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There is an electric field when you measure the Hall conductance. The electric field is not essential for the formation of the quantum Hall state itself: it is just a probe, and the Hall conductance i …

Confinement is usually discussed in the context of gauge theories. Here you do not have any specific information about the properties of the insulator, so it is not clear why you put "confined states" …

I don't know what is a "p+ip insulator". There is indeed p+ip superconductor, which belongs to the class D and characterized by an integer invariant, the Chern number.

The interaction term is not invariant under the gauge transformation. $b_k\rightarrow e^{i\theta_k}b_k$ is only a symmetry for the non-interacting Hamiltonian $H=\sum_k E(k)b_k^\dagger b_k$.

Here is a perhaps more direct argument: I suppose you define $|0\rangle$ to be the state with no fermion occupation, so it means that $n_I=\psi_I^\dagger\psi_I=0$ on this state. Define $N=\sum_I n_I$. …

You don't need the biquadratic interaction to realize the Haldane phase. The Heisenberg spin-1 chain already does the job. Adding the biquadratic term allows one to have a parent Hamiltonian for the e …

There is certainly a way to preserve the time-reversal symmetry with open boundary conditions: just put the two edge spins into a singlet state. You may say that this is cheating, since essentially we …

Q1: Consider the example of a Haldane phase protected by time-reversal symmetry. There is a spin-1/2 on the edge, and to polarize it (so the system is gapped and the ground state is unique) one must b …

The non-interacting classification was obtained in the seminal "periodic table" papers by Kitaev and Ryu/Snyder/Furusaki/Ludwig. The interacting classification of 2D fermionic SPT phases with time-rev …

To obtain the energy gap in the thermodynamic limit, one should take $N\rightarrow\infty, V\rightarrow\infty$ where $N$ is the number of atoms and $V$ the system size, but hold $N/V$ (i.e. density) fi …

It depends on what kind of pairing you are willing to include in the Hamiltonian. If only s-wave singlet pairing is present, and there is no spin-orbit coupling, the Hamiltonian has an additional $\ma …

Your argument is correct, in fact you can use it to show that the Hall conductivity is set by the density. However, there is an underlying assumption when you apply Galilean transformation, that is tr …

The relevant part of the sum as $\sum_{k^*,s_1,s_2}\delta_{k^*,s_1s_2}d_{s_1}d_{s_2}B^k_P$ Let me assume that the fusion category has no multiplicities, so $N_{ab}^c=0,1$, which I think Levin and We …

Just focusing on the $\cos^l\theta$ term is probably not going to get you anywhere, since $\cos^l\theta$, being a completely analytical function, is by no means singular (and following your argument y …

For 2D chiral p-wave, it has been shown that the Berry phase contribution vanishes and therefore the exchange statistics is entirely given by the monodromy. This was done in http://arxiv.org/abs/cond- …