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  Ground state of AKLT chain invariant under time-reversal?

+ 1 like - 0 dislike

The AKLT chain is an example of an SPT phase protected by time-reversal symmetry. The Hamiltonian of the system has time-reversal symmetry. The ground state wave function can be pictured as follows (from wikipedia https://en.wikipedia.org/wiki/AKLT_model): enter image description here

For the open boundary condition case, the edge spin 1/2s are unpaired. As a result, the ends of the chain behave like free spin 1/2 moments. Hence the ground state manifold has a four-fold degeneracy, and any particular ground state would break time-reversal symmetry. So it seems to me that the ground state wave function does not have a time-reversal symmetry (although the bulk wave function does).

Similar things happen for a fermionic analog of the Haldane chain, where I have a 1D fermionic SPT protected by $T^2=(-1)^F$ symmetry. I can imagine the only nontrivial phase as two copies of Kitaev chain. In the ground state, each end has two Majorana zero modes transforming as a Fermionic Kramers doublet. The two Majorana zero modes determines a complex fermion mode. So effectively there are two complex fermion modes living at each end of the chain. The ground state also has a four-fold degeneracy, corresponding to the occupying or un-occupying of the two complex fermion modes, which are time-reversal partners of each other. So any particular occupation of the complex fermion modes at the edge also breaks time-reversal symmetry.

However, by the definition of an SPT, it seems that the ground state wave function should preserve whatever symmetry the Hamiltonian of the system has, so I'm a little bit confused. Can anyone help?

This post imported from StackExchange Physics at 2015-09-23 09:33 (UTC), posted by SE-user Michael
asked Sep 21, 2015 in Theoretical Physics by Michael (5 points) [ no revision ]

"However, by the definition of an SPT, it seems that the ground state wave function should preserve whatever symmetry the Hamiltonian of the system has" -- why do you say so (for the OBC scenario you consider)?

This post imported from StackExchange Physics at 2015-09-23 09:33 (UTC), posted by SE-user Norbert Schuch

1 Answer

+ 2 like - 0 dislike

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 will have periodic boundary conditions by pairing the edge spins up. This depends on perspective, but indeed there are no other ways to have a completely symmetric ground state with open boundary conditions -- this should be seen as a feature, otherwise if there is a "local" way to make the ground state wavefunction symmetric, one should be able to tune the Hamiltonian, by adding local terms, in a symmetric way to favor that particular ground state, which would mean contradict the fact that this is a SPT.

The definition of SPT says that the Hamiltonian has the symmetry (symmetry is first defined at the Hamiltonian level), and the ground state does not break the symmetry spontaneously. The second statement is about the thermodynamic limit and boundary effects do not matter.

This post imported from StackExchange Physics at 2015-09-23 09:33 (UTC), posted by SE-user Meng Cheng
answered Sep 21, 2015 by Meng Cheng (40 points) [ no revision ]

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