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  • $\begingroup$ Hmmm. So, based on your comment, it seems to me then that my calculation in the way that I carried out is correct, due to the initial state of the particle being $\left|S_z^+\right>$ and the magnetic field being aligned with it. Classically, however, would not the direction of the moment change direction? The potential energy of the particle classically is: $U=-\vec{\mu}\cdot\mathbf{B}$, and to minimize this one would have $\mu$ align with $\mathbf{B}$. $\endgroup$ Commented Nov 1, 2018 at 6:03
  • $\begingroup$ @T.Zaborniak Yes, it's the fact that the field and the initial spin direction are aligned that leads to the spin not changing. (There is something a odd about the expression for the time-dependent wave function in the question though; the absolute value of the upper component of the spinor should be 1 for all time.) $\endgroup$ Commented Nov 1, 2018 at 6:08
  • $\begingroup$ @T.Zaborniak The classical torque is what I wrote in my answer, and if you take the dot product of $\vec{N}=d\vec{S}/dt$ with the direction $\hat{z}$ of $\vec{B}$, you can see that $S_{z}$ is independent of time. $\endgroup$ Commented Nov 1, 2018 at 6:12
  • $\begingroup$ I still don't get your classical explanation exactly. I'm thinking that in writing $S_z$ you mean $L_z$, where $L_z$ is the classical angular momentum in the $z$-direction, and by $\vec{S}$ you mean the classical angular momentum. That being the case, isn't it necessary that $L_z$ change when $t$ goes from $t<0$ to $t>0$ to have $|d\vec{L}/dt|$ remain a constant? The magnitude of $\vec{L}$ obviously would have to stay the same, but because it is $L_z\hat{\mathbf{z}}$, its direction would need to change... $\endgroup$ Commented Nov 1, 2018 at 6:31