Question:
What is $ [\hat{L_1}, \hat{r} \cdot \hat{p}] $
Solution:
With the identity and the experience deriving those, it is not complicated anymore.
\begin{eqnarray*} [\hat{L_1}, \hat{r} \cdot \hat{p}] &=& [\hat{L_1}, \hat{x_a}\hat{p_a}] \\ &=& [\hat{L_1}, \hat{x_a}]\hat{p_a} + \hat{x_a}[\hat{L_1}, \hat{p_a}] \\ &=& (\epsilon_{1ab}\hat{x_b})\hat{p_a} + \hat{x_a}(\epsilon_{1ab}\hat{p_b}) \\ &=& \epsilon_{1ab}(\hat{x_b}\hat{p_a} + \hat{x_a}\hat{p_b}) \\ &=& \hat{x_3}\hat{p_2} + \hat{x_2}\hat{p_3} - \hat{x_2}\hat{p_3} - \hat{x_3}\hat{p_2} \\ &=& 0 \end{eqnarray*}
What is $ [\hat{L_1}, \hat{r} \cdot \hat{p}] $
Solution:
With the identity and the experience deriving those, it is not complicated anymore.
\begin{eqnarray*} [\hat{L_1}, \hat{r} \cdot \hat{p}] &=& [\hat{L_1}, \hat{x_a}\hat{p_a}] \\ &=& [\hat{L_1}, \hat{x_a}]\hat{p_a} + \hat{x_a}[\hat{L_1}, \hat{p_a}] \\ &=& (\epsilon_{1ab}\hat{x_b})\hat{p_a} + \hat{x_a}(\epsilon_{1ab}\hat{p_b}) \\ &=& \epsilon_{1ab}(\hat{x_b}\hat{p_a} + \hat{x_a}\hat{p_b}) \\ &=& \hat{x_3}\hat{p_2} + \hat{x_2}\hat{p_3} - \hat{x_2}\hat{p_3} - \hat{x_3}\hat{p_2} \\ &=& 0 \end{eqnarray*}
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