Two-state systems and the Bloch sphere

A spin-½ in a magnetic field genuinely has a two-dimensional state space, and so does the polarisation of a photon in one momentum mode. Those are exact. Everything else is a truncation. A transmon carries a whole ladder of levels and is usable only because the ladder is anharmonic: the 0→1 transition sits near 5 GHz and the 1→2 transition some 200–300 MHz below it, so a drive tuned to one is detuned from the other. That detuning sets a speed limit — a pulse short enough to carry 200 MHz of bandwidth will drive the second transition too, which is why single-qubit gates run in tens of nanoseconds rather than fractions of one. Two more conditions hide in the same place. The splitting must beat the temperature: at 5 GHz and 20 mK, hf/k_BT = 12.0, so the thermal population of |1⟩ is about e⁻¹² = 6 × 10⁻⁶. And nothing in the experiment may push the system out of the chosen pair. Once it leaks, the whole construction below is simply wrong, because there is no third pole to draw.

Write |ψ⟩ = a|0⟩ + b|1⟩ with a and b complex: four real numbers. Normalisation |a|² + |b|² = 1 spends one. The global phase spends another, and it is worth seeing why rather than being told. Every prediction quantum mechanics makes is either a probability |⟨m|ψ⟩|² or an expectation ⟨ψ|A|ψ⟩. Replace |ψ⟩ by e^(iγ)|ψ⟩ and the factor enters once directly and once conjugated, so it cancels exactly — in every basis, for every observable. A quantity no experiment can reach is not part of the state. What does not cancel is the relative phase between a and b, because there the two amplitudes interfere with each other. |+⟩ = (|0⟩ + |1⟩)/√2 and |−⟩ = (|0⟩ − |1⟩)/√2 differ by that sign alone; both give P(0) = ½ in the z basis, yet they are orthogonal and an x measurement tells them apart every time.

Carry the two survivors as angles: |ψ⟩ = cos(θ/2)|0⟩ + e^(iφ) sin(θ/2)|1⟩. Then θ = 0 is |0⟩ at the north pole, θ = π is |1⟩ at the south, and θ = π/2 is the equator, where |+⟩, |−⟩ and the two circular states live. The half-angle carries the whole construction. It makes orthogonal states antipodal, 180° apart on a sphere on which nothing else is 180° apart, and in general |⟨χ|ψ⟩|² = cos²(Θ/2) for a Bloch angle Θ. The same factor of two says a physical rotation by 2π is only a half-turn for the ket: exp(−iπ n⋅σ) = −I, so a spin-½ comes back with its sign reversed while the sphere records nothing at all. That sign is not a bookkeeping ghost. Send one arm of a neutron interferometer through a 2π rotation and recombine it with an unrotated arm, and constructive interference becomes destructive — the 4π periodicity measured in 1975.

A measurement of n⋅σ has eigenvalues ±1 with projectors ½(I ± n⋅σ), so the Born rule collapses to P(±n) = ½(1 ± r⋅n) — one line covering every basis at once. Two readings follow. Along z, r⋅z = cos θ, giving P(0) = cos²(θ/2) and P(1) = sin²(θ/2), with φ gone: the relative phase is invisible in the very basis it is defined against. Along any axis with a transverse component φ returns in full. Take the pure state θ = 60°, φ = 0. Measuring z gives P(0) = cos² 30° = 0.750. Measuring x gives P(+) = ½(1 + sin 60°) = ½(1 + 0.866) = 0.933. Turn φ to 180° and leave θ alone: the z result does not move, while the x result drops to 0.067. Written as P = cos²(Θ/2) with Θ the angle between r and n, this is Malus's law for photons — the bench angle between two polarisers is half the Bloch angle between the states.

Unitary evolution preserves norms and inner products, so it cannot change |r|: every gate is a rotation of the ball about some axis, U = exp(−iα n⋅σ/2) turning r by α about n. X is a half-turn about x and swaps the poles. Z is a half-turn about z, adding π to φ and leaving every z probability untouched. The Hadamard is a half-turn about (x + z)/√2, which exchanges the z and x axes and so converts a phase into a population — the reason it bookends nearly every algorithm. In hardware these are pulses: a resonant drive of Rabi frequency Ω sweeps θ at rate Ω, so with Ω/2π = 25 MHz a π pulse takes 1/(2 × 25 MHz) = 20 ns to carry |0⟩ to |1⟩, and half of it prepares |+⟩. Because the motion is rigid, a point on the surface stays on the surface for ever. Nothing acting on the qubit alone can make it mixed, which tells you before any experiment that decoherence has to involve something outside it.

Prepare |+⟩ a thousand times, or flip a fair coin and prepare |0⟩ or |1⟩. Both give heads and tails in the z basis, and they are not the same state — no ket at all describes the second. The density matrix does: ρ = Σ pᵢ|ψᵢ⟩⟨ψᵢ|, and for one qubit every such object is ρ = ½(I + r⋅σ) with |r| ≤ 1. The superposition has r = (1, 0, 0) on the surface; the coin has r = 0 at the centre, ρ = ½ I. Measure x and they part company completely, 1 against ½. The diagonal of ρ holds the populations and the off-diagonal entries ρ₀₁ = ½(rₓ − i r_y) hold the coherences, which is why dephasing is written as those entries decaying: T₂ drives rₓ and r_y to zero while T₁ relaxes r_z toward its thermal value. Purity Tr ρ² = ½(1 + |r|²) runs from 1 on the surface to ½ at the centre, and measuring ⟨σₓ⟩, ⟨σ_y⟩ and ⟨σ_z⟩ fixes r, and so ρ, outright.