# Spin liquid states and conductors

1. Jun 18, 2012

### mavipranav

Hi,

All articles on spin liquids I've seen treat them as insulators. This is understandable in the context in which they were first introduced i.e. the resonating valence bond state in which every electron is singlet-ed with every other, and thus essentially blocking conduction.

Given that that the term spin liquids today, I believe, cover a much broader range of states, why are they still treated as insulators? Put otherwise, why cannot there be conducting spin liquid states?

Thanks,
Mavi

2. Jun 18, 2012

### Physics Monkey

Depending on what you mean, there is nothing wrong with having a conducting spin liquid. But it demands on what you mean. A Fermi liquid conducts and has no spin ordering so do you consider that a spin liquid? In the context of a purely spin model there is by definition no notion of charge and hence charge conductivity is meaningless. Of course, one can always embed these theories in larger frameworks that include explicit (but gapped) charge degrees of freedom. This leads, for example, to interesting predictions for the ac conductivity.

I would use a more general and less history dependent terminology to describe spin liquids and their more general conducting cousins: ideas like fractionalization and long range entanglement capture nicely the properties of spin liquids while easily generalizing to conducting states. They also make it clear that the "interesting" parts of spin liquids don't really require spin symmetry e.g. fractionalization can coexist with spin ordering.

3. Jun 18, 2012

### mavipranav

I guess my main problem was, to go into the specifics now, how it might be reasoned that a state with exotic properties like emergent gauge structure or topological order or fractionalisation of charge - as you point out - is equivalent to another standard definition of quantum SL (QSL) i.e. a Mott insulator with no magnetic ordering.

Therefore, I am skeptical that one can have spin ordering and call it a QSL. And no, a Fermi liquid would not be a QSL since such a Hamiltonian acts not on the spins but on the fermions; but the converse, as far as I see, need not hold i.e. a QSL might turn out to have Fermi-liquid like properties.

4. Jun 19, 2012

### Physics Monkey

One standard answer to this question is the folk theorem that one cannot have a gapped Mott insulator with one electron per unit cell (see Hastings version of LSM theorem for a precise statement, see also the work of Oshikawa http://arxiv.org/abs/cond-mat/0305505 http://arxiv.org/abs/cond-mat/0301338). One can get gapless states by symmetry breaking e.g. AF order (goldstone modes) or a valence bond crystal (degenerate ground states). However, if you rule this out then something "exotic" has to be going on, either topological order or some kind of gapless spin liquid.

Well, as I said above it depends on your definition. If your definition of QSL is that it isn't spin ordered then sure, but if your definition of QSL is that there are emergent spinons, say, then the two can clearly coexist. Indeed, you can break the spin symmmetry explicitly and not compromise the existence of spinons provided the perturbation is small. Of course their spin quantum number is mixed by the perturbation but one still has emergent fermions.

Also, if you demand a Hamiltonian that acts only on spins then again by definition you cannot have a charge conductivity since there are no charge degrees of freedom. If you want to to talk about charge in an electronic system then you're going to have to at least mention electrons even if some other bosonic mode is carrying the current. Certainly QSLs can have other Fermi liquid-like properties though e.g. thermal conductivity, and this is a major motivation for QSL proposals in the triangular lattice organic mott insulators.