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  1+1d TSC as $Z_2^f$ symmetry breaking topological order?

+ 3 like - 0 dislike

(This question was transfered from physics stack exchange by myself: http://physics.stackexchange.com/questions/275830/11d-tsc-as-z-2f-symmetry-breaking-topological-order)

I have been struggling recently with a comprehensive problem on the relationship between topological superconductor and topological order. My question originates from reading a work conducted by Prof. Wen http://arxiv.org/abs/1412.5985, also a previous answered question http://physics.stackexchange.com/questions/71351/do-topological-superconductors-exhibit-symmetry-enriched-topological-order/74821#74821,in which Prof. Wen's comment also deals a lot with my questions presented below.

It has been claimed that 1+1d topological superconductor is a case of 1+1d fermionic topological order (or more strictlly speaking, a symmetry-enriched topological order).

My understanding on topological order is that it is characterized by topological degeneracy, which is degenrate ground states roubust to any local perturbation. And in the paper 1412.5985, it is argued that 1+1d topological superconductor is such a fermionic topological order with fermion-parity $Z_2^f$ symmetry breaking. So far I have learned about one-dimensional (or maybe it's quasi-one-dimensional as you like) topological superconductor, the first case where we have detected the appearance of Majorana zero modes is 1d D-class topological superconductor. And the terminology on D-class is defined from a classification working with BdG symmetry class of Hamiltonian. And D-class means such 1d superconductor possesses only particle-hole symmetry (PHS: $\Xi^2=+1$), and based on Kitaev's K-theory jobs the topological invatiant is $\mathbb{Z}_2$. In non-trivial topological phase labeled with topological number $\nu=-1$, a Majorana zero mode appears at each end of 1d D-class topological superconductor. And under my comprehension, the two Majorana zero modes form a two-fold degenracte ground states $|0\rangle$,$|1\rangle$ with different fermion-parity, and in a fermionic system like superconductor, the Hamiltonian has a fermion-parity $Z_2^f$ symmetry, hence the actual ground state configuration just don't have such symmetry from Hamiltonian and therefore spontaneous breaks it.

However why it should be claimed that such $Z_2^f$ breaking just directly leads to the appearance of 1d fermionic topological order in this 1d topological superconductor? (Regarding to the original words in the paper "two-fold topological degeneracy is nothing but the two-fold degeneracy of the $Z_2^f$ symmetry breaking"). And Prof. Wen in that previous posted question also said that this kind of 1d fermionic topological order state is therefore the result of long-range entanglement(LRE), so Majorana chain is indeed LRE and Kitaev has just used another way with out local unitary transformation definition(LU) to describe topological order. So I want ask how should one treat such open-line 1d Majorana chain as a long-range entanglement quantum state? And what about describing it with LU definition? How should people be supposed to demonstrate that?

And the last but also the most important question which makes me quite confused is that, it seems like all 1d topological superconductors, regardless of what class or what symmetry they have, are 1d fermionic topological order state due to the appearance of topological degeneracy, which so far in here considered as Majorana zero modes. Well, at least in the paper mentioned above, it is about a two-fold topological degeneracy. Then let's consider another case, the 1d DIII-class topological superconductor with TRS $\Theta^2=-1$, and according to Kramer's theorem, each end has a pair of time-reverse counterpart of Majorana zero modes, i.e. Majorana doublet, which forms a four-fold degenracy. After we introduce TRS-breaking perturbation e.g. Zeeman field, the Majorana doublet is lifted and splits a finite energy level-------I have read the references PhysRevB.88.214514 as well as PhysRevLett.111.056402 , and I don't know whether I have got it correctly or not. So in these works, I found words claim that such four-fold degenracy is TRS topological protected. Does that imply this kind of 1d toplogical superconductor has actually no topological degenracy with Majorana zero modes since it needs symmetry protection and therefore should belongs to SPT rather then SET ? If it is a yes, then does it mean not all 1d topological superconductors are topological order but just those with two-fold degenarcy may belong to topological order ? If not, then where did I actually get wrong ?

I'd appreciate every helpful comments and replys. Thank you.

asked Sep 3, 2016 in Theoretical Physics by Tom Gao (15 points) [ revision history ]
recategorized Sep 3, 2016 by Dilaton

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