Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The fundamental question is

Why is Hall conductance quantized?

Let's start with the Hall bar, a 2D metal bar subject to a strong perpendicular magnetic field $B_0$. Let current $I$ flow in the x-direction, then the y-direction develops a voltage $V_H$. The Hall conductance is $\sigma_H = I/V_H$ Hall bar (http://members.home.nl/skoric/quantum/setup.gif)

To make Laughlin's charge pump, how should we wrap the Hall bar? Identify the left and right, or top and bottom sides?

Based on my understanding, we should paste top and bottom side together. (correct? Figure 1. left of the Paper maybe a little confusing.)

Laughlin assumes the Fermi level is in the middle of the gap, so that the ring is an insulator. But the changing flux will induce an current by taking "adiabatic derivative" of total energy w/r flux $$I = c\frac{\partial U}{\partial \Phi}$$ which flows in y-direction and where $c$ is speed of light.

Following Laughlin's calculations, as one threads one flux quantum, there will be $p$ (number of filled Landau levels) electrons transported. Then $$U=peV$$ where $V$ is the potential difference of two edges. From the current formula, we find the quantized Hall conductance.

The heart of the problem is

What is an adiabatic derivative? Why is ${\bf j} = \partial {\cal H}/\partial {\bf A}$ valid?

share|improve this question
The form of current operator is valid because $\hat H=(p + eA)^2/2m$ and $\delta \hat H/\delta A = e \delta H/\delta p = e\hat v$. Adiabatic derivative is meant to take derivative of $U=<H>$ without worrying about Boltzmann weights, so $I=c\delta U/\delta A$. –  ChenChao Dec 21 '12 at 7:21
It is the top and bottom sides that should be pasted (not the left and right), because they refer to $V_H$. –  Hai-Yao Deng Aug 15 at 22:08
The best presentation can be found in Laughlin's Nobel prize lecture. –  Hai-Yao Deng Aug 15 at 22:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.