Topological entanglement entropy (TEE, proposed by Levin, Wen, Kitaev, and Preskill) is a direct characterization of the topological order encoded in a wavefunction. Here I have some confusions, and let's take the spin-1/2 Kitaev model on the honeycomb lattice as an example.
The ground-state entanglement entropy of Kitaev model can be calculated exactly, where the TEE=$-ln2$ for both gapped phase and gapless phase. This is consistent with the 4-fold ground-state degeneracy on a torus for both gapped and gapless phases. [Although the ground-state degeneracy may be not well defined in the gapless phase.]
Question: The nonzero TEE of the gapless ground-state says that the gapless state has "topological order", but "topological order" is only defined for a gapped phase. How I understand this paradox ?
Remarks: I personally think that the concept of "topological order" for a gapped Hamiltonian and for a wavefunction may be different.
A related question is: whether a given state $\psi$ is gapped or not? One possible definition may be: If there exists a gapped Hamiltonian whose ground-state is $\psi$, then we say $\psi$ is a gapped state. But this definition seems to be not well defined, since there may exists another gappless Hamiltonian whose ground-state is also $\psi$. A simple example is a free fermion Hamiltonian $H(u)=\sum_k(k^2+u)C_k^\dagger C_k$, where the vacuum state $\left | 0 \right \rangle $ is a gappless ground-state of $H(u=0)$ while $\left | 0 \right \rangle $ is a gapped ground-state of $H(u>0)$, so the gap meaning of a given state (here $\left | 0 \right \rangle$) may be ambiguous.
So I personally think that the gapped and gapless ground-states in the Kitaev model are both topologically ordered wavefunctions (from the nonzero TEE), but only the gapped Kitaev Hamiltonian (rather than the gapless Kitaev Hamiltonian) has a well-defined topological order.
Thanks in advance!