 Solution

Use $[AB,C]=A[B,C]+[A,C]B$ and the canonical commutation relation. With $L_i={\epsilon }_{iab}{x}_a{p}_b$,Using these two transformation laws on $L_j={\epsilon }_{jab}{x}_a{p}_b$ givesThese are the orbital angular momentum commutation relations.

NowThe factor $[L_j,L_i]$ is antisymmetric in $j$ and its remaining product is symmetric after the two terms are combined, so the contraction vanishes:

The same commutators show that $p^2=p_j{p}_j$ and $r^2=x_j{x}_j$ are rotational scalars: their commutators with every $L_i$ vanish by contraction of the antisymmetric ${\epsilon }_{ijk}$ with a symmetric product. Therefore $U(r)$ also commutes with every $L_i$, and forwe have $[H,L_i]=0$, henceIn particular $H,L^2,L_3$ are pairwise commuting Hermitian operators. By simultaneous diagonalization, they admit a common eigenbasis, subject to the usual spectral-domain qualifications for unbounded operators. This is the rotational invariance of a central-potential Hamiltonian.