The GHZ Paradox

Bell's theorem is a statistical contradiction with local realism — it needs many runs to expose a violated inequality. The GHZ paradox is sharper: with three entangled qubits, local realism is refuted by a single run, with certainty. No inequalities, no statistics — an "all-or-nothing" nonlocality.

The state

Work with the three-qubit Greenberger–Horne–Zeilinger state

|\text{GHZ}\rangle = \frac{1}{\sqrt2}\big(|000\rangle + |111\rangle\big),

shared among Alice, Bob, and Charlie. Each measures either $X$ or $Y$ on their qubit.

Four deterministic quantum predictions

The GHZ state is a simultaneous $-1$ eigenstate of the commuting operators $XYY$ , $YXY$ , $YYX$ , and a $+1$ eigenstate of $XXX$ . Direct computation gives

XYY\,|\text{GHZ}\rangle = -|\text{GHZ}\rangle,\quad YXY\,|\text{GHZ}\rangle = -|\text{GHZ}\rangle,\quad YYX\,|\text{GHZ}\rangle = -|\text{GHZ}\rangle,

XXX\,|\text{GHZ}\rangle = +|\text{GHZ}\rangle.

So quantum mechanics predicts, with certainty, that the products of outcomes satisfy $x_A y_B y_C = -1$ , $y_A x_B y_C = -1$ , $y_A y_B x_C = -1$ , and $x_A x_B x_C = +1$ , where each $x_i,y_i\in\{+1,-1\}$ is the result that party $i$ would obtain.

The contradiction with local realism

A local-realist (hidden-variable) model must assign definite values $x_i, y_i \in\{\pm1\}$ to all six potential measurements before anyone chooses what to measure. Multiply the first three predictions:

(x_A y_B y_C)(y_A x_B y_C)(y_A y_B x_C) = x_A x_B x_C \,(y_A^2)(y_B^2)(y_C^2) = x_A x_B x_C,

since each $y_i^2 = 1$ . The left side equals $(-1)(-1)(-1) = -1$ , forcing $x_A x_B x_C = -1$ . But the fourth quantum prediction demands $x_A x_B x_C = +1$ . A predetermined value set therefore satisfies

-1 = x_A x_B x_C = +1,

a flat contradiction. No local hidden-variable assignment exists. Quantum mechanics, by contrast, has no trouble: the four products refer to non-commuting global measurements that cannot be assigned simultaneous definite values.

Mermin's operator viewpoint

The same fact reads cleanly at the operator level: the four operators $XYY, YXY, YYX, XXX$ pairwise commute on the relevant subspace, yet their product is $-I$ while any value assignment would force $+I$ . The Mermin–GHZ "magic" is exactly this incompatibility between the quantum algebra and a single global assignment — the seed of the contextuality story in the next lesson.

Try it

Build the GHZ state $\tfrac{1}{\sqrt2}(|000\rangle+|111\rangle)$ with a Hadamard and two CNOTs. The grader checks the full three-qubit statevector — the resource on which the entire paradox rests.

Run your code to see the quantum state.