Solution to the time-independent Schrödinger's equation for systems with spin

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: $$H|psirangle = E |psirangle quad , $$ 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 $|varphirangle in mathscr{H}_{mathrm{A}}$ and $|sigmarangle in mathscr{H}_{mathrm{B}}$ denote some eigenstates of $H_{mathrm{A}}$ and $H_{mathrm{B}}$, respectively. We compute

$$H left(|varphirangle otimes |sigmarangle right) = H_{mathrm{A}} |varphirangle otimes mathbb{I}_{mathrm{B}} |sigmarangle + mathbb{I}_{mathrm{A}} |varphirangle otimes H_{mathrm{B}} |sigmarangle quad , $$

which, using that $H_{mathrm{A}} |varphirangle = E_{varphi} |varphirangle$ and $H_{mathrm{B}}|sigmarangle = E_{sigma} |sigmarangle $, reduces to $$H left(|varphirangle otimes |sigmarangle right) = left(E_{varphi} + E_{sigma} right)|varphirangle otimes |sigmarangle quad. $$

Finally, if $H$ is time-independent, then $$ mathscr{H}ni|psi (t)rangle equiv e^{-i(E_{varphi} + E_{sigma})t} |varphirangle otimes |sigmarangle$$ 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 =|varphirangle otimes |sigmarangle $.