# Commutation relation of $J^2$ and $R(\alpha,\beta,\gamma)$

If $R(\alpha,\beta,\gamma)$ is the Rotation operator and $\alpha,\beta,\gamma$ are Euler angles and $J$ is the total angular momentum then how to get to this: $$[J^2,R]~=~0?$$

This is stated in Zettili's book Quantum Mechanics.

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You may just explicitly write $R$ as $R=\exp(i\alpha J_x/\hbar)\exp(i\beta J_z/\hbar)\exp(i\gamma J_x/\hbar)$, or some other convention with signs and axes, and because $J^2$ commutes both with $J_x$ and $J_z$, it also commutes with any and arbitrarily ordered function of them including $R$. –  Luboš Motl Dec 11 '12 at 18:33

The rotation operator $R$ is generated by the angular momentum operators $J_i$: $R$ can be written as a power series of $J_i$.
Since $[J^2,J_i]=0$ for $i=1,2,3$ the operator $J^2$ will commute with every term in the power series of $R$. By this we have $[R,J^2]=0$.