Building GHZ States
A GHZ state puts n qubits into an even superposition of all-zeros and all-ones. One Hadamard gate followed by a chain of CX gates builds it. The order of the CX gates matters. On five qubits, a straight chain needs four CX layers in a row, while a balanced tree needs only three. And a histogram with two peaks does not, by itself, prove the state is truly entangled.
What is a GHZ state?
A GHZ state is named after Greenberger, Horne and Zeilinger, who studied it in 1989. It is the n-qubit state (|00…0⟩ + |11…1⟩)/√2. In words, it is an even superposition of "all qubits read 0" and "all qubits read 1," with nothing in between.
It is the simplest example of multipartite entanglement. That means entanglement shared across more than two qubits. The whole group acts as one piece, not as a set of pairs. Measure any single qubit, and you know right away what every other qubit will read.
Picture a row of dominoes set up so that they all fall the same way. Push the first one either left or right, and the whole row copies that choice. The example stops working in one way: a domino row has a real direction the moment you push it. A GHZ state holds no hidden answer until it is measured.
The recipe grows from the Bell pair you already know. One H on q0 makes the two-branch superposition. Then CX gates fan out the branch choice. They copy the 0-or-1 choice, in the standard basis, from qubits that have it to qubits that don't. Each CX pulls one more qubit into the linked state.
- Worked example: how do the amplitudes move through the chain?
- Run it: GHZ-3INTERACTIVE
- Try this: what if I break the chain?INTERACTIVE
- Run it: GHZ-5, chain buildINTERACTIVE
- How does fan-out order change depth?
- Compare: GHZ-5, tree buildINTERACTIVE
- Does a two-peak histogram prove entanglement?
- What changes on real hardware?
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