Bell inequalities, CHSH and nonlocality

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 signs of a⋅λ and b⋅λ disagree on two antipodal lunes of total 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.

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.

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.

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, hence ‖S²‖ = 16 and ‖[A, A′]‖ ‖[B, B′]‖ = 12, while each factor is at most 2; 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.

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. The first two sit dozens of standard deviations above 2, Hensen's only 0.42/0.20 = 2.1, and none is above 2√2.

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 quantum-cryptography topic later in this unit.