The WP article on the density matrix has this remark:
It is now generally accepted that the description of quantum mechanics in which all self-adjoint operators represent observables is untenable.[17][18]
The first footnote is to the appendix in Mackey, Mathematical Foundations of Quantum Mechanics. I have a copy of Mackey, but I am unable to connect the material in the appendix to this statement. I also don't know anything about C* algebras — and Mackey doesn't seem to mention them either.
Can anyone explain this at an elementary level, maybe with a concrete example of a hermitian operator that wouldn't qualify as an observable?
I assume this issue only arises in infinite-dimensional spaces…? Intuitively, I don't see how it could be an issue in a finite-dimensional space.
Best Answer
The appendix of Mackey talks about superselection rules, and indeed superselection is the phenomenon where there are self-adjoint operators that are not observables. Whether this is obvious or not depends on how one defines "superselection".
The standard definition would be that the Hilbert space $H$ splits into the direct sum $H_1\oplus H_2$ such that for all $\lvert\psi\rangle\in H_1, \lvert \phi\rangle\in H_2$ and all observables $A$ we have that $\langle \psi \vert A\vert \phi \rangle = 0$, which implies that $AH_1 \subset H_1,AH_2\subset H_2$ for all observables, but which is clearly not true for all self-adjoint operators on $H_1\oplus H_2$.
An easy (albeit somewhat artificial) example of a superselected system is when we take $H_1$ to be the state space of a boson and $H_2$ the state space of a fermion, see also this answer of mine. Other examples can arise from theories with spontaneous symmetry breaking where states belonging to different VEVs cannot interact with each other and form superselection sectors.