Agreement between observers: a physical principle?
Patricia Contreras-Tejada, Giannicola Scarpa, Aleksander M. Kubicki, Adam Brandenburger, Pierfranceso La Mura
aa r X i v : . [ qu a n t - ph ] F e b Agreement between observers:a physical principle?
Patricia Contreras-Tejada ∗† Giannicola Scarpa ∗‡ Aleksander M. Kubicki § Adam Brandenburger ¶ Pierfrancesco La Mura †† Abstract
Is the world quantum? An active research line in quantum foundations is devotedto exploring what constraints can rule out the post-quantum theories that areconsistent with experimentally observed results. We explore this question in thecontext of epistemics, and ask whether agreement between observers can serve as aphysical principle that must hold for any theory of the world. Aumann’s seminalAgreement Theorem states that two (classical) agents cannot agree to disagree. Wepropose an extension of this theorem to no-signaling settings. In particular, weestablish an Agreement Theorem for quantum agents, while we construct examplesof (post-quantum) no-signaling boxes where agents can agree to disagree. The PRbox is an extremal instance of this phenomenon. These results make it plausiblethat agreement between observers might be a physical principle, while they alsoestablish links between the fields of epistemics and quantum information that seemworthy of further exploration.
Quantum mechanics famously made its creators uncomfortable. It is highlycounterintuitive and, almost a century after its introduction, it still sparks muchconceptual and philosophical discussion. Indeed, an active line of research inquantum foundations deals with the problem of singling out quantum theory fromother post-classical physical theories. This field is a delicate balance betweenproposals for new theories that are ‘tidier’ than quantum mechanics [28, 18]and proposals for desirable physical principles that such theories should obey[24, 9, 23, 29, 31].In this paper, we propose a new principle inspired by a famous result inepistemics, which is the formal study of knowledge and beliefs. In the domainof classical probability theory, Aumann proved that Bayesian agents cannot agreeto disagree [5]. A slightly more general restatement of Aumann’s theorem, which we ∗ These authors contributed equally to this work. † Instituto de Ciencias Matemáticas, 28049 Madrid, Spain, [email protected] ‡ Universidad Politécnica de Madrid, 28040 Madrid, Spain, [email protected] § Universidad Complutense de Madrid, 28040 Madrid, Spain, [email protected] ¶ †† HHL Leipzig Graduate School of Management, 04109 Leipzig, Germany, [email protected] ill refer to as the classical agreement theorem, states that, if Alice and Bob, basedon their partial information, assign probabilities q A , q B , respectively, to perfectlycorrelated events, and these probabilities are common certainty between them, then q A = q B . “Certainty” means assigning probability 1, and “common certainty” meansthat Alice is certain about q B ; Bob is certain about q A ; Alice is certain about Bobbeing certain about q A ; Bob is certain about Alice being certain about q A ; and soon infinitely.This result is considered a basic requirement in classical epistemics, and wecontend it should apply to all physical theories. The classical agreement theoremhas been used to show that two risk-neutral agents, starting from a common prior,cannot agree to bet with each other [27], to prove “no-trade” theorems for efficientmarkets [20], and to establish epistemic conditions for Nash equilibrium [6]. Theseapplications are all external to physics. Of course, the theorem holds equally in thephysical domain, provided that classical probability theory applies.But in the quantum domain, the classical model does not apply, and so wecannot assume that the same facts about agreement and disagreement betweenBayesian agents hold when they observe quantum phenomena. In particular, afundamental result of quantum mechanics is that no local hidden-variable theorycan model the results of all quantum experiments [8]. This implies that the classicalBayesian model does not apply, so the classical agreement theorem need not hold.The question then arises: can observers of quantum mechanical phenomena agreeto disagree? We address this question by exploring it in the broader no-signalingsetting.First, we establish that, in general, no-signaling agents can agree to disagreeabout perfectly correlated events, and we give explicit examples of disagreeing no-signaling distributions. In the particular case of two inputs and two outputs, wecharacterize the distributions that give rise to common certainty of disagreement.One might think that the fact that no-signaling agents can agree to disagree is adirect consequence of the multitude of uncertainty relations in quantum mechanics,all of which put a limit on the precision with which the values of incompatibleobservables can be measured and have even been linked to epistemic inconsistenciesin quantum mechanics [12]. Somewhat surprisingly, our next finding shows that thisis not the case. We find that disagreeing no-signaling distributions of two inputs andoutputs cannot be quantum—i.e., the agreement theorem holds for quantum agentsin this setting. Then, we go beyond this restriction and show that any disagreeingno-signaling distribution with more than two inputs or outputs induces a disagreeingdistribution with two inputs and outputs. Since the agreement theorem holds forquantum agents sharing distributions of two inputs and outputs, it does so for largerdistributions too. Thus, even if quantum mechanics features uncertainty relations,this does not apply to observers’ estimations of perfectly correlated events.Next, we ask if no-signaling and quantum agents can disagree in other ways.We define a new notion of disagreement, which we call singular disagreement , byremoving the requirement of common certainty and, instead, imposing q A = 1 ,q B = 0 , and we ask whether it holds for classical, quantum and no-signaling agents.We find the same pattern: singular disagreement does not hold for classical orquantum agents, but can occur in no-signaling settings, where we characterize thedistributions that feature it. We then put our two characterizations together andsearch for distributions that satisfy both common certainty of disagreement andsingular disagreement: we find that the PR box [24] is of this kind—i.e., it displaysextremal disagreement in the above sense. This is neat, as the PR box is known toexhibit the most extreme form of no-signaling correlations [7]. inally, we give further evidence that agreement between observers could be aconvenient principle for testing the consistency of new postquantum theories. Ourresults exhibit a clear parametrization of the set of the probability distributionsthat allow observer disagreement. This set is easy to work with, thanks to itsrestriction to two observations and two outcomes per observer. If a new theorycan be used to generate such a distribution, this might raise a red flag, as thistheory violates a reasonable and intuitive and, importantly, testable property thatquantum mechanics satisfies. As an example, we analyze through numerical searchthe “almost quantum correlations” [21], which were designed to obey all previouslyproposed physical principles. We find that some of these correlations can exhibitdisagreement. This new finding adds to the doubts that have been cast on theirusefulness [26, 14].Aumann’s theorem has appeared elsewhere in the physics literature. However, ithas been examined in a different context [16, 17], where agents are assumed to useBorn’s rule as their probability update rule. The authors conclude that Aumann’stheorem does not hold for this type of agents. Instead, our setting assumes that theagents are macroscopic and merely share a quantum state or a no-signaling box. Oursetting is appealing in quantum information for its applications to communicationcomplexity, cryptography, teleportation, and many other scenarios. In turn, Ref.[4] introduces a different notion of disagreement in a no-signaling context. Thedisagreement in that work concerns pieces of information about some variables, andagreement refers to consistency in the information provided about the variables.Hence, it is unrelated to the epistemic notion of disagreement that Aumann’stheorem defines, and that the present work revisits from a no-signaling perspective. Aumann’s theorem [5] is formulated on a probability space, and partial informationof the observers is represented by different partitions of the space. Each observerknows which of their partition elements obtains, and estimates the probabilityof an event of interest by Bayesian inference. We refer to a probability space,together with some given partitions, as a (classical) ontological model . Ontologicalmodels appearing in the literature (see, e.g., [11]) also contain a set of preparationsunderlying the distribution over the state space, and the partitions are usuallyphrased in terms of measurements and outcomes. However, we consider preparationsimplicit and use the language of partitions to bridge the gap between classicalprobability spaces and no-signaling boxes more smoothly.For the sake of simplicity and following Aumann, we will restrict our analysisto two observers, Alice and Bob. Aumann’s original theorem considers common knowledge about one single event of interest to both observers. We provide aslight generalization with common certainty about two perfectly correlated eventsof interest, one for each observer. This allows us to jump into the framework ofno-signaling boxes that we will use later. We refer to this generalization as theclassical agreement theorem. This nomenclature can be further motivated by thefact that, for purely classical situations, both statements—the original Aumann’stheorem and our formulation with perfectly correlated events—can be proven to be equivalent (as long as states of the world with null probability are ignored, as inRef. [1]).Consider a probability space (Ω , E , P ) where Ω is a set of possible states of the orld; E is its power set (i.e., the set of events); and P is a probability measureover Ω . We will consider two events E A , E B ∈ E of interest to Bob and Alice,respectively (the choice of subscripts will become clear later). We will assume thatthey are perfectly correlated : P ( E A \ E B ) = P ( E B \ E A ) = 0 . Fix partitions P A , P B of Ω for Alice and Bob, respectively. For convenience,assume that all members of the join (coarsest common refinement) of P A and P B are non-null. For a state ω ∈ Ω , P A,B ( ω ) is the partition element (of Alice’s orBob’s, respectively) that contains ω. For each n ∈ N , fix numbers q A , q B ∈ [0 , andconsider the following sets: A = { ω ∈ Ω : P ( E B |P A ( ω )) = q A } ,B = { ω ∈ Ω : P ( E A |P B ( ω )) = q B } ,A n +1 = { ω ∈ A n : P ( B n |P A ( ω )) = 1 } ,B n +1 = { ω ∈ B n : P ( A n |P B ( ω )) = 1 } . (1)Here, the set A is the set of states ω such that Alice assigns probability q A to event E B ; the set B is the set of states ω such that Bob assigns probability q B to event E A and probability 1 to the states in A —i.e., states where Bob assigns probability q B to E A and is certain that Alice assigned probability q A to E B ; and so on, andsimilarly for B , A , etc.It is common certainty at a state ω ∗ ∈ Ω that Alice assigns probability q A to E B and that Bob assigns probability q B to E A if ω ∗ ∈ A n ∩ B n ∀ n ∈ N . (2)If equation (2) does not hold for all n ∈ N , , but, instead, only for n ≤ N for acertain N ∈ N , then we talk about N th-order mutual certainty.We now state and prove the classical agreement theorem that will be the basisof our work: Theorem 1.
Fix a probability space (Ω , E , P ) , where E A and E B are perfectlycorrelated events. If it is common certainty at a state ω ∗ ∈ Ω that Alice assignsprobability q A to E B and Bob assigns probability q B to E A , then q A = q B . The full proof is given in the Supplementary Material. The main idea is to noticethat, since Ω is finite, there is a finite N ∈ N such that, for all n ≥ N, A n +1 = A n and B n +1 = B n . Using the definition of A N +1 , noticing that A N is a union ofAlice’s partition elements, and using convex combination arguments together withthe perfect correlations between E A and E B leads to P ( E A ∩ E B | A N ∩ B N ) = q A . (3)Running the parallel argument for Bob entails that the same expression is equal to q B , proving the claim. We now map the classical agreement theorem into the no-signaling framework, inorder to explore its applicability beyond the classical realm.We consider no-signaling distributions, or boxes [24], of the form { p ( ab | xy ) } a ∈A ,b ∈B ,x ∈X ,y ∈Y , (4)where x, y and a, b are Alice’s and Bob’s input and output, respectively, and A , B , X , Y are index sets, not necessarily of the same size. no-signaling box is local if, for each a, b, x, y,p ( ab | xy ) = X λ p A ( a | xλ ) p B ( b | yλ ) , (5)for some distributions { p A ( a | xλ ) } ( a,x,λ ) ∈A×X × Λ , { p B ( b | yλ ) } ( b,y,λ ) ∈B×Y× Λ where Λ isan index set. It is quantum if, for each x, y , there exist POVMs { E ax } a ∈A , { F by } b ∈B and a quantum state ρ such that, for each a, b , p ( ab | xy ) = tr (cid:16) E ax ⊗ F by ρ (cid:17) . (6)The set of local boxes is strictly included in the set of quantum boxes, which is, inturn, strictly included in the set of no-signaling boxes. No-signaling boxes that arenot quantum are termed post-quantum .We now show that we can associate a no-signaling box with any ontologicalmodel, and vice versa. Let A , B , X , Y be index sets. Let (Ω , F , P ) be a probabilityspace, and, for each x ∈ X , let A ax be a partition of the states ω ∈ Ω where a ∈ A denotes the partition elements. Similarly, for each y ∈ Y , let B by be anotherpartition of the states ω ∈ Ω , where b ∈ B denotes the partition elements. Accordingto that, we can understand labels x ∈ X , y ∈ Y as inputs —this information fixeswhat partition Alice and Bob look at—and a ∈ A , b ∈ B as outputs —this is theinformation that the agents gain by observing their corresponding partitions. Thisjargon will shortly become very natural.With all the above, (cid:8) (Ω , F , P ) , { A ax , B by } a,b,x,y (cid:9) is an ontological model that wenow want to associate to a no-signaling box that reproduces its statistics. In thisontological model, given inputs x ∈ X , y ∈ Y , the probability of obtaining outputs a ∈ A , b ∈ B is given by P ( A ax ∩ B by ) . This simple observation leads us to constructthe no-signaling box p ( a, b | x, y ) := P (cid:0) A ax ∩ B by (cid:1) , ∀ ( a, b, x, y ) ∈ A × B × X × Y . (7)It can be verified that the probabilities p are non-negative, normalized, and no-signaling.The converse process of finding an ontological model starting from a no-signalingbox can be also performed, as we show in the Supplementary Material. Remarkably,this can be accomplished even in the case in which the no-signaling box is non-local,obtaining an ontological model with a quasi-probability measure instead of standardpositive probabilities [2]. (The appearance of quasi-probabilities here should notsurprise the reader. In fact, one cannot hope to obtain ontological models withonly non-negative probabilities for post-classical no-signaling boxes, as this wouldprovide local hidden-variable models that contradict, for instance, Bell’s theorem.In any case, the use of this mathematical tool has been well rooted in the study ofquantum mechanics since its origins—see [11] for a nice review of this subject. Thismakes it possible to translate results from one framework to the other, somethingthat might be of interest in order to establish further connections between epistemicsand quantum theory. However, from now on, we focus on no-signaling boxes andleave this digression aside in the rest of the main text.With the mapping between ontological models and no-signaling boxes in mind,we next define common certainty of disagreement for no-signaling boxes. The ideais to reinterpret the definitions in Section 2.1 in this latter setting.We first provide meaning for the events of interest (previously identified as E A , E B ) in the present setting. Now, these events correspond to some set of outcomes,given that the no-signaling box was queried with some particular inputs. For the ake of concreteness, we fix these inputs to be x = 1 , y = 1 and the outcomesof interest to be a = 1 , b = 1 . This motivates us to consider the events F A = { (1 , b, , y ) } b ∈B ,y ∈Y (on Alice’s side) and F B = { ( a, , x, } a ∈A ,x ∈X (on Bob’s side).Then, we say that F A and F B are perfectly correlated when p ( a, b | x = 1 , y = 1) = 0 for all a = b. (8)Given this, we assume that the agents will actually conduct their measurementsaccording to some other partitions. Again, for concreteness, let us assume thatthose partitions are the ones associated with inputs x = 0 , y = 0 . These inputstake on the role of partitions P A , P B in the ontological model picture. The outputsobtained from these measurements are the no-signaling box analogue to the events P A ( ω ) , P B ( ω ) . In order to make the following expressions more concrete, we assume,when x = 0 , y = 0 are inputted, that the outputs obtained are a = 0 and b = 0 ,respectively.Therefore, given the perfectly correlated events F A , F B and numbers q A , q B ∈ [0 , , we define the sets α = { a : p ( b = 1 | a, x = 0 , y = 1) = q A } , (9) β = { b : p ( a = 1 | b, x = 1 , y = 0) = q B } , (10)and, for all n ≥ , α n +1 = { a ∈ α n : p ( B n | a, x = 0 , y = 0) = 1 } , (11) β n +1 = { b ∈ β n : p ( A n | b, x = 0 , y = 0) = 1 } , (12)where A n = α n × B × X × Y , (13) B n = A × β n × X × Y . (14)By analogy with the sets in equation (1), the set α is the set of Alice’s outcomessuch that she assigns probability q A to F B , having input x = 0 . The set β is theset of Bob’s outcomes such that he is certain that Alice assigned probability q A to F B , and so on, and similarly for β , α , etc.Suppose that Alice and Bob both input 0 and get output 0. Then, there is common certainty of disagreement about the event that Alice assigns probability q A to F B and Bob assigns probability q B to F A if q A = q B and ( a = 0 , b = 0 , x = 0 , y = 0) ∈ A n ∩ B n ∀ n ∈ N . (15)Notice the relationship between this definition and the previous one: the ω ∗ inequation (2), at which the disagreement occurred, fixed the partition elements thatAlice and Bob observed. Here, disagreement occurs at the inputs and outputs ( a = 0 , b = 0 , x = 0 , y = 0) that the agents obtain.The following is now a corollary of Theorem 1: Corollary 2.
Suppose that Alice and Bob share a local no-signaling box withunderlying probability distribution p . Let q A , q B ∈ [0 , , and let p ( b = 1 | a = 0 , x = 0 , y = 1) = q A ,p ( a = 1 | b = 0 , x = 1 , y = 0) = q B . (16) If q A and q B are common certainty between the agents, then q A = q B . In the Supplementary Material, we give a standalone proof of this result.Moreover, Theorems 1 and 2 can be shown to be equivalent. We now ask whetherthis theorem holds in quantum and no-signaling settings. .3 Quantum agents cannot agree to disagree Given the mapping exhibited above, as well as the restatement of the agreementtheorem for local boxes, it is now natural to ask whether the theorem holds whendropping the locality constraint. When we generalize the setting and allow theagents to share a generic no-signaling box, we find that the agreement theorem doesnot hold. That is, no-signaling observers can agree to disagree, and we characterizethe distributions that give rise to common certainty of disagreement. Then, we findthat no such distribution can be quantum—i.e., quantum observers cannot agree todisagree.We first present the following theorem in which the no-signaling box has twoinputs and two outputs, but we will show in Proposition 5 that the result is fullygeneral. In place of “common certainty of disagreement about the event that Aliceassigns probability q A to F B = { ( a, , x, } a ∈A ,x ∈X and Bob assigns probability q B to F A = { (1 , b, , y ) } b ∈B ,y ∈Y , at event (0 , , , ,” we simply say “common certaintyof disagreement.” Theorem 3.
A 2-input 2-output no-signaling box gives rise to common certainty ofdisagreement if and only if it takes the form of Table 1. xy \ ab
00 01 10 1100 r − r r − s s − r + t + s − t − s t − u u r − t + u − r − u t − t Table 1: Parametrization of 2-input 2-output no-signaling boxes withcommon certainty of disagreement. Here, r, s, t, u ∈ [0 , are such thatall the entries of the box are non-negative, and s − u = r − t. Here, we provide an outline of the proof; the full proof is provided in theSupplementary Material.To prove the direct implication, we first consider the case in which, for some n onwards, α n , β n each contain only one output, a = 0 , b = 0 , respectively. Bythe definition of α n +1 , this implies that p ( B n | a = 0 , x = 0 , y = 0) = 1 , and,thus, p (01 |
00) = 0 . Similarly, the definition of β n +1 gives p (10 |
00) = 0 . Perfectcorrelations in the inputs x = 1 , y = 1 imply that p (01 |
11) = p (10 |
11) = 0 , and therest of the table is deduced in terms of parameters r, s, t, u by using no-signalingand normalization constraints. Finally, q A = q B if and only if s − u = r − t, whichconcludes the proof of this case.If, for all n, one or both of α n , β n contain(s) both outputs, we find q A = q B , contradicting common certainty of disagreement.The converse implication is proved by writing q A , q B in terms of the parametersof the box. If α = { a = 0 } and β = { b = 0 } , then we find α = α and β = β , and common certainty of disagreement follows.If the parameters of the box are such that α = { a = 0 , a = 1 } but β = { b = 0 } ,then the definition of β implies p ( b = 0 | a = 0 , x = 0 , y = 0) = 1 ; therefore, (0 , , , ∈ A , and common certainty of disagreement follows. One can reasonsymmetrically if α = { a = 0 } but β = { b = 0 , b = 1 } . Finally, if both α and β are the full set of outcomes, then s − u = r − t, contradicting the statement of theTheorem. hile some no-signaling distributions can exhibit common certainty ofdisagreement, we find that probability distributions arising in quantum mechanicsdo satisfy the agreement theorem. Theorem 4.
No quantum box of 2 inputs and 2 outputs can give rise to commoncertainty of disagreement.Proof.
In order to give rise to common certainty of disagreement, the probabilitydistribution that the state and measurements generate must be of the form of Table1. By Tsirelson’s theorem [30], if there is a quantum realization of the box, thenthere exist real, unit vectors | w x i , | v y i (17)such that the correlations c xy := p ( a = b | xy ) − p ( a = b | xy ) (18)satisfy c xy = h w x | v y i (19)for each x, y . For the box in Theorem 3, this means, in particular, that h w | v i = 1 , h w | v i = 1 , (20)and, since the vectors have unit norm, this implies that | w i = | v i , | w i = | v i . (21)Then, we are left with c = h w | w i ,c = h w | w i . (22)Since the vectors are real, we find c = c , (23)but this implies that s − u = r − t, (24)which implies that q A = q B and, hence, impedes disagreement.We have seen that no 2-input 2-output quantum box can give rise to commoncertainty of disagreement. We now lift the restriction on the number of inputsand outputs and show that no quantum box can give rise to common certainty ofdisagreement.First, notice that the proof for 2 inputs and outputs did not require common certainty, but only first-order mutual certainty. Indeed, by observing the definitionsof the sets α n , β n , one can see that α n = α and β n = β for all n ≥ . This meansthat first-order mutual certainty implies common certainty, and, therefore, first-order certainty suffices to characterize the no-signaling box that displays commoncertainty of disagreement.As the number of outputs grows, first-order mutual certainty is no longersufficient. However, since the number of outputs is always finite, there exists an N ∈ N such that α n = α N and β n = β N for all n ≥ N . Since α n +1 ⊆ α n ∀ n , nd similarly for β , the sets α N , β N are the smallest sets of outputs for which thedisagreement occurs. Because of this, any ( a, b, x, y ) that belongs to A N ∩ B N will also belong to A n ∩ B n for all n ; that is, N th-order mutual certainty impliescommon certainty. So, for any finite no-signaling box, one needs only N th-ordermutual certainty to characterize it. As the number of outputs grows unboundedly,one needs common certainty to hold [13]. These observations will be relevant toextending Theorem 4 beyond two inputs and outputs. Theorem 5.
No quantum box can give rise to common certainty of disagreement.
We show that any no-signaling box with common certainty of disagreementinduces a 2-input 2-output no-signaling box with the same property. Thus, if thereexisted a quantum system that could generate the bigger box, it could also generatethe smaller box. Then, Theorem 3 implies that no quantum box can give rise tocommon certainty of disagreement.To show the reduction of the box, consider an n -input m -output no-signaling box displaying common certainty of disagreement. Let this box—referred to in the following as the original box —be denoted by the distribution { p ( ab | xy ) } a ∈A ,b ∈B ,x ∈X ,y ∈Y , where A = B = { , . . . , m } , X = Y = { , . . . , n } . Ouraim now is to construct a 2-input 2-output no-signaling box n ˜ p (˜ a ˜ b | ˜ x ˜ y ) o ˜ a, ˜ b, ˜ x, ˜ y ∈{ , } —referred to as the new box —also displaying common certainty of disagreement vialocal pre- and post-processing of the inputs and outputs of the original box.Without loss of generality, let us assume that the original box displays commoncertainty of disagreement at event (0 , , , about (1 , , , . We start by definingthe inputs ˜ x, ˜ y ∈ { , } of the new box. For that, we just consider the original boxwhen its inputs are restricted to or . This gives rise to the 2-input m -outputno-signaling box { p ( ab | ˜ x ˜ y ) } a ∈A ,b ∈B , ˜ x, ˜ y ∈{ , } . In the next paragraph, we refer to thisbox as intermediate .Given the above intermediate box, we now post-process its outputs a ∈ A , b ∈ B to obtain new dichotomic outcomes ˜ a, ˜ b ∈ { , } . Recalling the discussion precedingTheorem 5, for the intermediate no-signaling box { p ( ab | ˜ x ˜ y ) } a ∈A ,b ∈B , ˜ x, ˜ y ∈{ , } thereexists an N ∈ N such that α n = α N and β n = β N for all n ≥ N . Therefore, wefocus on sets α N , β N to group the outcomes a ∈ A , b ∈ B in such a way that theresulting box still displays common certainty of disagreement. For that, Alice, wheninputting ˜ x = 0 , assigns value ˜ a = 0 to outcomes a ∈ α N and ˜ a = 1 to the rest. Bobapplies a similar post-processing for his outcome based on β N instead of α N . WhenAlice inputs ˜ x = 1 , she assigns the value to ˜ a when output a = 1 is obtainedand value otherwise. Similarly, Bob, when inputting ˜ y = 1 , assigns the value to ˜ b when output b = 1 is obtained and value in all other cases. The conditionalprobability distribution of outcomes ˜ a , ˜ b , given inputs ˜ x , ˜ y , is the desired new box, n ˜ p (˜ a ˜ b | ˜ x ˜ y ) o ˜ a, ˜ b, ˜ x, ˜ y ∈{ , } .Finally, we check that the new box constructed in this way verifies the claimedproperties. First, since the reduction from the original box consists of local pre- andpost-processing, the resulting box ˜ p is well normalized and still no-signaling. Hence,what remains to be shown is that ˜ p displays common certainty of disagreement.We start by checking that inputs ˜ x = 1 , ˜ y = 1 have perfectly correlatedoutputs in ˜ p . That is straightforwardly done by considering how ˜ a was definedand that inputs x = 1 , y = 1 have perfectly correlated outputs in the originalbox. Accordingly, ˜ p (0 , | ,
1) = P a =1 p ( a, | ,
1) = 0 and, similarly, ˜ p (1 , | ,
1) = P b =1 p (1 , b | ,
1) = 0 . ext, we check that the corresponding probabilities ˜ q A , ˜ q B in ˜ p are different. Byhypothesis, q A = p ( b = 1 | a ∗ , x = 0 , y = 1) = p ( a = 1 | b ∗ , x = 1 , y = 0) = q B (25)holds for all a ∗ ∈ α and b ∗ ∈ β . Since α N ⊆ α and β N ⊆ β , equation (25) holdsin particular for all a ∗ ∈ α N and b ∗ ∈ β N . By using the definition of conditionalprobabilities to unpack q A , q B and summing over a ∗ ∈ α N on the left and b ∗ ∈ β N on the right, we arrive at the conclusion that ˜ p (˜ b = 1 | ˜ a, ˜ x = 0 , ˜ y = 1) = ˜ p (˜ a = 1 | ˜ b, ˜ x = 1 , ˜ y = 0) , (26)as required.To finish, we note that, for inputs ˜ x = 0 , ˜ y = 0 , outputs ˜ a , ˜ b were defined insuch a way that sets α n , β n for the new box are identically equal to { } for any n ∈ N . Recalling (15), this is enough to conclude the proof. See the SupplementaryInformation for further details omitted from this sketch of the proof. We explore other forms of disagreement that might arise about perfectly correlatedevents. Since common certainty is a strong requirement, we remove it and, instead,suppose that the agents assign probabilities that differ maximally. We find thatthis new notion of disagreement exhibits the same behaviour as common certaintyof disagreement.In a no-signaling box, there is singular disagreement about the probabilitiesassigned by Alice and Bob to perfectly correlated events F A = { (1 , b, , y ) } b ∈B ,y ∈Y and F B = { ( a, , x, } a ∈A ,x ∈X , respectively, at event (0 , , , if it holds that q A = 1 , q B = 0 . (27)This time, there is no notion of common certainty—we just require that Alice’s andBob’s assignments differ maximally.Similarly to the previous section, we refer to the above definition simply as“singular disagreement.”We restrict ourselves first to boxes of two inputs and outputs and show that localboxes cannot exhibit singular disagreement. Then, we characterize the no-signalingboxes that do satisfy singular disagreement and show they cannot be quantum.Finally, we generalize to boxes of any number of inputs and outputs. Theorem 6.
There is no local 2-input 2-output box that gives rise to singulardisagreement.
It turns out that singular disagreement induces a Hardy paradox in the system;therefore it cannot be local. The full proof is given in the Supplementary Material.We now lift the local restriction and characterize the no-signaling boxes in whichsingular disagreement occurs.
Theorem 7.
A 2-input 2-output no-signaling box gives rise to singular disagreementif and only if it takes the form of Table 2.Proof.
First, we show that singular disagreement implies that the no-signaling boxmust be of the above form. By construction, the inputs x = y = 1 have perfectlycorrelated outputs, so that p (01 |
11) = p (10 |
11) = 0 . (28) y \ ab
00 01 10 1100 s t − s − u − t u
01 0 s + t r − s − t − r − u − t u + t + r − − r r − r Table 2: Parametrization of 2-input 2-output no-signaling boxes withsingular disagreement. Here, r, s, t, u, ∈ [0 , are such that all theentries of the box are non-negative, s + t = 0 and u + t = 1 . Also, singular disagreement requires p ( b = 1 | a = 0 , x = 0 , y = 1) = 1 , (29) p ( a = 1 | b = 0 , x = 1 , y = 0) = 0 . (30)Equation (29) implies that p (00 |
01) = 0 and p (01 | = 0 , while Equation (30)implies that p (10 |
10) = 0 and p (00 | = 0 . The rest of the entries follow fromnormalization and no-signaling conditions. Therefore, any two-input two-outputno-signaling box that gives rise to singular disagreement must be of the above form.Proving the converse is straightforward, as it suffices to check that Equations(29) and (30) are satisfied for the parameters of the box.However, a box like the one in Table 2 cannot be quantum: Theorem 8.
No 2-input and 2-output quantum box can give rise to singulardisagreement.Proof.
Due to their form, the boxes in Theorem 7 are quantum voids [25]; i.e., theyare either local or post-quantum. This can be seen by observing that the mapping x x ⊕ , (31)which is a symmetry of the box, makes all four 0’s lie in entries p ( ab | xy ) such that a ⊕ b ⊕ xy . As stated in Sections III and V.B of Ref. [25], all boxes with four0’s in entries of the above form are quantum voids.Therefore, the box in Table 2 is either local, in which case it does not lead tosingular disagreement, or has no possible quantum realization, proving the claim.Finally, the above results can be generalized to any finite box: Theorem 9.
No quantum box can give rise to singular disagreement.
The proof is very similar to that of Theorem 5. The mapping from p to ˜ p forinputs x, y = 0 is as in Theorem 5 but substitutes α N , β N for α , β respectively. In this work, we have defined two notions of disagreement inspired by notions fromepistemics. These two definitions are compatible, and it is indeed possible to findexamples displaying both kinds of disagreement at once. Strikingly, a prime example f this is given by the PR box [24], proving that it is not only an extremal resourceas an extreme point of the polytope of no-signaling distributions, but also as adisagreeing distribution in the strongest possible sense.We now give an example of the usefulness and ease of application of theresults. We focus on almost-quantum correlations [21], a set of correlations strictlylarger than those achievable by measuring quantum states but that were designedto satisfy all physical principles previously proposed in the literature. We askwhether disagreement can help rule out almost-quantum correlations [21] from thelist of correlations compatible with physical principles. We conclude it can: thecorrelation with parameters r = t = 0 . , s = 0 . , u = 0 . in Theorem 3is almost-quantum and satisfies common certainty of disagreement. We also findan almost-quantum correlation that satisfies singular disagreement: the correlationwith parameters r = 0 . , s = 0 . , t = 0 . and u = 0 . in Theorem 7.In fact, there are no inclusion relations between almost-quantum correlations andobserver-agreeing correlations: the PR box is not almost- quantum and featuresboth kinds of disagreement; a noisy PR box [19] with visibility values above √ isnot almost-quantum and does not feature any disagreement; and one with visibilityvalues below √ is almost-quantum but still does not feature any disagreement.Almost-quantum correlations have already been found not to satisfy other physicalprinciples [26], and this new finding provides further evidence. In any case, this isa clear invitation for further research and application of our results.Another future direction for continuing this work concerns approximate notionsfor disagreement. In particular, it is natural to ask whether common certaintyof disagreement and singular disagreement are robust when some noise or error isconsidered. Such a result would be especially useful in designing experiments totest the behaviour of Nature. On a more speculative note, it would be also veryinteresting to explore the application of the notions introduced here to practicaltasks in which consensus between parties plays a role, such as, the coordination ofthe action of distributed agents or the verification of distributed computations. See[10] for some specific connections along these lines in the classical case.We conclude this paper with a brief comparison of the notions we presented.One might wonder: in testing new theories, is it better to focus on commoncertainty of disagreement or singular disagreement? We argue that both are equallyvaluable. First, they both induce an immediate test for new theories—i.e., thetables in Theorems 3 and 7. These tests are very general, in the sense that theyare based only on the capability of a theory to realize undesirable correlationsbetween non-communicating parties. Second, both principles have their rootsin epistemics, certainty of disagreement being closer to Aumann’s original idea,singular disagreement having a simpler description. Some limited evidence thatsingular disagreement might be a more powerful notion can be seen by trying to findcorrelations satisfying Theorems 3 and 7, which are approximately quantum, withthe NPA hierarchy [22]. Searching for a correlation featuring common certainty ofdisagreement fails at level 3 of the hierarchy, giving an alternative proof of Theorem4. However, the same task for singular disagreement continues finding solutions upto level 10, after which the computations become too complex for our machine. In asimilar manner, the proof of Theorem 9 relies on the recent and relatively complexresult of quantum voids, which can also be used to show Theorem 4. However,the latter follows more simply from Tsirelson’s theorem. Might it be that singulardisagreement, alone, singles out quantum mechanics? The question surely needsfurther investigation. cknowledgements. We thank Eduardo Zambrano for a helpfulcommunication. We also thank our audiences at Perimeter Institute, UniversidadComplutense de Madrid and QuSoft for their comments and illuminatingdiscussions. In particular, Elie Wolfe for bringing Ref [24] to our attention,Alex Pozas-Kerstjens for help with the numerical search, Serge Fehr, Peter vanEmde Boas and Ronald de Wolf. We acknowledge financial support from NYUStern School of Business, NYU Shanghai, and J.P. Valles (Brandenburger);MTM2017-88385-P (MINECO), QUITEMAD+CMS2013/ICE-2801 (Comunidadde Madrid) and ICMAT Severo Ochoa project SEV-2015-0554-16-3 (MINECO)(Contreras-Tejada); ERC Consolidator Grant GAPS (No. 648913) and MTM2014-57838-C2-2-P (MICINN) (Kubicki); Deutsche Bundesbank (La Mura); MTM2014-54240-P (MINECO), QUITEMAD+- CM Reference: S2013/ICE-2801 (Comunidadde Madrid), and ICMAT Severo Ochoa project SEV-2015-0554 (MINECO)(Scarpa); and NIH grant R01DA038063 (Steverson). Contreras-Tejada is gratefulfor the hospitality of Perimeter Institute, where part of this work was carried out.Research at Perimeter Institute is supported, in part, by the Government of Canadathrough the Department of Innovation, Science and Economic Development Canadaand by the Province of Ontario through the Ministry of Economic Development,Job Creation and Trade.
References [1] S. Aaronson. The complexity of agreement. In
Proceedings of the Thirty-Seventh Annual ACM Symposium on Theory of Computing , STOC ’05, pages634–643, New York, NY, USA, 2005. Association for Computing Machinery.[2] S. Abramsky and A. Brandenburger. The sheaf-theoretic structure of non-locality and contextuality.
New Journal of Physics , 13(11):113036, November2011.[3] S. Abramsky and A. Brandenburger. An Operational Interpretation of NegativeProbabilities and No-Signalling Models. In F. van Breugel, E. Kashefi,C. Palamidessi, and J. Rutten, editors,
Horizons of the Mind. A Tributeto Prakash Panangaden: Essays Dedicated to Prakash Panangaden on theOccasion of His 60th Birthday , Lecture Notes in Computer Science, pages 59–75. Springer International Publishing, Cham, 2014.[4] S. Abramsky and G. Carù. Non-locality, contextuality and valuation algebras: ageneral theory of disagreement.
Philosophical Transactions of the Royal SocietyA: Mathematical, Physical and Engineering Sciences , 377(2157):20190036,November 2019. arXiv: 1911.03521.[5] R. J. Aumann. Agreeing to Disagree.
The Annals of Statistics , 4(6):1236–1239,1976.[6] R. J. Aumann and A. Brandenburger. Epistemic conditions for nashequilibrium.
Econometrica , 63(5):1161–1180, 1995.[7] J. Barrett, N. Linden, S. Massar, S. Pironio, S. Popescu, and D. Roberts.Nonlocal correlations as an information-theoretic resource.
Phys. Rev. A ,71:022101, Feb 2005.[8] J. Bell. On the Einstein Podolsky Rosen paradox.
Physics Physique Fizika ,1(3):195–200, November 1964.
9] R. Clifton, J. Bub, and H. Halvorson. Characterizing quantum theory in termsof information-theoretic constraints.
Foundations of Physics , 33(11):1561–1591,2003.[10] R. Fagin, J. Halpern, Y. Moses, and M. Vardi.
Reasoning About Knowledge .MIT Press, 01 2003.[11] C. Ferrie. Quasi-probability representations of quantum theory withapplications to quantum information science.
Reports on Progress in Physics ,74(11):116001, 2011.[12] D. Frauchiger and R. Renner. Quantum theory cannot consistently describethe use of itself.
Nature Communications , 9:3711, 2018.[13] J. D. Geanakoplos and H. M. Polemarchakis. We can’t disagree forever.
Journalof Economic Theory , 28(1):192–200, 1982.[14] T. Gonda, R. Kunjwal, D. Schmid, E. Wolfe, and A. B. Sainz. Almost QuantumCorrelations are Inconsistent with Specker’s Principle.
Quantum , 2:87, August2018.[15] L. Hardy. Quantum mechanics, local realistic theories, and Lorentz-invariantrealistic theories.
Physical Review Letters , 68(20):2981–2984, May 1992.[16] A. Khrennikov. Quantum version of Aumann’s approach to commonknowledge: Sufficient conditions of impossibility to agree on disagree.
Journalof Mathematical Economics , 60:89–104, October 2015.[17] A. Khrennikov and I. Basieva. Possibility to agree on disagree from quantuminformation and decision making.
Journal of Mathematical Psychology , 62-63:1–15, October 2014.[18] J.-Å. Larsson. A contextual extension of Spekkens’ toy model.
AIP Conf.Proc. , 1424:211–220, 2012.[19] L. Masanes, A. Acín, and N. Gisin. General properties of nonsignaling theories.
Physical Review A , 73(1):012112, 2006.[20] P. Milgrom and N. Stokey. Information, trade and common knowledge.
Journalof Economic Theory , 26(1):17–27, 1982.[21] M. Navascués, Y. Guryanova, M. J. Hoban, and A. Acín. Almost quantumcorrelations.
Nature Communications , 6(1), 2015.[22] M. Navascués, S. Pironio, and A. Acín. A convergent hierarchy of semidefiniteprograms characterizing the set of quantum correlations.
New Journal ofPhysics , 10(7):073013, jul 2008.[23] M. Pawlowski, T. Paterek, D. Kaszlikowski, V. Scarani, A. Winter, andM. Zukowski. Information Causality as a Physical Principle.
Nature ,461(7267):1101–1104, October 2009.[24] S. Popescu and D. Rohrlich. Quantum nonlocality as an axiom.
Foundationsof Physics , 24(3):379–385, March 1994.[25] A. Rai, C. Duarte, S. Brito, and R. Chaves. Geometry of the quantum set onno-signaling faces.
Physical Review A , 99(3):032106, March 2019.[26] A. B. Sainz, Y. Guryanova, A. Acín, and M. Navascués. Almostquantum correlations violate the no-restriction hypothesis.
Phys. Rev. Lett. ,120(20):200402, May 2018.
27] J. K. Sebenius and J. Geanakoplos. Don’t bet on it: Contingent agreementswith asymmetric information.
Journal of the American Statistical Association ,78(382):424–426, 1983.[28] R. W. Spekkens. Evidence for the epistemic view of quantum states: A toytheory.
Phys. Rev. A , 75(3):032110, March 2007.[29] L.-L. Sun, X. Zhou, and S. Yu. No Disturbance Without Uncertainty as aPhysical Principle. arXiv:1906.11807 [quant-ph] , June 2019.[30] B. S. Tsirelson. Some results and problems on quantum Bell-type inequalities.
Hadronic Journal Supplement , 8(4):329–345, 1993.[31] B. Yan. Quantum Correlations are Tightly Bound by the Exclusivity Principle.
Phys. Rev. Lett. , 110(26):260406, June 2013. Supplementary material
We prove the results in the main text. We also provide the statements and somedefinitions in the interest of readability.
Theorem 1.
Fix a probability space (Ω , E , P ) where E A and E B are perfectlycorrelated events. If it is common certainty at a state ω ∗ ∈ Ω that Alice assignsprobability q A to E B and Bob assigns probability q B to E A , then q A = q B . Proof.
Since Ω is finite, there is a finite N ∈ N such that, for all n ≥ N, A n +1 = A n and B n +1 = B n . From the definition of A N +1 , we have that P ( B N |P A ( ω )) = 1 ∀ ω ∈ A N . (1)Now, A N is a union of partition elements of P A , i.e., A N = S i ∈ I π i where each π i ∈ P A and I is a finite index set. From Equation (1), we have P ( B N | π i ) = 1 ∀ i ∈ I. (2)Since P ( B N | A N ) is a convex combination of P ( B N | π i ) for i ∈ I, we must have P ( B N | A N ) = 1 . (3)Now, since A N ⊆ A , then P ( E B | π i ) = q A for all i ∈ I too. Using a convexcombination argument once more, this entails that P ( E B | A N ) = q A . (4)Equations (3) and (4) together imply that P ( E B | A N ∩ B N ) = q A . (5)But events E A and E B are perfectly correlated, so that P ( E A ∩ E B | A N ∩ B N ) = q A , (6)as well.Running the parallel argument with A and B interchanged, we obtain P ( E A ∩ E B | A N ∩ B N ) = q B , (7)which implies that q A = q B . Proposition A1.
Given any no-signaling box { p ( ab | xy ) } ( a,b,x,y ) ∈A×B×X ×Y , thereis a (non-unique) corresponding ontological model whose probabilities assigned to thestates of the world are not necessarily non-negative. This result was already derived in [2] from sheaf-theoretic concepts, however weprovide a much more direct proof that is more suitable for the purposes of this work.
Proof.
Let { p ( ab | xy ) } a,b,x,y be a no-signaling box. We construct its associatedontological model (cid:8) (Ω , F , P ) , { A ax , B by } a,b,x,y (cid:9) . We provide the proof for a, b, x, y ∈{ , } for ease of notation, but the generalization to more inputs and outputs isimmediate. o construct the ontological model, we postulate the existence of a set of states ω a a b b (8)with quasi-probabilities P a a b b ≡ P ( ω a a b b ) . (9)Each state corresponds to an instruction set [3], i.e., the state where Alice outputs a on input x = 0 and a on input x = 1 , and Bob outputs b on input y = 0 and b on input y = 1 . Then, each P a a b b is the quasi-probability of the correspondinginstruction set. Of course, if the given box is post-classical, not all of thesequasi-probabilities will be non-negative. In fact, in principle it need not even beguaranteed that one can find a quasi-probability distribution over these states. Butwe will use the probability distribution of the inputs and outputs of the given no-signaling box to derive a linear system of equations over the quasi-probabilities, andshow that it does have a solution.There are 16 states in total, as there are two possible outputs for each of the 4inputs ( |A| |X | · |B| |Y| in general). Then, each partition corresponds to a set of statesas follows: A ax = { ω a a b b : a x = a } B by = { ω a a b b : b y = b } . (10)We associate the probabilities p ( ab | xy ) of the no-signaling box to theprobabilities P ( A ax ∩ B by ) of each intersection of partitions, for each input pair ( x, y ) and output pair ( a, b ) . This gives rise to a set of equations for the probabilities P a a b b . Indeed, the probability of each intersection is given by P ( A ax ∩ B by ) = X a ¯ x ,b ¯ y P a x a ¯ x b y b ¯ y (11)where we denote the output corresponding to the input that is not x as a ¯ x , andsimilarly for b ¯ y , and so we have, for each a, b, x, y , X a ¯ x ,b ¯ y P a x a ¯ x b y b ¯ y = p ( ab | xy ) . (12)Since there are 16 values of p ( ab | xy ) in the 2-input 2-output no-signaling box, wearrive at 16 equations ( |A| × |B| × |X | × |Y| in general). Of course, there are somelinear dependencies between the equations, but we will show that the system stillhas a solution.The system of equations can be expressed as M P = C (13)where M is the matrix of coefficients, P is the vector of probabilities P a a b b and C is the vector of independent terms p ( ab | xy ) . The system has a solution (which isnot necessarily unique) if and only ifrank ( M ) = rank ( M | C ) . (14)Since the rank of a matrix is the number of linearly independent rows, it is triviallytrue that rank ( M ) ≤ rank ( M | C ) , (15)as including the independent terms can only remove some relations of lineardependence, not add more. Equivalently, the number of relations of linear ependence of M | C is always smaller than or equal to the number of relations oflinear dependence of M . Therefore, to show that their ranks are equal, it is sufficientto show that every relation of linear dependence that we find in M still holds in M | C .That is, for every relation of linear dependence between the probabilities P a a b b that is contained in M , it is sufficient to show that the relation still holds when thesums of probabilities are matched to the elements p ( ab | xy ) of the no-signaling boxin order to show that the system of equations has a solution.Observe that M contains only zeros and ones, as the equations (12) are justsums of probabilities. Moreover, each column of M corresponds to the probabilityof a state ω a a b b , while each row corresponds to an equation with independentterm p ( ab | xy ) . Because the equations (12) correspond to intersections of partitionsof the set of ω a a b b , we can observe that each row of M has a 1 in the columncorresponding to the states ω a a b b contained in the corresponding partition, anda 0 elsewhere. Put another way, in order to construct M one must first partitionthe set of ω a a b b in four different ways, corresponding to { A a } a , { A a } a , n B b o b , n B b o b (16)for Alice and Bob respectively. This gives partitions of the columns of M .
Then,the 16 possible ways of intersecting partitions of Alice’s with partitions of Bob’s givethe 16 equations with independent term p ( ab | xy ) . But notice now that the partitionstructure imposes a certain relation of linear dependence between the rows of M. Indeed, for each b, y, we have ([ a (cid:16) A a ∩ B by (cid:17)) = ([ a (cid:16) A a ∩ B by (cid:17)) , (17)as [ a (cid:16) A a ∩ B by (cid:17) = [ a A a ! ∩ B by = Ω ∩ B by = [ a A a ! ∩ B by = [ a (cid:16) A a ∩ B by (cid:17) , (18)and, similarly, for each a, x we have ([ b (cid:16) A ax ∩ B b (cid:17)) = ([ b (cid:16) A ax ∩ B b (cid:17)) . (19)Using the correspondence of these partitions with the partitions of the columns of M gives 8 relations of linear dependence between its rows. Now, noticing that aunion of columns of M corresponds to a sum of probabilities P a a b b , we find thatthese relations correspond exactly to the no-signaling conditions, as P [ a (cid:16) A ax ∩ B by (cid:17)! = X a P (cid:16) A ax ∩ B by (cid:17) (20)so X a P (cid:16) A a ∩ B by (cid:17) = X a P (cid:16) A a ∩ B by (cid:17) (21)and similarly for Bob. Of course, by definition of no-signaling box, these relationshold for the independent terms p ( ab | xy ) as well, since X a P (cid:16) A ax ∩ B by (cid:17) = X a p ( ab | xy ) , X b P (cid:16) A ax ∩ B by (cid:17) = X b p ( ab | xy ) (22) y construction of the linear system (see equations (11) and (12)). Therefore, everyrelation of linear dependence between the rows of M holds also between the rowsof M | C , as required.Notice also that the implication goes both ways: the linear system has a solution only if the set of probabilities p ( ab | xy ) is no-signaling. The coefficient matrix M incorporates the no-signaling conditions by construction of the states ω a a b b withprobabilities P a a b b . Therefore, if these conditions do not hold for the independentterms p ( ab | xy ) , then the rank of M | C must be larger than that of M , as M | C contains more linearly independent rows than M. We now state and prove the classical agreement theorem in the no-signalinglanguage, i.e. for local boxes. We restrict to boxes of two inputs and two outputssince, by Theorem 5, any larger box exhibiting disagreement can be reduced to a2-input 2-output box that also exhibits disagreement, while preserving its localityproperties.Recall that a distribution { p ( ab | xy ) } ( a,b,x,y ) ∈A×B×X ×Y is local if, for each a, b, x, y, p ( ab | xy ) = X λ p A ( a | xλ ) p B ( b | yλ ) , (23)for some distributions { p A ( a | xλ ) } ( a,x,λ ) ∈A×X × Λ , { p B ( b | yλ ) } ( b,y,λ ) ∈B×Y× Λ where Λ is an index set. Corollary 2.
Suppose Alice and Bob share a local no-signaling box with underlyingprobability distribution p . Let q A , q B ∈ [0 , , and let p ( b = 1 | a = 0 , x = 0 , y = 1) = q A ,p ( a = 1 | b = 0 , x = 1 , y = 0) = q B . (24) If q A and q B are common certainty between the agents, then q A = q B . Proof.
By definition of q A , q B , and using the fact that the shared distribution islocal and hence satisfies equation (23), we have q A X λ p λ p A (0 | λ ) = X λ p λ p A (0 | λ ) p B (1 | λ ) q B X λ p λ p B (0 | λ ) = X λ p λ p A (1 | λ ) p B (0 | λ ) . (25)From Theorem 3 we have that, if ∈ α n or ∈ β n for all n ∈ N then there isno common certainty of disagreement for any no-signaling distribution, and theseencompass local distributions. Hence there only remains to prove the claim for α n and β n , for some n ∈ N . This implies that p ( b = 0 | a = 0 , x = 0 , y = 0) = 1 p ( a = 0 | b = 0 , x = 0 , y = 0) = 1 (26)and hence X λ p λ p A (0 | λ ) = X λ p λ p A (0 | λ ) p B (0 | λ ) X λ p λ p B (0 | λ ) = X λ p λ p A (0 | λ ) p B (0 | λ ) , (27)which implies, on the one hand, that X λ p λ p A (0 | λ ) = X λ p λ p B (0 | λ ) (28) nd, on the other, that X λ p λ p A (0 | λ ) p B (1 | λ ) = X λ p λ p A (1 | λ ) p B (0 | λ ) = 0 , (29)that is, p A (0 | λ ) p B (1 | λ ) = p A (1 | λ ) p B (0 | λ ) = 0 (30)for all λ. Therefore, there remains to prove only that X λ p λ p A (0 | λ ) p B (1 | λ ) = X λ p λ p A (1 | λ ) p B (0 | λ ) . (31)Because the outputs for inputs x = 1 , y = 1 are perfectly correlated, we have p A (0 | λ ) p B (1 | λ ) = p A (1 | λ ) p B (0 | λ ) = 0 (32)for all λ and, since p A (0 | λ ) + p A (1 | λ ) = 1 and similarly for p B , this implies p A (1 | λ ) = p B (1 | λ ) . (33)Then we can prove (31) by simple manipulations of the probability distributions ofeach party: X λ p λ p A (0 | λ ) p B (1 | λ ) = X λ p λ p A (0 | λ ) p A (1 | λ ) ( p B (0 | λ ) + p B (1 | λ ))= X λ p λ p A (0 | λ ) p A (1 | λ ) p B (0 | λ )= X λ p λ ( p A (0 | λ ) + p A (1 | λ )) p A (1 | λ ) p B (0 | λ )= X λ p λ ( p A (0 | λ ) + p A (1 | λ )) p A (1 | λ ) p B (0 | λ )= X λ p λ p A (1 | λ ) p A (1 | λ ) p B (0 | λ )= X λ p λ p A (1 | λ ) p B (0 | λ ) (34)where we have used the fact that P b ∈B p B ( b | yλ ) = 1 for all y, λ in the first equality,(30) in the second and third, P a ∈A p A ( a | xλ ) = 1 for all x, λ in the fourth, (32)again in the fifth, and p A (1 | λ ) = p A (1 | λ ) for all λ (since p A ( a | xλ ) is either 1 or0 for every a, x, λ ) in the last.In the proof of Theorem 3, we will make use of the following Lemma: Lemma A2.
Consider a no-signaling box of 2 inputs and 2 outputs. Then, α = { , } if and only if q A = p ( b = 1 | y = 1) . Analogously, β = { , } if and only if q B = p ( a = 1 | x = 1) . Proof.
By hypothesis, q A = p ( b = 1 | a = 0 , x = 0 , y = 1) = p (01 | p ( a = 0 | x = 0)= p ( b = 1 | a = 1 , x = 0 , y = 1) = p (11 | p ( a = 1 | x = 0) . But now, we can write p ( b = 1 | y = 1) = p (01 |
01) + p (11 |
01) = p ( a = 0 | x = 0) q A + p ( a = 1 | x = 0) q A = q A . The reverse implication is trivial. The analogous statement can be proved byinterchanging the roles of Alice and Bob. heorem 3. A 2-input 2-output no-signaling box gives rise to common certainty ofdisagreement if and only if it takes the form of Table 1. xy \ ab
00 01 10 1100 r − r r − s s − r + t + s − t − s t − u u r − t + u − r − u t − t Table 1: Parametrization of no-signaling boxes with common certaintyof disagreement. Here, r, s, t, u ∈ [0 , are such that all the entries ofthe box are non-negative, and s − u = r − t. Proof.
We first prove that common certainty of disagreement imposes the claimedstructure for the no-signaling box. Therefore, we assume common certainty ofdisagreement, i.e., (0 , , , ∈ A n ∩ B n ∀ n ∈ N . (35)We split the proof into three cases based on the contents of the sets A n , B n : Case / ∈ α n , / ∈ β n for some n . From common certainty of disagreement(equation (35)), we have that p ( B n | a = 0 , x = 0 , y = 0) = 1 , p ( A n | b = 0 , x = 0 , y = 0) = 1 , which, together with / ∈ α n , / ∈ β n , translates into: p (01 |
00) = 0 , p (10 |
00) = 0 . The rest of the table is determined by no-signaling constraints in terms ofparameters r , s , t and u . Given the box in the statement of the theorem, q A = q B if and only if s − u = r − t , which concludes the proof of thiscase. Case α n = { , } , for all n ∈ N while / ∈ β m for some m . We show that thiscase implies q A = q B , so it contradicts common certainty of disagreement.Indeed, the definition of α m +1 enforces the conditions: p ( b = 0 | a = 0 , x = 0 , y = 0) = 1 = p ( b = 0 | a = 1 , x = 0 , y = 0) . This implies p ( b = 1 | a = 0 , x = 0 , y = 0) = p (01 | p ( a = 0 | x = 0) = 0 ⇒ p (01 |
00) = 0 ,p ( b = 1 | a = 1 , x = 0 , y = 0) = p (11 | p ( a = 1 | x = 0) = 0 ⇒ p (11 |
00) = 0 . Adding no-signaling conditions to these last equations, we also obtain p ( b = 1 | y = 0) = p (01 |
10) + p (11 | , (36)and so p (01 |
10) = 0 = p (11 | (37) This need not happen at the same stage, i.e., possibly / ∈ α m , for some m < n . However in thiscase, since the sets are nonempty by assumption, we have α n = α m . nd p ( b = 0 | y = 0) = 1 . (38)This allows us to identify q B with p ( a = 1 | x = 1) , since q B = p ( a = 1 | b = 0 , x = 1 , y = 0)= p (10 | p ( b = 0 | y = 0)= p (10 | p ( a = 1 | x = 1) − p (11 | p ( a = 1 | x = 1) , where the third and last equalities follow from equations (38) and(37) respectively. Now, taking into account Lemma A2 and perfectcorrelations, we have q A = p ( b = 1 | y = 1) = p ( a = 1 | x = 1) , which shows that q A = q B , as mentioned above. Case α n = { , } , β n = { , } for all n ∈ N . We now show that this case alsoimplies q A = q B , contradicting common certainty of disagreement. UsingLemma A2 we have q B = p ( a = 1 | x = 1) as well as q A = p ( b = 1 | y = 1) . Now, perfect correlations impose that p ( a = 1 | x = 1) = p ( b = 1 | y = 1) ,that is, q A = q B .Next, we prove the converse implication of the theorem. We show that any no-signaling box of the above form must exhibit common certainty of disagreement.Since s − u = r − t , we have that Alice and Bob assign different probabilities tooutput a, b = 1 on input x, y = 1 : q A = p ( b = 1 | a = 0 , x = 0 , y = 1) = s/r,q B = p ( a = 1 | b = 0 , x = 1 , y = 0) = ( r − t + u ) /r . (39)In the case that α , β , we also have that α = α and β = β , andcommon certainty of disagreement follows, because (0 , , , is in A n ∩ B n for all n . If the parameters are such that − t − s − r = sr , (40)but − r − u − r = r − t + ur , (41)then p ( b = 1 | a = 1 , x = 0 , y = 1) = q A , (42)as well, but p ( a = 1 | b = 1 , x = 1 , y = 0) = q B , (43)and so ∈ α , β . Since we have p ( b = 0 | a = 0 , x = 0 , y = 0) = 1 , (44) e find (0 , , , ∈ A , and hence all A n still contain (0 , , , , yielding commoncertainty of disagreement.Symmetric reasoning covers the case α , ∈ β , and only the case where α = { , } , β = { , } remains. This happens when p ( b = 1 | a = 1 , x = 0 , y = 1) = p ( b = 1 | a = 0 , x = 0 , y = 1) ,p ( a = 1 | b = 0 , x = 1 , y = 0) = p ( a = 1 | b = 1 , x = 1 , y = 0) (45)which, in terms of the parameters, is equivalent to − t − s − r = sr , (46) − r − u − r = r − t + ur . (47)However, these two conditions are satisfied simultaneously only when s − u = r − t ,as we now show. From Equation (46) we get s = r (1 − t ) , while from Equation (47) we obtain u = t (1 − r ) . This means that if Equations (46) and (47) are both satisfied, then s − u = r (1 − t ) − t (1 − r ) = r − t, which contradicts the statement of the Theorem. Theorem 5.
No quantum box can give rise to common certainty of disagreement.Proof.
We show that any no-signaling box with common certainty of disagreementinduces a 2-input 2-output no-signaling box with the same property. Thus, if thereexisted a quantum system that could generate the bigger box, it could also generatethe smaller box. Then, Theorem 3 implies that no quantum box can give rise tocommon certainty of disagreement.We define a mapping from a distribution { p ( ab | xy ) } a ∈A ,b ∈B ,x ∈X ,y ∈Y to an‘effective’ distribution n ˜ p (˜ a ˜ b | ˜ x ˜ y ) o ˜ a, ˜ b, ˜ x, ˜ y ∈{ , } such that the following conditions hold:(i) if { p ( ab | xy ) } is normalized and no-signaling, then so is n ˜ p (˜ a ˜ b | ˜ x ˜ y ) o , (ii) if { p ( ab | xy ) } satisfies common certainty of disagreement, then so does n ˜ p (˜ a ˜ b | ˜ x ˜ y ) o . First, notice that the number of inputs can be reduced to 2 without loss of generality,as common certainty of disagreement is always defined to be at an event (wlog, (0 , , , ) about another event (wlog, (1 , , , ). One can associate the inputs x = 0 , y = 0 with ˜ x = 0 , ˜ y = 0 , respectively, and x = 1 , y = 1 with ˜ x = 1 , ˜ y = 1 respectively, and ignore all other possible inputs in X , Y . The outputs, instead, mustbe grouped according to whether or not they belong in the sets A n , B n (for input0) and whether or not they correspond to the event obtaining, i.e. whether or notthey are equal to 1 (for input 1). Note (1 , , , A , though this does not affect the present proof. ( a,b | x,y ) abxy X α ˜ a | x ( a ) X β ˜ b | y ( b ) ˜ p (˜ a, ˜ b | ˜ x, ˜ y )˜ x ˜ y ˜ a ˜ b Figure 1: A diagrammatic representation of the construction of ˜ p . Since p satisfies common certainty of disagreement, we know that (0 , , , ∈ A n ∩ B n . Moreover, by the definitions of the sets α n , β n (and since we only considerfinite sets A , B , X , Y ) there exists an N ∈ N such that α n = α N and β n = β N forall n ≥ N .
Take such N, and define the following indicator functions: χ α | ( a ) = ( a α N a ∈ α N χ β | ( b ) = ( b β N b ∈ β N χ α | ( c ) = χ β | ( c ) = ( c = 11 c = 1 (48)(where c stands for output a, b for Alice and Bob, respectively), with χ α | x ( a ) = 1 − χ α | x ( a ) χ β | y ( b ) = 1 − χ β | y ( b ) (49)for each a, b, x, y . Then, the mapping from p to ˜ p is defined as follows: ˜ p (˜ a ˜ b | ˜ x ˜ y ) = X a,b δ x, ˜ x δ y, ˜ y χ α ˜ a | x ( a ) χ β ˜ b | y ( b ) p ( ab | xy ) (50)where δ s,t = ( s = t s = t . (51)We note that the distribution ˜ p is merely a local post-processing of p, and henceit is local if p is. Indeed, the function χ that defines ˜ p only relates the inputs andoutputs of each agent individually, and, if p is local, so is ˜ p, as ˜ p (˜ a ˜ b | ˜ x ˜ y ) = X a,b δ x, ˜ x δ y, ˜ y χ α ˜ a | x ( a ) χ β ˜ b | y ( b ) X λ p λ p A ( a | xλ ) p B ( b | yλ )= X λ p λ X a δ x, ˜ x χ α ˜ a | x ( a ) p A ( a | xλ ) ! X b δ y, ˜ y χ β ˜ b | y ( b ) p B ( b | yλ ) ! = X λ p λ ˜ p A (˜ a | ˜ xλ )˜ p B (˜ b | ˜ yλ ) (52) here ˜ p A (˜ a | ˜ xλ ) = X a δ x, ˜ x χ α ˜ a | x ( a ) p A ( a | xλ )˜ p B (˜ b | ˜ yλ ) = X b δ y, ˜ y χ β ˜ b | y ( b ) p B ( b | yλ ) (53)for each ˜ a, ˜ b, ˜ x, ˜ y, x, y, λ. To check that condition (i) holds, we assume that p is normalized and no-signaling. Normalization follows straightforwardly from the definition, since for eachinput, each output in p gets mapped to a unique output in ˜ p, and all of the outputs in p get mapped to some output in ˜ p (i.e. the map from p to ˜ p is a surjective function).Because the map is defined differently for each pair of inputs and outputs, the no-signaling conditions need to be checked for each line. However, the computationsall follow the same pattern, and we perform only one as an example: X ˜ a ˜ p (˜ a |
00) = X a ∈ α N b ∈ β N p ( ab |
00) + X a α N b ∈ β N p ( ab | X a ∈A b ∈ β N p ( ab | X a ∈A b ∈ β N p ( ab | X a =1 b ∈ β N p ( ab |
10) + X b ∈ β N p (1 b | X ˜ a ˜ p (˜ a | (54)where we have used the no-signaling property of p in the third line, and the restfollows from the definition of the map (50).To check condition (ii), let N be as in the definition of the map (50) and let a ∈ α N . Then, by definition of the set α N +1 , we have p ( β N | a, x = 0 , y = 0) = 1 (55)and, therefore, P b ∈ β N p ( ab | P b ∈B p ( ab |
00) = 1 , (56)which entails X b β N p ( ab |
00) = 0 . (57)Summing over a ∈ α N , we get X a ∈ α N b β N p ( ab |
00) = ˜ p (01 |
00) = 0 . (58)Similarly, we find ˜ p (10 |
00) = 0 . Since p satisfies common certainty of disagreement,its outputs on input x = 1 , y = 1 must be perfectly correlated. That is, p ( ab |
11) = 0 if a = b . Hence, ˜ p (01 |
11) = X a =1 p ( a |
11) = 0 (59) nd similarly for ˜ p (10 | . So far, the no-signaling box corresponding to ˜ p has twozeros in the first row and another two in the last. Using normalization and no-signaling conditions to fill in the rest of the table, we find it is of the form of theno-signaling box in Theorem 3. There remains to check for disagreement, i.e. thatif q A = p ( b = 1 | a = 0 , x = 0 , y = 1) = p ( a = 1 | b = 0 , x = 1 , y = 0) = q B (60)then ˜ p (˜ b = 1 | ˜ a, ˜ x = 0 , ˜ y = 1) = ˜ p (˜ a = 1 | ˜ b, ˜ x = 1 , ˜ y = 0) . (61)Since α N ⊆ α and β N ⊆ β , p ( b = 1 | a ∗ , x = 0 , y = 1) = p ( a = 1 | b ∗ , x = 1 , y = 0) holds in particular for all a ∗ ∈ α N , b ∗ ∈ β N . This means that, for a ∗ ∈ α N , b ∗ ∈ β N ,p ( a ∗ | P b ∈B p ( a ∗ b | = p (1 b ∗ | P a ∈A p ( ab ∗ | (62)and so p ( a ∗ | X a ∈A p ( ab ∗ | = p (1 b ∗ | X b ∈B p ( a ∗ b | . (63)Then, we can sum over α N and β N on both sides to find X a ∗ ∈ α N p ( a ∗ | X a ∈A b ∗ ∈ β N p ( ab ∗ | = X b ∗ ∈ β N p (1 b ∗ | X a ∗ ∈ α N b ∈B p ( a ∗ b | . (64)But in terms of ˜ p, this corresponds to ˜ p (01 | X ˜ a ∈{ , } ˜ p (˜ a | = ˜ p (10 | X ˜ b ∈{ , } ˜ p (0˜ b | (65)which implies ˜ p (˜ b = 1 | ˜ a = 0 , ˜ x = 0 , ˜ y = 1) = ˜ p (˜ a = 1 | ˜ b = 0 , ˜ x = 1 , ˜ y = 0) (66)and hence the disagreement occurs for the ˜ p distribution as well, which proves theresult.Notice that the sets ˜ α , ˜ β in the distribution ˜ p (defined analogously to α , β in the distribution p ) will correspond to outputs ˜ a, ˜ b = 0 , respectively. This is tobe expected, as the map p → ˜ p gives rise to a no-signaling box of the form of theone in Theorem 3, where the sets ˜ α , ˜ β contain a single element each. (In effect,this means we are ignoring the outputs a ∗ ∈ α \ α N and b ∗ ∈ β \ β N , but thoseoutputs lead to disagreement but not to common certainty of it, so they can besafely discarded.) Theorem 6.
There is no local 2-input 2-output box that gives rise to singulardisagreement.Proof.
Assume Alice and Bob measure x = y = 0 and obtain a = b = 0 . Thisimplies p (00 | > . (67)Alice assigns p ( b = 1 | a = 0 , x = 0 , y = 1) = 1 , (68) nd Bob assigns p ( a = 1 | b = 0 , x = 1 , y = 0) = 0 . (69)Further, the outputs for input ( x, y ) = (1 , are perfectly correlated, so, inparticular, p (01 |
11) = 0 . (70). Equations (68) and (69) imply, respectively, p (00 |
01) = 0 and p (10 |
10) = 0 . (71)However, equations (67), (70) and (71) make up a form of Hardy’s paradox [15],which is known not to hold for local distributions. Theorem 9.
No quantum box can give rise to singular disagreement.Proof.
Like in Theorem 5, we show that any no-signaling box with singulardisagreement induces a 2-input 2-output no-signaling box with the same property,and rely on Theorem 8 to deduce that no quantum system can give rise to singulardisagreement.Analogously to Theorem 5, to prove the Proposition for singular disagreementwe define a mapping from a distribution { p ( ab | xy ) } a ∈A ,b ∈B ,x ∈X ,y ∈Y to an ‘effective’distribution n ˜ p (˜ a ˜ b | ˜ x ˜ y ) o ˜ a, ˜ b, ˜ x, ˜ y ∈{ , } such that the following conditions hold:(i) if { p ( ab | xy ) } is normalized and no-signaling, then so is n ˜ p (˜ a ˜ b | ˜ x ˜ y ) o , (ii) if { p ( ab | xy ) } satisfies singular disagreement, then so does n ˜ p (˜ a ˜ b | ˜ x ˜ y ) o . Again, the number of inputs can be reduced to 2 without loss of generality . To groupthe outputs, we notice that the sets A , B also play a role in singular disagreement,as they group the outputs of each party which lead them to assign their respectiveprobabilities to the event. Then, we group the outputs according to whether or notthey belong in the sets α , β (for input 0) and whether or not they correspond tothe event obtaining, i.e. whether or not they are equal to 1 (for input 1). We obtainthe same mapping (50) as before, substituting α N for α and β N for β . With thisreplacement, condition (i) follows by the same proof as before. To check condition(ii), we know that, for all a ∗ ∈ α ,p ( b = 1 | a ∗ , x = 0 , y = 1) = 1 (72)and so p ( a ∗ |
01) = X b ∈B p ( a ∗ b | . (73)Summing over a ∗ ∈ α and rewriting the expression in terms of ˜ p, we find ˜ p (01 |
01) = X ˜ b ∈{ , } ˜ p (0˜ b | (74)which implies ˜ p (˜ b = 1 | ˜ a = 0 , ˜ x = 0 , ˜ y = 1) = 1 . (75)Similarly, for all b ∗ ∈ β we have p ( a = 1 | b ∗ , x = 1 , y = 0) = 0 , (76) ence p (1 b ∗ |
10) = 0 (77)and so, by adding over b ∗ ∈ β and mapping to ˜ p, we find ˜ p (10 |
10) = 0 (78)as required.(78)as required.