Hardy Paradox
Hardy’s paradox exhibits a positive-probability quantum event that is incompatible with three other quantum zero-probability events in a Bell-local model. “Paradox” names the conflict between quantum predictions and that model’s assumptions, not a contradiction within quantum theory. An explicit two-qubit construction makes every inference checkable.
Helpful background. Local Hidden Variables defines the local model contradicted by the four event probabilities.
A state and two settings per party
Section titled “A state and two settings per party”Use
For setting zero, both Alice and Bob label as outcome and as outcome . For setting one, label as outcome and as outcome . This outcome convention is essential: the setting-one plus label is the minus superposition vector.
The missing component gives . Destructive interference gives
so . But
The state is nonmaximally entangled. This convenient example is not the choice maximizing the Hardy success probability.
The local logical implication
Section titled “The local logical implication”In a deterministic local model consider a hidden state producing for both setting-one measurements. The forbidden event forces Bob’s setting-zero response to be . The forbidden event similarly forces Alice’s setting-zero response to be . The same hidden state must therefore contribute to , which is zero.
Setting independence and locality justify comparing these responses across alternative remote choices. Stochastic local models are convex mixtures of deterministic strategies, so the contradiction applies to them as well. No experiment simultaneously performs all four setting pairs on one copy; the model is what claims a common response assignment for the alternatives.
A probability bound replaces exact zeros
Section titled “A probability bound replaces exact zeros”For every deterministic strategy, success at setting pair implies at least one of the three forbidden events. The union bound therefore gives the Bell inequality
This is the practical form when measured probabilities are nonzero. If an experiment has success probability and the three unwanted probabilities are each at most , its ideal probability margin is at least . Finite counts require a statistical confidence statement for that margin and a protocol that handles losses and setting choices; point estimates alone do not establish the rejection strength.
What the example teaches
Section titled “What the example teaches”A maximally entangled state is not always optimal for a particular nonlocality witness. The success event relies on a specific pattern of constructive and destructive amplitudes, not simply “more entanglement.” The standard two-qubit Hardy construction does not work in the same form for maximally entangled states, although those states violate other Bell inequalities.
As with CHSH, no faster-than-light message follows. The remote setting affects correlations conditioned on both parties’ records, not the freely usable local marginal. See No-Signaling for that distinction.
Exercise: depolarizing noise
Section titled “Exercise: depolarizing noise”For , find when the Hardy inequality above is violated.
Solution
Every rank-one joint outcome has white-noise probability . Success is , while the three-event sum is . Violation requires , hence . The strong noise sensitivity is specific to this state, measurements and witness; it is not a general entanglement threshold.
References
Section titled “References”- Hardy, Lucien, “Nonlocality for two particles without inequalities for almost all entangled states,” Physical Review Letters 71, 1665–1668 (1993), doi:10.1103/PhysRevLett.71.1665.
- Brunner, Nicolas, Daniel Cavalcanti, Stefano Pironio, Valerio Scarani, and Stephanie Wehner, “Bell nonlocality,” Reviews of Modern Physics 86, 419–478 (2014), doi:10.1103/RevModPhys.86.419 — local polytopes and robust witnesses.