I'm studying quantum mechanic in particular tensor product and Hilbert space (for the first time). I have some doubts and I would like to check if I have understood correctly.
Factorization
The state of a particle can have many degrees of freedom and in order to describe the state of the particle respect one of them we use a vector in an appropriate Hilbert space.
When we want to describe the state respect more than one degree of freedom we use a vector that lives in a tensor product Hilbert space, for example when we describe spin and position of one particle: $$\mathcal{H}_{spin}\otimes \mathcal{H}_{position}$$ Or when we have two different independent particles: $$\mathcal{H}_{position1}\otimes \mathcal{H}_{position2}$$ Then if the degree of freedom are uncoupled (namely there aren't terms in the Hamiltonian that mix them) the eigenstates of the Hamiltonian can be factorized as a tensor product of two states that lives the proper Hilbert space, for example when we describe spin and position of one particle: $$S(spin)\otimes Ψ(r)$$ Or when we have two different independent particles: $$Ψ_a(r_1)\otimes Ψ_b(r_2)$$
Question
What I don't understand is why when we deal with one particle with spin and position the generic state is $S(spin)\oplus Ψ(r)$ and I have never seen something like: $$a\biggl(S_a(spin)\oplus Ψ_a(r)\biggr) +b\biggl(S_b(spin)\oplus Ψ_b(r)\biggr)$$ Instead when we deal with two particles systems writing $Ψ_a(r_1)\oplus Ψ_b(r_2)$ we mean a set of Eigenstates and the generic state is a linear combination of them like: $$a\biggl(Ψ_a(r_1)\otimes Ψ_b(r_2)\biggr) +b\biggl(Ψ_c(r_1)\otimes Ψ_d(r_2)\biggr)$$
Other rules
Eventually if the degree of freedom must respect some more rules we take as admissible eigenstates only a subset of the all possible eigenstates of the Hamiltonian. For example when we have two identical particle the states must be either symmetric or antisymmetric and so we have: symmetric eigenstates that span all the states of the bosons $$1/\sqrt{2}\biggl(Ψ_a(r_1)\otimes Ψ_b(r_2)\biggr) +1/\sqrt{2}\biggl(Ψ_a(r_2)\otimes Ψ_b(r_1)\biggr)$$ and antisymmetric eigenstates that span all the states of the fermions $$1/\sqrt{2}\biggl(Ψ_a(r_1)\otimes Ψ_b(r_2)\biggr) -1/\sqrt{2}\biggl(Ψ_a(r_2)\otimes Ψ_b(r_1)\biggr)$$