# Questions on gapless edge excitations in symmetry-protected topological state

I am studying a one-dimensional bosonic system with $U(1) \rtimes Z_2^T$ symmetry numerically, which might has a symmetry-proteced toplogical(SPT) phase. I have several questions about the symmetry-protected topological state, particularly, the gapless edge excitations in an SPT state.

On the first page of paper http://arxiv.org/abs/1209.4399 (PRB 87, 144421 (2013)) by Cenke Xu, he gave three criteria for SPT phases. In particular, the second criteria requires that the Hamiltonian with open boundary has either gapless edge excitations or gapped but degenerate ground states. (See also wiki)

Q1) What is the difference between the "gapless edge excitations" and "gapped but degenerate ground states"?

   I am confused about that because the bulk of SPT phase
is always gapped. Therefore, I think there is always
a gap above those gapless edge modes and these two
conditions should be equivalent in the thermodynamic limit.

Can someone give some examples to show the difference?
I prefer the examples in one dimension, for example,
the spin-1 Haldane chain with open boundary.


Q2) Since SPT phase is protected by some symmetry. How can I know the folds of degeneracy of the ground states from the symmetry without knowing the details of the Hamiltonian?

Q3) Is this a necessary condition for an SPT state? Or is this a necessary condiction in one dimension? If not, can you give several examples?

Q4) In one-dimensional bosonic systems, if an SPT phase is protected by $U(1)\rtimes Z_2^T$ symmetry, can I conclude that it is a Haldane phase?

## 1 Answer

Q1: Consider the example of a Haldane phase protected by time-reversal symmetry. There is a spin-1/2 on the edge, and to polarize it (so the system is gapped and the ground state is unique) one must break the symmetry. In this case, the ground state is always degenerate when the symmetry is not broken. Cenke Xu's statement is better understood for higher-dimensional system. For example, the edge of a 2D SPT has symmetry-protected gapless excitations. What this means is that in a finite-size system the gap between the ground state and the lowest excited state scales as 1/L where L is the (linear) size of the edge. It is possible that the gapless modes become gapped by spontaneously breaking the symmetry (if the symmetry is discrete), but then there is always a ground state degeneracy. The situation in 3D is more subtle: the surface can be gapless, or spontaneously break the symmetry (if the symmetry that is broken is discrete, there is a degeneracy. If it is a continuous symmetry that is broken, there is a gapless Goldstone mode), and very interestingly, the surface can be driven to a topologically ordered state which is fully gapped and preserves all symmetries. However, the way that the symmetry is manifested in the topologically ordered state is anomalous. In this case people often say that the surface is gapped but with topological degeneracy.

Q2: For 1D SPT, one can get the degeneracy by finding the minimal dimension of projective representations of the symmetry group.

Q3: I'm sure this is true for SPT states protected by on-site symmetry.

Q4: For $U(1)\rtimes Z_2^T$ there is only one bosonic SPT state in 1D, which is indeed the Haldane phase. The $U(1)$ here is inessential.