University Physics V · The Quantum Wavefunction · 3.8
Collapse & Superposition vs Mixture
Born's rule hands you the odds; this hands you the state you are left holding. Apply the projector update, watch a repeated measurement confirm it, then learn to read a density matrix — because once the diagonal is fixed, only the off-diagonal entries still know whether you have a superposition or a mixture.
Build the model
Connect the measurement to the mechanism.
Born's rule is half a postulate: it gives p(a) = ⟨ψ|Pₐ|ψ⟩ and stops, saying nothing about what you hold once the pointer has moved. The Lüders rule supplies the rest — take the projector belonging to the eigenvalue you got, apply it, renormalise, ρ → Pₐ ρ Pₐ / Tr(Pₐ ρ) — and everything characteristic of a measurement follows from Pₐ² = Pₐ: an immediate repeat returns the same value with certainty, and every observable commuting with A is left undisturbed. The cost is a second dynamical law, discontinuous and stochastic beside a unitary evolution that is neither, with no time of its own.
So be exact about what changed. Leave the outcome unread and the update becomes ρ → Σₐ Pₐ ρ Pₐ, an ordinary linear channel that touches no diagonal entry and erases every coherence between different eigenvalues. That is also, to the last matrix element, what an environment does: the reduced state carries ρ₀₁⟨E₁|E₀⟩ in place of ρ₀₁, and it is that overlap dying, not any collapse, that removes interference.
The gap between a superposition and a mixture therefore lives entirely off the diagonal, is measurable — ⟨σₓ⟩ = 1 against 0 for the two half-and-half qubit states — and is what decoherence destroys. What decoherence never does is select: the joint state stays pure throughout.
- Simple definition
- Collapse, in its Lüders form, is the postulate that a measurement of A returning eigenvalue a replaces the state by Pₐ|ψ⟩ divided by its norm, where Pₐ projects onto the whole eigenspace belonging to a.
- Example
- Measure σz on |+⟩ = (|0⟩ + |1⟩)/√2: each sign has probability ½, and the state left behind is |0⟩ or |1⟩, so an immediate repeat is certain — while ⟨σₓ⟩, equal to 1 beforehand, is now 0.
The denominator is √p(a), so the update is normalised and closed — the state is ready to be measured again.
Pₐ projects onto the entire eigenspace of A for eigenvalue a; Pₐ² = Pₐ = Pₐ†, dimensionless.
The only form that survives when the preparation is already a mixture, which every real preparation is.
ρ is the density operator, Tr ρ = 1, dimensionless. Valid for mixed input and for degenerate a.
Follows from Pₐ² = Pₐ alone. Repeatability is a theorem about projectors, not a separate postulate.
Immediate repeat: no evolution between the two measurements, so no time enters the statement.
Deletes every coherence between different eigenvalues while keeping those inside one degenerate eigenspace.
Sum over the distinct eigenvalues of A with the outcome unread; a linear, trace-preserving channel.
For (|0⟩ + |1⟩)/√2 and the half-and-half mixture: ⟨σz⟩ = 0 for both, ⟨σₓ⟩ = 1 against 0.
The two states share the diagonal; only ρ₀₁ differs, and ⟨σₓ⟩ = 2 Re ρ₀₁ is what reads it.
Orthogonal pointer states send ρ₀₁ to 0 and Tr ρ² to ½, while the system-plus-environment state stays pure.
a is the Bloch vector, |a| ≤ 1; |E₀⟩ and |E₁⟩ are the environment states the two branches drive.
Born's rule gives the odds and then stops
Born's rule answers one question: the probability of outcome a is p(a) = ⟨ψ|Pₐ|ψ⟩, or Tr(Pₐ ρ) in general. It says nothing about the state afterwards, and the Schrödinger equation cannot supply that either, because unitary evolution never turns a superposition into one of its terms. The Lüders rule is the extra postulate: apply the projector belonging to the eigenvalue you actually got and renormalise, |ψ⟩ → Pₐ|ψ⟩/√⟨ψ|Pₐ|ψ⟩, or ρ → Pₐ ρ Pₐ / Tr(Pₐ ρ). Two details deserve naming. Pₐ projects onto the eigenspace, so for a non-degenerate eigenvalue it is |a⟩⟨a| and the result is |a⟩ whatever you started from, while for a degenerate one it is Σₖ |a, k⟩⟨a, k| and the result keeps whatever structure |ψ⟩ had inside that subspace. And the denominator is exactly √p(a), which is what makes the rule closed. Take |ψ⟩ = 0.6|0⟩ + 0.8|1⟩ and measure σz: outcome +1 with probability 0.36, and the state afterwards is 0.6|0⟩ divided by 0.6, which is |0⟩.
Repeatability is a theorem about projectors
Ask what a second, immediate measurement of the same A returns. The state is ρₐ = Pₐ ρ Pₐ / Tr(Pₐ ρ), so the probability of a again is Tr(Pₐ ρₐ), and idempotence gives Pₐ Pₐ ρ Pₐ Pₐ = Pₐ ρ Pₐ, so that probability is 1. Nothing else was assumed: repeatability is Pₐ² = Pₐ wearing physical clothes. Three sharp edges. Immediate means no evolution in between — let H act for a time t with [H, A] ≠ 0 and the certainty decays. Not every measurement is of this kind: a photon absorbed at a detector has no post-measurement state at all, and such processes need the general instrument formalism, not a projector. And a continuous spectrum has no projector onto a point. Position offers only the projector for an interval [a, b], which multiplies ψ by that interval's indicator function; repeating it returns the same interval with probability 1, while the fabled collapse to δ(x − x₀) is not a vector in L², and a state squeezed into a narrow interval carries an energy growing as the inverse square of its width.
Degeneracy: Lüders against von Neumann
Take a three-level space where A has eigenvalue a twofold degenerate on |a₁⟩ and |a₂⟩ and eigenvalue b on |b⟩, in the state (|a₁⟩ + |a₂⟩ + |b⟩)/√3. Outcome a has probability 2/3. Lüders projects with the full Pₐ = |a₁⟩⟨a₁| + |a₂⟩⟨a₂| and leaves the pure state (|a₁⟩ + |a₂⟩)/√2. Von Neumann's original prescription instead summed over a chosen orthonormal basis inside the eigenspace, leaving the mixture ½|a₁⟩⟨a₁| + ½|a₂⟩⟨a₂|. These are different states and an experiment separates them: measure the observable whose eigenvector is (|a₁⟩ + |a₂⟩)/√2 and the Lüders state gives that outcome with probability 1, the von Neumann state with probability ½. Two reasons to prefer Lüders. It is basis-independent inside the eigenspace, where the older rule depends on a choice A itself never made. And it is minimally disturbing: read nothing and it leaves ⟨B⟩ unchanged for every B commuting with A, and it is the only update that does. A degenerate measurement asks a coarse question and should not take more than it asked for.
Superposition and mixture differ only off the diagonal
Write both qubit states as matrices. The superposition |+⟩ = (|0⟩ + |1⟩)/√2 gives ρₛ with all four entries ½. The half-and-half mixture gives ρₘ = ½I: the same diagonal, zeros off it. Every σz prediction is therefore identical, P(+1) = ρ₀₀ = ½ for both, and no quantity of data in that basis will separate them. Rotate the basis and they part at once. P(σₓ = +1) = ⟨+|ρ|+⟩ = ½(ρ₀₀ + ρ₁₁ + 2 Re ρ₀₁), which is 1 for ρₛ and ½ for ρₘ; equivalently ⟨σₓ⟩ = 2 Re ρ₀₁ = 1 against 0. One basis-free number says the same: Tr ρ² is 1 for the superposition and ½ for the mixture. Be careful with the word superposition — it is relative to a basis, since ρₛ is diagonal in the x basis and is an eigenstate of σₓ. What is not relative is purity. A pure state is an eigenstate of some observable, while the maximally mixed state is diagonal in every basis and an eigenstate of none, so no change of basis can carry one into the other.
Decoherence runs the same map and stops short
Leave the outcome unread and the update is non-selective: ρ → Σₐ Pₐ ρ Pₐ, summed over the distinct eigenvalues. This is an ordinary linear, trace-preserving channel. It leaves every diagonal entry alone, deletes every coherence between different eigenvalues, and keeps the coherences inside a degenerate eigenspace. Coupling to an environment performs the same arithmetic with no measurement made at all. Let (|0⟩ + |1⟩)/√2 drive a marker into |E₀⟩ or |E₁⟩: the joint state is (|0⟩|E₀⟩ + |1⟩|E₁⟩)/√2, and tracing out the marker leaves ρ₀₁ = ½⟨E₁|E₀⟩. Fringe visibility is |⟨E₁|E₀⟩|, so at overlap 0.6 the contrast is 0.6 and the purity is ½(1 + 0.36) = 0.68; orthogonal pointer states drive both to the mixture's values. Now what the argument does not deliver. The joint state is a unit vector throughout, so its purity is 1 at every step, and the map is linear and deterministic. Neither feature can produce one outcome out of two.
Which parts of the story an experiment can reach
Collapse is a computational rule with an odd status: discontinuous where the Schrödinger equation is continuous, stochastic where it is deterministic, and carrying no time of its own. Nobody has watched the transition; what is observed is that the statistics after a measurement match the projected state. Interpretations divide here, and it pays to know which differences are empirical. Everett keeps unitary evolution and reads the update as conditioning on a branch; de Broglie–Bohm gets effective collapse once the empty packets stop overlapping; textbook quantum mechanics simply postulates the projection. None of the three predicts a different number, so no experiment separates them. Dynamical-collapse models such as GRW and CSL do differ, adding a stochastic nonlinear term with a rate and a length scale that predicts spontaneous heating and X-ray emission, and experiments keep narrowing the allowed parameters. One check binds them all: tracing Alice's side out of Σₐ (Pₐ ⊗ I) ρ (Pₐ ⊗ I) returns ρB unchanged, which is why no collapse rule can signal.
Change one variable at a time
Make the relationship visible.
Start at γ = 1 with equal populations and drag the marker: the solid curve sweeps 0 to 1 while the dashed mixture lies flat at ½. Now pull γ to 0 — no diagonal entry moves, the θ = 0 prediction never changes, and the fringe flattens onto the mixture. Decoherence deletes ρ₀₁ and selects nothing.
COHERENCE ρ₀₁0.500
VISIBILITY |a|1.000
PURITY Tr ρ²1.000
P(+1) AT θ1.000
Live interpretationCOHERENCE ρ₀₁: 0.500. VISIBILITY |a|: 1.000. PURITY Tr ρ²: 1.000. P(+1) AT θ: 1.000
Catch the common trap
Explain before calculating.
An observable A on a three-dimensional space has eigenvalue a twofold degenerate on |a₁⟩ and |a₂⟩, and eigenvalue b on |b⟩. The state is |ψ⟩ = (|a₁⟩ + |a₂⟩ + |b⟩)/√3, and a measurement of A returns a. What is the state immediately afterwards?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA qubit is prepared in |ψ⟩ = 0.6|0⟩ + 0.8|1⟩. σz is measured and returns +1. Give the probability of that outcome, the state immediately afterwards, the probability that an immediate repeat of σz returns +1 again, and what happened to ⟨σₓ⟩.
- The projector for the outcome +1 is P₊ = |0⟩⟨0|, so p(+1) = ⟨ψ|P₊|ψ⟩ = |⟨0|ψ⟩|² = 0.6² = 0.36.
- Lüders: P₊|ψ⟩ = 0.6|0⟩, whose norm is 0.6, so the state afterwards is 0.6|0⟩/0.6 = |0⟩.
- Repeat immediately: p(+1) = |⟨0|0⟩|² = 1, exactly as P₊² = P₊ requires.
- Before: ρ₀₁ = c₀c₁* = 0.6 × 0.8 = 0.48, so ⟨σₓ⟩ = 2 Re ρ₀₁ = 0.96. After: the state is |0⟩, ρ₀₁ = 0 and ⟨σₓ⟩ = 0. The σz measurement bought certainty in z by spending the coherence that σₓ reads.
Answerp(+1) = 0.36; the state becomes |0⟩; an immediate repeat gives +1 with probability 1; ⟨σₓ⟩ falls from 0.96 to 0.
MediumTwo preparations of a qubit: (i) the pure state |+⟩ = (|0⟩ + |1⟩)/√2, and (ii) the mixture ρₘ that is |0⟩ half the time and |1⟩ half the time. Write each density matrix, give P(σz = +1), P(σₓ = +1) and Tr ρ² for both, and say which measurement separates them and how well.
- ρₛ = |+⟩⟨+| has every entry ½. ρₘ = ½|0⟩⟨0| + ½|1⟩⟨1| = ½I, the same diagonal with zeros off it. The whole difference is ρ₀₁ = ½ against 0.
- σz reads the diagonal only: P(+1) = ρ₀₀ = ½ for both. No number of runs in this basis will separate them.
- σₓ: P(+1) = ⟨+|ρ|+⟩ = ½(ρ₀₀ + ρ₁₁ + 2 Re ρ₀₁), giving ½(½ + ½ + 1) = 1 for ρₛ and ½(½ + ½ + 0) = ½ for ρₘ.
- Purity: ρₛ is a projector so Tr ρₛ² = 1; ρₘ² = ¼I so Tr ρₘ² = ½. Purity is basis-free and already separates them.
- How well, in one shot: ρₛ − ρₘ has entries 0 and ½ off-diagonal, eigenvalues ±½, so the trace distance is ½(|½| + |−½|) = ½ and the best possible single-run success rate is ½(1 + ½) = 0.75.
Answerρₛ = ½[[1,1],[1,1]], ρₘ = ½[[1,0],[0,1]]. P(σz = +1) = ½ for both; P(σₓ = +1) = 1 against ½; Tr ρ² = 1 against ½. One σₓ run identifies the state with probability 0.75.
HardA qubit in (|0⟩ + |1⟩)/√2 couples to a marker, giving the joint state |Ψ⟩ = (|0⟩|E₀⟩ + |1⟩|E₁⟩)/√2 with real overlap ⟨E₁|E₀⟩ = γ. Find the reduced ρS, its purity, P(σₓ = +1) and the fringe visibility as functions of γ; evaluate at γ = 0.6; then compare with a non-selective σz measurement of the qubit alone.
- Trace out the marker: ρS = ½(|0⟩⟨0| + ⟨E₁|E₀⟩|0⟩⟨1| + ⟨E₀|E₁⟩|1⟩⟨0| + |1⟩⟨1|), so ρS = ½[[1, γ],[γ, 1]] and ρ₀₁ = γ/2.
- In Bloch form ρS = ½(I + γσₓ), so |a| = γ and Tr ρS² = ½(1 + γ²).
- P(σₓ = +1) = ⟨+|ρS|+⟩ = ½(½ + γ/2 + γ/2 + ½) = ½(1 + γ). Sweeping the measurement axis in the x–z plane gives extremes ½(1 ± γ), so the visibility is γ: the marker's overlap is the fringe contrast.
- At γ = 0.6: ρS = [[0.5, 0.3],[0.3, 0.5]], Tr ρS² = ½(1 + 0.36) = 0.68, P(σₓ = +1) = 0.80, visibility 0.60.
- |Ψ⟩ is a unit vector for every γ, so the joint purity Tr ρSE² = 1 throughout. Only the marginal is mixed, and it is mixed because information sits in correlations, not because an outcome occurred.
- The non-selective map ρS → P₀ ρS P₀ + P₁ ρS P₁ = ½I gives purity ½, P(σₓ = +1) = ½ and visibility 0 — identical to γ = 0. The same arithmetic, and still no outcome selected.
AnswerρS = ½[[1, γ],[γ, 1]], Tr ρS² = ½(1 + γ²), P(σₓ = +1) = ½(1 + γ), visibility γ. At γ = 0.6: 0.68, 0.80 and 0.60. The joint state stays pure at every γ, and the non-selective σz map reproduces the γ = 0 marginal without selecting anything.