Commutation of the angular momentum operator How do I show that $\hat{L_i}$ commutes with $\mathbf{\hat{x}\cdot\hat{p}}$?
I know that I can write $\hat{L_i}=\epsilon_{ijk}\hat{x_j}\hat{p_k}$, but I don't know what to do from there.
 A: I assume  $\hbar=1$.
$$\hat{\bf x}\cdot \hat{\bf p}= \frac{1}{2}\hat{\bf x}\cdot \hat{\bf p}+
\frac{1}{2}\hat{\bf x}\cdot \hat{\bf p} = \frac{1}{2}\left(\hat{\bf x}\cdot \hat{\bf p} + \hat{\bf p}\cdot \hat{\bf x}\right) + 3i I = \frac{1}{2}\left[(\hat{\bf x} + \hat{\bf p})^2 - \hat{\bf x}^2 -  \hat{\bf p}^2\right] + 3 i I\:.$$
The rightmost term commutes with $\hat{L}_i$ because it is a sum of four scalars under rotations.
Another approach (which is not the direct computation described in the other answer) is
$$e^{-i \theta L_i} \hat{\bf x}\cdot \hat{\bf p} e^{i \theta L_i}= e^{-i \theta L_i} \hat{\bf x} e^{i \theta L_i} e^{-i \theta L_i}\cdot \hat{\bf p} e^{i \theta L_i} = 
(R_\theta \hat{\bf x}) \cdot (R_\theta \hat{\bf p})= \hat{\bf x}\cdot \hat{\bf p}$$
where $R_\theta$ is the rotation of $SO(3)$ around ${\bf e}_i$ of the angle $\theta$. Taking the derivative fort $\theta=0$ of both sides of $$e^{-i \theta L_i} \hat{\bf x}\cdot \hat{\bf p} e^{i \theta L_i}=
\hat{\bf x}\cdot \hat{\bf p}$$
we have $$[L_i,\hat{\bf x}\cdot \hat{\bf p} ]=0\:.$$
A: You can directly calculate the commutation relation $[L_i,x_jp_j]$. The Leibniz rule equates this to $[L_i,x_j]p_j+x_j[L_i,p_j]$.
Written out in full, this is $\epsilon_{ilm}[x_lp_m,x_j]p_j+x_j\epsilon_{ilm}[x_lp_m,p_j]$.
Now just use the Leibniz rule again ($[A,BC]=B[A,C]+[A,B]C$) and the canonical commutation relations ($[x_a,x_b]=[p_a,p_b]=0$ and $[x_a,p_b]=i \hbar \delta_{ab}$).
