---
title: Spin-½, the Pauli Matrices, and Stern–Gerlach
module: Spin
moduleNumber: 8
lessonNumber: 1
order: 801
summary: >
  A silver atom passing through an inhomogeneous magnetic field splits into two
  beams, not a smear. That single fact fixes the internal angular momentum of the
  electron to a two-valued quantity with no spatial wavefunction. We build the
  two-dimensional spin space, the Pauli matrices and their algebra, the spinor for
  measurement along an arbitrary axis, and the sequential Stern–Gerlach filters
  that expose measurement disturbance.
topics: [Spin]
sources:
  - book: Sakurai & Napolitano
    ref: "§1.1 The Stern–Gerlach Experiment; §3.2 Spin-½ Systems"
  - book: Griffiths & Schroeter
    ref: "§4.4.1 Spin ½"
  - book: Cohen-Tannoudji, Diu & Laloë
    ref: "Ch. IV — Application of the Postulates to Simple Cases"
draft: false
---

The [angular-momentum algebra](/quantum-mechanics/angular-momentum/the-angular-momentum-algebra)
admits half-integer quantum numbers that no wavefunction on the sphere can carry.
Orbital motion realizes only integer $\ell$, because $Y_\ell^m$ must be
single-valued under $\phi \to \phi + 2\pi$. The missing half-integers are physical,
and the electron carries the smallest one, $s = \tfrac12$. This degree of freedom
has no position-space representation: it is not the electron spinning on an axis
in any literal sense, and treating it that way gives a surface speed exceeding $c$.
Spin is an intrinsic two-valued observable, defined operationally by what a
magnet does to a beam of atoms.

## The Stern–Gerlach experiment

A neutral atom with magnetic moment $\vec\mu$ sitting in a field $\vec B$ has
energy $H = -\vec\mu\cdot\vec B$. If the field varies in space, the atom feels a
force

$$
\vec F = -\nabla H = \nabla(\vec\mu\cdot\vec B).
$$

Arrange the pole pieces so that near the beam $\vec B$ points along $z$ and its
magnitude changes rapidly with $z$ while the transverse variation is small. The
$z$-component of the force is then

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

proportional to the projection $\mu_z$ of the moment along the field. The
deflection at the screen measures $\mu_z$, and through $\vec\mu = \gamma\vec S$ it
measures $S_z$.

The 1922 experiment used silver. Silver's ground-state electron configuration
closes every shell except a single $5s$ electron, so the total orbital angular
momentum is zero and the atomic moment comes entirely from that one electron's
spin.[^sn-sg] Classically $\mu_z = |\vec\mu|\cos\theta$ takes every value in
$[-|\vec\mu|, +|\vec\mu|]$ as the moment's orientation ranges over the sphere, so
the screen should show a continuous vertical smear. It does not. The beam splits
into two sharp spots, symmetric about the undeflected line.

$$
% caption: A collimated beam of silver atoms is split by an inhomogeneous field
% into exactly two components, corresponding to the two eigenvalues of S_z; the
% classical prediction is a continuous smear.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% oven and collimator
\node[draw, minimum width=13mm, minimum height=9mm, fill=black!5] (oven) at (0,0) {oven};
\node[draw, minimum width=4mm, minimum height=9mm, fill=black!8] (col) at (2.0,0) {};
\node[font=\footnotesize, anchor=north] at (2.0,-0.7) {slit};
% magnet pole pieces
\draw[thick, fill=black!8] (3.6,0.55) -- (6.2,0.55) -- (6.2,1.15) -- (3.6,1.15) -- cycle;
\node[font=\footnotesize] at (4.9,0.85) {N};
\draw[thick, fill=black!8] (3.7,-0.95) -- (6.1,-0.95) -- (6.3,-0.55) -- (3.5,-0.55) -- cycle;
\node[font=\footnotesize] at (4.9,-0.78) {S};
% beam
\draw[->, black, thick] (oven) -- (col);
\draw[thick, black] (2.2,0) -- (3.6,0);
\draw[->, thick, acc] (3.6,0) .. controls (5.5,0) and (6.5,0.55) .. (8.4,0.75);
\draw[->, thick, acc] (3.6,0) .. controls (5.5,0) and (6.5,-0.55) .. (8.4,-0.75);
\draw[dashed, black] (3.6,0) -- (8.4,0);
% screen
\draw[very thick, black] (8.6,-1.3) -- (8.6,1.3);
\node[circle, fill=acc, inner sep=1.6pt] at (8.6,0.75) {};
\node[circle, fill=acc, inner sep=1.6pt] at (8.6,-0.75) {};
\node[font=\footnotesize, anchor=west] at (8.75,0.75) {spin up};
\node[font=\footnotesize, anchor=west] at (8.75,-0.75) {spin down};
\node[font=\footnotesize, anchor=west, text=black] at (8.75,0.0) {classical smear};
\draw[black, line width=2.2pt, opacity=0.4] (8.55,-0.6) -- (8.55,0.6);
\end{tikzpicture}
$$

Two spots means the observable $S_z$ takes exactly two values. Measuring them from
the deflection and the known field gradient gives $\pm\tfrac12\hbar$. The
apparatus is a two-outcome measuring device, and the atom's spin along $z$ is a
two-valued quantity. No continuous internal orientation survives contact with the
data.

## The two-dimensional state space

A single spin-½ has a two-dimensional state space. Take the $S_z$ eigenstates as
the basis,

$$
S_z\lvert +\rangle = +\tfrac{\hbar}{2}\lvert +\rangle,
\qquad
S_z\lvert -\rangle = -\tfrac{\hbar}{2}\lvert -\rangle,
$$

with $\langle +|-\rangle = 0$ and each normalized. A general spin state is a
superposition

$$
\lvert\chi\rangle = a\lvert +\rangle + b\lvert -\rangle,
\qquad
|a|^2 + |b|^2 = 1,
$$

written as a two-component column, the **spinor**

$$
\chi = \begin{pmatrix} a \\ b \end{pmatrix},
\qquad
\lvert +\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix},
\qquad
\lvert -\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.
$$

Born's rule reads off the outcome statistics: a measurement of $S_z$ yields
$+\hbar/2$ with probability $|a|^2$ and $-\hbar/2$ with probability $|b|^2$.
After the measurement the state collapses onto the corresponding eigenspinor.

The three spin components inherit the angular-momentum algebra
$[S_i, S_j] = i\hbar\,\varepsilon_{ijk}S_k$. In the $S_z$ basis they are
represented by $2\times 2$ Hermitian matrices. Writing $S_i = \tfrac{\hbar}{2}\sigma_i$
defines the **Pauli matrices** $\sigma_i$.

## The Pauli matrices

The matrix elements follow from the angular-momentum results for
$j = \tfrac12$.[^gs-spin] The ladder operators $S_\pm = S_x \pm iS_y$ act by
$S_+\lvert -\rangle = \hbar\lvert +\rangle$ and $S_-\lvert +\rangle = \hbar\lvert -\rangle$,
with the top and bottom rungs annihilated. Solving for $S_x, S_y$ gives

$$
\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\qquad
\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix},
\qquad
\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.
$$

These three matrices, with the identity $I$, span the real vector space of
$2\times 2$ Hermitian matrices. Their algebraic properties are the entire toolkit
for spin-½.

> **Definition (Pauli algebra).** The Pauli matrices are Hermitian and traceless,
> square to the identity, and satisfy
> $$
> \sigma_i\sigma_j = \delta_{ij}I + i\,\varepsilon_{ijk}\sigma_k .
> $$
> Splitting this into its symmetric and antisymmetric parts gives the
> anticommutator and commutator
> $$
> \{\sigma_i,\sigma_j\} = 2\delta_{ij}I,
> \qquad
> [\sigma_i,\sigma_j] = 2i\,\varepsilon_{ijk}\sigma_k .
> $$

Each $\sigma_i$ has eigenvalues $\pm 1$ (so each $S_i$ has eigenvalues
$\pm\hbar/2$), determinant $-1$, and trace $0$. The single product rule above
generates every identity one needs. Two consequences are used constantly. First,
for any two ordinary vectors $\vec a, \vec b$,

$$
(\vec\sigma\cdot\vec a)(\vec\sigma\cdot\vec b)
 = (\vec a\cdot\vec b)\,I + i\,\vec\sigma\cdot(\vec a\times\vec b).
$$

> **Proof.** Expand the left side over components:
> $(\vec\sigma\cdot\vec a)(\vec\sigma\cdot\vec b) = \sum_{i,j}\sigma_i\sigma_j\,a_i b_j$.
> Insert $\sigma_i\sigma_j = \delta_{ij}I + i\varepsilon_{ijk}\sigma_k$:
> $$
> \sum_{i,j}\big(\delta_{ij}I + i\varepsilon_{ijk}\sigma_k\big)a_i b_j
> = (\vec a\cdot\vec b)\,I + i\,\vec\sigma\cdot(\vec a\times\vec b),
> $$
> since $\sum_{ij}\delta_{ij}a_i b_j = \vec a\cdot\vec b$ and
> $\sum_{ijk}\varepsilon_{ijk}\sigma_k a_i b_j = \vec\sigma\cdot(\vec a\times\vec b)$.

Second, setting $\vec a = \vec b = \hat n$ a unit vector gives
$(\vec\sigma\cdot\hat n)^2 = I$, which lets any function of $\vec\sigma\cdot\hat n$
collapse to a linear expression. That fact drives spin rotations in the
[next lesson](/quantum-mechanics/spin/spin-in-a-magnetic-field-precession-and-resonance).

$$
% caption: The Pauli matrices close under multiplication in a cyclic pattern:
% sigma_x sigma_y = i sigma_z and its cyclic permutations, with a sign reversal
% for the reverse order. The nodes are labelled by axis; each solid arrow carries
% the factor i, the same su(2) structure as the cross product.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, circle, minimum size=11mm] (X) at (0,1.9) {$X$};
\node[draw, circle, minimum size=11mm] (Y) at (1.75,-1.0) {$Y$};
\node[draw, circle, minimum size=11mm] (Z) at (-1.75,-1.0) {$Z$};
\draw[->, thick, acc] (X) to[bend left=22] node[right, font=\footnotesize, black] {$i$} (Y);
\draw[->, thick, acc] (Y) to[bend left=22] node[below, font=\footnotesize, black] {$i$} (Z);
\draw[->, thick, acc] (Z) to[bend left=22] node[left, font=\footnotesize, black] {$i$} (X);
\node[font=\footnotesize, text=black, align=center] at (0,-0.1) {forward cycle};
\node[draw, fill=black!4, font=\footnotesize, align=left, anchor=west] at (3.0,0.4)
  {each square is $I$\\[2pt] Hermitian, trace $0$};
\end{tikzpicture}
$$

## Spin along an arbitrary axis

Point a Stern–Gerlach magnet along an arbitrary unit vector
$\hat n = (\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta)$. It
measures the component

$$
S_n = \vec S\cdot\hat n = \tfrac{\hbar}{2}\,\vec\sigma\cdot\hat n,
\qquad
\vec\sigma\cdot\hat n =
\begin{pmatrix}
 \cos\theta & \sin\theta\,e^{-i\varphi} \\
 \sin\theta\,e^{i\varphi} & -\cos\theta
\end{pmatrix}.
$$

Because $(\vec\sigma\cdot\hat n)^2 = I$, the eigenvalues are again $\pm 1$, so
every direction is a two-outcome measurement with results $\pm\hbar/2$. Any axis
is as good as $z$; nothing distinguishes a preferred direction. The eigenspinors
follow from diagonalizing the $2\times 2$ matrix.

> **Worked example.** Find the spin state pointing along $\hat n$, i.e. the
> $+\hbar/2$ eigenspinor of $S_n$. Writing $\chi_+ = (\alpha,\beta)^{\mathsf T}$,
> the eigenvalue equation $(\vec\sigma\cdot\hat n)\chi_+ = \chi_+$ gives the top row
> $$
> \cos\theta\,\alpha + \sin\theta\,e^{-i\varphi}\beta = \alpha
> \ \Longrightarrow\
> \frac{\beta}{\alpha} = \frac{1-\cos\theta}{\sin\theta}\,e^{i\varphi}
> = \tan\tfrac{\theta}{2}\,e^{i\varphi},
> $$
> using the half-angle identity $(1-\cos\theta)/\sin\theta = \tan(\theta/2)$.
> Normalizing $|\alpha|^2 + |\beta|^2 = 1$ and choosing the overall phase so that
> $\alpha$ is real,
> $$
> \chi_+(\hat n) =
> \begin{pmatrix} \cos\tfrac{\theta}{2} \\[2pt] \sin\tfrac{\theta}{2}\,e^{i\varphi} \end{pmatrix},
> \qquad
> \chi_-(\hat n) =
> \begin{pmatrix} \sin\tfrac{\theta}{2} \\[2pt] -\cos\tfrac{\theta}{2}\,e^{i\varphi} \end{pmatrix}.
> $$
> The polar angle enters as $\theta/2$: rotating the measurement axis by $2\pi$
> sends $\chi_+ \to -\chi_+$, the sign flip characteristic of half-integer spin.

The half-angle is the source of all the interesting statistics. Prepare an atom
in $\lvert +\rangle$ (spin up along $z$) and measure $S_n$. The amplitude for the
$+\hbar/2$ outcome is $\langle\chi_+(\hat n)\lvert +\rangle = \cos(\theta/2)$, so

$$
P_+(\theta) = \cos^2\tfrac{\theta}{2},
\qquad
P_-(\theta) = \sin^2\tfrac{\theta}{2},
\qquad
\langle S_n\rangle = \tfrac{\hbar}{2}\cos\theta .
$$

The expectation value tracks the classical projection $|\vec S|\cos\theta$, but
every individual measurement returns only $\pm\hbar/2$. The classical cosine
reappears as an average over the two quantized outcomes, not as a value any single
atom carries.

$$
% caption: Preparing spin up along z and measuring along an axis at polar angle
% theta gives outcome probabilities cos^2(theta/2) and sin^2(theta/2); the two
% curves cross at theta = pi/2, where the axes are mutually unbiased.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (0,0) -- (6.6,0) node[right, font=\footnotesize] {angle (deg)};
\draw[->, black] (0,0) -- (0,3.1) node[above, font=\footnotesize] {probability};
\node[font=\footnotesize, anchor=north] at (0,0) {$0$};
\node[font=\footnotesize, anchor=north] at (3.0,0) {$90$};
\node[font=\footnotesize, anchor=north] at (6.0,0) {$180$};
\node[font=\footnotesize, anchor=east] at (0,2.6) {$1$};
\node[font=\footnotesize, anchor=east] at (0,1.3) {$\tfrac{1}{2}$};
\draw[dashed, black] (0,2.6) -- (6.0,2.6);
\draw[dashed, black] (3.0,0) -- (3.0,1.3);
% cos^2(theta/2): from (0,1) down to (180deg,0), scaled y by 2.6
\draw[thick, acc] plot[domain=0:6.0, samples=60]
  ({\x}, {2.6*(cos(deg(\x/2)))^2});
% sin^2(theta/2): from (0,0) up to (180deg,1)
\draw[thick, dashed, black] plot[domain=0:6.0, samples=60]
  ({\x}, {2.6*(sin(deg(\x/2)))^2});
\node[text=acc, font=\footnotesize, anchor=south west, fill=white, inner sep=1pt] at (0.6,1.75) {up outcome};
\node[text=black, font=\footnotesize, anchor=south east, fill=white, inner sep=1pt] at (5.4,1.75) {down outcome};
\end{tikzpicture}
$$

## Sequential Stern–Gerlach filters

Cascading magnets exposes the collapse. Block one output of a first magnet and
feed the survivors into a second oriented along a different axis. The atoms have
already been filtered, yet the second magnet still splits them.

$$
% caption: Three Stern–Gerlach stages: a z-filter selects spin up, an x-filter
% selects spin up along x and discards the z information, and a final z-magnet
% again splits the beam into both z outcomes.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\tikzset{sg/.style={draw, minimum width=15mm, minimum height=13mm, fill=acc!8, align=center, font=\footnotesize}}
% source
\node[draw, fill=black!5, minimum width=11mm, minimum height=8mm, font=\footnotesize] (src) at (-1.7,0) {oven};
% stage 1: SG z
\node[sg] (z1) at (0.6,0) {SG\\$z$};
\draw[->, black] (src) -- (z1);
% two outputs, block lower
\draw[thick, acc, ->] (z1.east) -- ++(0.9,0.55) node[right, font=\footnotesize] {up};
\draw[thick, black, ->] (z1.east) -- ++(0.9,-0.55) node[right, font=\footnotesize, text=black] {blocked};
\draw[black, line width=2pt] (2.15,-0.65) -- (2.55,-0.25);
% stage 2: SG x
\node[sg] (x1) at (3.9,0.55) {SG\\$x$};
\draw[thick, acc, ->] (2.4,0.55) -- (x1.west);
\draw[thick, acc, ->] (x1.east) -- ++(0.9,0.5) node[right, font=\footnotesize] {up};
\draw[thick, black, ->] (x1.east) -- ++(0.9,-0.5) node[right, font=\footnotesize, text=black] {blocked};
\draw[black, line width=2pt] (5.35,-0.35) -- (5.75,0.05);
% stage 3: SG z
\node[sg] (z2) at (7.4,1.05) {SG\\$z$};
\draw[thick, acc, ->] (5.6,1.05) -- (z2.west);
\draw[thick, acc, ->] (z2.east) -- ++(0.9,0.45) node[right, font=\footnotesize] {up};
\draw[thick, acc, ->] (z2.east) -- ++(0.9,-0.45) node[right, font=\footnotesize] {down};
\node[font=\footnotesize, text=black, align=center, anchor=north] at (8.4,0.2) {both appear};
\end{tikzpicture}
$$

Read the cascade left to right. The first $z$-magnet passes only $\lvert +\rangle_z$.
The $x$-magnet decomposes that state in the $S_x$ basis,
$\lvert +\rangle_z = \tfrac{1}{\sqrt2}\big(\lvert +\rangle_x + \lvert -\rangle_x\big)$,
so it splits into two beams of equal intensity; keep $\lvert +\rangle_x$. That
survivor, expressed back in the $z$ basis, is
$\lvert +\rangle_x = \tfrac{1}{\sqrt2}\big(\lvert +\rangle_z + \lvert -\rangle_z\big)$.
The final $z$-magnet therefore splits it again, half into each $z$ outcome, even
though the very first magnet had already discarded $\lvert -\rangle_z$.

The intermediate $x$-measurement destroyed the $z$-information. $S_x$ and $S_z$ do
not commute, $[S_x, S_z] = -i\hbar S_y \neq 0$, so no state carries a definite
value of both. Filtering along $x$ prepares a definite $S_x$ and necessarily makes
$S_z$ maximally uncertain. Remove the middle magnet and the two $z$-magnets in
series pass every atom through the upper channel with no second split; insert it
and the beam is regenerated in both channels. Measurement is not passive readout;
it resets the state.

> **Worked example.** An atom leaves the first filter in $\lvert +\rangle_z$. It
> passes an $x$-filter (keep $+x$), then a magnet along $\hat n$ at polar angle
> $\theta$ from $z$ in the $xz$-plane (so $\varphi = 0$). What fraction of the
> original beam exits the $+\hat n$ channel? Each stage multiplies survival
> probabilities. The $x$-filter passes $|\langle +_x|+_z\rangle|^2 = \tfrac12$.
> The survivor $\lvert +\rangle_x$ has $\theta_x = \pi/2,\ \varphi = 0$; the final
> amplitude uses $\hat n$ against $\lvert +\rangle_x$:
> $$
> |\langle \chi_+(\hat n)\,|\,+_x\rangle|^2
> = \big|\cos\tfrac{\theta}{2}\cdot\tfrac{1}{\sqrt2} + \sin\tfrac{\theta}{2}\cdot\tfrac{1}{\sqrt2}\big|^2
> = \tfrac12\big(1+\sin\theta\big).
> $$
> The overall fraction is $\tfrac12\cdot\tfrac12(1+\sin\theta) = \tfrac14(1+\sin\theta)$,
> maximized at $\theta = \pi/2$ (the $+\hat n$ magnet aligned with the $x$-filter)
> where it reaches $\tfrac12$, and falling to $\tfrac14$ when $\hat n$ is along
> $\pm z$.

## The spinor and the density of a spin state

Any normalized spinor can be written, up to an overall phase, as
$\chi_+(\hat n)$ for a unique $\hat n$: solve
$\langle\chi|\vec\sigma|\chi\rangle = \hat n$. The vector
$\vec P = \langle\vec\sigma\rangle$ is the **polarization**, and for a pure state
$|\vec P| = 1$, so every pure spin-½ state points somewhere on the unit sphere.
This is the seed of the Bloch-sphere picture developed
[later in the module](/quantum-mechanics/spin/two-level-systems-and-the-bloch-sphere).
The map "pure state $\leftrightarrow$ direction $\hat n$" is two-to-one at the
level of $SU(2)$ versus $SO(3)$: the $\theta/2$ in the eigenspinor means a spatial
rotation by $2\pi$ multiplies the spinor by $-1$, and only a $4\pi$ rotation
returns it to itself. That sign is not a mathematical artifact; neutron
interferometry measures it directly by rotating a spin in one arm and reading the
interference against the other.[^ct-spinor]

$$
% caption: A pure spin-½ state corresponds to a direction on the unit sphere; the
% polar and azimuthal angles set the spinor components, and the polarization
% vector equals the expectation of the three Pauli operators.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% sphere
\draw[black] (0,0) circle (2.0);
\draw[black, dashed] (-2.0,0) arc (180:360:2.0 and 0.55);
\draw[black] (2.0,0) arc (0:180:2.0 and 0.55);
% axes
\draw[->, black] (0,0) -- (0,2.5) node[above, font=\footnotesize] {$z$};
\draw[->, black] (0,0) -- (2.55,-0.7) node[right, font=\footnotesize] {$x$};
\draw[->, black] (0,0) -- (0.95,1.0) node[above right, font=\footnotesize] {$y$};
% state vector
\draw[->, very thick, acc] (0,0) -- (1.15,1.45);
\node[circle, fill=acc, inner sep=1.5pt] at (1.15,1.45) {};
\node[text=acc, font=\footnotesize, anchor=south west] at (1.2,1.5) {$n$};
% angle marks (no greek in nodes -> label in caption)
\draw[thick] (0,0.85) arc (90:51:0.85);
\node[font=\footnotesize, text=black, anchor=west] at (0.35,0.72) {polar angle};
\node[draw, fill=black!4, font=\footnotesize, align=left, anchor=west] at (2.9,0.9)
  {pure state points\\[2pt] along the unit\\[2pt] vector $n$};
\node[font=\footnotesize, anchor=west, text=black] at (2.9,-0.6) {polarization length $1$};
\end{tikzpicture}
$$

## What spin adds

The electron's full state is a wavefunction times a spinor,
$\Psi(\vec r)\otimes\chi$, an element of $L^2(\mathbb R^3)\otimes\mathbb C^2$. When
the Hamiltonian ignores spin the two factors decouple and every spatial level
carries a two-fold spin degeneracy. Coupling appears through the magnetic moment
(the [Zeeman and spin–orbit terms](/quantum-mechanics/approximation-methods/fine-structure-and-the-real-hydrogen-atom))
and through the exchange symmetry of
[identical particles](/quantum-mechanics/identical-particles/identical-particles-and-exchange-symmetry),
where the antisymmetry of the total state ties the spatial and spin parts
together. The two-state structure isolated here — a qubit with the $SU(2)$ algebra
of the Pauli matrices — is also the smallest nontrivial quantum system and the
template for every two-level problem in the
[next two lessons](/quantum-mechanics/spin/spin-in-a-magnetic-field-precession-and-resonance).

[^sn-sg]: Sakurai & Napolitano, _Modern Quantum Mechanics_, 3rd ed. (Cambridge,
2021), §1.1. The choice of silver is not incidental: a hydrogen-like single
valence electron in an $\ell = 0$ orbital removes any orbital contribution, so the
deflection isolates spin.

[^gs-spin]: Griffiths & Schroeter, _Introduction to Quantum Mechanics_, 3rd ed.
(Cambridge, 2018), §4.4.1. The electron magnetic moment is
$\vec\mu = g_s\frac{-e}{2m_e}\vec S$ with $g_s \approx 2.00232$; the CODATA value
of the electron $g$-factor and the Bohr magneton are tabulated at
[physics.nist.gov/cuu/Constants](https://physics.nist.gov/cuu/Constants/).

[^ct-spinor]: Cohen-Tannoudji, Diu & Laloë, _Quantum Mechanics_, Vol. I (Wiley,
1977), Ch. IV and complements. The $2\pi$ sign change of a spinor was confirmed in
neutron-interferometry experiments (Rauch et al., 1975; Werner et al., 1975).