 Solution

orbital angular momentum generates spatial rotations and is $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. With $\mathbf{p}=-i\hslash \nabla$, its components satisfyThus $L^2$ and one component, conventionally $L_z$, have simultaneous eigenstates, while all three components cannot generally have sharp values. On ordinary single-valued spatial wavefunctions,The spherical harmonics form an orthonormal angular basis. The integer restriction is specific to orbital wavefunctions; half-integer spin is an intrinsic degree of freedom, not the angular dependence of a scalar spatial wavefunction.

The angular momentum ladder operators obeyThe ladder ends at $m=\pm \ell$, producing $2\ell +1$ states. A state with specified $m$ has a sharp projection $\hslash m$, but the angular momentum magnitude is $\hslash \sqrt{\ell (\ell +1)}$, not $\hslash \ell$. No classical picture of a vector with simultaneously fixed Cartesian components reproduces all these measurement properties.

In a central potential, rotational symmetry gives $[H,L_i]=0$. separation of variables yields $\psi (\mathbf{r})=R_{\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The radial Schrödinger equation includes a centrifugal potential ${\hslash }^2\ell (\ell +1)/(2m{r}^2)$, which suppresses small-radius amplitude for nonzero $\ell$. The energy is independent of $m$ when rotational symmetry is intact, but need not be independent of $\ell$. A magnetic field can break that degeneracy. Spatial inversion gives $Y_{\ell m}(-\hat{\mathbf{r}})=(-1{)}^{\ell }{Y}_{\ell m}(\hat{\mathbf{r}})$, connecting orbital angular momentum to parity and to the even/odd rotational restriction for the hydrogen spin isomers.