University Quantum Mechanics II · Composite Systems, Entanglement and Density Operators · 4.8

Singlet correlations and the CHSH inequality

Until now entanglement has been bookkeeping: tensor products, Schmidt coefficients, partial traces. Here it becomes a prediction a laboratory could refute. You compute one correlation from the Pauli algebra, prove a bound that every local hidden-variable model must obey, and watch the singlet exceed it by a factor of √2.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Two spin-½ particles in the singlet give individually random results — each side sees +1 and −1 equally often along every axis, because each spin's reduced state is I/2 — yet the product of the two results averages to E(a, b) = −a⋅b, which the Pauli algebra delivers in three lines because every component of σ₁ + σ₂ annihilates the singlet. Bell's question is whether that correlation could be an instruction set: a variable λ, fixed at the source, deciding each ±1 outcome from λ and the local setting alone. Any such model obeys a bound that needs no knowledge of λ.

For each λ the combination A(B − B′) + A′(B + B′) equals ±2, since one bracket vanishes and the other is ±2, so its average S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) satisfies |S| ≤ 2. The singlet at a, a′ = 0°, 90° and b, b′ = 45°, 135° gives S = −2√2 ≈ −2.83, and no quantum state does better (Tsirelson's bound, quoted). The argument stops at its assumptions: outcomes fixed by λ and the local setting, and settings chosen independently of λ.

A violation rules out that class of theories. It sends no signal, because Bob's reduced state stays I/2 whatever Alice does, and it leaves untouched theories that give up locality.

Simple definition: The CHSH inequality |S| ≤ 2 bounds a sum S of four spin correlations taken at two analyser settings per side; every local hidden-variable model obeys it, and the singlet state breaks it.
Example: With a, a′ at 0° and 90° and b, b′ at 45° and 135°, the singlet's E = −cos φ gives S = (−0.707) − (+0.707) + (−0.707) + (−0.707) = −2.83, so |S| beats the local bound 2 by 41%.

Singlet correlationE(a, b) = ⟨ψ⁻|(a⋅σ₁)(b⋅σ₂)|ψ⁻⟩ = −a⋅b = −cos φ

Derived from (σ₁ + σ₂)|ψ⁻⟩ = 0 and the Pauli product rule, for any a and b; a = b gives −1 and a ⊥ b gives 0.

a, b unit analyser directions, φ the angle between them; outcomes ±1 in units of ħ/2, so E is dimensionless

Joint outcome probabilitiesP(α, β) = (1 − αβ cos φ)/4, with α, β = ±1

What coincidence counters record: at φ = 45° the signs differ in 85.4% of pairs, while each side alone sees a fair coin.

α is Alice's result and β Bob's; the four P are dimensionless, sum to 1, and each marginal is ½

CHSH inequality for local hidden variables|S| ≤ 2, where S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′)

Derived: A(B − B′) + A′(B + B′) = ±2 for every λ, so its average over any setting-independent ρ(λ) lies between −2 and +2.

E = ∫ρ(λ)A(a, λ)B(b, λ)dλ with A, B = ±1 fixed by λ and the local setting alone; S is dimensionless

Singlet S at equally spaced settingsS(θ) = cos 3θ − 3 cos θ

Derived: stationary at θ = 45°, where S = −2√2 = −2.83; at θ = 0 it touches the local bound, S = −2.

θ the step between successive settings a, b, a′, b′ in one plane; the pair (a, b′) is 3θ apart

Tsirelson's bound (quoted)|S| ≤ 2√2 ≈ 2.83 for any quantum state and any ±1-valued observables

Quantum mechanics beats the local bound by at most √2, so a reported |S| above 2.83 would contradict quantum mechanics itself; in practice it flags an analysis error.

Quoted, not derived here (Tsirelson, 1980); S is dimensionless, and the singlet at θ = 45° attains the bound

No signalling through the singletTr_A[Σ_α (P_α ⊗ I) ρ (P_α ⊗ I)] = Tr_A ρ = I/2

Derived from Σ_α P_α = I and cyclicity inside the trace over A: Bob's counts stay 50:50 whatever Alice sets, or whether she measures at all.

P_α = (I + α a⋅σ)/2 are Alice's projectors, α = ±1; ρ = |ψ⁻⟩⟨ψ⁻|; the result is Bob's reduced density matrix

Three lines of Pauli algebra give −a⋅b

Write |ψ⁻⟩ = (|↑↓⟩ − |↓↑⟩)/√2, measure spin 1 along a and spin 2 along b, and record ±1 in units of ħ/2. The target is E(a, b) = ⟨ψ⁻|(a⋅σ₁)(b⋅σ₂)|ψ⁻⟩. The singlet has total spin zero, so every component of σ₁ + σ₂ annihilates it, and on this state (b⋅σ₂)|ψ⁻⟩ = −(b⋅σ₁)|ψ⁻⟩. Substituting, E = −⟨ψ⁻|(a⋅σ₁)(b⋅σ₁)|ψ⁻⟩. Quantum Mechanics I supplies (a⋅σ)(b⋅σ) = (a⋅b)I + i(a×b)⋅σ, so E = −a⋅b − i(a×b)⋅⟨σ₁⟩, and ⟨σ₁⟩ = Tr(ρ_A σ) = 0 because the partial trace gave ρ_A = I/2. Hence E = −a⋅b = −cos φ, derived for any two directions. Checks: a = b gives −1, the anticorrelation that total spin zero demands; a ⊥ b gives 0. In SI form ⟨(a⋅S₁)(b⋅S₂)⟩ = −(ħ²/4)a⋅b, in J² s² as a product of two angular momenta must be. No power of α enters: this is angular-momentum algebra, identical for electrons, protons and neutrons.

Individually random, jointly correlated

Correlation is only part of the statistics. Since ⟨a⋅σ₁⟩ = ⟨b⋅σ₂⟩ = 0, the four joint probabilities follow from E alone: P(α, β) = (1 + αβE)/4 = (1 − αβ cos φ)/4 for α, β = ±1. Summing over β gives P(α) = ½ at every φ, so each side alone sees a fair coin along every axis — the operational content of ρ_A = ρ_B = I/2. Everything of interest lives in the coincidences. At φ = 60° a pair shows the same sign with probability sin²30° = 0.25 and opposite signs with 0.75, so E = 0.25 − 0.75 = −0.50; at 90° the two sides are uncorrelated; at 180° they always agree. Most experiments use polarisation-entangled photons instead, where turning a polariser through χ moves the state through 2χ on the Bloch sphere (quoted from the photon polarisation qubit). The optimal 45° steps of spin analysers therefore become 22.5° steps of a polariser, and the bounds are unchanged.

What a local hidden-variable model is allowed to do

Bell asked whether −cos φ could be an instruction set carried from the source. Give both particles a variable λ drawn from ρ(λ), and let each outcome be fixed by λ and the local setting alone: A(a, λ) = ±1 and B(b, λ) = ±1. Two assumptions are built in and deserve names. Locality: A does not depend on b, nor B on a. Free settings: ρ(λ) is the same whichever settings are later chosen. Then E(a, b) = ∫ρ(λ)A(a, λ)B(b, λ)dλ. Such models are not straw men. Let λ be a unit vector uniform on the sphere, with A = sign(a⋅λ) and B = −sign(b⋅λ). The two signs differ only when the plane perpendicular to λ separates a from b, which happens with probability φ/π, so E_L = −(1 − 2φ/π), or −1 + φ/90° with φ in degrees. It agrees with the singlet exactly at 0°, 90° and 180°, and it misses in between: at 45° it gives −0.500 where the singlet gives −0.707.

Why every local model obeys |S| ≤ 2

Choose two settings per side, a or a′ for Alice and b or b′ for Bob, and form S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′). In a local model all four outcomes A, A′, B, B′ are defined for every λ, even though one run reveals only two, so S is the ρ(λ)-average of A(B − B′) + A′(B + B′). B and B′ are each ±1: either B = B′, so B − B′ = 0 and B + B′ = ±2, or B = −B′, so B + B′ = 0 and B − B′ = ±2. Exactly one bracket survives, multiplied by A or A′ = ±1, so the combination equals ±2 for every λ, and an average of numbers each equal to ±2 lies between −2 and +2. That is |S| ≤ 2, derived without knowing what λ is or how it is distributed. It survives stochastic local models, where λ fixes only local averages in [−1, 1], because the combination is linear in each and so takes its extremes at ±1. And it is tight: the sign model at equally spaced settings gives S = 3E_L(θ) − E_L(3θ) = −2 for every θ up to 60°.

The singlet reaches 2√2, and noise pulls it back

Put the settings in one plane, θ apart in the order a, b, a′, b′. Three pairs are θ apart and (a, b′) is 3θ apart, so the singlet gives S(θ) = −3 cos θ + cos 3θ. At θ = 0 it sits on the bound at −2; setting dS/dθ = 3 sin θ − 3 sin 3θ to zero gives θ = 45°, the settings 0°, 45°, 90°, 135°, and S = −3(0.7071) + cos 135° = −2.828 = −2√2. That the singlet reaches 2√2 is derived here; that no state and no ±1 observables can exceed it is Tsirelson's bound, quoted. Real sources add noise. Mix in white noise, ρ = V|ψ⁻⟩⟨ψ⁻| + (1 − V)I/4: the I/4 part contributes (1/4)Tr(a⋅σ)Tr(b⋅σ) = 0, so E = −V cos φ and |S| = 2√2 V, a violation only when V exceeds 1/√2 = 0.707. Aspect, Grangier and Roger measured |S| = 2.697 ± 0.015 in 1982, against 2.70 predicted for their apparatus, and the 2015 Delft test found |S| = 2.42 ± 0.20.

No signal crosses, and what a violation excludes

A violation is not a message. Let Alice measure along a and tell no one the result: the pair becomes Σ_α (P_α ⊗ I)ρ(P_α ⊗ I), with P_α = (I + α a⋅σ)/2. The partial trace over Alice's factor is cyclic in her operators, so each term traces to Tr_A[(P_α ⊗ I)ρ], and Σ_α P_α = I returns Tr_A ρ = I/2. Bob's counts are the same whether Alice chose a, a′ or nothing; the correlation appears only when the two lists are compared over an ordinary channel. What a violation excludes is the class above: outcomes fixed by λ and the local setting, with settings independent of λ. A real test must also close loopholes, named here only: locality (settings chosen and results recorded outside each other's light cones), detection (so many pairs lost that a local model hides in the missing ones), and the free choice of settings itself. Delft, Vienna and NIST closed the first two together in 2015. Theories that give up locality, as Bohmian mechanics does, are untouched.

Change one variable at a time

Make the relationship visible.

Interactive model

Setting step θ45 °

Visibility V1.00

At V = 1 drag θ from 0° to 90°: |S| leaves the local bound 2, peaks at 2.83 at θ = 45° and falls to 0. Then hold θ = 45° and lower V: the violation dies between V = 0.75 (|S| = 2.12) and V = 0.70 (|S| = 1.98).

S, NOISY SINGLET-2.828

S, LOCAL SIGN MODEL-2.000

E(θ) = −V cos θ-0.707

MARGIN |S| − 20.828

Live interpretationS, NOISY SINGLET: −2.828. S, LOCAL SIGN MODEL: −2.000. E(θ) = −V cos θ: −0.707. MARGIN |S| − 2: 0.828

Catch the common trap

Explain before calculating.

Misconception check

The tempting conclusion is that perfect anticorrelation is the quantum surprise: measure both spins along one axis, always get opposite results, so the first measurement must have reached across and fixed the second. It proves nothing. A pair of gloves does the same — one left, one right, decided at the source — and the sign model A = sign(a⋅λ), B = −sign(b⋅λ) reproduces E = −1 at φ = 0 and E = 0 at φ = 90° exactly. What no local model can match is the shape of −cos φ in between: at 45° the singlet gives −0.707 where the sign model gives −0.500, and CHSH turns that gap into an inequality, because a local model that matches the ends must pay somewhere else. The evidence against local hidden variables is |S| above 2 at intermediate angles, not the result at φ = 0.

Alice measures along a = 0° or a′ = 90°, Bob along b = 45° or b′ = 135°, all in one plane, on singlet pairs mixed with white noise so that E(φ) = −0.80 cos φ. What do they find for S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′), and what does it establish?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA source emits singlet pairs, and 8000 coincidences are recorded with Alice's and Bob's spin analysers 30° apart in one plane. Find E, the expected numbers of same-sign and opposite-sign coincidences, and compare the same-sign count with the local sign model E = −1 + φ/90°.

The singlet gives ⟨a⋅σ₁⟩ = ⟨b⋅σ₂⟩ = 0 and E = −cos φ, so P(α, β) = (1 + αβE)/4. Adding the two same-sign cases, P(same) = (1 + E)/2; likewise P(opposite) = (1 − E)/2.
E = −cos 30° = −0.8660, so P(same) = (1 − 0.8660)/2 = 0.0670 and P(opposite) = 0.9330.
Expected counts: 8000 × 0.0670 = 536 same-sign, 268 each of ++ and −−, and 8000 × 0.9330 = 7464 opposite-sign. Each side alone still records +1 in half of its 8000 events.
Local sign model: E_L = −1 + 30/90 = −0.667, so P(same) = (1 − 0.667)/2 = 0.167, about 1333 same-sign coincidences — two and a half times the singlet's.
Checks: every quantity is dimensionless; as φ → 0 both models give P(same) → 0, and at φ = 90° both give 4000 of each kind. The two models part company only at intermediate angles.

AnswerE = −0.866; about 536 same-sign coincidences (268 each of ++ and −−) and 7464 opposite-sign, against about 1333 same-sign for the local sign model.

MediumA singlet source is contaminated by white noise, ρ = V|ψ⁻⟩⟨ψ⁻| + (1 − V)I/4 with V = 0.85. Alice uses a = 0° or a′ = 90°, but Bob's mount has slipped, so he uses b = 20° and b′ = 110° instead of 45° and 135°, all in one plane. Find S, decide whether the local bound is violated, and find how far the mount could slip before the violation is lost.

The noise contributes Tr[(I/4)(a⋅σ ⊗ b⋅σ)] = (1/4)Tr(a⋅σ)Tr(b⋅σ) = 0, because each Pauli combination is traceless. So E(φ) = −V cos φ, with φ the angle between the two analysers.
Angles: (a, b) 20°, (a, b′) 110°, (a′, b) 70°, (a′, b′) 20°. With V = 0.85: E(a, b) = −0.799, E(a, b′) = +0.291, E(a′, b) = −0.291, E(a′, b′) = −0.799.
S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) = −0.799 − 0.291 − 0.291 − 0.799 = −2.18. So |S| = 2.18 exceeds 2: still a violation, though short of the 0.85 × 2√2 = 2.40 that the intended settings would give.
In general, with b = 45° − δ and b′ = 135° − δ, the four terms combine to S = −2V(cos β + sin β) with β = 45° − δ, which is −2√2 V cos δ. Check at δ = 25°: −2.828 × 0.85 × 0.906 = −2.18.
A violation needs 2√2 V cos δ to exceed 2, so cos δ must exceed 1/(√2 × 0.85) = 0.832, which holds for any slip δ up to 33.7°. At V = 1 the tolerance widens to 45°; once V falls to 1/√2 or below, no alignment rescues it.

AnswerS = −2.18, so |S| = 2.18 violates the local bound; the mount may slip up to 33.7° before |S| falls to 2.

HardAlice's two analysers are fixed 60° apart, at a = 0° and a′ = 60°, and Bob may choose any two directions. For singlet pairs, find the b and b′ that maximise |S|, the maximum itself, and the spacing Alice should have chosen instead. Then evaluate the local sign model E = −1 + φ/90° at Bob's optimal settings.

With E(x, y) = −x⋅y, S = −a⋅b + a⋅b′ − a′⋅b − a′⋅b′ = −(a + a′)⋅b − (a′ − a)⋅b′, so Bob's two choices enter separate terms.
A dot product with a unit vector is largest in size when that vector lies along the other factor, so take b along a + a′ and b′ along a′ − a: |S|max = |a + a′| + |a′ − a|.
For unit vectors γ apart, |a + a′| = 2 cos(γ/2) and |a′ − a| = 2 sin(γ/2). With γ = 60°, |S|max = √3 + 1 = 2.732, with b at 30° (the bisector) and b′ at 120°. Direct check: E = −0.866, +0.500, −0.866, −0.500 for (a, b), (a, b′), (a′, b), (a′, b′), so S = −0.866 − 0.500 − 0.866 − 0.500 = −2.732.
Over Alice's spacing, 2[cos(γ/2) + sin(γ/2)] = 2√2 sin(γ/2 + 45°), largest at γ = 90° with |S| = 2√2 = 2.83. This derives the ceiling for the singlet with spin measurements only; Tsirelson's statement for every state and observable stays quoted.
Local sign model at Bob's settings: the angles 30°, 120°, 30°, 60° give E_L = −0.667, +0.333, −0.667, −0.333, so S_L = −0.667 − 0.333 − 0.667 − 0.333 = −2.000, on the bound and no further.
Checks: γ → 0 gives |S|max = 2, since with a′ = a Alice has nothing to exploit; S is dimensionless and no power of α appears.

Answerb = 30° and b′ = 120° give |S|max = 1 + √3 = 2.73, a 37% violation; γ = 90° would give 2√2 = 2.83; the local sign model gives exactly |S| = 2.00 at the same settings.