Is quantum Hall current density local? (${\bf j}({\bf r}) = \sigma_H {\bf \hat n \times E}({\bf r})$)

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.

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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