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?


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.