Solution to the time-independent Schrödinger's equation for systems with spin I'm currently taking a course on quantum computing and we've just introduced the concept of spin in a not so very formal way and I'm no physicist, so please be gentle. Apparently, the solution to the time-independent Schrödinger's equation for, I guess, an electron-like particle has the following form:
$$\psi(\vec{x},s,t) = \psi(\vec{x})\otimes|s\rangle e^{-iEt/\hbar}$$
Where $s$ denotes the spin. My question is: on the right-hand side, why can we factorize the initial condition $\psi(\vec{x},s)$ as a tensor product of a spatial part and a spin part? I think the only assumption was that the spin contribution in the Hamiltonian is an additive term which depends only on $s$, and not on $\vec{x}$, e.g. in the case the particle is immersed in a (uniform?) magnetic field.
Possibly a duplicate of this, although I can't get much out of it.
 A: Let's say we consider a Hilbert space $\mathscr{H} \equiv \mathscr{H}_{\mathrm{A}} \otimes \mathscr{H}_{\mathrm{B}}$.
Here, $\mathscr{H}_{\mathrm{A}}$ and $\mathscr{H}_{\mathrm{B}}$ could be e.g. the Hilbert spaces of distinguishable particles; or if $ \mathscr{H}_{\mathrm{A}} = L^2(\mathbb{R}^3)$ and $ \mathscr{H}_{\mathrm{B}}= \mathbb{C}^2$, then $\mathscr{H}$ is the Hilbert space of a particle with spin $s=1/2$.
We want to obtain a solution of the time-independent Schrödinger equation:
$$H|\psi\rangle = E |\psi\rangle \quad , $$
where $H$ denotes the Hamiltonian. If it is of the form
$$H = H_{\mathrm{A}} \otimes \mathbb{I}_{\mathrm{B}} + \mathbb{I}_{\mathrm{A}} \otimes H_{\mathrm{B}} \quad ,$$
then an eigenstate of $H$ is given by the tensor product of eigenstates of $H_{\mathrm{A}}$ and $H_{\mathrm{B}}$, as we can easily verify: Let $|\varphi\rangle \in \mathscr{H}_{\mathrm{A}}$ and $|\sigma\rangle \in \mathscr{H}_{\mathrm{B}}$ denote some eigenstates of $H_{\mathrm{A}}$ and $H_{\mathrm{B}}$, respectively. We compute
$$H \left(|\varphi\rangle \otimes |\sigma\rangle \right) =  H_{\mathrm{A}} |\varphi\rangle \otimes \mathbb{I}_{\mathrm{B}} |\sigma\rangle + \mathbb{I}_{\mathrm{A}} |\varphi\rangle \otimes H_{\mathrm{B}} |\sigma\rangle \quad , $$
which, using that $H_{\mathrm{A}} |\varphi\rangle = E_{\varphi}  |\varphi\rangle$ and $H_{\mathrm{B}}|\sigma\rangle = E_{\sigma} |\sigma\rangle $, reduces to
$$H \left(|\varphi\rangle \otimes |\sigma\rangle \right) = \left(E_{\varphi} + E_{\sigma} \right)|\varphi\rangle \otimes |\sigma\rangle \quad.  $$
Finally, if $H$ is time-independent, then $$ \mathscr{H}\ni|\psi (t)\rangle \equiv e^{-i(E_{\varphi} + E_{\sigma})t} |\varphi\rangle \otimes |\sigma\rangle$$
solves the time-dependent Schrödinger equation, i.e. we find that
$$ i \, \partial_t |\psi(t) \rangle = H |\psi(t)\rangle $$
with $|\psi(0)\rangle =|\varphi\rangle \otimes |\sigma\rangle $.
