Field theory on the edge

In this section we will study the action (6.7). The space-time we work on is $\mathbbm{R}\times S^{1}$, infinite in the time direction and periodic in space. For the multi-valued field we will impose boundary conditions,

$$\phi(x+2\pi)=\phi(x)+w2\pi, \quad w\in\mathbbm{Z}.$$ (6.8)

In the rest of this chapter it will become clear why we choose these boundary conditions. The conjugate to $\phi$ for this action is obviously the same as for the action with vanishing v. Thus our Hamiltonian doesn't vanish anymore;

$$H=-\frac{mv}{4\pi}\int du \partial_{u} \phi \partial_{u} \phi.$$ (6.9)

For the action to be bounded from below, we have to choose $-mv>0$. The action is also invariant under translations, $u \to u +\delta u$, this symmetry of course implies a conserved quantity, momentum,

$$P = \frac{m}{4\pi} \int dx \vdiff{\mathcal{L}}{\partial_{t}\phi} \partial_{x} \phi = \int dx \partial_{x} \phi \partial_{x} \phi,$$ (6.10)

which is precisely $-\frac{1}{v}H$. Actually this action has much more symmetries, it is invariant under Lorentz transformations (which have a peculiar form in 1+1D) and the two dimensional conformal group. I will not go further into these matters. The equation of motion coming from $\phi$,

$$(\partial_{t}-v\partial_{u})\partial_{u}\phi=0,$$ (6.11)

gives us a relation between $\partial_{t}\phi$ and $\partial_{u}\phi$. Using Hamilton's equation,

$$\partial_{t} \phi(x) = i\comm{H}{\phi(x)} =-2v\;i \int dy \; \pi(y) \comm{\pi(y)}{\phi(x)},$$ (6.12)

we find a second equation. This gives us the commutation relation we must use to be in accordance with the equation of motion (6.11),

The commutation relations between $\pi$ and $\phi$ also imply commutation relations between for $\phi$ with itself and $\pi$ and itself.

In what follows we will work in units where $\abs{v}=1$. From (6.11) we know that $\pi$ is a function of $x_{+}=t+x$ and not of $x_{-}=t-x$. So the field $\pi$ propagates in only one direction, it is a chiral boson field. From the boundary conditions (6.8) and the equation of motion (6.11) we see that we can write the classical solution to the equation of motion as,

$$\phi(x)=\frac{1}{\sqrt{m}}\big(c_{0} + p_{0}(t+x) + i \sum_{k \neq 0} \frac{c_{k}}{k} e^{-i k (t+x)}\big).$$ (6.14)

Note that the condition on $\pi$ to be real classically, or hermitian in the quantum theory, yields the relation $c_{k}^{\dagger}=c_{-k}$ for the Fourier coefficients in the quantum theory. Plugging all this into the commutation relation (6.13c) we find the relations.

$$\comm{c_{k}}{c_{-l}} = k\delta_{kl},\qquad k \in \mathbbm{Z}_{+}$$ (6.15)

$$\comm{c_{0}}{p_{0}}=i$$ (6.16)

We will largely ignore the second equation, although it is clear it generates states labeled with a 'momentum', well know from quantum mechanics. The first equation describes a infinite set of harmonic oscillators. In terms of the operators $c_{k}$ the Hamiltonian, and thus the momentum operator, is diagonal

$$H=-\frac{1}{2} \sum_{k>0} c_{-k}c_{k}$$ (6.17)

where we defined the Hamiltonian as the normal ordered version of the classical Hamiltonian (6.9), so dropping an infinite constant. The ground state is defined as usual

$$\begin{split}c_{k}\ket{0}=0, \qquad k \in \mathbbm{Z}_{+}.\\ p_{0}\ket{0}=0 \end{split}$$ (6.18)

We thus have found a set of states $(c_{-k})^{n}\ket{0}$ with a momentum equal to minus their energy. These states are not the only states in this theory. The states generated by the creation operators $c_{-k}$ are all charge neutral with respect to the conserved current,

$$j^{\alpha}=\frac{1}{2\pi}\epsilon^{\alpha\beta}\partial_{\beta}\phi.$$ (6.19)

This is the electromagnetic charge current on the edge. To see this fact more clearly we begin with the charge current in the bulk.

$$\vdiff{S}{A_{\mu}}=\frac{1}{2\pi} \epsilon^{\mu\nu\lambda} \partial_{\nu} a_{\lambda}$$ (6.20)

We can find the current on the edge by averaging the bulk current in the direction perpendicular to the edge ($v$). After performing the average we send the width of the edge ($\lambda$) to zero. This has the result that only the terms that were derivatives in the direction perpendicular to the edge survive, the density becomes for instance,

$$\begin{split}j^{0}&=\lim_{\lambda \to 0} \frac{1}{2\pi} \int ... ...ac{1}{2\pi}a_{u} = \frac{1}{2\pi}\partial_{u} \phi, \end{split}$$ (6.21)

where we dropped the gauge field just outside the sample, because we can choose it it to vanish there.

From equation (6.21) we thus find a very nice interpretation as for the field $\pi(x)$. It is (up to a constant) the electron density on the edge. Because the bulk behaves as an incompressible fluid the only excitations possible are density waves. The equation of motion for $\pi$ (6.11) describes precisely that. It is a wave equation for waves that propagate with velocity $v$. We can now say more about the constant $v$ entering in the Lagrangian. If we study a quantum Hall fluid we need in addition to the perpendicular magnetic field a potential that keeps the fluid confined. Semi-classically this confining potential and the magnetic field will induce a current flowing along the edge,

$$\vec{j}=e\vec{v}=\vec{B} \wedge \vec{E},$$ (6.22)

hence also the density waves will travel with this velocity. For consistency reasons we thus have to choose $v=\frac{E}{B}$. The operators $c_{k}$ thus create propagating density waves.

We can also make charge minus one states which are charged with respect to the current (6.19). Such thus has to satisfy the following commutation relation.

$$\comm{\int dx j^{0}}{\psi(y)} = - \psi(y)$$ (6.23)

When we recognize the charge density as $\frac{2}{m}\pi$, up to a constant equal to the conjugate to the field $\phi$, we can immediately solve this equation yielding,

$$\psi(y)=e^{i \, m \phi(y)}.$$ (6.24)

We find this new state's equation of motion by calculating the commutator with the Hamiltonian,

$$\begin{split}\partial_{0}\psi&=i\comm{H}{\psi}=-imv \int dx \... ...{x} \phi(x) e^{im\phi(x)} = v \partial_{x} \psi(x), \end{split}$$ (6.25)

which states that also the field $\psi$ is chiral. By calculating the exchange of two such states we can determine their statistics,

$$\begin{split}e^{i m \phi(x)}e^{im \phi(y)}&=e^{im \phi(y)}e^{... ...im \phi(y)}e^{im \phi(x)} e^{-i\pi m\epsilon(x-y)}. \end{split}$$ (6.26)

For $m$ odd we have a fermionic operator, which we can identify with the electron operator. Now it is also clear why we choose boundary conditions (6.8). It assures us that the electron states on the edge are single-valued. The fundamental charged operator is the operator for $m=1$, it has charge $\frac{1}{m}$ and is anyonic with statistics $\frac{1}{m}$. These properties are exactly as we would expect for our quasiparticles.