Question

What is a continuous superselection sector?

Answer 1

I'm studying the terrible subject of continuous superselection rules and I faced with the following problem. Usually (continuous or discrete) superselection rules are defined involving a direct integral (with respect to some measure) of Hilbert spaces, each one to be interpreted as a "sector": an Hilbert space corresponding to the system for which some physical property (the one generating the superselection rules, e.g. the charge) is fixed. This interpretation is traditionally valid in the discrete case (direct sum) and I agree with it. But if the cardinality of the decomposition index set is the continuum one, I find some troubles trying to extend it. In general, indeed, a point of the index set can have null measure, hence the corresponding Hilbert space cannot be embedded in the global one, because the direct integral is defined with an "almost everywhere" construction. So what is a "sector" in this case? It seems to me that it would be more correct to consider again a direct sum, even if with a non countable index set. Now the sectors can be interpreted in the usual way, but the global space ceases to be separable. So my last question: could the Hilbert space associated with a physical system be non separable? Is there some physical (non mathematical) reason for the separability assumption in the quantum mechanics axioms

Answer 2

$<b><i><u>Superselection rules are postulated rules forbidding the preparation of quantum statesthat exhibit coherence between eigenstates of certain observables.[1] It was originally introduced by Wick, Wightman, and Wigner to impose additional restrictions to quantum theory beyond those of selection rules. Mathematically speaking, two quantum states {\displaystyle \psi _{1}} and {\displaystyle \psi _{2}}are separated by a selection rule if {\displaystyle \langle \psi _{1}|H|\psi _{2}\rangle =0} for any given Hamiltonian {\displaystyle H}, while they are separated by a superselection rule if {\displaystyle \langle \psi _{1}|A|\psi _{2}\rangle =0} for all physical observables {\displaystyle A}.$