# Quantum Stern-Gerlach Experiment
> The apparatus of the last lesson was built to measure a continuous quantity and it returned exactly
> two answers — the same two answers, no matter which way you turned it. This lesson takes that
> result seriously enough to follow it through five cascaded experiments, and comes out the other
> side holding a two-dimensional complex vector space and the Born rule.
## Learning Objectives
After this lesson you will be able to:
1. State what the 1922 plate actually showed, and identify precisely which classical assumption from the previous lesson each feature of the result violates.
2. Use the **Stern-Gerlach analyzer** abstraction — one input, two labelled outputs — to specify and reason about cascaded measurements.
3. Predict the outcome of the five canonical cascade experiments: repeated, reversed, perpendicular, three-stage, and rotated by an arbitrary angle.
4. Represent the atom's magnetic state as a vector in $\mathbb{C}^2$ and write down $\lvert{\pm z}\rangle$, $\lvert{\pm x}\rangle$, and $\lvert{+\hat{n}(\theta)}\rangle$.
5. Apply the **Born rule** to reproduce the empirical law $P(+,\theta) = \cos^2(\theta/2)$, and explain why the angle is halved.
6. Explain what it means for a measurement to *prepare* a state, and why the results cannot be explained by atoms that arrive with a pre-existing orientation.
---
## Intuition
Run the apparatus of Lesson 1 and you should get a filled band: uniform intensity from one edge to
the other, densest nowhere, populated everywhere. What Gerlach saw in February 1922 was **two
spots** — one deflected up, one deflected down, by equal amounts, with the entire middle of the band
empty.
Take a moment to appreciate how strange this is, because familiarity dulls it. The classical band is
not merely *wrong in detail*; the observed pattern is its exact photographic negative in the region
that matters. The only atoms that arrive are the ones at the two extreme values $\mu_z = \pm\mu$,
which classically are the rarest and least special — they require the moment to point exactly along
or exactly against the field, a set of measure zero on the sphere. Every atom in the beam behaves as
though it had been pre-aligned, and pre-aligned *both ways at once*.
Now the sentence that turns a surprise into a crisis. **The splitting is the same for every
orientation of the magnet.** Turn the apparatus on its side and you get two spots left and right,
same separation. Tilt it 30° and you get two spots along that tilt, same separation. If atoms left
the oven with fixed little arrows, some axis would have to be perpendicular to those arrows, and
measuring along it would give zero deflection — one undeflected spot. That never happens. There is
no arrangement of classical arrows, however cleverly chosen, that yields two equal and opposite
projections along *every* axis, and that impossibility is what Freericks (after Styer) calls the
**conundrum of projections** [Fre, §1.2]. Lesson 4 makes the argument airtight; this lesson gathers
the experimental facts that make it necessary.
---
## Theory
### What the experiment showed, and what it did not prove
Stern and Gerlach sent a collimated beam of silver atoms through the strongest field gradient they
could build, and collected the atoms on a glass plate. The beam speed was around 660 m/s, the magnet
about 5 cm long, the expected splitting a tenth of a millimetre — small enough that alignment
defeated them many times before the run that worked.
The immediate interpretation was that the result confirmed **space quantization**, the Bohr-Sommerfeld
doctrine that an atom's orbit can only take discrete orientations relative to a field. It was
celebrated as a triumph of the old quantum theory, and this is one of the great ironies of physics:
the theory was wrong, the atom's two-valuedness has nothing to do with orbital quantization, and the
correct explanation — the electron's **spin** — was not proposed until 1925 and not applied to this
experiment until 1927. The experiment was right; the reason given for it was not.
That the old explanation fails is worth seeing concretely, because it rules out an entire class of
patches. Orbital angular momentum with quantum number $\ell$ gives $2\ell + 1$ orientations, and
$2\ell + 1$ is always *odd* — one, three, five — never two. Worse, silver's 47 electrons form
closed shells plus a single valence electron in an $\ell = 0$ state, so silver's *orbital* moment is
zero and the beam should not split at all. Phipps and Taylor closed the argument in 1927 by running
the experiment on ground-state hydrogen, which has one electron and unambiguously zero orbital
angular momentum, and still saw two spots.
### The Stern-Gerlach analyzer
Since every orientation gives exactly two outcomes, we can stop drawing magnets and start drawing
boxes. A **Stern-Gerlach analyzer** is an idealized device with one input port and two output ports,
labelled $+$ and $-$, characterized entirely by the axis $\hat{n}$ its field defines: an atom
entering leaves by exactly one of the two ports, and the port it leaves by is what we mean by "the
value of the atom's magnetic projection along $\hat{n}$." Internally there are guide tubes that
restore the two deflected beams to parallel travel; externally it is a black box that sorts.
Fix a coordinate system once and use it for the rest of this course: the beam travels along $y$,
vertical is $z$, horizontal-transverse is $x$. An analyzer oriented along $\hat{z}$ we call SG $z$;
along $\hat{x}$, SG $x$; and one rotated by $\theta$ from vertical in the $x$–$z$ plane, SG $\theta$.
Because the analyzer's output ports are *spatially separated beams*, we can chain them: feed one
output port into the input of the next analyzer, and block the other. Five such cascades exhaust
what there is to learn [Fre, §1.2].
### Experiment 1 — repeatability
```mermaid
flowchart LR
SRC["Oven
(unpolarized)"] --> A["SG z"]
A -->|"plus"| B["SG z"]
A -->|"minus"| BLK["blocked"]
B -->|"plus"| D1["100% of input"]
B -->|"minus"| D2["0%"]
```
Take the $+$ output of an SG $z$ analyzer and send it into a second SG $z$ analyzer. Every atom
leaves by the second analyzer's $+$ port; the $-$ port stays dark. The same holds with $-$ in place
of $+$ throughout.
This is the single most important result in the set, and it is the reassuring one: measurement is
**repeatable**. It means the first analyzer did not merely reveal a random number — it left the atom
in a condition that the second analyzer reliably confirms. We say the first analyzer has performed
**state preparation**: whatever "having a positive projection along $z$" means, an atom emerging from
the $+$ port *has* it, definitely, and will keep having it until something else is done to it. Without
this fact there would be no such thing as a quantum state.
### Experiment 2 — reversing the analyzer
Now take the $-$ output of SG $z$ and feed it into an analyzer whose field points along $-\hat{z}$.
Everything comes out that analyzer's $+$ port. This is bookkeeping rather than physics: reversing
the axis exchanges the meaning of the labels, so an atom with a negative projection on $\hat{z}$ has a
positive projection on $-\hat{z}$. Results along a reversed axis are just as definite as along the
original one. Parallel and antiparallel analyzers are the two cases where the outcome is *certain*.
### Experiment 3 — perpendicular analyzers
```mermaid
flowchart LR
SRC["Oven
5000 atoms"] --> A["SG z"]
A -->|"plus (2500)"| B["SG x"]
A -->|"minus (2500)"| BLK["blocked"]
B -->|"plus"| D1["about 1250
(25% of source)"]
B -->|"minus"| D2["about 1250
(25% of source)"]
```
Feed the $+$ output of SG $z$ into SG $x$. The atoms split **50–50**. Each individual atom's choice
is unpredictable; the *proportions* are utterly reliable.
This is the first appearance of genuine quantum randomness, and it is worth being precise about what
kind of randomness it is. Every atom entering the second analyzer is in an identically prepared,
completely known condition — they all came out the same port of the same analyzer. There is no
hidden spread of initial conditions left to blame, as there was in the classical experiment where the
randomness came entirely from the oven's isotropic source. The randomness here is not ignorance about
which atom we have; it is irreducible. And "random" here does *not* mean "unpredictable": we cannot
say where the next atom goes, but we can predict the frequencies to arbitrary precision by running
long enough. Quantum mechanics is a theory of exactly this kind of prediction.
### Experiment 4 — three analyzers, and the memory test
Chain three: SG $z$, then SG $x$ on the $+$ port, then SG $z$ again on the $+$ port.
```mermaid
flowchart LR
A["SG z"] -->|"plus"| B["SG x"]
B -->|"plus"| C["SG z"]
C -->|"plus"| D1["50%"]
C -->|"minus"| D2["50%"]
```
Two plausible lines of reasoning give opposite answers, and it is worth having argued both before
seeing the result.
*Reasoning A.* The second analyzer prepared the atom along $x$. The third asks about $z$, which is
perpendicular to $x$, so by Experiment 3 we should get 50–50.
*Reasoning B.* The first analyzer certified the atom as $+z$. The second certified it as $+x$. So
entering the third, the atom is both $+z$ and $+x$; the third analyzer should confirm $+z$ with
certainty.
The answer is **A: a 50–50 split.** Half the atoms emerge with a *negative* projection along $z$ —
even though every one of them was certified $+z$ two stages earlier, and nothing in between touched
the $z$ axis except the act of measuring $x$. Reasoning B fails because its premise is false: the
atom is never "both $+z$ and $+x$". Measuring $x$ destroyed the $z$ information outright.
Confirm the diagnosis by making the third analyzer SG $x$ instead: then **100%** emerge from its $+$
port, exactly as Experiment 1 would say. So the atom does have a perfectly definite $x$ projection —
it simply no longer has a $z$ one. Freericks summarizes this memorably: atoms remember only the last
axis they were measured on, and nothing before it [Fre, §1.2].
Bookkeeping for the whole chain: of the original beam, half is blocked at stage one, half of the
remainder at stage two, and the survivors split evenly at stage three, so each final port receives
$\tfrac12\times\tfrac12\times\tfrac12 = \tfrac18$ of the source — or $\tfrac14$ of the atoms that
entered the second analyzer.
### Experiment 5 — the $\cos^2(\theta/2)$ law
Finally, keep the first analyzer vertical and rotate the second one by an angle $\theta$, recording
the fraction that emerges from its $+$ port. Three points are already known: $\theta = 0°$ gives 1
(Experiment 1), $\theta = 90°$ gives $\tfrac12$ (Experiment 3), $\theta = 180°$ gives 0
(Experiment 2). The measured curve through them is not the straight line joining them. At $45°$,
about 85 atoms in 100 emerge from the $+$ port, not 75. The empirical law is
$$
P(+,\theta) = \cos^2\!\left(\frac{\theta}{2}\right),
\qquad
P(-,\theta) = \sin^2\!\left(\frac{\theta}{2}\right),
$$
with $\theta$ the angle between the two analyzers, and the two probabilities summing to 1 as they
must. Check it: $\cos^2(22.5°) = 0.854$, so 85 out of 100 is right on the curve. Check the special
cases: $\cos^2 0 = 1$, $\cos^2 45° = \tfrac12$, $\cos^2 90° = 0$. Those three are worth memorizing;
the rest is a calculator.
Note the *half* angle. It is not a typo, and it is not the classical $\cos\theta$ projection law of
Lesson 1 in disguise. We are about to see where it comes from.
### Two outcomes force a two-dimensional space
Let us build the smallest mathematical object that reproduces all five experiments.
Along any axis there are exactly two outcomes, and Experiment 1 says each of them is a condition an
atom can definitely be in. Assign a **state** to each: $\lvert{+z}\rangle$ and $\lvert{-z}\rangle$
for the two exits of SG $z$. Experiment 2 says these are mutually exclusive — an atom certified
$+z$ never comes out the $-$ port — so we take them **orthogonal** and normalized:
$$
\langle{+z}|{+z}\rangle = \langle{-z}|{-z}\rangle = 1,
\qquad
\langle{+z}|{-z}\rangle = 0 .
$$
Any state of the atom's magnetic degree of freedom is a complex superposition
$\alpha\lvert{+z}\rangle + \beta\lvert{-z}\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$: the state space
is $\mathbb{C}^2$. Probabilities come from the **Born rule** — the probability of finding outcome
$\lvert{\phi}\rangle$ when the atom is in state $\lvert{\psi}\rangle$ is the squared modulus of their
overlap:
$$
P(\phi \mid \psi) = \bigl|\langle{\phi}|{\psi}\rangle\bigr|^2 .
$$
Now fit the data. The states of the $x$ analyzer must give $\tfrac12$ against both $z$ states
(Experiment 3) while being orthogonal to each other (Experiment 1 applied to SG $x$). The choice
$$
\lvert{\pm x}\rangle = \frac{1}{\sqrt{2}}\Bigl(\lvert{+z}\rangle \pm \lvert{-z}\rangle\Bigr)
$$
does both: $|\langle{\pm x}|{+z}\rangle|^2 = |1/\sqrt2|^2 = \tfrac12$ and
$\langle{+x}|{-x}\rangle = \tfrac12(1 - 1) = 0$. And it reproduces Experiment 4 with no further
input, since $|\langle{\pm z}|{+x}\rangle|^2 = \tfrac12$ — the $z$ information really is gone, as a
matter of arithmetic rather than of decree.
For a general angle $\theta$ in the $x$–$z$ plane, the state selected by SG $\theta$'s $+$ port is
$$
\lvert{+\hat{n}(\theta)}\rangle = \cos\frac{\theta}{2}\,\lvert{+z}\rangle + \sin\frac{\theta}{2}\,\lvert{-z}\rangle ,
$$
whose overlap with $\lvert{+z}\rangle$ is $\cos(\theta/2)$ — giving exactly the measured law
$P = \cos^2(\theta/2)$. At $\theta = 90°$ it reduces to $\lvert{+x}\rangle$; at $\theta = 180°$ it
becomes $\lvert{-z}\rangle$, as Experiment 2 demands. Five experiments, one two-component vector,
no free parameters left over.
### Why the angle is halved
The half-angle is the signature of a **spin-$\tfrac12$** system, and it encodes a real geometric
fact: rotating the *apparatus* by $\theta$ rotates the *state* by $\theta/2$. Turn the analyzer
through a full $360°$ and it is manifestly the same apparatus, but the state has only been taken
half way around:
$$
\lvert{+\hat{n}(360°)}\rangle = \cos 180°\,\lvert{+z}\rangle + \sin 180°\,\lvert{-z}\rangle = -\lvert{+z}\rangle .
$$
The state comes back with a minus sign. No measurement detects it — the Born rule squares moduli, so
a global phase is invisible — but it means the correspondence between rotations of space and
transformations of the state is two-to-one. This is the group $SU(2)$ double-covering $SO(3)$, and it
is why the natural picture of a qubit is a sphere on which physical angles appear halved
([1.2.2 The Bloch Sphere](../../term-1-quantum-mechanics/c02-qubits-bloch/02-bloch-sphere.md)). The
relative phase between $\lvert{+z}\rangle$ and $\lvert{-z}\rangle$, by contrast, is not invisible at
all: it is what distinguishes $\lvert{+x}\rangle$ from $\lvert{-x}\rangle$, and Lesson 3 will make it
do work.
### What is actually precessing
Lesson 1 argued that the apparatus only works for objects carrying angular momentum, so the atom's
moment must come with one. It is not orbital — silver and hydrogen both give two spots with
$\ell = 0$. It is **spin**: an intrinsic angular momentum of magnitude
$S_z = \pm\hbar/2$, with the associated moment $\boldsymbol{\mu} = -g_s\,(e/2m_e)\,\mathbf{S}$ and
$g_s \approx 2$, so that each spot deflects as if carrying one full Bohr magneton — quantitatively
what was measured.
Resist the picture of a spinning ball. Demanding that a sphere of the classical electron radius carry
$\hbar/2$ requires an equatorial speed of order $100c$, and no experiment has ever resolved any
spatial extent for the electron at all. Spin is a two-valued degree of freedom with the *algebra* of
angular momentum and no mechanical picture underneath — one of several places where the quantum world
supplies structure that classical language cannot narrate. The state space we just built, $\mathbb{C}^2$
with an orthonormal basis and the Born rule, *is* the qubit
([1.2.1 The Qubit](../../term-1-quantum-mechanics/c02-qubits-bloch/01-the-qubit.md)); rename
$\lvert{+z}\rangle \to \lvert{0}\rangle$ and $\lvert{-z}\rangle \to \lvert{1}\rangle$ and every
subsequent term of this program is about systems like the one on Gerlach's plate.
---
## Hands-on (Python)
Simulate the cascades and recover the $\cos^2(\theta/2)$ curve from shot noise.
```python
import numpy as np
rng = np.random.default_rng(7)
def state(theta):
"""+ output state of an analyzer rotated by theta (radians) from vertical."""
return np.array([np.cos(theta / 2), np.sin(theta / 2)], dtype=complex)
def analyze(psi, theta, rng):
"""One analyzer: return (+1 or -1, collapsed state) via the Born rule."""
plus, minus = state(theta), state(theta + np.pi)
p_plus = abs(np.vdot(plus, psi)) ** 2
if rng.random() < p_plus:
return +1, plus
return -1, minus
# --- Experiment 4: SG z (+) -> SG x (+) -> SG z ---------------------------
N, n_pass_x, n_final_plus = 100_000, 0, 0
for _ in range(N):
s, psi = analyze(state(0.0), np.pi / 2, rng) # z-prepared into SG x
if s > 0:
n_pass_x += 1
s2, _ = analyze(psi, 0.0, rng) # back into SG z
n_final_plus += (s2 > 0)
print(f"P(+x | +z) = {n_pass_x / N:.3f}") # ~0.500
print(f"P(+z | +x) = {n_final_plus / n_pass_x:.3f}") # ~0.500 <- z memory gone
print(f"P(+x then +z) = {n_final_plus / N:.3f}") # ~0.250
# --- Experiment 5: rotate the second analyzer ----------------------------
print("\n theta measured cos^2(theta/2)")
for deg in (0, 45, 90, 135, 180):
th = np.deg2rad(deg)
hits = sum(analyze(state(0.0), th, rng)[0] > 0 for _ in range(2000))
print(f" {deg:3d}° {hits / 2000:.3f} {np.cos(th / 2) ** 2:.3f}")
```
Run it and watch Reasoning B die: the final $z$ measurement in Experiment 4 comes out 50–50 every
time, no matter how definite the atom's $z$ projection was two stages earlier.
---
## Exercises
**E1 (easy).** Atoms leaving the $+$ port of an SG $z$ analyzer enter an analyzer oriented along
$-\hat{z}$. Which port do they leave by, and with what probability? Reconcile your answer with the
$\cos^2(\theta/2)$ law.
Solution
They leave by the $-$ port, with probability 1. The reversed analyzer is at $\theta = 180°$ relative
to the first, so $P(+,180°) = \cos^2 90° = 0$ and $P(-,180°) = \sin^2 90° = 1$. This is Experiment 2
read in the other direction: reversing the axis swaps the labels but destroys no information, and the
outcome stays certain. Parallel and antiparallel are the only two orientations with no randomness.
Solution
$P(+) = \cos^2 30° = 3/4$ and $P(-) = \sin^2 30° = 1/4$. Of 4000 atoms, about 3000 leave the $+$
port and about 1000 the $-$ port. Note this is *not* the classical projection $\cos 60° = \tfrac12$;
the half-angle law biases the outcome far more strongly toward the nearer pole than a naive
projection argument would suggest.
Solution
The relative angle is what matters, and it is $45° - (-45°) = 90°$, so
$P(\pm) = \cos^2 45° = \sin^2 45° = \tfrac12$ — an even split.
By vectors: the first analyzer prepares
$\lvert{+\hat n(45°)}\rangle = \cos 22.5°\lvert{+z}\rangle + \sin 22.5°\lvert{-z}\rangle$, and the
second's $+$ state is
$\lvert{+\hat n(-45°)}\rangle = \cos 22.5°\lvert{+z}\rangle - \sin 22.5°\lvert{-z}\rangle$. Their
overlap is $\cos^2 22.5° - \sin^2 22.5° = \cos 45° = 1/\sqrt2$, so the probability is $\tfrac12$. ✓
The lesson: only the *relative* orientation of consecutive analyzers enters, never the absolute one —
as it must be, since space has no preferred vertical.
Solution
Blocking the middle $-$ port, the surviving path is $+z \to +\hat n(\theta) \to +z$, so the fraction
is the product of two Born factors:
$$
P = \cos^2\!\frac{\theta}{2}\times\cos^2\!\frac{\theta}{2} = \cos^4\!\frac{\theta}{2} .
$$
It is largest at $\theta = 0$, where $P \to 1$ — but then the middle analyzer measures the same axis
as the others and does nothing at all (Experiment 1). At $\theta = 90°$ it gives
$\cos^4 45° = \tfrac14$, matching the three-analyzer bookkeeping in the text. The instructive case is
$\theta = 180°$, where $P = 0$: inserting a *reversed* middle analyzer and keeping its $+$ port
extinguishes the beam entirely, because the surviving atoms are exactly the ones that were $-z$.
Solution
**(a)** The prediction is $p = \cos^2 22.5° = 0.8536$, so the expected count in $N = 100$ shots is
$85.4$. Counts are binomial, so
$\sigma = \sqrt{Np(1-p)} = \sqrt{100 \times 0.8536 \times 0.1464} = 3.54$. The observed 85 sits
$|85 - 85.4|/3.54 = 0.10$ standard deviations from the prediction — as good agreement as one could
ask for. (It would be *suspicious* if every run landed exactly on the curve; genuine random data
scatters by about $\sigma$.)
**(b)** The rival predicts $P_{\text{lin}} = 1 - 45/180 = 0.75$, so the gap is
$\Delta = 0.8536 - 0.75 = 0.1036$. The standard error of an estimated probability from $N$ shots is
$\sigma_{\hat p} = \sqrt{p(1-p)/N} \approx 0.357/\sqrt{N}$. Requiring $\Delta \ge 3\sigma_{\hat p}$:
$$
0.1036 \ge \frac{3 \times 0.357}{\sqrt{N}}
\quad\Longrightarrow\quad
\sqrt{N} \ge 10.3
\quad\Longrightarrow\quad
N \gtrsim 107 .
$$
About 110 shots at $45°$ settle it — which is why the historical measurements at intermediate angles
were decisive despite modest statistics, and why $45°$ is the right place to look: it is where the two
laws differ most.
Answers
1. (i) The pattern is discrete — two spots rather than a filled band — which one orientation suffices to show. (ii) The splitting is the *same for every orientation*, which requires comparing runs at different angles and is the fact that rules out pre-existing classical arrows.
2. Orbital $\ell$ gives $2\ell+1$ orientations, always odd, never two; and silver's valence electron has $\ell = 0$, predicting no splitting at all. Phipps and Taylor's hydrogen run removed the last escape route — a one-electron atom with unambiguously zero orbital angular momentum still split in two.
3. (1) Repeating the same analyzer confirms the result with certainty. (2) Reversing the analyzer swaps the labels, still with certainty. (3) A perpendicular analyzer splits 50–50. (4) A third analyzer back along the original axis also splits 50–50 — the earlier result is erased — while a third along the middle axis confirms with certainty. (5) At relative angle $\theta$ the $+$ port receives $\cos^2(\theta/2)$.
4. $\lvert{\pm x}\rangle = (\lvert{+z}\rangle \pm \lvert{-z}\rangle)/\sqrt2$. Normalization: $\tfrac12(1+1) = 1$. Orthogonality: $\langle{+x}|{-x}\rangle = \tfrac12(1-1) = 0$. Born: $|\langle{\pm x}|{+z}\rangle|^2 = |1/\sqrt2|^2 = \tfrac12$. ✓
5. Rotating the apparatus by $\theta$ rotates the state by $\theta/2$ — the two-to-one correspondence between $SU(2)$ and $SO(3)$. A full $360°$ apparatus turn returns $-\lvert{+z}\rangle$: the same physical state, since global phase is unobservable, but a sign that shows the state space is a double cover of ordinary space.