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It seems clear that every observable in QM can be represented by some Hermitian operator in Hilbert space.

Does the reverse also hold? That is, for every Hermitian operator $O$ in Hilbert space, is there some observable represented by $O$?

Are there any arguments for or against a particular answer to this question?

Lory
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1 Answers1

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Hermitian Operators = Observables?

No, not in general.

It seems clear that every observable in QM can be represented by some Hermitian operator in Hilbert space.

An observable is a special kind of Hermitian operator. So it is a trivial statement that an observable is a Hermitian operator. It's "trivial" for the same reason that saying "a blue car is a car" is trivial.

Does the reverse also hold? That is, for every Hermitian operator $O$ in Hilbert space, is there some observable represented by $O$?

No.

Are there any arguments for or against a particular answer to this question?

Yes. The definition of an observable.

An observable is, by definition, a Hermitian operator that possesses a complete, orthonormal set of eigenfunctions. See, for example, Messiah "Quantum Mechanics Volume 1" (1961) at p. 188.

I will provide the same example that Messiah gives in his footnote on page 188.

Consider the operator $\hat O = -i\frac{d}{dx}$ acting on square-integrable functions defined on the semi-axis from $0$ to infinity, which vanish at $x=0$. For such functions $\hat O$ is Hermitian, since: $$ \langle\psi_1|\hat O\psi_2\rangle = \int_0^\infty dx \psi_1^*(x)\left(-i\frac{d\psi_2(x)}{dx}\right) $$ $$ =+i \int_0^\infty dx\frac{d\psi_1^*}{dx}\psi_2(x) = \int_0^\infty dx \left(-i\frac{d\psi_1(x)}{dx}\right)^*\psi_2(x) $$ $$ =\langle\hat O\psi_1|\psi_2\rangle\;. $$

But, unfortunately, the operator $\hat O$ has no eigenfunctions. The functions of the form $e^{ikx}$ that would be the eigenfunctions do not vanish at $x=0$. So, given the restriction to the semi-axis from $0$ to infinity and functions vanishing at $x=0$, there are no permissible eigenfunctions of the $\hat O$ operator. Thus, for this situation, $\hat O$ is a Hermitian operator, but not an observable.

hft
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  • Thanks, I think I should have clarified the meaning of 'observables.' Of course, one can define the term purely theoretically, but one could also use it to denote the observable or measurable quantities, which then cam be represented by, rather than identified with, Hermitian operators. – Lory Apr 04 '23 at 02:18
  • Then the question is whether for every Hermitian operator (or every Hermitian operator with a complete orthonormal set of eigenfunctions), there is an observable quantity represented by that operator, which doesn't get answered by the technical definition of observables. – Lory Apr 04 '23 at 02:21
  • The word "observable" has a specific meaning in the context of quantum mechanics. If you are going to use it in a different way then you should explain how you are using it in the body of your question. But, based on what you are saying here in the comments, it seems like the answer is still no. – hft Apr 04 '23 at 02:58
  • That's fair enough. So, again, by an 'observable' I meant any physical quantity that can be measured at least in principle (even if it's practically unmeasurable due to technical reasons). Given this definition, are you saying that there are some Hermitian operators such that (i) they have a complete orthonormal set of eigenfunctions and yet (ii) there are no associated physical quantities that are in principle measurable? – Lory Apr 05 '23 at 02:41