If $\hat{A}$ commutes with $\hat{r}^2$, does $\hat{A}$ commute with $\hat{r}$? Let $\hat{x}, \hat{y}, \hat{z}$ be the position operators in 3D space and $\hat{r}^2=\hat{x}^2+\hat{y}^2+\hat{z}^2$ as usual.
If one operator $\hat{A}$, commutes with $\hat{r}^2$, i.e.
$$\Big[\hat{A}, \hat{r}^2\Big] = 0$$
Does this imply that it also commutes with $\hat{r}$, $[\hat{A}, \hat{r}]=0$.
If the answer is no in general, what happens if we restrict $\hat{A}$ to some combination of $\hat{r}$ and $\hat{p}$ operators (with $\big[\hat{r}_i, \hat{p}_j\big] = i\delta_{ij}$)?
Before this comes up as a comment, I know that in general $\Big[\hat{A},\hat{B}^2\Big]=0$ does not imply $[\hat{A},\hat{B}]=0$, but I wonder if there is some property of the position operator to allow this.
The origin of my question is because in Quantum Physics sometimes one uses the argument that an operator (for example $\hat{L}$) commutes with $\hat{r}^2$, to justify that this operator commutes with any operator $V(\hat{r})$ in the Hamiltonian (for example, the Coulomb potential). And I want to know if this is really a good argument.
Thank you
 A: The point is that $\hat r$ is a positive operator and hence one has that $\hat r=\sqrt{\hat r^2}$.
Therefore, any operator commuting with $\hat r^2$ will also commute with $\hat r$.
The situation is entirely different with respect to a general self-adjoint operator, e.g. any $2\times 2$ matrix commutes with the square of $$A=\pmatrix{1&0\cr0&-1},$$
a.k.a. the identity matrix, but of course not all matrices commute with $A$!

EDIT:  Upon request let us show that, given two unbounded self-adjoint operators $A$ and $B$, if $B$ is positive and $A$
commutes with $B^2$, then $A$ also commutes with $B$.
First of all, recall that two unbounded self-adjoint operators $A$ and $B$ are said to "strongly commute" if all the projections in their associated projection-valued measures commute.  This is equivalent to saying that $e^{itA}$ commutes with
$e^{isB}$, for all $s, t\in {\mathbb R}$.
By contrast, simply saying that $[A, B]=0$ on the common domain is not a very useful condition and, in particular, this doesn't imply that $e^{iA}$ commutes with $e^{iB}$.
See the book by Reed and Simon, "Methods of mathematical physics" vol.1, chapter 8,  for a counter-example.
I am therefore referring to strong commutativity throughout.
This said, if $A$ and $B$ are self-adjoint operators, $B$ is positive (my term for positive semi-definite), and $A$
commutes (strongly of course) with $B^2$, then yes, it is true that $A$ commutes with $B$.
To prove it, denote the spectra of $B$ and $B^2$, respectively by $\Sigma _1$ and    $\Sigma _2$, and
consider the projection-valued measures
$P_1$ and $P_2$ for $B$ and $B^2$, respectively.
Thus $P_1$
(resp. $P_2$)
is defined on the Borel $\sigma $-algebra of  $\Sigma _1$
(resp. $\Sigma _2$), and
$$
B=\int _{\Sigma _1}\lambda \,dP_1(\lambda )$$$$
\Big (\text {resp}, \ B^2=\int _{\Sigma _2}\lambda \,dP_2 (\lambda )\Big ).
$$
Considering the "square" function
$$
  s:\lambda \in \Sigma _1\mapsto \lambda ^2\in \Sigma _2=\Sigma _1^2,
  $$
let $Q=s_*(P_1)$ be the push-forward measure of $P_1$ under $s$, namely  the projection-valued measure defined on $\Sigma _2$ by
$$
  Q(E) = P_1(s^{-1}(E)),
  $$
for every Borel subset $E\subseteq \Sigma _2$.  We then have that
$$
  \int _{\Sigma _2}\lambda \,dQ(\lambda ) =   \int _{\Sigma _2}\lambda \,ds_*(P_1)(\lambda ) =   \int _{\Sigma _1}s(\lambda )\,dP_1(\lambda ) = $$$$=
  \int _{\Sigma _1}\lambda ^2\,dP_1(\lambda ) =   \Big (\int _{\Sigma _1}\lambda \,dP_1(\lambda )\Big )^2 = B^2 = \int _{\Sigma _2}\lambda \,dP_2 (\lambda ).
  $$
By the uniqueness of the spectral decomposition we have that $Q=P_2$, meaning that
$$
  P_2(E) = P_1(s^{-1}(E)), \qquad (*)
  $$
for every Borel subset $E\subseteq \Sigma _2$.
The key hypothesis that $B$ is positive implies that $\Sigma _1\subseteq {\mathbb R}_+$, and hence that $s$ is a bijective function from $\Sigma _1$ to
$\Sigma _2$.  Plugging $E=s(F)$ in $(*)$, where $F$ is any Borel subset of $\Sigma _1$, we get
$$
  P_2(F^2) = P_1(F).
  $$
This says that all spectral projections for  $B$ (namely the $P_1(F)$) are found among the spectral projections of $B^2$.  Consequently, if a
spectral projection of $A$ commutes with all spectral projections of $B^2$, then it obviously commutes with all spectral
projections of $B$.
