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Can we use eigenstates of ANY observable as base of the Hilbert space? If we can, is this equal to the statement that those eigenstates are orthogonal to each other and normalizable?

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    $\begingroup$ Yes. (Assuming you know the correct definition of "observable") $\endgroup$ – ACuriousMind Jul 16 '14 at 13:38
  • $\begingroup$ As far as I know, observable is something that we are interested in and we want to measure, like position, momentum and energy. What do you you mean by "correct definition" ? did I miss something ? Sorry, I'm new in QM and a lot of things are unclear. :) $\endgroup$ – Mr. an Jul 16 '14 at 13:46
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An observable is a self-adjoint operator $\mathcal{O}$ on the Hilbert space of states $\mathcal{H}$.

The spectral theorem tells us that such an operator has an orthonormal basis of eigenvectors in $\mathcal{H}$, if it is compact.

If it is not compact, we have to "enlarge" the Hilbert space to something called rigged Hilbert space or Gelfand tripel. A good discussion of that is in Mathematical surprises and Dirac's formalism in quantum mechanics. Most introductory courses ignore this, however, and also assign "eigenvectors" to the non-compact operators like the position operator. It often works, but one must bear in mind that some non-sensical results coming from that really only occur because one has not been rigorous.

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  • $\begingroup$ You should read this answer or this other answer. For instance, the identity operator is self-adjoint, but does not correspond to any observable. $\endgroup$ – Trimok Jul 17 '14 at 10:35
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    $\begingroup$ @Trimok: I know. But I didn't say anything wrong, did I? That not every self-adjoint operator is an observable does not invalidate saying that every observable is a self-adjoint operator. The question here was not what the algebra of observables is, but whether we have an orthonormal basis for every observable. (though my comment above might have raised your expectation that I would discuss the $C^*$-algebras, I admit) $\endgroup$ – ACuriousMind Jul 17 '14 at 10:42
  • $\begingroup$ Yes, you are right, this was not really the question. $\endgroup$ – Trimok Jul 17 '14 at 10:51
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Unfortunately not every operator, which is used to get an observable gives you enough elements to make a base of the Hilbert space. As far as I know every eigenstate is orthogonal to each other. But the “length” could be arbitrary. Mostly used eigenstates in quantum mechanics are normalized for better usage.

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    $\begingroup$ I believe everything in this answer is correct, but could you elaborate with some technical considerations? Like, if a Hilbert space is a complete, normed, vector space, how can you be sure that observables don't form some kind of subHilbert space? Or, what is condition for observables to be pure states, that might help too. $\endgroup$ – levitopher Jul 16 '14 at 14:02
  • $\begingroup$ Every eigenstate is only orthogonal to another eigenstate if the eigenvalues of the two differ - which isn't guaranteed exactly because one operator isn't (usually) the complete set of commuting observables. $\endgroup$ – Luboš Motl Jul 17 '14 at 15:15

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