Spin ½, Pauli matrices and SU(2)

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, for the reason UPV-09.06 settled: on the sphere the l = ½ top state e^(iφ/2) √sin θ is annihilated by L₊, but lowering it twice gives −ħ² e^(−3iφ/2) sin^(−3/2) θ where the algebra demands zero, a function outside L² of the sphere, and no half-integer ladder closes there. 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)), with ψ_±(r) = ⟨r, ±|ψ⟩ in the |r⟩ ⊗ |±⟩ basis. Spin is an internal label, not a rotation of anything in the room.

Fix the basis by diagonalising S_z — 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 S_y = (S₊ − S₋)/2i = (ħ/2)[[0,−i],[i,0]], with S_z 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 + S_y² + S_z² = 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, S_y] − iħ S_z‖ comes back at machine epsilon.

A spin state is a normalised χ = a|↑⟩ + b|↓⟩ in C², with |a|² + |b|² = 1 and the global phase unobservable, which leaves two real parameters: exactly enough to name a direction. The Pauli algebra as a toolkit — commutators, anticommutators and the rule for (a⋅σ)(b⋅σ) — is worked in UPV-10.03; of its product identity, all that is needed here is (n̂⋅σ)² = I for every unit n̂. That makes the eigenvalues of n̂⋅σ ±1 and those of S⋅n̂ ±ħ/2 on every axis, so the representation singles out no direction and ẑ was only the axis chosen for the basis. Every normalised spinor is, up to phase, the +ħ/2 eigenstate of S⋅n̂ for exactly one n̂: χ₊(θ, φ) = (cos(θ/2), e^(iφ) sin(θ/2))ᵀ, with θ, φ the polar angles of n̂ = ⟨χ|σ|χ⟩. But the spinor is not that 3-vector. Orthogonal in C² means opposite in space: ⟨↑|↓⟩ = 0 while ẑ and −ẑ are antiparallel, and for two axes an angle γ apart |⟨χ₊(n̂)|χ₊(m̂)⟩|² = (1 + n̂⋅m̂)/2 = cos²(γ/2), a half-angle that no inner product of 3-vectors produces. That half-angle is the signature of the j = ½ representation, and the next section shows it is also what a rotation does to the state.

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 n_z)(c + i s n_z) + s²(nₓ² + n_y²) = 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.

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.

The electron's magnetic moment is μ = −gₛ μ_B S/ħ, with μ_B = 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 μ_B/2) σ⋅B; for B along ẑ the levels sit at ±gₛ μ_B B/2, a splitting of gₛ μ_B B ≈ 2 μ_B 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 F_z = μ_z ∂B_z/∂z, so the apparatus sorts on μ_z, not on S_z. The minus sign in μ, set by the electron's negative charge, flips them, so the sub-beam pulled towards strong field is the mₛ = −½ one.