I enclosed a computation from a well-known paper in the field of mathematical physics where the Chern number of the first Landau level is computed (the result claimed is $-1$) and the full paper can be found here click me. The Landau level is the first eigenstate of a Schrödinger operator with magnetic field.

The nice physics behind this computation is that it shows that the quantum Hall effect (which is the transverse current created as a response to an external current in a conductor that is placed in a magnetic field) is quantized and given in terms of a topological quantity, the Chern number.

Yet, I have difficulties understanding what happened in this computation.

The projection is given by the integral kernel

$$\Pi(x,y) = \frac{qB}{h} e^{-(qB/4\hbar)(x-y)^2-i(qB/2\hbar)x\wedge y}.$$

The authors compute a "derivative" of this expression and get an integral expression for the Chern character and claim it is equal to $-1$. It does not seem like a very sophisticated computation, but it is just very hard to understand so I would like to ask here about this since it is very likely that I am missing something.

  • $\begingroup$ It seems like you are asking for someone to write an exposition of that paper. I think that an exposition would be too long for the sort of short and specific answers that this web site was designed to encourage. $\endgroup$ – Ben McKay Jan 17 at 9:05
  • $\begingroup$ @BenMcKay Mhmm- I think my question: How is the Chern number of this Landau level computed refers to roughly half a page of this paper and I would like to ask for more details on that. Why do you say I ask for an exposition of the paper? $\endgroup$ – Ben Curnow Jan 17 at 15:53
  • $\begingroup$ It's tough to tell from your question where your sticking point is. Maybe you could give some more details about which parts you're having trouble understanding? $\endgroup$ – d_b Jan 17 at 20:46
  • $\begingroup$ my question is just: Given the Landau Hamiltonian, how do they end up with this integral expression for the chern number. So I wonder particularly about this integral $\endgroup$ – Ben Curnow Jan 17 at 20:50

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