# Doubts about the use of tensor product In quantum mechanics

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 $$1/\sqrt{2}\biggl(Ψ_a(r_1)\otimes Ψ_b(r_2)\biggr) -1/\sqrt{2}\biggl(Ψ_a(r_2)\otimes Ψ_b(r_1)\biggr)$$

## 1 Answer

Your judgement is correct: states such as $$a ( S_a(spin) \otimes \Psi_a(r) ) + b ( S_b(spin) \otimes \Psi_b(r) )$$ are perfectly well allowed and indeed they are commonly found in nature. The reason you haven't seen them yet is presumably because you are still at the introductory level, and indeed many textbooks are a little weak on this, in that they imply that spin state can always be factorised out, which is wrong.

The spin/space entangled state shown above describes the beam coming out of a Stern Gerlach magnet, for example. Also, it describes what really goes on amongst the electrons in any atom, although elementary treatments often keep this a bit hidden.