In the previous notebook, the hydrogen atom led us to a new question: can a particle have angular momentum that does not arise from orbiting through space? The Stern–Gerlach experiment gives us an experimental route toward an answer.
Imagine a narrow beam of neutral silver atoms passing through an inhomogeneous magnetic field. The atoms are not charged projectiles deflected by an electric force. Their magnetic moments interact with the field gradient. In our idealized spin-½ model, an analyzer aligned with an axis (first $z$) separates the beam into two outgoing channels, labelled $z+$ and $z-$. The labels refer to the measured spin projection $S_z=+\hbar/2$ or $S_z=-\hbar/2$; they do not by themselves specify which spot lies physically above the other.
Predict, before operating the instrument. If the incoming beam has no preferred spin direction, would you expect one spot, two distinct spots, or a continuous distribution? If you send only a small number of atoms, must the two spot counts be identical?
Virtual experiment — one analyzer. Start with the unpolarized source and a terminal $z$ analyzer. Choose the number of launched atoms and perform an acquisition. Inspect the two detector counts, not just the model's probability readout. Increase the sample size or acquire another reproducible sample. Which result is stable as an expected pattern? Which numerical details fluctuate?
The last active stage is always the terminal detector; earlier stages are selectors. Set Stages to 1 and Stage 1 axis to z. Change Launched N_0 if desired. A control change updates the current acquisition; Acquire again requests a new acquisition, which can coincidentally have the same integers.
Observe and interpret. Two possible detector channels appear in the ideal spin-½ model. For an unpolarized ensemble, each has theoretical probability $1/2$. Individual atoms are recorded in one channel, and finite detector counts need not be equal. A simple classical picture with continuously distributed projection values does not predict just these two ideal spin-projection values.
Keep the categories separate: the detector-like readout is the integer channel count; $1/2$ is a model probability; a count divided by the number actually detected is a sample fraction. An unpolarized beam is a mixed ensemble, not a single coherent two-component state with a secretly chosen direction.
The apparatus drawing is qualitative flow only: it does not calibrate fields or deflections, and paths are not recombined. Signs label spin projections, not magnetic-moment orientation.
An analyzer does more than separate two possible results: if we keep only one outgoing path, it becomes a state-preparation device.
Predict. Send an unpolarized beam into a $z$ analyzer, transmit only the $z+$ path, and then analyze those transmitted atoms along $z$ again. Will the final detector split its incoming beam equally, or will one channel be certain? Will every originally launched atom arrive there?
Virtual experiment — repeat the same axis. Return to the instrument above; set Stages to 2, Stage 1 axis to z, Stage 1 select to +, and Stage 2 axis to z. Only active controls describe this experiment.
Configure a first $z$ analyzer as a selector that transmits $z+$ and blocks $z-$. Follow it with a terminal $z$ analyzer. Inspect the launched count $N_0$, the number reaching the second analyzer, the blocked number, and the final detector channels. Do not interpret intermediate branch assignments as measurements made by a destructive detector placed inside the beamline.
Observe and interpret. A selected $z+$ state has a definite $S_z$ result. Every particle that reaches an ideal second $z$ analyzer exits through $z+$; however, the first selector typically transmits only about half of an initially unpolarized beam. The statement “100% in the final $z+$ channel” is conditional on arrival at the final analyzer, not 100% of $N_0$.
Mathematical Window — a prepared spin state. We name the two normalized $z$-basis states $|+z\rangle$ and $|-z\rangle$. A $z+$ selector prepares $|+z\rangle$ for every atom it transmits. For this ideal prepared state,
$$ P(z+\mid z+\ \mathrm{prepared})=1,\qquad P(z-\mid z+\ \mathrm{prepared})=0. $$
The ideal expected terminal shares relative to $N_0$ are $(1/2,0)$. This relation predicts the conditional detector result; it does not undo atoms removed by the selector.
A quantum state prepared to give a definite answer about $S_z$ need not give a definite answer about a different component.
Predict. Keep the first selector at $z+$, but set the final analyzer to $x$. If the first measurement established the spin, why might the second measurement again produce two outcomes? Is the final split measured relative to the original beam or to particles entering that analyzer?
Virtual experiment — orthogonal axes. Return to the instrument above; set Stages to 2, Stage 1 axis to z, Stage 1 select to +, and Stage 2 axis to x.
Compare the configurations $z+\to z$ and $z+\to x$ while keeping the launched count and acquisition settings controlled. Track final detector counts, their measured sample fractions, and the model probabilities separately.
Observe and interpret. After preparing $|+z\rangle$, an ideal $x$ measurement has two equally probable outcomes:
$$ P(x+\mid z+)=P(x-\mid z+)=\frac12. $$
The ideal expected terminal shares relative to $N_0$ are $(1/4,1/4)$. This is not a loss of instrumental accuracy. It is a property of the prepared quantum state and the measurement basis. An exactly known $S_z$ projection does not imply an exactly known $S_x$ projection.
Spin uncertainty, interpreting the model. For $|+z\rangle$, $\Delta S_z=0$, while $\Delta S_x=\hbar/2$. The first value expresses certainty about $S_z$; the second represents the spread of possible $S_x$ outcomes, which are $\pm\hbar/2$ with equal probabilities. These are model-defined standard deviations of spin-component outcomes, not widths of the beam spots or errors of the detector.
Now add a third analyzer. The purpose is not to make the instrument more complicated; it is to test whether the result of a later $z$ measurement depends on what we did in between.
Predict. Compare two paths:
Does the last $z$ analyzer behave the same way in both cases? What happens to the number of particles that reach it?
Virtual experiment — sequential incompatible selection. Return to the instrument above; set Stages to 3, Stage 1 axis to z, Stage 1 select to +, Stage 2 axis to x, Stage 2 select to +, and Stage 3 axis to z.
Run the three-stage sequence
$$ \text{unpolarized source} \longrightarrow \mathrm{SG}_z\ (z+\ \text{selected}) \longrightarrow \mathrm{SG}_x\ (x+\ \text{selected}) \longrightarrow \mathrm{SG}_z\ (\text{terminal}). $$
Compare it with the two-stage compatible repetition. At every selector, check entering, selected/transmitted, and blocked counts. At the final detector, inspect $z+$ and $z-$ counts. If no atoms survive an earlier filter during a small finite acquisition, the later detector has no observations: an empirical fraction is not zero, but undefined for that acquisition.
Observe and interpret. The intermediate $x+$ selection prepares the transmitted atoms in a new state. A subsequent $z$ measurement has probabilities
$$ P(z+\mid x+)=P(z-\mid x+)=\frac12. $$
The initial $z$ certainty has not survived the incompatible intermediate preparation. The sequence therefore demonstrates why preparation and the order of spin measurements matter.
Mathematical Window — never lose the denominator. For a large ensemble, the ideal predictions for this particular unpolarized-source sequence are:
| Step | Probability conditional on reaching the step | Expected fraction of original $N_0$ |
|---|---|---|
| Transmit $z+$ through first selector | $1/2$ | $1/2$ |
| Transmit $x+$ through second selector | $1/2$ | $1/4$ |
| Terminal $z+$ detection | $1/2$ | $1/8$ |
| Terminal $z-$ detection | $1/2$ | $1/8$ |
Here $n_{\mathrm{enter}}$ is the actual number entering a stage and is the denominator of its empirical conditional fractions; $N_0$ is the launched-beam denominator for the model fractions. If $n_{\mathrm{enter}}=0$, terminal counts are zero and empirical conditional fractions are undefined, not 0%. Hypothetical conditional model probabilities remain meaningful but are not observations. Intermediate assignments are simulated bookkeeping, not destructive detector readings.
The fractions in the right column are theoretical expectations, not exact identities for small randomly sampled counts. At every stage, ask: What population is in the denominator?
Reflect. Why would displaying only a final “50%/50%” result, without the transmitted counts, give an incomplete account of the experiment?
We have seen that a $z+$ preparation predicts certain $z$ outcomes but equal $x$ outcomes. We now introduce the smallest mathematical model that explains both.
Mathematical Window — spinors and projections. In the $z$ basis a normalized pure spin-½ state can be written
$$ |\psi\rangle=\alpha|+z\rangle+\beta|-z\rangle, \qquad |\alpha|^2+|\beta|^2=1. $$
The coefficients are probability amplitudes, not detector signals or classical fractions of two beam paths. If we measure $S_z$ on this pure state, the model predicts $P(z+)=|\alpha|^2$ and $P(z-)=|\beta|^2$.
The two eigenstates are represented in the $z$ basis by
$$ |+z\rangle=\begin{pmatrix}1\\0\end{pmatrix}, \qquad |-z\rangle=\begin{pmatrix}0\\1\end{pmatrix}. $$
The equally normalized $x$ eigenstates are
$$ |+x\rangle=\frac{1}{\sqrt2} \begin{pmatrix}1\\1\end{pmatrix}, \qquad |-x\rangle=\frac{1}{\sqrt2} \begin{pmatrix}1\\-1\end{pmatrix}. $$
The overlap gives the probability for measuring $x+$ after preparing $z+$:
$$ P(x+\mid z+)=|\langle +x|+z\rangle|^2=\frac12. $$
Mathematical Window — operators and values. The component of a spin-½ angular momentum along axis $i$ is represented by
$$ S_i=\frac{\hbar}{2}\sigma_i, \qquad i\in\{x,y,z\}, $$
where the Pauli matrices in the $z$ basis are (the $y$ matrix is theory, not an instrument control)
$$ \sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad \sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\quad \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$
Every $S_i$ has the two possible measured eigenvalues $\pm\hbar/2$. However, different component operators are incompatible: their order matters ($S_xS_z\ne S_zS_x$). This concise mathematical statement is consistent with the earlier $z\to x\to z$ experiment. We do not need to derive a general uncertainty theorem here.
Generalization, not another required control. If an atom is prepared in the $+$ state along an axis $\mathbf a$ and measured along axis $\mathbf b$, separated by angle $\theta$, the ideal model predicts
$$ P(+_{\mathbf b}\mid +_{\mathbf a})=\cos^2\left(\frac{\theta}{2}\right), \quad P(-_{\mathbf b}\mid +_{\mathbf a})=\sin^2\left(\frac{\theta}{2}\right). $$
The same-axis case gives certainty, and perpendicular axes give $1/2$–$1/2$. The angle refers to measurement directions; it does not mean that a tiny classical ball simply points along a hidden axis.
Concept check. If a pure $|+x\rangle$ state and an unpolarized ensemble both yield 50:50 results when analyzed along $z$, are they the same physical state? What other measurement would distinguish them?
The silver-atom Stern–Gerlach experiment is not an MRI scanner. It illustrates something more basic: a quantum spin has discrete measurement outcomes, its state can be prepared, and the probabilities depend on the measurement axis.
In ordinary clinical proton MRI, the important spins are primarily nuclear spins of hydrogen nuclei, not the electron spin responsible for the simplified silver-atom demonstration. Magnetic fields, resonant driving, and detection of collective magnetic signals connect those spins to measurable imaging data. We will study the needed dynamics and interactions in Notebook 08, and their application to NMR/MRI in Notebook 11. This simulation does not model NMR driving, resonance signals, or MRI imaging. The present notebook establishes the preparation-and-measurement logic that those later models build upon.
Reflect. Which quantities have been directly represented as detector readings in our virtual Stern–Gerlach experiment, and which are predictions of its ideal quantum model? Why would a measurement of magnetic resonance involve new physics beyond this stationary analyzer?
Return to the Question of the Day: Why do spin measurements give discrete outcomes? In the spin-½ model, an analyzer of any chosen spin component has just two eigenvalues, $+\hbar/2$ and $-\hbar/2$. A given particle contributes one outcome, while a prepared state predicts probabilities for repeated measurements. Selecting a beam branch prepares a new state and changes both later probabilities and transmitted intensity.
Self-check (without changing the physics implementation):
So far we have changed what we measure, not how a spin state evolves in time. What happens if a magnetic interaction couples two levels or a field drives transitions between them? That is the next question.