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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)

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  • $\begingroup$ what do you mean with an hamiltonian like that? is there a direct sum instead of a tensor product? $\endgroup$ Mar 20, 2021 at 13:44
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    $\begingroup$ it was supposed to be the tensor product of two operators acting in two different hilbert spaces, the momentum space and the spin space $\endgroup$ Mar 20, 2021 at 13:48
  • $\begingroup$ I don't understand the meaning of that hamiltonian, that is a kinetic energy multiplied by another term, that's odd. Do you mean the following hamiltonian? $$ H = \frac{p^2}{2m} \otimes \mathbb{1}_{momentum} + \hbar \omega(|\uparrow\rangle\langle\uparrow|-| \downarrow\rangle\langle\downarrow|) \otimes \mathbb{1}_{spin} $$ $\endgroup$ Mar 20, 2021 at 13:58
  • $\begingroup$ nope it's as above, we had many examples in lecture and in specific the relation $(\hat{A} \otimes \hat{B}) | \psi_1 \rangle | \psi_2 \rangle = (\hat{A} | \psi_1 \rangle) (\hat{B} | \psi_2 \rangle)$ holds, the hamiltonian was given to us in the meaning that the operators in the product act in the different hilbert spaces. $\endgroup$ Mar 20, 2021 at 14:25
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    $\begingroup$ Your Hamiltonian is malformed! The Hamiltonian is a generator of the evolution group, so it is a coproduct so it should present as a sum of the two tensor factors, but @Matteo crossed the labels of the identities. It is the group generators, the evolution operators which tensor multiply plainly. See this for angular momentum. $\endgroup$ Mar 20, 2021 at 15:21

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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.

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  • $\begingroup$ Thank you for clearing up my missconception about the hamiltonian. This answer has certainly helped me definitly! $\endgroup$ Mar 20, 2021 at 17:39

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