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]≠0\end{equation}

TheLis no long a quantum number anymore.

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Hamiltonian with position spin coupling

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

Loading...

**Physics Forums | Science Articles, Homework Help, Discussion**