I Representation of Spin 1/2 quantum state

[PeterDonis]

It should mean that $$(I \otimes \sigma_z)(\vec{x} \otimes I) \ket{\psi} = (\vec{x} \otimes I) (I \otimes \sigma_z) \ket{\psi}, \forall \ket{\psi}$$

Yes.

We can check that it is true using the definition of tensor product operators acting on a generic tensor product of type $\ket{\psi} = \ket{\alpha} \otimes \ket {\beta}$.

Yes.

Since both $I \otimes \sigma_z$ and $\vec{x} \otimes I$ are hermitian and commute each other, they have at least a common eigenbasis -- see also Commuting Operators Have the Same Eigenvectors.

Yes.

However any eigenbasis of the first operator is an eigenbasis for the second (and the other way around) if and only if the commutating hermitian operators have both nondegerate eigenvalues -- see Common eigenfunctions of commuting operators.

Yes.

 

[cianfa72]

Ok, so coming back to the case of entangled position/momentum plus spin state, such a state can be written as linear combination of tensor products basis $\{v_i \otimes w_j\}$ that will be an eigenbasis of the hermitian/self-adjoint tensor product operator $\vec{x} \otimes \sigma_z$. Note that $\{v_i\}$ and $\{w_j\}$ are eigenbasis of operator $\vec{x}$ and $\sigma_z$ respectively.