Measurement and CHSH

The Born rule, and what a second measurement does

Measuring in an orthonormal basis {|e_k〉} gives outcome k with probability |〈e_k|ψ〉|². These sum to 1 because the basis is complete. The state afterwards is the normalised projection P_k|ψ〉/‖P_k|ψ〉‖, so measuring again returns the same outcome with probability 1. For an observable A with A² = I (an ±1 observable such as a Pauli) the expectation is 〈ψ|A|ψ〉 and the variance is 1 − 〈A〉².

Uncertainty (Robertson). For any two observables, Var(A) Var(B) ≥ |〈[A,B]〉|²/4. With A = Z, B = X the commutator is [Z,X] = 2iY, so Var(Z) Var(X) ≥ 〈Y〉²: Z and X cannot both be sharp.

Schmidt decomposition and the singlet

Every two-qubit state has a Schmidt form √λ₁|a₁〉|b₁〉 + √λ₂|a₂〉|b₂〉, where λ_1,2 are the eigenvalues of M M^† for the 2×2 coefficient matrix M. The entanglement entropy is −Σ λ_i log₂ λ_i: zero for a product state and exactly one bit for the singlet (|01〉 − |10〉)/√2, whose λ₁ = λ₂ = 1/2.

Two parties, four settings, and a number that separates theories

Alice measures ±1 observables A₀ or A₁, Bob B₀ or B₁. Define S = 〈A₀B₀〉 + 〈A₀B₁〉 + 〈A₁B₀〉 − 〈A₁B₁〉. Local models. If the outcomes are fixed in advance, each of A₀, A₁, B₀, B₁ is ±1 and S = A₀(B₀ + B₁) + A₁(B₀ − B₁); one of the brackets is 0 and the other ±2, so |S| ≤ 2, and every mixture of such assignments obeys it too.

Quantum. On the singlet, for observables cosa Z + sina X and cosb Z + sinb X the correlator is −cos(a − b). At a₀ = 0, a₁ = π/2, b₀ = π/4, b₁ = −π/4 this gives |S| = 2√2 ≈ 2.828.

Tsirelson's bound, algebraically. For any A₀, A₁, B₀, B₁ with squares equal to I, let S = A₀⊗(B₀ + B₁) + A₁⊗(B₀ − B₁). Expanding S² and using the squares, the cross terms collect into a commutator:

S^2  =  4 I  -  [A0, A1] (x) [B0, B1]

Each commutator has norm at most 2, so ‖S‖² ≤ 4 + 4 = 8 and ‖S‖ ≤ 2√2. The identity needs only A² = B² = I, not Hermiticity, a point the checker's control found out the hard way (see below).

The same number as a game. A referee sends bits x, y to Alice and Bob, who may not communicate; they answer a, b and win when a ⊕ b = x·y. A win at xy = 0 means equal answers and at xy = 1 different answers, so the win probability is 1/2 + S/8. Consider the family of no-signalling boxes with uniform marginals and P(win) = (1+ε)/2: every correlator is ε(−1)^xy and S = 4ε. Deterministic local strategies win at most 3/4 (8 of the 16 reach it), and mixing those 8 gives exactly the ε = 1/2 box, so ε > 1/2 is outside every local model. The singlet at the settings above, with Bob's observables negated to match the −cos convention, gives P(win) = cos²(π/8) ≈ 0.854, that is ε = 1/√2. The Popescu–Rohrlich box, ε = 1, wins always and has S = 4, so no-signalling alone does not give Tsirelson's bound. Barandes, Hasan and Kagan (Phys. Rev. A 114, 022208, 2026) re-derive 2√2 in their stochastic–quantum framework, where causal locality forces the unitary to factorise as U_A⊗U_B; the bound then follows from the same operator inequality used above. That framework is an interpretation, not consensus, and my reading of their paper is from the text supplied, not machine-checked.

Computed from matrices, not from the formula

Born rule: probabilities sum to 1 (worst 1.1e-15); repeat measurement -> same outcome (6.7e-16)
Robertson: Var(Z)Var(X) >= |<[Z,X]>|^2/4 on 500 random states, minimum slack +9.8e-07
Schmidt: singlet entropy exactly 1 bit; product state 0
Correlator -cos(a-b) vs explicit kron(A(a),B(b)) on the singlet: 25 x 25 grid, worst 4.4e-16
CHSH local:   all 16 deterministic assignments: max |S| = 2
CHSH quantum: |S| = 2.828427124746190 = 2 sqrt 2;  16^4 angle choices: max 2.828427124746
Tsirelson identity S^2 = 4I - [A0,A1](x)[B0,B1]: 60 random observable quadruples, worst 6.2e-15
spectral norm of S over those: max 2.818011 <= 2 sqrt 2
Game: box family eps in [0,1] is no-signalling; win = (1+eps)/2, S = 4 eps (101 values, worst 4.4e-16)
Local: 16 deterministic strategies, best win 3/4 (8 reach it); uniform mixture of those 8 = the eps = 1/2 box exactly
Quantum: singlet box wins cos^2(pi/8) = 0.8535533906 = (1+1/sqrt 2)/2; PR box (eps = 1) wins 1, S = 4

One game check was vacuous on first writing: my no-signalling test compared Alice's marginal across her own input, which may vary, and passed only because every marginal in the family is uniform. A box where Alice's answer equals Bob's input exposed it (the test scored 0.00 on a box that signals); it now compares across the other party's input and scores 1.00 there. A second control confirms the sign convention: with Bob's observables un-negated the singlet wins only 0.146.

Controls, each must fail and does: −cos(2(a−b)) in place of −cos(a−b) against the explicit singlet; a product state on the same grid stays at max |S| = 1.755, inside the local bound; and a non-involutory A₀ breaks the identity. The checker's first control for that last point used the matrix [[1,1],[0,−1]], which is non-Hermitian but does square to I, so the identity held and the control did not fail. It was replaced with [[1,1],[0,1]], whose square is not I.

Run it: python3 quantum-maths/qm2-verify.py (standard library only; the sourcing block says SKIP, never PASS, if the source PDF is not in ~/Downloads).