The spin-half as a qubit

The computational basis is a choice, and for a spin-half the natural one is the S_z eigenbasis: |0⟩ ≡ |↑⟩ with S_z = +ħ/2 and |1⟩ ≡ |↓⟩ with S_z = −ħ/2. In that basis Z = σ_z is S_z in units of ħ/2, X = σₓ swaps the two, and the general normalised state is |ψ⟩ = cos(ϑ/2)|0⟩ + e^(iφ) sin(ϑ/2)|1⟩ — the eigenspinor |n̂,+⟩ of topic 10.4, with (ϑ, φ) the polar angles of the Bloch vector a = ⟨σ⟩ = (sin ϑ cos φ, sin ϑ sin φ, cos ϑ). Two real parameters survive normalisation and the unobservable global phase, so a pure qubit state is a point on the unit sphere and nothing more. H|0⟩ = (|0⟩ + |1⟩)/√2 is the point ϑ = π/2, φ = 0 on the +x axis. Everything that follows is about moving that point — and about what the point fails to record.

Put the spin in a field B along n̂ for a time t. The Hamiltonian is H = −γ B n̂⋅S = −(γBħ/2) n̂⋅σ, so U(t) = exp(−iHt/ħ) = exp(+iγBt n̂⋅σ/2). Because (n̂⋅σ)² = I, the even powers of the exponential series collect into a cosine and the odd powers into a sine: exp(−iθ n̂⋅σ/2) = cos(θ/2) I − i sin(θ/2) n̂⋅σ, and by the adjoint action this turns the Bloch vector through θ about n̂. The pulse supplies θ = −γBt, so the sense of the turn is set by the sign of γ — the Larmor precession of topic 10.8 read as a gate. For a proton, γ/2π = 42.58 MHz T⁻¹, so a 1.0 mT field in the rotating frame gives γB = 2.68 × 10⁵ rad s⁻¹: a π/2 pulse lasts 5.87 μs and a π pulse 11.7 μs. The pulse table of an NMR spectrometer and the native gate set of a spin-qubit processor are the same list.

Set θ = π: Rₙ̂(π) = −i n̂⋅σ, so n̂⋅σ = i Rₙ̂(π). With n̂ = x̂, ŷ, ẑ that reads X = i Rₓ(π), Y = i R_y(π), Z = i R_z(π); with n̂ = (x̂ + ẑ)/√2 it reads (X + Z)/√2 = H. Four gates, four half-turns. Why the factor i? det exp(A) = exp(tr A), and n̂⋅σ is traceless, so every Rₙ̂(θ) has determinant +1 — it lives in SU(2). But det X = det H = −1. No pulse can equal X exactly; the closest it comes is −iX, and (−i)² = −1 repairs the determinant. On the qubit alone the i is invisible, since |⟨a|e^(iα)ψ⟩|² = |⟨a|ψ⟩|² for every outcome a. Check the Bloch action instead: X sends (aₓ, a_y, a_z) to (aₓ, −a_y, −a_z), and H sends it to (a_z, −a_y, aₓ), a swap of x and z. Both are 3×3 rotations of determinant +1 — the −1 never reaches the sphere.

A unitary on C² has four real parameters, and the Euler form U = e^(iα) R_z(β) R_y(γ) R_z(δ) supplies them: one global phase and three angles. Any single-qubit gate is therefore at most three pulses about two axes, which is how a compiler maps circuit gates onto a machine's native pulses. For Hadamard, compute R_y(π/2) R_z(π). R_z(π) = −iZ = diag(−i, i), and R_y(π/2) = (I − iY)/√2 has rows (1, −1) and (1, 1) over √2; the product has rows (−i, −i) and (−i, i) over √2, which is −iH. Hence H = i R_y(π/2) R_z(π): a π pulse about z followed by a π/2 pulse about y, with α = π/2, β = 0, γ = π/2, δ = π. Numerically, scipy.linalg.expm(−1j θ/2 (n⋅σ)) agrees with cos(θ/2) I − i sin(θ/2) n⋅σ to 10⁻¹⁵, and U†U = I checks unitarity — the first cell of any qubit notebook.

Set θ = 2π: Rₙ̂(2π) = cos π I = −I for every axis. The Bloch vector has come all the way round — the induced 3×3 rotation has period 2π — so no measurement on the rotated qubit changes; the spinor needs 4π. To see the sign, give it something to be relative to. Split a beam, rotate the spin in arm A only, and recombine on a 50:50 splitter: the state (|A⟩R|s⟩ + |B⟩|s⟩)/√2 leaves the bright port with amplitude (R + I)|s⟩/2 and probability ½[1 + Re⟨s|R|s⟩]. For |s⟩ = |0⟩, ⟨0|Rₙ̂(θ)|0⟩ = cos(θ/2) − i n_z sin(θ/2), so the port fraction is ½[1 + cos(θ/2)] whatever the axis: bright at 0, dark at 2π, bright again at 4π. Neutron interferometers measured this period in 1975 and found roughly 720°, not 360°. The −1 was a global phase only because nothing else was in the experiment.

Count what a single-qubit computation contains. The state is a Bloch vector, three real numbers; a gate is a 3×3 rotation; a circuit of k gates is a product of k such matrices, which is one rotation. A laptop follows it in microseconds, and ħ never enters the cost — the interference and the 4π sign are real physics, not hard arithmetic. The count changes with more qubits. n qubits in a product state |ψ₁⟩ ⊗ … ⊗ |ψₙ⟩ still need only 2n Bloch angles, but a general state on the 2ⁿ-dimensional product space needs 2ⁿ⁺¹ − 2 real numbers: for n = 50 that is about 2.3 × 10¹⁵ against 100. Single-qubit gates never leave the product manifold; an entangling gate such as CNOT = |0⟩⟨0| ⊗ I + |1⟩⟨1| ⊗ X does, exactly as the Stern–Gerlach field of topic 10.5 entangled spin with position. The spin-half makes an excellent qubit. Its power lies in the coupling, which is why the course's closing unit builds its state spaces by tensor product.