PeterDonis said:
Again, you're using a classical description, but we're not talking about classical physics, we're talking about quantum physics. In quantum physics, i.e., in actual experiments when they are actually run, the SG device does not change the amplitudes of the spin components at all. It just entangles them with linear momentum. Indeed, it would make no sense to use SG devices to measure the spin components (by measuring the relative frequencies of particles in the output beams) if the device changed them.
You have an oversimplified picture of the SGE, but let's discuss this idealized picture first, because as I tried to clarify in my previous posting, you can realize it at very good approximation. My arguments were indeed semiclassical, but since the equations of motion in the discussed approximation are linear, it's a very good picture.
Quantum mechanically, the indealized SGE is indeed an example for a von Neumann measurement. Let's assume that the (large homogeneous part of the) ##\vec{B}## field is along the ##z##-direction, i.e., we look at an SGE that measures the ##z##-component of the spin:
$$\vec{B}=(B_0 + \beta z) \vec{e}_z - \beta y \vec{e}_y.$$
Solving the Pauli equation for the approximate Hamiltonian,
$$\hat{H}_0=\frac{1}{2m} \hat{\vec{p}}^2 + \mu_B g_s (B_0+\beta z) \hat{s}_z$$
shows that you indeed entangle the ##s_z##-component with the position of the atom, i.e., if you arrange the initial conditions and the magnetic field properly you can achieve that one partial beam is prepared in a ##\sigma_1=+1/2## and the other partial beam in a ##\sigma_z=-1/2## eigenstate of ##\hat{s}_z##. Using the position as a pointer variable you have thus a perfect measurement of the spin-##z## component and by blocking one of the partial beams you even have a perfect preparation procedure for eigenstates of ##\hat{s}_z##, i.e., a "von Neumann filter measurement".
Now what do I mean "you change the spin" with this experiment. It, of course means, this preparation procedure, i.e., independent from which initial (mixed or pure) spin state you start, in each of the partial beams you have prepared an eigenstate of ##\hat{s}_z##. Only if the original particle was in such an eigenstate, this state remains unchanged. So in general "you change the spin state" with this experiment. That's the only meaning to say "the spin changes" that makes sense within QT.
You can also look at the expectation values ##\langle \vec{s} \rangle##. As long as the magnetic field is acting this expecation value precesses around the ##z## axis. Also this shows that of course within a magnetic field the spin changes, i.e., it's not a conserved quantity of the dynamics (which also is reflected in the operator equations of motion, ##\mathring{\hat{\vec{s}}}=-\mathrm{i} [\hat{\vec{s}},\hat{H}] \neq 0##.
You couldn't measure or prepare the spin-##z##-component if spin were conserved!
PeterDonis said:
Yes, a probability of a spin flip. This flip would be a "change in spin" since in general it changes the spin components (although if a ##z## oriented device is being used to measure particles prepared in the spin ##x## up state, the flip actually does not change the relative weights of the spin components), but, as you note, it is considered an error and the devices are constructed to make this probability as small as possible. So the vast majority of the time, this spin flip does not happen and the spin does not change. And that vast majority of the time, when the SG device is functioning properly as part of a spin measurement, is what the post I originally responded to was talking about. In that case, the spin does not change, which was my point.
As detailed above, what does not change in the "vast majority of time" is the spin-##z## component, because the approximate Hamiltonian ##\hat{H}_0## commutes with ##\hat{s}_z##, but it doesn't commute with the other spin components, i.e., the spin changes due to the interaction with the magnetic field even in this approximation of the Hamiltonian describing the ideal SGE, and that's the curcial dynamics that makes the SG magnet a measurement device for the spin-##z## component in the here discussed setup.
It's of course also important that the perturbation due to the "rest Hamiltonian"
$$\hat{H}_1=-\mu_B g_s \beta y \hat{s}_y$$
is indeed "small". This can be checked in first-order time-dependent perturbation theory, which also shows, how the magnetic field must be chosen to get a nearly ideal SGE for measurement of the spin-##z## component. What's shown by this calculation is the correctness of the semiclassical heuristic argument, usually given in the textbooks.