GHZ Paradox

Three observers — Alice, Bob, Carol — each receive one qubit from a shared GHZ state (|000⟩ + |111⟩)/√2. Each independently chooses to measure their qubit in basis X or Y. The product of their three results is not just statistically different in the quantum and classical worlds — it's deterministically opposite. No probability, no inequality, no statistical loophole. Just contradiction.

Alice

measure X
—

Bob

measure X
—

Carol

measure X
—
A · B · C = — predicted (quantum): +1 predicted (classical): −1
scenario quantum predicts any local hidden-variable predicts
XXX (all measure X)+1−1 (impossible to satisfy)
XYY−1±1 needed
YXY−1±1 needed
YYX−1±1 needed
why this is more dramatic than CHSH
CHSH (the two-party version) shows quantum and classical differ statistically: 85% vs 75% win rates. GHZ shows they differ deterministically.

For the GHZ state (|000⟩ + |111⟩)/√2, quantum mechanics predicts with certainty:
• A_x · B_y · C_y = −1 every time
• A_y · B_x · C_y = −1 every time
• A_y · B_y · C_x = −1 every time
• A_x · B_x · C_x = +1 every time

Any local hidden-variable theory assigns each party a pre-determined value for X (call them α, β, γ ∈ ±1) and for Y (α', β', γ' ∈ ±1). Multiplying the first three predictions: (α · β' · γ') · (α' · β · γ') · (α' · β' · γ) = α α' β β' γ γ' · (α'β'γ'). After cancellation, this equals α' β' γ' · (α' β' γ') wait let me redo. The trick is: multiplying the LHS of all three "−1 results" gives α β'γ' · α'β γ' · α'β'γ = α·α'·α' · β'·β·β' · γ'·γ'·γ = α · β · γ (since each prime appears twice and squares to 1).

So classical predicts α β γ = (−1) × (−1) × (−1) = −1. But the fourth measurement A_x B_x C_x = α β γ is predicted by quantum to be +1. Contradiction. No probability needed. This is the Mermin-Greenberger-Horne-Zeilinger paradox in its sharpest form: local realism predicts the opposite of what quantum mechanics — and experiment — confirm.