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The edge of a fractional quantum Hall state is a chiral conformal field theory. In the Laughlin case it corresponds to the chiral boson,

$$ S = \frac{1}{4\pi} \int dt dx \left[\partial_t\phi\partial_x\phi - v (\partial_x\phi)^2\right]$$

Here the field $\phi$ is identified with the charge density operator:

$$ \rho(x) = \frac{\sqrt{\nu}}{2\pi} \partial_x\phi$$

Now, you expect this operator to be the zero'th component of a two-vector, $J^\mu = (\rho, j)$ with $j$ the edge current. So how does $j$ relate to $\phi$ ? The reason I'm asking is that the literature gives different answers for this.

  1. This paper states at the beginning of Chapter III to use the continuity equation to obtain $j = -v\frac{\sqrt{\nu}}{2\pi} \partial_x\phi$.
  2. This paper (pdf warning) has equation (2.12) relating $j = -\frac{\sqrt{\nu}}{2\pi} \partial_t\phi$. No motivation though.
  3. This paper first couples the theory to an electromagnetic potential with components $(a_0, a_x)$ at the boundary ($D_\mu = \partial_\mu + \sqrt{\nu}a_\mu$), $$ S = \int dt dx \frac{1}{4\pi}\left[D_t\phi D_x\phi - v_c (D_x\phi)^2\right]+\frac{\sqrt{\nu}}{2\pi}\epsilon^{\mu\lambda}a_{\mu}\partial_\lambda\phi$$ and defines the current through $J^\mu = \frac{\delta S}{\delta a_{\mu}}$ giving $J^\mu=(\frac{\sqrt{\nu}}{2\pi} D_x\phi, -v\frac{\sqrt{\nu}}{2\pi} D_x\phi)$.

So I notice that case 3 reduces to case 1 when $a_x = 0$. Furthermore, case 3 is gauge invariant, so this operator seems like a logical choice.

My problem with this choice is as follows: Consider a system with two edges (infininte quantum Hall bar) and suppose you have a non-zero potential along one edge, $a_t = U$ and $a_x = 0$ along this edge. You therefore expect a current to run through the system, because the edges are held at different potentials (and the current runs perpendicular to the potential difference, because of the quantum Hall relation). But: $\langle \partial_x \phi\rangle = 0$, so for case 1 and 3 there's no current.... Does that make case 2 the correct choice?

So perhaps the question comes down to: What operator represents the edge current? What operator is 'measured' in an experiment where the current is probed?

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1 Answer 1

All the three forms agree with each other since, on the chiral edge, $\phi$ has a form $\phi(x,t)=\phi(x-vt)$ (as a time dependent operator).

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