# Why we call the ground state of Kitaev model a Spin Liquid?

Now we always talk about the so-called Kitaev spin liquid. One important property of spin liquid is global spin rotation symmetry. Let $\Psi$ represents a spin ground state, if $\Psi$ has global spin rotation symmetry, then it's easy to show this simple identity $< \Psi \mid S_i^xS_j^x\mid \Psi >=< \Psi \mid S_i^yS_j^y\mid \Psi >=< \Psi \mid S_i^zS_j^z\mid \Psi >$. But Baskaran's exact calculation of spin dynamics in Kitaev model shows that only the components of spin-spin correlations(nearest neighbour sites) matching the bond type are nonzero, which violates the above identity, further means that the ground state of Kitaev model does not have global spin rotation symmetry.

So why we still call the ground state of Kitaev model a Spin Liquid ?

• Maybe the term "Spin Liquid" here means that there is no symmetry breaking in the ground state of Kitaev model. May 25, 2013 at 21:52
• @Xiao-Gang Wen Thank you Prof.Wen. How to see "there is no symmetry breaking in the ground state of Kitaev model" under periodic boundary conditions(PBC)? For example, under open boundary conditions, the ground state is unique and hence preserve all the symmetries of the Hamiltonian; while under PBC, there is a 4-fold ground state degeneracy due to the $Z_2$ gauge structure, and how to understand "no symmetry breaking" in this case. Thank you very much. Mar 12, 2015 at 9:15

The defining property of the spin liquid is the intrinsic topological order (or the quantum order for gapless spin liquid). The Kitaev spin liquid possesses the $$\mathbb{Z}_2$$ topological order, which makes it a spin liquid, although the spin rotation symmetry is explicitly broken on the Hamiltonian level. In the current classification scheme, the Kitaev spin liquid is an SET state with $$\mathbb{Z}_2$$ topological order enriched by space group (translation, $$C_6$$ rotation and reflection) and time-reversal symmetry, and therefore satisfies the modern SET definition of the quantum spin liquid.
• @ Everett You, ok... But when the spin liquid is in a gapless phase, say $H=-J_K\sum S_i^\gamma S_j^\gamma$ where $J_x=J_y=J_z=J_K$ , is the intrinsic topological order still well defined? May 25, 2013 at 21:28
• Hi, I realize that I still don't quite understand why we say spin-SO(3) symmetry? For example, consider the simplest spin rotational symmetric Heisenberg model with N spin 1/2, then how to define the symmetry group G of the model? If G is the set of all the global spin-rotation operators $e^{i\alpha S_z}e^{i\beta S_y}e^{i\gamma S_z}$, then G=SU(2) (but when the number of spins N is even, G=SU(2)/Z2=SO(3)? Am I wrong?), where $S_{y,z}$ are the total spin operators; Mar 3 at 21:41