Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?]

share|cite|improve this question
up vote 1 down vote accepted

The definition of a spin liquid as a spin system "with no spontaneously broken symmetries" is out of date and no longer used, partially for the reason you describe.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.

share|cite|improve this answer
    
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? – Kai Li May 11 at 9:29
1  
@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. – tparker May 11 at 22:50
1  
@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. – tparker May 11 at 22:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.