The Stern–Gerlach experiment, in depth

The previous section handed spin a crisp set of mathematics: measure along any direction and there are only two answers, $\pm\tfrac\hbar2$. That sounds far too tidy to be how the real world behaves. This section turns to experiment — the best case in the history of quantum mechanics of “one apparatus teaching an entire course”.

1922: a beam of silver atoms and two traces

Here is what Otto Stern and Walther Gerlach did in Frankfurt: heat silver in an oven until it evaporates, let the silver atoms stream through a slit to form a beam, fly it past a magnet with a knife-edge pole above and a flat pole below (the key point: the field is inhomogeneous, stronger near the sharp pole), and let the atoms land on a glass plate to be developed.

Why inhomogeneous? The energy of a magnetic moment $\vec\mu$ in a field is $U=-\vec\mu\cdot\vec B$. A uniform field only twists the moment (a torque) and exerts no net force; only a non-uniform field pulls:

$$F_z=-\frac{\partial U}{\partial z}=\mu_z\frac{\partial B_z}{\partial z}\tag{5.4.1}$$

The direction of the force is set by the sign of $\mu_z$, its size by the value of $\mu_z$. So the landing position on the glass plate is a readout of $\mu_z$ — and a silver atom’s magnetic moment is proportional to the spin of its lone valence electron (of silver’s 47 electrons, 46 pair off and cancel; the remaining one happens to sit in an $l=0$ orbital, so orbital angular momentum can neither help nor interfere — silver is nature’s gift of a pure spin sample). Hence: this apparatus is an $\hat S_z$ meter, with the landing spot as its needle.

Now compare two predictions:

• Classical prediction: atoms leave the oven with randomly oriented moments, $\mu_z$ distributed continuously from $-\mu$ to $+\mu$, so the glass plate should show one continuous vertical smear.
• Actual result: no smear. The silver atoms split neatly into two clusters, with empty space between.

Two clusters — no more, no fewer. Compare with the previous section: $2s+1=2$ gives $s=\tfrac12$, each atom’s $S_z$ can only be $+\tfrac\hbar2$ or $-\tfrac\hbar2$, with nothing in between. Orientation in space is quantised — which is why the phenomenon was christened “space quantisation”.

Chaining the devices: three configurations

A single device is merely evidence of quantisation. The genuine textbook-grade shock comes from cascading: feed one output beam of an SG device into the next. Notation: SGz is a device with its field along z, SGx along x; each has two exits, labelled $+$ and $-$.

The three cascaded configurations, computed one by onebasic~10 minexpand

Only three tools are needed: the $\hat S_z$ basis $\ket{\uparrow},\ket{\downarrow}$; the $\hat S_x$ eigenstates solved in the previous section,

$$\ket{+x}=\frac{\ket{\uparrow}+\ket{\downarrow}}{\sqrt2},\qquad \ket{-x}=\frac{\ket{\uparrow}-\ket{\downarrow}}{\sqrt2}\tag{5.4.2}$$

(set $\theta=\tfrac\pi2$ and $\phi=0$ or $\pi$ in the arbitrary-direction formula); and the projection postulate of section 3.8: probabilities are squared amplitudes, and the post-measurement state is the corresponding eigenstate.

Configuration 1: SGz → SGz. A consistency check on the instrument.

Atoms from the oven enter the first SGz and split randomly into two beams; keep only the $+$ exit (the state is projected to $\ket{\uparrow}$) and feed it into a second SGz. Probabilities:

$$P(+)=|\braket{\uparrow}{\uparrow}|^2=1,\qquad P(-)=|\braket{\downarrow}{\uparrow}|^2=0\tag{5.4.3}$$

Output: $100\%$ from the $+$ exit. Repeating a measurement of the same observable repeats the result — the “repeatability” clause of the projection postulate. The instrument is self-consistent, not a random-number generator.

Configuration 2: SGz → SGx. Ask the question along a different axis.

Take the same $\ket{\uparrow}$ beam and send it into SGx. Rewrite it in the $\hat S_x$ basis:

$$\ket{\uparrow}=\frac{\ket{+x}+\ket{-x}}{\sqrt2}\tag{5.4.4}$$

(add the two defining equations of $\ket{\pm x}$ to verify). Then

$$P(+x)=\left|\braket{+x}{\uparrow}\right|^2=\left|\tfrac1{\sqrt2}\right|^2=\frac12, \qquad P(-x)=\frac12\tag{5.4.5}$$

Output: $50\%:50\%$, completely random atom by atom. A state with definite $S_z$ has no answer to the question “what is $S_x$” — only probabilities — because $[\hat S_z,\hat S_x]\ne0$: the two share no common eigenstates.

Configuration 3: SGz → SGx → SGz. The knockout punch.

Continue from configuration 2, keeping only the $+$ exit of SGx — this beam’s state has been projected to $\ket{+x}$. Now send it into a third device, SGz. The naive expectation: these atoms were already “certified” back at the first device as $S_z=+\tfrac\hbar2$; in between we merely measured an unrelated $S_x$, so surely $100\%$ should exit through $+$.

Compute it. The current state is $\ket{+x}=(\ket{\uparrow}+\ket{\downarrow})/\sqrt2$:

$$P(+)=\left|\braket{\uparrow}{+x}\right|^2=\frac12,\qquad P(-)=\left|\braket{\downarrow}{+x}\right|^2=\frac12\tag{5.4.6}$$

Output: $50\%:50\%$. Half the atoms have $S_z=-\tfrac\hbar2$ — even though every atom in this beam was once confirmed by the first device to have $S_z=+\tfrac\hbar2$, and nothing touched their z direction along the way.

Tallying the survival fractions across all three stages: oven → SGz keep $+$ ($\tfrac12$) → SGx keep $+$ ($\tfrac12$) → SGz exit $-$ ($\tfrac12$): in the end $\tfrac18$ of the atoms complete this “impossible” journey.

Configuration 3 is non-commutativity made flesh: the very act of measuring $S_x$ erases the memory of $S_z$. After projection onto $\ket{+x}$, not one bit of “used to be $\ket{\uparrow}$” survives in the state — $\ket{+x}$ is $\ket{+x}$, regardless of which beam it was projected from. The definiteness of $S_z$ and $S_x$ works like a seesaw: press one end down and the other must rise.

“Maybe the atoms just carry labels?” — the rout of naive hidden variables

Faced with configuration 3, a natural line of resistance is: perhaps the randomness is merely our ignorance. Imagine each atom leaves the factory with a label listing its answer to every possible direction, say “if asked z answer $+$, if asked x answer $-$”. Measurement merely reads the label, and the 50:50 of configuration 3 just reflects the statistics of the labels. No “measurement changes the state” required.

This model can explain the two clusters of a single measurement, and can even fake the 50:50 of configuration 2. But there is one upgraded experiment it cannot survive.

Models of this kind — the atom carrying a pre-written answer sheet for every question, i.e. non-contextual hidden variables — are already limping badly here; chapter 9’s Bell inequalities will give them, together with all their local cousins, a formal burial. For now, the experimental fact is enough: the values of non-commuting observables do not pre-exist, waiting to be read off.

∎Key formulas

• The SG force

  $$F_z=\mu_z\,\frac{\partial B_z}{\partial z}$$

  The gradient translates the value of μ_z into a deflection — the landing spot is the readout

• Basis change

  $$\ket{\pm x}=\tfrac{1}{\sqrt2}\big(\ket{\uparrow}\pm\ket{\downarrow}\big)$$

  The complete toolkit for every cascade calculation

• Cascade probability

  $$P=\lvert\braket{\text{exit}}{\text{incident}}\rvert^2$$

  z+ → x: 50/50; z+ → x+ → z: 50/50 — the z memory is erased

• Recombination identity

  $$\tfrac1{\sqrt2}\ket{+x}+\tfrac1{\sqrt2}\ket{-x}=\ket{\uparrow}$$

  With no path record the amplitudes fold back coherently: 100% recovery of the original state

?Self-check4 questions

1. 1.

   Why must the SG experiment use an inhomogeneous magnetic field?

   • A uniform field gives the moment a torque but no net force; only an inhomogeneous field pulls atoms with different μ_z to different positions
   • A uniform field would accelerate the atoms too much
   • Only an inhomogeneous field can quantise the magnetic moment
   • A uniform field would erase the spin

   Submit
2. 2.

   In configuration 3 (SGz keep + → SGx keep + → SGz), the third device finds half the atoms with S_z down. The reason is:

   • Measuring S_x projects the state onto |+x⟩, which retains no information whatsoever about "once had S_z up"
   • Collisional disturbance by the SGx field changed S_z
   • The first device gets it wrong half the time
   • The atoms depolarise naturally in flight

   Submit
3. 3.

   The recombination experiment (SGx splits the beam, then merges it with no measurement) outputs 100% |↑⟩. This shows: (Select all that apply.)

   Select all that apply

   • Splitting a beam is not in itself a measurement; with no path information recorded there is no projection
   • The amplitudes of the two paths superpose coherently and interfere back into the original state
   • The naive hidden-variable model of "each atom carries an answer sheet for every direction" contradicts experiment
   • The magnetic field has no effect on the spin at all

   Submit
4. 4.

   A beam of S_z = +ħ/2 atoms passes first through SGx (keeping the +x exit), then through SGz. What fraction of the incident beam finally exits through the "−" port of SGz?

   1% relative tolerance

   Submit

What comes next

So far, the SG apparatus has only answered “which way does the spin point at this instant”. But between two measurements, what is the spin doing?

Put the atom in a uniform magnetic field (no gradient needed this time — we are not measuring, only evolving), and the spin’s expectation value circles the field like a gyroscope. This rotation can be calculated, can be measured, and is the entire working principle of nuclear magnetic resonance and hospital MRI — the subject of the next section.