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The good old Ohm's law $${\bf j}({\bf r}) = \sigma_O {\bf E}({\bf r})$$ if translated into words would be "the local current density is proportional to a local electric field."

In a quantum Hall state (QHS), the linear response to an external electric field is given by the following $${\bf j} = \sigma_H \bf \hat n \times E$$ where $\bf \hat n$ is the surface normal.

Is it also a local relation between applied field ${\bf E}({\bf r})$ and response current ${\bf j}({\bf r})$?

If the electric field is non-zero inside the bulk, it means there must be current propagating inside the gapped bulk. I tend to understand it in an intuitive way. As you see in the picture, for a uniform electric field in $x$-direction, the chemical potential is raised up linearly in $x$ and it intersects levels in the bulk. The current in the bulk comes from the chiral edge modes and sink into edge back again. I don't know whether my picture is correct or not.

As you see, for a uniform electric field in $x$-direction, the chemical potential is raised up linearly in $x$ and it intersects levels in the bulk.

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yes, your picture is correct inside the material, but the current doesn't flow outside, it just collects charge on the surface, which changes the direction of the electric field until the current is parallel to the wire. The equation you write for the Hall conductivity is completely local, but it predicts the Hall voltage, which is tantamount to the charge buildup on the two sides of the wire when the current is running, which is what you measure in an experiment.

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Thank you! In graphene, the Landau levels are no longer equally spaced, but response equation remains the same. So a uniform $E$ will give non-uniform current density if interpreted in terms of the chemical potential picture! –  ChenChao Sep 21 '12 at 12:19
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