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?
 A: A short answer: Why not?
HQ states do not have time reversal symmetry. So the right moving excitations and left moving excitations may behave differently -- thus chiral.
The edge states of most FQH states are very chiral, in the sense that
even the numbers of left moving modes and right moving modes are different.
Topological insulator and topological superconducting have time reversal symmetry and their edges, exactly speaking, are non-chiral, such as the right movers and the left movers have exactly the same velocity. Certainly the
the numbers of left moving modes and right moving modes are the same.
A: 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.
A: At fundamental level, the quantum hall physics describes non-commutative, interacting, guiding centers which satisfy the non-commutative algebra
$[R^a_i, R^b_j] = -i\delta_{ij}\epsilon^{ab}l_B^2$ where $l_B$ is magnetic length, $i,j$ labels particles, $a,b$ labels 2D direction.
This fundamental algebra is time-reversal odd.
