Three-qubit entanglement that refutes local realism
How does three-qubit entanglement differ from two-qubit entanglement, and why does it provide a stronger test of quantum mechanics?
The Greenberger-Horne-Zeilinger (GHZ) state extends Bell states to three qubits and reveals an even more dramatic departure from classical physics. While Bell states require statistical analysis over many measurements to show non-classicality, the GHZ state provides a single-shot contradiction with classical local realism. GHZ states are used in quantum error correction, multiparty quantum communication, and as benchmarks for quantum hardware.
The GHZ state is like a Bell state scaled up: all three qubits are either all 0 or all 1, with nothing in between. Each qubit individually looks completely random, but measuring any one instantly tells you what the other two will be. The state is genuinely three-way entangled — it cannot be decomposed into any pair being entangled while the third is independent.
GHZ states demonstrate genuine multipartite entanglement. Unlike W states (where pairs remain entangled if one qubit is lost), losing any single qubit of a GHZ state completely destroys the entanglement between the remaining two.
All three qubits start in . No entanglement.
Put the first qubit in superposition. The state is now .
Entangle qubit 0 with qubit 1. The branch flips qubit 1.
Extend the entanglement to qubit 2. The branch flips qubit 2. Now all three qubits are entangled.
You get 000 or 111, each with 50% probability. No other outcomes ever appear. All three qubits are perfectly correlated.
Outcomes 000 and 111 with equal probability (never 001, 010, 011, 100, 101, 110) confirm the GHZ state. Any deviation indicates noise or decoherence.
The GHZ state is a three-qubit state with a remarkable fragility. If you mathematically remove any one qubit from the description (called tracing out), the remaining two qubits are left in a state that is classically correlated (both same) but not entangled. This means the entanglement is genuinely three-way -- it cannot survive the loss of any part. The GHZ argument shows that for certain measurement axis choices, quantum mechanics makes deterministic predictions that no classical pre-agreement (local hidden variable theory) can reproduce. Unlike Bell inequality violations, which are statistical, the GHZ contradiction is all-or-nothing.
Classically, you could prepare three coins to be all-heads or all-tails, but they would have definite values. The GHZ argument shows that no classical pre-agreement can reproduce all quantum predictions.
The GHZ state has no definite values before measurement. Certain measurement combinations give deterministic outcomes that contradict any local hidden variable assignment — not statistically, but with certainty.
It is tempting to think the GHZ state is just a bigger version of a Bell state. The entanglement is qualitatively different. A Bell pair's entanglement survives if you only look at the two qubits involved. GHZ entanglement is fragile: losing any single qubit destroys all entanglement between the remaining two. This fragility is also what gives GHZ its unique power as a test of quantum mechanics.
GHZ states generalize to n-qubit cat states (|00…0⟩+|11…1⟩)/, which are central to quantum error correction and quantum metrology.
If you lose (trace out) one qubit from a GHZ state, what happens to the entanglement between the remaining two?
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