DarMM said:
Yes, but what about the experiment where you take particles of definite ##S_z## and then perform an ##S_x## measurement? Afterward they've gone from a state with definite ##S_z## to one of definite ##S_x## in a way that cannot be described by unitary evolution.
Again there's no problem. You just use a magnetic field with its large homogeneous part in ##x## direction. Then a silver-atom prepared to have ##S_z=1/2## is randomly deflected (with probability 50%) in the one or the other direction. Now the position of the particle is entangled with ##S_x## rather than with ##S_z##, and you have prepared a particle with ##S_x=-1/2## or ##S_x=+1/2##, depending of the direction it went when running through the ##\text{SG}_x## magnet. You cannot decide beforehand, which value of ##S_z## you get, you only know that in about half the cases you get ##S_x=+1/2## and the other half of the cases you get ##S_x=-1/2##, but after the particle has run through the magnet, it's clear which of the two possible ##S_x## values it has.
You can understand this intuitively nearly by thinking in classical terms (being a bit sloppy in thinking about spin, which is not a classical observable to begin with): If you have the particle prepared in a ##S_z=1/2## eigenstate, the ##S_x## and ##S_y## components are indetermined. This implies that the magnetic moment vector has a determined ##\mu_z## component but indetermined ##\mu_x## and ##\mu_y## components. If the particle now enters an inhomogeneous magnetic field with a large homogeneous part in ##x##-direction, the magnetic moment starts to precess rapidly around the ##x## direction. In moving through the magnetic field, which must also have some inhomogeneous part, the particle is deflected according to the force ##\vec{F}=-\vec{\nabla} (\vec{\mu} \cdot \vec{B})##. Now the ##\mu_y## and ##\mu_z## components are rapidly oscillating since ##\vec{\mu}## rapidly precesses around the ##x## direction (with the Larmor frequency ##\omega=g e B/(2m) \simeq g e B_0/(2m)## (for a silver atom ##g=g_{\text{electron}} \simeq 2##). Thus it is a good approximation to assume that the deflection is given by ##\vec{F} \simeq -\vec{\nabla} (\mu_x B_x)##. The other components ##\mu_y B_y## and ##\mu_z B_z## can be assumed tobe so rapidly oscillating that they average out to 0 over the typical much longer time scales the silver atom moves inside the magnet. Let the magnet be along the ##y## axis and the particle's momentum well peaked around the ##y## axis too. Then we can assume that in the time it's inside the magnet it's not too far refleced from the ##y## axis, and we can approximate the ##B## field as
$$\vec{B}=(B_0 + \beta x)\vec{e}_x -\beta y \vec{e}_y.$$
Note that the last term must be there, because we need to fulfill ##\vec{\nabla} \cdot \vec{B}=0##. Nevertheless since we can approximate due to the above argument
$$\vec{F} \simeq -\vec{\nabla} (\mu_x B_x)=-\mu_x \beta \vec{e}_x.$$
The particle gets deflected in ##x## direction in two opposite directions depending on the two possible signs ##\mu_x## can take.
You can make this fully quantum by reading everywhere operators instead of usual c-numbers and use time-dependent perturbation theory with
$$\hat{H}_0=\frac{\vec{p}^2}{2m} + \hat{\mu}_x (B_0+\beta \hat{x}), \quad \hat{H}_1=-\hat{\mu}_y \beta \hat{y}.$$
Then you can calculate the time evolution of a Pauli wave function, initially given as an apprpriate Gaussian wave packet with only a component referring to ##\sigma_z=+1/2##. The time evolution with ##\hat{H}_0## gives a two-bump wave packet, where particles in one bump have ##\sigma_x=+1/2## and ##\sigma_x=-1/2##. Then you can use first-order time-dependent perturbation theory to show that the correction due to the perturbation ##\hat{H}_1## is indeed small due to the rapid Larmor oscillation of the ##\mu_y##. Thus you get (nearly) perfect entanglement between ##\sigma_x## and the particle's position coordinate ##x##, i.e., blocking one of the two partial beams leads to a (nearly perfect) beam of particles with ##\sigma_x=+1/2##.
