Bloch sphere, density operators and mixed states

A ket α|0⟩ + β|1⟩ holds four real numbers. Normalisation |α|² + |β|² = 1 removes one; the global phase, which no measurement can see, removes another. Spend the two that remain as α = cos(θ/2) and β = e^(iφ) sin(θ/2) and compute the three Pauli expectation values: ⟨σ_z⟩ = |α|² − |β|² = cos θ, ⟨σₓ⟩ = 2 Re(α*β) = sin θ cos φ, ⟨σ_y⟩ = 2 Im(α*β) = sin θ sin φ. Those are the Cartesian components of a unit vector at polar angle θ and azimuth φ — the Bloch vector r. The half-angle matters: |0⟩ sits at θ = 0 and |1⟩ at θ = π, so states orthogonal in C² are antipodal on the sphere, and |+⟩ = (|0⟩ + |1⟩)/√2 lands on the equator at r = (1, 0, 0). A σ_z measurement of |+⟩ gives ±1 with equal probability, but that is a fact about the z axis, not a sign that the state is uncertain: along x it reads +1 every time.

Take a source that emits |0⟩ on half its runs and |1⟩ on the rest. No vector in C² reproduces its statistics: any α|0⟩ + β|1⟩ with |α|² = |β|² = ½ has a definite axis along which it reads +1 with certainty, while this source gives 50/50 along every axis. The description that works is the density operator ρ = Σ pᵢ|ψᵢ⟩⟨ψᵢ|, the preparation-weighted sum of projectors. Averaging ⟨ψᵢ|A|ψᵢ⟩ over the pᵢ gives ⟨A⟩ = Σ pᵢ Tr(|ψᵢ⟩⟨ψᵢ|A) = Tr(ρA), so ρ carries every expectation value. It is Hermitian because each projector is, positive because each pᵢ ≥ 0, and of unit trace because Σ pᵢ = 1. For the 50/50 source, ρ = ½|0⟩⟨0| + ½|1⟩⟨1| = I/2. For |+⟩ prepared on every run, ρ = |+⟩⟨+| = ½[[1, 1], [1, 1]] = (I + σₓ)/2. The difference between the two lives entirely in the off-diagonal entries — the coherences — and ρ² = ρ holds only for the second.

The 2 × 2 Hermitian matrices form a four-dimensional real vector space, and (I, σₓ, σ_y, σ_z) is an orthogonal basis for it under the Hilbert–Schmidt product: Tr(σᵢσⱼ) = 2δᵢⱼ and Tr σᵢ = 0. Expand ρ = (a I + r⋅σ)/2; unit trace forces a = 1, and taking Tr(ρσᵢ) picks out rᵢ = ⟨σᵢ⟩. So ρ = (I + r⋅σ)/2 for a real three-vector r, and positivity decides how long r may be. Because (r⋅σ)² = |r|² I, the operator r⋅σ has eigenvalues ±|r| and ρ has eigenvalues (1 ± |r|)/2; the smaller one is non-negative only for |r| ≤ 1. Every state is a point of the closed unit ball: the surface |r| = 1 is exactly the pure states of the previous section, the centre is I/2. The purity Tr ρ² = ¼ Tr(I + 2 r⋅σ + |r|² I) = (1 + |r|²)/2 rises from ½ at the centre to 1 on the surface. For ρ = diag(0.8, 0.2), r = (0, 0, 0.6), the eigenvalues are 0.8 and 0.2, the purity is 0.68, and the von Neumann entropy −0.8 log₂ 0.8 − 0.2 log₂ 0.2 = 0.72 bits says how much is unknown.

Measuring the Pauli observable n⋅σ along a unit vector n uses the projectors P± = (I ± n⋅σ)/2, and Tr(ρP±) = ½ Tr ρ ± ½ Tr(ρ n⋅σ) = (1 ± r⋅n)/2. The whole predictive content of a qubit is one dot product, so two preparations with the same r are indistinguishable by any measurement, however many copies you are given. That is the cost of the description. The centre I/2 is ½|0⟩⟨0| + ½|1⟩⟨1|, and equally ½|+⟩⟨+| + ½|−⟩⟨−|, and equally a pure state drawn uniformly from the whole sphere. The state r = (0, 0, 0.6) is 0.8|0⟩⟨0| + 0.2|1⟩⟨1| and also an equal mixture of the two pure states at r = (±0.8, 0, 0.6), whose average is the same arrow. Any convex combination of surface points that lands on r is a legitimate ensemble for it. The same ρ even arises with no ensemble at all, as the partial trace of an entangled pair — the improper mixture of Topic 20.4 — and no local experiment tells the cases apart. ρ fixes statistics, not its origin.

A unitary acts as ρ → UρU†, and for U = exp(−iα n⋅σ/2) = cos(α/2) I − i sin(α/2) n⋅σ the Pauli algebra gives U(r⋅σ)U† = (R r)⋅σ with R the SO(3) rotation about n by α. Every gate is a rotation of the arrow. X = i⋅exp(−iπσₓ/2) is a half-turn about x, sending (rₓ, r_y, r_z) to (rₓ, −r_y, −r_z); the Hadamard is a half-turn about (x̂ + ẑ)/√2, sending (rₓ, r_y, r_z) to (r_z, −r_y, rₓ); S and T are quarter- and eighth-turns about z. Rotations preserve |r|, so no gate changes purity: a qubit at r = (0, 0, 0.6) leaves a Hadamard at (0.6, 0, 0) with purity still 0.68. The map U → R is two-to-one, since U and −U conjugate ρ identically, and α = 2π gives U = −I — a full turn returns the arrow but flips the ket's sign, a sign only interference can see, as neutron interferometry confirmed. Shortening the arrow needs non-unitary evolution: pure dephasing multiplies rₓ and r_y by e^(−t/T₂) while leaving r_z, so |+⟩ after one T₂ sits at (0.37, 0, 0) with purity 0.57, sliding into the ball along a path no gate can reverse.

In NumPy the whole construction is four lines: build ρ as np.outer(ψ, psi.conj()) or as a weighted sum of projectors, read r off as [np.trace(ρ @ s).real for s in (sx, sy, sz)], take np.linalg.eigvalsh(ρ) for the eigenvalues and np.trace(ρ @ ρ).real for the purity — then check that |r| ≤ 1 and that the eigenvalues agree with (1 ± |r|)/2 to floating precision. The picture, though, is a two-dimensional accident. A qutrit's ρ has eight real parameters, expanding on the Gell-Mann matrices, and positivity does not carve out a ball: the pure states form a four-dimensional subset of the seven-sphere, and most points of that sphere are not states at all. The Bloch ball is the exact geometry of one qubit and nothing larger — which is why this unit builds every protocol from qubits and tensor products of them, where each factor still has its arrow.