According to different sources: Tong lectures on IQHE (Tong), MIT Open courses (MIT) etc, when calculating the number of states in each Landau Level all of them impose (in the Landau gauge) periodic boundary conditions in the traslational invariance direction (lets supose is the y direction). However if you impose this conditions somehow you are suposing that your sample "behaves" like an infinite cylinder in the x direction, which in real samples is not the case. This boundary conditions $k_y=n\frac{2\pi}{L_y}$ give the degeneracy expected of $N=AB/\phi_0$

However in a more realistic calculation you should impose fixed boundary conditions where the wavefunction is zero in the edges of the sample, that is to impose $k_y= n\frac{\pi}{L_y}$. In the MIT courses they said that this 2 conditions for large enough samples are the same but I dont see that. Why appling periodic boundary conditions instead of fixed boundary conditions gives the correct degeneracy that it is seen in experiments?


1 Answer 1


in general, the basic assumption (which is not always true) is that boundary conditions are not important for macroscopic quantities, and we can choose them as we want, assuming we take the large $L$ limit of the system size. The intuition is that deep inside the bulk the effects of the boundary are not felt, and as the boundary only contributes $O(1)$ vs. $O(L)$ for the rest of the system, we can ignore its specific nature and choose boundary conditions that are convenient to us.

The IQH is a nice example for where this assumption works and where it (kind of) fails. The degeneracy of the Landau levels is a macroscopic quantity, and we expect it to be more or less the same no matter what the boundary conditions are. If we look at something like the density of states, a free particle in a box $L$ can have periodic boundary conditions with $k_n=2\pi n/L$ or infinite-well boundary conditions $k_n=\pi n / L$. In the first case, the allowed $n$ are in the range $(-\infty, \infty)$ while for the second case we are only in the range $[1, \infty)$, as changing $n$ to $-n$ leaves us with the same function. So indeed the two cases are identical in terms of number of states in the interval of $(|k|, |k|+dk)$, up to the single state $n=0$ which exists only for the periodic boundary conditions. But we don't care about a single state anyway when looking at a macroscopic system. Another way to see that - when taking the limit $L\to \infty$, summation over $k$ becomes in the first case $\frac{1}{L}\sum_{n=-\infty}^{\infty} \to \int_{-\infty}^{\infty}\frac{dk}{2\pi}$ while for the other case $\frac{1}{L}\sum_{n=1}^{\infty} \to \int_{0}^{\infty}\frac{dk}{\pi}$.

You can see that for the specific question of counting the degeneracy the factor $2$ is balanced between the periodic boundary conditions and the fixed boundary conditions with zero by the fact that we allow or disallow negative $n$.

The IQH is interesting in that it does have a very important boundary contribution. For periodic boundary conditions the system is completely insulating. But when we have a boundary a perfect conduction channel appears there. So looking at the conduction of the system, the boundary matters a lot (notice that this is not conductivity - as this is still zero in the large $L$ limit). This is a signature of topological states of matter, where the boundary plays a very important role in the phenomena that interests us. However, for the degeneracy level, it still doesn't matter.

  • $\begingroup$ Perfectly explained, however still have a question. In Tong's lectures he derived the degeneracy saying that if our sample has dimensions of $L_x \times L_y$ and the oscillators are centered in the position $x_0=-l_B^2k_y$, then the maximum value of $k_y$ is the one that makes (ignoring the sign) $x_0=L_x$, which gives the $k_y$ range $[\frac{-L_x}{l_B^2},0]$ and then as you did changing from the summatory to the integral $N=\Delta k_y \int^0_\frac{-L_x}{l_B^2} dk$. I dont see how to calculate the degeneracy the same way but with the fixed boundary conditions. $\endgroup$ Commented Sep 9, 2022 at 16:30
  • $\begingroup$ I'm not sure. I think it wouldn't be so simple. The ansatz in solving the Landau level for a periodic boundary conditions is to choose a gauge such that we can exactly solve in the direction with the pbc, leaving the other direction as harmonic oscillator with a shifted potential, where the shift is proportional to $p_y$. For the solution with infinite-well bc, $p_y$ is not a good quantum number. So the method fails from the start. The solution for a harmonic quantum oscillator inside an infinite well is by itself complicated. See here physics.stackexchange.com/a/92072/275556 $\endgroup$
    – user275556
    Commented Sep 12, 2022 at 7:52

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