# Commutation between angular momentum and Hamiltonian

Consider the following Hamiltonian of a 3-dimensional system:

$$H=\frac{p^2}{2m}+V(r)$$

If the components of the angular momentum, $L_i$, commute with $H$, then:

$$[H,L_i]=0$$

This condition can be satisfied if:

$$[p^2,L_i]=0\hspace{10pt}(1),\hspace{15pt}[V(r),L_i]=0\hspace{10pt}(2)$$

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.

• I agree with you on that. But how could I prove it to odd powers of $r$? – Élio Pereira Nov 26 '15 at 20:41
• Just think at $r$ as $r=\sqrt{r^2}$ and expand the square root in power series. – gented Nov 26 '15 at 21:02
• I know that $[L_i,r^2]=0\Rightarrow [L_i,r^{2n}]=0$, because $[A,BC]=B[A,C]+[A,B]C$. If $A=L_i$ and $B,C=r^2$, then $[A,BC]=[L_i,r^4]=r^2[L_i,r^2]+[L_i,r^2]r^2=0$, and we can extend this to $r^{2n}$. But we can't say that $[L_i,r^2]=0\Rightarrow [L_i,r]=0$ because $[L_i,r^2]=0\Leftrightarrow r[L_i,r]+[L_i,r]r=0$, which doesn't imply that $[L_i,r]=0$ although the inverse does. – Élio Pereira Nov 26 '15 at 21:21
• just would like to point out, that from the extract of the book the potential is $V(\sqrt{r^2})$. the only reason I see for doing so, is to have a potential that is proportional to $|r|$. – PinkFloyd Nov 27 '15 at 7:17