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Is it possible to have topological degeneracy in 1D ?

+ 4 like - 0 dislike
75 views

I mean to have q-fold degenerate ground states on a ring which could not be lifted by local perturbation.

If the answer is no, then what is the physical (or mathematical) reason against having such a state in 1D ?

This post imported from StackExchange Physics at 2014-04-11 15:50 (UCT), posted by SE-user user23765
asked May 1, 2013 in Theoretical Physics by user23765 (50 points) [ no revision ]
Perhaps if you have $N$ segments of the wire which are topological superconductors ($p$-wave), all of these segments have non-topological segments between them, and the boundaries between the non-topological and topological sections are well separated then you would have a topologically protected $2^N$-fold degenerate ground state. However, it won't have topological order (i.e. long-range entanglement); I'm not 100% sure why. In a recent talk Kitaev mentioned that the 1-D $p$-wave chain does not have topological order.

This post imported from StackExchange Physics at 2014-04-11 15:50 (UCT), posted by SE-user NanoPhys

1 Answer

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It is not possible to have topological degeneracy on a 1D ring, because $d$-dimensional topologically ordered states on $S^d$ have no degeneracy.

It is possible to have topological degeneracy on a 1D segment, but only for fermion systems, since for bosonic system, there is no 1D topological order.

answered Aug 7, 2015 by Xiao-Gang Wen (3,319 points) [ no revision ]

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