Magnetic Moments, Stern–Gerlach & Spin

An orbiting charge is a tiny magnet, so a magnetic field reads out angular momentum — and in 1922 Stern and Gerlach pointed one at a beam of silver atoms expecting a smear and got two clean spots. The culprit is a brand-new kind of angular momentum, spin, with the half-integer value the ladder algebra always allowed and orbital motion never used. This final lesson works through the experiment that discovered the qubit, then hands the whole program over to Term 1. The wave-mechanics door and the axiomatic door open into the same room; curiosity, for once, rewards the cat.

Learning Objectives

After this lesson you will be able to:

1. Derive the orbital magnetic moment $\boldsymbol\mu = -\frac{e}{2m_e}\mathbf L$ and quantify moments in Bohr magnetons.
2. Compute normal Zeeman splittings, Larmor precession frequencies, and the deflecting force $F_z = \mu_z\,\partial B_z/\partial z$ in an inhomogeneous field.
3. Explain why the Stern–Gerlach result — two spots — is incompatible with classical physics and with orbital angular momentum, and how spin $s = \tfrac12$ resolves it.
4. Analyze sequential Stern–Gerlach experiments with two-component states, computing outcome probabilities via inner products.
5. Argue quantitatively that spin is not literal rotation, and state the values of $S^2$, $S_z$, and $g_s$ for the electron.
6. Identify the spin-$\tfrac12$ system as a physical qubit, connecting every Stern–Gerlach observation to the postulates of Term 1.

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Intuition

Hydrogen's stationary states are labeled $(n,\ell,m)$, but the energy $-13.6\ \text{eV}/n^2$ hears only $n$. To see $\ell$ and $m$ you must break the spherical symmetry — apply a magnetic field that picks out a direction. A circulating electron is a current loop, hence a magnetic dipole proportional to $\mathbf L$; a field then shifts energies by $m$ (the Zeeman effect), twists the dipole into precession, and — if the field is inhomogeneous — pulls the atom up or down by an amount set by $L_z$. Magnetic fields are nature's $L_z$-meters.

So here is a clean experimental question: send atoms through a field gradient and watch the beam. Classical physics predicts a continuous smear (dipoles point every which way). Quantum mechanics predicts $2\ell+1$ discrete spots — an odd number. Nature, asked politely with silver atoms in 1922, answered: two. Neither theory on the table could produce an even number. The resolution is the last, strangest angular momentum — and the first qubit.

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Theory

The orbital magnetic moment

Model the electron (charge $-e$, mass $m_e$) as a classical circular orbit of radius $r$ and speed $v$. It passes any point once per period $T = 2\pi r/v$, so it constitutes a current $I = -e/T = -ev/2\pi r$. A current loop of area $A = \pi r^2$ has magnetic moment

$$\mu = IA = -\frac{ev}{2\pi r}\,\pi r^2 = -\frac{evr}{2} = -\frac{e}{2m_e}\,(m_e v r) ,$$

and $m_e v r$ is exactly the orbital angular momentum $L$. Directions agree (both $\boldsymbol\mu$ and $\mathbf L$ are normal to the orbit; the minus sign flips $\boldsymbol\mu$ antiparallel because the charge is negative), and the shape of the orbit drops out for any planar loop, so

$$\boldsymbol\mu = -\frac{e}{2m_e}\,\mathbf L .$$

The ratio $\gamma = -e/2m_e$ is the gyromagnetic ratio; we promote the relation directly to an operator identity. Quantization of $L_z$ then quantizes the moment:

$$\mu_z = -\frac{e}{2m_e}\,\hbar m = -m\,\mu_B, \qquad \mu_B \equiv \frac{e\hbar}{2m_e} = 9.274\times10^{-24}\ \text{J/T} = 5.788\times10^{-5}\ \text{eV/T},$$

the Bohr magneton — the natural unit of atomic magnetism.

Energy in a field: the normal Zeeman effect

A dipole in a field $\mathbf B$ has interaction energy $U = -\boldsymbol\mu\cdot\mathbf B$ (least energy when aligned). Take $\mathbf B = B\hat z$, uniform. Then

$$U = -\mu_z B = +\,\mu_B B\, m ,$$

so the $2\ell+1$ states of given $(n,\ell)$, degenerate in zero field, fan out into equally spaced levels: $E_{n\ell m} = E_n + \mu_B B\,m$. Between adjacent $m$-levels,

$$\Delta E = \mu_B B\,\Delta m = \mu_B B , \qquad \Delta\nu = \frac{\mu_B B}{h} = \frac{eB}{4\pi m_e} ,$$

the normal Zeeman splitting — this is why $m$ is called the magnetic quantum number. (Many real spectral lines split in "anomalous" patterns that this formula cannot produce; that anomaly was an early fingerprint of the spin we are about to meet.)

Larmor precession

A uniform field exerts no net force, but it does exert a torque $\boldsymbol\tau = \boldsymbol\mu\times\mathbf B$. Then

$$\frac{d\mathbf L}{dt} = \boldsymbol\tau = -\frac{e}{2m_e}\,\mathbf L\times\mathbf B .$$

The change is always perpendicular to both $\mathbf L$ and $\mathbf B$, so $|\mathbf L|$ and the angle $\alpha$ to $\mathbf B$ are constant: the tip of $\mathbf L$ sweeps a circle of radius $L\sin\alpha$ at speed $|d\mathbf L/dt| = \frac{e}{2m_e}LB\sin\alpha$. The angular rate is therefore

$$\omega_L = \frac{|d\mathbf L/dt|}{L\sin\alpha} = \frac{eB}{2m_e} ,$$

the Larmor frequency — independent of $\alpha$. Quantum mechanically the same statement holds for expectation values, and for a spin-$\tfrac12$ it becomes something you will meet again almost immediately: uniform precession of $\langle\mathbf S\rangle$ about $\mathbf B$ is exactly rotation about the $z$-axis of the Bloch sphere (1.2.2 The Bloch Sphere).

Force in an inhomogeneous field

If $\mathbf B$ varies in space, the energy $U = -\boldsymbol\mu\cdot\mathbf B(\mathbf r)$ does too, and the atom feels $\mathbf F = -\nabla U = \nabla(\boldsymbol\mu\cdot\mathbf B)$. Let the field point mainly along $z$ with a strong gradient $\partial B_z/\partial z$. Larmor precession spins the transverse components of $\boldsymbol\mu$ rapidly about $z$, averaging their force to zero, while $\mu_z$ is constant. The surviving force is

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

An inhomogeneous magnet is thus a $\mu_z$-meter: it translates the value of $\mu_z$ — hence of the angular momentum projection — into a measurable deflection.

The Stern–Gerlach experiment (1922)

Stern and Gerlach vaporized silver in an oven, collimated the escaping atoms into a beam ($v \approx 550$ m/s), passed it between the asymmetric pole pieces of a magnet ($\partial B_z/\partial z \sim 10^3$ T/m over a few cm), and collected the atoms on a glass plate.

• Classical prediction. Thermal atoms have randomly oriented moments: $\mu_z = \mu\cos\alpha$ fills $[-\mu, +\mu]$ continuously, so the deposit should be one continuous smear.
• Observation. Two discrete spots, symmetric about the axis, and nothing in between.

Discreteness itself was celebrated as "space quantization" — $L_z$ really does take only discrete values. But look closer and the triumph collapses:

1. Orbital angular momentum gives $2\ell+1$ spots, and $2\ell+1$ is always odd. No value of $\ell$ yields two.
2. Worse: the silver atom's 47 electrons form closed shells plus a single valence electron in a 5s state — $\ell = 0$. Its orbital moment is zero. The beam should not split at all.

Something in the atom carries a magnetic moment that is not orbital motion — with exactly two projections.

Spin: the resolution

Goudsmit and Uhlenbeck (1925) proposed that the electron carries an intrinsic angular momentum, spin, with fixed quantum number $s = \tfrac12$:

$$S^2 = s(s+1)\hbar^2 = \frac34\hbar^2 , \qquad S_z = m_s\hbar = \pm\frac\hbar2 .$$

This is precisely the half-integer representation that the ladder algebra of P.6.1 permitted and single-valuedness forbade for orbital motion — spin is no one's $e^{im\varphi}$; the loophole closes here. Two values of $m_s$ mean two spots. The magnetic moment needs one correction: experiment (and later Dirac's relativistic theory) gives a g-factor $g_s \approx 2$ (measured: $2.00232\dots$),

$$\boldsymbol\mu_s = -g_s\frac{e}{2m_e}\mathbf S \quad\Longrightarrow\quad \mu_z = -g_s\frac{e}{2m_e}\Bigl(\pm\frac\hbar2\Bigr) = \mp\frac{g_s}{2}\mu_B \approx \mp\mu_B ,$$

so each spot deflects as if carrying one full Bohr magneton — quantitatively what Stern and Gerlach measured.

Do not picture a spinning ball. Take the classical electron radius $r_e = \frac{e^2}{4\pi\epsilon_0 m_ec^2} = 2.82\times10^{-15}$ m and demand that a uniform sphere spinning with $L = I\omega = \frac25 m_e r_e^2\,\omega$ carry $\hbar/2$. The equatorial speed would be

$$v = \omega r_e = \frac{5\hbar}{4m_e r_e} = \frac{5\times1.055\times10^{-34}}{4\times9.11\times10^{-31}\times2.82\times10^{-15}} \approx 5.1\times10^{10}\ \text{m/s} \approx 170\,c .$$

No rotation of matter can do this. Spin is an irreducibly quantum degree of freedom: a two-valued label with the algebra of angular momentum and no mechanical picture underneath.

Sequential Stern–Gerlach experiments

Since spin has exactly two outcomes per axis, describe it by two-component states: an orthonormal basis $\lvert{+z}\rangle, \lvert{-z}\rangle$ (the two exit ports of a $z$-oriented apparatus, "SG$\hat z$"), and for the $x$-axis

$$\lvert{\pm x}\rangle = \frac{1}{\sqrt2}\bigl(\lvert{+z}\rangle \pm \lvert{-z}\rangle\bigr),$$

(orthonormal, and symmetric between $\pm z$ as they must be). The Born rule gives outcome probabilities as squared overlaps. Now chain apparatuses, Sakurai-style:

Experiment 1: SG$\hat z$ → SG$\hat z$. Select the $+z$ beam, measure $S_z$ again: $P(+z) = |\langle{+z}|{+z}\rangle|^2 = 1$. Measurement is repeatable — the first apparatus prepared a definite $S_z$.

Experiment 2: SG$\hat z$ → SG$\hat x$. Feed the $+z$ beam into an $x$-apparatus:

$$P(\pm x) = \bigl|\langle{\pm x}|{+z}\rangle\bigr|^2 = \left|\frac{1}{\sqrt2}\bigl(\langle{+z}\rvert \pm \langle{-z}\rvert\bigr)\lvert{+z}\rangle\right|^2 = \frac12 \quad\text{each.}$$

Experiment 3: SG$\hat z$ → SG$\hat x$ → SG$\hat z$. Select $+x$ from Experiment 2 and measure $S_z$ once more. The state is now $\lvert{+x}\rangle$, so

$$P(\pm z) = \bigl|\langle{\pm z}|{+x}\rangle\bigr|^2 = \frac12 \quad\text{each.}$$

Half the atoms come out $-z$ — even though every atom entering the $x$-apparatus had been certified $+z$! Measuring $S_x$ erased the $S_z$ value. Of the original $+z$ beam, a fraction $\tfrac12\cdot\tfrac12 = \tfrac14$ lands in each final port. $S_z$ and $S_x$ are incompatible observables: no state has both sharp, and measuring one collapses the state into an eigenstate of it, randomizing the other. This tabletop cascade is the entire content of the projective measurement formalism — 1.3.1 Projective Measurement and 1.1.2 Observables & the Measurement Postulate — made experimental, three years before anyone wrote it down.

Caution. "Spin up" does not mean the spin vector points along $+z$. Its length is $|\mathbf S| = \sqrt{s(s+1)}\,\hbar = \frac{\sqrt3}{2}\hbar > \frac\hbar2$, so even the $+z$ eigenstate leans off-axis, with $\langle S_x^2 + S_y^2\rangle = \hbar^2/2 > 0$: the transverse components remain indeterminate. Same cone, same warning as P.6.1 — now for the smallest ladder there is.

The handoff: the qubit was here all along

Step back and inventory what the electron's spin actually is. Its state space is spanned by two basis states, $\lvert{+z}\rangle$ and $\lvert{-z}\rangle$; a general state is a normalized complex combination $\alpha\lvert{+z}\rangle + \beta\lvert{-z}\rangle$ — the space is $\mathbb C^2$. Rename the basis

$$\lvert{+z}\rangle \equiv \lvert 0\rangle, \qquad \lvert{-z}\rangle \equiv \lvert 1\rangle,$$

and you are holding the qubit (1.2.1 The Qubit). Everything the main program builds from Term 1 onward is the physics of systems like this one: the state postulate (1.1.1) axiomatizes the two-component states you just used; the Bloch sphere is the geometry of spin directions, with Larmor precession as its native rotation; projective measurement is a Stern–Gerlach magnet; entanglement is what happens when two such spins share a state. Even the hardware road returns here — superconducting circuits are engineered artificial two-level atoms (Term 4.4). The wave-mechanics road ends exactly where the program's axiomatic road begins; from here on the convention $\hbar = 1$ takes over. Walk through the other door.

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Worked Examples

Example 1 — Zeeman splitting at 1 tesla

A hydrogen 2p level ($\ell = 1$) in $B = 1$ T splits into three levels spaced by $\Delta E = \mu_B B = 5.79\times10^{-5}$ eV, i.e. $\Delta\nu = \mu_B B/h = 14.0$ GHz. Compare scales: the 2p → 1s photon is $10.2$ eV, so the splitting is a fractional shift of only $\sim6\times10^{-6}$ — resolvable, but it takes a good spectrometer, which is why the Zeeman effect waited until 1896. Thermal energy at room temperature ($k_BT \approx 0.025$ eV) exceeds the splitting 400-fold: field-on level populations barely notice.

Example 2 — The deflection that made history

Silver atoms ($M = 1.79\times10^{-25}$ kg, $v = 550$ m/s) cross an $\ell = 3.5$ cm magnet with $\partial B_z/\partial z = 1400$ T/m, and $\mu_z = \pm\mu_B$. Force and kinematics:

$$F_z = \mu_B\frac{\partial B_z}{\partial z} = 9.274\times10^{-24}\times1400 = 1.30\times10^{-20}\ \text{N}, \qquad a = \frac{F_z}{M} = 7.3\times10^{4}\ \text{m/s}^2 .$$

Transit time $t = 0.035/550 = 6.4\times10^{-5}$ s, so each beam deflects $z = \tfrac12 a t^2 \approx 1.5\times10^{-4}$ m — the two spots sit $\approx 0.3$ mm apart. Tiny, but resolvable on a glass plate in 1922 (legend says Stern's cheap-cigar sulfur helped develop the faint silver deposit). One quantum of angular momentum, visible by eye.

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Hands-on (Python)

Classical smear vs quantum spots — simulate the detector plate:

import numpy as np
import matplotlib.pyplot as plt

rng = np.random.default_rng(42)
N = 200_000
# Deflection in units of z0 = (mu_B dB/dz) L^2 / (2 M v^2) ~ 0.15 mm (Example 2).

# (a) Classical: isotropic dipoles -> mu_z/mu_B = cos(alpha), uniform in [-1, 1].
z_cl = rng.uniform(-1, 1, N)         + 0.15*rng.normal(size=N)   # + beam width
# (b) Quantum: mu_z = ±mu_B, equal probability.
z_qm = rng.choice([-1.0, 1.0], N)    + 0.15*rng.normal(size=N)

fig, ax = plt.subplots(1, 2, figsize=(10, 4), sharey=True)
ax[0].hist(z_cl, bins=200, density=True); ax[0].set_title("classical prediction: smear")
ax[1].hist(z_qm, bins=200, density=True); ax[1].set_title("observed (quantum): two spots")
for a in ax: a.set_xlabel("deflection $z/z_0$")
plt.tight_layout(); plt.show()
# Expect: left, a flat-topped band filling [-1, 1]; right, two sharp peaks at ±1
# with an empty middle -- the 1922 photographic plate, in histogram form.

Sequential Stern–Gerlach with spinors — a NumPy-only preview of Term 1 measurement:

hbar = 1.0
sigma_x = np.array([[0, 1], [1, 0]], dtype=complex)
sigma_z = np.array([[1, 0], [0, -1]], dtype=complex)
Sx, Sz = hbar/2*sigma_x, hbar/2*sigma_z          # S = (hbar/2) sigma

def measure(state, S):
    """Projective measurement of observable S: returns (outcome, collapsed state)."""
    evals, evecs = np.linalg.eigh(S)
    probs = np.abs(evecs.conj().T @ state)**2     # Born rule
    k = rng.choice(len(evals), p=probs/probs.sum())
    post = evecs[:, k]                            # projector + renormalize (nondegenerate)
    return evals[k], post

trials, n_px, n_px_then_pz = 100_000, 0, 0
plus_z = np.array([1, 0], dtype=complex)          # certified +z beam from SGz
for _ in range(trials):
    sx, state = measure(plus_z, Sx)               # ... into SGx
    if sx > 0:
        n_px += 1
        sz, _ = measure(state, Sz)                # +x beam into SGz
        if sz > 0:
            n_px_then_pz += 1

print(f"P(+x | +z)        = {n_px/trials:.3f}")          # ~0.500
print(f"P(+x then +z)     = {n_px_then_pz/trials:.3f}")  # ~0.250
print(f"P(+z | +x)        = {n_px_then_pz/n_px:.3f}")    # ~0.500  <- Sz was erased

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Exercises

E1 (easy). From $e = 1.602\times10^{-19}$ C, $\hbar = 1.055\times10^{-34}$ J·s, and $m_e = 9.109\times10^{-31}$ kg, compute $\mu_B$ in J/T and eV/T, and the Larmor frequency $f_L = \omega_L/2\pi$ in a $0.5$ T field.

Solution

$\mu_B = e\hbar/2m_e = (1.602\times10^{-19}\times1.055\times10^{-34})/(2\times9.109\times10^{-31}) = 9.27\times10^{-24}$J/T; dividing by$e$:$5.79\times10^{-5}$ eV/T. $\omega_L = eB/2m_e = (1.602\times10^{-19}\times0.5)/(2\times9.109\times10^{-31}) = 4.40\times10^{10}$ rad/s, so $f_L = 7.0$ GHz — microwave territory, the working band of spin resonance (and, not coincidentally, of superconducting qubits).

E2 (easy). Into how many beams does a Stern–Gerlach magnet split (a) atoms with $\ell = 1$ and no spin effects, (b) hydrogen atoms in their ground state? What did (b) show when Phipps and Taylor did it in 1927?

Solution

(a) $2\ell+1 = 3$ beams ($m = -1, 0, +1$). (b) Ground-state hydrogen has $\ell = 0$: one undeflected beam if only orbital moments existed. Phipps and Taylor observed two — with a single electron and definitely zero orbital angular momentum, this pinned the moment on electron spin itself, removing any doubt that silver's complexity was to blame.

E3 (medium). Define $\lvert{\pm y}\rangle = \frac{1}{\sqrt2}(\lvert{+z}\rangle \pm i\lvert{-z}\rangle)$. Verify orthonormality and compute $|\langle{+y}|{+z}\rangle|^2$ and $|\langle{+y}|{+x}\rangle|^2$. What does the pattern tell you?

Solution

$\langle{+y}|{+y}\rangle = \tfrac12(1 + 1) = 1$; $\langle{+y}|{-y}\rangle = \tfrac12(1 \cdot 1 + (-i)(i)(-1)) = \tfrac12(1 - 1) = 0$. ✓ $|\langle{+y}|{+z}\rangle|^2 = |\tfrac{1}{\sqrt2}|^2 = \tfrac12$; $\langle{+y}|{+x}\rangle = \tfrac12(1\cdot1 + (-i)\cdot1) = \tfrac{1-i}{2}$, so $|\langle{+y}|{+x}\rangle|^2 = \tfrac{2}{4} = \tfrac12$. All three axes are pairwise "maximally incompatible": knowing the spin along any one axis makes the other two perfect coin flips. The three bases are mutually unbiased — in Term 1 language, the eigenbases of the three Pauli operators.

E4 (medium). In Experiment 3, suppose the SG$\hat x$ stage separates the beams but both are recombined coherently (nothing measured, nothing blocked) before the final SG$\hat z$. What does the final measurement give? Compare with blocking the $-x$ beam, and explain the difference.

Solution

Recombining without measuring undoes the separation: the state entering the last magnet is still $\lvert{+z}\rangle = \tfrac{1}{\sqrt2}(\lvert{+x}\rangle + \lvert{-x}\rangle)$ — the two paths' amplitudes re-add — so $P(+z) = 1$: all atoms exit $+z$, as if the $x$-stage weren't there. Blocking $-x$ instead leaves the (renormalized) state $\lvert{+x}\rangle$, giving $P(\pm z) = \tfrac12$ and only $\tfrac14$ of the original beam in each port. The difference between "split and recombined" and "split and known" is interference of amplitudes versus collapse — the double-slit lesson of Course P.3, replayed with spins.

E5 (hard). Let $\hat n = (\sin\theta, 0, \cos\theta)$. Using $S_n = \frac\hbar2(\sigma_z\cos\theta + \sigma_x\sin\theta)$, find the $+\hbar/2$ eigenstate $\lvert{+n}\rangle$ and show that a $+z$-prepared atom passes an SG$\hat n$ filter with probability $\cos^2(\theta/2)$. Check $\theta = \pi/2$ against Experiment 2.

Solution

$S_n = \frac\hbar2\begin{pmatrix}\cos\theta & \sin\theta\\ \sin\theta & -\cos\theta\end{pmatrix}$. For eigenvalue $+\frac\hbar2$: $(\cos\theta - 1)a + \sin\theta\, b = 0$, so $b/a = \frac{1-\cos\theta}{\sin\theta} = \tan(\theta/2)$, giving the normalized eigenstate $\lvert{+n}\rangle = \cos\frac\theta2\,\lvert{+z}\rangle + \sin\frac\theta2\,\lvert{-z}\rangle$. Then $P = |\langle{+n}|{+z}\rangle|^2 = \cos^2(\theta/2)$. At $\theta = \pi/2$: $\lvert{+n}\rangle = \lvert{+x}\rangle$ and $P = \tfrac12$ ✓. The half-angle is the signature of spin-$\tfrac12$: rotating the apparatus by $\theta$ rotates the state by $\theta/2$ — the geometry that becomes the Bloch sphere in 1.2.2.

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Checkpoint

1. Why does a Stern–Gerlach magnet need an inhomogeneous field, and what quantity does the deflection measure?
2. Give both reasons why orbital angular momentum cannot explain silver's two spots.
3. State $S^2$ and $S_z$ for an electron. What is $g_s$, and where does the "spinning ball exceeds $c$" argument leave the meaning of spin?
4. In the chain SG$\hat z(+) \to$ SG$\hat x(+) \to$ SG$\hat z$, what fraction of the original $+z$ atoms exits each final port, and why?
5. In what precise sense is the spin-$\tfrac12$ electron a qubit?

Answers

1. A uniform field only torques a dipole ($U$ independent of position); a gradient makes $U = -\boldsymbol\mu\cdot\mathbf B$ position-dependent, giving $F_z = \mu_z\,\partial B_z/\partial z$. The deflection measures $\mu_z$, hence the angular momentum projection along $z$.
2. (i) Orbital multiplets have $2\ell+1$ (odd) orientations — never two. (ii) Silver's valence electron is 5s, $\ell = 0$: zero orbital moment, so the beam shouldn't split at all.
3. $S^2 = \frac34\hbar^2$, $S_z = \pm\frac\hbar2$; $g_s \approx 2$ ($2.00232\dots$). A classical sphere would need equatorial speeds $\sim170c$ to carry $\hbar/2$ — spin is intrinsic angular momentum with no rotating-matter picture.
4. $\tfrac14$ in $+z$ and $\tfrac14$ in $-z$ (of the original beam): each stage is a Born-rule coin flip, $\tfrac12\times\tfrac12$, because measuring $S_x$ collapses the state to $\lvert{+x}\rangle$ and erases the previously sharp $S_z$.
5. Its state space is $\mathbb C^2$ with basis $\{\lvert{+z}\rangle, \lvert{-z}\rangle\}$; relabeling these $\lvert 0\rangle, \lvert 1\rangle$ satisfies the Term 1 state postulate exactly — superpositions, Born-rule measurement, unitary (Larmor) evolution and all.

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Further Reading

• [Sak] Sakurai & Napolitano, §1.1 — the celebrated Stern–Gerlach opening; sequential experiments as the definition of quantum mechanics.
• [Gri] Griffiths & Schroeter, Ch. 4 (§4.4) — spin, magnetic interactions, and Larmor precession in wave-mechanics language.
• [NC] Nielsen & Chuang, §1.5.1 — Stern–Gerlach retold from the quantum-information side: the prototype qubit.
• [ER] Eisberg & Resnick, Ch. 8 — spin and magnetic moments with the full experimental story, including Phipps–Taylor.

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