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In fine structure calculation we use the perturbation theory. The basic Hamiltonian $H_0$ is perturbed as:

$H = H_0 + W$

First, the basic problem assume that $H_0$ is an observable. That allows to consider the set of linearly independent eigenfunctions as a complete basis of the wave-functions' space.

The Hamiltonian $H$ of the complete problem (perturbed problem) should normally be an observable too.

Should the perturbations $W$ be an observable too (in math. view)?

Note:

In C. C. Tanoudji, an observable is defined mathematically as a hermitian operator such that there is a complete basis of eigenfunctions for the wave-space. Let's just call it a "complete operator" to avoid confusion.

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  • $\begingroup$ Crazy people can only down score important questions to witch they can't answer or they can't understand. :( $\endgroup$ – aayyachi Aug 16 '15 at 18:37
  • $\begingroup$ In C. C. Tanoudji, an observable is defined mathematically as a hermitian operator such that there is a complete basis of eigenfunctions for the wave-space. $\endgroup$ – aayyachi Aug 16 '15 at 18:55
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Only the full hamiltonian is observable, as in "corresponds to a physical quantity that can be observed."

The "free" and "perturbed" parts are a convenient split when we do a calculation but are not separately observable. In fact it would be surprising if they were, since the split between free and perturbed was arbitrary, it is our choice how to make the split.

This conceptual distinction is important, it's always important to keep in mind what quantities are unambiguous and physical and which contain some arbitrariness, usually to facilitate a calculation. As you progress in physics it becomes increasingly important to be aware of the differences, for example the fact that the free and perturbing Hamiltonian are not separately measurable becomes extremely important in quantum field theory when you discuss renormalization.

Having said all of that I should point out that, almost always, the free and perturbed Hamiltonian (as well as the full Hamiltonian) are hermitian operators. But I wouldn't say this makes them observable quantities.

I suppose if you wanted to be somewhat perverse and say all hermitian operators were "Observables" then technically the free and perturbed hamiltonians are "Observables," but I think this is a misleading use of language.

(All oberservable quantities are represented by hermitian operators of course, but typically when people talk about Observables they also mean a quantity that you can really measure, which is a more restrictive condition than just hermiticity.)

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Yes. If $A$ and $B$ are observables, then so is $A+B$. In this case, since $H_0$ and $H$ are observables, so is $H-H_0 = W$.

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