# What is the minimal symmetry required for a spin Hamiltonian to describe a spin-liquid ground state?

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?]

• @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 '16 at 22:50
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.