Hamiltonian with position spin coupling

  • #1
I am solving a Hamiltonian including a term \begin{equation}(x\cdot S)^2\end{equation}


The Hamiltonian is like this form:
\begin{equation}
H=L\cdot S+(x\cdot S)^2
\end{equation}
where L is angular momentum operator and S is spin operator. The eigenvalue for \begin{equation}L^2 , S^2\end{equation} are \begin{equation}l(l+1), s(s+1)\end{equation}

If the Hamiltonian only has the first term, it is just spin orbital coupling and it is easy to solve. The total J=L+S, L2 and S2 are quantum number. However, when we consider the second term position and spin coupling: \begin{equation}(x\cdot S)^2\end{equation} it becomes much harder. The total J is still a quantum number. We have \begin{equation}[(x\cdot S)^2, J]=0\end{equation}. However, \begin{equation}[(x\cdot S)^2,L^2]≠0\end{equation}
The L is no long a quantum number anymore.

Anybody have ideas on how to solve this kind of Hamiltonian?
 

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