I got some naive questions on the ground states of honeycomb Kitaev model (with open boundary conditions):

(1) Consider a simple case that $J_x=J_y=0$, then the model reduces to $$H=J_z\sum_{z\text{ }\text{links}}S_i^zS_j^z.$$ It's obvious that $H$ has highly degenerate GSs (degeneracy$=2^N$, where $N$ is the number of unit cells), and in each GS configuration, every two spins connected by a $z$ link are aligned parallel (if $J_z<0$) or antiparallel (if $J_z>0$). Thus, each of these GSs corresponds to a spin configuration and is not a spin liquid (SL).

On the other hand, the Majorana exact solution tells us that the GS of $H$ is a gapped SL.

So, does the exact solution just single out one of the highly degenerate GSs, which happens to be a SL (some certain manner of superposition of the above $2^N$ spin configurations) ??

(2) Following question (1), if the exact solution does single out just one of the highly degenerate GSs of $H$, and how could this be?

According to Kitaev's discussion on P.19 in his paper, the GS is achieved by the vortex-free field configuration, which follows from a theorem proved by Lieb.

But shouldn't an exact solution gives all the $2^N$ GSs ? Why only single out one which strongly contradicts the obviously correct GS degeneracy$=2^N$ ? Is there something wrong with applying the Lieb's theorem here ? I'm confused....


I think I can answer my question now.

The Lieb's theorem here is a statement about which flux configuration (or flux configurations) minimizes the ground state energy of the quadratic Hamiltonians with each corresponding flux configuration/sector. In the case that $J_x=J_y=0$ (or more generic cases that $J_xJ_yJ_z=0$), the corresponding $x,y$ link terms are absence in the quadratic Hamiltonians, and the flux (and hence the flux configuration) through each elementary plaquette is even not well defined. Therefore, the Lieb's theorem does not work in the case that $J_x=J_y=0$ (or more generic cases that $J_xJ_yJ_z=0$) while the theorem itself is still correct. And the Kitaev model at $J_x=J_y=0$ does have highly degenerate GSs.

On the other hand, in the cases that $J_xJ_yJ_z\neq 0$, the Lieb's theorem works, which states that only one flux configuration (vortex-free field configuration here) minimizes the ground state energy. Thus, the GS of Kitaev model is always unique (non-degenerate) as long as $J_xJ_yJ_z\neq 0$. Due to this uniqueness, the GS must preserve all the symmetries of the Kitaev Hamiltonian. And this should be one of the reasons for the name of spin liquid GS of Kitaev model. (And see Prof.Wen's comments there.)

Note that for the Kitaev model defined on a decorated honeycomb lattice, there are two distinct flux configurations (which are related by time-reversal) that minimize the ground state energy, and hence there are two degenerate chiral spin liquid GSs which breaks time-reversal symmetry while preserves lattice translational and spin symmetries due to the 'uniform' flux configuration instead of the 'uniqueness' of flux configuration in the honeycomb case.

Note: Note that for the honeycomb Kitaev model, as we take the thermodynamic limit, we have to specify a certain sequence of sizes $\rightarrow \infty $, e.g., let the number of unit cells$\rightarrow \infty $ rather than the number of lattice sites$\rightarrow \infty $ , which ensure that the number of lattice sites is always even. Otherwise, for the odd number of lattice sites, there would be Kramers degeneracy due to the time-reversal symmetry.

BTW: I get another naive question. As we know, if an eigenstate of a Hamiltonian is non-degenerate, then this eigenstate must preserve all the symmetries of the Hamiltonian. While, is there any possibility that this eigenstate has additional symmetry which is absence in the Hamiltonian ? Does someone know any example of this kind?

  • 1
    $\begingroup$ Yes ground states can have additional symmetries. I suppose the most 'obvious' examples (that is, obvious once someone has explicitly pointed it out) is the emergent conformal invariance for critical Hamiltonians. But there are examples of gapped phases of as well, in fact I am currently working on some 1D phases where the hamiltonian has a U(1) symmetry but the ground state has a SO(3) symmetry. (I can give you more exact references if you are interested.) $\endgroup$ – Ruben Verresen Feb 16 '16 at 15:00
  • $\begingroup$ @Ruben Verresen Thanks for your comment. Sure, I would be interested in your suggested references. $\endgroup$ – Kai Li Feb 17 '16 at 16:32

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.