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

7 events

when toggle format	what		by	license	comment
Dec 7, 2017 at 6:37	vote	accept	Saúl Pilatowsky-Cameo
Nov 28, 2017 at 14:48	answer	added	John Donne		timeline score: 0
Nov 26, 2017 at 21:39	comment	added	Saúl Pilatowsky-Cameo		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 21:33	comment	added	Saúl Pilatowsky-Cameo		$\textbf{L} \times \textbf{L}=i \hbar \textbf{L}$ 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 21:19	comment	added	M111		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.
Nov 26, 2017 at 20:40	review	First posts
Nov 26, 2017 at 22:22
Nov 26, 2017 at 20:35	history	asked	Saúl Pilatowsky-Cameo	CC BY-SA 3.0