# Eigenbasis of a complete set of commuting observables

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?

• The question is somewhat circular, since, in order to have two operators commuting, you need to define them on the same basis. Even though I do agree that asking under which conditions a basis can be represented as a product basis makes sense. Oct 6, 2020 at 11:58
• I agree with @Vadim that there is a problem with the question. Neither A nor B acts on a product space, so their common basis can't be written as a product. Oct 6, 2020 at 12:02
• Note that you don't need to explicitly reference edits to the post. There is an edit history available for those who are interested. Oct 6, 2020 at 12:46

$$A$$ 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.
• Ok, I understand what you say from a mathematics point of view, but in general what is the ''common'' method in QM ? Suppose I have a quantum isotropic harmonic oscillator in two dimension. If I choose $(n_x, n_y)$ as quantum numbers I say that the complete set of observables are the hamiltonians $(\hat{H}_x, \hat{H}_y)$. Now, there is an intrinsic reason for why do consider usually the direct product of the eigenbasis of the two operator, or is just a choice ? Oct 6, 2020 at 12:24