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

+ 3 like - 0 dislike
1320 views

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 ?

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user K-boy
asked May 25, 2013 in Theoretical Physics by Kai Li (980 points) [ no revision ]
@ Brandon Enright, thanks for your edit.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user K-boy
Maybe the term "Spin Liquid" here means that there is no symmetry breaking in the ground state of Kitaev model.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user Xiao-Gang Wen
@ Xiao-Gang Wen, Yeah, the term here is likely to be what you mean. And does the ground state of Kitaev model have "more" symmetries than the Kitaev Hamiltonian, like emergent symmetries ? Thanks a lot.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user K-boy

1 Answer

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Unfortunately, the global spin rotation symmetry is not essential for the spin liquid 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 even on the model level. Of cause, you may restrict the discussion to the spin rotational symmetric cases, i.e. the spin-SU(2) 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-SU(2) symmetric, and the resulting ground state is a rotational symmetric spin liquid.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user Everett You
answered May 25, 2013 by Everett You (785 points) [ no revision ]
@ 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 ?

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user K-boy
@K-boy The short-range correlation is also not essential. Almost all spin systems have short-range correlations.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user Everett You
@ 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?

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user K-boy
@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.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user Everett You

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