Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


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?

share|cite|improve this question
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
up vote 17 down vote accepted

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

share|cite|improve this answer
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

Without superselection rules to restrict the observables, any Hermitian operator is an admissible observable. The case of multiple identical systems is very important. Indeed, if the systems are really identical, only observables that are symmetric under the exchange of the systems are admissible. In such a case, technically speaking you should only consider observables that commute with all possible permutation operators (i.e., with the elements of the representation of the permutation group on the Hilbert space of the systems).

share|cite|improve this answer
Thank you for the answer. Indeed, for identical particles one takes as the Hilbert space as the quotient of the tensor product by the appropriate ideal. Do you know some bibliographic references showing that the only restrictions to a Hermitian operator to be observable are only these? – Cristi Stoica Feb 14 '12 at 9:34
You can impose other superselection rules, typically based on some symmetry, not necessarily the exchange symmetry (that applies only for multi-particle systems). I cannot recommend a reference based on my direct knowledge, but I'd say you might start from – Marco Feb 14 '12 at 18:51
Thank you, this helps. – Cristi Stoica Feb 14 '12 at 19:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.