Entanglement

GHZ State

Equal superposition of all-zero and all-one. The n-qubit form of a Bell pair.

Qubits 3 Cost O(n) Default 3 qubits

The Greenberger-Horne-Zeilinger state is (|0...0⟩ + |1...1⟩)/√2. Any single-qubit Z measurement collapses the rest. Tracing out one qubit leaves a classical mixture, so GHZ entanglement is fragile under loss.

Build it with H on q0 and a CX chain q0→q1→...→q(n-1). The default below is n = 3.

How it works

  1. 1
    Seed superposition

    H 0 prepares (|000⟩ + |100⟩)/√2 in |q2 q1 q0⟩.

  2. 2
    CX cascade

    CX [0,1] then CX [1,2] copies the bit along the chain. |100⟩ becomes |111⟩.

  3. 3
    Readout

    Shots are 000 or 111 only. Measuring any qubit in Z predicts the other two.

Default Qubi (static)
H 0
CX [0,1]
CX [1,2]

Function form: GHZ(0..2)

Circuit
q0q1q2H
Expected result

State. |GHZ⟩ = (|000⟩ + |111⟩)/√2

P(000) = P(111) = 1/2.

BasisAmplitudeProbability
|000⟩ 1/√2
50%
|111⟩ 1/√2
50%

Notes

  • W states are a different entanglement class; you cannot convert GHZ to W with local operations and classical communication.
  • Losing one qubit of a GHZ state destroys the remaining entanglement.

Change parameters

The write-up above is for the default circuit. Use this control to generate other variants and load them in the simulator.

See alsoBell StateW State
← Bell State All algorithms W State →
Open default circuit in Simulator