Separable Hamiltonians evolve independently

When we have a Separable Hamiltonian $H(x_1,p_1,x_2,p_2)=H_1(x_1,p_1)+H_2(x_2,p_2)$, we can solve the 2 sub-systems independently to obtain the joint solution.

Proof

We have that $[H_1,H_2]=0$ since $H_1$ and $H_2$ act on different components of the Tensor product. As such we can find their simultaneous eigenstates, which are given by

$$\begin{array}{cll}H|E⟩& =(H_1+H_2)(|E_1⟩\otimes |E_2⟩)& \text{since addition distributes over the tensor product}\\ & =H_1(|E_1⟩\otimes |E_2⟩)+H_2(|E_1⟩\otimes |E_2⟩)& \text{Since Hi only operates on it’s "component"}\\ & =E_1|E_1⟩\otimes |E_2⟩+|E_1⟩\otimes E_2|E_2⟩& \\ & =E_1(|E_1⟩\otimes |E_2⟩)+E_2(|E_1⟩\otimes |E_2⟩)& \\ & =(E_1+E_2)(|E_1⟩\otimes |E_2⟩)& \\ & =E|E⟩& \text{where E:=E1+E2 and ∣E⟩:=∣E1⟩⊗∣E2⟩=∣E1E2⟩ }\end{array}$$