Schrodinger equation, commutative operators, and Symmetry

Question

When solving Schrodinger's equation in 3D with a spherical laplacian you reach a point at which you introduce a separation constant and can see that the same eigenvalue satisfies the radial and angular equations

$$\frac{d}{dr}\big(r^2\frac{dR}{dr}\big) - \frac{2mr^2}{\hbar^2}[V(r) - E]R = l(l + 1)R$$

and

$$\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\big(\sin \theta\frac{\partial Y}{\partial \theta}\big) + \frac{1}{\sin^2 \theta}\frac{\partial^2Y}{\partial\phi^2} = -l(l + 1)Y$$

This suggests that two operators, $\hat{R}$ and $\hat{\Omega}$ share the same eigenvalue. What is the signifigance of this? The pieces I am trying to fit together with this is commutativity of operators, symmetry, and separation of variables. I can't seem to get it all together and make enough sense of it. Is there a simple way of seeing how commutativity, symmetry and sep. of vars. work in this? Some have suggested Lie Algebra and groups but I have not studied this. Any help would be appreciated.

Answer 1

Except that they don't have the same eigenvalue. To understand it in terms of differential operators; divide the first equation by $1/r^2$, this gives you an eigenvalue equation for the total hamiltonian operator $H = -h^2\nabla^2/2m + V$, whose eigenvalue is clearly $E$. Then we divide $\nabla^2 = \hat{R} + 1/r^2 \hat{\Omega}$ in your notation (into radial and angular differentiation operators). Then its clear that as long as $V$ is isotropic $[H,\hat{\Omega}]=[\hat{R},\hat{\Omega}]=0$ while $[\hat{R},H]\neq 0$ because of the $1/r^2$ in front of $\hat{\Omega}$ as well as $V(r)$. So we can find simultaneously eigenvalue of $\hat{\Omega}$ which is $l(l+1)$ and of $H$ which is $E$, but not the eigenvalue of $\hat{R}$ because it doesn't commute with H.