Can I see separation of variables as a tensor product?

Question

Can I see separation of variables as a tensor product? For example, in a radial potential, the separation of variables brings to the solution $R(r)\Theta(\theta) \Phi (\phi)$. This sounds like an element of the space spanned by $|r\rangle\otimes|\theta\rangle\otimes |\phi\rangle$ where $R(r)$ lives in the space spanned by $|R\rangle$ and so on.

Answer 1

Yes, that is exactly what separation of variables is in terms of the Hilbert space - generally, we have that $L^2(X\times Y) = L^2(X)\otimes L^2(Y)$, i.e. the space of square-integrable functions on a Cartesian product is the tensor product of the square-integrable functions on the factors of the Cartesian product. In particular, $L^2(\mathbb{R}^n) = \bigotimes_{i=1}^nL^2(\mathbb{R})$, which corresponds to writing a function $f(x,y,z)$ as linear combinations of functions $f_x(x)f_y(y)f_z(z)$. The ansatz of "separation of variables" in these terms is nothing but the assumption that the solution is a simple tensor, i.e. the linear combination has only a single non-zero summand.

There is a slight subtlely for polar coordinates since they don't really cover all of $\mathbb{R}^3$, but the points where they are fishy is a zero measure set, so it doesn't really matter.