= Solution
The product rule for a <commutator> gives $[A,B^2]=[A,B]B+B[A,B]$. Using the permitted cyclic <angular momentum commutation relations>,
$$
\begin{aligned}
[L_1,L_2^2]&=i\hbar(L_3L_2+L_2L_3),\\
[L_1,L_3^2]&=-i\hbar(L_2L_3+L_3L_2),\\
[L_1,L_1^2]&=0.
\end{aligned}
$$
The first two expressions cancel, even though the factors need not commute. Hence
$$
\boxed{[\widehat L_1,\widehat{\mathbf L}^{\,2}]=0.}
$$
The squared <angular momentum operator> commutes with each of its components by the same cyclic calculation.
Back to article page
