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In mathematical words, an observable is an operator that a set of linearly independent eigenfunctions constitutes a complete basis of the wave-functions' space.

Now, let's consider some observables: $O_i, i \in I$ (ex.: $P^2, L^2, S_z, ...$) and a multi-variable function $f$. Let $O = f(O_i, i \in I)$ (ex.: $O = L.M_S$).

Is $O$ an observable?

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Edit:

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I'm interested to the property that a hermitian operator has a complete set of eigenfunctions (that generate the whole wave-space : basis). Some authors (C. C. Tanoudji) call it the observability. Let's just call that property "completeness", to avoid confusion.

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  • $\begingroup$ Comment to the question (v2): OP's definition is ambiguous/incomplete as no operator ordering prescription is specified. E.g. if $f(x, y) = xy$, does $f(X,Y)$ stand for $XY$, $YX$, $\frac{1}{2}(XY+YX)$, or something else (which for commuting elements $x$ and $y$ reduce back to $xy$)? Hence the question Is $f(X,Y)$ an observable? cannot be uniquely answered. $\endgroup$ – Qmechanic♦ Aug 16 '15 at 7:27
  • $\begingroup$ Let's say that you need to define a proper quantization in order to translate classical real-valued functions into observables. Using Weyl quantization, for example, it is possible to do it for a large class of functions. $\endgroup$ – yuggib Aug 16 '15 at 7:32
  • $\begingroup$ It doesn't matter. The question is general. $\endgroup$ – aayyachi Aug 16 '15 at 7:32
  • $\begingroup$ @aayyachi The question suggested by the comments is how general an answer do you want... $\endgroup$ – yuggib Aug 16 '15 at 7:40
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    $\begingroup$ @aayyachi: How am I supposed to edit the English mistake if the English sentence doesn't make sense to me? And I wasn't critizing your background knowledge. $\endgroup$ – jjack Aug 18 '15 at 5:20
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No. For example, $f(x, y) = xy$ gives $f(x, p) = xp$ which is not an observable since $(xp)^\dagger = px \neq xp$.

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  • $\begingroup$ In fine structure calculation, we use $H=m_e c^2 + H_0 + W_{mv} + W_{SO} + ...$. $H_0$ is an observable. Should the perturbations be observables too? $\endgroup$ – aayyachi Aug 16 '15 at 7:41
  • $\begingroup$ @aayyachi you are making the common mistake of putting the cart in front of the horse. An observable is something that can be experimented with and measured, the horse. . The successful quantum mechanical model is successful for that reason, that there exists a, by hand, defined/postulated correspondence between measurable quantities and the mathematical operators. The mathematics is the cart. $\endgroup$ – anna v Aug 16 '15 at 7:51
  • $\begingroup$ @annav My skills in quantum mechanics are still fresh but my question is clear. $\endgroup$ – aayyachi Aug 16 '15 at 8:08
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Lets put it another way.

Every physical observable corresponds to a quantum mechanical (mathematical) operator, i.e. a (usually) differential that acts on the wave function. It is part of the postulates of quantum mechanics (page 2 in link) as wave mechanics given by the solutions of the Schrodinger equation. It is necessary that every physical observable corresponds to an operator. But being an operator is not a sufficient condition to generate an observable; mathematics is a tool to model nature, not a creator/generator of nature.

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  • $\begingroup$ You didn't understand my question. Some authors (C. C. Tanoudji) define the observability as the property that a hermitian operator has a complete set of eigenfunctions (that generate the whole wave-space : basis). Let's just call that property "completeness", to avoid confusion. $\endgroup$ – aayyachi Aug 16 '15 at 22:24
  • $\begingroup$ I am answering with "observables are defined by hermitian operators but hermiticity is not sufficient to define an observable", "necessary but not sufficient" , which last was a concept ingrained in my mathematical studies but seems to be missing in a lot of background to questions here, $\endgroup$ – anna v Aug 17 '15 at 3:39
  • $\begingroup$ What do you mean by: "seems to be missing in a lot of background to questions here"? Of course I know that "hermiticity" doesn't imply "observability" $\endgroup$ – aayyachi Aug 17 '15 at 20:33

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