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Quantum constant of motion in a magnetic field

  1. Oct 28, 2017 #1
    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.)
     
  2. jcsd
  3. Oct 30, 2017 #2

    DrClaude

    User Avatar

    Staff: Mentor

    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}##?
     
  4. Oct 30, 2017 #3
    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.

    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 wich 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.
     
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