# 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.
---
## 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.
---
## 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](../../term-1-quantum-mechanics/c02-qubits-bloch/02-bloch-sphere.md)).
### 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](01-central-potentials-angular-momentum.md) 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](../../term-1-quantum-mechanics/c03-measurement/01-projective-measurement.md)
and [1.1.2 Observables & the Measurement Postulate](../../term-1-quantum-mechanics/c01-postulates/02-observables-measurement-postulate.md)
— 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](01-central-potentials-angular-momentum.md) — 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](../../term-1-quantum-mechanics/c02-qubits-bloch/01-the-qubit.md)).
Everything the main program builds from Term 1 onward is the physics of systems like this one: the
state postulate ([1.1.1](../../term-1-quantum-mechanics/c01-postulates/01-state-postulate.md))
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.
---
## 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.
---
## Hands-on (Python)
**Classical smear vs quantum spots** — simulate the detector plate:
```python
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:
```python
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
```
---
## 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).
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.
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.
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.
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](../../term-1-quantum-mechanics/c02-qubits-bloch/02-bloch-sphere.md).
Solution
(a) $\boldsymbol\mu_s = -g_s\frac{e}{2m_e}\mathbf S = -\frac{e}{m_e}\cdot\frac\hbar2\boldsymbol\sigma = -\mu_B\boldsymbol\sigma$,
so $H = -\boldsymbol\mu_s\cdot\mathbf B = \mu_BB\,\sigma_x$. Write $\Omega = \mu_BB/\hbar$.
(b) Since $\sigma_x^2 = I$, $e^{-i\Omega t\sigma_x} = \cos(\Omega t)\,I - i\sin(\Omega t)\,\sigma_x$, and
$\sigma_x$ swaps $\lvert{+z}\rangle \leftrightarrow \lvert{-z}\rangle$:
$\lvert\psi(t)\rangle = \cos(\Omega t)\lvert{+z}\rangle - i\sin(\Omega t)\lvert{-z}\rangle$. Hence
$P(-z) = \sin^2(\Omega t) = \tfrac12\big(1 - \cos 2\Omega t\big)$.
(c) The flip is certain when $\Omega t = \pi/2$:
$t = \pi\hbar/2\mu_BB = \pi m_e/eB = \pi\times1.055\times10^{-34}/(2\times9.27\times10^{-24}\times10^{-3}) = 17.9$ ns. The probability
oscillates at $2\Omega/2\pi = \mu_BB/\pi\hbar = 28.0$ MHz, the electron's spin precession frequency,
28 GHz per tesla. (d) The Bloch vector starts at the north pole and precesses about the $x$-axis at
angular rate $2\Omega = eB/m_e$, passing through $-y$ to the south pole at $t = 17.9$ ns. That rate is
twice the orbital Larmor frequency $eB/2m_e$, because $g_s = 2$. This is the quantum version of the
Larmor precession above, and it is the same physics as the Rabi oscillations of
[1.6.2](../../term-1-quantum-mechanics/c06-dynamics/02-two-level-rabi.md): a timed pulse of field is a
quantum gate, here a NOT up to phase.
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.
6. With $H = \mu_BB\,\sigma_x$ the state is $\cos(\Omega t)\lvert{+z}\rangle - i\sin(\Omega t)\lvert{-z}\rangle$, $\Omega = \mu_BB/\hbar$, so $P(-z) = \sin^2\Omega t$: the spin precesses about $x$ and flips completely at $t = \pi\hbar/2\mu_BB$ (about 18 ns at 1 mT).