Wave–particle duality and complementarity

The claim that a single quantum interferes with itself has to be earned, because a dim classical wave also makes fringes. Two things earn it. First, the pattern is built one detection at a time. In an electron biprism interferometer at 50 kV the electrons travel at 1.24 × 10⁸ m s⁻¹ and arrive about a thousand per second, so successive electrons sit some 124 km apart along a column 1.5 m long: the mean number in flight together is 1.2 × 10⁻⁵. The first ten counts look like random dust, a few thousand hint at structure, and a hundred thousand are a clean fringe pattern — a histogram of single events, not a wave washing over the screen. Second, for light you must show the arrivals are single quanta at all. Split the beam at a beam splitter and measure the anticorrelation parameter of the two detectors: a classical field of any intensity cannot push it below 1, while a heralded single photon takes it to about 0.2. Only then does 'one particle in the apparatus' mean anything.

A Mach–Zehnder interferometer splits the input at one 50:50 beam splitter, carries the two amplitudes along separate arms, and recombines them at a second. The arms differ by a phase φ, and the detection probabilities at the two output ports are P± = ½(1 ± cos φ) — complementary, always summing to 1, because the quanta have to go somewhere. Phase is cheap to control: at 633 nm a path difference of 158 nm gives φ = 2π(158/633) = 1.568 rad, a hair under π/2, putting one port at P = 0.501, midway between a maximum and a minimum. What you actually record is counts, and the number you extract from them is the visibility V = (Pₘₐₓ − Pₘᵢₙ)/(Pₘₐₓ + Pₘᵢₙ). Sweep φ, find 8900 counts at the peak and 1100 in the trough, and V = 7800/10000 = 0.78. That single number is the entire 'wave' side of the ledger.

Now put a marker in the arms — the cleanest version is polarisation. Leave arm 1 at |H⟩ and rotate arm 2's polarisation by α, so |m₂⟩ = cos α|H⟩ + sin α|V⟩. The state leaving the arms is |ψ⟩ = (|1⟩|m₁⟩ + e^(iφ)|2⟩|m₂⟩)/√2, and the detector at the output port is blind to polarisation, so you trace the marker out. The diagonal terms are untouched; the surviving off-diagonal coherence is ρ₂₁ = ½e^(iφ)⟨m₁|m₂⟩. Everything follows: V = |⟨m₁|m₂⟩| = cos α. At α = 0 the marker records nothing and V = 1; at α = 90° the marker states are orthogonal, the record is perfect, and the fringes are gone entirely. Notice what never appeared in that calculation: momentum. A wave plate rewrites an internal label and leaves the transverse motion alone, transferring no measurable recoil. The fringes died of correlation.

The 'particle' side of the ledger needs a number too, and the honest definition is operational. Hand the marker to someone who must name the arm; the best strategy succeeds with probability P_guess = ½(1 + D), which defines the distinguishability D between 0 (a coin flip) and 1 (certainty). For pure markers and a balanced splitter, D = √(1 − |⟨m₁|m₂⟩|²). Take an overlap of 0.80: V = 0.80, D = √0.36 = 0.60, and the guess succeeds 80% of the time. Then D² + V² = 0.36 + 0.64 = 1 exactly. That is the general result — Englert's relation D² + V² ≤ 1 — and a pure state with an optimal marker measurement sits precisely on the quarter-circle. Two things push a state inside it: a marker that is itself noisy, and information that leaks somewhere you cannot reach. Both cost contrast without giving you the knowledge you paid for.

Set α = 90°, so the markers are |H⟩ and |V⟩ and V = 0. The output port shows a flat 20 000 counts with no fringes at all. Now measure the marker in the diagonal basis |±⟩ = (|H⟩ ± |V⟩)/√2 instead of the H/V basis. Each outcome occurs half the time, and the path state conditioned on it is (|1⟩ ± e^(iφ)|2⟩)/√2 — full visibility, V = 1, in each subensemble, with the two fringe patterns exactly π out of phase. Sort your 20 000 counts by the marker result and you get 10 000 counts of clean fringes plus 10 000 counts of the complementary fringes; add them back together and you recover the flat line you started with. Nothing at the interferometer changed, and nothing has to happen before the particle is detected — the basis may be chosen afterwards. That is also why erasure cannot signal: the unconditioned pattern is flat whatever the distant experimenter chooses, and the fringes live only in the coincidence record.

D² + V² ≤ 1 is not an extra postulate about observers or clumsy instruments. It is a theorem about inner products in the joint state space: unitarity fixes the interference term as an overlap, and the same overlap fixes how well two marker states can be told apart. Nothing in the derivation mentions the size of a detector or the momentum it transfers, which is why it holds identically for photons, electrons, neutrons, atoms and molecules as heavy as C₆₀. It is a separate theorem from the Fourier bound of the previous topic — one counts distinguishable paths, the other bounds the spread of a single packet — though the two agree wherever both apply. In a double slit of separation d, locating the particle's transverse position well enough to name the slit costs a momentum spread of order ħ/d, and that smears the pattern by an angle of the same order as the fringe spacing λ/d, so Bohr's recoiling-slit estimate lands in the right place. But the polarisation marker shows the estimate is not the mechanism: there, no momentum moves at all, and the fringes still go.