# Integer Quantum Hall effect, conductivity & edge states

I'm confused about the conductivity and the edge states in the IQHE. On the plateaux, we zero the longitudinal conductivity and resistivity, right? So is it really true, that on the plateaux, there is no current flowing in the longitudinal direction, only in transverse?

From this image, it looks to me, that the edge states carry the current in longitudinal direction and that there is no current flowing in transverse direction. What did I miss understand?

Greetings

The two dimensional conductivity ($$\sigma$$) and resistivity ($$\rho$$)tensors are defined by $$j_a= \sigma_{ab}E_b,\\ E_b= \rho_{ab}j_b$$ respectively. Here $$a,b$$ stand for the $$x,y$$ directions. This means that $$\sigma_{ab}$$ is the inverse matrix to $$\rho_{ab}$$. On an IQHE plateau $$\sigma_{ab}= \frac{ne^2}{h}\left(\matrix{0&1\cr -1&0}\right)_{ab}$$ and the inverse matrix is $$\rho_{ab}=\frac{h}{ne^2}\left(\matrix{0&-1\cr 1&0}\right)_{ab}.$$ We see that both longitudinal conductiites $$\sigma_{xx}$$ and $$\sigma_{yy}$$ are zero as are the longitudinal resitivities $$\rho_{xx}$$ and $$\rho_{yy}$$. There can certainly be non-zero currents and voltages however. It is just that the current and voltage must be perpendicular to each other. You can have a current in the $$x$$ direction and an $${\bf E}$$ field in the $$y$$ direction. In your pictured Hall bar there is net left-to-right current, but only a top-to-bottom potential drop: $$j_x= \frac{ne^2}{h} E_y.$$
• No. As I explained, it just means that the voltage drop is pependicular to the current. The situation is similar to a superconductor in that ${\bf j}\cdot {\bf E}=0$ means that there is no dissipation, but it is different from superconductivity in many ways. – mike stone Apr 29 at 15:05
• In superconductor we can have a current with no electric field. On a QHE plateau a current always has an electric field at right-angles to it. Zero longitudinal conductivity means that given an ${\bf E}$ field there is no curent parallel to ${\bf E}$. There can be a current caused by ${\bf E}$ and at right angles to ${\bf E}$. The matrix expressions I wrote in my answer make this clearer than words can. – mike stone Apr 29 at 15:53