# In a spherically symmetric central potential why do we look for eigenfunctions of the angular momentum operator?

Well, to put it shortly, it's just because $$L^2$$ and $$L_z$$ are two observables that have no $$r$$ dependence. Since the kinetic terms of $$H$$ can be written as functions of $$r$$ and $$L^2$$, and the potential depends only on $$r$$, it is clear that the Hamiltonian commutes with the angular momenta. As such, it makes sense to write down $$Y$$ in terms of these two.

But if we'd like to go a little deeper, it is a bit related to representation theory. The fact that $$L^2$$ and $$L_z$$ form a complete description of vector rotations (they represent the Lie algebra $$\mathfrak{so}(3)$$ of the Lie group $$\mathrm{SO}(3)$$) and therefore are the most natural way to express the angular dependence of a rotation-invariant space in terms of orthonormalizable functions.

After reading @ZeroTheHero 's comment, i figured it would be instructive to provide some intuition as to why angular momentum does not depend on radius. Remember the classical definition

$$\mathbf L=\mathbf r\times\mathbf p$$

Let's then give and estimate of $$L$$. Suppose an atom or something has some characteristic radius $$R$$, and the electron moves with some characteristic momentum $$p$$. Then we estimate

$$L\sim Rp$$ But the de Broigle relation gives us an estimate of momentum in terms of the characteristic length

$$p\sim\frac{h}{R}$$

Where $$h$$ is Planck's constant. As such, out estimate for angular momentum amounts to

$$L\sim h,$$

Which has no explicit dependence on characteristic lengths of the system. As such it is reasonable to say it has no $$r$$ dependence, even though I've not given a rigorous proof.

One objective of quantum mechanics is to fully labelled states of the system. In some cases energy is not enough: for instance in the hydrogen atom some energy levels occur more than once; the same happens in the 3d harmonic oscillator. In such cases, simply giving the energy is not enough to completely identify the state so we look for other quantum numbers'' to refine the identification.

Of course one would prefer to label states with constants of motion, since they are constant so what we call state $$n\ell m$$ at the start we can use the same name at the end. Hence the idea of using a complete set of commuting operators - the Hamiltonian and others that commute with it - so as to uniquely identify the quantum states with constants of motion.

Of course when the potential is spherical $$V(r)$$, what enters in the radial part of the Schrödinger equation is actually the effective potential $$V_{eff }=V(r)+\frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2} \tag{1}$$ and so you see that the radial Schrödinger equation actually explicitly contains an angular momentum part so it is quite natural to look for functions $$\psi(r,\theta,\phi)$$ which are also eigenfunctions $$Y^\ell_m(\theta,\phi)$$ of $$\hat L^2$$ as these will provide the correct extra factor in Eq.(1) when separating the variables. Since there are many functions $$Y^\ell_m(\theta,\phi)$$ with the same $$\hbar^2\ell(\ell+1)$$, one uses the eigenvalue $$\hbar m$$ of $$\hat L_z$$ to distinguish them.