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Let $\vec{X}=(X_1,X_2,X_3)^T$ and $\vec{P}=(P_1,P_2,P_3)^T$. Define $\vec{L}=\vec{X}\times\vec{P}$. Then, I can calculate $\vec{L}=(X_2P_3-X_3P_2,\,X_3P_2-X_2P_3,\,X_1P_2-P_1X_2)^t$. For all compononents of $L$, I want to compute $[\vec{L}_a,\vec{X}\cdot\vec{P}]$ for $a=1,2,3$. I know that $[X_i,P_j]=i\hbar\delta_{ij}$, $[x_i,x_j]=[P_i,P_j]=0$. For $a=1$:

$[X_2P_3-X_3P_2,X_1P_1+X_2P_2+X_3P_3]=[X_2P_3-X_3P_2,X_2P_2+X_3P_3]=[X_2P_3,X_2P_2]+[X_2P_3,X_3P_3]-[X_3P_2,X_2P_2]-[X_3P_2,X_3P_3]=$ $$ [X_2,X_2P_2]P_3+X_2[P_3,X_3P_3]-X_3[P_2,X_2P_2]-[X_3,X_3P_3]P_2= $$ $$ X_2[X_2,P_2]P_3+X_2[P_3,X_3]P_3-X_3[P_2,X_2]P_2-X_3[X_3,P_3]P_2=0 $$ and likewise for $a2$ and $a=3$. Is this the correct approach? And what does this physically tell us?

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    $\begingroup$ Yes. There are more compact ways to organize it, but you got it. One says that X.P is a rotational invariant. $\endgroup$ – Cosmas Zachos Jan 29 at 20:19

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