The Hall conductivity is given by

$$\sigma_{xy}^{2D}=\frac{e^2}{\hbar} \int \frac{d\vec{k}}{(2\pi)^d} f(\epsilon(\vec{k}))\Omega_{k_xk_y} \tag{1} $$ in which $f$ is the Fermi distribution function and $\Omega$ the Berry curvature. If we write the Berry curvature in terms of the Berry vector potential and integrate Eq.$(1)$ by part, one finds $$\sigma_{xy}^{2D}=\frac{e^2}{\hbar} \int \frac{d\vec{k}}{(2\pi)^d} \left( \frac{\partial f}{\partial k_y} \mathcal{A}_{k_x}-\frac{\partial f}{\partial k_x} \mathcal{A}_{k_y} \right) \tag{2}.$$ Note that the Fermi distribution function $f$ is a step function at the Fermi energy. If we assume the Fermi surface is a closed loop in the Brillouin zone, then we have: $$\sigma_{xy}^{2D}=\frac{e^2}{2\pi\hbar}\oint d\vec{k}\cdot\vec{\mathcal{A}} \tag{3}.$$ The integral is nothing but the Berry phase along the Fermi circle in the Brillouin zone.

My question is how to derive the formula Eq.$(3)$ from Eq.$(2)$? One can get the Eq.$(3)$ from Eq.$(1)$ by Stokes theorem.


1 Answer 1


If the distribution function has a step discontinuity at the Fermi surface, its derivatives are Dirac delta functions which reduce the integral to one over the Fermi surface. I will illustrate this for the simple case of a 2D system with a rectangular Fermi surface, but the full argument for 2D and 3D can be found in Haldane's Phys. Rev. Lett. 93, 206602 (2004) [arXiv version]. A related paper is X. Wang, D. Vanderbilt, J. R. Yates and I. Souza's Phys. Rev. B 76, 195109 (2009) [arXiv version]. (The rectangle is topologically equivalent to the circle, but makes it easy to work with an explicit coordinate system, thus making the problem concrete.)

We start with your Eq. (2), $$\sigma_{xy}^{2D}=\frac{e^2}{\hbar} \int \frac{d\vec{k}}{(2\pi)^2} \left( \frac{\partial f}{\partial k_y} \mathcal{A}_{k_x}-\frac{\partial f}{\partial k_x} \mathcal{A}_{k_y} \right) \tag{4}.$$ Now, assume a rectangular distribution function. A simple way to write one is $$ f = \left[ \theta \left( k_x \right) - \theta \left( k_x -k_{\mathrm{F},x} \right) \right] \left[ \theta \left( k_y \right) - \theta \left( k_y -k_{\mathrm{F},y} \right) \right], \tag{5} $$ which is constant and equal to 1 in the crosshatched area of the figure below, and vanishes outside it.

Rectangular Fermi surface and integral contours

It of course follows that $$ \frac{\partial f}{\partial k_x} = \left[ \delta \left( k_x \right) - \delta \left( k_x -k_{\mathrm{F},x} \right) \right] \left[ \theta \left( k_y \right) - \theta \left( k_y -k_{\mathrm{F},y} \right) \right], \tag{6} $$ $$ \frac{\partial f}{\partial k_y} = \left[ \theta\left( k_x \right) - \theta\left( k_x -k_{\mathrm{F},x} \right) \right] \left[ \delta\left( k_y \right) - \delta \left( k_y -k_{\mathrm{F},y} \right) \right]. \tag{7} $$ Substituting these expressions into (4) and carrying out integrals over the Delta functions we obtain $$\sigma_{xy}^{2D}=\frac{e^2}{\hbar} \int \frac{dk_x}{\left( 2\pi\right)^2} \left[ \theta \left( k_x \right) - \theta \left( k_x -k_{\mathrm{F},x} \right) \right] \left[ \mathcal{A}_{k_x}|_{k_y=0} - \mathcal{A}_{k_x}|_{k_y=k_{\mathrm{F},y}} \right]\\ - \frac{e^2}{\hbar} \int \frac{dk_y}{\left( 2\pi\right)^2} \left[ \theta \left( k_y \right) - \theta \left( k_y -k_{\mathrm{F},y} \right) \right] \left[ \mathcal{A}_{k_y}|_{k_x=0} - \mathcal{A}_{k_y}|_{k_x=k_{\mathrm{F},x}} \right], \tag{8}$$ where the notation $f|_{x=0}$ indicates that $f$ is evaluated at $x=0$. Now it is just a matter of identifying the different terms with different paths in the figure, which yields $$ \sigma_{xy}^{2D}=\frac{e^2}{\hbar} \left[ \int_{\Gamma_1} \mathcal{A}_{k_x} dk_x + \int_{\Gamma_2} \mathcal{A}_{k_y} dk_y + \int_{\Gamma_3} \mathcal{A}_{k_x} dk_x + \int_{\Gamma_4} \mathcal{A}_{k_y} dk_y \right]. \tag{9} $$ Since $\Gamma=\Gamma_1+\Gamma_2+\Gamma_3+\Gamma_4$ is a closed contour we have $$ \sigma_{xy}^{2D}=\frac{e^2}{\hbar \left( 2\pi\right)^2} \oint_\Gamma \vec{\mathcal{A}} \cdot \vec{dk}, \tag{10}$$ which matches your Eq. (3) up to a factor of $2\pi$, and agrees with Haldane's result in the article mentioned above. Finding the source of this discrepancy is left as an exercise.

  • $\begingroup$ Wow, an amazing answer! Thanks! $\endgroup$ Jun 28 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.