# Commutation of abstract o(3) generators and vectors.

I've been given the following problem, and I'm quite lost with it - any help would be fantastic!!

Let $L_1$, $L_2$, and $L_3$ denote the abstract o(3) algebras. You are given that $\vec{A} = (A_1, A_2, A_3)$ and $\vec{B} = (B_1, B_2, B_3)$ transform as vector operators of o(3).

Show that $[L_j, \vec{A} \cdot \vec{B}] = 0$

I know that $L_j = \varepsilon_{jlm} q_l p_m$, and I can obviously determine the dot product, but I'm not sure where to go from there.

I do, however, know that $\vec{A} = \frac{1}{Ze^{2}\mu}(\vec{L} \times \vec{p}) + (\frac{1}{r})\vec{r}$, but I'm not sure ho to integrate that into this problem.

Any help would be fantastic!! :)

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The definition of a vector operator under ${\frak o} (3)$ is $$[L_i, A_j] = i \sum_k \epsilon_{ijk} A_k.$$ Using this we compute $$[L_i, \sum_k A_k B_k]=$$ $$= \sum_k \left( [L_i, A_k] B_k + A_k [L_i, B_k] \right)=$$ $$=i\sum_{k,j} \left (\epsilon_{ikj} A_j B_k + \epsilon_{ikj} A_k B_j \right).$$ This expression is both symmetric and antisymmetric in $j,k$ indices and therefore equals zero.
The intuition for this result is that since both $\vec A$ and $\vec B$ transform as vectors $\vec A \cdot \vec B$ must transform as scalar. But that means that it is invariant under rotations and so commutes with the generators of the symmetry.