Commutator $\vec{L}$ with $\vec{X}\cdot\vec{P}$

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?

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