Common eigenkets of spherically symmetric Hamiltonian

In a QM text it states: "Consider a spinless particle subjected to a spherical symmetrical potential. The wave equation is known to be separable coordinates, and the energy eigenfunctions can be written as $$\langle \mathbf{x'}| n,l,m \rangle = R_{nl}(r)Y_{l}^{m}(\theta, \phi)$$ where the position vector $\mathbf{x}'$ is specified by the spherical coordinates $r, \theta$ and $\phi$."

Can we consider that we obtain this by the following steps: $$|n,l,m \rangle = |n,l \rangle \otimes |l,m \rangle$$ hence $$\langle x'|n, l, m \rangle = (\langle \vec{r}| \otimes \langle \theta, \phi|)(|n, l \rangle \otimes |l, m \rangle) = (\langle \vec{r}| n, l \rangle)\otimes(\langle \theta, \phi | l, m \rangle) = R_{nl}(r)Y_{l}^{m}(\theta, \phi)$$

Is what I wrote nonsense or is there something there? The question was motivated by the fact that $|l,m\rangle = Y_{l}^{m}(\theta, \phi)$...hence what we have left is $R_{nl}(r)$.

Thanks.

• How did you define $\lvert n,l\rangle$ and $\lvert l,m\rangle$? In what spaces do these states live? – ACuriousMind May 10 '17 at 15:56
• @ACuriousMind $|l, m \rangle$ is the common eigenstate of $J_{z}$ and $J^{2}$ and $|n l \rangle$ is the leftover part from the radial equation :) What do you think, does that make any sense? – user100411 May 10 '17 at 17:03

• Can anything more be said about the vector $|n, l \rangle$ – user100411 May 15 '17 at 12:28