# Sequential Stern-Gerlach experiment

Consider the following diagram:

(Sakurai)

In the first lecture of MIT OCW Quantum Physics 1, 2013 (https://www.youtube.com/watch?v=lZ3bPUKo5zc), Allan Adams implies that if we remove the barrier on the Sx- beam, thus allowing the Sx+ beam and Sx- beam to merge together and pass through the next SG measurement setup, the output of that final measurement will be only Sz+ atoms, rather than both Sz+ and Sz-. Why does this happen? Why do the atoms not care about being collapsed into Sx eigenstates in this scenario?

• Can you point out the timestamp in which this point is made? It doesn't make sense to me that the removal of the Sx- block would produce such a result. Instead, the proposed conclusion would only make sense if you remove the entire SGx machine altogether – talrefae Aug 5 '19 at 1:39
• @talrefae actually, removal of the block and allowing the beams to merge back is equivalent to removal of the whole SGx machine. – Ruslan Aug 5 '19 at 15:13

Let us assume that the lecturer is correct. My analysis would be as follows: Consider an initial state $$|1 \rangle = \frac{1}{\sqrt{2}} (|S_z + \rangle + |S_z -\rangle).$$ Now it passes through the first SG experiment, and all down spins are filtered out, so you have a collapsed state $$|2 \rangle = |S_z + \rangle = \frac{1}{\sqrt{2}} (|S_x + \rangle + |S_x -\rangle).$$ If a measurement is made after the second SG experiment, you again collapse into a state $$S_x+$$ $$|3 \rangle = \frac{1}{\sqrt{2}} (|S_z +\rangle + |S_z - \rangle) = |1\rangle.$$ However if no measurement is made, and the beams are merged, your system is still described by $$|2 \rangle$$, so a measurement of $$S_z$$ will only yield $$|S_z +\rangle$$.
The key point is that splitting out into the two $$S_x$$ components and merging again without ever measuring will not collapse the state. Superposition can persist over distance as long as no measurement is made, see e.g. the quantum erasure experiment with spacelike separation for practical examples of this.