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Sakurai (Modern Quantum Mechanics, by J.J. Sakurai) states in the section on compatible operators:

Let us first consider the case of compatible observables A and B. As usual, we assume that the ket space is spanned by the eigenkets of A. We may also regard the same ket space as being spanned by the eigenkets of B

Under what conditions can we say that the space spanned by the eigenkets of A can be spanned by that of B? If this is true only for commuting operators (compatible observables), how can we derive the truth of this fact from the definition of compatible observables, i.e. [$A,B] = \hat{0}$?

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    $\begingroup$ Assuming the observables are defined on the same space, why would you need commutativity to say that their eigenkets span the same space? You can just apply the spectral theorem to each separately, no? $\endgroup$
    – ACuriousMind ♦
    Commented Feb 4, 2017 at 14:29
  • $\begingroup$ Yes, but are we assuming here that the operators are defined on the same domain space? They should, I guess.. $\endgroup$
    – Arkya
    Commented Feb 4, 2017 at 14:31
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    $\begingroup$ If you're not already assuming that the operators are defined on the same domain, vanishing commutator won't help you much because the commutator would only be defined on some type of an intersection of the domains and its vanishing nature would only be applicable in that intersection. So, I don't see how a vanishing commutator can help us claim that the operators share the same domain. $\endgroup$
    – user87745
    Commented May 17, 2020 at 23:11
  • $\begingroup$ Thanks. After revisiting this question, I realize I had misunderstood what was being said in the quoted excerpt. $\endgroup$
    – Arkya
    Commented May 19, 2020 at 0:15

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I don't think Sakurai is talking about the domains of the operators, because claims about their commutator wouldn't help us determine if they share the same domain or not.

However, assuming they share the same domain, the eigenstates of each operator would span the domain because observables are Hermitian operators and the eigenstates of a Hermitian operator form a complete basis. Again, compatibility is irrelevant here, Hermiticity is the relevant thing.

What would be special when the observables are compatible is that they can share the same set of eigenstates that span the domain. In other words, they can be simultaneously diagonalized.

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This is how I think about it.

There is only one space in question here, the Hilbert space. My system's state depends entirely on the state vector it is in, and this state vector belongs in the Hilbert space.

Now, how do I define this space? I define it by saying that it is the basis formed by the eigenkets of an operator(observable), any operator.

After all, when I try to measure the value of that observable, it causes the system to land in another state vector (which we know should be an eigenket), therefore it makes sense that if two operators are being used on the same system, they should span the same space.

This is therefore not at all related to the compatibility.

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    $\begingroup$ A priori, therefore, we need to know what observable(s), we're interested in? If that is the case, what if there's more internal structure to the space? For example, spin was hypothesized initially because hydrogen spectrum was seen to have fine structures which couldn't be explained within the original Hilbert space. On a similar vein, does our hypothesis of a Hilbert space depend on experiment? This seems slightly unsettling to me. $\endgroup$
    – Arkya
    Commented Dec 10, 2017 at 6:04

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