Can spin operator expectation value be time-independent while commutator with Hamiltonian is non-zero? Considering the following (magnetic field) hamiltonian: $\hat{H}=-\gamma B_z \hat{S}_z$ ($\gamma$ is a constant). Suppose an electron is in an eigenstate of $S_x$, and we ask ourselves the question whether the expected value of $S_z$ changes with time. This is not the case, as $[\hat{H}, \hat{S}_z]=0$.
Now suppose we are in an eigenstate of $S_z$, does the $S_x$ expectation value change with time? Logically I would say it does because of the answer to the previous question. However, $[\hat{H}, \hat{S}_x]\neq 0$ indicating time dependence. So, will the expectation value of $S_x$ change with time or not?
 A: The spin is undergoing Larmor precession about the magnetic field, so yes the $x$ and $y$ components will change while the component parallel to the field remains fixed..
A: The Ehrenfest theorem uses the expected value of the commutator. i.e. for some operator $\hat{A}$ the Ehrenfest theorem says
$$\frac{d}{dt} \langle\hat{A}\rangle = \frac{1}{i\hbar} \langle[\hat{A}, \hat{H}]\rangle + \langle\frac{\partial\hat{A}}{\partial t}\rangle.$$
In your case while $[\hat{S}_x, \hat{H}] \neq 0,$ its expectation value will still be zero thus the expectation value of $\hat{S}_x$ will not depend on time. Should be straightforward to show that $\langle[\hat{S}_x, \hat{S}_z]\rangle = 0$ with the spin commutator relations.
Edit: For your first point, if the only Hamiltonian is the one you've given then the particle won't be in an eigenstate of $\hat{S}_x$. If it was in such an eigenstate before the magnetic field was turned on the wavefunction will collapse into one of the possible $\hat{S}_z$ eigenstates once the field is turned on.
