# Quantum Vector Operators. Showing $\textbf{r}(\textbf{L} \cdot\textbf{p})-\textbf{p}(\textbf{L} \cdot\textbf{r})=0$

I'm asked to prove this relation: $$\left [\textbf{A},\textbf{L} \cdot \textbf{B} \right]=-i \hbar \textbf{A} \times \textbf{B} + L_i \left [\textbf{A},B_i \right].$$ Where $\textbf{A}$ and $\textbf{B}$ are vector operators that satisfy $$\left[L_i,A_j \right]=i \hbar A_k\varepsilon_{ijk},$$ and $\textbf{L}$ is the angular momentum operator.

This is easy using $\left[A_k,L_j B_j \right]=L_j \left[A_k, B_j \right]+ \left[A_k, L_j \right]B_j=L_j \left[A_k, B_j \right]-i \hbar A_l\varepsilon_{jkl}Bj$. Then I'm asked to use this relation to show that $$\textbf{r}(\textbf{L} \cdot\textbf{p})-\textbf{p}(\textbf{L} \cdot\textbf{r})=0.$$ Here is where I'm stuck. Setting $\textbf{A}=\textbf{r}$ and $\textbf{B}=\textbf{p}$ I am able to show that $\textbf{r}(\textbf{L} \cdot\textbf{p})-(\textbf{L} \cdot\textbf{p})\textbf{r}=0.$ With $\textbf{B}=\textbf{r}$ and $\textbf{A}=\textbf{p}$ I get $\textbf{p}(\textbf{L} \cdot\textbf{r})-(\textbf{L} \cdot\textbf{r})\textbf{p}=0.$ But this gets me nowhere.

Am I missing something? The formula is similar to an identity involving $\textbf{L} \times (\textbf{r} \times \textbf{p})$. But I think that the result has to be more or less immediate as it is stated as a corollary to the first relation.

• If you subtract the two terms you get $Req + (L.r)p - (L.p)r = 0$. Where Req is the quantity you have to show to be equal to zero. The remaining part can be expressed as $L \times (r \times p)$ which is zero, since $L = r \times p$. I think this process is fine.
– M111
Nov 26, 2017 at 21:19
• $\textbf{L} \times \textbf{L}=i \hbar \textbf{L}$ is not zero. And Req is not equal to that, it is missing a commutator. $a × (b × c) = b (a \cdot c) − (a \cdot b) c + [ a_j , b ] c_j$ Nov 26, 2017 at 21:33
• Still, I have the feeling that using that identity is too much work. I would have to prove it, together with the fact that $\textbf{L} \times \textbf{L}=i \hbar \textbf{L}$. This seems more work than simply proving $\textbf{r}(\textbf{L} \cdot\textbf{p})-\textbf{p}(\textbf{L} \cdot\textbf{r})=0$ by hand. I was hoping there was some kind of neat trick here. Nov 26, 2017 at 21:39

If you're allowed to assume that $\textbf{L}=\textbf{r} \times \textbf{p}$ then the equation you're trying to prove is in some sense totally trivial, because both terms vanish identically: $$\textbf{L} \cdot \textbf{r} = \textbf{L} \cdot \textbf{p}=0$$ This is very easy to prove by noting that the usual vector equality $\textbf{L}=\textbf{r} \times \textbf{p}=-\textbf{p} \times \textbf{r}$ still holds for those operators.
However if you're not allowed to assume that, it is not true in general that the dot product of $\textbf{L}$ and an arbitrary vector operator vanishes (the obvious counterxample is $\textbf{L}\cdot \textbf{L}\neq 0)$. Therefore, to avoid being blinded by preconceived notions, one should try to prove the following: $$\textbf{A}(\textbf{L} \cdot\textbf{B})-\textbf{B}(\textbf{L} \cdot\textbf{A})=0$$ for any two vector operators $\textbf{A}$ and $\textbf{B}$. However this is not true in general, the counterexample given by taking $\textbf{A}=\textbf{r}$ and $\textbf{B}=\textbf{L}$, the usual $\textbf{L}$. Then $$\textbf{A}(\textbf{L} \cdot\textbf{B})-\textbf{B}(\textbf{L} \cdot\textbf{A})=\textbf{r}\textbf{L}^2\neq 0$$ Therefore I suspect that the first part of the question is indeed what they were looking for. The result is simple enough to be immediate even without resorting to any lemmas.