# Constants of motion in quantum mechanics

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1. Aug 20, 2017

### Dario SLC

1. The problem statement, all variables and given/known data
A particle of mass m and spin s, it's subject at next central potential:
$\begin{equation*} V(\mathbf{r})= \begin{cases} 0\text{ r<a}\\ V_0\text{ a<r<b}\\ 0\text{ r>b} \end{cases} \end{equation*}$
Find the constants of motion of the system and the set of observables that commute.

2. Relevant equations
Ehrenfest's theorem
$\frac{\partial \left<\hat A\right>}{\partial t}=\frac{i}{\hbar}\left<[\hat H,\hat A]\right>+\left<\frac{\partial\hat A}{\partial t}\right>$ when $\hat A$ it is an any observable.

3. The attempt at a solution
Because of that it's a problem of central potential, like hydrogen atom, $\hat{L}_z$ and $\hat{L}^2$ commute with hamiltonian $\hat{H}$, and not depend explicity of the time, them are constant of motion (for the Ehrenfest's theorem), similar the hamiltonian because only depend of r coordinate and $[\hat{H},\hat{H}]=0$. But the system has spin, therefore it add other degree of freedom, ie the system has four degree of freedom (if have'nt spin, only three degree of freedom), then the wave function will be:
$$\psi(r,\theta,\phi,m_s)=R(r)Y_{lm}(\theta,\phi)g(m_s)$$
when $g(m_s)$ will be $\alpha$ or $\beta$ corresponding at spin up or spin down respectly.

Like the operator $S_z$ corresponding to the observable relative to proyection about z-axis for the momentum of spin, it commute with hamiltonian, because it do not depends of $m_s$.

Therefore, the constants of motion will be:
$$\hat{L}_z\text{, }\hat{L}^2\text{, }\hat{H}\text{ and } \hat{S_z}$$

On the other hand, these constants they are a set of observables that commute.

It is fine?

In addition, why isn't $\hat{S}^2$ a constant of motion?, this will have to do with the fact of that s=1/2, and no option to choose, therefore $S^2$ it is not a constant of motion?

2. Aug 20, 2017

### blue_leaf77

It should be fine.
$\hat S^2$ is also a constant of motion. It's, however, usually omitted in the notation of the eigenfunction because one typically only deals with one type of particle in such type of problem.