For those spin liquids with SU(2) spin-rotation symmetry or time-reversal(TR) symmetry , the Spin Density Wave (SDW) order parameters are always zero, say $\left \langle \mathbf{S}_i \right \rangle=\mathbf{0}$, due to the spin-rotation symmetry or TR symmetry.

But if a spin-liquid state is neither spin-rotation symmetric nor TR symmetric, e.g. the exact chiral spin liquid ground state of the generalized Kitaev model on the decorated honeycomb lattice, is there any possibility that the SDW order parameters $\left \langle \mathbf{S}_i \right \rangle\neq \mathbf{0}$ ? Therefore, if a spin-liquid state has nonzero order parameters $\left \langle \mathbf{S}_i \right \rangle$, why we still call it a "spin liquid" rather than a "SDW phase"?

Remarks: The spin-rotation mentioned here should be understood as the continuous $SU(2)$ or $SO(3)$ one, say all the spin-rotation transformations. Although the Kitaev model(on the decorated honeycomb lattice) and its ground state break this continuous spin-rotation symmetry, they still possess the $\pi$ spin-rotation symmetry about $S_x,S_y$ and $S_z$ spin-axes, and this is enough to ensure $\left \langle \mathbf{S}_i \right \rangle=\mathbf{0}$.

Thanks in advance.

  • $\begingroup$ For a $N$ spin-$S$ system, for any spin state $\psi$, we have $\left | \left \langle \mathbf{S}_i \right \rangle_\psi \right |\leqslant S$. This inequality can be proved by using Shwinger-boson or Holstein-Primakoff boson representations. Note that for a spin coherent state $\mid \mathbf{n} \rangle = \mid \mathbf{n}_1,\mathbf{n}_2,...,\mathbf{n}_N\rangle$, we have $ \left \langle \mathbf{n}\left | \mathbf{S}_i \right | \mathbf{n}\right \rangle=S\mathbf{n}_i$, where $\mathbf{n}_i$ are unit vectors. $\endgroup$
    – Kai Li
    May 19, 2014 at 9:54

1 Answer 1


The "old" definition of spin liquid is a state that breaks no symmetry, e.g. the ground state is like a fluctuating liquid of spins. This is still very useful intuition for designing models and in experiments. Nevertheless, it is a fundamentally bad definition since it defines a spin liquid by what it isn't and misses the key physics.

The key physics of the spin liquid is the presence of long-range entanglement. This is the "new" definition of a spin liquid. Long range entanglement typically manifests itself in terms of emergent gauge fields, fractionalization of quantum numbers, emergent anyons, etc.

To understand what this means, take two examples.

First take a conventional SDW, say an AF state in the square lattice Heisenberg model. Break all symmetries by imposing external fields. Then the system will be gapped and can be smoothly deformed into a product state. Hence without symmetry the system is short-range entangled (can be deformed smoothly to a product state).

On the other hand, what we really mean by a spin liquid is a long range entangled state. Take such a spin liquid and let's assume its gapped as in your example. Break all symmetries including discrete symmetries. The state cannot be deformed into a product state without encountering a phase transition.

So to summarize, a spin liquid, by which I mean a long-range entangled state, can always have some magnetic order or other broken symmetry sitting on top of it. This symmetry breaking can be spontaneous, or if the Hamiltonian lacks the relevant symmetry, such magnetic order will be present on symmetry grounds. However, as long as you have long-range entanglement the physics will be much richer.

This is why we still call a long-range entangled state with broken symmetry a "spin liquid" and not a SDW, for example. But you're absolutely right that the terminology is confusing and could be improved.

  • 1
    $\begingroup$ @ Physics Monkey, thanks for your comment. I want to know whether there exists any example for spin-liquid with nonzero SDW order parameters? Or do you know some papers or books for this situation? $\endgroup$
    – Kai Li
    Oct 6, 2013 at 15:32

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