Eigenbasis of a complete set of commuting observables

Question

Suppose I have a complete set of two commuting observables $hat{A}, hat{B}$ for which:
$$ hat{A} |{a} rangle = a |arangle $$
$$ hat{B} |b rangle = b |brangle $$
Now, I can find a common eigenstate in the Hilbert space where $hat{A}$ and $hat{B}$ operates, or I can define a new Hilbert space which is the direct product of the space spanned by $|{a} rangle$ and $|{b} rangle$:
$${ |a,b rangle }_{H} = { |a rangle }_{H_1} otimes { |b rangle }_{H_2}$$
Where $H$ are Hilbert spaces, and with the operators extension $A to A  otimes  1$ ; $B to 1  otimes  B$.
In the new Hilbert space I can then choose the common eigenbasis as $|{a} rangle otimes|{b} rangle$.
Now there is a rationale in Quantum Mechanics to choose one method or another, or is a matter choice ? And if indeed is a matter of choice, what are the situation in which one may prefer a method over another?

Groggy · Answer

That depends on the Hilbertspace:
$A$ is defined only on the first Hilbertspace spanned by the $|arangle$ vectors.
$B$ is similarly defined on a second Hilbertspace spanned by the $|brangle$ vectors.
You ask about vectors in the tensor product of the two spaces, where neither $A$ nor $B$ is defined.
If you extend them by $A mapsto Aotimes 1$, than your statement is of course true, since the $A$ only "cares" about the first part of the tensor product.
If you want $A$ and $B$ to akt on the same hilbertspace, the vectors $|arangle otimes |brangle$ are not in the correct space and there is no logical answer.

Eigenbasis of a complete set of commuting observables

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