University Quantum Mechanics II · Composite Systems, Entanglement and Density Operators · 4.4

Different ensembles, one density operator

This is the lesson where the density operator stops being bookkeeping and becomes the state. We build one beam two ways, prove that no experiment on it, nor any chain of experiments, can say which way, quote the exact condition for two recipes to give one ρ and check it on that beam, and ask what a recipe still means once nothing can test it.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum Mechanics I left one beam unwritten. An oven's silver atoms split 1:1 in a Stern–Gerlach magnet turned to any axis, yet every spinor is spin-up with certainty along some axis, so no spinor describes them. A density operator does, and more than one recipe produces it: an equal mixture of |+z⟩ and |−z⟩ gives ½(|+z⟩⟨+z| + |−z⟩⟨−z|) = I/2 by completeness, and an equal mixture of |+x⟩ and |−x⟩ gives I/2 by completeness in the other basis.

Every outcome probability is Tr(ρΠ), every post-measurement ensemble is ΠρΠ/Tr(ρΠ) and every evolution is UρU†, so the recipe drops out of every experiment and every chain of experiments: the two are one beam. The general condition is quoted: two recipes give one ρ exactly when their vectors √pₖ|ψₖ⟩, the shorter list padded with zeros, are related by a unitary matrix, which for the z and x recipes is the Hadamard matrix. So ρ is the state, the recipe is history, and a mixed ρ has infinitely many decompositions into pure states.

The freedom has edges. A pure state has exactly one recipe. And a recipe becomes testable again wherever a trace of it survives outside the one-particle ρ, in a preparer's record, a partner particle, or a source that emits its atoms in a fixed order rather than at random; each of those is a larger system with a different density operator of its own.

Simple definition: Two ensembles of pure states that produce the same density operator make the same predictions for every measurement, so they are one physical state, however differently they were prepared.
Example: A 50:50 mixture of |+z⟩ and |−z⟩ and a 50:50 mixture of |+x⟩ and |−x⟩ both give ρ = I/2, so a Stern–Gerlach analyser tilted 30° from z sends exactly 0.500 of either beam into its up channel.

The unpolarised beam, two ways½|+z⟩⟨+z| + ½|−z⟩⟨−z| = ½|+x⟩⟨+x| + ½|−x⟩⟨−x| = I/2

Spin-up along any axis n then has probability Tr(Πₙ)/2 = ½. No spinor does that, since every spinor is spin-up with certainty along some axis.

Derived here from completeness in each basis. ρ is a dimensionless 2 × 2 matrix with trace 1; I is the identity.

Unitary freedom between two recipesΣₖ pₖ|ψₖ⟩⟨ψₖ| = Σⱼ qⱼ|φⱼ⟩⟨φⱼ| ⟺ √pₖ|ψₖ⟩ = Σⱼ uₖⱼ √qⱼ|φⱼ⟩, with u unitary

Whether two recipes make one beam becomes a question about matrices: exhibit a unitary u linking the weighted lists, or show that none exists.

pₖ, qⱼ dimensionless weights summing to 1; pad the shorter list with zero vectors so u is square. Quoted; the ⟸ half is derived.

The unitary linking the z and x recipes|±z⟩/√2 = (1/√2)(|+x⟩/√2 ± |−x⟩/√2), so u = (1/√2)[1 11 −1]

The quoted criterion checked on the headline case. For any two orthonormal bases taken with equal weights, u is simply the change-of-basis matrix.

Derived here: u is the Hadamard matrix, real and dimensionless, with u†u = I. Rows label the z list, columns the x list.

The ensemble after outcome aρₐ = Πₐ ρ Πₐ / Tr(ρ Πₐ)

The update sees the recipe only through ρ, as evolution UρU† does, so no chain of measurements and waits, however long, splits two recipes that share ρ.

Derived here by Bayes' rule on the members. Πₐ is the projector for outcome a and Tr(ρΠₐ) its probability; all are dimensionless.

Mixing averages the spin axesΣₖ pₖ ½(I + nₖ⋅σ) = ½(I + r⋅σ), with r = Σₖ pₖ nₖ

Two spin-½ recipes share ρ exactly when their weighted axes average to one r, so every chord through r is a two-state recipe, weighted by the lever rule.

nₖ the unit axis along which |ψₖ⟩ is spin-up, σ the Pauli matrices, r dimensionless. Derived from QM I's projector ½(I + n⋅σ).

A pure state has one recipe|χ⟩⟨χ| = Σₖ pₖ|ψₖ⟩⟨ψₖ| with every pₖ ≠ 0 ⟹ |ψₖ⟩ = e^(iδₖ)|χ⟩ for every k

The limiting case: the freedom to re-describe a beam belongs to mixed states alone, and it shrinks to nothing as the beam becomes fully polarised.

Derived from ⟨χ⊥|ρ|χ⊥⟩ = Σₖ pₖ|⟨χ⊥|ψₖ⟩|² = 0 for any |χ⊥⟩ orthogonal to |χ⟩; δₖ is a real phase in rad.

A beam no spinor can describe

Quantum Mechanics I showed that every spinor a|+z⟩ + b|−z⟩ is spin-up with certainty along some unit vector n. An oven's beam of silver atoms has no such axis: it splits 1:1 in a Stern–Gerlach magnet, and turned to any other axis the magnet still splits it 1:1, so no spinor describes it. A density operator does, in more than one way. Emit |+z⟩ and |−z⟩ with probability ½ each and ρ = ½(|+z⟩⟨+z| + |−z⟩⟨−z|) = I/2 by completeness; emit |+x⟩ and |−x⟩ instead and completeness in the x basis gives I/2 again. For an analyser along any n the up probability is Tr(ρΠₙ) = ½Tr(Πₙ) = ½, because a rank-one projector has trace 1. Atom by atom the recipes differ. With n tilted 30° from z towards x, the z recipe's members pass with probabilities cos²15° = 0.933 and cos²75° = 0.067, the x recipe's with cos²30° = 0.750 and cos²60° = 0.250. Both pairs average to 0.500, which is the measured 1:1. Light obeys the same algebra: a polariser turned to any angle passes the same fraction of direct sunlight, half for an ideal one.

No experiment, and no chain of experiments, tells them apart

The last lesson derived P(a) = Tr(ρΠₐ), which settles single measurements. A sceptic can measure, wait and measure again, hoping a later result keeps a trace of the recipe. It cannot. A member prepared in |ψₖ⟩ that gives outcome a is left in Πₐ|ψₖ⟩, normalised, and by Bayes' rule the members that gave a carry weights p(k|a) = pₖ⟨ψₖ|Πₐ|ψₖ⟩/P(a). Summing their projectors, the factors ⟨ψₖ|Πₐ|ψₖ⟩ cancel and the survivors form ρₐ = ΠₐρΠₐ/Tr(ρΠₐ). Waiting a time t sends each |ψₖ⟩ to U|ψₖ⟩, so ρ becomes UρU†. Every step of any protocol, a probability, an update or an evolution, takes ρ in and hands ρ on, so two recipes with one ρ agree on the joint statistics of every sequence of tests. Checks: ρ, Πₐ and every probability are pure numbers, and Tr ρₐ = 1 by cyclicity of the trace and Πₐ² = Πₐ. No coupling constant enters, so there is no order in the fine-structure constant α to count: the result is exact for any spin in any field.

The criterion: a unitary matrix between the lists

Write a recipe as a list of unnormalised vectors |ψ̃ₖ⟩ = √pₖ|ψₖ⟩, so that ρ = Σₖ|ψ̃ₖ⟩⟨ψ̃ₖ|. Two lists give one ρ exactly when |ψ̃ₖ⟩ = Σⱼ uₖⱼ|φ̃ⱼ⟩ for some unitary matrix u, the shorter list padded with zero vectors. The theorem, Schrödinger's in 1936 and rediscovered by Hughston, Jozsa and Wootters in 1993, is quoted: its 'only if' half builds u from ρ's eigenvectors and extends it to a square unitary. The 'if' half is one line: Σₖ|ψ̃ₖ⟩⟨ψ̃ₖ| = Σⱼⱼ′(Σₖ uₖⱼu*ₖⱼ′)|φ̃ⱼ⟩⟨φ̃ⱼ′|, and the bracket is (u†u)ⱼ′ⱼ = δⱼⱼ′. Check it on the headline case. With |φ̃±⟩ = |±x⟩/√2 = (|+z⟩ ± |−z⟩)/2, their sum and difference give |+z⟩/√2 = (|φ̃₊⟩ + |φ̃₋⟩)/√2 and |−z⟩/√2 = (|φ̃₊⟩ − |φ̃₋⟩)/√2, so u = (1/√2)[1 1; 1 −1], which is unitary. Lists of different lengths work too: three states 120° apart on the x–z circle, weight ⅓ each, also give I/2, and a 3 × 3 real orthogonal u links them to the z list plus one zero vector.

Every chord through r is a recipe

For spin ½ the criterion has a picture. Quantum Mechanics I wrote the projector onto spin-up along a unit vector n as ½(I + n⋅σ), and ρ is linear in the projectors, so a recipe with weights pₖ and axes nₖ gives ρ = ½(I + r⋅σ) with r = Σₖpₖnₖ: mixing averages the axes. Because ρ fixes r and r fixes ρ, two recipes share ρ exactly when their weighted axes average to the same r. A two-state recipe is then a chord of the unit sphere through r, and each end's weight is the fraction of the chord lying on the far side of r, the lever rule. Take r = 0.5 along z, so ρ = diag(0.75, 0.25). The vertical chord gives |+z⟩ and |−z⟩ with weights 0.75 and 0.25. The horizontal chord gives two states at ±60° from z, weight 0.5 each. The chord through +x gives |+x⟩ with weight 0.375 and a state at −36.9° with weight 0.625. At r = 0 every diameter works with weights ½, which is the unpolarised beam's whole two-state freedom. The next lesson fills in the rest of this ball.

The pure limit, and what a recipe is worth

Push r to 1 and the freedom disappears. Suppose |χ⟩⟨χ| = Σₖpₖ|ψₖ⟩⟨ψₖ| with every pₖ ≠ 0, and take any |χ⊥⟩ orthogonal to |χ⟩. Sandwiching both sides gives 0 on the left and Σₖpₖ|⟨χ⊥|ψₖ⟩|² on the right, a sum of non-negative terms, so every ⟨χ⊥|ψₖ⟩ vanishes and every |ψₖ⟩ is |χ⟩ up to a phase. A pure state has one recipe; only mixed states have many, and in the chord picture the freedom grows from a single point at r = 1 to every diameter at r = 0. The operational conclusion follows. Every step of every protocol passes through ρ alone, so nothing done to the beam tests its recipe: ρ is the state and the recipe is history. A mixed ρ carries no fact about which pure states its members are really in. Even the eigen-decomposition, the one recipe with orthogonal members when its two weights differ, is singled out by convenience rather than by nature: for I/2, every basis is an eigenbasis.

A record makes a larger system, and a singlet keeps it honest

A recipe matters again when its record survives. If the preparer writes down k for every member, spin and notebook form a joint system in Σₖpₖ|k⟩⟨k| ⊗ |ψₖ⟩⟨ψₖ|, and that joint state does differ between recipes: under the z recipe the note predicts every S_z outcome, while under the x recipe note and S_z outcome are uncorrelated. The story became testable because the note became part of what is measured. A source that emits |+z⟩, |−z⟩, |+z⟩, … in strict alternation is the same case, its order a record held in the many-atom state. The singlet shows why the one-atom rule must hold. The state (|+z,−z⟩ − |−z,+z⟩)/√2 equals (|−x,+x⟩ − |+x,−x⟩)/√2, so Alice measuring her spin along z leaves Bob's spin as the z recipe, and measuring along x leaves the x recipe. Were those distinguishable, Bob could read Alice's choice of axis instantly, at any distance. They share ρ = I/2, so he reads nothing; the partial-trace lesson derives his I/2 without mentioning Alice at all.

Change one variable at a time

Make the relationship visible.

Interactive model

Beam polarisation r0.50

Axis of state 1, θ₁ from z90 °

Analyser axis α from z60 °

At r = 0 state 2 sits opposite state 1 and both weights are ½: θ₁ = 0° is the |±z⟩ mix, 90° the |±x⟩ mix. At r = 0.5 sweep θ₁: each member's own P swings widely, yet the weighted P never moves; θ₁ = 90° puts |+x⟩ in at weight 0.375.

WEIGHT p₁ OF STATE 10.375

P(UP ALONG a), STATE 10.933

P(UP ALONG a), STATE 20.440

½(1 + r cos α), FROM ρ0.625

Live interpretationWEIGHT p₁ OF STATE 1: 0.375. P(UP ALONG a), STATE 1: 0.933. P(UP ALONG a), STATE 2: 0.440. ½(1 + r cos α), FROM ρ: 0.625

Catch the common trap

Explain before calculating.

Misconception check

The tempting argument is that measuring S_z unmasks the z recipe: there every atom already has a definite S_z, so its outcomes should come out less random than the x recipe's, or should follow the preparation. They cannot. To anyone without the preparer's list, each S_z outcome in either beam is +ħ/2 or −ħ/2 with probability Tr(ρΠ) = ½, independently from atom to atom, and a second S_z measurement on the same atom repeats the first in both beams, since with Π₊ = |+z⟩⟨+z| the update Π₊ρΠ₊/½ returns |+z⟩⟨+z| whatever the recipe. The only difference is a correlation between outcomes and the list, and the list is not in the beam. A second slip runs the other way and writes the x recipe as the superposition (|+z⟩ + |−z⟩)/√2. That is |+x⟩ itself, a pure state that an x analyser passes every time; the mixture passes it half the time.

A beam is an equal mixture of |ψ₊⟩ = cos(θ/2)|+z⟩ + sin(θ/2)|−z⟩ and |ψ₋⟩ = cos(θ/2)|+z⟩ − sin(θ/2)|−z⟩. For which θ is it the same physical state as a beam that is 80% |+z⟩ and 20% |−z⟩?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA sender flips one fair coin to pick the z or the x basis and a second to pick the sign, then emits |+z⟩, |−z⟩, |+x⟩ or |−x⟩. An eavesdropper hoping to learn the basis measures every spin along an axis n tilted 50° from z towards x. Find the density operator of the z-basis emissions and of the x-basis emissions, and the eavesdropper's probability of spin-up in each case.

z basis: ½|+z⟩⟨+z| + ½|−z⟩⟨−z| = I/2 by completeness. x basis: ½|+x⟩⟨+x| + ½|−x⟩⟨−x| = I/2 by completeness in that basis. The two recipes share one ρ.
Member by member, Quantum Mechanics I gives P(up along n) = cos²(γ/2), with γ the angle between n and the member's spin axis. For the z basis γ = 50° or 130°: cos²25° = 0.821 and cos²65° = 0.179.
For the x basis γ = 40° or 140°: cos²20° = 0.883 and cos²70° = 0.117.
Average over each basis's two members: ½(0.821 + 0.179) = 0.500 and ½(0.883 + 0.117) = 0.500, as Tr(Πₙ)/2 = ½ demands for any n.
Limiting case: turn n to z itself. The z-basis members then give 1 and 0, the x-basis members ½ and ½, and both bases still average to ½. However sharply the members differ, the average never does.

AnswerBoth bases give ρ = I/2 and P(up along n) = 0.500. The members differ, 0.821 and 0.179 against 0.883 and 0.117, but no statistic of the spins alone reveals the basis.

MediumA lab notebook says a spin source emits |+z⟩ with probability 0.60 and |+x⟩ with probability 0.40. Find ρ in the z basis and its eigenvalues and eigenvectors, which form an orthogonal recipe for the same beam. Then find the probability of spin-up along the eigenvector with the larger eigenvalue, once from the notebook's recipe and once from the orthogonal one.

|+z⟩⟨+z| = [1 0; 0 0] and |+x⟩⟨+x| = ½[1 1; 1 1], so ρ = 0.60[1 0; 0 0] + 0.20[1 1; 1 1] = [0.80 0.20; 0.20 0.20], with Tr ρ = 1.00 as required.
Eigenvalues: λ = 0.50 ± √(0.30² + 0.20²) = 0.50 ± 0.361, so λ₁ = 0.861 and λ₂ = 0.139.
For λ₁, (0.80 − 0.861)a + 0.20b = 0 gives b/a = 0.303, so |e₁⟩ = 0.957|+z⟩ + 0.290|−z⟩, spin-up along an axis 2 arctan(0.303) = 33.7° from z towards x. |e₂⟩ is spin-down along that axis, and the orthogonal recipe is 0.861 of |e₁⟩ with 0.139 of |e₂⟩.
From the orthogonal recipe: P(up along e₁) = 0.861 × 1 + 0.139 × 0 = 0.861.
From the notebook: |+z⟩ lies 33.7° from that axis and |+x⟩ lies 56.3° from it, so P = 0.60 cos²16.85° + 0.40 cos²28.15° = 0.60 × 0.916 + 0.40 × 0.777 = 0.550 + 0.311 = 0.861.
Check by averaging axes: the notebook's r has (x, z) components (0.40, 0.60), length √(0.16 + 0.36) = 0.721 at arctan(0.40/0.60) = 33.7° from z, and (1 ± 0.721)/2 returns 0.861 and 0.139.

Answerρ = [0.80 0.20; 0.20 0.20], with eigenvalues 0.861 and 0.139 on an axis 33.7° from z towards x. Both recipes give P = 0.861 along that axis: 0.60 and 0.40 of non-orthogonal states, and 0.861 and 0.139 of orthogonal ones, describe one beam.

HardA beam has ρ = 0.75|+z⟩⟨+z| + 0.25|−z⟩⟨−z|. A colleague claims it is really a mixture of |+x⟩ and one other pure state |χ⟩. Find the weight q of |+x⟩ and the state |χ⟩ that make the claim consistent, then find the unitary u linking this recipe to the recipe of 0.75 |+z⟩ and 0.25 |−z⟩, and confirm that u is unitary.

ρ − q|+x⟩⟨+x| must equal (1 − q)|χ⟩⟨χ|, a matrix of rank one, so its determinant vanishes. It is [0.75 − q/2, −q/2; −q/2, 0.25 − q/2], whose determinant is 0.1875 − 0.5q, so q = 0.375.
The remainder is [0.5625 −0.1875; −0.1875 0.0625] = 0.625 × [0.9 −0.3; −0.3 0.1]. So 1 − q = 0.625 and |χ⟩ = (3|+z⟩ − |−z⟩)/√10 = 0.949|+z⟩ − 0.316|−z⟩, spin-up along an axis 36.9° from z towards −x.
Weighted vectors as (|+z⟩, |−z⟩) components: |ψ̃₁⟩ = √0.75|+z⟩ = (0.866, 0) and |ψ̃₂⟩ = √0.25|−z⟩ = (0, 0.500); |φ̃₁⟩ = √0.625|χ⟩ = (0.750, −0.250) and |φ̃₂⟩ = √0.375|+x⟩ = (0.433, 0.433).
Try a rotation: 0.866|φ̃₁⟩ + 0.500|φ̃₂⟩ = (0.650 + 0.217, −0.217 + 0.217) = (0.866, 0) = |ψ̃₁⟩, and −0.500|φ̃₁⟩ + 0.866|φ̃₂⟩ = (−0.375 + 0.375, 0.125 + 0.375) = (0, 0.500) = |ψ̃₂⟩. So u = [0.866 0.500; −0.500 0.866].
Unitarity: each row of u has length √(0.750 + 0.250) = 1 and the rows' dot product is −0.433 + 0.433 = 0, so u†u = I. u is a rotation by 30°, and the quoted criterion confirms that the two recipes are one beam.
Checks: the weighted spin axes average to 0.625 × (−0.600, 0.800) + 0.375 × (1, 0) = (0, 0.500) in (x, z) components, the beam's r; and no two-state recipe can give |+x⟩ any other weight, since the chord from +x through r meets the circle again at one point only.

Answerq = 0.375, with |χ⟩ = (3|+z⟩ − |−z⟩)/√10 at weight 0.625, linked to the z recipe by u = [0.866 0.500; −0.500 0.866], a 30° rotation. The claim is consistent and untestable: every recipe for this ρ predicts the same.