I've seen in numerous places the argument: if the hamiltonian $H$ is rotationally invariant, then it commutes with the components of the angular momentum: $$[H,J_i]=0$$

I'm not completely sure why that is. My guess is that, since $H$ is rotationally invariant, it commutes with all the rotation operators $R(\vec{\alpha})=\exp(-i\vec{\alpha}\cdot J)$. In particular, $[H,\exp(-i\vec{\alpha}\cdot J_i)]=0$, for any $i=1,2,3$. Since $H$ commutes with a function of $J_i$, it must commute with $J_i$.

I haven't seen this argument written explicitly, so I want to confirm it.

  • $\begingroup$ Hint: what happens to the rotation operator if you differentiate w.r.t. the angle of rotation? $\endgroup$ – Emilio Pisanty Jan 27 '17 at 11:50

You are on the right way. Consider the Hamiltonian under an infinitesimal rotation $\delta \vec{\alpha}$. Then, the Hamiltonian transforms as

$H \mapsto R(\delta \vec{\alpha})HR^\dagger(\delta \vec{\alpha})$

under this infinitesimal rotation. Finally, use Taylor expansion to first order in $\delta \vec{\alpha}$ and the fact that $H$ is unchanged after an infinitesimal rotation, i.e.

$H = R(\delta \vec{\alpha})HR^\dagger(\delta \vec{\alpha})$.

  • $\begingroup$ Ok, but why is my reasoning not sufficient? $\endgroup$ – Soap Jan 27 '17 at 12:23

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