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I'm reading the proof of "nonlocality without inequality" presented in (Hardy 1992). In this protocol, we consider two particles (say, an electron and a positron) evolving almost independently: they both pass through one of two beamsplitters independently from the other's path, but can cross paths. If that happens, they annihilate each other with unit probability, and therefore are not measured at the end. Here is the setup as given in the paper:

Here, $C^\pm$ and $D^\pm$ are the possible output events, and $P$ is the point at which $e_+$ and $e_-$ interact if they both pass through it. Note how, if there was no point of interaction $P$, the two particles would evolve independently, and the beamsplitters cancel each other out.

We focus on the output events in four scenarios: 1) when all BSs are as in the above figure, 2) when BS2$^+$ is removed, 3) when BS2$^-$ is removed, 4) when both BS2$^+$ and BS2$^-$ are removed.

Using the convention for beamsplitter evolutions in which a phase $i$ is introduced upon reflection, $a_1\to a_1 + i a_2$ and $a_2 \to ia_1 + a_2$, we can compute the output states in the four cases, and find the following

\begin{align} &\,\,\,\,\,\text{both BS}\,: &&-3|cc\rangle + \,\,\,i |cd\rangle + \,\,\,i |dc\rangle - |dd\rangle, \\ &\text{only BS2$^+$}: &&\phantom3-|cc\rangle + 2i |cd\rangle + \,\,\,i|dc\rangle, \\ &\text{only BS2$^-$}: &&\phantom3-|cc\rangle + \,\,\,i| cd\rangle + 2i |dc\rangle, \\ &\,\,\,\,\,\text{no BS2$^\pm$}: &&\phantom1 \qquad\qquad\,\,\, i| cd\rangle + \,\,\,i |dc\rangle + |dd\rangle. \end{align}

We now consider what happens when the output $|dd\rangle$ is observed when both BSs are used. This happens with probability $1/16$. Then, the author remarks that

  1. From the only BS2$^+$ case, we conclude that observing $d_+$ (i.e. $d$ for the first particle) implies that we should observe $c_-$ when no BS2$^-$ is used.
  2. From the only BS2$^-$ case, we conclude that observing $d_-$ (i.e. $d$ for the second particle) implies that we should observe $c_+$ when no BS2$^+$ is used.
  3. The above two observations tell us that, when neither BS2$^+$ nor BS2$^-$ are used, we should observe the output $|cc\rangle$. But this is incompatible with the output state we know we have in this instance, which assigns zero probability to such event.

We then conclude that observing $|dd\rangle$ contradicts local realistic explanations.

The latter conclusion is what I'm trying to understand. I follow all the calculations, but fail to see clearly the connection with the local realistic hypothesis. Why can we mix the output states in the four different scenarios (yes/no BS$^+$/BS$^-$) like this? Why does it matter what we would observe when one of the two BSs is removed, when both of them are present in the considered experimental run?

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Answering your precise question: mixing the four scenarios is not particular to Hardy's argument, it is done in all nonlocality proofs. The fundamental assumption is that the distribution of the hidden variables doesn't depend on the measurement setting, i.e., whether BS2$^+$ or BS2$^-$ are there, so that we can actually use the different measurements to probe the state of the hidden variables. This assumption is known as "no superdeterminism" or "no conspiracy".

More generally, keep in mind that Hardy wrote his paper in 1992. The language and mentality at this time were very different. Nowadays this kind of proof is known as "possibilistic nonlocality". I find this paper about it quite clear.

Don't worry, though, Hardy's proof is completely kosher, and can be rephrased in the usual language. The idea of a proof of nonlocality is to show that a correlation $p(ab|xy)$, obtainable from quantum mechanics, cannot be reproduced with local hidden variable models (LHVs). One can always do that via Bell inequalities, as they completely delimitate the polytope of correlations that can be produced via LHVs. Hardy's insight was that in some cases you don't need to know all the probabilities, and in fact not even the precise values, but only that for some $a,b,x,y$ we have $p(ab|xy) > 0$ and for some other $p(ab|xy) = 0$.

For convenience, let's rename the settings: if BS2$^+$ is there Alice has setting 1, and if it's not there it has setting 0. Same for Bob. The quantum correlations are then \begin{align*} p(cc|00) = 0 \\ p(dd|01) = 0 \\ p(dd|10) = 0 \\ p(dd|11) > 0 \end{align*} Hardy's argument implies that there is no LHV model that can reproduce them (and in particular they violate the CHSH inequality). If you find it unpersuasive, perhaps it's more enlightening to try to build a LHV model for these correlations via brute force. It will be a mixture of all 16 possible deterministic assignments for the results of Alice and Bob when they use or not BS2. Writing it in the order Alice-0, Alice-1, Bob-0, and Bob-1, one possible assignment is $(c,d,c,d)$, which means that Alice gets result $c$ when she doesn't use BS2, result $d$ when she does use it, and the same for Bob.

Now, from the condition $p(dd|11) > 0$, we see that the LHV model must have nonzero weight in one of the four assignments $(c,d,c,d)$, $(c,d,d,d)$, $(d,d,c,d)$, and $(d,d,d,d)$, as these are the only ones with output $d,d$ for settings $1,1$. But the fact that $p(cc|00) = 0$ implies that $(c,d,c,d)$ is forbidden. Likewise, $p(dd|01) = 0$ implies that $(d,d,c,d)$, and $(d,d,d,d)$ are forbidden, and $p(dd|10) = 0$ implies that $(c,d,d,d)$ and $(d,d,d,d)$ are forbidden. Therefore, all four possibilities must have zero weight, and no LHV can reproduce those four characteristics of the quantum correlation.

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