Euro Jnl Phil Sci (2014) 4:361–368 DOI 10.1007/s13194-014-0089-2 ORIGINAL PAPER IN PHILOSOPHY OF PHYSICS On a conjecture by San Pedro Leszek Wroński Received: 11 September 2013 / Accepted: 4 April 2014 / Published online: 7 June 2014 © The Author(s) 2014. This article is published with open access at Springerlink.com Abstract In a recent paper in this Journal Iñaki San Pedro put forward a conjec- ture regarding the relationship between no-conspiracy and parameter independence in EPR scenarios; namely, that violation of the former implies violation of the latter. He also offered an argument supporting the conjecture. In this short note I present a method of constructing counterexamples to the conjecture and point to a mistake in the argument. Keywords EPR · Parameter independence · No conspiracy · Measurement independence 1 Introduction Since the EPR correlations and the philosophical issues concerning them have been described by (San Pedro 2012), to which paper this note is a technical follow-up, I will be very brief here. In the setup with which we are concerned two electrons are emitted in opposite directions. Each electron from a given pair proceeds towards a detector which can be set to measure the electron’s spin along an axis chosen from a previously determined set of directions (typically from 2 or 3 options); the results of the measurement are binary: either “spin-up” or “spin-down”. Very roughly, the EPR correlations consist in the fact that while the probability of each result for any measurement direction in any of the two wings of the experiment is one-half, the probabilities of obtaining pairs of results (one from each wing) typically differ from one-fourth. This is puzzling due to the spatiotemporal features of the setup which L. Wroński Leszek Wroński Institute of Philosophy, Jagiellonian University, Grodzka 52, 31-044 Kraków, Poland e-mail: leszek.wronski@uj.edu.pl mailto:leszek.wronski@uj.edu.pl 362 Euro Jnl Phil Sci (2014) 4:361–368 suggest various independencies: the outcome at one wing cannot influence the out- come in the other wing, and the choice of measurement setting at one wing cannot influence the outcome in the other wing. The famous 1964 result of Bell shows that one cannot account for the correlations by positing a hidden variable subject to certain formal conditions which are supposed to formulate the natural independencies like the ones described above. There have been many attempts to formulate alternative explanations by means of various con- structs subject to various independence relations; one usual relation of this sort is that of no conspiracy, or measure independence: the posited hidden variable should be statistically independent from the measurement settings. It is to this field that San Pedro makes a contribution. Apart from philosophical points about backwards causation, which will not concern me here, he offers a conjecture; namely, that when- ever measurement independence is violated, the so called parameter independence condition—which says that if we fix the value of the posited hidden factor the chance for obtaining a given result in one wing should not depend on the measurement setting in the other wing—is also violated. San Pedro offers a mathematical argument sup- porting this conjecture. In the following sections I will present a method of construct- ing counterexamples to the conjecture and point to a flaw in San Pedro’s argument. 2 The formal setup San Pedro’s notational conventions dictate that we use: • Li for the event of the detector in the left wing being set to the axis i; • Rj for the event of the detector in the right wing being set to the axis j; • La i for the event of the detector in the left wing, set to the axis i, displaying the result a; • Rb j for the event of the detector in the right wing, set to the axis j, displaying the result b; where i, j ∈ {1, 2, 3} and a, b ∈ {+, −}. For the hidden causal factors San Pedro proposes to use the notation Cab ij , with i, j, a, b as before. Now, due to the difference between “common common causes” and “separate common causes” (see e.g. Hofer-Szabó et al. 2013) it is currently standard to require different causal factors screening off correlations at different mea- surement settings. However, San Pedro aims to go one step further, and require that at the measurement setting Li ∧ Rj , the event Cabij should screen off the correla- tion between La i and Rb j and none of the other three (anti-)correlations. That is, for example, C++13 should screen off L + 1 from R + 3 , but not L + 1 from R − 3 . 1 This seems to be a mistake. Any event screening off one correlation at a given combination of measurement settings screens off all the other correlations at the same pair of measurement settings. Let me show an example of this (the argument is 1 The occurrence of C++13 does not in general entail the occurence of L + 1 ∧ R+3 ; i.e., we are not assuming the Cab ij ’s generally act as deterministic screening-off factors. Euro Jnl Phil Sci (2014) 4:361–368 363 similar in all other cases): if C++13 screens off L + 1 from R + 3 , then it also screens off L + 1 from R − 3 . Assume that (*) P(L+1 ∧ R+3 |L1 ∧ R3 ∧ C++13 ) = P(L+1 |L1 ∧ R3 ∧ C++13 ) · P(R + 3 |L1 ∧ R3 ∧ C++13 ). Now, P(L+1 ∧ R−3 |L1 ∧ R3 ∧ C++13 ) = P(L+1 |L1 ∧ R3 ∧ C ++ 13 ) − P(L+1 ∧ R+3 |L1 ∧ R3 ∧ C++13 ). Due to (*), this is equal to P(L+1 |L1 ∧ R3 ∧ C ++ 13 ) − P(L+1 |L1 ∧ R3 ∧ C++13 ) · P(R+3 |L1 ∧ R3 ∧ C++13 ), which in turn equals P(L + 1 |L1 ∧ R3 ∧ C++13 ) · ( 1 − P(R+3 |L1 ∧ R3 ∧ C++13 ) ) . The last expression is just P(L+1 |L1 ∧ R3 ∧ C++13 ) · P(R−3 |L1 ∧ R3 ∧ C++13 ). Putting all this together, we get the result that P(L+1 ∧ R−3 |L1 ∧ R3 ∧ C++13 ) = P(L+1 |L1 ∧ R3 ∧ C ++ 13 )·P(R−3 |L1 ∧R3 ∧C++13 ). And so the screening off we wanted to establish holds. San Pedro’s notation seems to be redundant, then. There is no great harm in keep- ing it, though; we will simply consider the existence of various screening-off factors for the correlations at a given pair of measurement settings. We can now formulate the well-known independence conditions. The screening off condition is also called outcome independence (OI, p. 147): P(L a i ∧ Rbj |Li ∧ Rj ∧ Cabij ) = P(Lai |Li ∧ Rj ∧ Cabij ) · P(Rbj |Li ∧ Rj ∧ Cabij ) (OI) The other two requirements are parameter independence (PI, p. 153) and measure- ment independence (MI, p. 147; also called no conspiracy), each of which consists of a symmetric pair of conditions: P(L a i |Li ∧ Cabij ) = P(Lai |Li ∧ Rj ∧ Cabij ) (PI) P(R b j |Rj ∧ Cabij ) = P(Rbj |Li ∧ Rj ∧ Cabij ) (PI) P(C ab ij ∧ Li) = P(Cabij ) · P(Li) (MI) P(C ab ij ∧ Rj ) = P(Cabij ) · P(Rj ) (MI) Notice that the notion of measurement independence used by San Pedro is relatively weak: it does not require the independence of Cab ij ’s and pairs of measure- ment settings, and neither the independence of measurement settings and Boolean combinations of Cab ij ’s (which are non-trivial for different pairs of measurement settings). Since I am commenting on a conjecture by San Pedro, I will keep his nomenclature. Apart from the disjointness assumption, i.e. that with fixed i and j, for a �= c, b �= d, Cab ij ∩ Cad ij = Cab ij ∩ Ccb ij = ∅, on p. 146 San Pedro sets down another requirement for Cab ij , from which it follows that C ab ij ⊂ Li ∩ Rj (SC) which I will call the subset condition.2 I will not discuss the philosophical advantages or disadvantages of the subset con- dition, but will notice that, by mere mathematics, it renders San Pedro’s conjecture 2 Following San Pedro, when writing about probability of the conjunction of two events I will use the “∧” sign, but when the context is clearly set-theoretical I will use “∩” instead. 364 Euro Jnl Phil Sci (2014) 4:361–368 less interesting that it (I think) would be if the condition was removed (I consider this option in Sections 3.1 and 3.2). First, San Pedro himself notices (p. 147) that his Cab ij ’s violate MI. Note that it is due to the subset condition and nothing else. It is elementary that if A ⊂ B and P(B) �= 1, then P(A) · P(B) �= P(A ∧ B). The subset condition gives us Cab ij ⊂ Li , and since there is more than one possible measurement setting in the left wing of the experiment, P(Li) �= 1. Therefore P(Cabij ∧ Li) �= P(Cabij ) · P(Li), that is, MI is violated. Second, notice that due to the subset condition (and, again, nothing else) PI uni- versally holds. This is because it follows immediately from the subset condition that Li ∩ Cabij = Li ∩ Rj ∩ Cabij (as both sides are identical to Cabij ). Therefore P(La i |Li ∧ Cabij ) = P(Lai |Li ∧ Rj ∧ Cabij ); similarly for the symmetric condition. Therefore PI always holds. 3 The conjecture and the argument What is the conjecture, then? To quote San Pedro (p. 152): Conjecture 1 If measurement independence is violated then parameter indepen- dence is also violated. By the light of the previous section, we have to say that the conjecture is simply false. If we use San Pedro’s definition, we cannot even contemplate a violation of PI. At this point one may be tempted to consider an alternative formulation of parameter independence; namely, for j �= k and h �= i: P(L a i |Li ∧ Rk ∧ Cabij ) = P(Lai |Li ∧ Rj ∧ Cabij ) (PI’) P(R b j |Lh ∧ Rj ∧ Cabij ) = P(Rbj |Li ∧ Rj ∧ Cabij ) (PI’) Notice, though, that due to the subset condition (and the fact that j �= k and h �= i) Li ∩ Rk ∩ Cabij = Lh ∩ Rj ∩ Cabij = ∅, so P(Li ∧ Rk ∧ Cabij ) = P(Lh ∧Rj ∧Cabij ) = 0, and thus the left-hand side probabilities in the PI’ conditions are undefined. So PI’ universally fails. On this formulation of parameter indepen- dence, the conjecture is true, but not interesting, akin to an implication with a previously known consequent. And again, the subset condition is the culprit. (From now on I will stick to PI as opposed to PI’, since it is the formulation used by San Pedro.) I suspect San Pedro himself had second thoughts about the subset condition, since he offered an argument in support of the conjecture, the falsity of which we have seen to follow directly from the subset condition. It might be interesting, then, to consider the conjecture without assuming SC. Euro Jnl Phil Sci (2014) 4:361–368 365 3.1 Without the subset condition: flaw in the argument Before we tackle the conjecture itself, let us look at the argument for it given by San Pedro on p. 153. Consider the following excerpt: A violation of measurement independence entails that (...) P(C ab ij ∧ Li) �= P(Cabij ) · P(Li), (18) P(C ab ij ∧ Rj ) �= P(Cabij ) · P(Rj ), (19) which entails that P(C ab ij ∧ Li ∧ Rj ) �= P(Cabij ) · P(Li ∧ Rj ), (20) as long as we assume that Li and Rj are probabilistically independent. Notice again that in the presence of SC (20) would need no argument. If we forget about SC, though, the entailment does not hold: it is possible to make (18) and (19) true while making (20) false, while keeping Li and Rj probabilistically independent. Consider the probability space with � = [0, 1]×[0, 1], the event algebra F being the smallest σ−algebra containing all “rectangles” of the form “a Borel subset of [0, 1] × a Borel subset of [0, 1]”, and the measure being the unique extension to the whole F of the function Pr defined on the rectangles so that Pr(A × B) = L(A) · L(B), where L is the Lebesgue measure. (In other words, consider a space defined over a square of area 1, with the probability of its measurable subsets being their area.) Now define the following:3 L1 = [0, 1] × [1/2, 1]; L2 = � \ L1; R1 = [0, 1/2] × [0, 1]; R2 = � \ R3; C ++ 12 = [0, 1/2] × [1/4, 1/2] ∪ [0, 1/2] × [7/8, 1] ∪ ∪ (1/2, 1] × [1/8, 1/2] ∪ (1/2, 1] × [3/8, 1]. The Reader may check that in this situation, after substituting L1 for Li , R2 for Rj , and C++12 for C ab ij , inequalities (18) and (19) hold, “Li and Rj are probabilistically independent”, but inequality (20) is false. Of course, San Pedro only intended the argument to inspire more confidence in the truth of the conjecture, without giving an exact proof (he would not call it a “conjecture” in that case!); nonetheless, I think it is fitting when arguing against a conjecture supported by an argument to also point to a flaw in the argument. I can now turn to presenting the method of constructing counterexamples to San Pedro’s conjecture. 3 For brevity I am defining the space for two possible measurement settings at each detector (San Pedro considers three) but it should be obvious how to extend the example to cover more settings. 366 Euro Jnl Phil Sci (2014) 4:361–368 3.2 Without the subset condition: falsity of the conjecture In this section I will show how, given a probability space modelling some set of EPR correlations, that is, containing the events Li , Rj , L a i and Rb j , to add to it events Cab ij so that MI is violated while PI is satisfied. The construction will also satisfy OI. For an easier argument (saving us from an additional argument by cases), assume the (anti-)correlations are not perfect, i.e. that any combination of La i ∧ Rb j occurs with a non-zero probability. We can without loss of generality assume the probability space is atomless; that is, given an event A with P(A) > 0, we can always find an event B such that B ⊆ A and P(A) > P(B) > 0. We do not lose any generality because any classical probability space can be embedded in an atomless classical probability space (see chapter 4 of Wroński 2010 for a short proof). I will also again assume that we have two possible settings at each of the detectors. This is just for brevity, the reasoning goes through for any finite number of settings. I will label the directions for the left wing “1” and “2”, and the other two “3” and “4”. Therefore the events of selecting the measurement directions are L1, L2, R3 and R4. We can think of the sample space as divided into four “quadrants” of the form Li ∩ Rj . So to define each Cabij we need to set its intersection with each quadrant, making sure the resulting event satisfies OI and PI. For example, for any a, b ∈ {+, −}, Cab13 should screen off all the (anti-)correlations between Lc1 and R d 3 (for any c, d ∈ {+, −}); also, the chance of Lc1 should be the same conditional on L1 ∧ Cab13 as conditional on L1 ∧ R3 ∧ Cab13 , and so on. The key observation is the following: for any Cab ij , making sure it satisfies OI and PI only requires paying attention to its intersection with three quadrants. For example, in the case of C++13 , we need to specify its intersection with L1 ∩ R3 (for OI to work) and with L1 ∩ R4 and L2 ∩ R3 (for PI to work). Its intersection with the last remaining quadrant, L2 ∩ R4, is irrelevant for the purposes of establishing OI and PI. Once the intersections of C++13 with three quadrants are fixed, MI might already be violated. But if it is not, then there is at most one num- ber fit for the probability of P(C++13 ∧ L2 ∧ R4) if MI is to be sustained; label the number r. But since we are (due to atomlessness) at our complete liberty when it comes to choosing the probability of P(C++13 ∧ L2 ∧ R4), we can choose the event so that its probability does not equal r, therefore violating MI. (In all but some very exceptional cases,4 setting the intersection C++13 ∩ L2 ∩ R4 to be the empty set will be enough to the trick.) And so what is really needed is only the way of defin- ing C++13 ’s intersection with the three abovementioned quadrants so that OI and PI are satisfied. This we will achieve in the following way. For each pair of detector settings i, j we will consider four events: C++ ij , C−+ ij , C+− ij , and C−− ij . The intuitive meaning is 4 Exactly: if the numerator of the fraction at the right-hand side of the identity (**) below is not equal to 0. Euro Jnl Phil Sci (2014) 4:361–368 367 that Cab ij should guarantee5, if we set the left detector to i and the right detector to j, the occurence of La i ∧ Rb j . In the particular case of C++13 we will achieve this (which means that OI is satisfied), as well as the satisfaction of PI, by setting the intersection of C++13 with the three relevant quadrants in the following way: C ++ 13 ∩ L1 := ( L + 1 ∩ R+3 ) ∪ (L+1 ∩ R+4 ) , C ++ 13 ∩ L2 ∩ R3 := L+2 ∩ R+3 . With C+−13 we set the intersections as follows: C +− 13 ∩ L1 := ( L + 1 ∩ R−3 ) ∪ (L+1 ∩ R−4 ) , C +− 13 ∩ L2 ∩ R3 := L+2 ∩ R−3 , with the remaining two cases dealt with analogously. It is elementary that the Cab13 ’s defined in this way satisfy OI and PI no matter what their intersection with L2 ∩ R4 is. We now need to ensure the violation of MI. Notice that MI requires that P(C ++ 13 ) · P(L2) = P(C++13 ∧ L2 ∧ R3) + P(C++13 ∧ L2 ∧ R4), which we can transform into P(C ++ 13 ∧ L2 ∧ R4) = ( P(C ++ 13 ∧ R3) + P(C++13 ∧ L1 ∧ R4) ) · P(L2) − P(C++13 ∧ L2 ∧ R3) P(L1) (**) Notice that the values of all the terms on the right hand side of the equality are known. Label the value of the fraction r. We now know that if we set the intersection C ++ 13 ∩ L2 ∩ R4 so that its probability is different than r, MI will be violated. Thanks to atomlessness, there are infinitely many ways of doing so.6 At this point we have secured the violation of MI. As I have already stated, all Cab13 ’s satisfy OI and PI. Now all that needs to be done is to repeat the above construction for Cab14 ’s, C ab 23 , and C ab 24 , changing the role of the quadrants accordingly. Notice also that, for any i and j, nothing prevents us from setting the events in the “irrelevant” quadrant so that the Cab ij ’s form a partition of the sample space. We now know that if we remove the subset condition, the conjecture is false. I have presented the construction for imperfect (anti-)correlations, but it should be evident how to transform it into one fit for that special case. 5 Please notice that this interpretaion of Cab ij is intended only in my construction and not in San Pedro’s approach (though it of course permits it); in my counterexamples the Cab ij act as “deterministic” screening- off factors. San Pedro does not require this in his framework. It is just that I think using screening-off factors which are deterministic is the simplest way to find counterexamples. 6 If we do not assume atomlessness, we can just set the intersection to be the empty set or, in the unlikely case in which r = 0, to be the full quadrant L2 ∩ R4; then we would set the intersection of C+−13 , C−+13 , and C−−13 with L2 ∩ R4 to be empty. 368 Euro Jnl Phil Sci (2014) 4:361–368 4 Conclusions In the previous section I argued that San Pedro’s conjecture, taken as a mathematical statement (since its proponent himself gives a mathematical argument in its support) is false. I showed that if we keep San Pedro’s definitions, the conjecture fails imme- diately; while if we remove one of the conditions, it is still possible to provide a method for obtaining (infinitely many) counterexamples. Now, the counterexamples are mathematical constructions. It would be interesting if there were philosophical reasons for rejecting them, especially if some of the reasons could arguably result in additional formal requirements which my construction fails to satisfy. I would welcome such a development; as it stands, though, San Pedro’s conjecture is false. Acknowledgments I would like to thank Juliusz Doboszewski, Michał Marczyk and Tomasz Placek for commenting on the initial version of the manuscript and also the two anonymous reviewers for their helpful comments. My research has been aided by the Foundation for Polish Science “START” Fellowship and also by the Foundation for Polish Science “MISTRZ” Fellowship received by Tomasz Placek. Open Access This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited. References Hofer-Szabó, G., Rédei, M., Szabó, L.E. (2013). The common cause principle. Cambridge University Press. San Pedro, I. (2012). Causation, measurement relevance and no-conspiracy in EPR. European Journal for Philosophy of Science, 2, 137–156. Wroński, L. (2010). The common cause principle. Explanation via screening off. PhD thesis, Jagiellonian University, Kraków. Archived at http://jagiellonian.academia.edu/LeszekWroński. Forthcoming as a book from De Gruyter Open. http://jagiellonian.academia.edu/LeszekWro%C5%84ski On a conjecture by San Pedro Abstract Introduction The formal setup The conjecture and the argument Without the subset condition: flaw in the argument Without the subset condition: falsity of the conjecture Conclusions Acknowledgments Open Access References