University Physics V · Foundations of Quantum Mechanics · 2.7

Duality Made Quantitative on a Two-State Space

Two path alternatives are a qubit, so "wave or particle" becomes arithmetic on a 2×2 density matrix. This topic replaces the slogan with two perpendicular components of one Bloch vector, proves their Pythagorean tradeoff, then sharpens it into an equality every which-path marker must obey.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Complementarity stops being a slogan the moment you notice that two path alternatives span a two-dimensional Hilbert space. Any preparation of that path degree of freedom — pure, mixed, or entangled with a marker — is a 2×2 density matrix, ρ = ½(I + a⋅σ) in the Pauli basis, and the whole debate lives in the real vector a. Predictability P = |a_z| is the bias of the arm populations; visibility V = √(aₓ² + a_y²) is twice the coherence |ρ₁₂| and equals the contrast of the fringes p(φ) = ½[1 + V cos(φ + α)].

Pythagoras then does the physics: P² + V² = |a|² = 2Tr(ρ²) − 1 ≤ 1, with equality exactly when the path qubit is pure. Couple a marker and the story tightens: the pure joint state √w₁|1⟩|m₁⟩ + √w₂|2⟩|m₂⟩ leaves V = 2√(w₁w₂)|⟨m₁|m₂⟩|, while Englert's distinguishability D — the trace norm Tr|w₁ρ₁ᴹ − w₂ρ₂ᴹ|, the best possible read of the marker — obeys D² + V² = 1 exactly. What the theorem costs is a favourite mechanism: nothing in it mentions momentum kicks or disturbance.

The tradeoff is enforced by the geometry of the joint state space, and it holds whether or not anyone ever looks at the marker.

Simple definition: Wave-particle duality on a two-path system is the statement that predictability P and fringe visibility V are perpendicular components of one Bloch vector, so P² + V² ≤ 1, with a which-path marker's distinguishability D tightening this to D² + V² = 1 on any pure joint state.
Example: A balanced interferometer whose pure marker states overlap by ⟨m₁|m₂⟩ = 0.80 shows fringes of contrast V = 0.80, while the best marker measurement identifies the path with D = √(1 − 0.80²) = 0.60: 0.60² + 0.80² = 1 exactly.

Path qubit in the Pauli basisρ = ½(I + a⋅σ), ρ₁₁ = ½(1 + a_z), ρ₁₂ = ½(aₓ − i a_y)

One 3-vector carries every measurable fact about the path — populations along z, coherence in the equator.

a is the real, dimensionless Bloch vector, |a| ≤ 1; σ = (σx, σy, σz) the Pauli matrices

Predictability and visibilityP = |Tr(ρ σ_z)| = |a_z| · V = 2|ρ₁₂| = √(aₓ² + a_y²)

P and V are perpendicular components of the same vector — that is the entire geometry of duality.

both dimensionless, 0 to 1; P = |w₁ − w₂| is the bias of the arm populations

Fringe lawp(φ) = ½[1 + V cos(φ + α)]

Ties V to the laboratory: the contrast of a measured interferogram is the coherence, twice |ρ₁₂|.

φ the interferometer phase, α = arg ρ₁₂; contrast (pₘₐₓ − pₘᵢₙ)/(pₘₐₓ + pₘᵢₙ) = V

Wave-particle boundP² + V² = |a|² = 2 Tr(ρ²) − 1 ≤ 1

A mixed path qubit is one entangled with something traced out — the slack 1 − P² − V² meters that leak.

equality iff the path qubit is pure; both sides dimensionless

Marking the pathV = 2√(w₁w₂) |⟨m₁|m₂⟩| for |Ψ⟩ = √w₁|1⟩|m₁⟩ + √w₂|2⟩|m₂⟩

Coherence becomes an overlap: fringes fade because the marker could name the path, read or not.

wᵢ path weights, |mᵢ⟩ normalised marker states; V read from ρ after tracing the marker out

Englert's distinguishabilityD = Tr|w₁ρ₁ᴹ − w₂ρ₂ᴹ|, D² + V² ≤ 1, = 1 for pure |Ψ⟩

D ≥ P: the marker only adds to the priors. The equality is the exact exchange rate between fringes and which-path knowledge.

ρᵢᴹ the marker state given path i; the optimal guess of the path succeeds with ½(1 + D)

Two paths span a qubit

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.

P and V are components of one vector

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.

Pythagoras is the inequality

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 experimental sloppiness. A mixed path state is precisely one entangled with degrees of freedom the trace discarded — a stray photon, a vibrating mirror, a which-path marker — so 1 − P² − V² is a meter reading for how much coherence has leaked into correlations with the rest of the world. The interesting physics is in what saturates the bound, and that is where the marker enters.

A marker turns coherence into an overlap

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. Project the marker onto the normalised (|m₁⟩ ± |m₂⟩) states and each outcome's conditional fringe pattern returns at full contrast, the two in antiphase, summing back to none. A classical kick, once delivered, could not be sorted out again by a later choice of measurement basis; an entanglement can.

Distinguishability closes the inequality

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 absolute eigenvalues sum to give D² = (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.

No kicks required — a laboratory verdict

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Marker overlap |⟨m₁|m₂⟩|0.80

Path weight w₁0.50

Drop the overlap from 1 and watch the filled dot ride the arc while the fringes flatten between the dashed lines; then raise w₁ and watch P climb toward D. At overlap 1 the two dots merge — a marker whose states are identical adds nothing to the priors.

VISIBILITY V0.80

PREDICTABILITY P0.00

DISTINGUISHABILITY D0.60

P² + V²0.64

Live interpretationVISIBILITY V: 0.80. PREDICTABILITY P: 0.00. DISTINGUISHABILITY D: 0.60. P² + V²: 0.64

Catch the common trap

Explain before calculating.

Misconception check

Students carry a disturbance story: the marker kicks the particle, jittering the phase until the fringes wash out — and if the marker is never read, the fringes should return. Both halves are wrong. V = 2√(w₁w₂)|⟨m₁|m₂⟩| is fixed by the overlap the coupling unitary leaves between the marker states: the fringes are lost the moment the correlation exists, whether or not any observer reads the marker, and in the micromaser and atom-interferometer versions the momentum transferred is orders of magnitude too small to scramble the phase. What is true is subtler — an unread marker means the fringes are hidden, not gone. Sorting runs by a later marker measurement in the (|m₁⟩ ± |m₂⟩) basis recovers two conditional patterns in antiphase. A random kick could never be sorted back out; an entangling unitary can.

A balanced two-path interferometer carries a pure which-path marker whose conditional states obey ⟨m₁|m₂⟩ = 0.60. What are the fringe visibility V and the distinguishability D?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn the path basis (|1⟩, |2⟩) an interferometer's state is ρ = [[0.70, 0.30], [0.30, 0.30]]. Find P, V and the Bloch vector, test the duality bound, and decide whether the path qubit is pure.

Populations: w₁ = 0.70 and w₂ = 0.30, so the predictability is P = |w₁ − w₂| = 0.40 — a bare which-arm bet succeeds with probability ½(1 + P) = 0.70.
Coherence: ρ₁₂ = 0.30, real, so V = 2|ρ₁₂| = 0.60 with α = 0; the fringes p(φ) = ½[1 + 0.60 cos φ] swing from 0.20 to 0.80.
Bloch vector: a_z = w₁ − w₂ = 0.40 and aₓ − i a_y = 2ρ₁₂ = 0.60, so a = (0.60, 0, 0.40) and |a|² = 0.36 + 0.16 = 0.52.
Bound: P² + V² = 0.16 + 0.36 = 0.52 ≤ 1, and Tr(ρ²) = ½(1 + |a|²) = 0.76 < 1: the qubit is mixed — entangled with something the trace discarded.

AnswerP = 0.40, V = 0.60, a = (0.60, 0, 0.40); P² + V² = 0.52 < 1, so the path qubit is mixed (purity 0.76).

MediumA balanced interferometer (w₁ = w₂ = ½) carries a which-path marker with pure conditional states obeying ⟨m₁|m₂⟩ = 0.80. Find V and D, verify Englert's relation, and give the optimal probability of guessing the path from the marker.

Trace out the marker: ρ₁₂ = √(w₁w₂)⟨m₂|m₁⟩ = ½ × 0.80 = 0.40, so V = 2|ρ₁₂| = 0.80. Balanced arms mean P = |w₁ − w₂| = 0.
Distinguishability of two pure states with equal priors: D = √(1 − |⟨m₁|m₂⟩|²) = √(1 − 0.64) = √0.36 = 0.60.
Englert: D² + V² = 0.36 + 0.64 = 1.00 exactly, because the joint path-marker state is pure.
Helstrom: the best marker measurement identifies the path with probability ½(1 + D) = ½ × 1.60 = 0.80 — while the screen still shows 80% fringe contrast.

AnswerV = 0.80, D = 0.60, D² + V² = 1 exactly; the optimal which-path guess succeeds with probability 0.80.

HardAn unbalanced interferometer has w₁ = 0.64, w₂ = 0.36, and a pure marker with |⟨m₁|m₂⟩| = 0.50. Compute P, V and D, verify D ≥ P and D² + V² = 1, and compare guessing the path from the priors alone with guessing after the optimal marker measurement.

Predictability from the weights alone: P = |0.64 − 0.36| = 0.28.
Visibility: V = 2√(w₁w₂)|c| = 2√(0.64 × 0.36) × 0.50 = 2 × 0.48 × 0.50 = 0.48.
Distinguishability: D² = (w₁ − w₂)² + 4w₁w₂(1 − |c|²) = 0.0784 + 4 × 0.2304 × 0.75 = 0.0784 + 0.6912 = 0.7696, so D = √0.7696 ≈ 0.877.
Checks: D = 0.877 ≥ P = 0.28, and D² + V² = 0.7696 + 0.2304 = 1.0000 exactly — the joint state is pure, so the relation saturates.
Betting: the priors alone succeed with ½(1 + P) = 0.64; the optimal marker measurement succeeds with ½(1 + D) = ½ × 1.877 ≈ 0.94.

AnswerP = 0.28, V = 0.48, D ≈ 0.877; D² + V² = 1 exactly, and the marker lifts the which-path guess from 64% to 94%.