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The edge of a fractional quantum Hall state is an example of a chiral Luttinger liquid. Take, for the sake of simplicity, the edge of the Laughlin state. The Hamiltonian is:

$$H = \frac{2\pi}{\nu}\frac{v_c}{2} \int_{\textrm{edge}} dx \rho(x)^2 $$

Here $\nu$ is the filling fraction, which is constant, $v_c$ is the velocity of the edge mode and $\rho$ is the charge density operator. You can think of this Hamiltonian as a delta-function interaction $V(x,x') = \delta(x-x')$.

Together with this Hamiltonian there is also the commutation relation of the field $\rho$:

$$ [\rho(x),\rho(x')] = i\frac{\nu}{2\pi} \partial_x \delta(x-x')$$

I haven't gone through the exercise myself, but I presume this is derived by going to momentum space, obtaining the canonical momenta through Hamilton's equation of motion and performing a canonical quantization. These equal-time commutation relations together with the Heisenberg equations of motion lead to:

$$\partial_t \rho(x,t) = i[H,\rho(x,t)] = v_c \partial_x\rho(x,t) $$

This demonstrates that the edge is chiral, since $(\partial_t - v_c\partial_x)\rho(x,t) = 0$ and therefore the correlator involving $\rho(x,t)$ (or any other correlator) is a function of $t+x/v_c$ alone (hence the name "chiral" and "left-moving").

There are also particle excitations (e.g. the electron) which are generated through vertex operator $\Psi(x,t)$ (these can be motivated through bosonization and/or conformal field theory, but I won't go into that). In any case these field operators have the following equal-time commutation relations with the current operator:

$$[\rho(x),\Psi(x')] = Q \Psi(x')\delta(x-x')$$

Here $Q$ is the charge of the operator $\Psi$ with respect to the charge density operator $\rho$. This charge is then of course the electric charge.

My question is now: how do you generalize the commutation relations to non-equal time? What is:

$$[\rho(x,t),\rho(x',t')] = \ldots $$ $$[\rho(x,t),\Psi(x',t')] = \ldots $$


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up vote 2 down vote accepted

You may simply replace the arguments $ x \to x+vt $ in all of those commutators. The way to prove that is to go to momentum space, where $\rho (k, t) = \rho (k,0) e^{i k v t} $.

Proving that about the commutator $[\rho (x,t),\psi (x',t')]$ is a little trickier, because we don't know the explicit form of $\psi (t)$, but the way I did it was to use $$ [\rho (x,t),\psi (x',t')] = U^\dagger (t') [\rho (x,t-t'),\psi (x',0)] U (t') $$ Where U is the time evolution operator. This way only the time evolution of the density is needed.

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Fantastic. It's so elegant I wonder why I couldnt come up with it. I went through the whole thing and it works out likes you said ($x\rightarrow x+vt$). For what it's worth: the field operators take on the form $\Psi(x) \sim exp(i \alpha \int_{-\infty}^x \rho(x))$, so the relation is carried over automatically. – Olaf Aug 26 '11 at 14:53

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