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In the literature I read that the Hall conductivity is quantized because the Hall conductivity is actually the winding number associated with the mapping from the brillouin zone (a torus) to the space of Hamiltoninans. The calculation assumes that system has an energy gap above the ground state. But doesn't that mean that the system is an insulator ? If the last statement is true how does the system conduct ?

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Yes. According to the classification scheme, the quantum Hall insulator is first an insulator, then a band insulator, then a topological band insulator (or abbreviated as a topological insulator), and finally a 2D topological insulator in symmetry class A.

But quantum Hall insulator can conduct Hall current! This statement is not a contradiction.

You need to understand the precise definition of insulators. A state of matter is classified as an insulator if the state does not conduct when it is placed on a closed manifold (manifold with no boundary). So the definition does not rule out the possibility that an insulator can conduct when it is placed on an open manifold (manifold that has boundary). In fact, the key feature of a topological insulator is insulating in the bulk and conducting on the boundary.

Quantum Hall insulator is indeed insulating and has no Hall conductivity if there is no boundary. One you cut open the boundary, gapless edge modes will emerge on the boundary, which then carry the Hall current and contribute to the Hall conductivity.

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  • $\begingroup$ Thanks a lot for the answer. The other thing that confuses me and I think is related to the above question is the Laughlin's argument. In Laughlin's argument he starts with the ground state and then talks about changing the magnetic flux adiabatically from zero to one full flux quanta. At the end of the process the Hamiltonian for the system returns to its original form so that the spectrum becomes the same as the one he started with. $\endgroup$ – Tuhin Subhra Mukherjee Apr 25 '16 at 12:21
  • $\begingroup$ But then he argues that in this process the the system might end up in an excited state of the Hamiltonian. This means there must be a level crossing during the process ? $\endgroup$ – Tuhin Subhra Mukherjee Apr 25 '16 at 12:21
  • $\begingroup$ @TuhinSubhraMukherjee Yes, there is a level crossing during the process. But at the level crossing point, the two degenerated states are differed from each other by one unit of charge transfer, but since the bulk is insulating, it is exponentially unlikely to resonance between the two degenerated states to avoid the level cross, or in other words, the crossing is topologically protected. So after the flux threading, the final state is inevitably an excited state. $\endgroup$ – Everett You Apr 27 '16 at 5:09
  • $\begingroup$ Oh ok. That actually cleared a lot of confusions. Thank you. $\endgroup$ – Tuhin Subhra Mukherjee Apr 27 '16 at 9:33
  • $\begingroup$ Sorry. So the level crossing does not occur in the bulk ? $\endgroup$ – Tuhin Subhra Mukherjee Apr 27 '16 at 10:10

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