi(k)|\nabla_{k_y}|\psi(k)>$.

I assume $\nabla_{k_x}|\psi(k)>$ is simply $\frac{\partial}{\partial x} \psi(k)$.

Doing for $E_+$, I got $A(k_x) = \frac{-ik_x^2}{\sqrt{k_x^2+k_y^2+m^2}}-i\sqrt{k_x^2+k_y^2+m^2}-2ik_x$

and $A(k_y)=\frac{-2ik_xk_y}{\sqrt{k_x^2+k_y^2+m^2}}-2m-4ik_y$


$\nabla_k \times A(k) = \frac{\partial A(k_y)}{\partial k_x} - \frac{\partial A(k_x)}{\partial k_y} = \frac{-ik_y(k_y^2+m^2)}{(k_x^2+k_y^2+m^2)^{3/2}}$

Lastly, it's $Q = \frac{1}{2\pi}\int d^2k (\nabla_k\times A(k))$. Which I don't know what boundaries I'm supposed to integrate around.

  • $\begingroup$ What's the trouble? It looks very straightforward. $\endgroup$ – Meng Cheng Jan 11 '17 at 13:40
  • $\begingroup$ @MengCheng I'm not sure how to get $|\psi(k)>$, I'm assuming it is simply the eigenvector of the Hamiltonian. When I worked it out to the integral, I'm unable to get an integer for the Chern number despite trying different boundaries for the integral. I guess something went wrong when calcuating $A(k)$, probably with $\nabla_k|\psi(k)>$. $\endgroup$ – SataSata Jan 11 '17 at 13:52
  • $\begingroup$ Then to enhance the probability of getting meaningful response, you may want to detail what you have actually done and where exactly you got stuck. $\endgroup$ – Meng Cheng Jan 11 '17 at 14:17
  • $\begingroup$ If I suppose you did everything correct, then $Q=0$ because the field is odd in $k_y$. Some remarks : $A$ is a vector, usually one writes its components as $A_x$ and $A_y$, not $A(x)$ or $A(y)$ as you did. Are you sure your $\Psi$ are normalised ? Are you sure there is only one eigenvector ? Are you sure it is correct ? Perhaps before trying to calculate some topological things, you should review the basics of linear algebra ! $\endgroup$ – FraSchelle Jan 11 '17 at 16:15

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