GHZ theorem

The GHZ theorem contradicts predetermined local outcomes using perfect quantum correlations. For the GHZ state $(|000⟩+|111⟩)/\sqrt{2}$, the products of Pauli operators obeywith certainty, while $X_1{X}_2{X}_3=+1$ with certainty.

Suppose each site has predetermined outcomes $x_i,y_i\in \{-1,+1\}$, independent of which observable is chosen at the other sites. Multiplying the first three constraints cancels every $y_i^2$, forcing $x_1{x}_2{x}_3=-1$, in contradiction with the quantum measurement prediction. The Pauli operators evade that classical multiplication rule because $XY=-YX$ at the same site. The displayed signs use the plus-GHZ state; changing its relative sign changes the individual correlations consistently. ^[ref]

A three-party nonlocal game with binary questions of even Hamming weight. Its winning output parity is half the question weight modulo two. For the four possible input triples, a hypothetical deterministic classical response would give four parity equations whose left sides cancel under exclusive or, while their right sides have odd total parity. Thus no classical strategy succeeds on every allowed triple, even with pre-shared randomness.

On the GHZ state, measure the Pauli X gate as an observable for question zero and the Pauli Y gate for question one, and encode the measured eigenvalue as $(-1{)}^x$. The joint observables XXX, XYY, YXY and YYX have respective GHZ eigenvalues $+1,-1,-1,-1$. Their products of local outcomes enforce the required parity with certainty. The strategy uses pre-shared quantum entanglement and local measurements during the game, without communication. It realizes the GHZ theorem as an operational task. If the four questions are uniform, the best classical success is $3/4$: the parity contradiction bounds it above, and the deterministic rule $x=a$, $y=z=0$ wins three triples and loses only 011. The uniformity assumption is necessary for that numerical score, not for the impossibility of worst-case certainty.

If each of $XXY$, $XYX$ and $YXX$ has product $-1$ except with probability $ϵ$, predetermined local X/Y values imply that a positive YYY product requires at least one of those constraints to fail. The union bound gives the displayed result without assuming independent errors. For the mixture of $(|000⟩-i|111⟩)/\sqrt{2}$ and its phase-flipped partner with weights $1-ϵ,ϵ$, quantum mechanics instead gives $P(YYY=+1)=1-ϵ$. The violation survives for $ϵ<1/4$. A noisy local outcome histogram is not uniquely determined by these parity statistics alone.

For $|G_N^{-}⟩=(|0{⟩}^{\otimes N}-|1{⟩}^{\otimes N})/\sqrt{2}$, let $O_j$ have one Pauli X gate at site $j$ and Pauli Y gates elsewhere, and let $X_N=X^{\otimes N}$. The relations $Y|0⟩=i|1⟩$, $Y|1⟩=-i|0⟩$ show that $O_j|G_N^{-}⟩$ is proportional to $|G_N^{-}⟩$ exactly when $N$ is odd. In that case its eigenvalue is ${\lambda }_N=(-1{)}^{(N+1)/2}$, while $X_N$ always has eigenvalue $-1$.

Under the EPR criterion of reality and locality, each site's $X$ and $Y$ outcomes have predetermined values $x_j,y_j\in \{\pm 1\}$. Multiplying the $N$ certain equations $x_j{\prod }_{k\ne j}{y}_k={\lambda }_N$ gives ${\prod }_j{x}_j={\lambda }_N$ when $N$ is odd, since every $y_k$ appears an even number of times. This contradicts the certain quantum value $-1$ precisely when $N\equiv 3(\mathrm{mod}4)$. Thus the displayed GHZ state gives an infinite family of perfect-correlation contradictions with local hidden-variable theory, without an inequality.