Unproven claim in Weinberg's textbook

[dEdt]

In his Lectures on Quantum Mechanics, Weinberg makes the following claim about solutions to the Schrodinger equation for a central potential: Suppose \psi(\mathbf{x},t) is an eigenfunction of H, \mathbf{L}^2, and L_z. According to Weinberg, "since \mathbf{L}^2 acts only on angles, such a wavefunction must be proportional to a function only of angles, with a coefficient of proportionality R that can depend only on r. That is, for all r,
\psi(\mathbf{x})=R(r)Y(\theta,\phi)."

He does not elaborate further on this, in my view, non-trivial statement. Can someone here provide a proof of his claim?

 

[Nugatory]

He does not elaborate further on this, in my view, non-trivial statement. Can someone here provide a proof of his claim?

I'd be inclined to work on the intuition behind that claim instead of looking for a rigorous proof, because the general idea is useful in so many other problems.

But first, which part of the statement are you asking about? The part about $L^2$ acting only on $\theta$ and $\phi$, or the claim that this implies the $R(r)Y(\theta,\phi)$ form of the solution?

 

[dEdt]

I'm talking about the claim that \psi=R(r)Y(\theta,\phi) given that \mathbf{L}^2 acts only on angles.

 

[dextercioby]

This is trivial once you consider the uniparticle Hilbert space for the 'dummy' particle 'moving' aroung the CoM. Its Hilbert space is L^2(R^3) = L^2 (R) X L^2 (R) X L^2 (R) and after performing a unitary transformation of the whole space you have that L^2(R^3) ~ L^((0,infinity),dr) X L^2 (S^2,dOmega).

In the position space the wavefunction of the 'dummy' particle would be then a product of a (sq integrable) function of r times a product of a function of spherical angles.