University Physics IV · Photons and Matter Waves · 4.9
Wave–Particle Duality & Complementarity
Duality stops being a slogan the moment you can measure both halves of it. Here you learn to read a fringe pattern as one number, a path record as another, and the quarter-circle that ties them together: the first scraps of which-path knowledge cost almost no contrast, and the last ones cost nearly all of it.
Build the model
Connect the measurement to the mechanism.
Complementarity is usually recited as a slogan — wave or particle, never both — and the slogan is a poor summary of what the formalism says. Take a two-path interferometer and let the path become correlated with the state of anything else: a polarisation, a spin, an atom's internal level, a stray photon. The joint state is then |ψ⟩ = (|1⟩|m₁⟩ + e(iφ)|2⟩|m₂⟩)/√2, and since the pattern at the output depends on the path alone, you trace the marker out.
Tracing leaves the interference term multiplied by ⟨m₁|m₂⟩ and touches nothing else. That one line does all the work: the fringe visibility equals the modulus of the marker overlap, while the best possible guess of the path from that same marker improves exactly as the overlap shrinks, and the two are locked together as D² + V² ≤ 1. So duality is not a switch between two pictures but a continuous trade along one curve, and its currency is correlation, not disturbance.
What it costs you is the comfortable idea that the particle had a path and the interference was a separate wave effect layered on top. However much path information the record holds, exactly that much coherence has left the reduced state — and it comes back only by sorting the data on what the marker is later found to say.
- Simple definition
- Complementarity is the statement that in a two-path interferometer the contrast of the fringes and the which-path information available about those same quanta trade against each other, with distinguishability squared plus visibility squared never exceeding one.
- Example
- Put a marker in the arms whose two states overlap by 0.28: the fringe visibility is exactly 0.28, the distinguishability is √(1 − 0.0784) = 0.96, and the best possible guess of which arm the particle took succeeds 98% of the time.
Turns a picture of fringes into one number a fit can return — and it is the only feature of the pattern the duality bound constrains.
P is the detection probability at one output port; V is dimensionless, 0 ≤ V ≤ 1
The marker overlap multiplies the interference term and nothing else, so V = |⟨m₁|m₂⟩| for a balanced interferometer.
φ is the arm phase difference in rad; χ = arg⟨m₁|m₂⟩; the two ports always sum to 1
Tracing out the marker leaves the path coherence scaled by the overlap: the fringes leave through entanglement, not through recoil.
|1⟩, |2⟩ are the arm states, |m₁⟩, |m₂⟩ the marker states they become correlated with; ρ₁₂ is the complex conjugate of ρ₂₁
Defines which-path knowledge operationally, as the best single-shot success rate, so it can be put into an inequality.
D is dimensionless; the second form holds for pure markers and equal path weights
Reads as a quarter-circle: path knowledge costs contrast quadratically, and only total ignorance of the path buys V = 1.
equality for a pure state read with an optimal marker measurement; strict once the marker is mixed or partly inaccessible
Path knowledge you hold before the run counts too: a 70:30 splitter alone caps visibility at 2√0.21 = 0.917, with no marker anywhere.
p₁, p₂ are the intensity fractions in the two arms, with p₁ + p₂ = 1; D falls back to P when the overlap is 1
One quantum in the apparatus at a time
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.
What a fringe is, and what visibility measures
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.
Marking a path entangles it; it does not kick it
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.
Turning both halves into numbers: D, V and the quarter-circle
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 Pguess = 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.
Erasure: the fringes come back, but only in coincidence
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.
Where the bound comes from, and how far it reaches
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.
Change one variable at a time
Make the relationship visible.
Pull the marker overlap from 1 down to 0: the fringes flatten as the point climbs the arc to D = 1, and it never leaves the arc. Then restore the overlap and unbalance the splitter instead — the contrast falls again, this time paid for by predictability, with no marker present at all.
VISIBILITY V0.80
DISTINGUISHABILITY D0.60
BEST PATH GUESS80.0 %
D squared + V squared1.00
Live interpretationVISIBILITY V: 0.80. DISTINGUISHABILITY D: 0.60. BEST PATH GUESS: 80.0 %. D squared + V squared: 1.00
Catch the common trap
Explain before calculating.
A balanced Mach–Zehnder interferometer carries single photons, and a path marker leaves its two states overlapping by |⟨m₁|m₂⟩| = 0.60. The marker is filed away and never measured. What are the fringe visibility and the which-path distinguishability?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron biprism interferometer runs at 50 kV through a column 1.5 m long and detects about 1000 electrons per second. Show that the fringes it builds up cannot be interference between different electrons.
- Speed first. The kinetic energy is 50 keV against a rest energy of 511 keV, so γ = 1 + 50/511 = 1.0978.
- β = √(1 − 1/γ²) = √(1 − 1/1.2053) = √0.1703 = 0.4127, giving v = 0.4127 × 3.00 × 10⁸ = 1.24 × 10⁸ m s⁻¹.
- At a rate R = 1000 s⁻¹ the mean spacing between successive electrons is v/R = 1.24 × 10⁸ / 1000 = 1.24 × 10⁵ m, or about 124 km.
- The mean number inside the 1.5 m column at any instant is 1.5 / (1.24 × 10⁵) = 1.2 × 10⁻⁵. Roughly one transit in 10⁵ has a second electron anywhere in the instrument, and the fringes appear anyway.
AnswerAbout 1.2 × 10⁻⁵ electrons occupy the column at once, with 124 km of vacuum between neighbours. Each electron interferes only with itself; the pattern is a histogram of single events.
MediumA balanced Mach–Zehnder carries single photons. Arm 1 leaves the polarisation at |H⟩; arm 2 contains a wave plate that rotates it to cos 40°|H⟩ + sin 40°|V⟩. Find the visibility, the distinguishability, the best path-guessing rate, and the largest and smallest detection probabilities at one output port.
- Marker overlap: ⟨m₁|m₂⟩ = ⟨H|(cos 40°|H⟩ + sin 40°|V⟩) = cos 40° = 0.766. The arms are balanced, so V = |⟨m₁|m₂⟩| = 0.766.
- Distinguishability: D = √(1 − 0.766²) = √(1 − 0.5868) = √0.4132 = 0.643, which is just sin 40°.
- Best guess: Pguess = ½(1 + D) = ½(1.643) = 0.821, so an optimal marker measurement names the arm correctly 82.1% of the time.
- Port probabilities: Pₘₐₓ = ½(1 + V) = 0.883 and Pₘᵢₙ = ½(1 − V) = 0.117. Check: (0.883 − 0.117)/(0.883 + 0.117) = 0.766 = V ✓.
- Bound: D² + V² = 0.413 + 0.587 = 1.000. A pure marker state on a balanced interferometer saturates the relation exactly.
AnswerV = 0.766, D = 0.643, best guess 82.1% correct, and P runs between 0.117 and 0.883. D² + V² = 1.000.
HardThe first beam splitter is now unbalanced, sending 70% of the intensity into arm 1 and 30% into arm 2, and a marker of overlap 0.80 sits in the arms. Find the visibility and the distinguishability, check the duality bound, and say which of the two effects costs more contrast.
- Unbalanced two-path interference: V = 2√(p₁p₂) · s = 2√(0.7 × 0.3) × 0.80 = 2(0.4583)(0.80) = 0.733.
- The two costs multiply. Imbalance alone (s = 1) caps V at 2√0.21 = 0.917; the marker alone on a balanced splitter would give V = 0.80. Their product is 0.917 × 0.80 = 0.733 ✓, and the marker is the more expensive of the two.
- Distinguishability combines the prior imbalance with the marker: D = √[(p₁ − p₂)² + 4p₁p₂(1 − s²)] = √(0.16 + 0.84 × 0.36) = √(0.16 + 0.3024) = √0.4624 = 0.680.
- Best guess: Pguess = ½(1 + 0.680) = 0.840, so 84.0% correct — better than a coin even before the marker's contribution, because 70:30 already favours arm 1.
- Bound: D² + V² = 0.4624 + 0.5376 = 1.000. Algebraically (p₁ − p₂)² + 4p₁p₂(1 − s²) + 4p₁p₂s² = (p₁ + p₂)² = 1, so any pure state sits exactly on the circle however you split and mark it.
AnswerV = 0.733, D = 0.680, best guess 84.0%, and D² + V² = 1.000. The marker costs more contrast than the 70:30 imbalance: factors of 0.80 against 0.917.