Stern Gerlach and interference I recently came across this experiment: a beam of spin 1/2 particles pass through a Stern Gerlach apparatus oriented in the z direction. After passing through it and splitting, the beams are again merged into one with another magnetic field.First part of the experiment in the z direction.

This beam then passes through another Stern Gerlach apparatus this time oriented in the x direction. Now, according to Quantum Mechanics (Auletta, Fortunato, Parisi) in the absence of a detector, these two beams will interfere constructively in the +x direction and destructively in -x direction. So there will be no particles in the -x direction. Why is this? I understand how to get this result by calculating probabilities, but I don't understand why there's interference in the x and not z direction and why it's constructive in one and completely destructive in the other direction. Thanks!
 A: The evolution for a particle with a well defined spin along the $z$-axis is :
$$\begin{array}c
I && III && V \\
|\phi_I\rangle\otimes|+\rangle_z & \longrightarrow & |\psi_\uparrow\rangle\otimes |+\rangle_z & \longrightarrow & |\psi_V\rangle\otimes|+\rangle_z \\
|\phi_I\rangle\otimes|-\rangle_z & \longrightarrow & |\psi_\downarrow\rangle\otimes |-\rangle_z & \longrightarrow & |\psi_V\rangle\otimes|-\rangle_z \\
\end{array}$$
where $|\phi_I\rangle, |\phi_V\rangle$ are wave-packets localized in region I and V respectively, and $|\psi_\uparrow\rangle$, $|\psi_\downarrow\rangle$ are localized in region III on the up and down path respectively.
In the usual Stern-Gerlach experiment, the position of the particle in region III is measured, which allows to tell $|\psi_\uparrow\rangle$ from $|\psi_\downarrow\rangle$ and in turns gives us the spin of the particle.
Now, consider any particle at the start of the experiment. Its state is :
$$|\Psi_I\rangle = |\phi_I\rangle \otimes |u\rangle$$
where  $|u\rangle  =\alpha |+\rangle_z + \beta|-\rangle_z$ with $|\alpha|^2  +|\beta|^2=  1$. By linearity, it evolves as :
$$\begin{array}c
I && III && V \\
|\phi_I\rangle \otimes |u\rangle & \longrightarrow & \alpha |\psi_\uparrow\rangle\otimes |+\rangle_z + \beta  |\psi_\downarrow\rangle\otimes |-\rangle_z & \longrightarrow &|\psi_V\rangle\otimes |u\rangle
\end{array}$$
ie, the spin of the particle is unchanged. Looking only at $III\rightarrow V$, the superposition interferes just in the right way to give the initial spin state.
(In the book, they choose $\alpha = \beta = 1/\sqrt 2$, ie $|u\rangle = |+\rangle_x$, but this works for any initial spin state)
