Duality made quantitative on a two-state space

An interferometer's two arms |1⟩ and |2⟩ are an orthonormal basis of C² — a qubit made of alternatives rather than spin. Once everything else is traced out, the most general path state is a 2×2 density matrix, and every Hermitian, unit-trace, positive 2×2 matrix can be written ρ = ½(I + a⋅σ) with a a real vector, |a| ≤ 1. The dictionary is exact: the populations sit on the diagonal as w₁ = ½(1 + a_z) and w₂ = ½(1 − a_z), and the coherence between the arms is ρ₁₂ = ½(aₓ − i a_y). So ρ = [[0.70, 0.30], [0.30, 0.30]] is the vector a = (0.60, 0, 0.40), with eigenvalues ½(1 ± |a|) — positivity is precisely the condition |a| ≤ 1. The Pauli basis is not decoration: it turns every claim about wave-versus-particle behaviour into a claim about the components of one vector, which is what lets the whole topic end in a single Pythagorean statement.

Ask the particle question: which arm? That is a measurement of σ_z, outcomes ±1 with probabilities w₁ and w₂, and the best bet succeeds with probability ½(1 + P), where P = |⟨σ_z⟩| = |w₁ − w₂| = |a_z| is the predictability. Ask the wave question instead: recombine the arms with a phase φ, which measures in the basis (|1⟩ ± e^(iφ)|2⟩)/√2. The click probability is p(φ) = ½[1 + V cos(φ + α)] with V = 2|ρ₁₂| = √(aₓ² + a_y²) and α = arg ρ₁₂ — sinusoidal fringes whose contrast (pₘₐₓ − pₘᵢₙ)/(pₘₐₓ + pₘᵢₙ) is exactly V. For a = (0.60, 0, 0.40): P = 0.40, and the fringes swing between 0.20 and 0.80, so V = 0.60. Nothing metaphysical has happened yet — P is the z-component of a and V its equatorial radius, two perpendicular projections of one object, extracted by two incompatible experiments.

Now add the squares: P² + V² = a_z² + (aₓ² + a_y²) = |a|², and |a|² = 2Tr(ρ²) − 1 ≤ 1 because the eigenvalues ½(1 ± |a|) may not go negative. That is the entire proof of the duality inequality — one line of linear algebra, with equality exactly when the path qubit is pure. For the state above, P² + V² = 0.16 + 0.36 = 0.52 and the purity is Tr(ρ²) = 0.76: the qubit is mixed, and 0.48 of the budget is simply unclaimed. The slack is not a flaw in the bound. A mixed path state records either entanglement with degrees of freedom the trace discarded — a stray photon, a which-path marker — or classical noise in the preparation, such as a vibrating mirror's random phase, so 1 − P² − V² is a meter reading for how much purity has been lost to correlations with the rest of the world or to noise. The interesting physics is in what saturates the bound, and that is where the marker enters.

Couple a marker so the joint state is |Ψ⟩ = √w₁|1⟩|m₁⟩ + √w₂|2⟩|m₂⟩ with |m₁⟩, |m₂⟩ normalised. Tracing the marker out leaves the populations untouched but multiplies the coherence by the overlap: ρ₁₂ = √(w₁w₂)⟨m₂|m₁⟩, so V = 2√(w₁w₂)|⟨m₁|m₂⟩|. Orthogonal marker states kill the fringes outright; identical ones leave them alone; anything between scales them linearly in the overlap. Notice what is absent: nobody has read the marker, and no random phase has been stamped on the arms — a single entangling unitary did all the damage. The quantum-eraser corollary proves the point. Take the overlap ⟨m₁|m₂⟩ real and project the marker onto the normalised (|m₁⟩ ± |m₂⟩) states: each outcome's conditional fringe pattern returns at the unmarked contrast 2√(w₁w₂), the two in antiphase, and weighted by their probabilities ½(1 ± ⟨m₁|m₂⟩) they sum back to the faded V — to no fringes at all when the marker states are orthogonal. A classical kick, once delivered, could not be sorted out again by a later choice of measurement basis; an entanglement can.

How well could the marker answer the path question? Given path i, the marker holds ρᵢᴹ with prior wᵢ, and Helstrom's bound says the optimal measurement guesses the path with probability ½(1 + D), where D = Tr|w₁ρ₁ᴹ − w₂ρ₂ᴹ| — the trace norm of the weighted difference, Englert's distinguishability. For pure marker states the operator w₁|m₁⟩⟨m₁| − w₂|m₂⟩⟨m₂| has trace w₁ − w₂ and determinant −w₁w₂(1 − |c|²), with c = ⟨m₁|m₂⟩, so its two eigenvalues have opposite signs, their moduli sum to D, and D² = (trace)² − 4 det = (w₁ − w₂)² + 4w₁w₂(1 − |c|²) = 1 − V². Hence D² + V² = 1 on every pure joint state — an equality, not a bound — and D ≥ P always, since the marker can only add to the priors. Numbers: w₁ = 0.64 and |c| = 0.50 give V = 0.48 and D = 0.877, a which-path guess that succeeds 94% of the time while the screen still shows 48% fringe contrast.

The older story said the marker's back-action — Einstein's recoiling slit, a momentum kick of order h/d — jitters the relative phase until the fringes wash out. The theorem just proved never mentions momentum. Scully, Englert and Walther made the point sharp with a micromaser marker: an excited atom leaves one photon in whichever cavity it traverses, storing perfect which-path information while transferring momentum far too small to displace a fringe. Dürr, Nonn and Rempe ran the test in 1998 in a rubidium atom interferometer, writing the path label into an internal hyperfine state: fringes vanished as D rose, the measured points obeyed the duality relation, and the phase diffusion available from the microwave coupling was orders of magnitude too weak to explain the loss. Complementarity is enforced by which correlations exist in the joint Hilbert space, not by the clumsiness of the apparatus.