I am solving a Hamiltonian including a term \begin{equation}(x\cdot S)^2\end{equation}(adsbygoogle = window.adsbygoogle || []).push({});

The Hamiltonian is like this form:

\begin{equation}

H=L\cdot S+(x\cdot S)^2

\end{equation}

whereLis angular momentum operator andSis 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 totalJ=L+S,L^{2}andS^{2}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 totalJis 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]â‰ 0\end{equation}

TheLis no long a quantum number anymore.

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

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# Hamiltonian with position spin coupling

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