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GHZ State

Extend entanglement to three qubits

intermediate3 qubits·~2 min

The question

How does three-qubit entanglement differ from two-qubit?

Before you start

This experiment builds on the Bell state. You already know how H + CNOT entangles two qubits. Here you will add a second CNOT to bring a third qubit into the entangled group. The result is a GHZ state, named after physicists Greenberger, Horne, and Zeilinger.

What you will see

Start with the Bell state recipe: H on qubit 0, then CNOT to entangle qubit 1. Now add a second CNOT that links qubit 2 to qubit 0. The $∣000 ⟩$ branch stays $∣000 ⟩$ , and the $∣110 ⟩$ branch becomes $∣111 ⟩$ (the second CNOT flips qubit 2 when qubit 0 is $∣1 ⟩$ . The final state has all three qubits either all 0 or all 1, with nothing else. Each qubit alone looks completely random, but measuring all three always gives a perfectly correlated result.

The circuit

Circuit

Open in simulator →

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Step-by-step walkthrough

Apply Hadamard to qubit 0

H puts qubit 0 into superposition. Qubits 1 and 2 remain in $∣0 ⟩$ . The histogram shows equal weight on $∣000 ⟩$ and $∣100 ⟩$ . This is the same starting point as the Bell state experiment.

\frac{∣000 ⟩ + ∣100 ⟩}{2}

Apply CNOT (control: q0, target: q1)

The first CNOT entangles qubits 0 and 1, creating $(∣000 ⟩ + ∣110 ⟩) / 2$ . The histogram now shows only $∣000 ⟩$ and $∣110 ⟩$ . Qubit 2 is still unentangled at this point.

\frac{∣000 ⟩ + ∣110 ⟩}{2}

Apply CNOT (control: q0, target: q2)

The second CNOT extends the entanglement to qubit 2. The $∣110 ⟩$ branch becomes $∣111 ⟩$ (qubit 0 is $∣1 ⟩$ , so qubit 2 flips). Now all three qubits are entangled. The histogram shows only $∣000 ⟩$ and $∣111 ⟩$ with equal weight.

\frac{∣000 ⟩ + ∣111 ⟩}{2} = ∣ GHZ ⟩

What to notice

Only the two extreme outcomes (000 and 111) appear in the histogram. All six other combinations have zero weight.
All three single-qubit Bloch spheres shrink inward, showing that each qubit alone is mixed.
Compare with the Bell state: the pattern is the same (all match or none match) but extended to three qubits.
If you could trace out any one qubit, the remaining two would lose their entanglement entirely. GHZ entanglement is fragile in this specific way.

Tempting but wrong

It is tempting to think the GHZ state is just a bigger Bell state with the same properties. The entanglement is qualitatively different. In a Bell pair, the two qubits stay entangled no matter what. In a GHZ state, losing any single qubit destroys all entanglement between the remaining two. This fragility is what makes three-qubit entanglement a genuinely new phenomenon.

Expected result

You see only 000 and 111 in the output, each with roughly 50% frequency. You never see any of the six other possible three-bit strings. Over 1000 shots, the counts should split near 500 each.

Connection to the theory

This experiment extends the Entanglement lesson from two qubits to three. The lesson explains why GHZ entanglement cannot survive the loss of a qubit and why it provides a stronger test of quantum mechanics than Bell states. Running this experiment gives you direct experience with multipartite entanglement before studying the theory.

Read the full lesson →

Test your understanding

If you lose one qubit from a GHZ state, what happens to the entanglement between the remaining two?

▶ Load in simulator ↗ Nielsen and Chuang, Quantum Computation and Quantum Information ↗ MIT OCW 8.06: entanglement notes

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