The systems I have in mind are for example Kitaev's toric code model (arXiv:quant-ph/9707021) and Kitaev's honeycomb model (arXiv:cond-mat/0506438). What I'm looking for is a mathematically rigorous definition of superselection sector, also known as quasiparticle type, in such anyon systems, which I've been unable to find in the literature.

(Actually I was able to make a rigorous definition of superselection sector that works basically for any model whose fusion rules are always of the form $x\times y = z$, i.e. there's only one term in the right-hand side. This encompasses the toric code but unfortunately not the honeycomb model, so it doesn't seem illuminating to spell out my definition here.)

Kitaev said in the latter paper that they "do not care about rigorous definition" of superselection sector, but it wasn't clear to me that an appropriate, rigorous definition can be made either. Granted, we can get away with many physical concepts (e.g. thermodynamic limit) without rigorously defining them, but the situation with superselection sector in anyon systems seems slightly different. This is because the way we approach such systems is that we'll first cook up a unitary braided fusion category (UBFC) out of the physical systems, and then apply the mathematical theorems concerning UBFC back to the physical systems. UBFC is itself rigorously defined and intensively studied in the mathematical literature. I'd have reservation about the applicability of the theorems unless I'm confident that everything is on solid ground.

What bothered me was that superselection sector is pretty much the only thing that has not been rigorously defined, yet it's supposed to be the starting point of the construction of a UBFC. Everything that follows, e.g. splitting/fusion spaces, can be made well-defined. One can then construct, as far as I can envision, everything we need in a UBFC: the objects, the hom-spaces, the composition map, the biproduct, the monoidal structure, the conjugation, the dual objects, the pivotal structure, etc.

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An anyonic superselection sector is defined to be an equivalence class of local excitations under local operations. A non-trivial anyon (i.e. a non-trivial superselection sector) is a "local excitation which cannot be created locally".

Here is a precise statement of this notion which works for systems with zero correlation length, where anyons can be exactly localized into a finite region. The main point is that to make the notion of superselection sector precise, you have to consider an infinite system. In a finite system with periodic boundary conditions, the overall state always has trivial anyonic charge.

Let $\mathcal{L}$ be the algebra of operators acting on finite sets of spins within an infinite spin lattice $\mathbb{Z}^d$. A state is defined to be a function mapping operators to their expectation values, i.e. a linear map $\omega : \mathcal{L} \to \mathbb{C}$, such that $\omega(A) \in \mathbb{R}$ if $A = A^{\dagger}$, $\omega(A) \geq 0$ if $A$ is positive-semidefinite, and $\omega(\mathbb{I}) = 1$. (If the system were finite, we could define $\omega(A) = \mathrm{Tr}(A\rho)$, where $\rho$ is the density matrix representing the state of the system, and the three conditions stated would correspond to $\rho = \rho^{\dagger}$, $\rho \geq 0$ and $\mathrm{Tr} \rho = 1$ respectively. However, it is not possible to define a sensible notion of density matrix in an infinite system.)

A pure state is a state $\omega$ which cannot be expressed as a convex linear combination of other states: $$\omega = p\omega_1 + (1-p) \omega_2, p \in [0,1] \implies \omega_1 = \omega_2 = \omega.$$ Again, in a finite system this would be equivalent to saying that the density matrix is a pure state $\rho = |\Psi\rangle \langle \Psi|$.

Let us fix one particular pure state $\omega_{GS}$ which represents the ground state of our topologically ordered system. A local excitation is any pure state $\omega$ which "looks the same as the ground state far away", that is there exists some finite region $R \subseteq \mathbb{Z}^d$ such that $\omega(A) = \omega_{\mathrm{GS}}(A)$ for any operator $A$ that acts trivially inside of $R$. EDIT: We also have to require that the support of $A$ does not enclose $R$ -- otherwise we could detect the presence of an anyonic charge through a Wilson loop.

We can define an equivalence relation on local excitations: $\omega \sim \omega^{\prime} $ iff $\omega^{\prime}$ can be created from $\omega$ locally, by acting with a unitary $U$ on a finite region, so that $\omega^{\prime}(A) = \omega(U^{\dagger} A U)$ for all $A$. The superselection sectors are precisely the equivalence classes of local excitaions under this equivalence relation.

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    Thank you! This is an excellent answer. I have two follow-up questions. I assume this is how you would define $\omega_{GS}$. Consider an exhaustion by $\mathbb Z^d$ by a sequence of finite subsets $L_1 \subset L_2 \subset L_3 \subset \ldots$. Define $\omega_{GS}^{L_i}(A)={\rm Tr}(A\rho_{GS}^{L_i})$ for any operator $A$ on $L_i$, where $\rho_{GS}^{L_i}$ is the density matrix for the ground state on $L_i$. Then given compactly supported $A$ on $\mathbb Z^d$, we define $\omega_{GS}(A)=\lim_{i\rightarrow\infty}\omega_{GS}^{L_i}(A)$. This is well-defined because ${\rm supp}(A)$ is in some $L_i$. – user46652 May 29 '15 at 8:07
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    (cont'd) I can even visualize how taking the limit somehow gives us nontrivial superselection sectors. Imagine a particle-antiparticle pair. As we take larger and larger lattices, we keep the particle at a fixed location but the antiparticle farther and farther away. It's easy to see the limit of the corresponding $\omega^{L_i}$ exists. Now, at each stage we can apply a unitary operator to annihilate the pair, but the support of the operator has to grow. Thus in the limit of infinite lattice it's impossible to annihilate the pair. – user46652 May 29 '15 at 8:22
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    (cont'd) Question 1: Do we need some condition on the "purity" of the state $\omega$ on $\mathbb Z^d$? One way to formulate this would be to require $\omega$ to be the limit of a sequence $\omega^{L_i}$ defined by pure density matrices. Without the purity condition, I think we would have infinitely many superselection sectors even on a finite lattice, because the transformation $U\rho U^\dagger$ doesn't affect the eigenvalues of $\rho$. I imagine if we take the limit of a sequence of mixed states, the limit won't be related to one that's coming from pure states. – user46652 May 29 '15 at 8:26
  • (cont'd) Question 2: What part of your answer is well-known and what part is your personal definition? For the well-known part, are there any references where I can learn more about the formulation? – user46652 May 29 '15 at 8:29
  • @user46652: 1. Yes, I think you're right that we need to require local excitations to be pure states. I edited my post to state a definition of pure state. I think the one you give is too strong: a limit of mixed states can still be pure. In particular, if we construct the state by exhaustion as you suggested, then we ought to take $\rho^{L_i}$ to be the reduced density matrix on $L_i$ starting from an infinite (or sufficiently larger than $L_i$, it should give the same result) lattice. In that case the $\rho^{L_i}$ will clearly not be pure, but the infinite-system state will be. – Dominic Else May 29 '15 at 18:47

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