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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 ?

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    $\begingroup$ Maybe the term "Spin Liquid" here means that there is no symmetry breaking in the ground state of Kitaev model. $\endgroup$ May 25, 2013 at 21:52
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    $\begingroup$ @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. $\endgroup$
    – Kai Li
    Mar 12, 2015 at 9:15

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The previous understanding of the quantum spin liquid as a ground state of spin systems with spin rotation symmetry is not only out-of-date but also misleading. In modern language, quantum spin liquids are classified as symmetry enriched topological(SET) states, which possess anyon excitations carrying fractionalized symmetry charges, meaning that the anyons transform projectively under symmetry actions. The symmetry does not need to include the global spin rotation symmetry SO(3). Therefore, a quantum spin liquid does not need to preserve the spin-SO(3) symmetry in the general sense.

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.

Of cause, you may restrict the discussion of spin liquid to the spin rotational symmetric cases, i.e. the spin-SO(3) symmetric spin liquid, which is just a subclass of all spin liquids, and indeed Kitaev spin liquid does not belong to this subclass. However, it is possible to write down a variation of the Kitaev model which is spin-SO(3) symmetric, and the resulting ground state is a spin-SO(3) symmetric spin liquid.

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  • $\begingroup$ @ Everett You, thanks for your answer. What about the short-ranged spin-spin correlations ? Is this feature essential for spin liquid in the general sense ? $\endgroup$
    – Kai Li
    May 25, 2013 at 17:40
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    $\begingroup$ @K-boy The short-range correlation is also not essential. Almost all spin systems have short-range correlations. $\endgroup$ May 25, 2013 at 19:49
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    $\begingroup$ @ 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? $\endgroup$
    – Kai Li
    May 25, 2013 at 21:28
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    $\begingroup$ @K-boy In the gapless case, the topological order is not defined, but the quantum order is still defined, manifested by the deconfined spinon and vison excitations. Thanks for reminding me. I have added this point to the answer. $\endgroup$ May 26, 2013 at 4:39
  • $\begingroup$ 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; $\endgroup$
    – Kai Li
    Mar 3 at 21:41

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