Sequential Stern–Gerlach

The Stern–Gerlach apparatus is not just a demonstration of spin quantization — it is a filter. By blocking one of its output beams you can prepare a beam in a definite spin eigenstate along a chosen axis. Sending that prepared beam through a second, differently oriented apparatus reveals something striking about quantum measurement: measuring along a new axis completely destroys the information about the previous axis.

One apparatus: a quick recap

A single SG-z apparatus splits a beam into two sub-beams:

$|{+z}\rangle$ atoms, deflected upward (detected at the upper spot), and
$|{-z}\rangle$ atoms, deflected downward (detected at the lower spot).

If you block the lower beam and let only the $|{+z}\rangle$ atoms continue, you have a state preparation device: every atom leaving the open port is guaranteed to have $S_z = +\hbar/2$ .

Sequential measurement: z then x

Now pass the $|{+z}\rangle$ beam through a second apparatus oriented along $x$ (an SG-x device). To predict the outcome we express $|{+z}\rangle$ in the eigenbasis of $S_x$ .

The eigenstates of $S_x$ are

|{-x}\rangle = \frac{1}{\sqrt{2}}\bigl(|{+z}\rangle - |{-z}\rangle\bigr).$$ Inverting these relations: $$|{+z}\rangle = \frac{1}{\sqrt{2}}\bigl(|{+x}\rangle + |{-x}\rangle\bigr).$$ The Born rule then gives the probability of each $S_x$ outcome when the state is $|{+z}\rangle$: $$P(+x \mid +z) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}, \qquad P(-x \mid +z) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}.$$ The beam splits **equally** regardless of which $z$-eigenstate you started with. ## The key experiment: z then x then z again The most striking variant uses **three** apparatuses in sequence: SG-z, then SG-x, then SG-z again. 1. **SG-z (first):** Prepare $|{+z}\rangle$ by blocking the lower beam. 2. **SG-x (middle):** Let both outputs through — or block one and keep only $|{+x}\rangle$. 3. **SG-z (last):** Measure $S_z$ one more time. If you let both SG-x outputs pass into the final SG-z, interference between $|{+x}\rangle$ and $|{-x}\rangle$ reconstitutes $|{+z}\rangle$, and every atom exits the final SG-z as spin-up. The intermediate measurement is "erased" because both paths are preserved. If instead you **block** the $|{-x}\rangle$ beam, the situation changes completely. The atoms that pass are now in the state $|{+x}\rangle$. Expanding $|{+x}\rangle$ in the $z$-basis: $$|{+x}\rangle = \frac{1}{\sqrt{2}}\bigl(|{+z}\rangle + |{-z}\rangle\bigr).$$ The final SG-z apparatus therefore finds: $$P(+z \mid +x) = \frac{1}{2}, \qquad P(-z \mid +x) = \frac{1}{2}.$$ **Half** the atoms exit as spin-up along $z$ and half as spin-down — even though the original beam had $S_z = +\hbar/2$ with certainty. Measuring $S_x$ in between erased the previous information about $S_z$. <Callout type="tip"> This is not a matter of disturbing the atom mechanically. The information is lost because $|{+z}\rangle$ and $|{-z}\rangle$ are not eigenstates of $S_x$; projecting onto $|{+x}\rangle$ puts the atom into a genuine superposition of $z$-outcomes. There is no hidden variable that kept track of the original $S_z$. </Callout> ## Fraction reaching the final spin-up port Starting with a beam of $N$ atoms all in $|{+z}\rangle$: | Stage | Atoms remaining | State | |---|---|---| | After SG-z preparation | $N$ | $|{+z}\rangle$ | | After SG-x (keep $|{+x}\rangle$) | $N/2$ | $|{+x}\rangle$ | | After final SG-z (keep $|{+z}\rangle$) | $N/4$ | $|{+z}\rangle$ | The fraction of the original beam that reaches the spin-up port of the last SG-z apparatus is $$\frac{N/4}{N} = \frac{1}{4} = 0.25,$$ derived as the product $P(+x \mid +z) \times P(+z \mid +x) = \tfrac{1}{2} \times \tfrac{1}{2}$. ## Non-commutativity of spin operators The sequential experiment is a direct consequence of the fact that $S_z$ and $S_x$ **do not commute**. Their commutator is $$[S_z, S_x] = i\hbar S_y \neq 0.$$ Non-commuting observables cannot share a common set of eigenstates; knowing $S_z$ precisely means $S_x$ is maximally uncertain, and measuring $S_x$ completely scrambles any previously known value of $S_z$. The Heisenberg uncertainty relation for spin gives $$\Delta S_z \,\Delta S_x \geq \frac{\hbar}{2}\,|\langle S_y \rangle|.$$ The sequential Stern–Gerlach arrangement makes this non-commutativity tangible with a visible, macroscopic spot on a detector plate. ## Try it This is a numerical exercise — `return` a number. A beam is prepared in $|{+z}\rangle$, passed through SG-x with only the $|{+x}\rangle$ output kept, and then passed through a final SG-z apparatus. What fraction of the original beam exits as spin-up from the final SG-z? <RunnableExample />