University Physics V · Introduction to Quantum Information · 20.5

Bell Inequalities, CHSH & Nonlocality

The singlet's correlations are a number you can compute, a bound you can prove, and an experiment you can run. Do all three here: derive |S| ≤ 2 from nothing but locality, take the operator expectation that gives −cos(a − b), find the analyser angles that push it to 2√2, then say what the violation kills and what it leaves standing.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Bell asked whether the singlet's correlations could come from a model in which each particle leaves the source carrying an answer to every setting it might meet. Write that model down: a variable λ with density ρ(λ), outcomes A(a, λ) = ±1 and B(b, λ) = ±1 that each depend on their own analyser only, and a correlation E(a, b) = ∫ρ A B dλ. Nothing in that sentence is quantum, and the CHSH bound follows from it in three lines: for every λ the combination A(B − B′) + A′(B + B′) is ±2, so its average S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) cannot exceed 2.

Quantum mechanics replaces the integral with a trace. For |ψ⁻⟩ the operator (a⋅σ) ⊗ (b⋅σ) has expectation −a⋅b = −cos(a − b), and the cosine bows below every straight line: at 45° it gives −0.707 where the best local model manages −0.5, and four such pairs stack to |S| = 2√2. The bound returns on the quantum side as Tsirelson's: S² = 4I + [A, A′] ⊗ [B, B′] caps ‖S‖ at 2√2 because a commutator of ±1 operators has norm at most 2.

What a violation costs is the conjunction it tested. Local causality, pre-existing values and free setting choice cannot all hold; experiment does not say which, and Bohmian mechanics keeps the values by giving up locality. What it does not buy is a signal: Bob's reduced state is I/2 whichever setting Alice chooses, and the correlation lives only in the record the two compare over a classical channel.

Simple definition: A Bell inequality is a bound on a combination of two-particle correlations that every local hidden-variable model must obey; CHSH is |E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′)| ≤ 2, and quantum mechanics predicts values up to 2√2.
Example: For a spin singlet with Alice at 0° or 90° and Bob at 45° or 135°, E = −cos(a − b) gives −0.707 for three pairs and +0.707 for (0°, 135°), so S = −3(0.707) − 0.707 = −2.83: above 2 in magnitude, and 0.83 more than any local model can reach.

CHSH combinationS = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′)

Four correlation experiments on four sub-ensembles of pairs, folded into one number whose bound separates the two theories.

E(a, b) = ⟨Aₐ B_b⟩, the mean product of two ±1 outcomes, dimensionless in [−1, 1]; a, a′, b, b′ are analyser directions

Local boundE(a, b) = ∫ ρ(λ) A(a, λ) B(b, λ) dλ ⇒ |S| ≤ 2

A(B − B′) + A′(B + B′) = ±2 for every λ, so the average cannot pass 2. Locality and pre-existing values are the only inputs.

ρ(λ) ≥ 0 with ∫ρ dλ = 1; A, B = ±1, each a function of its own setting and λ only

Singlet correlationE(a, b) = ⟨ψ⁻|(a⋅σ) ⊗ (b⋅σ)|ψ⁻⟩ = −a⋅b = −cos(a − b)

The whole quantum side is one dot product. The cosine bows below any straight line through (0°, −1) and (180°, +1), and that bow is the violation.

|ψ⁻⟩ = (|01⟩ − |10⟩)/√2; a, b unit vectors and a − b the angle between them; photon polarisation uses −cos 2(a − b)

Tsirelson boundS² = 4I + [A, A′] ⊗ [B, B′] ⇒ ‖S‖ ≤ 2√2

Quantum mechanics beats 2 only because A and A′ need not commute, and cannot beat 2√2 because a commutator of ±1 operators has norm at most 2.

for S = A ⊗ (B − B′) + A′ ⊗ (B + B′) with A² = A′² = B² = B′² = I; the other CHSH sign convention flips + to −

Evenly spaced analysersa = 0, b = δ, a′ = 2δ, b′ = 3δ: S = −3 cos δ + cos 3δ → |S| = 2√2 at δ = 45°

Turns CHSH into one function of one angle: dS/dδ = 0 at sin δ = sin 3δ, so δ = 45°. This is the curve the figure below plots.

spin-½ convention; the same set-up at δ = 22.5° gives only 2.39, and 22.5° is the optimum for the photon −cos 2(a − b) rule

Werner state and visibilityρ_W = V|ψ⁻⟩⟨ψ⁻| + (1 − V) I/4 ⇒ E = −V cos(a − b), |S|ₘₐₓ = 2√2 V

Separates entangled from Bell-violating: between V = 1/3 and 0.707 the state is entangled yet passes every CHSH test, so entanglement is necessary, not sufficient.

V ∈ [0, 1] is the dimensionless visibility; CHSH violated only for V > 1/√2 = 0.707; entangled for V > 1/3

Write the local model down before you attack it

Bell's target is not vague 'classical physics' but a precise class of theories. A hidden variable λ, of any kind, is set at the source with density ρ(λ) ≥ 0 and ∫ρ dλ = 1. Alice's outcome is a function A(a, λ) = ±1 of her own setting and λ; Bob's is B(b, λ) = ±1 of his setting and λ. Three assumptions are buried there: locality (A does not see b), realism (a value exists for the setting not chosen, since A(a′, λ) is defined on the same λ), and measurement independence (ρ does not depend on a or b). The correlation is E(a, b) = ∫ρ A B dλ, and the stochastic version P(A, B|a, b, λ) = P(A|a, λ)P(B|b, λ) changes nothing below. Bell's own example shows the class is not empty: take λ a unit vector uniform on the sphere, A = sign(a⋅λ), B = −sign(b⋅λ). The two signs disagree on a lune of solid-angle fraction θ/π, so E_L(θ) = −(1 − 2θ/π) = −1 + 2θ/π. It reproduces perfect anticorrelation at θ = 0 and +1 at 180°, and it is the dashed line in the figure.

The CHSH bound is three lines of arithmetic

Fix λ and write A = A(a, λ), A′ = A(a′, λ), B = B(b, λ), B′ = B(b′, λ), four numbers each ±1. Then A(B − B′) + A′(B + B′) has one bracket equal to 0 and the other to ±2, so it equals ±2 for every λ. Integrate against ρ: S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) = ∫ρ [A(B − B′) + A′(B + B′)] dλ, whose magnitude cannot exceed ∫ρ × 2 = 2. That is the whole proof; nothing in it is quantum, and nothing in it is about spin. Notice what was used. The same λ supplied both A and A′, although any one particle meets only one setting — that is realism doing work, not locality. And the four terms are measured on four different sub-ensembles of pairs; the bound assumes those sub-ensembles sample the same ρ, which is measurement independence. Bell's sawtooth saturates the bound: with evenly spaced analysers, 3E_L(δ) − E_L(3δ) = 3(−1 + 2δ/π) − (−1 + 6δ/π) = −2 for every δ up to 60°. A local model can reach 2; the theorem says none can pass it.

Compute the quantum side as an operator expectation

Name the state, the operators and the basis. State: |ψ⁻⟩ = (|01⟩ − |10⟩)/√2 in the σ_z ⊗ σ_z product basis. Observables: Aₐ = a⋅σ ⊗ I and B_b = I ⊗ b⋅σ, each with eigenvalues ±1, and their product (a⋅σ) ⊗ (b⋅σ). The singlet is rotationally invariant with ⟨σᵢ ⊗ σⱼ⟩ = −δᵢⱼ, so E(a, b) = Σ aᵢ bⱼ (−δᵢⱼ) = −a⋅b = −cos(a − b). Its joint probabilities follow from the Born rule: P(±,±) = (1 − cos θ)/4 and P(±,∓) = (1 + cos θ)/4. In NumPy this is six lines: build sx and sz, ψ = np.array([0, 1, −1, 0])/np.√(2), na = np.sin(a)*sx + np.cos(a)*sz, and E = ψ @ np.kron(na, nb) @ psi. At the canonical settings a = 0°, a′ = 90°, b = 45°, b′ = 135°, three pairs differ by 45° and give −0.7071; (a, b′) differs by 135° and gives +0.7071. So S = −0.7071 − 0.7071 − 0.7071 − 0.7071 = −2.8284 = −2√2. The sign is the singlet's: |Φ⁺⟩ = (|00⟩ + |11⟩)/√2 has E = +cos(a − b) in the x–z plane and gives +2√2 at the same angles.

Why 2√2 and not 4: Tsirelson's bound from a commutator

S is also an operator, S = A ⊗ (B − B′) + A′ ⊗ (B + B′), and ⟨S⟩ is bounded by its norm. Square it using A² = A′² = B² = B′² = I: the diagonal terms give 4I and the cross terms give [A, A′] ⊗ [B, B′], so S² = 4I + [A, A′] ⊗ [B, B′]. Each commutator of two ±1 operators has norm at most 2, so ‖S²‖ ≤ 8 and |⟨S⟩| ≤ 2√2. Two things are now visible. If A and A′ commute — one shared hidden variable is exactly that — the commutator term vanishes and the bound drops to 2, which is the local bound recovered from the quantum side. And the algebraic maximum 4 would need all four correlations at ±1 with the CHSH signs, forcing the commutator's norm to 4; no pair of operators with square I manages it. At the canonical angles [σ_z, σₓ] = 2iσ_y and [B, B′] = 2iσ_y, so S² = 4I − 4 σ_y ⊗ σ_y, whose eigenvalues are 8 (on |ψ⁻⟩ and |Φ⁺⟩) and 0. The singlet is an exact eigenvector of S with eigenvalue −2√2, which is why it saturates the bound rather than merely approaching it.

Angles, conventions and the recorded numbers

With Alice at 0° and 2δ and Bob at δ and 3δ, three pairs differ by δ and one by 3δ, so S(δ) = −3 cos δ + cos 3δ. Its extremum is at sin δ = sin 3δ, so δ = 45°, where |S| = 3(0.7071) + 0.7071 = 2√2; at δ = 0 the value is exactly 2, and at 60° it is 2.5. Photon experiments quote 22.5°, and the reason is the operator, not the physics: linear polarisation at angle a is the projector onto cos a|H⟩ + sin a|V⟩, an object with period 180°, so E = −cos 2(a − b) and every angle halves. Set a spin-½ experiment at 22.5° and you get 3 cos 22.5° − cos 67.5° = 2.772 − 0.383 = 2.39: a violation, but not the maximum. The recorded values sit below 2√2 for physical reasons. Aspect, Grangier and Roger (1982) measured S = 2.697 ± 0.015 with cascade photons; Weihs et al. (1998) found 2.73 ± 0.02 with settings switched after the photons left the source; Hensen et al. (2015), closing the locality and detection loopholes together with electron spins 1.3 km apart, reported 2.42 ± 0.20 from 245 trials. Each is many standard deviations above 2 and none is above 2√2.

What the violation excludes, and the message it does not send

The experiment refutes a conjunction: local causality, pre-existing values for unmeasured settings, and setting choices independent of λ. It does not name the culprit. Bohmian mechanics keeps definite values and drops locality; Everett keeps locality and drops single outcomes; superdeterminism keeps both by denying free choice, at the price of explaining every random-number generator ever used to pick a setting. Nor does a violation carry a signal. Alice's measurement acts on ρ = |ψ⁻⟩⟨ψ⁻| as Σₖ (Pₖ ⊗ I) ρ (Pₖ ⊗ I), and tracing out her qubit gives Tr_A ρ = I/2 whatever axis her Pₖ belong to: Bob's marginals are ½ and ½ for both of her settings, and for no measurement at all. The correlation exists only in the joint record, which needs a classical channel to compare, so the violation respects relativity exactly as the no-signalling theorem says it must. What the violation does buy is a certificate. If Alice and Bob measure |S| > 2 on the pairs they share, no third party holds a local copy of their outcomes, and that is the resource E91 turns into a key in the next topic.

Change one variable at a time

Make the relationship visible.

Interactive model

Analyser spacing δ45 °

Visibility V (Werner weight)1.00

Keep V = 1 and sweep δ: the bar leaves the local line as soon as the analysers separate, touches 2√2 at δ = 45°, and falls back to 2.5 at 60°, because the cosine bows below the local line at δ and above it at 3δ. Then lower V at 45°: the bar drops through 2 at V = 0.71, the 1/√2 visibility floor.

E(δ) = −V cos δ-0.707

E(3δ) = −V cos 3δ0.707

S = 3E(δ) − E(3δ)-2.828

V NEEDED FOR |S| > 20.707

Live interpretationE(δ) = −V cos δ: −0.707. E(3δ) = −V cos 3δ: 0.707. S = 3E(δ) − E(3δ): −2.828. V NEEDED FOR |S| > 2: 0.707

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that perfect anticorrelation at equal settings is the nonlocal part — Alice reads +1, so Bob's particle 'instantly becomes' −1. It is not the evidence. Bell's sawtooth model, E = −1 + 2θ/π, gives exactly −1 at θ = 0 with a local hidden variable fixed at the source, like a pair of gloves in two boxes. The evidence sits at intermediate angles: at 45° the singlet gives −0.707 where any local model fitting the endpoints manages only −0.5, and at 135° it gives +0.707 against +0.5. The cosine bows past the line, and four such pairs push |S| to 2.83. Gloves cannot do that, and no local set of pre-drawn answers can either. The second error follows from the first: if Alice's reading 'changed' Bob's particle, she could signal. She cannot — his statistics are ½, ½ for every setting she picks — because what she changes is the joint description, not his reduced state.

A spin singlet is shared. Alice measures along a = 0° or a′ = 90° and Bob along b = 45° or b′ = 135°. With E(a, b) = −cos(a − b) and S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′), what is S, and what does the result establish?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo spin-½ particles are in the singlet |ψ⁻⟩ = (|01⟩ − |10⟩)/√2. Find the four joint probabilities and the correlation E when the analysers differ by 60°, then evaluate S for a = 0°, a′ = 90°, b = 45°, b′ = 135° and compare it with the local bound.

Operators and state: the product observable is (a⋅σ) ⊗ (b⋅σ), and the singlet's rotational invariance gives ⟨σᵢ ⊗ σⱼ⟩ = −δᵢⱼ, so E(a, b) = −a⋅b = −cos θ with θ the angle between the analysers.
Joint probabilities from the Born rule: P(+,+) = P(−,−) = (1 − cos θ)/4 and P(+,−) = P(−,+) = (1 + cos θ)/4. At θ = 60°: P(+,+) = (1 − 0.5)/4 = 0.125 and P(+,−) = 0.375; the four sum to 1, and each marginal is 0.125 + 0.375 = 0.5.
Correlation: E = P(+,+) + P(−,−) − P(+,−) − P(−,+) = 0.25 − 0.75 = −0.50, which is −cos 60° as required.
CHSH: the pairs (0°, 45°), (90°, 45°) and (90°, 135°) differ by 45° and give E = −cos 45° = −0.7071; the pair (0°, 135°) differs by 135° and gives −cos 135° = +0.7071. S = (−0.7071) − (+0.7071) + (−0.7071) + (−0.7071) = −2.8284.
|S| = 2.83 > 2, so the singlet's correlations lie outside every local hidden-variable model; and 2.83 = 2√2 exactly, so these angles also saturate Tsirelson's bound.

AnswerAt 60°: P(+,+) = P(−,−) = 0.125, P(+,−) = P(−,+) = 0.375, E = −0.50. At the canonical angles S = −2√2 ≈ −2.83, exceeding the local bound of 2 by 0.83.

MediumA NumPy simulation of the singlet records 2000 pairs at each of the four CHSH settings. Counts (N₊₊, N₊₋, N₋₊, N₋₋) are: (a, b) = (148, 852, 850, 150); (a, b′) = (855, 145, 149, 851); (a′, b) = (143, 857, 853, 147); (a′, b′) = (152, 848, 846, 154). Estimate each E with its standard error, then S with its uncertainty, and judge the run against both bounds.

Each E is the mean of ±1 products: E = (N₊₊ + N₋₋ − N₊₋ − N₋₊)/N. (a, b): (148 + 150 − 852 − 850)/2000 = −1404/2000 = −0.702. (a, b′): (855 + 851 − 145 − 149)/2000 = +0.706. (a′, b): (143 + 147 − 857 − 853)/2000 = −0.710. (a′, b′): (152 + 154 − 848 − 846)/2000 = −0.694.
A product of ±1 outcomes has variance 1 − E², so each estimate carries σ_E = √((1 − E²)/N): 0.0159, 0.0158, 0.0157 and 0.0161 for the four settings — all close to 1/√(2N) = 0.0158 because E² ≈ ½.
S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) = −0.702 − 0.706 − 0.710 − 0.694 = −2.812. The four runs are independent, so the errors add in quadrature: σ_S = √(0.0159² + 0.0158² + 0.0157² + 0.0161²) = 0.0318.
Against the bounds: |S| − 2 = 0.812, which is 0.812/0.0318 = 25.5 standard errors above the local bound, so no local model survives this run. Against Tsirelson: 2.828 − 2.812 = 0.016, half a standard error below 2√2, consistent with a perfect singlet and no loss of visibility.
Implied visibility: V = |S|/(2√2) = 2.812/2.828 = 0.994 ± 0.011, so at two standard errors the run bounds the source at V > 0.97. A local simulation confirms the algebra, not nature: it draws outcomes from the Born rule it was given, and a real test also needs spacelike-separated setting choices and closed detection.

AnswerE = −0.702, +0.706, −0.710, −0.694 (each ± 0.016); S = −2.812 ± 0.032, 25 standard errors beyond |S| = 2 and 0.5 standard errors inside 2√2; implied visibility V = 0.994 ± 0.011.

HardFor A = σ_z, A′ = σₓ, B = (σ_z + σₓ)/√2 and B′ = (σₓ − σ_z)/√2, build the CHSH operator S = A ⊗ (B − B′) + A′ ⊗ (B + B′), find S² and its spectrum, and show |ψ⁻⟩ is an eigenvector. Then find the range of visibility V for which the Werner state ρ_W = V|ψ⁻⟩⟨ψ⁻| + (1 − V) I/4 violates CHSH, and the range for which it is entangled.

Square using A² = A′² = B² = B′² = I. The diagonal terms give A² ⊗ (B − B′)² + A′² ⊗ (B + B′)² = I ⊗ (2I − (B, B′)) + I ⊗ (2I + (B, B′)) = 4I. The cross terms give AA′ ⊗ (B − B′)(B + B′) + A′A ⊗ (B + B′)(B − B′) = AA′ ⊗ [B, B′] − A′A ⊗ [B, B′] = [A, A′] ⊗ [B, B′]. So S² = 4I + [A, A′] ⊗ [B, B′].
Evaluate the commutators with σ_z σₓ = iσ_y: [A, A′] = [σ_z, σₓ] = 2iσ_y, and [B, B′] = ½[σ_z + σₓ, σₓ − σ_z] = ½([σ_z, σₓ] − [σₓ, σ_z]) = ½(2iσ_y + 2iσ_y) = 2iσ_y. Hence S² = 4I + (2i)(2i) σ_y ⊗ σ_y = 4I − 4 σ_y ⊗ σ_y.
σ_y ⊗ σ_y has eigenvalues ±1, each twice, so S² has eigenvalues 8 and 0, each twice, and S has spectrum (−2√2, 0, 0, +2√2): the norm is 2√2 with nothing above it. Since σ_y|0⟩ = i|1⟩ and σ_y|1⟩ = −i|0⟩, σ_y ⊗ σ_y|ψ⁻⟩ = (|10⟩ − |01⟩)/√2 = −|ψ⁻⟩, so S²|ψ⁻⟩ = 8|ψ⁻⟩; with ⟨ψ⁻|S|ψ⁻⟩ = −3 cos 45° + cos 135° = −2√2, the minimum eigenvalue, S|ψ⁻⟩ = −2√2|ψ⁻⟩ exactly.
Werner state: Tr(ρ_W S) = V⟨ψ⁻|S|ψ⁻⟩ + (1 − V) Tr S/4, and S is traceless because its spectrum is symmetric, so ⟨S⟩_W = −2√2 V. CHSH is violated when 2√2 V > 2, that is V > 1/√2 = 0.7071; at V = 0.8 the value is −2.263, at V = 0.7 it is −1.980 and the state passes the test.
Entanglement by the Peres–Horodecki test: partial transposition sends |ψ⁻⟩⟨ψ⁻| to ½I − |Φ⁺⟩⟨Φ⁺|, with eigenvalues ½, ½, ½, −½, so ρ_W^(T_B) = ((1 + V)/4) I − V|Φ⁺⟩⟨Φ⁺| has eigenvalues (1 + V)/4 three times and (1 − 3V)/4 once. That is negative, and ρ_W entangled, exactly when V > 1/3; for two qubits the test is necessary and sufficient.
Read the two thresholds together: for 1/3 < V ≤ 1/√2 the Werner state is entangled but its CHSH value lies inside the local bound — at V = 0.6 it is entangled and |S| = 1.70. Entanglement is necessary for a violation, not sufficient, and a source is characterised by its visibility, not by whether it is entangled.

AnswerS² = 4I − 4 σ_y ⊗ σ_y, the spectrum of S is (−2√2, 0, 0, +2√2), and S|ψ⁻⟩ = −2√2|ψ⁻⟩. The Werner state violates CHSH for V > 1/√2 ≈ 0.707 and is entangled for V > 1/3, so 0.333 < V ≤ 0.707 is entangled without violating.