I Separability of a Hamiltonian with spin

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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?
 
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