Landau level edge states as simple harmonic oscillator with boundary

Question

Halperin 1982 investigated the Landau level edge state spectrum by solving the harmonic oscillator on a half-infinite line $x \in (-\infty,s]$, where varying $s$ is equivalent to varying the momentum parallel to the edge:

$$[-\frac{d^2}{dx^2} + x^2] g_{\nu,s}(x) = E_\nu g_{\nu,s}(x)$$

where I've set all physical constants to 1, and the boundary conditions are

$$\lim_{x\to s} g_{\nu,s}= \lim_{x\to-\infty} g_{\nu,s} = 0.$$

He claims:

• When $s \to \infty$, $E_{\nu} \to \nu+\frac12$, which makes sense since this is the usual harmonic oscillator
• As $s\to -\infty$, $E_\nu$ passes through $2\nu+\frac32$ at $s=0$ and then grows as $s^2$.

This leads to the well-known energy levels for the quantum Hall edge states and shows they are chiral. But how does one derive the second claim? Also, is there a simple solution for $g_{\nu,s}$?

Answer 1

For the $s = 0$ case we just have to look at the eigenmodes of the usual quantum harmonic oscillator (the $s \to \infty$ case) which have a node at $x = 0$. They are precisely the ones with odd $\nu$ energies $E_\nu = \nu + \frac{1}{2}$, so writing $\nu = 2 n + 1$ we get $E_n = 2n + \frac{3}{2}$.

For the case $s < 0$ I only have a crappy guess to why it scales as $\sim s^2$. The harmonic solutions $|\Psi_\nu \rangle$ for the $s \to \infty$ still span the Hilbert space, so we can expand the eigenstates $ | \Phi_n \rangle$ of any $s < 0$ problem into the $| \Psi_\nu \rangle$ basis. Then, my guess is that the $| \Psi_\nu \rangle$ state which will most contribute to the expansion of the $s < 0$ ground state $| \Phi_0 \rangle$ will be the lowest energy one that has a node close to $s$

Image by AllenMcC, distributed under a CC BY-SA 3.0 license.

Looking at this image, I suppose that the location of the outermost node of the $ |\Psi_\nu \rangle$ state goes as $ r_\nu = - \sqrt{E_\nu} $, the classical turning point for the harmonic oscillator. Thus, the ground state energy should go as $E \sim E_\nu = r_\nu^2 \sim s^2$.