Separability of a Hamiltonian with spin

[Salmone]

I'd like to know if this Hamiltonian $\hat{H}=\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2+\frac{A}{\hbar^2}(J^2-L^2-S^2)$ is separable into two parts $H_1=\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2$ and $H_2=\frac{A}{\hbar^2}(J^2-L^2-S^2)$ and $[H_1,H_2]=0$. Here A is a constant. I did so:

$[H_1,H_2]=[\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2,\frac{A}{\hbar^2}(J^2-L^2-S^2)]=$
$=[\frac{p^2}{2m},\frac{A}{\hbar^2}(J^2-L^2-S^2)]+[\frac{1}{2}m\omega^2r^2,\frac{A}{\hbar^2}(J^2-L^2-S^2)]=$
$=\frac{A}{2m \hbar^2}([p^2,J^2]-[p^2,L^2]-[p^2,S^2])+\frac{a}{2 \hbar^2}([r^2,J^2]-[r^2,L^2]-[r^2,S^2])=$ $=\frac{A}{2m \hbar^2}([p^2,L^2]+[p^2,S^2]+2[p^2,L_xS_x]+2[p^2,L_yS_y]+2[p^2,L_zS_z]-[p^2,L_x^2]-[p^2,L_y^2]-[p^2,L_z^2]-$ $-[p^2,S^2])+\frac{A}{2 \hbar^2}([r^2,J^2]-[r^2,L^2]-[r^2,S^2])=\frac{A}{2m \hbar^2}([p^2,2L_xS_x]+[p^2,2L_yS_y]+[p^2,2L_zS_z])+\frac{A}{2 \hbar^2}([r^2,J^2]-[r^2,L^2]-[r^2,S^2])$ now, since $p^2$ is a scalar and it's a function of spatial coordinates, it commutes with the component $L_i$ of the angular moment and with Spin operators and the same can be said about $r^2$ so the first commutator is $0$ and the hamiltonian is separable in $H_1+H_2$. Am I right?

 

[vanhees71]

That's right!