Spin-half states in C²

The ladder argument of the previous lesson gives 2s + 1 = 2 states at s = ½, so the spin factor of the state space is C². What it does not give is a basis. S_z is Hermitian, so its eigenvectors span C² and can be chosen orthonormal; call them |↑⟩ = |½, +½⟩ and |↓⟩ = |½, −½⟩, with S_z|↑⟩ = (ħ/2)|↑⟩ and S_z|↓⟩ = −(ħ/2)|↓⟩. That is a choice of coordinates, not a fact about spin — diagonalising Sₓ instead gives an equally good pair. The completeness relation |↑⟩⟨↑| + |↓⟩⟨↓| = I is the promise that two vectors are enough. Once the basis is fixed, every ket is a column (α, β)ᵀ = (⟨↑|ψ⟩, ⟨↓|ψ⟩)ᵀ, every bra the conjugate row (α*, β*), and every operator the matrix Aᵢⱼ = ⟨i|A|j⟩, so S_z = (ħ/2) diag(1, −1). Inner products become arithmetic: for |φ⟩ = (1, 1)/√2 and |ψ⟩ = (1, i)/√2, ⟨φ|ψ⟩ = (1⋅1 + 1⋅i)/2 = (1 + i)/2, and |⟨φ|ψ⟩|² = ½. Conjugate the bra, not the ket: ⟨ψ|φ⟩ = (1 − i)/2 is the complex conjugate, and forgetting that flips the sign of every relative phase you compute.

The Born rule says P(S_z = +ħ/2) = |⟨↑|ψ⟩|² = |α|² and P(S_z = −ħ/2) = |β|², so ⟨ψ|ψ⟩ = |α|² + |β|² = 1 is not decoration: it is the two outcomes exhausting the possibilities. Take the un-normalised column (2, 1 + i). Its norm squared is |2|² + |1 + i|² = 4 + 2 = 6 — the modulus of 1 + i is √2, not 2 — so the normalised state is (2, 1 + i)/√6. Then P(↑) = 4/6 = 0.667, P(↓) = 2/6 = 0.333, and ⟨S_z⟩ = (ħ/2)(|α|² − |β|²) = (ħ/2)(1/3) = ħ/6 ≈ 1.76 × 10⁻³⁵ J s. Two cautions follow. First, the moduli speak about S_z only: the same state has ⟨Sₓ⟩ = ħ Re(α*β) = ħ Re[2(1 + i)]/6 = ħ/3, hence P(Sₓ = +ħ/2) = ½(1 + 2⟨Sₓ⟩/ħ) = ½(1 + 2/3) = 5/6, a number no amount of staring at |α|² would reveal. Second, normalisation is a convention on the representative: (2, 1 + i) and (2, 1 + i)/√6 are the same physical state, and the second is merely the version whose components can be read as amplitudes without a further division.

α and β are four real numbers. Normalisation spends one. The global phase spends another: replace |ψ⟩ by e^(iγ)|ψ⟩ and every probability |⟨a|ψ⟩|² picks up |e^(iγ)|² = 1, every expectation ⟨ψ|A|ψ⟩ picks up e^(−iγ)e^(iγ) = 1, so no measurement on the spin alone can see γ. A physical state is therefore a ray — a vector up to complex scale — and two real parameters remain. Spend the phase freedom making α real and non-negative; normalisation then lets you write α = cos(θ/2), β = e^(iφ) sin(θ/2) with θ ∈ [0, π] and φ ∈ [0, 2π). The two survivors are the polar and azimuthal angles of a direction, and the space of rays in C² is the Bloch sphere. The relative phase φ is not a global phase and is fully physical: (1, 1)/√2 and (1, −1)/√2 are orthogonal, with ⟨Sₓ⟩ = +ħ/2 for one and −ħ/2 for the other. Worked: (√3/2, i/2) has cos(θ/2) = √3/2, so θ/2 = 30° and θ = 60°, and arg β − arg α = 90° = φ. Its Bloch vector (sin 60° cos 90°, sin 60° sin 90°, cos 60°) = (0, 0.866, 0.5) gives ⟨S⟩ = (ħ/2)(0, 0.866, 0.5): tilted 60° from z toward +y.

With Sᵢ = (ħ/2)σᵢ and σᵢ² = I, S² = Sₓ² + S_y² + S_z² = 3(ħ/2)² I = (3/4)ħ² I, which is s(s + 1)ħ² at s = ½. Every vector in C² is an eigenvector with the same eigenvalue, so S² has expectation 3ħ²/4 and zero variance in every state, commutes with every 2 × 2 matrix, and a measurement of it returns only what the dimension already told you. Contrast orbital angular momentum, where L² takes different values l(l + 1)ħ² on different subspaces of L²(S²) and genuinely sorts states. The identity is still worth one look. A pure spinor has |⟨S⟩| = ħ/2, so |⟨S⟩|² = ħ²/4 while ⟨S²⟩ = 3ħ²/4: the missing ħ²/2 is quantum spread. For |↑⟩ the pieces are ⟨S_z²⟩ = ħ²/4 and ⟨Sₓ²⟩ = ⟨S_y²⟩ = ħ²/4 each, even though ⟨Sₓ⟩ = ⟨S_y⟩ = 0 — the transverse components are never zero, only zero on average. So ⟨S⟩ never reaches the full length √3ħ/2: the ratio (ħ/2)/(√3ħ/2) = 1/√3 is cos 54.7°, the cone half-angle in every vector-model picture.

An electron's state lives in L²(R³) ⊗ C², so in full it is a two-component wavefunction Ψ(r) = ψ↑(r)|↑⟩ + ψ↓(r)|↓⟩ normalised by ∫(|ψ↑|² + |ψ↓|²) d³r = 1. A lone spinor (α, β) describes the spin exactly only when Ψ factorises, Ψ = φ(r) ⊗ (α|↑⟩ + β|↓⟩), which means ψ↑ and ψ↓ are the same spatial function up to the constants α and β. Hamiltonians of the form H_space ⊗ I + I ⊗ Hₛₚᵢₙ keep products as products: a uniform field precesses the spinor and leaves φ alone. A gradient does not. H = −γₑ S_z B′z, with γₑ the gyromagnetic ratio, couples the two factors, and the Stern-Gerlach state α|↑⟩⊗|χ₊⟩ + β|↓⟩⊗|χ₋⟩ with ⟨χ₊|χ₋⟩ → 0 has no spinor at all. What survives is the reduced density matrix ρᵢⱼ = ∫ψᵢ ψⱼ* d³r, a 2 × 2 Hermitian matrix with unit trace; for that state it is diag(|α|², |β|²), whose purity Tr ρ² = |α|⁴ + |β|⁴ falls to ½ at α = β = 1/√2. Every result in this unit that quotes a single (α, β) is silently assuming the product form.

Represent kets as complex arrays and let the machine police the conjugates. With up = np.array([1, 0], complex) and dn = np.array([0, 1], complex), the completeness check is np.allclose(np.outer(up, up.conj()) + np.outer(dn, dn.conj()), np.eye(2)). For ψ = np.array([2, 1 + 1j]) / np.√(6), the inner product is np.vdot(φ, ψ) — vdot conjugates its first argument, np.dot does not, and that one letter is the difference between ⟨φ|ψ⟩ and a meaningless bilinear form. The S_z probabilities are np.abs(ψ)**2, which returns [0.667, 0.333]. Build S2 = sum((0.5 * s) @ (0.5 * s) for s in (sx, sy, sz)) in units of ħ and assert np.allclose(S2, 0.75 * np.eye(2)). The Bloch vector is [np.vdot(ψ, s @ ψ).real for s in (sx, sy, sz)], which gives (0.667, 0.667, 0.333) with norm 1.000 — unit length is the numerical signature of a pure state — and multiplying psi by np.exp(0.7j) first leaves all three components unchanged to machine precision, which is the global phase dropping out in code.