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Is $Tr(\mathbb{x})$ supposed to represent the trace of that messy tensor, $\mathbb{x}$? If so, then the result should be a scalar by definition, I believe.

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First of all, the SSH model does not have particle-hole symmetry. Particle-hole symmetry is an exact symmetry (at mean field level) reserved for superconductors and is an anti-unitary symmetry. A symmetric spectrum does not mean particle-hole symmetry. SSH model and end states Let me first explain a simple way to understand the topological end states. The ...

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This is a very interesting point and it is somehow related to universality classes. A typical examples are the universality classes defined by time reversal symmetry. This is at the basis of Random Matrix Theory and has application for the analysis of the atomic emission spectra and in mesoscopic physics as well. There is a direct connection between the ...

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First you need to bring it into the following form: $H=\Psi^\dagger h \Psi$ Here $\Psi$ is a big column vector: $\Psi=(\dots, c_{m,n}, \dots, c_{m,n}^\dagger, \dots)^T$ Basically, the first half of $\Psi$ are all annihilation operators, and the second half are all creation ones. If the number of sites is $N$, the size of $\Psi$ is $2N$. So $h$ is a ...

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So: I assume you want to diagonalize this problem by rewriting the Hamiltonian as $H=\sum E_id_i^\dagger d_i$, where $d_i$ are quasiparticle operators which obey the Fermionic commutation relations. If we only had $c^\dagger c$ terms, we would be able to write H as $$H=H_{ij}c_i^\dagger c_j$$ We could then prove that if $\{c_i\}$ obey the Fermion ...

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Clearly the "TMI" and the slave-rotor mean-field state are very different, because the TMI, as you assume, has no topological degeneracy while the other state is topologically ordered. However, I feel this answer is not very meaningful without seeing more details of the slave-rotor mean-field state. I'm afraid this is not a very well-known (or even ...

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