# Exotic quantum observables

Quantum physics inherited its physical quantities from classical mechanics (energy, momentum, etc). Each quantity is associated with Hermitian operator, but there seems to be a lot more Hermitian operators than classic physical quantities! Granted, density operator can be viewed as an observable, however, it obeys restriction on its spectrum bounded between 0 and 1. Are there quantities besides classic ones?

• The statement of QM is "Every observable is Hermitian (or, rather, self-adjoint)", not "Every Hermitian/self-adjoint operator is an observable". Jan 24 '15 at 22:40
• @ACuriousMind : you can answer this question better than I. Of course there are observable properties with no classic counterpart: parity. But I am no good at this. Classical objects are distinguishable. But all sort of properties under rotation or reflection in space, in charge, and others, are purely quantum. Jan 24 '15 at 22:52

Let us first note that, although all observables are self-adjoint, not all self-adjoint operators are observables. The obvious counterexample is the identity - something that "observes" the identity is just a silly thing like an instrument whose dial has been fixed to point at a "1", it has no physical meaning.

Another instance where non-observable self-adjoint operators appear is whenever we have superselection sectors1, which, by assumption, are invariant under observables. Any self-adjoint operator which does not have the sectors as invariant subspaces is not an observable, yet, such operators certainly exist.

Also, the usual quantization schemes, regardless of their rigor, do not assign to any self-adjoint operator a physical meaning. Rather, they take the classical phase space observables and "embed" them into the self-adjoint operators of the quantum theory.

Nevertheless, there are non-classical observables, and there is one that everyone knows: Spin. Spin does not exist in a classical theory, but emerges more or less naturally in the course of quantization when we are faced with constructing representations of the Lorentz group (or, non-relativistically, the rotation group).

1Superselection sectors occur when the representation of the algebra of observables is not irreducible, and are exactly the irreducible summands.

• Isn't a spin merely a fancy angular momentum? Parity would be a better example. BTW, parity is not covered in introductory QM texts, what is its spectrum? Jan 24 '15 at 23:09
• It would be worth also noting that it is a theorem of Wightman, Wick and Wigner that, whenever superselection rules hold, there exist self-adjoint operators that are not observable. Jan 24 '15 at 23:10
• @TegiriNenashi: Spin is part of the total angular momentum of a quantum theory - that is, it is part of the Noether charge conserved under rotation - , but it is not the quantization of the classical angular momentum $L = \vec x \times \vec p$, which we usually call the orbital angular momentum. Jan 24 '15 at 23:12
• @ACuriousMind : spin is not a good example. It has no classical explanation, but he existence of a magnetic momentum of elementary particles is known for a long time (1922). Jan 24 '15 at 23:18
• @Sofia: Parity certainly is observable, I just don't think it is "non-classical". It's just the behaviour of stuff if we reverse all spatial directions. Where parity becomes quantumly interesting is when it is not conserved, but that is connected deeply with the concept of chirality, and hence essentially spin again (or better, representations of the Lorentz group again). Jan 24 '15 at 23:40

Every hermitian operator that appears in quantum mechanics (in some dynamics or symmetry of the system) is an observable in principle. Some of this operator is more easy to measure, and anothers is more (very more) difficult. When I say measure, I think in some mechanism that can correlate the hermitian operator of the system with the measure apparatus.

Operators that do not appears in dynamics and symmetry, are illness. How we can talk about things that don't have any connection to any thing.

Think about measure of the spin (electron). We can proceed with this because exist some sort of interaction that connect spins and magnetic field. Actually, for any generator of symmetry is expected that some interaction access this operator. Example, the momentum of a particle is a generator of symmetry, and we expect that this operator do something within, at least, a momentum of other particle. If our expectation failure we may have a very boring theory that do anything.