Separable Hamiltonian systems in quantum mechanics Why is it that in separable Hamiltonian problems the total eigenfunction is equal to the product of the individual eigenfunctions, but the individual Hamiltonians must commute?
In mathematics, when the method of separating variables is used, for example for some PDEs, it is assumed that the total solution is product of functions of the individual variables.
But why, in quantum mechanics, must the individual Hamiltonians commute with each other in order to use this method of resolution?
I refer for example to the case of the hydrogen atom:
after the change of variables, one can write the total Hamiltonian as the sum of two Hamiltonians that commute with each other and the eigenfunction of $H$ is the product of the individual eigenfunctions.
In this case (relative coordinates hamiltonian of a 3D hydrogen atom in spherical coordinates) my book only states: "Since a central hamiltonian commutes with $L^2$ and $L_z$, we can write the solutions of TISE as: $\psi(r,\theta,\phi)=F(\theta,\phi)R(r)."$
 A: I think the other answer is (at least partially) mistaken. Let me first answer your question, and then explain why I have issues with the other answer.
Suppose you have a Hamiltonian of the form $H = H_1 + H_2$, where $[H_1, H_2] = 0$. Then, since $H_1$ and $H_2$ commute, they can be simultaneously diagonalized. That is, there exists an eigenbasis of the form $| \varepsilon_1, \varepsilon_2 \rangle$ where $H_1 | \varepsilon_1, \varepsilon_2 \rangle = \varepsilon_1 | \varepsilon_1, \varepsilon_2 \rangle$ and $H_2 | \varepsilon_1, \varepsilon_2 \rangle = \varepsilon_2 |\varepsilon_1, \varepsilon_2 \rangle$. An arbitrary state in your Hilbert space can be written in the form
$$
| \psi \rangle = \sum_{\varepsilon_1, \varepsilon_2} c_{\varepsilon_1,\varepsilon_2} |\varepsilon_1, \varepsilon_2 \rangle ,
$$
but if your goal is only to find energy eigenstates, you are allowed to assume that you are working in a common eigenstate of $H_1$ and $H_2$ separately. This is the complete answer to your question. Note that if $H_1$ and $H_2$ didn't commute, then there would not exist a common eigenbasis, and it would not be possible to factorize your wavefunction.
Now, here is the problem with the other answer: there is no reason to assume that $H_1$ and $H_2$ have tensor product structure. Separation of variables works even when the Hamiltonian is NOT of the form $H = H_1 \otimes 1 + 1 \otimes H_2$. The difference is that when $H$ has this tensor product structure, the eigenvalues $\varepsilon_1$ and $\varepsilon_2$ can be chosen independently, but this doesn't have to be the case; indeed, this is not the case for the hydrogen atom!! In the hydrogen atom, the allowed quantum numbers $(n,\ell,m)$ cannot all be chosen independently, for example $\ell$ is not allowed to exceed $n$. This is indicative of the fact that, while the Hamiltonian splits into two commuting pieces, it does not have the tensor product structure indicated by the other answer.
A: I think this is a mistake on the language of the textbook. The important part is that the Hamiltonian splits into a radial and angular part $H=H_r+H_\Omega$ in a way such that for separable wavefunctions $\psi(r,\Omega)=F(\Omega)R(r)$
$$H_r(\psi)=FH_r(R)\text{ and } H_\Omega(\psi)=H_\Omega(F)R.$$
In particular, this implies the two Hamiltonians commute. However, the converse is not true.
A mathematical language that is well suited for this is that of tensor products. What is done in the Hydrogen atom is to find a decomposition of the Hilbert space into a tensor product $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$ in which the Hamiltonian decomposes into something of the form $H=H_1\otimes \mathbb{1}_2+\mathbb{1}_1\otimes H_2$. Then a basis of eigenvectors of $H$ can be found by taking the tensor product of eigenvectors of $H_1$ and eigenvectors of $H_2$. It is a side remark of this that $H_1\otimes \mathbb{1}_2$ and $\mathbb{1}_1\otimes H_2$ commute and certainly the fact that they commute does not guarantee the tensor product structure before hand. For example, there are commuting operators in $\mathbb{C}^3$ but the latter does not admit a non-trivial tensor decomposition. To see this, note that the dimension of a tensor product is the product of the dimensions of its factors and 3 is a prime number.
