Greenberger–Horne–Zeilinger state

The three-qubit GHZ state is

A postselection scheme in which two polarization-entangled photon pairs pass through polarizing beam splitters, a half-wave plate and a nonpolarizing beam splitter. One photon at each of a trigger and three output detectors selects a GHZ state in the three output polarizations, provided optical mode matching removes distinguishable pair labels. All four detections are needed for the selected event; the trigger alone does not guarantee the three other output occupations.

The 1999 experiment used an approximately $200\mathrm{fs}$, $394\mathrm{nm}$ pump pulse and a beta barium borate crystal. Two polarization-entangled photon pairs supplied four photons. A trigger and three other photodetectors selected a GHZ state through fourfold coincidence counting. Narrow interference filters lengthened the photon coherence time to approximately $500\mathrm{fs}$, exceeding the pump duration and suppressing timing information about which pair supplied which photon. The optical mode matching was essential to coherence, not merely to routing photons.^[ref]

The observed state was selected by all four detections, which also measured the photons. A trigger alone did not demonstrate a guaranteed surviving three-photon output. This distinction connects postselection to what an apparatus actually supplies, rather than identifying a fourfold coincidence with an unconditional source.^[ref]

For a source with exactly two photons in each input arm, let $q$ be the probability of one horizontal and one vertical photon in the trigger arm. This sector alone supplies exactly one trigger photon. Splitting the other arm across its two output ports has probability $1/2$; the final polarizing beam splitter selects equal polarizations with probability $1/2$. The fourfold success probability is therefore $q/4$. The two-pair component of polarization-entangled down-conversion has $q=1/3$, giving $1/12$; independent pair bookkeeping has $q=1/2$, giving $1/8$. Pair-label distinguishability can also destroy the conditional GHZ state coherence and must be checked separately.

Charlie measures his qubit of a GHZ state in the $X$ basis. The two equally likely outcomes leave Alice and Bob in the Bell states ${\Phi }^{+}$ or ${\Phi }^{-}$, respectively. Charlie sends the outcome bit $c$ to Bob, who applies the Pauli operator $X{Z}^c$, giving ${\Psi }^{+}$ in either case. This is deterministic LOCC entanglement localization. If Charlie's record is ignored, Alice and Bob instead have the separable mixture $(|00⟩⟨00|+|11⟩⟨11|)/2$. The classical outcome is necessary to turn the conditional entanglement into a known usable Bell pair.

For $n\ge 2$, apply a Hadamard gate to one qubit, then controlled-NOTs from it to all others. Replacing each controlled-NOT gate by a target Hadamard, a Controlled-Z gate, and another target Hadamard uses $1+3(n-1)$ allowed gates. Each one-qubit reduced state is $I/2$, proving entanglement across each qubit-versus-rest bipartition.

For $|\alpha ⟩=a|0⟩+b|1⟩$, a Bell measurement on $|\alpha ⟩$ and the first leg of a GHZ state leaves the other two legs in $a|00⟩\pm b|11⟩$, after applying $X$ to both legs for the two $\Psi$ outcomes.

Starting from $|000⟩$, apply a Hadamard gate to the first qubit and then controlled-NOT gates from the first qubit to each of the other two qubits.

If two senders share a GHZ state with a receiver, one sender can encode two bits with $X^x{Z}^z$ and the other one bit with $X^b$. Sending both qubits to the receiver produces one of eight orthogonal GHZ-basis states, so a joint measurement recovers all three bits.

The GHZ theorem uses perfect multipartite quantum correlations to contradict local predetermined values without a statistical inequality. For the three-qubit GHZ state, three certain Pauli-product relations force a fourth product to have the opposite sign from its quantum-mechanical value.

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.