# Commutation of abstract $O(3)$ generators and vectors

I've been given the following problem, and I'm quite lost with it.

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 how to integrate that into this problem.

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You don't need to know the specific form of $\vec{A}$ or representation of $\vec{L}$, only that $\vec{A},\vec{B}$ transform like vectors. That tells you the commutators $[L_j, A_i]$ and $[L_j, B_i]$, and from that you can compute $[L_j, \vec{A}\cdot\vec{B}]$. Also note the homework tag applies even if this is not assigned coursework - it's close enough. :) –  Michael Brown Sep 18 '13 at 8:31
@MichaelBrown I just saw this; I probably shouldn't have written my response given this comment. My apologies. –  joshphysics Sep 18 '13 at 8:42
@joshphysics No worries. The only reason I didn't post the commutators is that I can never remember the sign. :) –  Michael Brown Sep 18 '13 at 9:03

A collection $\{V_1, V_2, V_3\}$ of operators on a vector space $V$ is called an $\mathfrak{o}(3)$ vector operator with respect to a representation $\rho$ of $\mathfrak{o}(3)$ acting on $V$ provided \begin{align} [V_i, L_j] = i\epsilon_{ijk}V_k \end{align} where $L_1, L_2, L_3$ are the representatives of the standard basis on $\mathfrak{o}(3)$. This means, in your case, that merely knowing that $A_i$ and $B_i$ are the components of vector operators tells you that \begin{align} [A_i, L_j]=i\epsilon_{ijk}A_k, \qquad [B_i, L_j]=i\epsilon_{ijk}B_k \end{align} These commutation relations alone are sufficient to demonstrate the desired result using commutator identities.
I know this is probably really stretching my luck, but how do I go about applying those commutation relations?? Should I be using $[A, BC] = [A, B]C + B[A, C]$?? –  Jack Sep 18 '13 at 10:07