Spin — intrinsic angular momentum: Worked examples

The scenario matrix

Every spin problem you'll meet lives in one of these cells. The examples below are labelled with the cell they cover.

Cell	What makes it different	Covered by
A. Magnitude, any $s$	depends on $s (s + 1)$ only, never on $m_{s}$	Ex 1
B. Projection, $m_{s} > 0$ (spin up)	positive $S_{z}$ , force one way	Ex 2
C. Projection, $m_{s} < 0$ (spin down)	negative $S_{z}$ , force the other way	Ex 2, Ex 5
D. Tilt angle from $z$	why $θ \neq = 0$ ever; all allowed $θ$	Ex 3
E. Degenerate input $s = 0$	zero magnitude, one spot, no splitting	Ex 4
F. Bigger ladder $s = 1$ , $s = \frac{3}{2}$	$2 s + 1$ spots, all $m_{s}$ steps	Ex 6
G. Limiting / classical check	"spinning ball" would break light-speed	Ex 7
H. Real-world word problem	Stern–Gerlach with actual numbers/units	Ex 8
I. Exam-style twist	energy splitting $U = - μ_{z} B$ , spin-flip photon	Ex 9

We use these constants throughout (memorise the symbols, not the digits):

Cell A — magnitude for any spin

Worked example Ex 1 — Magnitude of an electron's spin (Cell A)

Find $∣ S ∣$ for an electron ( $s = \frac{1}{2}$ ) as a multiple of $ℏ$ and in SI units.

Forecast: guess before reading — is it $\frac{1}{2} ℏ$ ? Write your guess down. (It is not; that's the projection, not the magnitude.)

Write the magnitude rule. $∣ S ∣ = ℏ s (s + 1)$ . Why this step? Magnitude never uses $m_{s}$ — it depends only on the total-spin number $s$ . This is the "size" of the vector, not its shadow on the $z$ -axis.
Substitute $s = \frac{1}{2}$ . $s (s + 1) = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$ . Why this step? Plugging in now keeps the algebra honest — we haven't touched $m_{s}$ , proving the answer is the same for spin-up and spin-down.
Take the root. $∣ S ∣ = ℏ \frac{3}{4} = \frac{3}{2} ℏ \approx 0.8660 ℏ$ .
Convert to SI. $0.8660 \times 1.0546 \times 1 0^{- 34} \approx 9.13 \times 1 0^{- 35} J\cdots$ .

Verify: $(\frac{3}{2})^{2} = \frac{3}{4} = s (s + 1)$ ✔. Units: $ℏ$ is in J·s, and angular momentum has units J·s ✔.

Cells B & C — projection, both signs

Worked example Ex 2 — Both

$S_{z}$ values and their forces (Cells B, C) For an electron, list both allowed $S_{z}$ and the sign of the Stern–Gerlach force each feels in a gradient $\partial B_{z} / \partial z > 0$ .

Forecast: how many values, and are the two forces equal-and-opposite or unrelated?

Allowed $m_{s}$ . For $s = \frac{1}{2}$ : $m_{s} = + \frac{1}{2}$ (up) and $m_{s} = - \frac{1}{2}$ (down). Why this step? $m_{s}$ runs $- s, \dots, + s$ in integer steps, so only two rungs exist.
Projection. $S_{z} = m_{s} ℏ = + \frac{1}{2} ℏ$ or $- \frac{1}{2} ℏ$ .
Magnetic moment. $μ_{z} = - g_{s} \frac{e}{2 m _{e}} S_{z} = - g_{s} μ_{B} m_{s}$ . With $g_{s} \approx 2$ : $μ_{z} = - 2 μ_{B} (\pm \frac{1}{2}) = \mp μ_{B}$ . Why the minus? The electron charge is negative, so $μ$ points opposite to $S$ . Spin-up ( $m_{s} = + \frac{1}{2}$ ) gives $μ_{z} = - μ_{B}$ ; spin-down gives $μ_{z} = + μ_{B}$ .
Force. $F_{z} = μ_{z} \frac{\partial B _{z}}{\partial z}$ . The two moments $\pm μ_{B}$ give two opposite forces → two spots.

Verify: $μ_{z}^{u p} + μ_{z}^{d o w n} = (- μ_{B}) + (+ μ_{B}) = 0$ : forces are exactly equal and opposite ✔. Number of spots $= 2 s + 1 = 2$ ✔.

Figure — Spin — intrinsic angular momentum

Cell D — the tilt angle

The vector $S$ can never lie flat along $z$ . Here is why, and all the angles it is allowed to make.

Worked example Ex 3 — Angle of

$S$ from the $z$ -axis (Cell D) Find the angle $θ$ between $S$ and the $+ z$ axis for an electron, for both spin states.

Forecast: can $θ = 0^{\circ}$ (perfect alignment)? Guess yes/no first.

Set up the cosine. For a vector, the cosine of its angle from $z$ is (its $z$ -shadow)÷(its length): $cos θ = \frac{S _{z}}{∣ S ∣} = \frac{m _{s} ℏ}{ℏ s ( s + 1 )} = \frac{m _{s}}{s ( s + 1 )} .$ Why cosine and not sine? Because $S_{z} = ∣ S ∣ cos θ$ is the adjacent side of the right triangle formed by the vector and its $z$ -projection — cosine is "adjacent over hypotenuse."
Spin up. $cos θ = \frac{1/2}{3 /2} = \frac{1}{3} \approx 0.5774$ , so $θ = arccos (0.5774) \approx 54. 7^{\circ}$ .
Spin down. $m_{s} = - \frac{1}{2}$ gives $cos θ = - \frac{1}{3}$ , so $θ \approx 125. 3^{\circ}$ (i.e. $18 0^{\circ} - 54. 7^{\circ}$ ). Why this step? The down-state is the mirror image through the $x y$ -plane; its tilt from $+ z$ is the supplement.
Why never $0^{\circ}$ ? Perfect alignment would need $S_{z} = ∣ S ∣$ , i.e. $\frac{1}{2} ℏ = \frac{3}{2} ℏ$ — false. If $S$ pointed exactly along $z$ , we'd know $S_{x} = S_{y} = 0$ too, violating the uncertainty built into the spin algebra. So $S$ always tilts.

Verify: $cos (54.735 6^{\circ}) = 1/ 3 \approx 0.5774$ ✔; $54.735 6^{\circ} + 125.264 4^{\circ} = 18 0^{\circ}$ ✔.

Cell E — the degenerate case $s = 0$

Worked example Ex 4 — A spin-0 particle (Cell E)

A silver atom in Stern–Gerlach behaves as if $L = 0$ ; suppose additionally a hypothetical particle has total spin $s = 0$ . Find $∣ S ∣$ , the allowed $m_{s}$ , and the number of spots.

Forecast: how many spots — zero, one, or two?

Magnitude. $∣ S ∣ = ℏ 0 (0 + 1) = ℏ 0 = 0$ . Why this step? With $s = 0$ the vector has zero length — no spin angular momentum at all.
Allowed $m_{s}$ . Range $- s \dots + s$ collapses to the single value $m_{s} = 0$ .
Number of spots. $2 s + 1 = 2 (0) + 1 = 1$ → one undeflected spot. Why this matters: this is exactly what physicists expected for silver before 1922. Getting two spots instead is what forced spin to exist (see Stern-Gerlach Experiment).

Verify: $∣ S ∣ = 0$ and spot count $= 1$ ✔; a spin-0 particle has no magnetic moment ( $μ_{z} = - g_{s} μ_{B} \cdot 0 = 0$ ) so zero force ✔.

Cell C again — a proton (positive charge flips a sign)

Worked example Ex 5 — Sign care for a positive charge (Cell C)

A proton also has $s = \frac{1}{2}$ . For its spin-up state, in which direction does the magnetic moment point relative to $S$ ? (Do not worry about the proton's own g-factor here; just track the charge sign with a generic positive charge $+ e$ .)

Forecast: same direction as the electron's, or opposite?

General moment. For charge $q$ : $μ = g \frac{q}{2 m} S$ . Why this step? The classical loop formula carries the actual sign of the charge. Electrons had $q = - e$ (hence the minus in the parent note); a proton has $q = + e$ .
Substitute $q = + e$ . $μ_{z} = + g \frac{e}{2 m _{p}} S_{z}$ — no minus. Why this step? A positive charge makes $μ$ point the same way as $S$ , unlike the electron.
Spin up ⇒ $S_{z} = + \frac{1}{2} ℏ > 0$ ⇒ $μ_{z} > 0$ : moment points along $S$ (up).

Verify: flipping the charge sign flips the moment direction — electron up-state has $μ_{z} < 0$ , proton up-state has $μ_{z} > 0$ ; they are opposite ✔. Case coverage: we've now shown both charge signs.

Cell F — bigger ladders

Cell G — the classical limit that fails

Worked example Ex 7 — Would a spinning ball break light-speed? (Cell G)

Model the electron as a uniform sphere of "classical electron radius" $r_{e} = 2.82 \times 1 0^{- 15} m$ with moment of inertia $I = \frac{2}{5} m_{e} r_{e}^{2}$ . If its spin angular momentum equals $∣ S ∣ = \frac{3}{2} ℏ$ , find the required equatorial speed $v = ω r_{e}$ and compare to $c$ .

Forecast: a fraction of $c$ , roughly $c$ , or many times $c$ ?

Angular speed from $L = I ω$ . $ω = \frac{∣ S ∣}{I} = \frac{\frac{3}{2} ℏ}{\frac{2}{5} m _{e} r _{e}^{2}}$ . Why this step? For a rigid body angular momentum is $I ω$ ; solving gives the spin rate the "ball" would need.
Equatorial speed. $v = ω r_{e} = \frac{\frac{3}{2} ℏ}{\frac{2}{5} m _{e} r _{e}^{2}} r_{e} = \frac{5 3 ℏ}{4 m _{e} r _{e}}$ .
Plug in. $v = \frac{5 \cdot 1.732 \cdot 1.0546 \times 1 0 ^{- 34}}{4 \cdot 9.109 \times 1 0 ^{- 31} \cdot 2.82 \times 1 0 ^{- 15}} \approx 8.9 \times 1 0^{10} m/s$ .
Compare to $c = 3.0 \times 1 0^{8} m/s$ . Ratio $v / c \approx 296$ — nearly 300 times the speed of light, impossible.

Verify: $v / c \approx 2.96 \times 1 0^{2} ≫ 1$ ✔ — confirms the parent note's claim that spin is not literal rotation. The "tiny spinning ball" picture is only a mnemonic.

Cell H — real-world Stern–Gerlach numbers

Cell I — exam-style twist (spin-flip energy)

Worked example Ex 9 — Photon energy to flip a spin (Cell I)

In a magnetic field $B = 1.0 T$ along $z$ , the two electron spin states have energy $U = - μ_{z} B$ . Find the energy gap $Δ U$ between them and the frequency $f$ of a photon that flips the spin ( $Δ U = h f$ , with $h = 2 π ℏ$ ). This underlies the Zeeman Effect.

Forecast: microwave ( $\sim 1 0^{10}$ Hz), infrared, or visible?

Energies of each state. $U = - μ_{z} B$ . Up-state $μ_{z} = - μ_{B} \Rightarrow U_{u p} = + μ_{B} B$ ; down-state $μ_{z} = + μ_{B} \Rightarrow U_{d o w n} = - μ_{B} B$ . Why signs matter: the lower energy is the state whose moment aligns with $B$ . Getting the sign wrong flips which state is the ground state.
Energy gap. $Δ U = U_{u p} - U_{d o w n} = 2 μ_{B} B$ .
Plug in. $Δ U = 2 \times 9.274 \times 1 0^{- 24} \times 1.0 = 1.855 \times 1 0^{- 23} J$ .
Photon frequency. $f = \frac{Δ U}{h} = \frac{Δ U}{2 π ℏ} = \frac{1.855 \times 1 0 ^{- 23}}{6.626 \times 1 0 ^{- 34}} \approx 2.80 \times 1 0^{10} Hz$ — a microwave (~28 GHz).

Verify: $2 μ_{B} B = 1.855 \times 1 0^{- 23}$ J ✔; $f \approx 2.80 \times 1 0^{10}$ Hz, in the microwave band ✔. (This is exactly electron spin resonance.)

Recall

Connections

Parent: Spin — intrinsic angular momentum
Stern-Gerlach Experiment — Ex 2, 4, 8 are its arithmetic.
Orbital Angular Momentum — same $ℓ (ℓ + 1)$ magnitude rule, integer only.
Quantum Numbers — where $m_{s}$ sits in the full state label.
Bohr Magneton — the unit doing the work in Ex 8, 9.
Zeeman Effect & Fine Structure — Ex 9's energy splitting.
Pauli Exclusion Principle — needs the two $m_{s}$ of Ex 2.
Dirac Equation — the source of $g_{s} \approx 2$ used everywhere above.