# 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!

• Can you link the experiment you are talking about? Commented Aug 21, 2021 at 16:00
• I'm not sure what the question here is - the fact that the probability amplitudes in one direction cancel out and not in the other is precisely what is meant by there being "destructive interference" in one direction and "constructive interference" in the other. Commented Aug 21, 2021 at 16:18
• @Bill Alsept I hope this link works: books.google.hr/… - part of the experiment I'm asking about is explained on page 303 (link takes ypu to page 301) Commented Aug 21, 2021 at 22:17
• @ACuriousMind oh, yeah, you're right. I have a wrong image in my mind when I think about interference. But why is there interference in the x direction? In this part of the experiment, the beam is merged back into one and there are no detectors - the beam should be in the same state as before the first Stern - Gerlach apparatus (in the z direction). If we had only one apparatus in the x direction (so none in z), we would get two beams and no interference. Am I right? – Commented Aug 21, 2021 at 22:41
• I see no mystery in the Stern Gerlach experiment. Its important to remember that the experiment does more than just measure and compare. Each apparatus has a magnetic field that influences the spins of the particles. When you send a beam of individual particles with random spins, the apparatus will physically (In real life) rotate each and every particle up or down depending on the hemisphere the dipole's were already pointing to. So when the particles reach the second apparatus orientated left and right it doesn't matter if spins are up and down. Now they will all be rotated left and right. Commented Aug 22, 2021 at 17:08

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)