# What Hermitian operators can be observables?

We can construct a Hermitian operator $O$ in the following general way:

1. find a complete set of projectors $P_\lambda$ which commute,
2. assign to each projector a unique real number $\lambda\in\mathbb R$.

By this, each projector defines an eigenspace of the operator $O$, and the corresponding eigenvalues are the real numbers $\lambda$. In the particular case in which the eigenvalues are non-degenerate, the operator $O$ has the form $$O=\sum_\lambda\lambda|\lambda\rangle\langle\lambda|$$

Question: what restrictions which prevent $O$ from being an observable are known?

For example, we can't admit as observables the Hermitian operators having as eigenstates superpositions forbidden by the superselection rules.

a) Where can I find an exhaustive list of the superselection rules?

b) Are there other rules?

Update:

c) Is the particular case when the Hilbert space is the tensor product of two Hilbert spaces (representing two quantum systems), special from this viewpoint?

• A superselection rule is when no local operator can link between two states. It's exactly like when in statistical mechanics, you have zero probability of making a macroscopic motion. There is no exhaustive list, as any vaccum condensate which breaks an exact symmetry or makes a SUSY modulus is automatically a superselection sector maker. – Ron Maimon May 8 '12 at 15:24
• As an observation (pardon the pun), the identity operator is Hermitian with a degenerate eigenvalue of 1. Every vector is an eigenvector, and any chosen one could be included in an orthonormal basis. So without the restrictions you are looking for it would seem that every state would be observable (I think ?) – Tom Collinge Sep 5 '15 at 11:52

You might ask where do we get the algebra from. Well, this already should be supplied by the particular model. For a quantum mechanical particle moving on a manifold $M$, the C*-algebra consists of all bounded operators on $L^2(\hat{M})$ commuting with $\pi_1(M)$ where $\hat{M}$ is the universal cover of $M$. The superselection sectors correspond to irreducible representations of $\pi_1(M)$. For QFTs the problem of constructing the algebra of observables is in general open however certain cases (such as free QFT and I believe rational CFT as well) were solved. An approach emphasizing the algebra point of view is Haag-Kastler axiomatic QFT