Consider the following Hamiltonian of a 3-dimensional system:
If the components of the angular momentum, $L_i$, commute with $H$, then:
This condition can be satisfied if:
I've seen on Griffiths' book that, it's enough to have (1) and $[r^2,L_i]=0$ satisfied:
My doubt is:
Why is correct to say: $[r^2,L_i]=0\Rightarrow [V(r),L_i]=0$?
$V(r)$ could be a function of any power of $r$, wich couldn't let this statement be a general truth.