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!


A very good question. First of all, topological order strictly speaking is only defined for gapped states. But to some extent it can coexist with gapless degrees of freedom. A rather trivial example is just adding something gapless decoupled from the topological order (i.e. phonons). The example of the Kitaev model is quite different though, since the gapless part are fermionic spinons and the gapped part are visons ($Z_2$ gauge fields). The TEE says that the wavefunction of the $Z_2$ gauge field has non-local constraints(i.e. electric field lines must be closed, or only end on the spinons), which is also reflected on the mutual braiding statistics between spinons and visons. But on the other hand, the gapless spinons do have significant effect: for example, if one asks for the topological degeneracy (on top of the $1/L$ low-energy spinon excitations) on a torus, my thought is that the ground state degeneracy is reduced from $4$ to $1$, because spinons going around the large cycles can measure the vison fluxes and such a process has an amplitude $\sim 1/L$.

So indeed the distinction you draw is an important one. On the other hand, this gapless state is not robust: there is nothing preventing the spinons from opening a mass gap unless additional symmetries are imposed. So near the gapless phase, there are gapped A and B phases, both of which have the same value of TEE. So it can be considered as the "critical point" between the A and B phase.

I recommend a very insightful paper by Bonderson and Nayak, http://arxiv.org/abs/1212.6395, which discussed in great depth how one can define topological order in the presence of gapless degrees of freedom, and how ground state degeneracy and braiding statistics are affected.

For the second question, a gapped state should probably be defined as a state where all correlation functions (of physical, local operators) are short-ranged. It seems to me that if a state is gapless, then some correlation function should detect the gaplessness: for example, in your example of Kitaev honeycomb model, although the spin correlation functions are short-ranged, the bond energy correlations are algebraic. I have not seen a counterexample to this criteria, but I don't think it has been proved rigorously either. You can look at http://arxiv.org/abs/math-ph/0507008 for a proof of spectral gap implying short-ranged correlations (also need to be careful about whether one should use connected correlation functions, so many subtle details...).

  • 1
    $\begingroup$ Thanks a lot for your insightful comments and suggested references. The former part in your answer seems deep to me and I may spend more time to understand it. However, I learned from you that the definition of a gapped state $\psi$ is via the correlation length rather than some parent Hamiltonian $H$. So is there a possibility or an example that a gapless Hamiltonian possesses a gapped ground-state? $\endgroup$ – Kai Li Mar 15 '15 at 6:41
  • 2
    $\begingroup$ There may be other simpler examples, but here is one: journals.aps.org/prl/abstract/10.1103/PhysRevLett.109.260401 $\endgroup$ – Meng Cheng Mar 15 '15 at 6:48
  • $\begingroup$ As to the honeycomb Kitaev model, for the gapless phase (e.g., gapless spin Hamiltonian at $J_x=J_y=J_z$), is the corresponding ground-state also gapless (in the sense that some kind of correlation-length is large as compared to the system size)? If the ground-state is gapless, then is the TEE(=$-ln2$) still well defined? $\endgroup$ – Kai Li Mar 15 '15 at 6:51
  • 1
    $\begingroup$ Yes, as I mentioned in the answer, the bond energy correlation function should be algebraic. For TEE, first one needs to have an area law for the entanglement entropy, which is true for gapless systems with linear dispersion (or emergent Lorentz invariance). Then TEE is just the correction to the area law. $\endgroup$ – Meng Cheng Mar 15 '15 at 6:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.