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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.

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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.

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

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