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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?

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    $\begingroup$ But... look at the m=1/2 eigenstate of $S_z$. It gives $\langle S_x\rangle=0$. $\endgroup$ – Cosmas Zachos Oct 8 '20 at 21:05
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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..

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    $\begingroup$ ...unless both x and y expectation values vanish. $\endgroup$ – Cosmas Zachos Oct 8 '20 at 21:09
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    $\begingroup$ Alright, so x and y components ofcourse change (i.e. are time dependent), but nevertheless the expectation values for x and y are constant ($=0$) (makes sense as the precession creates a perfect circle). I guess I should not be too quick to make assumptions solely on the Ehrenfest theorem. $\endgroup$ – 6thsense Oct 8 '20 at 21:16
  • $\begingroup$ @CosmosZachos. Only if $J=0$. $\endgroup$ – mike stone Oct 8 '20 at 23:31
  • $\begingroup$ What do you mean by "expectation of $x$ $y$" do you mean $S_x$, $S_y$? $\endgroup$ – mike stone Oct 8 '20 at 23:32
  • $\begingroup$ Yes, consider J=1/2, and $\langle S_x\rangle=\langle S_y\rangle=0$. $\endgroup$ – Cosmas Zachos Oct 9 '20 at 0:17
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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.

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