University Quantum Mechanics II · Composite Systems, Entanglement and Density Operators · 4.5
The spin-half density matrix and the Bloch ball
Topic 4.3 gave you the density operator in general; for one spin-half it becomes an object you can draw and measure. We derive the ball from positivity, read spin, purity and every analyser count off a single arrow, rebuild that arrow from three Stern–Gerlach settings, and find out why no larger system gets a ball.
Build the model
Connect the measurement to the mechanism.
Topic 4.3 made ρ a Hermitian, positive, unit-trace operator with ⟨A⟩ = Tr(ρA); for a spin-half those conditions can be solved completely, and the solution is a picture. The 2 × 2 Hermitian matrices form a four-dimensional real space with I, σₓ, σy, σz as an orthogonal basis under Tr(AB), so unit trace leaves exactly three real numbers: ρ = (I + r⋅σ)/2 with rᵢ = Tr(ρσᵢ). Positivity is the only condition left, and because (r⋅σ)² = |r|² I it becomes a statement about length: the eigenvalues are (1 ± |r|)/2, so |r| ≤ 1 and the states fill a solid ball.
Everything about the ensemble is then geometry. Its mean spin is (ħ/2) r, so |r| is the polarisation a polarimeter reports; its purity is (1 + |r|²)/2, so the pure states are the surface and the unpolarised beam the centre; mixing is linear, so a blend of two ensembles sits on the chord between them; and an analyser along n̂ passes (1 + r⋅n̂)/2, so three non-coplanar settings reconstruct r. All of it is derived here and exact, with no small parameter and no power of α.
What ends it is dimension. For d levels ρ has d² − 1 parameters but a pure state only 2(d − 1), the pure states fill the sphere of unit purity only when d = 2, and already for spin 1 the states form a lopsided body in which the reflection of a pure state through the centre is not a state at all.
- Simple definition
- The Bloch vector of a spin-half ensemble is the real vector r with components rᵢ = Tr(ρσᵢ) = 2⟨Sᵢ⟩/ħ; it fixes ρ = (I + r⋅σ)/2 completely, and positivity confines it to the unit ball |r| ≤ 1.
- Example
- An electron beam polarised 0.86 along +z with no transverse part has r = (0, 0, 0.86): ρ = diag(0.93, 0.07), ⟨Sz⟩ = 0.43ħ = 4.5 × 10⁻³⁵ J s, and purity (1 + 0.86²)/2 = 0.87.
Derived from Tr(σᵢσⱼ) = 2δᵢⱼ and Tr ρ = 1. Three real numbers are the whole state: rz is the population difference, rₓ and ry the coherence.
rᵢ = Tr(ρσᵢ), real and dimensionless; I the 2 × 2 identity; σ the Pauli matrices; ρ itself dimensionless
Derived from (r⋅σ)² = |r|² I. Positivity, the one condition the expansion does not build in, confines r to the closed unit ball.
λ± dimensionless, with the spin states along ±r̂ as eigenvectors; det ρ = (1 − |r|²)/4 gives the same bound
Derived from S = (ħ/2)σ. The ball is the set of mean spins an ensemble can have, of radius ħ/2, and the length of r is the polarisation a polarimeter measures.
⟨S⟩ in J s; ħ = 1.055 × 10⁻³⁴ J s; |r| = (N₊ − N₋)/(N₊ + N₋), the polarisation along r̂
Derived by squaring the expansion. ρ² = ρ only at |r| = 1, so the pure states are exactly the sphere Quantum Mechanics I built from kets.
dimensionless; ½ for the unpolarised centre, 1 on the surface, and no state lies outside that range
Derived as P± = Tr(ρΠ±). Each setting returns one component, r⋅n̂ = P₊ − P₋, so three non-coplanar settings fix r; as n̂ turns, P₊ runs from λ₋ to λ₊.
n̂ the analyser's unit axis; probabilities dimensionless; the projector Π± = (I ± n̂⋅σ)/2 is itself the pure state at ±n̂
Derived from the linearity of rᵢ = Tr(ρσᵢ). The ball is strictly convex, so a blend of two different states never reaches the surface: no such mixture is pure.
p the mixing weight, 0 ≤ p ≤ 1, dimensionless; r₁ and r₂ the Bloch vectors of the two ensembles
Four Hermitian matrices, three real numbers
Quoted from Quantum Mechanics I: the Pauli matrices are Hermitian and traceless with σᵢσⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ, so Tr(σᵢσⱼ) = 2δᵢⱼ. A general Hermitian 2 × 2 matrix [[a, c − id], [c + id, b]] holds four real numbers, and I, σₓ, σy, σz are four mutually orthogonal Hermitian matrices under the product Tr(A†B), hence a basis. Expand ρ = ½(r₀ I + r⋅σ) and take traces: r₀ = Tr ρ = 1 and rᵢ = Tr(ρσᵢ), real because ρ and σᵢ are both Hermitian. So a spin-half density matrix is three real numbers, derived from nothing beyond Topic 4.3's conditions. Reading them off is mechanical: rz = ρ₁₁ − ρ₂₂ is the population difference, while rₓ = 2 Re ρ₁₂ and ry = −2 Im ρ₁₂ carry the coherence. For ρ = [[0.7, 0.2 − 0.1i], [0.2 + 0.1i, 0.3]] that gives r = (0.4, 0.2, 0.4). The expansion belongs to any two-level system; for light the same three numbers, up to a relabelling of axes, are the normalised Stokes parameters, quoted here from optics.
Positivity is a statement about length
Hermiticity and unit trace are now built in; positivity is not, and it is what makes the diagonal of ρ in every basis a list of probabilities. Because the Pauli matrices anticommute, (r⋅σ)² = |r|² I, so r⋅σ has eigenvalues ±|r| with the spin states along ±r̂ as eigenvectors, and ρ shares them with eigenvalues λ± = (1 ± |r|)/2. Hence λ₋ ≥ 0 exactly when |r| ≤ 1: of all of R³ that Hermiticity and trace allow, the states are the closed unit ball. A second route is handy for checking numbers: det ρ = ¼[(1 − rz²) − (rₓ² + ry²)] = (1 − |r|²)/4, and a 2 × 2 matrix of trace 1 has both eigenvalues non-negative exactly when its determinant is. For the matrix above, |r| = 0.6, λ = 0.8 and 0.2, and det ρ = 0.21 − 0.05 = 0.16 = (1 − 0.36)/4. The limits check the geometry: at |r| = 1, λ = 1 and 0 and ρ is a projector; at r = 0, ρ = I/2. The spectral form also says something physical: every spin-half ensemble acts as a fraction λ₊ up and λ₋ down along its own axis r̂.
The arrow is the measured spin
With S = (ħ/2)σ, ⟨S⟩ = Tr(ρS) = (ħ/2) r, derived in one line. Dimensions: ħ/2 carries J s and r is a pure number, so ⟨S⟩ is an angular momentum, as it must be, and the ball is the set of mean spins any ensemble can have, of radius ħ/2. Its length is the polarisation P = (N₊ − N₋)/(N₊ + N₋) along r̂, the number a Mott polarimeter or a Stern–Gerlach magnet reports. No power of α enters: this is exact kinematics of a two-dimensional space, true whatever the Hamiltonian. Measured beams span the ball. Silver atoms leaving Stern and Gerlach's oven sit at the centre. Electrons photo-emitted from bulk GaAs by circularly polarised light are capped at |r| = (3 − 1)/(3 + 1) = 0.5 by a 3 : 1 ratio of transition strengths, quoted here, not derived, from the 1 × ½ Clebsch–Gordan table that Unit 5 builds for the j = 3/2 valence band; real cathodes give about 0.3 to 0.4. Strained GaAs structures, which split the two hole levels so only one transition is driven, deliver measured polarisations near 0.85 to 0.9.
Surface, centre and chord
Squaring the expansion gives Tr ρ² = ¼ Tr[(1 + |r|²) I + 2 r⋅σ] = (1 + |r|²)/2, and ρ² = ρ only at |r| = 1: the pure states are exactly the surface, the Bloch sphere Quantum Mechanics I built from kets, and the unpolarised ensemble I/2 is the centre. Because ρ and r are linearly related, mixing ensembles mixes arrows: ρ = pρ₁ + (1 − p)ρ₂ has r = p r₁ + (1 − p) r₂, the point a fraction p of the way along the chord from r₂ to r₁. For two pure states at angle γ, |r|² = 1 − 2p(1 − p)(1 − cos γ), which returns 1 at γ = 0 and (2p − 1)² for opposite states. An equal blend of spin-up along z and spin-up along an axis 60° away has |r|² = 0.75, so |r| = 0.866, purity 0.875, and eigenvalues 0.933 and 0.067 along the bisector, 30° from each. The ball is strictly convex, so the chord between two distinct points never touches the surface: no mixture of different states is pure, and the pure states are exactly the extreme points. Every other chord through the same r is another recipe for the same ρ, which is Topic 4.4's result seen as geometry.
Three analyser settings fix the state
The projector for +ħ/2 along n̂ is Π₊ = (I + n̂⋅σ)/2, itself the pure state at the surface point n̂, so the probabilities are P₊ = Tr(ρΠ₊) = (1 + r⋅n̂)/2 and P₋ = Tr(ρΠ₋) = (1 − r⋅n̂)/2 (derived). Every Stern–Gerlach statistic is a projection of one arrow, running from λ₋ to λ₊ as n̂ turns, and each setting delivers one component, r⋅n̂ = P₊ − P₋. Three non-coplanar settings give three independent linear equations for r, and x, y, z give each rᵢ directly. Two are never enough: the states (rₓ, ±ry, rz) agree along x and z. Counting sets the precision. Each atom gives ±1 with mean rᵢ and variance 1 − rᵢ², so N atoms per setting fix rᵢ to √((1 − rᵢ²)/N): 2500 atoms at rᵢ = 0.5 give ±0.017. Near the surface that noise can carry the raw estimate out of the ball. For a pure state along z, the z counts return exactly 1 while x and y scatter by 1/√N, so the raw |r| exceeds 1 on essentially every run and must be pulled back before it is a density matrix.
Why the ball belongs to two levels only
Count parameters. For d levels a Hermitian unit-trace ρ has d² − 1 real parameters; write ρ = I/d + T with T traceless, and Tr ρ² = 1/d + Tr T² puts every pure state on one sphere of dimension d² − 2. A pure state, though, is a ket: 2d real numbers less normalisation and phase, 2(d − 1). The counts agree only when d² − 2 = 2d − 2, that is d = 2, so only there do the pure states fill the sphere and the states form a ball; for spin 1 they are a four-dimensional sliver of a seven-sphere. The diagonal slice makes it concrete. For d = 2 the diagonal matrices are the diameter −1 ≤ rz ≤ 1, as a ball requires; for d = 3 they are the triangle of probability triples (p₁, p₂, p₃), and a triangle is no disc. Reflect the pure state diag(1, 0, 0) through the centre I/3 and you get diag(−1/3, 2/3, 2/3), with a negative eigenvalue. Physically, the count says a spin-1 ensemble needs 8 − 3 = 5 alignment numbers beyond ⟨S⟩: the tensor polarisation quoted for polarised deuteron targets.
Change one variable at a time
Make the relationship visible.
Raise P(+y) from 0.70 to 0.85: no single probability is odd, yet the slice shrinks past the tip, λ₋ turns negative and the purity exceeds 1, so no ensemble gives those counts. Then lower P(+x) to 0.75: the tip sits just inside the rim with λ₋ ≈ 0.003.
BLOCH LENGTH |r|0.877
SMALLER EIGENVALUE λ₋0.061
PURITY Tr ρ²0.885
MEAN SPIN |⟨S⟩|0.439 ħ
Live interpretationBLOCH LENGTH |r|: 0.877. SMALLER EIGENVALUE λ₋: 0.061. PURITY Tr ρ²: 0.885. MEAN SPIN |⟨S⟩|: 0.439 ħ
Catch the common trap
Explain before calculating.
Equal intensities of an atomic beam fully polarised along +z and one fully polarised along +x are merged into one beam. What is the largest fraction of the merged beam that any single Stern–Gerlach analyser setting can pass as spin-up, and along which axis?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTomography of a spin-half ensemble returns ρ = [[0.62, 0.15 + 0.20i], [0.15 − 0.20i, 0.38]] in the Sz basis. Find its Bloch vector, check that it is a legitimate state, and give its purity and the size of its mean spin ⟨S⟩.
- Read r off the entries: rz = ρ₁₁ − ρ₂₂ = 0.62 − 0.38 = 0.24, rₓ = 2 Re ρ₁₂ = 0.30, and ry = −2 Im ρ₁₂ = −0.40.
- Length: |r|² = 0.30² + 0.40² + 0.24² = 0.0900 + 0.1600 + 0.0576 = 0.3076, so |r| = 0.555, inside the ball.
- Eigenvalues: λ± = (1 ± 0.555)/2 = 0.777 and 0.223, both positive. Cross-check with the determinant: 0.62 × 0.38 − |0.15 + 0.20i|² = 0.2356 − 0.0625 = 0.1731, and (1 − 0.3076)/4 = 0.1731.
- Purity: Tr ρ² = (1 + 0.3076)/2 = 0.654, between ½ and 1, so this is a partially polarised mixture.
- Mean spin: ⟨S⟩ = (ħ/2) r = ħ(0.150, −0.200, 0.120), of magnitude 0.277ħ = 0.277 × 1.055 × 10⁻³⁴ J s = 2.92 × 10⁻³⁵ J s.
Answerr = (0.30, −0.40, 0.24) with |r| = 0.555: a valid state with eigenvalues 0.777 and 0.223, purity 0.654, and |⟨S⟩| = 0.277ħ = 2.92 × 10⁻³⁵ J s.
MediumA source emits spin-half atoms prepared spin-up along +z with probability 0.70, and spin-up along an axis 120° from +z, tilted towards +x, with probability 0.30. Find the Bloch vector and purity, the analyser axis that passes the largest fraction and that fraction, and the fraction passed by an analyser along +x.
- Pure states sit on the surface: r₁ = (0, 0, 1) and r₂ = (sin 120°, 0, cos 120°) = (0.866, 0, −0.500).
- Mixing is linear: r = 0.70 r₁ + 0.30 r₂ = (0.260, 0, 0.700 − 0.150) = (0.260, 0, 0.550), on the chord between the two surface points.
- |r|² = 0.0675 + 0.3025 = 0.370, so |r| = 0.608. The chord formula agrees: 1 − 2(0.70)(0.30)(1 − cos 120°) = 1 − 0.42 × 1.5 = 0.370. Purity is (1 + 0.370)/2 = 0.685.
- The best analyser lies along r̂, at arctan(0.260/0.550) = 25.3° from +z towards +x, and passes λ₊ = (1 + 0.608)/2 = 0.804; the worst axis, −r̂, passes λ₋ = 0.196.
- Along +x: P₊ = (1 + rₓ)/2 = (1 + 0.260)/2 = 0.630.
Answerr = (0.260, 0, 0.550), |r| = 0.608, purity 0.685. The best axis is 25.3° from +z towards +x and passes 0.804; an analyser along +x passes 0.630.
HardA spin-half atomic beam is analysed with 2000 atoms at each of three Stern–Gerlach settings, giving spin-up counts of 1830 along +x, 1018 along +y and 1590 along +z. Reconstruct r, show that the raw estimate is not a density matrix, decide from the counting statistics whether the data are consistent with a pure state, and give the nearest physical Bloch vector with its angles.
- Each setting gives one component, rᵢ = (N₊ − N₋)/N = (2N₊ − N)/N: rₓ = (3660 − 2000)/2000 = 0.830, ry = (2036 − 2000)/2000 = 0.018, rz = (3180 − 2000)/2000 = 0.590.
- |r|² = 0.6889 + 0.0003 + 0.3481 = 1.0373, so |r| = 1.0185. Then λ₋ = (1 − 1.0185)/2 = −0.00925 and Tr ρ² = (1 + 1.0373)/2 = 1.019, above 1: the raw estimate is not a density matrix.
- Counting errors: each atom gives ±1 with variance 1 − rᵢ², so σᵢ = √((1 − rᵢ²)/N): σₓ = √(0.3111/2000) = 0.0125, σy = √(0.9997/2000) = 0.0224, σz = √(0.6519/2000) = 0.0181.
- Propagate to the length: σ(|r|) = √[Σ(rᵢσᵢ)²]/|r| = √(1.07 × 10⁻⁴ + 1.6 × 10⁻⁷ + 1.13 × 10⁻⁴)/1.0185 = 0.0149/1.0185 = 0.0146. The excess 0.0185 is 1.3σ, so the counts are consistent with a pure state; a raw estimator near the surface routinely overshoots by this much.
- Nearest physical state: the Hilbert–Schmidt distance between (I + r⋅σ)/2 and (I + r′⋅σ)/2 is |r − r′|/√2, so the nearest point of the ball keeps the direction and sets the length to 1: r = (0.830, 0.018, 0.590)/1.0185 = (0.815, 0.018, 0.579).
- Angles: θ = arccos 0.579 = 54.6° and φ = arctan(0.018/0.830) = 1.2°, which is 0.8σ from zero since σφ ≈ σy/rₓ = 1.5°. As a ket, cos 27.3° |↑⟩ + e(iφ) sin 27.3° |↓⟩ = 0.889|↑⟩ + e(iφ) 0.459|↓⟩.
AnswerRaw |r| = 1.019 ± 0.015 with λ₋ = −0.009, so not a state but only 1.3σ outside the ball and consistent with a pure state. Nearest physical r = (0.815, 0.018, 0.579): θ = 54.6°, φ = 1.2°.