 Solution

Keep one normalized spin function fixed and write $A=1/\sqrt{24}$ and $f_m(\theta ,\varphi )=\sin \theta e^{im\varphi }$, with $m=\pm 1$. Overall constants and the spin factor pass unchanged through the spatial operators.

First, in atomic units, ${\hat{l}}_z=-i{\partial }_{\varphi }$, soFor the squared orbital angular momentum, ${\partial }_{\varphi }^2{f}_m=-f_m$ andTherefore the two terms of the angular operator giveusing $1-{\cos }^2\theta ={\sin }^2\theta$. Thus ${\hat{l}}^2\Psi =2\Psi$.

Finally, the spatial hydrogen Hamiltonian isFor $R=Ar{e}^{-r/2}$,Substitute the already established angular eigenvalue $l(l+1)=2$:Hence $E=-1/8$ hartree, consistently with $n=2$. The angular calculation must precede the radial substitution if one wants to avoid differentiating the entire product repeatedly.