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I have a spin 1/2 chain with open boundary conditions described by Hamiltonian $H=\sum_i \sigma_{2i}^z \sigma_{2i+1}^z$. From $H$ it's clear that boundary sites are decoupled from the rest of the system. Thus, there are free spin 1/2s at the edges and G.S. will be 4-fold degenerate. These edge modes will be destroyed if I add a perturbation which breaks time-reversal symmetry (TRS) (e.g. If we add a magnetic field then spins will be polarized and degeneracy will be lifted). Its clear that it's a topological phase but how can it be shown rigorously that if we add any perturbation respecting TRS, edge modes will survive?

Is there a way to show that through the projective representation of TRS group. Since these free spins will transform in a non-trivial way under TRS ($\mathcal(T)^2=-1$).

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  • $\begingroup$ As it is written, the Hamiltonian describes system of decoupled pairs of spins because of spin $2i$ interacts only with spin $2i+1$. So the degeneration here is much more then 4-fold. $\endgroup$
    – Gec
    Commented Nov 29, 2023 at 14:58
  • $\begingroup$ Yeah, but I want to focus on degeneracy coming explicitly because of dangling spin-1/2 at the end of the chain, which will vanish in topologically trivial case (dimerization of 2i-1 and 2i site spins). $\endgroup$
    – Barry
    Commented Nov 29, 2023 at 16:18

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This is not a topological (SPT) phase with stable edge modes.

To show this, notice that the local perturbation $$ \Delta H = \lambda \sum_i \sigma^z_{2i-1} \sigma^z_{2i} $$ leaves the system with a two-fold ground state degeneracy. This is the ground state degeneracy associated to the ferromagnet and is a bulk property. There is no boundary mode.

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  • $\begingroup$ But if you treat this perturbation via perturbation theory, it will not allow scattering between states of degenerate ground state manifold at any order. So degeneracy seems to be robust. On the other hand, if we add perturbation like $\sum_i \sigma_i^x \sigma_{i+1}^x + \sigma_i^y \sigma_{i+1}^y$, these allow scattering between degenerate states at L-th order of perturbation theory (L=system size), but these matrix elements will be exponentially suppressed so again degeneracy seems to be robust in thermodynamic limit ($L\rightarrow \infty$). $\endgroup$
    – Barry
    Commented Dec 1, 2023 at 9:34
  • $\begingroup$ In the Z basis, you are correct that it does not have any off diagonal processes (even at L-th order in perturbation theory!) because the Hamiltonian commutes with every S_z. But that’s not the only way you can open up a gap in degenerate perturbation theory (DPT), as in this case there are nontrivial diagonal terms that open up the gap. The degeneracy is not robust in the thermodynamic limit. Try doing DPT for a two level system with a sigma_z perturbation if that makes it clear… $\endgroup$
    – Nandagopal Manoj
    Commented Dec 1, 2023 at 18:46

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