Bell made Simple - Hardy result

[Eye_in_the_Sky]

I am starting this B thread for people to discuss and learn from.

In particular, I am hoping it will help to clarify some of the confusing issues which have come up in the "CFD - Counterfactual Definiteness" thread.

"Well wishings to ALL, for a complete and successful understanding!"
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The example below is due to Hardy.
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Each of the instruments of Alice and Bob (which together perform a joint-entanglement-measurement at spacelike separation) have two settings, 1 and 2, for which the outcomes can be YES or NO.

There is a quantum-mechanical entanglement scenario for which each of the following statements is true:

(0) For the configuration <a₁,b₁>: the outcome (NO,NO) is sometimes obtained.

(1) For the configuration <a₁,b₂>: if a₁ gives NO, then b₂ gives YES with certainty.

(2) For the configuration <a₂,b₁>: if b₁ gives NO, then a₂ gives YES with certainty.

(3) For the configuration <a₂,b₂>: the outcome (YES,YES) is forbidden.
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I have taken this example (but adapted the notation) from Penrose's "Road to Reality", §23.5, p. 590. There, he mentions that the (NO,NO) outcome for the setting in (0) is obtained with a probability of 1/12.

 

[Strilanc]

Here's a set of slides discussing the Hardy experiment. Unfortunately the original paper is behind a paywall.

I consider this inequality more complicated than the CHSH one. You do best at it under partial entanglement.

 

[Demystifier]

For a simple explanation of the Hardy result see e.g.
http://arxiv.org/abs/quant-ph/0609163
around Eqs. (43)-(46).

 

[Strilanc]

Here's a circuit that meets the desired criteria:

The first gate is just making a 1/3-2/3 split. Its matrix is {{√⅔,-√⅓},{√⅓,√⅔}}. Then the controlled-Hadamard finishes the setup of the shared partially-entangled state $|00\rangle + |01\rangle + |10\rangle$. Later, the referees make their choice of which case to query via a Hadamard+measure. Then the players can apply operations that depend on the choice of their referee; they both do nothing in the ON case and apply H+X in the OFF case.

Note the grid of outputs to the right. Imagine splitting it up into four 2x2 blocks. The top-left block is the A1,B1 case. Top-right is A1,B2. Bottom-left is A2,B1. Bottom-right is A2,B2. Within each 2x2 block the top-left is NO,NO and the bottom-right is YES,YES.

There's holes over the cells corresponding to (A2,B1,NO,NO), (A1,B2,NO,NO), and (A2,B2,YES,YES). But the cell for (A1,B1,NO,NO) isn't empty. Those are the requirements.