Eigenvalue for a Hamiltonian

[yuanyuan5220]

Homework Statement

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

Homework Equations

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}

The Attempt at a Solution

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 \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 Hamiltonian?

 

[mark726]

I would suggest considering using perturbation theory to solve this Hamiltonian. Since the first term is easy to solve, we can treat it as the unperturbed Hamiltonian and the second term as the perturbation. This will allow us to find an approximate solution for the total energy and the corresponding wavefunction.

Another approach could be to use the ladder operator method, where we can use the commutation relations between the operators to find the eigenstates and eigenvalues. This may be more complicated, but it could potentially give a more accurate solution.

Additionally, we can also try to use numerical methods, such as the variational method or Monte Carlo simulations, to solve this Hamiltonian and obtain an approximate solution.

It may also be helpful to look for similar Hamiltonians that have been studied before and see if there are any known techniques or methods that can be applied to this problem. Collaborating with other scientists or seeking guidance from a mentor or advisor could also be beneficial in finding a solution to this Hamiltonian.