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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.

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]

In a local hidden-variable description of a Bell experiment, shared hidden data have a distribution independent of the measurement settings. Each outcome depends on the local setting and those shared data, without dependence on the remote setting. Local randomness can be included in the hidden data. Perfect correlations then constrain simultaneous assignments to the possible local measurements. The GHZ theorem shows an all-versus-nothing inconsistency between such assignments and certain three-qubit correlations.

For deterministic classical responses, denote the two possible local answers by $x_0,x_1$, $y_0,y_1$ and $z_0,z_1$. Certainty in the GHZ game would requireTaking the exclusive or of all four equations gives zero on the left, since every answer occurs twice, but one on the right. This contradiction rules out deterministic certainty. Pre-shared randomness cannot cure it: conditioning on the random seed fixes a deterministic response table, and none succeeds on every allowed input. Thus noncommunicating classical players cannot win with certainty. No uniform distribution on the referee's questions is needed for this worst-case claim.

For the quantum strategy, share the GHZ state and measure the Pauli X gate observable when the local question is 0, or the Pauli Y gate observable when it is 1. Return bit 0 for measurement eigenvalue $+1$ and bit 1 for eigenvalue $-1$. The four relevant operator identities areFor the first identity, the three bit flips exchange $|000⟩$ and $|111⟩$. For each remaining identity, the two Y actions contribute $i^2=-1$ on $|000⟩$ and $(-i{)}^2=-1$ on $|111⟩$, so they exchange the terms with a minus sign. Local measurements on separate systems commute, and the product of their observed eigenvalues is therefore fixed by the displayed joint eigenvalue. That product equals $(-1{)}^{x+y+z}$, giving exactly the required output parity for each promised input. Hence the quantum strategy wins with certainty without communication. This is the GHZ theorem in nonlocal-game form; it does not let a player control another player's individual outcome.

Encode $|0⟩=|H⟩$ and $|1⟩=|V⟩$ in two orthogonal linear polarizations. Then $|\pm ⟩=(|H⟩\pm |V⟩)/\sqrt{2}$ are linear polarizations at opposite forty-five-degree angles, while $(|H⟩\pm i|V⟩)/\sqrt{2}$ are the two circular polarizations. Their right/left names depend on the viewing convention. A half-wave plate with its axis at $22.5^{\circ }$ to H implements H up to a common phase, since its Jones matrix has entries $\cos 2\theta ,\sin 2\theta$ and $-\cos 2\theta$.

In principle, a nonlinear interaction producing $e^{i\pi n_{1V}{n}_{2V}}$ supplies a deterministic Controlled-Z gate: only the joint VV component acquires a minus sign. A strongly coupled matter mediator can also provide the required conditional interaction. At optical single-photon energies the direct nonlinear interaction is weak; loss, unwanted noise and maintaining indistinguishable spatial and temporal modes make a high-fidelity deterministic gate difficult. Passive linear optics by itself does not supply this deterministic entangling interaction; measurement-based optical gates introduce additional resources and success conditions.

For the printed circuit, the first three H gates create $|+{⟩}^{\otimes 3}$. The two controlled-Z operations from the first qubit to the others giveThe final H gates on the last two qubits therefore produce the GHZ stateDefine $|s_X⟩=(|0⟩+s|1⟩)/\sqrt{2}$ and $|s_Y⟩=(|0⟩+is|1⟩)/\sqrt{2}$ for $s=\pm 1$. In any of the XYY, YXY and YYX bases, direct overlaps givewith the same coefficient formula when the X position is permuted. Hence the three explicit expansions can be written compactly asIn each sum the four sign triples are $(+,+,-),(+,-,+),(-,+,+),(-,-,-)$. They have equal probabilities $1/4$. Thus two Y outcomes determine the X outcome with certainty: their product times the X outcome is $-1$.

In the XXX basis,The four allowed triples are $(+,+,+),(+,-,-),(-,+,-),(-,-,+)$, so the XXX outcome product is $+1$.

The GHZ theorem follows without a statistical inequality. If local predetermined values $X_i,Y_i\in \{\pm 1\}$ reproduced the three mixed-basis correlations, multiplying them would givebecause every Y value is squared. Quantum mechanics instead predicts the XXX product $+1$. This is an all-versus-nothing contradiction with local realism using ideal perfect correlations. Standard two-qubit Bell-state tests compare statistical correlations through a Bell inequality; the GHZ construction makes the inconsistency visible already at the level of these deterministic parity constraints.

In a local hidden-variable description, assign predetermined values $X_j,Y_j\in \{+1,-1\}$ to the two possible measurements at photon $j$. The perfect correlations of b requireMultiplying these identities cancels each squared X value and gives $\boxed{Y_1{Y}_2{Y}_3=-1}$. Thus local realism predicts only the YYY outcomes $++-,$ $+-+,$ $-++,$ and $---$. Combining this parity rule with the measured unbiased single-Y marginals forces all four frequencies to be $1/4$: four weights subject to normalization and the three zero-mean conditions are uniquely uniform.

Quantum mechanics instead has $k=3$ in the amplitude formula of b, giving ${\mathcal{A}}_3=(1+abc)/4$. ThereforeQuantum mechanics predicts each positive-product outcome with probability $1/4$ and every negative-product outcome with probability zero. The local and quantum allowed sets are disjoint.

This is the GHZ theorem: in the ideal case the contradiction follows from deterministic product constraints, an all-versus-nothing argument. A usual two-party Bell-state test instead violates a statistical Bell inequality assembled from several settings. Both require appropriate locality and measurement-choice assumptions and careful treatment of experimental imperfections; the distinction is the deterministic form of the ideal three-party argument.

The EPR criterion of reality says that if a physical quantity can be predicted with certainty without disturbing the system, then it represents an element of physical reality. For separated particles, the EPR locality assumption regards choices of measurements on the other particles as nondisturbing. The perfect correlations below therefore imply predetermined local values for both $X={\sigma }_x$ and $Y={\sigma }_y$, independent of which distant measurements are chosen.

Put $|0⟩=|\uparrow ⟩$ and $|1⟩=|\downarrow ⟩$. The Pauli matrices satisfyFor three particles, each operator with one $X$ and two $Y$ takes $|000⟩$ to $-|111⟩$ and $|111⟩$ to $-|000⟩$, whereas $X\otimes X\otimes X$ interchanges the two strings without a minus sign. Thus the GHZ state in the question has the certain outcomesEach local $X$ or $Y$ value can be predicted from measurements at the other two sites in one of these contexts. If the EPR criterion of reality and locality assign values $x_j,y_j\in \{\pm 1\}$, the first three equalities requireMultiplying these ordinary numbers and using $y_j^2=1$ yields $x_1{x}_2{x}_3=1$, contradicting the certain quantum outcome $-1$. This derives the GHZ theorem explicitly; the multiplication is of hypothetical local values, not of mutually incompatible local observables.

For general $N$, let $O_j=X_j{⨂}_{k\ne j}{Y}_k$, with the tensor factors ordered by their site labels, and let $X_N=X^{\otimes N}$. Directly,For the difference of the two strings to be an eigenstate, these phases must agree, which happens exactly when $N-1$ is even. For odd $N$ the common phase is $(-1{)}^{(N-1)/2}$, and interchanging the strings negates their difference. ConsequentlyFor even $N$, $X_N$ still has eigenvalue $-1$, but every $O_j$ takes the difference state to a phase times the sum state, which is orthogonal to it. Thus it is not an eigenstate of any $O_j$; those operators each have outcomes $\pm 1$ with equal probabilities. Among the stated $N\ge 4$, simultaneous eigenstates therefore occur precisely at $N=5,7,9,\dots$.

For odd $N$, certainty of each $O_j$ again permits remote prediction of both local $X$ and $Y$. Writing ${\lambda }_N=(-1{)}^{(N+1)/2}$, the local hidden-variable theory would require $x_j{\prod }_{k\ne j}{y}_k={\lambda }_N$ for all $j$. Their product ishence ${\prod }_j{x}_j={\lambda }_N$, since $N-1$ is even and $N$ is odd. Quantum theory requires the same product to be $-1$. The contradiction occurs when ${\lambda }_N=+1$, giving the infinite familyFor $N\equiv 1(\mathrm{mod}4)$, these particular certain product constraints are consistent; for example $x_j=-1$, $y_j=+1$ satisfies them. This distinguishes the GHZ contradiction for N congruent to 3 modulo 4 from the larger set of odd $N$ for which simultaneous eigenstates exist.

A putative simultaneous assignment of classical values ${\tilde{s}}_x^{(i)},{\tilde{s}}_y^{(i)}\in \{\pm a\}$ reproducing the first three certain quantum outcomes must obey, almost surely,Multiply the three equalities. Each $y$ value occurs twice and its square is $a^2$, leaving $a^6{\tilde{s}}_x^{(1)}{\tilde{s}}_x^{(2)}{\tilde{s}}_x^{(3)}=a^9$. Thuscontrary to the quantum prediction $-a^3$.

This is the GHZ theorem against predetermined local measurement values. Experimentally prepare many copies of the three-spin state and choose independently among the four measurement settings on separated subsystems. The three mixed correlations force the classical prediction for the fourth, whereas quantum mechanics predicts the opposite sign. Since $s_x,s_y$ on the same subsystem do not commute, the four settings are measured on separate copies; the classical contradiction relies precisely on the common assignment of values across settings.

The Einstein–Podolsky–Rosen criterion of reality says that if a physical quantity can be predicted with certainty without disturbing a system, then an element of physical reality corresponds to that quantity.

Let $X_i,Y_i$ be Pauli measurements on particle $i$. The GHZ state in the question obeyswith certainty. Each local outcome can therefore be predicted by spacelike-separated measurements on the other two particles. The EPR criterion assigns predetermined values $x_i,y_i\in \{-1,1\}$ satisfying those three equations. Multiplying them and using $y_i^2=1$ givesQuantum theory instead predictsso a joint $X$ measurement has product $-1$ with certainty. The EPR elements of reality therefore contradict the quantum prediction, which is the GHZ theorem form of Bell theorem.

For tensor products of Pauli factors from $\{I,X,Y,Z\}$, let $k$ count the sites where both factors are nonidentity and different. Distinct nonidentity Pauli matrices anticommute, and equal or identity factors commute. Reversing the product site by site therefore gives $PQ=(-1{)}^{k}QP$. Thus two Pauli strings commute exactly when $k$ is even. The criterion applies to the perfect-correlation observables in the GHZ theorem.