The question is the one of the title, let $\hat{O}_1$ and $\hat{O}_2$ two commuting operators: $[\hat{O}_1,\hat{O}_2]=0$, there is an orthonormal basis formed by their simultaneous eigenstates.
These eigenstates could be written as $|\epsilon_1,\epsilon_2\rangle$ so that $\hat{O}_1|\epsilon_1,\epsilon_2\rangle=\epsilon_1|\epsilon_1,\epsilon_2\rangle$ and $\hat{O}_2|\epsilon_1,\epsilon_2\rangle=\epsilon_2|\epsilon_1,\epsilon_2\rangle$. Are $|\epsilon_1,\epsilon_2\rangle=|\epsilon_1\rangle\otimes|\epsilon_2\rangle?$
Quantum Mechanics – Are Simultaneous Eigenstates Always a Tensorial Product of Two Eigenstates?
eigenvectorsquantum mechanics
Best Answer
No. A simple counterexample to this is taking a spin-1/2 representation of the rotation group $H_{1/2}$ and a spin-1 representation $H_1$ - the direct sum $H_{1/2}\oplus H_1$ is a five-dimensional Hilbert space in which $S^2$ and e.g. $S_z$ commute, and so it has a basis of common eigenstates $\lvert 1/2,1/2\rangle, \lvert 1/2,-1/2\rangle,\lvert 1,1\rangle, \lvert 1,0\rangle, \lvert 1,-1\rangle$. But since it is five-dimensional and 5 is prime, it is not the tensor product of any other space except in the trivial way (tensoring with the one-dimensional Hilbert space).
More explicitly, if we had $\lvert 1/2,-1/2\rangle = \lvert 1/2\rangle_{S^2}\otimes\lvert -1/2\rangle_{S_z}$ and $\lvert 1,1\rangle = \lvert 1\rangle_{S^2}\otimes\lvert 1\rangle_{S_z}$, then we would also have to have a state like $\lvert 1\rangle_{S^2}\otimes \lvert 1/2\rangle_{S_z}$, i.e. with eigenvalue 1 for $S^2$ and $1/2$ for $S_z$, but of course there is no such state in $H_{1/2}\oplus H_1$.