Quantum constant of motion in a magnetic field

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
Dario SLC
Hello, I have a doubt about the Complete Set of commuting observables (CSCO) in the cases when there are a magnetic field ##B## in z.
The statement is find the constant of motion and CSCO for a particle of mass m and spin 1/2, not necessary a electron or any atomic particle.

I know that the CSCO is ##\{\hat{H}, \hat{L^2},\hat{L_z},\hat{S_z},\hat{S^2}\}## if not active the magnetic field, and them are constants of motion.
If not consider the coupling ##\hat{L}\hat{S}##, I think that the CSCO do not change, ie, conserve the constants of motion. But if I consider ##L\cdot S##, the new CSCO is ##\{\hat{H}, \hat{L^2},\hat{J_z},\hat{J^2},\hat{S^2}\}## (in absent of field ##B##)

That is true?

(The potential ##V##, is a spherical potential, only depends of ##r## coordinate.)
 
on Phys.org
Dario SLC said:
The statement is find the constant of motion and CSCO for a particle of mass m and spin 1/2, not necessary a electron or any atomic particle.
Dario SLC said:
(The potential ##V##, is a spherical potential, only depends of ##r## coordinate.)
This is why we require that the homework template be filled, including the full statement of the problem. Is there a potential or not?

The answer to your question depends on what's included in the Hamiltonian. So, what is ##\hat{H}##?
 
DrClaude said:
This is why we require that the homework template be filled, including the full statement of the problem. Is there a potential or not?
Hello, yes I don't put it, this potencial is a spherical step of height ##V_0##.
Stament
A particle of mass ##m## and spin 1/2, is subject a potential spherical step of height ##V_0##. If a external magnetic field constant ##B_0## is activated in the direction of ##z##. Find the Hamiltonian and CSCO.

DrClaude said:
The answer to your question depends on what's included in the Hamiltonian. So, what is ##\hat{H}##?
I think that the Hamiltonian is:
$$
\hat{H_0}=\left(\frac{\hat{p}_0^2}{2m}+\hat{V_0}\right)+\frac{\hat{L^2}}{2mr}
$$
and the Hamiltonian due to external magnetic field ##B=B_0\hat{z}##
$$
\hat{H_1}=-\frac{\mu_B}{\hbar}\left(g_l\hat{L}+g_s\hat{S}\right)B_0
$$
when ##g_l## and ##g_s## there are de gyromagnetic factor (orbital and spin respectly), and because I don't know which class of particle is, I don't use the values 1 and 2 for ##g_l## and ##g_s##
then ##\hat{H}=\hat{H_0}+\hat{H_1}##

I don't include the interaction ##L\cdot S## because it's a particle of mass ##m##, I don't know if the mass generates a magnetic field like the electron. By this I think that the CSCO is the same independient of the external magnetic field, all conmutations remain valid.