GHZ state

The three-qubit GHZ state is

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.