Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 2 / 17B / Solution

The total angular momentum operator isUsing the cyclic orbital angular momentum commutation relations,while $[L_z,L_z^2]=0$; hence $\boxed{[L_z,{\mathbf{L}}^2]=0}$. In Cartesian coordinates,Because $L_z(x+iy)=\hslash (x+iy)$ and it annihilates $z$ and every radial function,Replacing $i$ by $-i$ gives eigenvalue $-m\hslash$.

The six-dimensional space splits into the five trace-free quadratic spherical harmonics with $\ell =2$ and one radial state with $\ell =0$. A simultaneous eigenbasis for $L_z$ and ${\mathbf{L}}^2$ iswhere the eigenvalue $\ell (\ell +1){\hslash }^2$ of ${\mathbf{L}}^2$ was used. Every displayed state is a complex linear combination of the original six energy eigenstates, so it is still an energy eigenstate at the same level.