Hamiltonian with position spin coupling

[yuanyuan5220]

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?

 

[azntoon]

Any suggestions would be greatly appreciated. One possible approach is to use the Wigner-Eckart theorem to simplify the problem. This theorem states that the expectation values for operators in a certain state obey certain symmetry relations. By using this theorem, you can reduce the problem to one involving only the spin operators and the total angular momentum operator, which can then be solved more easily.