# Landau level edge states as simple harmonic oscillator with boundary

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 $$g_{\nu,s}$$?

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 $$s$$
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 $$E \sim E_\nu = r_\nu^2 \sim s^2$$.