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 
 A: 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:
$$
j_x= \frac{ne^2}{h} E_y.
$$
A: The essential reason for the simultaneous zero longitudinal conductance and resistance is that the edge states are ballistic because they are unidirectional at each edge providing no backscattering states. This ballistic nature has an important implication that electrons on these edge states are in equilibrium with electrons in the electrode where the edge electrons are coming from, i.e., they have the same chemical potential.
Then if you imagine applying a current $I_x$ between the source and drain, the electron chemical potential at any point on the top edge is all equal to the chemical potential of the source, meaning the voltage drop $V_x$ across any two points on the edge is all zero, hence $R_\text{L}=V_x/I_x=0$.
On the other hand, consider applying a voltage $V_x$ between the source and drain and measure the transverse current $I_y$ and longitudinal current $I_x$. This case actually got me think a lot. Below I will state my thought, but I'm not 100% sure. I think the most important thing in this case is that if you want to measure (or allow) $I_y$, you need to short one of the top electrode to one of the bottom electrode. The voltage source should initially inject some electrons onto the left edge state (going up along the source/2DEG interface) to increase its chemical potential. This leads to a larger current on the left edge state than on the right edge state. But this excess current on the left edge state will not flow to the right edge state, because it will flow directly through the shorted top and bottom electrodes. So you will measure a finite current $I_y$. But $I_x$ through the source and drain will be zero, because after the initial electron injection from the source into the left edge state, the current on all the edge states including the shorting connection is steady. Hence $\sigma_{xx}=I_x/V_x=0$.
