University Physics V · Quantum Spin · 10.6

Superposition, Relative Phase & Mixed States

Two ensembles can give identical up/down counts on a z analyser and still be different physical objects. This is where you learn to separate them: turn the analyser off axis, read the relative phase out of ⟨Sₓ⟩ and ⟨S_y⟩, and let ρ = ½(I + a⋅σ) carry both cases in one object.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A spin-half pure state carries four real numbers in α and β, and physics keeps two: normalisation removes one, and the global phase removes another, since e^(iγ) cancels in every |⟨a|ψ⟩|². What survives is a population difference and one relative phase, and both are measurable — ⟨S_z⟩ = (ħ/2)(|α|² − |β|²) reads the first, ⟨Sₓ⟩ = ħ Re(α*β) and ⟨S_y⟩ = ħ Im(α*β) read the second. Package them as the Bloch vector a = ⟨σ⟩ and the state becomes ρ = ½(I + a⋅σ), whose every prediction is P_±(n̂) = ½(1 ± a⋅n̂).

The pay-off is that ρ also describes what no ket can: a classical mixture, where a is the probability-weighted average of the members' Bloch vectors and so falls strictly inside the unit ball. Purity Tr(ρ²) = (1 + |a|²)/2 then carries the whole distinction — on the surface means coherent, inside means the coin was flipped before you looked. The cost is twofold. ρ is complete for predictions but silent on provenance: I/2 is an equal mix of |↑⟩ and |↓⟩ and equally one of |→⟩ and |←⟩.

And obtaining a at all takes three incompatible measurements on three sub-ensembles, because σₓ, σ_y and σ_z cannot be read off a single copy.

Simple definition: A superposition is one vector α|↑⟩ + β|↓⟩ whose relative phase arg(β/α) is physical, a mixture is a classical probability list over states, and the density operator ρ = ½(I + a⋅σ) is the single object that encodes exactly what either can predict.
Example: (|↑⟩ + |↓⟩)/√2 and a 50/50 mix of |↑⟩ and |↓⟩ both give 50% up along z. Along x the first gives 100% and the second 50%: Bloch vectors (1, 0, 0) and (0, 0, 0), purities 1 and 0.5.

Bloch vector of a spinora = (2Re(α*β), 2Im(α*β), |α|² − |β|²)

⟨S⟩ = (ħ/2)a, so these three numbers are the entire measurable content of the state.

α, β dimensionless amplitudes with |α|² + |β|² = 1; a is dimensionless, and |a| = 1 for any ket.

Relative phase in the transverse spin⟨Sₓ⟩ = ħ Re(α*β) · ⟨S_y⟩ = ħ Im(α*β)

φ is invisible to a z analyser and fully visible to an x or y one — that is how you measure a phase.

Both in J s. With α = cos(θ/2), β = e^(iφ)sin(θ/2): ⟨Sₓ⟩ = (ħ/2)sinθ cosφ, ⟨S_y⟩ = (ħ/2)sinθ sinφ.

Density operator in the Pauli basisρ = ½(I + a⋅σ), with a = Tr(ρσ) = ⟨σ⟩

Three real numbers replace the ket, and mixing preparations becomes ordinary averaging of a.

ρ is Hermitian, Tr ρ = 1, ρ ≥ 0; all entries dimensionless. Holds for pure and mixed states alike.

Born rule for any analyser axisP_±(n̂) = Tr(ρ P_±) = ½(1 ± a⋅n̂)

One line covers superposition and mixture, and the contrast of P₊ against angle is exactly |a|.

n̂ a unit vector, P_± = ½(I ± n̂⋅σ) the projectors; P_± dimensionless and summing to 1.

Purity and the Bloch ballTr(ρ²) = (1 + |a|²)/2 · eigenvalues (1 ± |a|)/2

A single scalar separates a coherent state from a coin that was flipped before you looked.

|a| = 1 on the surface gives Tr(ρ²) = 1; |a| = 0 at the centre gives ½, the floor in two dimensions.

Mixing averages the Bloch vectorsρ = Σ pₖ ρₖ ⟹ a = Σ pₖ aₖ

Explains why every mixture sits strictly inside the ball, and why the recipe cannot be read back out.

pₖ ≥ 0 with Σ pₖ = 1, all dimensionless. Follows from ρ being linear in the preparation probabilities.

Four real numbers go in, two survive

Write |ψ⟩ = α|↑⟩ + β|↓⟩ and count: two complex amplitudes are four real numbers. Normalisation |α|² + |β|² = 1 spends one. The global phase spends another, because replacing |ψ⟩ by e^(iγ)|ψ⟩ leaves every probability |⟨a|ψ⟩|² and every expectation ⟨ψ|A|ψ⟩ untouched — γ cancels between the bra and the ket. Two real parameters survive, and the standard choice is α = cos(θ/2), β = e^(iφ)sin(θ/2) with θ in [0, π] and φ in [0, 2π). So the ratio β/α is physical while α and β separately are not. Concretely, (|↑⟩ + |↓⟩)/√2 and (|↑⟩ + i|↓⟩)/√2 share |α| = |β| = 1/√2 and differ only in φ, 0 against 90° — and they are genuinely different states, eigenstates of Sₓ and of S_y respectively.

Where the phase shows up: compute ⟨Sₓ⟩ and ⟨S_y⟩

Do the matrix arithmetic rather than quoting the result. With |ψ⟩ = (α, β)ᵀ, σₓ|ψ⟩ = (β, α)ᵀ, so ⟨σₓ⟩ = α*β + β*α = 2Re(α*β) and ⟨Sₓ⟩ = ħ Re(α*β). Likewise σ_y|ψ⟩ = (−iβ, iα)ᵀ gives ⟨σ_y⟩ = 2Im(α*β) and ⟨S_y⟩ = ħ Im(α*β), while σ_z is diagonal and gives ⟨S_z⟩ = (ħ/2)(|α|² − |β|²). Substituting the standard parametrisation, ⟨Sₓ⟩ = (ħ/2)sinθ cosφ and ⟨S_y⟩ = (ħ/2)sinθ sinφ. Now fix θ = 90°. At φ = 0 the state is spin-up along x; at φ = 90° it is spin-up along y; at φ = 180° it is spin-down along x — and all three give P(up along z) = 0.500 exactly. A z analyser is blind to φ, while one Sₓ reading and one S_y reading pin it down completely.

The density operator holds what a ket cannot

For a pure state ρ = |ψ⟩⟨ψ| has entries [[|α|², αβ*], [α*β, |β|²]]: the diagonal is the pair of z populations, the off-diagonal is the coherence and carries the phase. For a mixture ρ = Σ pₖ |ψₖ⟩⟨ψₖ|. Both are Hermitian, unit-trace and positive, and since I together with σₓ, σ_y and σ_z spans the Hermitian 2×2 matrices, every such ρ is ½(I + a⋅σ) with a = ⟨σ⟩. Predictions follow from P_±(n̂) = Tr(ρ P_±) = ½(1 ± a⋅n̂), which is linear in ρ — so mixing preparations means averaging their Bloch vectors, nothing more. Compare the two states with the same z data: (|↑⟩ + |↓⟩)/√2 gives [[0.5, 0.5], [0.5, 0.5]] with a = (1, 0, 0), while the coin-flip mixture gives [[0.5, 0], [0, 0.5]] with a = 0. Identical diagonals, different off-diagonals; that gap is the entire physical difference.

Purity, and the difference between the surface and the inside

Purity turns the picture into a number: Tr(ρ²) = (1 + |a|²)/2, and the eigenvalues of ρ are (1 ± |a|)/2. If |a| = 1 the eigenvalues are 1 and 0, ρ has rank one, and the state is pure — and then a is a unit vector n̂, so |ψ⟩ = |n, +⟩. That deserves stating plainly: every pure spinor is spin-up along some axis, so no pure state is unpolarised, however random its z counts look. If |a| = 0 then ρ = I/2, every analyser returns 50/50, and Tr(ρ²) = 0.5, the floor in two dimensions. In between, |a| = 0.4 gives eigenvalues 0.7 and 0.3 and Tr(ρ²) = 0.58. The surface of the ball is coherence, the interior is classical ignorance, and the radius says how much of each you have.

Dephasing walks the Bloch vector inward

Take an environment that couples only to S_z — a fluctuating B_z, say. It never moves the populations, but it randomises the accumulated phase φ. Averaging e^(iφ) over a distribution that broadens with time gives ρ_↑↓(t) = ρ_↑↓(0)e^(−t/T₂), so aₓ and a_y shrink by the factor e^(−t/T₂) while a_z is untouched. A state prepared as |→⟩ with a = (1, 0, 0) therefore walks radially inward: at t = T₂ the Bloch length is 0.368 and the x-analyser transmission has fallen from 1.000 to 0.684; at t = 3T₂ they are 0.050 and 0.525. Every z count is unchanged throughout. That is the operational content of the phrase 'a superposition becomes a mixture'. Add the T₁ processes that do flip populations and a_z decays as well, so the state ends at thermal equilibrium rather than at the centre of the ball.

Tomography: three sub-ensembles and shot noise

You cannot read a off one atom: σₓ, σ_y and σ_z pairwise fail to commute, and a projective measurement along one axis destroys the information about the other two. So split the ensemble in three, measure one Pauli on each sub-ensemble of N copies, and estimate aᵢ = (N₊ − N₋)/N = 2fᵢ − 1. Each is the mean of ±1 outcomes with variance 1 − aᵢ², so u(aᵢ) = √((1 − aᵢ²)/N): about 0.02 at N = 2500, about 0.01 at N = 10⁴. Rebuild ρ = ½(I + a⋅σ) and check |a| ≤ 1, because noise can push a raw estimate outside the ball and return a ρ with a negative eigenvalue — which is why real experiments fit by maximum likelihood. Note what tomography does not deliver: ρ fixes every probability, but I/2 is an equal mix of |↑⟩ and |↓⟩ and equally an equal mix of |→⟩ and |←⟩, and no data can choose between those recipes.

Change one variable at a time

Make the relationship visible.

Interactive model

State polar angle θ90 °

Coherence factor c1.00

Analyser angle Θ from +z90 °

Set θ = 90° and slide c from 1 down to 0: the fringe collapses from full contrast onto the flat dashed 0.5 line, while the reading at Θ = 0° never moves. That is a superposition turning into a mixture with identical z counts throughout.

P+ AT ANALYSER1.000

BLOCH LENGTH |a|1.000

PURITY Tr(ρ²)1.000

P+ ALONG z0.500

Live interpretationP+ AT ANALYSER: 1.000. BLOCH LENGTH |a|: 1.000. PURITY Tr(ρ²): 1.000. P+ ALONG z: 0.500

Catch the common trap

Explain before calculating.

Misconception check

The error is reading α|↑⟩ + β|↓⟩ as "the spin is already up or already down and I merely do not know which". That reading is a 50/50 mixture, and it is not a different interpretation of the same state — it is a different state with different numbers. Both give 50% up on a z analyser, so z data cannot separate them; rotate to x and (|↑⟩ + |↓⟩)/√2 transmits with probability 1 while the mixture transmits 0.5. In ρ the two differ in the off-diagonal entries, 0.5 against 0; on the Bloch ball they sit at (1, 0, 0) and at the centre; their purities are 1 and 0.5. The converse error is just as common: a mixture is not "a superposition nobody has bothered to write down". There is no pair α, β for which |ψ⟩⟨ψ| comes out diagonal with two nonzero entries, because a rank-one projector cannot have rank two.

One ensemble is prepared by flipping a fair coin and sending each atom in |↑⟩ or |↓⟩ accordingly. A second has every atom in (|↑⟩ + i|↓⟩)/√2. Which single Stern-Gerlach setting separates them, and what does it give?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spin-half is prepared in |ψ⟩ = 0.6|↑⟩ + 0.8i|↓⟩. Confirm it is normalised, find its Bloch vector a, show the state is pure, and give the probability that an analyser along +y returns +ħ/2.

Normalisation: |α|² + |β|² = 0.36 + 0.64 = 1, so no rescaling is needed. Here α = 0.6 and β = 0.8i.
The phase sits in the product α*β = 0.6 × 0.8i = 0.48i, so Re(α*β) = 0 and Im(α*β) = 0.48.
Bloch components: aₓ = 2Re(α*β) = 0, a_y = 2Im(α*β) = 0.96, a_z = |α|² − |β|² = 0.36 − 0.64 = −0.28.
|a|² = 0 + 0.9216 + 0.0784 = 1.0000, so |a| = 1 and Tr(ρ²) = (1 + 1)/2 = 1 — pure, as any single ket must be.
Born rule with n̂ = ŷ: P₊ = ½(1 + a⋅ŷ) = ½(1 + 0.96) = 0.98. Along z the same state gives ½(1 + a_z) = ½(0.72) = 0.36.

Answera = (0, 0.96, −0.28) with |a| = 1, so the state is pure; P(S_y = +ħ/2) = 0.98 against P(S_z = +ħ/2) = 0.36.

MediumAn ensemble is 70% in |→⟩ = (|↑⟩ + |↓⟩)/√2 and 30% in |←⟩ = (|↑⟩ − |↓⟩)/√2. Find ρ, its Bloch vector, purity and eigenvalues, and the probabilities along x and z. Then produce a completely different preparation recipe with the identical ρ.

Bloch vectors of the members: |→⟩ has a = (1, 0, 0) and |←⟩ has a = (−1, 0, 0). Mixing is linear in ρ and therefore linear in a.
a = 0.7(1, 0, 0) + 0.3(−1, 0, 0) = (0.4, 0, 0), so |a| = 0.4 — inside the ball, not on its surface.
ρ = ½(I + 0.4σₓ) = [[0.5, 0.2], [0.2, 0.5]]. The trace is 1, and the off-diagonal 0.2 is the coherence that survived the mixing.
Purity Tr(ρ²) = (1 + 0.16)/2 = 0.58, and the eigenvalues are (1 ± 0.4)/2 = 0.7 and 0.3 — which simply hands back the recipe you started with.
Probabilities: P₊(x̂) = ½(1 + 0.4) = 0.7 and P₊(ẑ) = ½(1 + 0) = 0.5. The z analyser cannot see the 0.4 at all.
A different recipe, same ρ: mix 50/50 two pure states tilted ±Θ from x̂ in the x–z plane, with a = (cosΘ, 0, ±sinΘ). Their average is (cosΘ, 0, 0), which equals (0.4, 0, 0) when cosΘ = 0.4, i.e. Θ = 66.4°. No measurement whatever distinguishes the two recipes.

Answera = (0.4, 0, 0), ρ = [[0.5, 0.2], [0.2, 0.5]], Tr(ρ²) = 0.58, eigenvalues 0.7 and 0.3; P₊(x̂) = 0.7 and P₊(ẑ) = 0.5. A 50/50 mix of pure states at ±66.4° to x̂ gives exactly the same ρ.

HardTomography on 2500 copies per axis returns up-fractions fₓ = 0.62, f_y = 0.46, f_z = 0.80. Reconstruct a and ρ, get the purity and eigenvalues, quote the shot-noise uncertainty on each component, and decide whether the data are consistent with a pure state.

Each Pauli needs its own sub-ensemble because σₓ, σ_y and σ_z do not commute. From aᵢ = 2fᵢ − 1: a = (0.24, −0.08, 0.60).
|a|² = 0.0576 + 0.0064 + 0.3600 = 0.4240, so |a| = 0.651 and Tr(ρ²) = (1 + 0.424)/2 = 0.712.
ρ = ½(I + a⋅σ) = [[0.80, 0.12 + 0.04i], [0.12 − 0.04i, 0.20]]. Trace 1; det = 0.16 − 0.016 = 0.144; eigenvalues (1 ± 0.651)/2 = 0.826 and 0.174.
Shot noise on a ±1 observable is u(aᵢ) = √((1 − aᵢ²)/N). At N = 2500 that is 0.019 for x, 0.020 for y and 0.016 for z.
Propagate into |a| = √(Σ aᵢ²) using ∂|a|/∂aᵢ = aᵢ/|a|: u(|a|) = (1/0.651)√(0.24²×0.019² + 0.08²×0.020² + 0.60²×0.016²) = 0.017.
Purity test: 1 − |a| = 0.349, about 21 standard errors, so the state is decisively mixed. Its eigen-decomposition is 0.826|n,+⟩⟨n,+| + 0.174|n,−⟩⟨n,−| with n̂ = a/|a| = (0.369, −0.123, 0.921).

Answera = (0.24, −0.08, 0.60), |a| = 0.651 ± 0.017, Tr(ρ²) = 0.712, eigenvalues 0.826 and 0.174. The state sits about 21σ inside the Bloch sphere, so it is mixed, not pure.