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If a scalar operator $\hat{S}$ is defined as an operator that is invariant under rotations, i.e $$U^\dagger S U = S,\,\,\,\,\,\,\, U=e^{-i\theta\hat{\mathbf{J}}\cdot{\mathbf{n}}}$$ which is equivalent to $$[J_i,S]=0$$

Then, the potential energy $V(\mathbf{r})$ is not a scalar operator unless $V(\mathbf{r})= V(r), r=|\mathbf{r}|$ which seems to contradict this wikipedia page (https://en.wikipedia.org/wiki/Tensor_operator#Scalar_operators) and whole lot of standard textbooks.

What am I misunderstanding?

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    $\begingroup$ I think that since usually potential energy only depends on $r$ they are making this assumption. $\endgroup$ – valerio Jun 5 '17 at 19:46
  • $\begingroup$ @valerio92, as a follow up, if it is not a scalar (or a vector or a tensor) operator, what is it then? $\endgroup$ – stupidquestions Jun 5 '17 at 20:49
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    $\begingroup$ I think we can safely say that it is a scalar operator in most physical situations. $\endgroup$ – valerio Jun 5 '17 at 20:56
  • $\begingroup$ It depends on the interaction responsible for the potential energy. Sometimes it is a scalar, other times it is not. $\endgroup$ – Lewis Miller Jun 6 '17 at 2:35
  • $\begingroup$ The is just the definition of scalar operator. For example, a potential $v(x,y,z) = z$ is the $0$th component of a rank $1$ spherical tensor operator. Section 3.11 of Sakurai's Modern Quantum Mechanics is a good introduction on this topic. $\endgroup$ – Eric Yang Jun 6 '17 at 9:12

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