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It is stated that commuting Hermitian (linear + self-adjoint) operators have 'simultaneous eigenstates'? Which of the following is most correct, and why?

2 commuting operators share AN eigenstate

2 commuting operators share SOME eigenstates

2 commuting operators share THE SET of all possible eigenstates of the operator

My intuition would be that 2 commuting operators have to share the EXACT SAME FULL SET of all possible eigenstates, but the Quantum Mechanics textbook I am reading from is not sufficiently specific. If I am correct, any relatively accessible mathematical proof, or at least conceptual explanation, would be much appreciated.

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If this is a physics book rather than a mathematics book, it is quite likely that (in the case of continuous spectrum) what it calls an "eigenstate" is not what a mathematician would call an eigenstate. Moreover, if the operators involved are unbounded (and thus not everywhere defined), you have to be careful with what you mean by "commuting": it is possible that the intersection of the range of operator $A$ and the domain of operator $B$ is $\{0\}$, so that $BA$ doesn't make sense on any nonzero vector. One sensible definition is this: (Reed and Simon, Methods of Modern Mathematical Physics I: Functional Analysis, sec. VIII.5)

Two (possibly unbounded) self-adjoint operators are said to commute if all the projections in their associated projection-valued measures commute.

It's certainly not true that the set of eigenstates is the same. For example, the identity operator $I$ commutes with any other operator $B$, and all nonzero vectors are eigenvectors of $I$, but that won't generally be true for $B$.

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