# 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?

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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

I think the fundamental object is quantum mechanics is not the Hilbert space and operators on it but the C*-algebra of observables. In this picture the Hilbert space appears as a representation of the algebra. Different irreducible representations are different superselection sectors. The answer to "which operator is observable" is thus simple: the observable operators are those that come from the algebra. Indeed its better to think of observables as self-adjoint elements of the algebra rather than as operators

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

From the point of view of deformation quantization the quantum observable algebra is a non-commutative deformation of the algebra of continuous (say) functions on the classical phase space. This point of view is not fundamental but it's useful. For example it allows to understand different values of spin and different quantum statistics as superselection sectors

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Thank you, this is very useful. –  Cristi Stoica Feb 14 '12 at 19:13
-1: This is useless. It doesn't answer the question, and is just a collection of impressive sounding trivial gibberish. You didn't say which observables could be measured, you defined "superselection sector" in the maximally unenlightening way, and your example of "different spin values" is wrong and the stuff you talk about is not useful. This is just terrible. –  Ron Maimon May 8 '12 at 15:19