Question regarding product state and Schrödinger equation Assume we look at the state $|\Psi(t) \rangle = |\psi(t)\rangle |m(t)\rangle$ with the Hamiltonian acting in the two Hilbert spaces for example: $$\hat{H} = \frac{\hat{p}^2}{2M} \otimes (| \uparrow \rangle \langle \uparrow | - | \downarrow \rangle \langle \downarrow |)$$ then the Schrödinger equation after applying the product rule looks something like: $$i \hbar (\frac{\partial}{\partial t}| \psi(t) \rangle  | m(t) \rangle + | \psi(t) \rangle  \frac{\partial}{\partial t} | m(t) \rangle) = \frac{\hat{p}^2}{2M} |\psi(t) \rangle  \otimes  (| \uparrow \rangle \langle \uparrow | m(t) \rangle - | \downarrow \rangle \langle \downarrow | m(t) \rangle)$$ which leads me to the question of how u would solve this Schrödinger equation with the product states for any given initial value for the amplitudes? (assuming that the first state is a continuum state and the second one is discrete)
 A: You are confusing a group with its generator, cf this answer for rotations versus angular momentum operators. So your hamiltonian is badly malformed as you wrote it. Your A and B comment example holds for evolution operators in the respective spaces, and therefore, distinctly not Hamiltonians, their generators!
Let me write down the correct expressions first, and you could reassure yourself how they fit.
$$
H_1=\hat p ^2/2M, \qquad H_2= \sigma_z, \\
|\Psi(t)\rangle = |\psi(t)\rangle \otimes |m(t)\rangle =e^{-iH_1 t/\hbar}|\psi(0)\rangle \otimes e^{-iH_2 t/\hbar}|m(0)\rangle\\
=(e^{-iH_1 t/\hbar}\otimes e^{-iH_2 t/\hbar}  ) (    |\psi(0)\rangle \otimes |m(0)\rangle\\ =e^{-i(H_1  \otimes {\mathbb 1}_2 +  {\mathbb 1}_1\otimes  H_2 )t/\hbar}  ~~ (    |\psi(0)\rangle \otimes |m(0)\rangle) \leadsto \\
H= H_1  \otimes {\mathbb 1}_2 +  {\mathbb 1}_1\otimes  H_2 , \\ e^{-iH t/\hbar} |\Psi(0)\rangle= |\Psi(t)\rangle.
$$
The total hamiltonian is a coproduct, and you may differentiate the third line w.r.t. t, to confirm the correct Leibniz rule in the Schrodinger equation.
