I've looked through several papers that talk about the anomalous integer quantum Hall effect of graphene (such as http://journals.aps.org/prl/pdf/10.1103/PhysRevLett.95.146801), and they all state that the Hall effect is anomalous for graphene because the n=0 Landau level has half the degeneracy of the n>1 levels, with little to no explanation and without citing any (helpful) sources.

One of the papers (http://journals.aps.org/prb/pdf/10.1103/PhysRevB.75.165411) states that "the Dirac fermions have acquired an effective gap in a form of a “relativistic mass” due to the Coulomb interaction. Such a gap reduces the degeneracy of only the zeroth LL." Could someone explain this in more depth, specifically why the Coulomb interaction only breaks the degeneracy of the zeroth LL?

The 0th Landau level also has a degeneracy of four as do other LLs. The significance of 0th LL is that it is shared by conduction and valence band with equal weight. That is to say: two of 0th LLs are electron like and two of them are hole-like (because the degeneracy is four, I imagine there are four "seperated" non-degenerate LLs at the same energy for pedagogical simplicity).

When an electron-like Landau-level is fully occupied by electrons, it contributes a unit conductance; when a hole-like Landau-level is fully occupied by holes, it contributes a negative unit conductance. (Assuming that the landau level has no degeneracy)

When an electron-like Landau-level is empty, it contributes no conductance (of course!); when a hole-like Landau-level is filled with electrons, it contributes no conductance.

Now let's see how the anomalous quantum hall effect arises:

When the 0th LL is fully occupied, the two electron-like LLs are filled with electron and two hole-like LLs are also filled with electron. Thus the conductance of the system is 2 unit conductance.

When the 0th LL is empty, the two electron-like LLs are empty and two hole-like LLs are filled with holes. Thus contribute to the conductance of -2 unit conductance.

Similarly, when both 0th and 1th LLs are also occupied with electrons, the total conductance is 4+2.

The origin of the anomalous Hall conductance

The anomalous Hall conductance arises because of the particle hole symmetry and the linear dispersion of graphene. For the 0th LL, because its energy is just 0, so it must be shared by the valence and conduction band.

However, for the parabolic dispersion, even if the original band have particle-hole symmetry. When forming LLs, the energy of lowest LLs formed by conduction band is 1/2, While the energy of lowest LL formed by valence band is -1/2. They do not mix.

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    But why is the zeroth LL shared by the conduction and valence band with equal weight in graphene, but not in other materials? – Izzhov May 5 '14 at 20:01
  • This is only for dirac like electrons. – buzhidao May 6 '14 at 0:11
  • And why is that? – Izzhov May 6 '14 at 0:40
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    Because for Dirac electrons there's no band gap — $E(k)$ functions for electrons and holes touch each other at $k=0$. – Ruslan May 6 '14 at 10:08

Because without magnetic field, the chemical potential is at the Dirac point. In other words all the valence bands are occupied. The system is particle-hole symmetric. With an external magnetic field added, the energy levels are quantized into Landau levels(LLs). Hlaf of the zero-th LL comes from the original valence band and the other half from the original conduction band. Since the number of electrons is unchanged, the 0-th LL should be half-filled so that the whole system is still particle-hole symmetric.

Regarding to your 2nd question, I do not know the details. But I think the logic is: if we assume the Coulomb interaction energy scale is small compared with the LL spacing, the problem basically reduces to the 0th LL problem. Namely we only need to consider electrons within the 0th LL, interacting via Coulomb interactions. We can ignore the Landau level mixing since this Coulomb interaction is weak. Therefore degeneracies of other LLs do not change.

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