Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hamiltonian with position spin coupling

  1. Nov 23, 2011 #1
    I am solving a Hamiltonian including a term \begin{equation}(x\cdot S)^2\end{equation}

    The Hamiltonian is like this form:
    H=L\cdot S+(x\cdot S)^2
    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?
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted

Similar Discussions: Hamiltonian with position spin coupling
  1. Spin Hamiltonian (Replies: 4)

  2. Coupled Spin operators (Replies: 7)

  3. Spin-Orbit Coupling (Replies: 10)