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:

$$ \hat{A} |{a} \rangle = a |a\rangle $$

$$ \hat{B} |b \rangle = b |b\rangle $$

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?

Answer 1

That depends on the Hilbertspace:

$A$ is defined only on the first Hilbertspace spanned by the $|a\rangle$ vectors. $B$ is similarly defined on a second Hilbertspace spanned by the $|b\rangle$ 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 $A \mapsto A\otimes 1$, 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 $|a\rangle \otimes |b\rangle$ are not in the correct space and there is no logical answer.