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  Difference between "ordinary" quantum Hall effect and quantum anomalous Hall effect

+ 2 like - 0 dislike

I am reading a review article on Weyl semimetal by Burkov where he writes, top of page 5:

A 3D quantum anomalous Hall insulator may be obtained by making a stack of 2D quantum Hall insulators [Ref. 23].

Ref. 23 in his paper is the generalization of the 2D TKNN invariant to the 3D case.

I am a little confused about what Burkov meant here. The TKNN invariant in 3D was derived in the presence of a nonzero external magnetic field. I understand that, from Haldane's model of a Chern insulator, with Streda's formula, the quantum anomalous Hall conductance is the limit (for a 2D system):

\begin{equation} \lim_{B_k\to 0} \sigma_{ij} = \lim_{B_k\to 0} \epsilon_{ijk} \frac{\partial \rho}{\partial B_k} \neq 0, \end{equation}

where $\rho$ is the electric-charge density, $B_k$ is the external magnetic field, and $\{i,j,k\}$ are spatial indices. So, it seems that Burkov is implying that the above limit exists in 3D. If such a nontrivial limit indeed exists in 3D (which I think is true after reading Sec. III of Ref. 23), doesn't this imply all time-reversal broken quantum Hall systems in 3D are also quantum anomalous Hall systems? This seems a little odd to me. What is then the difference between an "ordinary" quantum Hall system and a quantum anomalous Hall system? Do they belong to the same topological phase (i.e. connected by a continuous adiabatic transformation)?

This post imported from StackExchange Physics at 2020-09-11 18:14 (UTC), posted by SE-user Congmiao
asked Sep 8, 2020 in Theoretical Physics by Waterfall (30 points) [ no revision ]

1 Answer

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I think that I may have an answer to my question here.

Perhaps, the right question to ask is not whether an "ordinary" quantum Hall system (QH) in 3D and quantum anomalous Hall system (QAH) is connected through some adiabatic path. When the 3D TKNN invariant is valid, we see that Hall conductance is independent of the external magnetic field:

\begin{equation} \lim_{\vec{B}\to0} \sigma_{ij} = \lim_{\vec{B}\to0}\epsilon_{ijk}\frac{\partial \rho}{\partial B_k} = \frac{e^2}{2\pi h} \epsilon_{ijk} G_k, \end{equation}

for some reciprocal lattice vector $\vec{G}$ from Ref. 23 in the question. Then, perhaps, all we can conclude QAH is a special case of 3D QH.

In general, QAH may be formed by a variety of physics (such as ferromagnetism and others as discussed here: http://10.1103/RevModPhys.83.1057), QAH may not necessarily be related to QH, and QAH may not need to be a limit of QH at all.

Moreover, Streda's formula may be regarded more as a theoretic "shortcut" to Hall conductance. Experimentally, the distinction between QAH and QH is that chiral modes exist for QAH at zero magnetic fields, which can be observed by supplying an external electric field. The bulk charge density $\rho$ is not meaningful in the sense that it does not couple to experimental probes.

This post imported from StackExchange Physics at 2020-09-11 18:14 (UTC), posted by SE-user Congmiao
answered Sep 10, 2020 by Waterfall (30 points) [ no revision ]

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