Is the Hilbert space $S$ of states for a particle the tensor product of spaces generated by each observable eigenvectors?

Question

There is a nice answer to this question by @joshphysics but it applies to multiple quantum systems combination. I was wondering whether this could also be applied to a single quantum system? For example, the Hilbert space $S$ of states for a single particle (the space abstraction including its spin, position, momentum, charge, energy, ...) is the tensor product of the Hilbert spaces generated by the basis (eigen-) vectors of each observable considered in the $S$ space.

Answer 1

The answer is negative if I correctly understand the question. There are observables which do not commute, whereas if each of them could be viewed in its own Hilbert space and the whole Hilbert space were the tensor product of all those Hilbert spaces, these observables would commute.

What it is true is that there is a maximal set of pairwise commuting observables whose common eigenvectors span the whole Hilbert space (the rigorous mathematical statement is a bit different and uses the direct integral, but the physical substance is that). This fact can be reformulated in terms of tensor product if necessary, even if the use of the tensor product is not necessary...