# Quantum constant of motion in a magnetic field

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1. Oct 28, 2017

### 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.)

2. Oct 30, 2017

### 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 $\hat{H}$?

3. Oct 30, 2017

### Dario SLC

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.