Abstract

The split property of a pure state for a certain cut of a quantum spin system can be understood as the entanglement between the two subsystems being weak. From this point of view, we may say that if it is not possible to transform a state omega via sufficiently local automorphisms (in a sense that we will make precise) into a state satisfying the split property, then the state omega has a long-range entanglement. It is well known that in 1D, gapped ground states have the split property with respect to cutting the system into left and right half-chains. In 2D, however, the split property fails to hold for interesting models such as Kitaev’s toric code. Here we show that this failure is the reason that anyons can exist in that model. There is a folklore saying that the existence of anyons, like in the toric code model, implies long-range entanglement of the state. In this paper, we prove this folklore in an infinite dimensional setting. More precisely, we show that long-range entanglement, in a way that we will define precisely, is a necessary condition to have non-trivial superselection sectors. Anyons in particular give rise to such non-trivial sectors. States with the split property for cones, on the other hand, do not admit non-trivial sectors. A key technical ingredient of our proof is that under suitable assumptions on locality, the automorphisms generated by local interactions can be “approximately factorized.” That is, they can be written as the tensor product of automorphisms localized in a cone and its complement respectively, followed by an automorphism acting near the “boundary” of Lambda , and conjugation with a unitary. This result may be of independent interest. This technique also allows us to prove that the approximate split property, a weaker version of the split property that is satisfied in e.g. the toric code, is stable under applying such automorphisms.

Full Text
Published version (Free)

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call