# Ground state of AKLT chain invariant under time-reversal?

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):

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?

• "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)? – Norbert Schuch Sep 21 '15 at 7:24