# Commutation relation for a rotated vector operator

It is well known that a set of operators $(V_x,V_y,V_z)$ is called a vector operator if, given the total angular momentum $(J_x,J_y,J_z)$ of a system, it satisfies the following commutation relations \begin{equation} \left[V_a,J_b\right]=i\epsilon_{abc}V_c \end{equation} Let us assume now that we consider a new set of operators obtained by applying a rotation to $\vec{V}$ \begin{equation} V'_{a}=\sum_b R_{ab}V_b \end{equation} where $R$ is the corresponding rotation matrix (equivalently one could use the unitary transformation associated to $R$). My qustion is, can we show that the $\vec{V}'$ is again a vector operator? That is, does the commutation relation \begin{equation} \left[V'_a,J_b\right]=i\epsilon_{abc}V'_c \end{equation} hold? To my intuition it should, but i haven't found an easy way to show that.

If you "go finite", note first that your definition of a vector operator is equivalent to $$R(\omega)V_mR^{-1}(\omega)=\sum_k V_kD^1_{km}(\omega)\, ,$$ where $R(\omega)$ is a rotation operator and $D^1_{km}(\omega)$ a rotation matrix. In its most intuitive form $$R(\omega)= \exp\left(-i \vec{\omega} \cdot {\vec J} \right)$$ and you can recover $$[V_m,J_b]=i\frac{\partial}{\partial \omega_b} R(\omega)V_m R^{-1}(\omega)\Bigl\vert_{\vec{\omega} =0}\, .\tag{1}$$ In this way you can then verify that $V'_a=R(\Omega)V_aR^{-1}(\Omega)$ transforms properly because \begin{align} R(\omega) V_a'R^{-1}(\omega)&= R(\omega)R(\Omega)V_aR^{-1}(\Omega)R^{-1}(\omega)\, ,\\ &=R(\omega\circ\Omega)V_a R^{-1}(\omega\circ\Omega)\, ,\\ &=\sum_k V_k D^1_{ka}(\omega\circ\Omega)\, ,\\ &=\sum_k V_k \left[\sum_m D^1_{km}(\omega)D^1_{ma}(\Omega)\right]. \tag{2} \end{align} as it should. Here, $R(\omega\circ\Omega)=R(\omega)R(\Omega)$ is the product of rotatons; since rotations form a group it a rotation parametrized by three angles denoted as $\omega\circ\Omega$ since the rotation parameters $\omega$ and $\Omega$ are not additive (rotations are not Abelian). Alternatively, you can multiply the $D$'s in (2) and use this to infer the angles $\omega\circ\Omega$. Either way you can then recover the commutator version using (1).
Alternatively, you can start with $V'_a$ just an infinitesimal rotation away from $V_a$ and work entirely with commutators.