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As part of an exercise we have been asked to consider a free particle in a uniform magnetic field $\vec B = (0,0,B)$, bounded by a strong confining potential:

$$H = \frac {(-i\hbar\nabla^ -e \vec A)^2}{2m} + \frac {m\omega^2y^2}{2}$$

I found the eigenvalues of the Hamiltonian to be

$$E_{n,k} = \hbar \omega_H(n+\frac 12)+\frac {\hbar^2k^2}{2m}\frac{\omega^2}{\omega_H^2}$$

Where $\omega_H = \sqrt{\omega^2 + \omega_c^2}$ and $\omega_c$ is the cyclotron frequency $\omega_c = \frac {eB}{m}$. At $n = 0$ this simplifies to:

$$E_{0,k} = \frac {\hbar \omega_H}{2} + \frac {\hbar^2k^2}{2m}\frac{\omega^2}{\omega_H^2}$$

Next I have been asked to find the maximal and minimal values of $k$ assuming only the $n=0$ Landau level is occupied. I'm not sure what this means but on a hunch I found the roots of:

$$E_{0,k}-\frac {\hbar \omega_H}{2} + \frac {\hbar^2k^2}{2m}\frac{\omega^2}{\omega_H^2}=0$$

This yielded:

$$k = i\frac {\hbar}{\sqrt{m\hbar\omega_H-2mE_{0,k}}}$$

Which doesn't seem very useful. I also have to find "the positions in the y-direction" corresponding to the maximal and minimal values of $k$ and the velocities of these states. I know these refer to "edge states" but other than their name I don't really know what they are. Could someone please explain what these edge states are and how I might find their positions and velocities from $k$-values? Thank you!

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  • $\begingroup$ This is an interesting question, however please correct some typoes, like the Hamiltonian, where the square is missing in the kinetic part, and perhaps $\omega_0$ should be relabeled $\omega$. Also, are you considering a uniform magnetic field along the $z$ axis? $\endgroup$
    – Matteo
    Apr 3, 2022 at 23:42
  • $\begingroup$ Thank you for pointing out the typoes @Matteo, I jumped between notations a few times as I was working on the problem, should be fixed now. And yes its a uniform field along z-axis, I've updated the question to reflect this. $\endgroup$
    – Allod
    Apr 4, 2022 at 9:55
  • $\begingroup$ Don't worry! There is still a typo, possibly related to your problem: the kinetic term should be $(-i\hbar\nabla - e \vec{A} )^2$ (all the thing is squared!). Please consider if you have computed the eigenvalues of the right Hamiltonian $\endgroup$
    – Matteo
    Apr 4, 2022 at 9:58
  • $\begingroup$ Damn, I really am all over the place with this @Matteo, fixed again. Yes I've compared with some colleagues and we all calculated these to be the eigenvalues. It's really the lack of understanding about the relationship between $k$ and the position of the edge states that's holding me back I think. $\endgroup$
    – Allod
    Apr 4, 2022 at 10:04

1 Answer 1

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If only the first Landau level is occupied, then we can say that the Fermi level is $\mu \leq 3 \hbar \omega_H / 2$. This value is nothing but the lowest energy level of the second Landau band, corresponding to $n=1$ and $k=0$. If you assume a larger value of the Fermi level, then the second Landau level is partially filled.

The largest available value of $k$ can be determined by solving $E_{0,k}=\mu$, which yields $$ \hbar \omega_H = \frac{\hbar^2 k^2}{2m} \frac{\omega^2}{\omega_H^2} \;\;\;\; \to \;\;\;\; k_{\pm} = \pm \sqrt{\frac{2m \omega_H^3}{\hbar\omega^2}}. $$

Now if you look back at the Schrodinger equation that you have solved to find the eigenvalues, probably at some point you have used the fact that the wave function can be factorized (in the Landau gauge) as $\psi(x,y) = e^{ikx} \varphi(y)$, and you have realized that $\varphi(y)$ is the eigenstate of an effective harmonic oscilator in the $y$ direction centered about the point $y_0 \propto k $. For example, in the lowest Landau band, for $n=0$, $\varphi(y)$ is a gaussian centered about $y_0\propto k$. If you consider the largest possible values of $k$ obtained above ($k_{\pm}$), they correspond to the largest possible values of $y_0$ ($y_0^{\pm}$) around which the eigenstate is centered. The states with largest $k$ thus correspond to states that are localized at the outermost positions in the $y$ direction of the system, hence the name "edge states". Furthermore, the state at $y_0^+ > 0$ has $k_+ > 0$, while the state located at $y_0^- <0$ has $k_-<0$: this provides a physical interpretation in terms of electrons localized at the $y$-edges that are counterpropagating in opposite directions along the $x$ axis. This circumstance can be understood semi-classically by the concept of skipping-orbits, that you might have heard.

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