# Why are there chiral edge states in the quantum hall effect?

The most popular explanation for the existence of chiral edge states is probably the following: in a magnetic field, electrons move in cyclotron orbits, and such such cyclotron orbits ensure electrons to move in a single direction at the edge. That is why the edge state is chiral.

I think this picture is too classical. Could anyone provide me with other explanations for this phenomenon? Does it have a relation with time-reversal symmetry breaking, Chern numbers, or some other topological phenomenon?

Here is an explanation that's purely quantum.

A charged quantum particle in a magnetic field is subject to Landau quantization. Taking the magnetic field in the $z$ direction, we can choose the Landau gauge for the vector potential: $$\mathbf{A} = B x \hat{y} ~~ \Rightarrow ~~ \mathbf{B} = B \hat{z}.$$ The Hamiltonian in the coordinates $xy$, ignoring (for now) the edges of the sample:

$$H = \frac{1}{2m} \left( \mathbf{p} - \frac{e \mathbf{A}}{c}\right)^2 = \frac{1}{2m} \left[ p_x^2 + \left(p_y- m \omega_c x\right)^2\right],$$

where $\omega_c = eB/mc$ is the cyclotron frequency.
After separation of variables we get the wavefunctions:

$$\psi(x,y) = f_n ( x- k_y / m \omega_c ) e^{i k_y y},$$

where $f_n$ are the eigenfunctions of the simple harmonic oscillator ($n=0,1,2...$). The expectation values of $p_y$ and $x$ for this wavefunction are $\langle p_y \rangle =k_y$ and $\langle x \rangle =k_y / m \omega_c$, and the current along the $y$ direction is proportional to the generalized momentum in that direction:

$$\langle I_y \rangle = \frac{-e}{m} \langle p_y - m \omega_c x\rangle = \frac{-e}{m} (k_y - m \omega_c \frac{k_y}{ m \omega_c} )=0.$$

As expected, we get zero current in the bulk of the sample.
Now let's imagine we are near the edge of sample on the negative side of the $x$ axis. This means the particle will feel a confining potential $U(x)$ that looks roughly like:
This potential will deform the wavefunction $f_n$ to a wavefunction that has more weight in the positive direction of $x$ than before, and then we'll get $\langle x \rangle > k_y / m \omega_c$, leading to: $$\langle I_y \rangle > 0,$$ i.e. edge current in the positive $y$ direction. Notice that this is the same direction predicted classically.

• Thanks very much for your answer. This is quite a good point of view. Feb 2, 2013 at 6:25
• I know that quantum spin hall effect and topological superconducting state also has chiral state. Is chiral state a result of bulk topological properity? Feb 2, 2013 at 6:27
• When you have a system in a non-trivial topological phase (such as a topological superconductor), then it will generally have edge states. This is because the bulk of the system is gaped, and the outside is gaped (say outside the sample there's just vacuum), but the gap in the sample is inverted. When you consider how the spectrum changes between the sample and the outside, the gap has to close at the edge because of the inversion - and this means there's a state at the edge. I'm not sure if you can determine whether this edge state is chiral or not from this consideration.
– Joe
Feb 2, 2013 at 8:45
• Thanks. Yes, I know closing gap leads to quantum phase transition, which will change the topological invariant; and this is the reason surface states are always gapless. Feb 2, 2013 at 10:13
• Well this is a topic for a whole new question, but in short: The invariant is protected from being changed by the gap. During band inversion the gap closes, and the invariant momentarily loses its protection and can be changed.
– Joe
Feb 3, 2013 at 15:36

$[R^a_i, R^b_j] = -i\delta_{ij}\epsilon^{ab}l_B^2$ where $l_B$ is magnetic length, $i,j$ labels particles, $a,b$ labels 2D direction.