I was reading about identical particles and i came across this example:
Consider two electrons with spin 1/2. The Hamiltonian for this system is:
$$Η=\frac{p_1^2}{2m}+\frac{p_2^2}{2m}+\frac{1}{2}m\omega x_1^2+\frac{1}{2}m\omega x_2^2+g\vec{S_1}\cdot\vec{S_2}(x_1-x_2)^2$$
Im not sure what the last term means exactly, is it correct writing in this form?
$$Η=1\otimes1\otimes\frac{p_1^2}{2m}+1\otimes1\otimes\frac{p_2^2}{2m}+1\otimes1\otimes\frac{1}{2}m\omega x_1^2+1\otimes1\otimes\frac{1}{2}m\omega x_2^2+g\vec{S_1}\otimes\vec{S_2}\otimes(x_1-x_2)^2$$
If this is correct i tried changing the coordinates by using this transformation:
$$R=\frac{x_1+x_2}{m},r=x_1-x_2, \mu=\frac{m}{2}, M=2m$$
and then the new Hamiltonian becomes:
$$Η=1\otimes1\otimes\frac{p_r^2}{2\mu}+1\otimes1\otimes\frac{p_R^2}{2M}+1\otimes1\otimes\frac{1}{2}M\omega R^2+1\otimes1\otimes\frac{1}{2}\mu\omega r^2+g\vec{S_1}\otimes\vec{S_2}\otimes(x_1-x_2)^2$$
Does the spin part transform with the new coordinates i set? If so how? Another problem i have is to find the eigenstates of this hamiltonian, the only way i can think of is to consider: $$Η_0=1\otimes1\otimes\frac{p_r^2}{2\mu}+1\otimes1\otimes\frac{p_R^2}{2M}+1\otimes1\otimes\frac{1}{2}M\omega R^2+1\otimes1\otimes\frac{1}{2}\mu\omega r^2$$ as my original hamiltonian and
$$H_p=g\vec{S_1}\otimes\vec{S_2}\otimes(x_1-x_2)^2$$ but does this hold for any value of $g$?
Last question is there any way to compute the exact solutions to this hamiltonian?