advanced31 min studyLesson 365 in Physics

Density matrices and mixed preparations

Combine preparation probabilities at the operator level and distinguish statistical mixtures from coherent superpositions in a chosen basis.

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What you’ll understand

  • Construct a density operator from a probabilistic preparation recipe.
  • Check positivity, trace, and purity of a finite density matrix.
  • Explain why one density operator can have several ensemble decompositions.

A preparation device may select among pure states using a classical random record that is unavailable to the observer. Predicting outcomes then requires averaging probabilities over that record, rather than adding amplitudes across distinct preparations. The density operator collects these predictions into one positive unit-trace object. Its matrix depends on the chosen basis, but the operator's spectrum, purity, and predictions for specified measurements do not depend on that representation.

Average projectors instead of state vectors

If preparation j occurs with probability wj and produces a normalized ket, average the rank-one projectors weighted by those probabilities. Averaging kets directly is incorrect: independent preparation labels are classical alternatives, so no controlled relative phase links them. The resulting operator is positive because every vector has a nonnegative quadratic form against each projector. Its trace is one because the recipe weights sum to one. These properties characterize finite-dimensional density operators.

For an observable A, expectation is the trace of rho times A. For a projective outcome, replace A by its projector to obtain a Born probability. In infinite dimension, rho must be a positive trace-class unit-trace operator, and unbounded-observable moments need additional finiteness conditions. A matrix that is Hermitian with trace one but has a negative eigenvalue is not a valid density matrix: some possible measurement would receive a negative probability.

ρ=∑jwj∣ψj⟩⟨ψj∣,wj≥0,∑jwj=1,⟨A⟩=tr⁡(ρA)\rho=\sum_jw_j|\psi_j\rangle\langle\psi_j|,\qquad w_j\ge0,\quad\sum_jw_j=1,\qquad \langle A\rangle=\operatorname{tr}(\rho A)
ρ and the weights are dimensionless. A and its expectation carry the observable's units. The ensemble formula uses normalized pure states and classical preparation probabilities; it is not a coherent sum of kets. The trace expectation assumes A is bounded or the required moment exists.↔ Scroll wide equations horizontally.

Coherence is basis dependent, purity is not

Off-diagonal matrix entries encode phase-sensitive predictions relative to the selected basis. A coherent superposition can have nonzero off-diagonal entries even when its diagonal probabilities match those of an incoherent mixture. Diagonalizing any density operator removes those entries in its eigenbasis, so having a diagonal matrix does not by itself mean a preparation lacks coherence relative to every physically relevant basis. State properties and basis-specific resources must be described separately.

Purity is the trace of rho squared. For a finite d-dimensional normalized state it lies between one divided by d and one. Equality to one means a rank-one projector, which describes a pure state ray; smaller purity means a mixed state. In a qubit, the Bloch-vector length provides the same test. Pure states lie on the unit sphere, mixed states inside it, and the maximally mixed state sits at its centre.

ρ=12(I+r⋅σ),∣r∣≤1,tr⁡(ρ2)=1+∣r∣22\rho=\frac12(I+\mathbf r\cdot\boldsymbol\sigma),\qquad |\mathbf r|\le1,\qquad \operatorname{tr}(\rho^2)=\frac{1+|\mathbf r|^2}{2}
For a qubit, σ denotes the three dimensionless Pauli matrices and r is a real dimensionless Bloch vector. Purity equals one only on the sphere. This is a density-operator description, unlike a lab restricted to pure-state surface points.↔ Scroll wide equations horizontally.

The recipe is not uniquely recoverable from rho

A maximally mixed qubit can be prepared by selecting the two computational states equally, or by selecting the plus and minus X eigenstates equally. Both recipes produce the same operator and therefore identical statistics for every measurement on the qubit alone. Their discarded classical labels are different records, but those records are not recoverable from rho without additional access to a preparation register. This is ensemble nonuniqueness rather than an uncertainty about which matrix entries to use.

The same mathematical density operator can also describe a subsystem entangled with another system. Locally, its measurement predictions match those of an appropriate classical preparation mixture, even though the global correlations differ. Calling every mixed state merely ignorance of a unique pre-existing pure state therefore adds information that the operator does not contain. The next lessons introduce the tensor-product description needed to distinguish a local state from its possible global extensions.

A nonorthogonal preparation mixture

A device prepares |0⟩ with probability 3/5 and ∣+⟩=(∣0⟩+∣1⟩)/2|+\rangle=(|0\rangle+|1\rangle)/\sqrt2 with probability 2/5, discarding the label. Find ρ, the zero and plus outcome probabilities, and its purity.

  1. Use projectors: ∣0⟩⟨0∣=(1000)|0\rangle\langle0|=\begin{pmatrix}1&0\\0&0\end{pmatrix} and ∣+⟩⟨+∣=12(1111)|+\rangle\langle+|=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}. The weighted sum is ρ=(4/51/51/51/5)\rho=\begin{pmatrix}4/5&1/5\\1/5&1/5\end{pmatrix}, with unit trace and positive determinant 3/25.

  2. The computational zero probability is ρ00=4/5. Projecting onto plus gives p(+)=12(ρ00+ρ01+ρ10+ρ11)=7/10p(+)=\tfrac12(\rho_{00}+\rho_{01}+\rho_{10}+\rho_{11})=7/10. This differs from using the diagonal entries alone because the X measurement is sensitive to off-diagonal coherence.

  3. Square the matrix and take its trace: (4/5)2+(1/5)2+2(1/5)2=19/25=0.76(4/5)^2+(1/5)^2+2(1/5)^2=19/25=0.76. Its Bloch vector is (2/5,0,3/5), whose squared length 13/25 gives the same purity.

The result

The mixture predicts p(0)=0.80 and p(+)=0.70, with purity 0.76. It is mixed despite retaining off-diagonal entries in the computational basis.

Explore the model

Interactive lab

Interactive lab · quantum

Two pure components do not display their mixture

Reset and set both controls to θ=90° and φ=0°; predict Bloch (1,0,0). For a fresh comparison, Reset and set both controls to θ=90° and φ=180°; predict Bloch (−1,0,0). Both runs have equal computational probabilities. An equally weighted preparation mixture has average Bloch vector zero, which must be calculated separately and is not displayed by either pure-state run.

Two pure components do not display their mixtureA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe instrument plots pure two-level geometry only. It has no mixture-weight control, interior mixed-state point, random preparation sampling, or purity readout.

Check your understanding

Which statement about a maximally mixed qubit is correct?

Try it yourself

Prepare |+⟩ and |−⟩ with equal classical probabilities, where ∣±⟩=(∣0⟩±∣1⟩)/2|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2. Find the density matrix, its purity, and probabilities for computational zero and X-plus.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Add the two projectors with weights one half. Their off-diagonal entries have opposite signs.

Compare with the solution

The weighted projectors sum to ρ=I/2\rho=I/2. Hence tr⁡(ρ2)=tr⁡(I/4)=1/2\operatorname{tr}(\rho^2)=\operatorname{tr}(I/4)=1/2. Both p(0) and p(+) equal 1/2. This is different from adding the kets coherently, whose normalized sum would be |0⟩ and would give p(0)=1. The mixed recipe is operationally identical on the qubit to an equal mixture of |0⟩ and |1⟩, although its preparation labels describe different pure components.

Take it with you

  • Classical preparation probabilities weight projectors, rather than coherent amplitudes.
  • A valid density operator is positive and unit trace; purity detects a rank-one state.
  • Ensemble decompositions are nonunique even though all specified local predictions are fixed.

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