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Let's restrict to the case of spin-1/2 system. As we know, a spin-liquid (SL) state is the ground state of a lattice spin Hamiltonian with no spontaneous broken symmetries (sometime it may spontaneously break time-reversal symmetry and is called a chiral SL), where two essential symmetries of a SL state are lattice translation and spin-rotation symmetries.

Since, traditionally, we usually describe a SL state by using a spin Hamiltonian with the full $SU(2)$ spin-rotation symmetry (e.g., Heisenberg model), and the corresponding SL state is hence also $SU(2)$ symmetric, i.e., a RVB type SL. While, the honeycomb Kitaev model provides us an exact SL ground state with $Q_8$ spin-rotation symmetry, where $Q_8$ is a finite subgroup of $SU(2)$, indicating that the Kitaev SL does NOT belong to the RVB type.

Thus, my question is: Generally speaking, what is the minimal spin-rotation symmetry required for a spin Hamiltonian to describe a SL ground state? Is $Q_8$ group the minimal one? Thank you very much.

[My motivation for this question is that for a spin Hamiltonian without any spin-rotation symmetry, whether or not can it possess a SL ground state? And does the existence of a SL state with some spin-rotation symmetry imply the occurrence of emergent symmetries?]

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The definition of a spin liquid as a spin system "with no spontaneously broken symmetries" is out of date and no longer used, partially for the reason you describe. If you perturb as spin-liquid Hamiltonian by adding small terms that break all the symmetries, then the ground state will still be a spin liquid even though there are no longer any symmetries that could possibly be broken. Moreover, a spin liquid actually can spontaneously break symmetries; see the third paragraph of http://arxiv.org/abs/1112.2241.

The more modern definition of a spin liquid is a spin system with "intrinsic topological order." This can be defined in many equivalent ways (at least for a gapped system - the gapless case raises more subtle issues): (a) the inability to be deformed into a product state by local unitary operations, (b) nonzero topological entanglement entropy, (c) low-energy physics that can be described by a topological quantum field theory, (d) excitations with anyonic statistics, etc.

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  • $\begingroup$ thanks for your comments. Part of the reasons I asked this question is: If a spin liquid state has no symmetries, then can the local moments exist? I mean can a spin liquid state host a magnetic order? $\endgroup$ – Kai Li May 11 '16 at 9:29
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    $\begingroup$ @KaiLi If I understand your question correctly, you are asking "If you add a term that breaks the SU(2) symmetry, can the ground state be a spin liquid with local magnetic moments?" The answer is yes: for example, if you apply a field $H \approx J/3$ to the nearest-neighbor Kagome antiferromagnet, you get a gapped state that is probably a $\mathbb{Z}_3$ spin liquid, in which the spins have a magnetic moment parallel to the field. See nature.com/ncomms/2013/130805/ncomms3287/full/ncomms3287.html. $\endgroup$ – tparker May 11 '16 at 22:50
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    $\begingroup$ @KaiLi In general, a system which has both (a) intrinsic topological order (i.e. a spin liquid, if it's a spin system) and (b) a global symmetry is called a system with "symmetry enriched topological order" (search that phrase for more information). If you take a system with SET order and add terms that explicitly break the symmetry, then you can remove the SET order while preserving the intrinsic topological order. $\endgroup$ – tparker May 11 '16 at 22:56
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tparker's answer is absolutely correct, but it's worth noting why the "old fashioned" definition was still useful. According to the higher dimensional extension of the Lieb-Schultz-Mattis theorem by Hastings and others, a gapped system with (a) translational invariance (b) an odd number of S=1/2 moments per magnetic unit cell, and (c) unbroken SO(3) symmetry must be a spin-liquid in the sense he described (anyons). So the old definition was a sufficient, but not necessary, condition for spin-liquid physics.

One way to rephrase your question is: how much symmetry is required for a Lieb-Schultz-Mattis theorem to still hold? For example, Oshikawa and Hastings showed that you can break down $SO(3) \to U(1)$ (rotation invariance about one axis), and the theorem still holds at zero-magnetization. Later work showed you can breakdown $SO(3) \to \mathbb{Z}_2 \times \mathbb{Z}_2$, or you even break $SO(3)$ completely if you keep time-reversal invariance. These two are probably the minimal cases in the sense you're asking.

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