University Physics IV · Angular Momentum and Spin · 9.6
Intrinsic Spin, Spinors & Pauli Matrices
Spin is the angular momentum that survives when you switch off position. Nothing on the sphere carries it, so the state shrinks to a column of two complex numbers and every observable becomes a 2×2 matrix. That small piece of algebra predicts every analyser chain you will build — and it kills the spinning-ball picture.
Build the model
Connect the measurement to the mechanism.
The ladder argument never forbade half-integers; the sphere did. From [Sₓ, Sy] = iħSz alone the spectrum is s(s+1)ħ² and mħ with s integer or half-integer, and orbital motion lost the half-integers only because the l = ½ tower cannot be built from normalisable functions of θ and φ. So the algebra leaves room for a degree of freedom with no position representation at all, and Stern-Gerlach, the alkali doublets and the anomalous Zeeman patterns say nature uses one.
Take s = ½: the state space has 2s + 1 = 2 dimensions, the state is a spinor χ = (a, b)ᵀ with |a|² + |b|² = 1, and every observable is a 2×2 Hermitian matrix, with S = (ħ/2)σ. The payoff is that the algebra closes completely. σᵢσⱼ = δᵢⱼ + iεᵢⱼₖσₖ forces σᵢ² = 1, so a measurement along any axis whatsoever returns ±ħ/2; S² = ¾ħ² times the identity, so s = ½ labels the particle and is not a quantum number the state can change; and the eigenspinor along any direction is written down in one line of half-angles. The cost is the picture. ⟨S⟩ has length ħ/2 while |S| = √3ħ/2, so no arrow represents the state, and a 2π rotation multiplies a spinor by −1, so its period is 4π.
Whatever spin is, it is not a small ball turning.
- Simple definition
- Intrinsic spin is an angular momentum a particle carries as an internal property: it obeys the same commutation relations as orbital angular momentum but depends on no coordinate, so its state is a finite column of complex numbers rather than a function of space.
- Example
- An electron has s = ½, so S² = s(s+1)ħ² = ¾ħ² and |S| = (√3/2)ħ = 9.13 × 10⁻³⁵ J s — yet any single component you measure, along any axis at all, is only ±ħ/2 = ±5.27 × 10⁻³⁵ J s.
Identical to L's algebra, so the ladder result carries over — but with no ∂/∂φ available to veto half-integers.
S carries J s; s = ½ for the electron, proton, neutron and quarks
Turns operator algebra into 2×2 arithmetic: every spin question becomes a matrix multiplication you can do by hand.
σ is dimensionless, Hermitian, traceless, det = −1, eigenvalues ±1
Its symmetric part gives σᵢ² = 1, hence ±ħ/2 on every axis; its antisymmetric part regenerates the commutators.
1 is the 2×2 identity; a and b are ordinary c-number vectors
Any analyser direction without solving a new eigenvalue problem — and it gives ⟨S⟩ = (ħ/2)n̂ exactly.
θ, φ are the polar angles of n̂; χ₋ = (−e(−iφ)sin(θ/2), cos(θ/2))ᵀ
The measurable content of the half-angle: this is what a Stern-Gerlach chain actually counts.
Θ is the angle between the two analyser axes; Θ = 90° gives ½
The 4π period. The minus sign is global and hidden — until only one interferometer arm is rotated.
α = 2π gives U = −1; only α = 4π gives U = +1
Half-integers were never ruled out by the algebra
Run the ladder argument on any triple obeying [Jₓ, Jy] = iħJz. Demanding that the ladder terminate at both ends forces the eigenvalues j(j+1)ħ² and mħ with 2j + 1 rungs, and j comes out integer or half-integer. Orbital angular momentum kept only the integers, and not because e(imφ) must be single-valued — a half-integer m only flips an overall sign, which is unobservable. It kept them because the l = ½ tower fails: L₋ does not annihilate the supposed bottom rung, and lowering again leaves the normalisable functions on the sphere altogether. That veto is a property of the position representation, not of the algebra. A degree of freedom that has no position representation is free to take j = ½ — and the evidence says one exists: two Stern-Gerlach beams from an l = 0 silver atom, doublet lines in sodium, and Zeeman patterns with the wrong number of components.
Building the Pauli matrices from the ladder
With s = ½ the space has 2s + 1 = 2 dimensions, so take |↑⟩ = (1,0)ᵀ and |↓⟩ = (0,1)ᵀ as the Sz eigenvectors with eigenvalues ±ħ/2. That alone fixes Sz = (ħ/2)diag(1, −1). The other two come from the ladder: S₊|s, m⟩ = ħ√(s(s+1) − m(m+1))|s, m+1⟩, and for s = ½, m = −½ the root is √(¾ + ¼) = 1, so S₊|↓⟩ = ħ|↑⟩ while S₊|↑⟩ = 0. In matrix form S₊ = ħ[[0,1],[0,0]] and S₋ = ħ[[0,0],[1,0]]. Then Sₓ = (S₊ + S₋)/2 = (ħ/2)[[0,1],[1,0]] and Sy = (S₊ − S₋)/2i = (ħ/2)[[0,−i],[i,0]]. Nothing was assumed except the commutators and the dimension: the Pauli matrices are the output of the derivation, not an input to it. Writing S = (ħ/2)σ strips off the units and leaves three dimensionless matrices to work with.
One identity does the work of a chapter
Everything about a spin-half follows from σᵢσⱼ = δᵢⱼ1 + iεᵢⱼₖσₖ. Set j = i: σᵢ² = 1, so each Pauli matrix has eigenvalues ±1 and every spin component, along every axis, measures ±ħ/2 — no state and no direction gives anything else. Symmetrise: (σᵢ, σⱼ) = 2δᵢⱼ1, so different components anticommute, the sharpest possible incompatibility. Antisymmetrise: [σᵢ, σⱼ] = 2iεᵢⱼₖσₖ, which is [Sₓ, Sy] = iħSz restored. Sum the squares: S² = (ħ²/4)(σₓ² + σy² + σz²) = ¾ħ²1, proportional to the identity, so every spinor is an S² eigenstate and s = ½ is a property of the electron rather than a quantum number of its state. Finally 1, σₓ, σy and σz span all 2×2 Hermitian matrices, so every observable of a spin-half is a1 + b⋅σ and there is nothing else left to find.
Any axis in closed form, and why the angle is halved
Write n̂ = (sinθ cosφ, sinθ sinφ, cosθ). Then S⋅n̂ = (ħ/2)[[cosθ, sinθ e(−iφ)], [sinθ e(iφ), −cosθ]], whose trace is 0 and determinant −ħ²/4 for every θ and φ — so the eigenvalues are ±ħ/2 on every axis and no direction in space is privileged. Solving for the + eigenvector gives χ₊(n̂) = (cos(θ/2), e(iφ)sin(θ/2))ᵀ. Check it: θ = 90°, φ = 0 returns (1/√2)(1, 1)ᵀ, the +x spinor. Two features matter. First, ⟨S⟩ = χ₊†Sχ₊ = (ħ/2)n̂ exactly, so the mean vector has length ħ/2 and not √3ħ/2; the shortfall is transverse variance, not a shorter arrow. Second, the angle is halved: turn n̂ through θ and the components turn through θ/2, so χ₊ at θ = 2π is −χ₊ at θ = 0. The half-angle is not a convention — it is the whole reason the period is 4π.
cos²(half the angle) is what an analyser chain counts
Prepare |↑⟩ and analyse along n̂ at angle Θ from z. The amplitude is ⟨χ₊(n̂)|↑⟩ = cos(Θ/2), so P(+) = cos²(Θ/2) = (1 + cosΘ)/2 and P(−) = sin²(Θ/2). In numbers: Θ = 0 gives 1, Θ = 60° gives 0.750, Θ = 90° gives 0.500, Θ = 120° gives 0.250, Θ = 180° gives 0. Now chain three analysers, z then x then z. The x stage splits the beam 50/50, and its +x output — a state that no longer knows anything about z — splits 50/50 again, so a quarter of an initially pure +z beam ends in the −z channel although the x stage removed nothing along z. No assignment of definite values to every axis, fixed in advance, reproduces that. Be precise about what it refutes, though: it kills values pre-existing along all axes at once, but it is a single-particle result, not a Bell test of local hidden variables.
The 2π sign, the 4π period, and the end of the ball
Expand the rotation operator using (n̂⋅σ)² = 1 and it collapses to U(α, n̂) = cos(α/2)1 − i sin(α/2)(n̂⋅σ). At α = 2π that is −1: every spinor comes back reversed, and only α = 4π restores it. A global sign cannot be measured, but a relative one can — split a neutron beam, rotate the spin in one arm only, and the interference intensity is periodic in 4π of rotation, which is what the 1975 interferometry experiments found. No classical picture survives alongside that. Model the electron as a uniform sphere of the classical electron radius 2.82 × 10⁻¹⁵ m carrying L = ħ/2: the equatorial speed is 5ħ/(4mr) = 5.1 × 10¹⁰ m s⁻¹, about 170c, and keeping it below c would demand a radius above 4.8 × 10⁻¹³ m when experiment bounds the electron below 10⁻¹⁸ m. Spin says how a state transforms under rotations, not how fast anything turns.
Change one variable at a time
Make the relationship visible.
Set Θ₀ = 0 and drag Θ from 0 to 720°. The solid probability curve is back to 1 at 360°, but the dashed amplitude is −1 there and only returns to +1 at 720° — the 4π period. Then move Θ₀: only the difference Θ − Θ₀ matters.
AMPLITUDE cos((Θ−Θ₀)/2)0.707
PROBABILITY P(+n̂)0.500
P(−n̂)0.500
RELATIVE ANGLE Θ − Θ₀90 °
Live interpretationAMPLITUDE cos((Θ−Θ₀)/2): 0.707. PROBABILITY P(+n̂): 0.500. P(−n̂): 0.500. RELATIVE ANGLE Θ − Θ₀: 90 °
Catch the common trap
Explain before calculating.
A beam of silver atoms is prepared in the Sz = +ħ/2 state. It passes an analyser whose axis lies 45° from z in the x–z plane, and only that analyser's +ħ/2 output is kept. Those atoms then enter a second analyser aligned with z. What fraction of the original beam leaves the second analyser in the Sz = −ħ/2 channel?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is in the spinor state χ = (1/√5)(1, 2i)ᵀ written in the Sz basis. Check that it is normalised, find the probabilities of the two Sz outcomes, then compute ⟨Sz⟩ and ⟨Sₓ⟩.
- Normalisation: |a|² + |b|² = (1/5)(|1|² + |2i|²) = (1/5)(1 + 4) = 1 ✓. The i does not survive the modulus squared.
- P(Sz = +ħ/2) = |a|² = 1/5 = 0.20 and P(Sz = −ħ/2) = |b|² = 4/5 = 0.80. These exhaust the possibilities because σz has only two eigenvalues.
- ⟨Sz⟩ = (+ħ/2)(0.20) + (−ħ/2)(0.80) = −0.30ħ = −3.16 × 10⁻³⁵ J s. It lies between the two allowed values without ever being measured as one.
- ⟨Sₓ⟩ = (ħ/2)χ†σₓχ. Here σₓχ = (1/√5)(2i, 1)ᵀ, so χ†σₓχ = (1/5)[(1)(2i) + (−2i)(1)] = 0, giving ⟨Sₓ⟩ = 0 — though a measurement of Sₓ still returns ±ħ/2 every time.
AnswerNormalised; P(+ħ/2) = 0.20 and P(−ħ/2) = 0.80; ⟨Sz⟩ = −0.30ħ = −3.16 × 10⁻³⁵ J s; ⟨Sₓ⟩ = 0.
MediumA beam prepared with Sz = +ħ/2 meets an analyser tilted 60° from z in the x–z plane. Find the probabilities of its two outputs, then send the + output into a z analyser and find the probability of Sz = −ħ/2. What fraction of the original beam ends in that channel, and what is ⟨Sz⟩ for the atoms leaving the tilted analyser's + port?
- The prepared state is |↑⟩ and the analyser projects onto χ₊(n̂) = (cos30°, sin30°)ᵀ for θ = 60°, φ = 0. The amplitude is ⟨χ₊|↑⟩ = cos30° = 0.8660.
- P(+n̂) = cos²30° = 0.750 and P(−n̂) = sin²30° = 0.250, and the two sum to 1 as they must.
- The surviving atoms are now in the state χ₊(n̂) = (0.8660, 0.5000)ᵀ, whatever they were before. Read that in the z basis: P(Sz = +ħ/2) = 0.8660² = 0.750 and P(Sz = −ħ/2) = 0.5000² = 0.250.
- Fraction of the original beam reaching the final −z channel = 0.750 × 0.250 = 0.1875, i.e. 18.75%. A beam that was purely +z has acquired a −z component from a filter that never measured z.
- ⟨Sz⟩ at the + port = (+ħ/2)(0.750) + (−ħ/2)(0.250) = ħ/4, which is (ħ/2)cos60° — the z projection of ⟨S⟩ = (ħ/2)n̂.
AnswerP(+n̂) = 0.750 and P(−n̂) = 0.250; the following z analyser gives −ħ/2 with probability 0.250, so 18.75% of the original beam ends there; ⟨Sz⟩ = ħ/4 at the + port.
HardModel the electron as a uniform solid sphere of the classical electron radius r = 2.82 × 10⁻¹⁵ m and mass m = 9.11 × 10⁻³¹ kg, spinning with angular momentum ħ/2. Find the equatorial speed and compare it with c, then find the radius the sphere would need for that speed to fall below c.
- A uniform sphere has I = (2/5)mr², and the equatorial speed is v = ωr, so L = Iω = (2/5)mrv and therefore v = 5L/(2mr) = 5ħ/(4mr) with L = ħ/2.
- Numerator: 5ħ = 5 × 1.055 × 10⁻³⁴ = 5.27 × 10⁻³⁴ J s. Denominator: 4mr = 4 × 9.11 × 10⁻³¹ × 2.82 × 10⁻¹⁵ = 1.03 × 10⁻⁴⁴ kg m.
- v = 5.27 × 10⁻³⁴ / 1.03 × 10⁻⁴⁴ = 5.1 × 10¹⁰ m s⁻¹, which is 1.7 × 10² times c. The model is dead on arrival, and no choice of moment-of-inertia factor rescues a factor of 171.
- Set v = c and solve for the radius: r = 5ħ/(4mc) = 5.27 × 10⁻³⁴ / (4 × 9.11 × 10⁻³¹ × 3.00 × 10⁸) = 5.27 × 10⁻³⁴ / 1.09 × 10⁻²¹ = 4.8 × 10⁻¹³ m.
- That is 171 times the classical electron radius and some 600 proton radii, while electron scattering bounds any electron substructure below 10⁻¹⁸ m — five orders of magnitude smaller still.
- So ħ/2 cannot be stored as rotation of matter. It is a label for how the two-component state transforms, which is why S has a matrix representation but no differential one.
Answerv ≈ 5.1 × 10¹⁰ m s⁻¹ ≈ 1.7 × 10² c; sub-luminal rotation would need r ≳ 4.8 × 10⁻¹³ m, more than five orders of magnitude above the experimental bound on the electron's size.