The classical bit and the qubit

A classical bit takes one of two values, so its state space is the set {0, 1} and there is nothing between them. The qubit keeps the two labels but changes what they are: |0⟩ = (1, 0)ᵀ and |1⟩ = (0, 1)ᵀ are an orthonormal basis of C², chosen as the eigenvectors of σ_z = diag(1, −1), so that reading the bit means measuring σ_z and translating +1 → 0 and −1 → 1. Every unit vector α|0⟩ + β|1⟩ is now a legitimate state, and almost all of them are neither basis vector. The field has to be C, not R. Over the reals the unit vectors form a circle and (|0⟩ + i|1⟩)/√2 does not exist, yet that vector is an eigenstate of σ_y, an observable as real as σ_z; complex amplitudes are what let one space carry every direction of a spin-½. In NumPy the object is simply ψ = np.array([α, beta], dtype=complex), and the inner product ⟨φ|ψ⟩ is np.vdot(φ, ψ), which conjugates its first argument exactly as the bra demands.

The Born rule gives P(0) = |⟨0|ψ⟩|² = |α|² and P(1) = |β|², so |α|² + |β|² = 1 is the statement that one of the two outcomes must occur. A vector handed to you unnormalised, say (1, 2)ᵀ, is not yet a state: divide by √(1 + 4) = √5 to get P(0) = 0.20 and P(1) = 0.80. The expectation ⟨σ_z⟩ = (+1)P(0) + (−1)P(1) = |α|² − |β|² is then −0.60. None of this is visible in one shot. A single σ_z measurement returns 0 or 1, and the amplitudes only show up as frequencies: in 1000 shots on that state you expect 200 zeros with a binomial spread of √(1000 × 0.2 × 0.8) ≈ 13, so a count of 187 or 213 is unremarkable. Code the readout as p0 = abs(ψ[0])**2 followed by rng.choice([0, 1], size=1000, p=[p0, 1 - p0]) with rng = np.random.default_rng(); the histogram converges on the amplitudes, the individual shots never do.

Two complex amplitudes are four real numbers. Normalisation spends one. The second goes because the vector e^(iγ)|ψ⟩ predicts exactly what |ψ⟩ predicts: for any Hermitian A, ⟨ψ|e^(−iγ) A e^(iγ)|ψ⟩ = ⟨ψ|A|ψ⟩, and every probability is the expectation value of a projector. So −|ψ⟩, i|ψ⟩ and e^(iπ/4)|ψ⟩ are the same state as |ψ⟩; the physical object is the ray, the one-dimensional subspace the vector spans. Two real parameters remain, and the standard way to spend them is to make α real and non-negative: |ψ⟩ = cos(θ/2)|0⟩ + e^(iφ) sin(θ/2)|1⟩ with 0 ≤ θ ≤ π and 0 ≤ φ < 2π. The half-angle is not decoration: θ = 0 gives |0⟩ and θ = π gives |1⟩, and those two are orthogonal in C² even though the next lesson will draw them at opposite poles of a sphere. The phase that survives, φ, sits between the two amplitudes rather than on the whole vector, and that is exactly why it cannot be dropped.

A σ_z readout is blind to φ: P(0) = cos²(θ/2) contains no φ at all, so (|0⟩ + |1⟩)/√2 and (|0⟩ − |1⟩)/√2 both return 0 and 1 half the time each. They are still different states, and the way to show it is to measure a different operator. σₓ has eigenvectors |±⟩ = (|0⟩ ± |1⟩)/√2, and ⟨+|ψ⟩ = (cos(θ/2) + e^(iφ) sin(θ/2))/√2 gives P(+) = (1 + sin θ cos φ)/2. At θ = π/2 the two states above give P(+) = 1 and P(+) = 0: a phase invisible in one basis is a certainty in another. On hardware nobody rebuilds the detector; the Hadamard H = (1/√2)[[1, 1], [1, −1]] maps |+⟩ → |0⟩ and |−⟩ → |1⟩, so apply H, then read σ_z, is the σₓ measurement, and in NumPy pₚₗᵤₛ = abs((H @ ψ)[0])**2. Compare a random classical bit, |0⟩ or |1⟩ with probability ½ each. It matches the superposition on P(0) = ½, but in the new basis it gives P(+) = ½|⟨+|0⟩|² + ½|⟨+|1⟩|² = ½, because it has no phase to carry. The qubit's second parameter is what separates them.

It is tempting to read θ and φ as an infinite amount of stored information. They are not retrievable that way. One measurement of any Hermitian operator on C² has at most two outcomes, so it returns at most one bit, and by the projection postulate the state afterwards is the eigenvector it landed on: the original amplitudes are gone. Learning θ and φ means preparing many identical copies and estimating frequencies. With N copies read in σ_z, the estimate of P(0) has standard deviation √(P(0)(1 − P(0))/N), which for P(0) ≈ ½ and N = 10 000 is 0.005, and the same again in the σₓ and σ_y bases to pin down φ. That is state tomography, and it is expensive because the resource is the number of copies, not the number of parameters. The Holevo bound makes it a theorem: a single qubit sent on its own, with no entangled partner shared in advance, can transmit at most one classical bit, however finely its amplitudes were prepared. Superposition is a resource for interference and for correlation with other qubits, not a way of packing a real number into a two-level system.

C² is exact for a spin-½ magnetic moment with its position frozen: the operator algebra is su(2) and there is no third level to leak into. Every other qubit is a truncation. A transmon is a weakly anharmonic oscillator with ω₀₁/2π ≈ 5 GHz and a next transition ω₁₂ lower by roughly 200 to 300 MHz; a trapped-ion qubit is two hyperfine levels picked out of many; a photonic qubit is one photon in two modes, with the vacuum one loss away. In each case the true state lives in a larger space, and writing α|0⟩ + β|1⟩ asserts that the population outside span(|0⟩, |1⟩) is negligible. A drive pulse of duration τ has spectral width of order 1/τ, so a 20 ns pulse spreads over about 50 MHz, comfortably inside a 250 MHz anharmonicity, which is why it can flip the qubit without lifting it to |2⟩. Shorten the pulse to 2 ns and its 500 MHz bandwidth reaches the next level, and the two-level algebra of this lesson quietly stops describing the device. Leakage is measured, budgeted and corrected precisely because the model is an approximation.