Superposition, relative phase and mixed states

Write |ψ⟩ = α|↑⟩ + β|↓⟩ and count: two complex amplitudes are four real numbers. Normalisation |α|² + |β|² = 1 spends one. The global phase spends another, because replacing |ψ⟩ by e^(iγ)|ψ⟩ leaves every probability |⟨a|ψ⟩|² and every expectation ⟨ψ|A|ψ⟩ untouched — γ cancels between the bra and the ket. Two real parameters survive, and the standard choice is α = cos(θ/2), β = e^(iφ)sin(θ/2) with θ in [0, π] and φ in [0, 2π). So the ratio β/α is physical while α and β separately are not. Concretely, (|↑⟩ + |↓⟩)/√2 and (|↑⟩ + i|↓⟩)/√2 share |α| = |β| = 1/√2 and differ only in φ, 0 against 90° — and they are genuinely different states, eigenstates of Sₓ and of S_y respectively.

Do the matrix arithmetic rather than quoting the result. With |ψ⟩ = (α, β)ᵀ, σₓ|ψ⟩ = (β, α)ᵀ, so ⟨σₓ⟩ = α*β + β*α = 2Re(α*β) and ⟨Sₓ⟩ = ħ Re(α*β). Likewise σ_y|ψ⟩ = (−iβ, iα)ᵀ gives ⟨σ_y⟩ = 2Im(α*β) and ⟨S_y⟩ = ħ Im(α*β), while σ_z is diagonal and gives ⟨S_z⟩ = (ħ/2)(|α|² − |β|²). Substituting the standard parametrisation, ⟨Sₓ⟩ = (ħ/2)sinθ cosφ and ⟨S_y⟩ = (ħ/2)sinθ sinφ. Now fix θ = 90°. At φ = 0 the state is spin-up along x; at φ = 90° it is spin-up along y; at φ = 180° it is spin-down along x — and all three give P(up along z) = 0.500 exactly. A z analyser is blind to φ, while the two ensemble averages ⟨Sₓ⟩ and ⟨S_y⟩ pin it down completely.

For a pure state ρ = |ψ⟩⟨ψ| has entries [[|α|², αβ*], [α*β, |β|²]]: the diagonal is the pair of z populations, the off-diagonal is the coherence and carries the phase. For a mixture ρ = Σ pₖ |ψₖ⟩⟨ψₖ|. Both are Hermitian, unit-trace and positive, and since I together with σₓ, σ_y and σ_z spans the Hermitian 2×2 matrices, every such ρ is ½(I + a⋅σ) with a = ⟨σ⟩. Predictions follow from P_±(n̂) = Tr(ρ P̂_±) = ½(1 ± a⋅n̂), where P̂_± = ½(I ± n̂⋅σ) are the projectors along n̂; the rule is linear in ρ — so mixing preparations means averaging their Bloch vectors, nothing more. Compare the two states with the same z data: (|↑⟩ + |↓⟩)/√2 gives [[0.5, 0.5], [0.5, 0.5]] with a = (1, 0, 0), while the coin-flip mixture gives [[0.5, 0], [0, 0.5]] with a = 0. Identical diagonals, different off-diagonals; that gap is the entire physical difference.

Purity turns the picture into a number: Tr(ρ²) = (1 + |a|²)/2, and the eigenvalues of ρ are (1 ± |a|)/2. If |a| = 1 the eigenvalues are 1 and 0, ρ has rank one, and the state is pure — and then a is a unit vector n̂, so |ψ⟩ = |n, +⟩. That deserves stating plainly: every pure spinor is spin-up along some axis, so no pure state is unpolarised, however random its z counts look. If |a| = 0 then ρ = I/2, every analyser returns 50/50, and Tr(ρ²) = 0.5, the floor in two dimensions. In between, |a| = 0.4 gives eigenvalues 0.7 and 0.3 and Tr(ρ²) = 0.58. The surface of the ball is coherence; the interior is mixedness, whether from classical ignorance of a coin-flip preparation or from entanglement with a system you are not measuring; and the radius says how much of each you have.

Take an environment that couples only to S_z — a fluctuating B_z, say. It never moves the populations, but it randomises the accumulated phase φ. If the noise is fast, φ random-walks: its spread is Gaussian with variance ⟨φ²⟩ = 2t/T₂ growing linearly in time, so averaging gives ⟨e^(iφ)⟩ = e^{(−⟨φ²⟩/2)} and ρ_↑↓(t) = ρ_↑↓(0)e^(−t/T₂). Thus aₓ and a_y shrink by the factor e^(−t/T₂) while a_z is untouched; a static spread of fields would instead give a Gaussian decay in t. A state prepared as |→⟩ with a = (1, 0, 0) therefore walks radially inward: at t = T₂ the Bloch length is 0.368 and the x-analyser transmission has fallen from 1.000 to 0.684; at t = 3T₂ they are 0.050 and 0.525. Every z count is unchanged throughout. That is the operational content of the phrase 'a superposition becomes a mixture'. Add the T₁ processes that do flip populations and a_z relaxes too, towards its thermal value, so the state ends at thermal equilibrium on the z axis rather than at the centre of the ball.

You cannot read a off one atom: σₓ, σ_y and σ_z pairwise fail to commute, and a projective measurement along one axis destroys the information about the other two. So split the ensemble in three, measure one Pauli on each sub-ensemble of N copies, and estimate aᵢ = (N₊ − N₋)/N = 2fᵢ − 1. Each is the mean of ±1 outcomes with variance 1 − aᵢ², so u(aᵢ) = √((1 − aᵢ²)/N): about 0.02 at N = 2500, about 0.01 at N = 10⁴. Rebuild ρ = ½(I + a⋅σ) and check |a| ≤ 1, because noise can push a raw estimate outside the ball and return a ρ with a negative eigenvalue — which is why real experiments fit by maximum likelihood. Note what tomography does not deliver: ρ fixes every probability, but I/2 is an equal mix of |↑⟩ and |↓⟩ and equally an equal mix of |→⟩ and |←⟩, and no data can choose between those recipes.