Does every hermitian operator represent a measurable quantity?

In Quantum mechanics, observables are represented by hermitian operator. But does every hermitian operator represent a observable? If not , how do we know that whether a hermitian operator represent observable or not? What is the precise definition of the term "observable"?

-
Possible duplicates: physics.stackexchange.com/q/27038/2451 and links therein. –  Qmechanic Aug 27 '13 at 6:14

Given a quantum system with associated Hilbert space $\mathcal H$, the set of all self-adjoint bounded operators is $\newcommand{\bh}{\mathcal B(\mathcal H)_\text{sa}}\bh$. In general, only a small subset of $\bh$ will represent physically observable operators. For infinite-dimensional systems, $\bh$ is huge and there's no hope ever finding experiments for all its members; even in finite-dimensional systems it is very challenging to find experimental schemes sensitive to even a vector-space basis for $\bh$.

The physical approach to this is to begin with a finite set of operators which you know you can measure. For a single free particle, for example, you'd take position and momentum; for a finite set of spins you'd take all their Pauli matrices. You then form the set $\mathcal A$ of all operators that can be formed from them via products and linear combinations, which has the structure of a $\mathcal C^\ast$ algebra, and that is your set of physical observables. The $\mathcal C^\ast$algebra itself is the really fundamental description of the system; the Hilbert space is simply one possible representation.

In this formalism, states are functionals on $\mathcal A$: they are functions $$\rho:\mathcal A\rightarrow \mathbb C$$ that take an observable and give its measured value (or probable measured value, etc.) in that state. (In a Hilbert space representation, each such functional is associated with a density matrix $\hat\rho$, a trace-class positive operator such that $\rho(A)=\text{Tr}(\hat\rho\hat A)$ for $\hat A$ the Hilbert space operator associated with an arbitrary $A\in\mathcal A$.

A good reference for this is

'What is a Thing?': Topos Theory in the Foundations of Physics. A. Döring and C. Isham. In New Structures for Physics, ed. B. Coecke, Lecture Notes in Physics, vol. 813, 2011, pp 753-937. arXiv:0803.0417 [quant-ph].

Blatant advert: A. Döring is currently lecturing about this at the QuICC 2013 Summer School on Quantum Information, Computing and Control; stay tuned for the lecture videos.

-
While researching my own answer, I didn't quite understand one thing: how does one rule out hidden observables in this scheme? Is there any definitive proof or experiment that can be performed to finally say 'Yes our commuting set is complete'? Does the existence or non-existence of these hidden observables have anything to do with the EPR paradox? –  dj_mummy Aug 27 '13 at 14:35
@dj_mummy I should think the possibility of extra observables can never be conclusively ruled out. However, as they would be fully quantum observables, they would not help in resolving EPR paradoxes: they are either local, and thus within Bell's treatment, or entangling, in which case the locality assumption is broken. –  Emilio Pisanty Aug 27 '13 at 15:06
Thanks, now things are a lot clearer for me. –  dj_mummy Aug 27 '13 at 15:07
@EmilioPisanty I'm confused. In standard courses on quantum mechanics, we usually consider certain unbounded, self-adjoint operators to be observables, but presumably $\mathcal B(\mathcal H)_\mathrm{sa}$ does not contain these beasts. What am I missing? –  joshphysics Aug 27 '13 at 18:13
Emilio and @joshphysics . It would be great to get a reference: am I missing something (probably!)? Isn't resolving an operator into its unbounded-weight sum of always bounded projectors spectral decomposition itself - so your comment seems a bit like question begging. I don't doubt you're likely to be right (I've seen a good number of your posts!) - it's just that my (and probably joshphysics's) conception of spectral theory is of something that gets lots of fiddly bits when we wander into unbounded operator lands. I would love to see better versions and it sounds like you've glimpsed them! –  WetSavannaAnimal aka Rod Vance Oct 4 '13 at 0:29

Edit: My examples had consistency problems:

1) I considered unity to be a counter example of the OP's statement.

2) My second example was originally meant to be an incomplete commuting set of observables {O}, amended with an Hermitian operator K that represented the missing observable to close it. I expressed it incorrectly.

3) In any case the existence of such K to close {O} requires it to not be a dummy by definition of a complete quantum measurement.

4) Now that I understand the OP's question, my answer is:

We start with our experiments and write out all the observables known and from experiments we also decide whether they commute or not and associated structures. From here we construct C*-algebra of observables with an associated binary operation (composition) and a unit element. So every operator is hermitian and represents a measurable quantity.

If we discover newer observables and their associated properties through experiment, we will define a new set of all possible Hermitian observables and their binary operations such that they satisfy the C*-algebra.

-
What is the precise definition of the term observable? How do you know a hermitian operator is observable or not? –  user774025 Aug 27 '13 at 6:29
-1: I don't know the answer to this question, but judging by the multitude of responses to similar questions on physics.SE and the nuances they demonstrate, I don't think it's this simple. –  joshphysics Aug 27 '13 at 7:16
@joshphysics I agree that it is far more nuanced than I have shown. But a lot of the nuances are more axiomatic in nature. In that sense my answer is outdated. I took the idea from Dirac's 'Principles of Quantum Mechanics'. This early work had little use for too many axioms and QM was tailor-made to suit the needs of the hour. I have added a note to my answer to prevent misunderstanding –  dj_mummy Aug 27 '13 at 7:48
I think the question is legitimate (although duplicate). The one who asked did not made the confusion you claim. The first example you give, the identity operator, is not an example of a hermitian operator that doesn't have physical correspondent. It corresponds to "no measurement", and the observed system is unaffected by it, because the corresponding projection operator is the identity. The other example is actually the same, since, by definition, if the set of observables is complete, you can't add another operator, except the identity. So I see no counterexample here. –  Cristi Stoica Aug 27 '13 at 9:52
I made major edits to the answer. Hopefully that will be satisfactory. I will remove my first comment to prevent further confusion. –  dj_mummy Aug 27 '13 at 11:59