Question about L.S component of hamiltonian

[Leandro Tabak]

Homework Statement

I need to find the energies of the lowest level of the following hamiltonian

H = p^2/2m + (mw^2r^2)/2 + L.S

(spin 1/2 particle of mass m)

The tridimensional harmonic oscilator with a coupling term

Homework Equations

The first two terms are solved, my question is about the term L.S, which l correspond to the lowest energy? l=0 or l= 1? First, I thought that it should be l=0, but in my class they stated l=1, is it related to the Heisenberg principle?

The Attempt at a Solution

If the hamiltonian was the free particle with just the term L.S i think l = 0 would follow to a fundamental state of 0 energy and it is wrong (i think)...could you help me please?

 

[mark726]

Thank you for your question. The term L.S in the given Hamiltonian represents the interaction between the orbital angular momentum (L) and the spin angular momentum (S) of the particle. In this case, the value of l would depend on the total angular momentum of the system, which is given by the addition of L and S.

In general, for a spin 1/2 particle, the possible values of l can range from 0 to 1. However, in this specific case of a tridimensional harmonic oscillator, the term L.S would only contribute to the energy if the total angular momentum is non-zero. Therefore, the lowest energy level would correspond to l=0, as you initially thought. This is not related to the Heisenberg principle, but rather to the specific form of the Hamiltonian and the system being studied.

I hope this clarifies your doubts. If you need further assistance, please do not hesitate to ask. Good luck with your research!