University Physics V · Angular Momentum in Quantum Mechanics · 9.8
Spin ½, Pauli Matrices & SU(2)
Every other angular momentum in this course lives on a wavefunction; this one has none. Here you build the two-dimensional representation the algebra insists on, learn the single identity that generates all Pauli algebra, and find out why a spinor needs two full turns to come home.
Build the model
Connect the measurement to the mechanism.
The commutator algebra [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ knows nothing about wavefunctions, and the ladder argument that follows from it permits j = ½ as freely as it permits j = 1. Orbital motion cannot cash that permission in: realised on the sphere as Lz = −iħ ∂/∂φ, the lowering operator applied to a would-be l = ½ bottom state refuses to vanish, the ladder never closes, and the functions it generates leave L² of the sphere. So the j = ½ representation exists only abstractly — a Hilbert space C² carried along with the particle, with no position-space realisation at all.
Fix the basis by diagonalising Sz and the ladder matrix elements hand you three 2 × 2 Hermitian matrices, S = (ħ/2)σ. Everything else is consequence. One identity, σᵢ σⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ, generates the commutators, the anticommutators, and (n̂⋅σ)² = I, and that last fact collapses the rotation exponential into the exact closed form cos(θ/2) I − i sin(θ/2) n̂⋅σ.
The cost is the half-angle. A rotation through 2π returns every vector in the room to itself but multiplies the spinor by −1; the map SU(2) → SO(3) is two-to-one, and the state only comes home after 4π. That sign is invisible on a state as a whole and perfectly visible in an interferometer where only one arm is turned.
- Simple definition
- Spin one-half is the two-dimensional irreducible representation of the angular-momentum algebra: S = (ħ/2)σ acting on C², an internal space the particle carries rather than a function of where it is.
- Example
- Sz = (ħ/2) diag(1, −1), so a measurement along any axis returns ±ħ/2 = ±5.27 × 10⁻³⁵ J s and never anything between; along an axis 60° from the preparation axis the plus outcome appears cos²30° = 75% of the time.
j = ½ gives 2j + 1 = 2 states, so every spin observable here is a 2 × 2 matrix.
ħ/2 = 5.27 × 10⁻³⁵ J s; the σᵢ are dimensionless, so S carries the unit of action
This basis is the one diagonalising Sz; naming ẑ is a convention, not physics.
Hermitian, traceless, det = −1, eigenvalues ±1, and σᵢ² = I
Gives (a⋅σ)(b⋅σ) = (a⋅b) I + i(a×b)⋅σ, hence (n̂⋅σ)² = I for every unit n̂.
Antisymmetric part [σᵢ, σⱼ] = 2i εᵢⱼₖ σₖ; symmetric part (σᵢ, σⱼ) = 2 δᵢⱼ I
(n̂⋅σ)² = I collapses the series exactly — no truncation, no matrix exponential.
θ in radians about the unit vector n̂; U†U = I and det U = cos²(θ/2) + sin²(θ/2) = 1
SU(2) → SO(3) is two-to-one: vectors close after 2π, spinors only after 4π.
R(n̂, θ) ∈ SO(3) is the rotation by θ about n̂; U and −U give the same R
Half-angle in the amplitude: θ = 90° gives 0.500, θ = 180° gives exactly 0.
θ, φ are the polar angles of n̂; P₊ is for +ħ/2 along n̂ given the state |↑z⟩
A representation the sphere cannot hold
The ladder argument used only [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ and the positivity of norms, and it permits half-integer j. Orbital motion cannot cash that permission in. Realised on the sphere, Lz = −iħ ∂/∂φ has eigenfunctions e(imφ), and the honest reason integer l wins is not single-valuedness but closure: construct the would-be l = ½ bottom state and L₋ applied to it fails to vanish, so the ladder runs on forever and the functions it generates leave L² of the sphere. So j = ½ has no position-space realisation whatsoever. It exists as an abstract two-dimensional representation of su(2), carried on a Hilbert space C² that travels with the particle. The full state space of one electron is L²(R³) ⊗ C², and its wavefunction becomes a two-component object ψ(r) = (ψ₊(r), ψ₋(r)). Spin is an internal label, not a rotation of anything in the room.
Building S from the ladder matrix elements
Fix the basis by diagonalising Sz — nothing in the algebra prefers ẑ, so this is a convention — and label the two states |↑⟩ = |½, +½⟩ and |↓⟩ = |½, −½⟩. The ladder matrix element ⟨j, m±1| J_± |j, m⟩ = ħ√(j(j+1) − m(m±1)) supplies the rest. For j = ½ and m = −½ it gives ħ√(3/4 + 1/4) = ħ, so S₊|↓⟩ = ħ|↑⟩ while S₊|↑⟩ = 0: 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]], with Sz already diagonal at (ħ/2) diag(1, −1). Factor ħ/2 out of all three and the Pauli matrices are what remains. As a check, S² = Sₓ² + Sy² + Sz² = 3(ħ/2)² I = (3/4) ħ² I, which is ħ² j(j+1) at j = ½. Four lines of NumPy build these, and the residual ‖[Sₓ, Sy] − iħ Sz‖ comes back at machine epsilon.
One identity finishes the algebra
Everything algebraic about spin-½ follows from σᵢ σⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ. Its symmetric part is the anticommutator (σᵢ, σⱼ) = 2 δᵢⱼ I, which says σᵢ² = I and that distinct Pauli matrices anticommute; its antisymmetric part is [σᵢ, σⱼ] = 2i εᵢⱼₖ σₖ, the su(2) algebra again. Contract it with two vectors and you get (a⋅σ)(b⋅σ) = (a⋅b) I + i(a×b)⋅σ. Put a = b = n̂ and the cross product dies: (n̂⋅σ)² = I for every unit vector, so n̂⋅σ has eigenvalues ±1 and S⋅n̂ has eigenvalues ±ħ/2 along every axis, with no preferred direction anywhere in sight. The identity also shows that (I, σₓ, σy, σz) spans the real space of 2 × 2 Hermitian matrices: any spin observable is A = a₀ I + a⋅σ, and any state, pure or mixed, is ρ = ½(I + s⋅σ) with |s| ≤ 1 — the Bloch ball.
Exponentiating: SU(2) in closed form
Because (n̂⋅σ)² = I, the exponential series splits cleanly. Even powers of −iθ(n̂⋅σ)/2 are multiples of I and sum to cos(θ/2); odd powers are multiples of n̂⋅σ and sum to −i sin(θ/2). So U(n̂, θ) = exp(−iθ n̂⋅σ/2) = cos(θ/2) I − i sin(θ/2) n̂⋅σ exactly. Verify it: U†U = (cos² + sin²) I = I, and writing c = cos(θ/2), s = sin(θ/2), det U = (c − i s nz)(c + i s nz) + s²(nₓ² + ny²) = c² + s² = 1. Unitary with unit determinant is the definition of SU(2), and every element of SU(2) has this form; written as U = [[a, −b̄], [b, ā]] with |a|² + |b|² = 1, the group is the unit 3-sphere. The simplest case is the whole lesson in one matrix: about ẑ, U = diag(e(−iθ/2), e(+iθ/2)), the two components picking up opposite half-angle phases.
The double cover, and where the minus sign becomes visible
Conjugation turns SU(2) into rotations of ordinary vectors: U(n̂, θ)(a⋅σ)U† = (R(n̂, θ) a)⋅σ, where R is the honest 3 × 3 SO(3) rotation by θ about n̂. The map U ↦ R is a homomorphism, and it is two-to-one, because U and −U conjugate identically. Feed it θ = 2π: U = cos(π) I − i sin(π)(n̂⋅σ) = −I for every axis, while R is the identity. Only θ = 4π brings U back to +I. Is that sign physical? On the state as a whole, no: ⟨ψ|A|ψ⟩ and |⟨φ|ψ⟩|² are both blind to an overall phase. It becomes physical the moment it is relative. Split a neutron beam, rotate the spin in one arm only through α, recombine, and the interference term carries a factor cos(α/2), so the count rate is periodic in α with period 4π rather than 2π. Rauch's and Werner's 1975 interferometers measured exactly that.
What Stern–Gerlach measures, and why gₛ ≈ 2 matters
The electron's magnetic moment is μ = −gₛ μB S/ħ, with μB = 9.274 × 10⁻²⁴ J T⁻¹ and gₛ = 2.00232. That g-factor is input at this level, not derived: Stern–Gerlach and spectroscopy measure it, the Dirac equation later explains the 2, and QED supplies the 0.00232. Put it into H = −μ⋅B = (gₛ μB/2) σ⋅B; for B along ẑ the levels sit at ±gₛ μB B/2, a splitting of gₛ μB B ≈ 2 μB B — twice what one step in mₗ gives in the same field, which is why the anomalous Zeeman effect was unexplainable without spin. At B = 1.00 T that is 1.86 × 10⁻²³ J, or 28.0 GHz. And note what the magnet does: the force is Fz = μz ∂Bz/∂z, so the apparatus sorts on μz, not on Sz. The negative g-factor flips them, so the sub-beam pulled towards strong field is the mₛ = −½ one.
Change one variable at a time
Make the relationship visible.
Leave δ at 0 and drag θ to 360°: the dashed probability is back at 1.000 while the solid amplitude has reached −1.000 — that is U(2π) = −I. Only at 720° do both return. Then set δ = 90° and read the probability at θ = 0: exactly 0.500.
AMPLITUDE ⟨+n̂|ψ⟩0.707
PROBABILITY P(+ħ/2)0.500
⟨S⋅n̂⟩ ÷ (ħ/2)0.000
½ Tr U(θ)0.707
Live interpretationAMPLITUDE ⟨+n̂|ψ⟩: 0.707. PROBABILITY P(+ħ/2): 0.500. ⟨S⋅n̂⟩ ÷ (ħ/2): 0.000. ½ Tr U(θ): 0.707
Catch the common trap
Explain before calculating.
A spin-½ state is rotated through θ = 2π about an arbitrary unit axis n̂ using U(n̂, θ) = exp(−iθ n̂⋅σ/2). Which statement about U and its observable consequences is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyMultiply out σₓ σy and σy σₓ using the 2 × 2 matrices, then use the results to write down the commutator [σₓ, σy] and the anticommutator (σₓ, σy). Finally convert the commutator into a statement about Sₓ and Sy.
- σₓ σy = [[0,1],[1,0]] · [[0,−i],[i,0]]. Top row: (0⋅0 + 1⋅i, 0⋅(−i) + 1⋅0) = (i, 0). Bottom row: (1⋅0 + 0⋅i, 1⋅(−i) + 0⋅0) = (0, −i). So σₓ σy = [[i,0],[0,−i]] = i σz.
- Reverse the order: σy σₓ top row is (0⋅0 + (−i)⋅1, 0⋅1 + (−i)⋅0) = (−i, 0), and bottom row is (i⋅0 + 0⋅1, i⋅1 + 0⋅0) = (0, i). So σy σₓ = −i σz.
- Subtract: [σₓ, σy] = i σz − (−i σz) = 2i σz. Add: (σₓ, σy) = i σz + (−i σz) = 0. Both are the i = x, j = y case of σᵢ σⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ, with δxy = 0 and εxyz = +1.
- Since Sᵢ = (ħ/2) σᵢ, [Sₓ, Sy] = (ħ/2)² [σₓ, σy] = (ħ²/4)(2i σz) = iħ (ħ/2) σz = iħ Sz — the su(2) algebra, reproduced by 2 × 2 matrices with no differential operator anywhere in sight.
Answerσₓ σy = i σz and σy σₓ = −i σz, so [σₓ, σy] = 2i σz and (σₓ, σy) = 0; scaling by (ħ/2)² gives [Sₓ, Sy] = iħ Sz.
MediumA spin-½ prepared as |↑z⟩ is rotated through θ = 120° about ŷ by U = exp(−iθ σy/2). Write U as an explicit 2 × 2 matrix, apply it, and find the probability that a subsequent Sz measurement returns +ħ/2, together with ⟨Sz⟩ in the rotated state.
- (n̂⋅σ)² = I for a unit n̂, so the series collapses: U = cos(θ/2) I − i sin(θ/2) σy. With θ = 120°, θ/2 = 60°, cos 60° = 1/2 and sin 60° = √3/2 = 0.8660.
- Evaluate the second term: −i(√3/2) σy = −i(√3/2)[[0,−i],[i,0]] = [[0, −√3/2], [√3/2, 0]], since (−i)(−i) = −1 and (−i)(i) = +1. Hence U = [[1/2, −√3/2], [√3/2, 1/2]].
- Act on |↑z⟩ = (1, 0)ᵀ: |ψ⟩ = (1/2, √3/2)ᵀ = cos 60° |↑⟩ + sin 60° |↓⟩. That is exactly χ₊(θ = 120°, φ = 0), the eigenspinor of n̂⋅σ for n̂ tilted 120° from ẑ — the rotation moved the spin axis by the full 120° while the amplitudes moved by the half-angle.
- Born rule: P(+ħ/2) = |⟨↑|ψ⟩|² = (1/2)² = 0.250, and P(−ħ/2) = 3/4 = 0.750. Note this is cos²(θ/2), not cos θ.
- ⟨Sz⟩ = (ħ/2)(0.250 − 0.750) = −ħ/4 = −2.64 × 10⁻³⁵ J s, which equals (ħ/2) cos 120°. The mean follows the full angle even though the amplitudes follow the half-angle.
AnswerU = [[½, −√3/2], [√3/2, ½]] and |ψ⟩ = (½, √3/2)ᵀ. P(+ħ/2) = 0.250, P(−ħ/2) = 0.750, and ⟨Sz⟩ = −ħ/4 = −2.64 × 10⁻³⁵ J s.
HardIn a neutron interferometer one arm passes 2.00 cm through a uniform field B, which rotates the spin in that arm only. The neutrons have λ = 1.80 Å and γₙ = 1.832 × 10⁸ rad s⁻¹ T⁻¹. Find the field that rotates the spin through 2π, say what it does to the fringes, and find the field that restores them.
- Speed from de Broglie: v = h/(mₙ λ) = 6.626 × 10⁻³⁴ / (1.675 × 10⁻²⁷ × 1.80 × 10⁻¹⁰) = 6.626 × 10⁻³⁴ / 3.015 × 10⁻³⁷ = 2.20 × 10³ m s⁻¹. Time in the field: t = 0.0200 / 2198 = 9.10 × 10⁻⁶ s.
- H = −μ⋅B = (ħ ωL/2)(n̂⋅σ) with ωL = γₙ B, so U = exp(−i(ωL t/2) n̂⋅σ) is precisely an SU(2) rotation through α = γₙ B t about the field direction.
- Set α = 2π: B = 2π / (γₙ t) = 6.2832 / (1.832 × 10⁸ × 9.10 × 10⁻⁶) = 6.2832 / 1667 = 3.77 × 10⁻³ T.
- At α = 2π, U = cos(π) I − i sin(π)(n̂⋅σ) = −I. The rotated arm's amplitude flips sign, so ψ₁ + ψ₂ becomes −ψ₁ + ψ₂ and a constructive fringe goes destructive. A global −1 would be invisible; this one is relative, so it is not.
- Restoration needs α = 4π, i.e. B = 7.54 × 10⁻³ T, where U = +I. The count rate is therefore periodic in B with period 7.54 mT, and a 3.77 mT step inverts rather than repeats it.
AnswerB(2π) = 3.77 mT inverts the fringes; B(4π) = 7.54 mT restores them. The interferogram is periodic in B with period 7.54 mT, not 3.77 mT — the double cover, measured.