# How to show that $[L_i, v_j]=i\hbar\sum_k \epsilon_{ijk}v_k$ for any vector $\textbf{v}$ constructed from $\textbf{x}$ and/or $\nabla$?

It follows from the fact that $$\hat L_i$$ are generators of the rotations.

Generally, the vector operator $$\hat V_i$$ is not just "a list" of operators. It has to transform like a vector under rotations $$\hat V_i' = R_{ij} \hat V_j$$ (summation implied).

This identity can be proven by considering an observable in a rotated frame of reference. The frame rotation can be carried out through a transformation $$R$$ acting on $$\hat V$$ or a unitary operator $$\hat{U}(R)$$: $$$$\left._R\right. \langle\alpha|\hat{V}_i|\beta\rangle_R = \langle\alpha|\hat U^\dagger \hat V_i \hat U|\beta\rangle = R_{ij}\langle\alpha| \hat V_j |\beta\rangle$$$$

Thus, we have: $$\hat U^\dagger \hat V_i \hat U = R_{ij} \hat V_j$$. By considering infinitesimal rotation $$\hat U = 1 - \frac{i \epsilon \bf{J}\cdot \bf{n}}{\hbar}$$ and corresponding $$R = \left(\begin{matrix}1& -\epsilon & 0\\ \epsilon & 1 &0 \\ 0 & 0 & 1\end{matrix}\right)$$ we find at first order in $$\epsilon$$: $$$$[V_i, J_j] = i \hbar \epsilon_{ijk} V_k$$$$

A nice proof of this identity can be found here: https://www.oulu.fi/tf/kvmIII/english/2004/09_tensop.pdf.

Similar question: Angular and linear momentum operators' commutation

Another useful reference: https://www.wikiwand.com/en/Tensor_operator#/Vector_operators