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1-In QM every observable is described mathematically by a linear Hermitian operator. Does that mean every Hermitian linear operator can represent an observable?

2-What are the criteria to say whether some quantity can be considered as an observable or not?

3-An observable is represented by an operator via a recipe called quantization if it has an analog in classical mechanics. If not, such as spin since it has no classical analog, do we use data from experiment to guess what this operator could look like? are there any other methods for finding that?

4-Are there other observables, besides spin, which also have no classical analog?

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I don`t think it is practical for physicists to postulate one - to - one correspondence between physical observables and self - adjoint operators, there would be too many observables !! In practice most applications of quantum mechanics involves only a finite number of observales such as energy, position, etc –  Serifo Blade Feb 16 '12 at 13:50
@Revo: If you allow external perturbations, you can measure any observable. On a quantum computer, you can measure any observable. In real life, it is very hard to measure an arbitrary observable. If someone presents QM as some set of "postulates", instead of by a series of examples of by physical reasoning, this person doesn't know what they're talking about and you should read another book. –  Ron Maimon May 8 '12 at 14:54
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3 Answers

up vote 4 down vote accepted

Questions 1,2:

An observable is an element that is obtained from experiments. You can take this as the definition of an observable. The fact that we make an operator and give it some properties does not change/influence the outcome of an experiment. It just so happens that the theory we have ascribes linear, hermitian operators to explain experiments. With this in mind, it is easy to say that not all linear, hermitian operators we cook up describe observables.

Question 3

Initially, the classical-quantum correspondence was used, but people quickly realized that it was of limited use. The modern view is that nature can be described by Group Theory (especially the Poincare Group) and everything that is observed follows from there. With this in mind, you don't have to guess about the existence of the Spin Operator, it comes up naturally. What is more important though, is the representations of the operator. When you relate theory and experiments, remember that you are dealing with the representations of an operator. An operator cannot be measured and is useless by itself unless you specify the basis.

Question 4

I don't know the answer to this, but I can tell you that we never measure spin by itself, but the interaction of a spin with something else. Why? In my view, that is the definition of a measurement.

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Don't know why this was downvoted. The final judge about a theory is by experiment. Physics is about science, not maths. Or let's say it with Feynman: No matter how smart your theory is or how smart you are: if it doesn’t agree with experiment it’s wrong –  Alex1167623 Feb 29 '12 at 10:31
I'm interested in the claim that "nature can be described by Group Theory". I guess that's only true if you use Group Theory in a very broad sense. I'd say it's a tool once you introduce space and time, etc., but things (space and time, spin) don't follow from the Poincare Group itself. There is the possibility for spin if you use group theory, but what does "you don't have to guess about the existence of the Spin Operator" really hold? –  NiftyKitty95 Mar 9 '12 at 11:31
@RonMaimon: Why do you ask? Did I make statements about what's more non-classical or more counterintuitive? I wasn't even talking about QM here in particular. I was using the spin observable as an example for something which is described by group theory and was asking why he says they would necessarily follow to exist in nature. His statement seems to imply all possible realizations would have to be present in nature. –  NiftyKitty95 May 8 '12 at 15:06
@NickKidman: Because I'm stupid, erased. –  Ron Maimon May 8 '12 at 15:08
+1: I don't know why I said this answer is wrong--- it's fine. –  Ron Maimon May 8 '12 at 22:28
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In principle, every Hermition operator can be an observable in the sense this term is used in quantum mechanics. For finite systems (edit: i.e., those with a finite-dimensional Hilbert space) I have seen theoretical results that prove measurability of any Hermitian operators, though I couldn't locate a reference.

But measuring an observable becomes very difficult when it is a contrived operator rather than one of those you find typically discussed.

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Does "finite system" here mean that there is a number of operators $O_1,O_2,\dots,O_n$ and if you know their spectrum, lets denote it $S^o$, then you can show that the spectra of all operators in the system can really be written down a functions of elements of $S^o$? –  NiftyKitty95 Mar 9 '12 at 11:26
finite system = finite-dimensional Hilbert space. –  Arnold Neumaier Mar 9 '12 at 12:17
@NickKidman: An operator is a matrix, so in a finite dimensional Hilbert space you can do this with linear combinations of a finite basis. –  Ron Maimon May 8 '12 at 14:50
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The parity operator, the unitary operator which implements reflections on the wavefunction is both next to impossible to measure, and has no classical analog. This operator is unitary, but its real and imaginary parts are Hermitian.

The spin is not nonclassical, is the spinning particle. It's not like parity and other discrete symmetry observables.

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