# 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)

• what do you mean with an hamiltonian like that? is there a direct sum instead of a tensor product? Mar 20, 2021 at 13:44
• it was supposed to be the tensor product of two operators acting in two different hilbert spaces, the momentum space and the spin space Mar 20, 2021 at 13:48
• 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? $$H = \frac{p^2}{2m} \otimes \mathbb{1}_{momentum} + \hbar \omega(|\uparrow\rangle\langle\uparrow|-| \downarrow\rangle\langle\downarrow|) \otimes \mathbb{1}_{spin}$$ Mar 20, 2021 at 13:58
• 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. Mar 20, 2021 at 14:25
• 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. Mar 20, 2021 at 15:21

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.