Allen Stairs and Jeffrey Bub Correlations, Contexts and Quantum Logic

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1 Allen Stairs and Jeffrey Bub Correlations, Contexts and Quantum Logic ABSTRACT: Quantum theory is a probabilistic theory whose states include correlations that are stronger than any that classical theories allow but not as strong as those of hypothetical "superquantum" theories. This raises the question "Why the quantum?" whether we can articulate principles that account for the character of quantum probability. We ask what quantum-logical notions to this investigation. This project isn't meant to compete with the many beautiful results that information-theoretic approaches have yielded but rather aims to complement that work. KEY WORDS: probability, polytope, convex sets, PR box, quantum logic, quantum foundations Probabilities and Polytopes Our goal in this paper is to ask what quantum logical approaches can add to the attempt to answer the question "Why the quantum?" but we begin with a highly general way of thinking about quantum correlations, most easily introduced through an example, Consider two experimenters, Alice and Bob, performing experiments on a pair of systems. Each chooses between two mutually exclusive inputs for Alice, for Bob with, {0,1}. Each input has two possible outcomes or outputs, for Alice and for Bob, with, {0,1}. Assuming that Alice and Bob each perform a measurement on each run of the experiment, the probabilities that interest us are of the form p( ). This is the joint probability for the outcomes and in the measurement context. Measurement contexts here are understood operationally. A context simply represents a pair of "measurements" or experiments or operations that Alice and Bob might perform. A state is a complete assignment of probabilities to each of the four contexts 0 0, 0 1, 1 0 and 1 1. The set of states is convex: mixtures of states in are also in. We will depict states in with diagrams such as those in (Jones and Masanes 2005) p(00 00) p(10 00) p(00 10) p(01 00) p(11 00) p(01 10) 1 p(00 01) p(10 01) p(00 11) p(01 01) p(11 01) p(01 11) p(10 10) p(11 10) p(10 11) p(11 11) Probabilities add to 1 within each cell, and we can label each cell by its input 00, 01, 10 or 11. Among the states of is the set of so-called "deterministic" states, i.e., states that

2 assign probability one to a single outcome in each cell. Since there are four ways of assigning probability one to one element of a cell, there are 4 4 = 256 members of. Let us abbreviate this table a bit further. A general state has this form: p p p p S: p 00 p 20 p 10 p p p 32 1 p 01 p 21 p 11 p p p 33 Within each cell, the 0th entry is the upper left, the 1st the lower left, and so on, picked out by the first index. Similarly, the 0th cell is the upper left, the 1st is the lower left, and so on, picked out by the second index. Deterministic state members of are fixed by the entries to which they assign probability 1. We can denote members of by expressions such as E The quadruple index tells us which elements of cells 0, 1, 2 and 3, in that order are assigned probability 1. In the case of E 0132, for example, we have E 0132 = Any state in can be expressed as a mixture of such states. One way to see this: given a state S, with entries p 00, etc. as above, assign the weight w ijkl = p i0 p j1 p k2 p l3 to the state E ijkl. The set of all such weights is the set of 256 terms in the expansion of (p 00 + p 10 + p 20 + p 30 )( p 01 + p 11 + p 21 + p 31 )( p 02 + p 12 + p 22 + p 32 )( p 03 + p 13 + p 23 + p 33 ) Since each bracketed sum equals 1, the product also equals 1 and hence so does the sum of the w ijkl. We then combine the weighted members of into a mixture W W = ijkl ijkl E ijkl, We have, for example, p W (00 00) = ijkl ijkl p(00 00,E ijkl )

3 Here p(00 00,E ijkl ) is the probability of outputs 00 given inputs 00 and state E ijkl. Note, however, that p(00 00,E ijkl ) equals 1 iff i = 0 and equals 0 otherwise. Thus we have p W (00 00) = jkl 0jkl p(00 00,0 0jkl ) = jkl 0jkl = 0jkl p 00 p j1 p k2 p l3 = p 00 Similar remarks apply to the other terms in S. This shows that any member of can be written as a mixture of deterministic states. is a polytope; in general, decompositions into mixtures of members of are not unique. Consider S': /2 0 1/ /2 0 1/2 1 1/2 0 1/ /2 0 1/2 We can express S' as an equal mixture of these states: or as an equal mixture of these states: It may seem puzzling that states in have non-unique decompositions into mixtures of deterministic states. To understand this, consider a simpler example: Alice alone, choosing between two inputs with two possible outputs for each input. We will call this example the mini-

4 qubit, but we don't assume the specifically quantum restrictions on the probabilities. In general, mini-qubit states take this form p( 1 ) p(+1 ) p( 1 ) p(+1 ) This set of states has four deterministic extremal points: Now consider the state Q, in which all entries equal 1/2. We can represent Q as an equal mixture of D 1 and D 4, or of D 2 and D 3. Or we could represent it as this mixture: Q = 3/8(D 1 ) + 1/8(D 2 ) + 1/8(D 3 ) + 3/8(D 4 ) In fact, there are infinitely many decompositions of Q into deterministic states. This seems odd because the D i pick out the points in a classical outcome set that we can write as: O = { x =+1 & y =+1, =+1 & y = 1, x = 1 & y =+1, x = 1 & y = 1} In a classical probability space, every measure has a unique decomposition into measures concentrated on the points. However since Q provides no information about joint x y probabilities it doesn't determine a unique measure on the subsets of O. The polytope representation makes no assumptions about the relationship of one cell to another. The nonuniqueness is a consequence of this informational modesty. Signaling Modesty may be a virtue, but in this case it has a striking consequence. Consider these two states:

5 The one on the left is unremarkable: Alice and Bob each get the result 0, regardless of their input. The one on the right is more peculiar. Compare the cells in the left column. Alice's input is 0 in both cells; Bob's input is 0 in the upper cell and 1 in the lower. But this difference in Bob's input goes with a difference in Alice's output. The second state is a signaling state; Bob could send a message to Alice simply by having her keep her input set to 0 and choosing between 0 or 1 as input. A deterministic state is effectively a value assignment to each of and in each cell. Deterministic states are signaling if an -value differs from one cell to the cell above or below it, or a -value differs from one cell to a cell to its left or right. A deterministic non-signaling (DNS) member of must fix local outputs as a function of local inputs alone, and so a deterministic non-signaling state must associate an unambiguous 0 or a 1 with each of 0, 1, 0, 1. That yields 2 4 = 16 possible DNS states. The remaining 240 members of are signaling. More generally, signaling states are the ones that violate Non-Signaling: For any and ', b p( ) = b p( ') = p( ) For any and ', a p( ) = a p( ' ) = p( ) If the marginal probability of differs within two cells of a column of cells or the marginal probability of differs within two cells of a row, then the state deterministic or not is a signaling state. Thus, non-signaling states determine well-defined probabilities p( ) and p( ). The state S, recall, was one in which Alice and Bob's results always agree, though results 00 and 11 were equally likely. We noted that S = ½ E ½E It's easily checked that E 0303 and E 3030 are signaling state, but we also noted that S = ½ E ½ E E 0000 and E 3333 are both DNS states. Thus, while we can represent S by signaling states, we needn't. However, consider the state PR, first introduced by Popescu and Rohrlich (Popescu and Rohrlich 1994). PR: /2 0 1/ /2 0 1/2 1 1/ /2 0 1/2 1/2 0

6 Alice and Bob's results always agree except when they both select input 1. In that case, their results disagree. We can characterize a PR box by two constraints: PR Boxes: = (where " " is addition mod 2) Probabilities must satisfy Non-Signaling Taken together, these two requirements determine the probabilities above. In spite of the requirement that the probabilities satisfy Non-Signaling, however, any mixture of -states that reproduces PR must be composed entirely of signaling states. To see this, suppose E is a nonsignaling member of and assume that E is a non-zero element of a mixture representing PR. The upper-left cell requires that 0 = 0 in E. The upper-right cell requires 0 = 1. Further, the lower left cell requires 0 = 1. Thus, E must provide unambiguous values satisfying 1) 0 = 0, 0 = 1, 0 = 1 It follows that 1 = 1. But PR assigns probability 0 to this possibility, as the lower right cell makes clear. Thus E can't have non-zero weight in the mixture. The CHSH Inequality The argument just rehearsed leads to a well-known result. First, re-label 0 outcomes as 1. This doesn't change anything substantive, but it makes some things easier to see. Now let < i, j > be the expected value of the product of Alice and Bob's outcomes for the input pair. With our re-labeling, i and j { 1,+1} and so the product on any run is ±1. Thus, 1 < i, j > +1. We have 2) 4 < 0, 0 > + < 0, 1 > + < 1, 0 > + < 1, 1 > +4 It's clear that DNS states can yield ±4 for example, the state E 0000 returns a value of 4 for this sum. But now consider the sum < 0, 0 > + < 0, 1 > + < 1, 0 > < 1, 1 > As a matter of arithmetic, we still have 3) 4 < 0, 0 > + < 0, 1 > + < 1, 0 > < 1, 1 > +4. However, if we restrict ourselves to non-signaling states in, the minimum and maximum are out of reach. Consider the maximum. To reach it, we need +1 = < 0, 0 > = < 0, 1 > = < 1, 0 >. But < 0, 0 > = +1 requires 0 = 0, < 0, 1 > requires 0 = 1, and < 1, 0 > requires 1 = 0. This amounts to imposing the series of equalities 1) above and entails 1 = 1, hence < 1, 1 > = +1. Thus, if we restrict ourselves to DNS states, we have

7 CHSH C 2 < 0, 0 > + < 0, 1 > + < 1, 0 > < 1, 1 > +2. CHSH C is the Clauser-Horne-Shimony-Holt inequality for classical systems (Clauser et al. 1969). We have shown that it holds for members of, but it governs a wider class of states. Call a state factorizable if for each entry, p( ) = p( )p( ); DNS states are a special case. If a state is factorizable, then it must obey CHSH C. To see this, suppose Alice and Bob each possess a mini-qubit. Each mini-qubit state can be written as a weighted sum of the states D i. In general, We can introduce a minimal product of states in a straightforward way: Note that the state on the right has the general form of a factorizable state, and that it is nonsignaling. In the special case where Alice and Bob's states are both deterministic, the product is a DNS state for the pair. Now put these facts together: any factorizable state is a product of a state for Alice and a state for Bob. Further, each of Alice's and Bob's states can be written as weighted sums of deterministic states. Taking the product of those sums yields a representation of any factorizable state as a weighted sum of DNS states. Since DNS states satisfy CHSH C, so will any factorizable state or any weighted sum of factorizable states.

8 Classical, Quantum, Super-Quantum The sum < 0, 0 > + < 0, 1 > + < 1, 0 > < 1, 1 > is a useful tool for considering strength of correlations. The arithmetical upper and lower bounds are 4 and +4, but as we just saw, for theories whose states can be written as mixtures of factorizable states, the lower and upper bounds are 2 and +2. Quantum theory falls in between. Tsirelson demonstrated that in the quantum case, the lower and upper bounds are 2 2 and (Tsireslon 1980) Thus we have the following three inequalities: CHSH C 2 < 0, 0 > + < 0, 1 > + < 1, 0 > < 1, 1 > +2 CHSH Q 2 2 < 0, 0 > + < 0, 1 > + < 1, 0 > < 1, 1 > +2 2 CHSH 4 < 0, 0 > + < 0, 1 > + < 1, 0 > < 1, 1 > +4 The relationship between these inequalities and signaling is not straightforward. A state can be signaling even though it doesn't exceed the bounds of CHSH C, as this case demonstrates: The value of CHSH for this state is 0. However, we have p( ) + p( ) = 1/2 but p( ) + p( ) = 7/8 and so this is a signaling state. Furthermore, a state can exceed the classical CHSH bound and still be non-signaling. Quantum mechanics contains no signaling states but contains many states that exceed CHSH C. PR itself is non-signaling, even though it saturates CHSH. Indeed, this was the point when Popescu and Rohrlich introduced PR: to demonstrate that stronger-than-quantum correlations could still be non-signaling. What we can say is this: I Any state that can be represented as a mixture of DNS states does not violate CHSH C. II Any state that violates CHSH C can be represented as a mixture of deterministic states, but only if at least one of those states is signaling. III Any state that saturates CHSH can be represented as a mixture of deterministic states, but only if all non-zero elements of the mixture are signaling states.

9 II is what underlies the temptation to call non-factorizable states non-local. An intuitive though inadequate way to put it: any attempt to build a mechanical model of states that exceed the CHSH C bound requires non-local causality. The further temptation is to think that indeterminism just amounts to "loosening up" deterministic non-signaling states, and that any such loosening-up shouldn't allow us to perform feats that the unloosened states would forbid. What quantum theory suggests, however, is that this isn't so: indeterminism isn't just determinism with its belt unbuckled and its shoes undone. Foundational research typically sets aside the signaling elements of, restricting itself to the non-signaling polytope. has two important characteristics. First, p( ) and p( ) are well-defined for all values of the variables. Second, the extremal states of are no longer all deterministic. Since PR is non-signaling, PR is included in. However, PR can't be written as a mixture of DNS states, nor into any other set of states. PR is an extremal state of, as are other states that saturate CHSH. (See Barrett and Pironio 2005, Barrett et al. 2007, and Jones and Masanes 2005) In particular, consider a Bell-type quantum state and a choice of angles that yields a violation of CHSH C. The result will be a state of that is non-signaling, nonfactorizable and non-extremal. As II notes, it can be decomposed into a mixture of extremal states of, but only if at least one of those states is a non-deterministic state like PR. The non-signaling polytope has proved useful object for research, but from one point of view, both and have a classical character. Consider the form of p( ). This is the probability of a pair of outcome events conditional on the choice of an experimental arrangement. and represent four mutually exclusive experimental arrangements, 00, 01, 10 and 11, and one could embed in a classical probability space whose point events were =0& =0& =0& =0, =0& =1& =0& =0, =1& =0& =0& =0, =1& =1& =0& =0 =0& =0& =1& =1, =0& =1& =1& =1, =1& =0& =1& =1, =1& =1& =1& =1 This would allow us to model all the probabilities of in a classical probability space. Since there is little motive for assigning probabilities to experimental contexts =0& =0, etc., there is little reason to pursue this approach, but the possibility points something out: contexts have a classical character in the polytope representation; distinct contexts are mutually exclusive in a straightforwardly classical way.

10 Quantum mechanics treats contexts differently; it doesn't represent them by classically exclusive elements. The mark of distinct contexts in quantum mechanics is non-commutativity. Thus, for example, spin-x and spin-y are represented by non-commuting operators x and y. In typical quantum logical approaches, commutativity corresponds to membership in a common Boolean algebra of the logical structure, and conversely, non-commutativity corresponds to the absence of a common Boolean algebra. We will examine two quantum-logical approaches more closely: partial Boolean algebras and lattices, in particular proposition systems in the sense of Jauch and Piron. What they have in common is that contexts are associated with Boolean algebras, and when contexts are distinct, there is no Boolean algebra that includes both. We begin with partial Boolean algebras (PBAs). Following Kochen and Specker (Kochen and Specker 1967), a partial Boolean algebra is a family of Boolean algebras i, i I, such that (a) for every i, j I, there is a k I such that i j = k (b) if each pair from 1,.., n lies in some common i, then there is a k I such that 1,.., n k. If i and j belong to a common i, we say that they are compatible, and write i j. Thus, (b) says that if a set of elements is pairwise compatible, then the elements of the set are mutually compatible. Note that if i and j are not compatible, then neither i j nor i j is defined, hence the term partial Boolean algebra. Boolean algebras are trivial examples of PBAs. However, each quantum mechanical observable is associated with a Boolean algebra, and the set of such algebras for a quantum system is also a partial Boolean algebra. In this case, we have i j just in case the projectors associated with i and j commute. It will be instructive to consider a much simpler example, however: the "mini-qubit" introduced above. Each context is associated with a four-element Boolean algebra. If we add the trivial Boolean algebra {I,0}, the three Boolean algebras form a PBA. We will label the elements of Alice's PBA as follows: 0, ~ 0, 1, ~ 1, 0 A, I A The non-trivial Boolean subalgebras of Alice's PBA are A0, with elements { 0, ~ 0, 0 A, I A }, and A1. with elements { 1, ~ 1, 0 A, I A }. Parallel comments apply to Bob's mini-qubit. Note that " 0 1," for example, is not defined. A function from a PBA into [0,1] is a probability measure iff its restriction to each i is a probability measure. We will call such measures states.

11 Suppose Alice and Bob are carrying out joint experiments. What is the appropriate "tensor product" of the two PBAs, suitable for representing the joint experiments? If the PBAs were both simply Boolean algebras if there were no non-trivial contexts the appropriate object would be the Boolean product of A and B. The Boolean product is a Boolean algebra generated by isomorphic, algebraically independent images of A and B. For non-trivial PBAs, here are some plausible constraints. First, the identities of Alice's and Bob's systems should be preserved. It should be possible to talk meaningfully of Alice's system, of Bob's, and of Alice's and Bob's together. This is a requirement not of causal locality but of separability; it reflects Einstein's notion that spatially distinct systems should have their own "being thus." (Einstein 1948) Second, if Alice is performing an experiment associated with Ai and Bob is performing an experiment associated with Bj, then the Boolean product of Finally, the images of the factor PBAs should generate the product PBA. The motivation for this constraint is that if the product represents the "logical space" of the pair, then it shouldn't contain elements that aren't, as it were, logical functions of the elements in the factor spaces. In the case of classical systems, the Boolean product is the appropriate object. This suggests the following definition. Denote the product of two PBAs 1 and 2 by 1 2 and let the product satisfy: (i) 1 2 is a PBA with zero 0 and unit I (ii) 1 and 2 map 1 and 2 isomorphically into 1 2, with 1 (0 1 ) = 0, 1 (I 1 ) = I, and similarly for B. (iii) For any Boolean subalgebras 1i of 1 and 2j of 2, 1 ( i ) 2 ( ii ) generates a Boolean product of 1i and 2j (iv) 1 ( 1 ) 2 ( 2 ) generates 1 2. As it turns out, this definition does not pick out a unique product; more on that below. However, we can identify a minimal product that is a substructure of any product fitting the definition. Until further notice, it is this minimal product that we will denote by 1 2. We can illustrate this minimal product with the case of Alice and Bob and their miniqubits. A B contains four maximal Boolean subalgebras, one for each of A0 B0, A1 B0, A0 B1, A1 B1. A B contains 16 atoms (minimal non-zero elements), which correspond to the Cartesian product of the set of atoms of A with the set of atoms of B. States on A B automatically satisfy Non-Signaling: each element of A and each element of B has a unique representative in

12 A B ; hence each is assigned a univocal probability. Turning to the factors, A and B each permit deterministic states, and all other states A and on B can be written as mixtures of these exactly on analogy with the states D i above. Thus, if we restricted our attention to either miniqubit, the structure of the PBA would give us no to conclude that it is indeterministic. A B itself also permits deterministic states which, like all states on A B, all of which will be nonsignaling, since all states on A B are non-signaling. However, A B also admits nondeterministic extremal states. PR, for example, is extremal. It can be faithfully reproduced in A B, but not as a mixture of any other set of states. A B admits the full range of states that violate CHSH C, including states whose probabilities match the quantum mechanical Belltype correlations of states that violate CHSH C. These states, likewise, are not mixtures of deterministic states on A B, but they are also not extremal. (See Jones and Masanes 2005 and Barrett et al. and Gisin 2005). PBAs hint as some insight into the origins of indeterminism. When we combine local contextuality PBAs rather than Boolean algebras as the representatives of the subsystems with separability as embodied in the definition of A B, the possibility of deeply indeterministic states emerges - states whose probabilities can't be understood as mixtures of the deterministic states that the structure admits. It might be complained that PBAs offer nothing new: the states on A B are exactly the states of the non-signaling polytope. Indeed, we might say that simply is A B in a different guise. There is some justice to this claim. Nonetheless, the PBA approach provides a motivation for non-signaling that doesn't treat it simply as a brute empirical constraint, and it also provides insight into the role of contextuality in shaping the structure of the probabilities. The fact that the product permits states that violate CHSH C is a consequence of the contextual structure of the subsystems, together with the conditions on the product. That said, while PBAs might help us understand how there could be stronger-than-classical correlations, it has nothing to offer on the question of why we don't have stronger-than-quantum correlations. This is where proposition systems have a contribution to make. We should make clear that we aren't simply proposing a "quantum logical" solution to this question. On the contrary, some of the deepest insights into the question of "Why the quantum?" may well come from considerations of a very different sort. In a recent paper, Pawlowski and several collaborators propose a condition that generalizes No-Signaling and promises to tell us a great deal about why quantum probabilities behave as they do. (Pawlowski

13 et al. 2009) The condition is called information causality and it can be stated informally as follows: Information Causality (IC) The transmission of classical bits can cause an information gain of at most bits. A thorough discussion of IC is beyond the scope of the present paper, but we get a better feel for the condition by noting the simplest case where it is violated. Suppose that Alice is given two classical bits 0, 1 {0,1}. Bob will try to guess the value of 0 or of 1, though Alice doesn't know which. Alice is allowed to send one classical bit {0,1} to Bob, with determined by some method on which she and Bob have agreed. If Alice and Bob share a PR box, there is a protocol that will let Bob guess either of Alice's bits, as he chooses. The shared information between Alice's transmitted bit and Bob's guess (i.e., H( ) + H( ) H(, )) is 2, even though Bob has received only one classical bit. Pawlowski et al show that no shared quantum state permits a violation of IC, and more generally, violations of IC are possible exactly for probabilities that violate CHSH Q. This is a remarkable result and moves us much further toward an information-theoretic answer to the "Why the quantum?" question. What follows is not intended to replace such results, but nonetheless cast a different kind light on why quantum correlations can't saturate CHSH. We turn to proposition systems as defined by Jauch and Piron (Jauch and Piron 1969). A proposition systems is a lattice that satisfies the orthomodular law = ( ), and the atomicity axiom A(i) For every element of there exists an atom (minimal non-zero element) such that. A(i) If is an atom, then for all and, ( ) [ = or = ] If is finite-dimensional, then also satisfies the modular law ( ) = ( ) Since we will be concerned exclusively with finite-dimensional cases, proposition systems can be thought of simply as orthocomplemented modular lattices. The Hasse diagram for Alice's mini-qubit looks like this:

14 Once again, we can ask what the appropriate product of two proposition systems should be. One of us considered this question some years ago (Stairs 1983) and proposed the following answer. (i) 1 2 is a proposition system with zero 0 and unit I (ii) 1 and 2 map 1 and 2 isomorphically into 1 2, with 1 (0 1 ) = 0, 1 (I 1 ) = I, and similarly for 2. (iii) For any Boolean sublattices 1i of A and 2j of B, 1 ( 1i ) 2 ( ) generates a Boolean product of 1i and 2j (iv) 1 ( 1 ) 2 ( 2 ) generates 1 2. Here 1 and 2 are any two proposition systems. Conditions (i) (iv) are simply the conditions on products of PBAs, modified for the case of proposition systems. With this definition in hand, we turn to a purely lattice-theoretic result. Restrict attention for the moment to the case of A and B, our mini-qubits, with lattice structures as in the Hasse diagram above. Suppose that A B exists. Then purely lattice-theoretic calculations show that A B contains atoms other than the sixteen atoms 0 0, 0 ~ 0,, ~ 1 ~ 1. That is, purely lattice-theoretic calculations ensure that even in this minimal case, there are lattice analogues of vectors in a tensor product space that can't be written as simple tensors. Note that this is quite unlike the minimal product of PBAs. It depends on the fact that in a lattice, and are always defined, whereas in the case of a PBA, they are defined only if and belong to a common Boolean algebra. The proof (see the appendix of Stairs 1983 for details) proceeds by considering ( 0 0 ) (~ 0 ~ 0 ) and ( 1 1 ) (~ 1 ~ 1 ) In the minimal PBA product, there is no Boolean algebra that contains both of these elements. Consequently, their intersection is not defined. This has a further consequence: the PBA approach provides no constraints on the relationship between these elements. In the lattice case,

15 however, the intersection of these two elements must exist. What the lattice calculations of the proof show is that this intersection is an atom, hence is non-zero, and that it is distinct from the sixteen simple atoms noted above. One might think that the existence of these "extra" atoms would provide a richer set of states on A B than are permitted by A B, but roughly the opposite is true. It is the facts about these extra atoms that help us see why we end up with fewer rather than more states. Roughly, the "extra atoms" don't exist in the abundance that saturation of CHSH requires. This point requires explaining and qualifiying. Begin by considering the "novel" atom = [( 0 0 ) (~ 0 ~ 0 )] [( 1 1 ) (~ 1 ~ 1 )]. Suppose is assigned probability one. In any such state, both "conjuncts" in the lattice polynomial on the right will be assigned probability one. Thus, provides a support in the lattice for a state that requires perfect correlation of outputs when Alice and Bob both input 0, and when they both input 1. Likewise, we can find an element in A B that provides a support for perfect anti-correlation between Alice's and Bob's outcomes in these cases. There are also supports for perfect correlations between Alice's 0 and Bob's 1 results, and vice-versa. But return to PRboxes. They require the following correlations: 0 0, 0 1, 1 0, 1 ~ 1 Each of these correlations, taken one-by-one, has support in A B. However, when we try to account for more than one pair of correlations at once, we run into difficulties. Consider the first two correlations, 0 0 and 0 1. Expressed in lattice terms, they are ( 0 0 ) (~ 0 ~ 0 ) and ( 0 1 ) (~ 0 ~ 1 ) Joint support for these two correlations in A B requires [( 0 0 ) (~ 0 ~ 0 )] [( 0 1 ) (~ 0 ~ 1 )] However, straightforward lattice calculations show that this intersection must be. If the intersection is non-zero, then there must be an atom such that ( 0 0 ) (~ 0 ~ 0 ) and ( 0 1 ) (~ 0 ~ 1 ).

16 However, it can be shown that if is under ( 0 1 ) (~ 0 ~ 1 ) then is not under ( 0 0 ) (~ 0 ~ 0 ). Hence the intersection is zero. We need to be clear about what this point does and does not amount to. First, the point is not that no proposition system could ever provide support for two correlations of the form ( 0 0 ) & ( 0 1 ). On the contrary: if 0, 0 and 1 all belonged to a common Boolean algebra in some proposition system, such support would straightforwardly exist. However, we are considering the quantummechanically motivated case in which these elements have the compatibility relations found in the lattice A B. The point is that proposition systems permit such cases, the proposition systems associated with qubits call for them, and in that case, the lattice does not provide the requisite element. The account of this fact assumes only the handful of relations in the lattices A and B, together with the definition of the product set A B. The product of proposition systems introduces constraints that aren't found in the product of PBAs. It's also important to understand that simply from the lack of a common support for ( 0 0 ) and ( 0 1 ), it doesn't follow that no state on A B could assign probability one to both correlations. To see why, consider something much simpler: the lattice A. This lattice does not provide a common support for 0 and 1, but there is a measure on that assigns probability one to 0 and to 1. The same is true if we consider the full lattice associated with a full qubit. Lattice theoretic considerations by themselves don't guarantee that two lattice elements can be assigned probability one if and only if their intersection is assigned probability one. This may seem to rob what has been said of its point. To see why this is not so, we need to say more about the product A B. We began with the case of two mini-qubits because of its conceptual simplicity. In fact, when considering proposition systems, the mini-qubit turns out to be too simple: it isn't clear that the product of the two 6-element proposition systems exists. The reason is that the product would have to be a complemented modular lattice of four dimensions, and it would have to have a trivial center (that is, only the 0 and the I would belong to all Boolean sublattices). The only elements that belong to all Boolean subalgebras would have to be and 1. This means that the product would have to be a geometry over the reals, the complex numbers or the quaternions (see Varadarajan 1968 for details) and hence would have to have uncountably many elements. This makes it doubtful that the generation condition (iv) can be satisfied for the product of these small

17 finite lattices. However, the mini-qubit is an unrealistic toy example. If instead of considering the mini-qubit, we consider two real qubits, hence the lattices of subspaces of two-dimensional Hilbert spaces, products do exist. In particular, ( 2 2 ) satisfies the definition of the product. All of the points made above, however, remain intact. In particular, there will be no common support for ( 0 0 ) and ( 0 1 ). In the case of ( 2 ) ( 2 ) = ( 2 2 ), the consequences of Gleason's theorem (Gleason 1957) apply. And one of those consequences is that in the case of Hilbert space lattices of dimension 3, what we might call the intersection principle for probability one applies: for any and in the lattice, if p( ) = 1 and p( ), then p( ) = 1. Thus, the fact that ( 0 0 ) and ( 0 1 ) have no common support means that they can't both be assigned probability one. But that they have no common support is a consequence of straightforward lattice-theoretic considerations. It's worth making a comparison with PBAs. On the one hand, there is an obvious PBA associated with 2 2, and it satisfies the definition of the PBA product for two qubits. However, it is not a minimal product. All that the basic constraints on PBA products require is a structure with essentially the same character as A B, the minimal PBA product for two miniqubits. The atoms of the product still correspond to the elements of the Cartesian product of the atoms of the two subsystems, but two elements of the product PBA can both be assigned probability one whether or not their intersection is defined. In general, the PBA product leaves things loose and separate, as Hume might say; loose enough and separate enough to permit the saturation of CHSH. We posed a narrow question when we turned our attention to proposition systems: why don't quantum correlations saturate CHSH? In contrast to the PBA framework, the lattice approach requires meets and joins to be everywhere defined. This imposes more constraints on the relations among the events over which probabilities are defined. In particular, it entails that the correlations ( 0 0 ) and ( 0 1 ) lack a common support and hence, given deeper theorems about the relevant lattices, cannot both obtain. In fact, there is a simple corollary that is nonetheless intuitively striking. It's not just that the lattice product for two qubits doesn't permit the CHSH-saturating correlations that we originally set out to examine. For exactly the same reason, it also doesn't allow saturation of this classically-trivial inequality: 4 < 0, 0 > + < 0, 1 > + < 1, 0 > + < 1, 1 > +4 The reason is simple: just as with CHSH, saturation requires ( 0 0 ) and ( 0 1 ).

18 Part of what a lattice-theoretic approach provides, then, is a particular kind of insight into the question of why quantum systems can't saturate CHSH. However, part of what makes that question interesting is an obvious contrast with the classical case. Classical systems, which combine factorizability and Non-Signaling, can't even approach the Tsirelson bound, let alone the values ±4. This is because even though classical systems can display probabilistic behavior, it's always mathematically possible to represent their probabilities as mixtures of deterministic non-signaling state. Quantum systems aren't like that. Their behavior is probabilistic, but the probabilistic behavior can't be modeled by deterministic non-signaling states. The latticetheoretic approach suggests that the source of this fact lies in the nature of quantum contextuality. Different contexts don't correspond to classically-exclusive propositions. Contextuality has a distinctive logical character; propositions that correspond to different contexts do not admit of classical logical combination. Partial Boolean algebras and proposition systems both embody this idea, but the structural constraints of partial Boolean algebras are too weak. They ignore the fact that even though different contexts don't fit into a common Boolean algebra, they are not simply unrelated. If we assume that the structure underlying quantum probability is a proposition system rather than a PBA, then something interesting emerges. Even in the simplest realistic case (the qubit), the most natural way of combining proposition systems entails that quantum systems are deeply stochastic that none of their states are DNS states. While the PBA approach permits deeply stochastic states, the looser constraints have two notable consequences. First, they permit deterministic states that the lattice approach rules out. Second, they permit non-deterministic states that the lattice approach also excludes. Because PBAs don't require joins and meets to be everywhere defined, they doesn't impose systematic relations among correlations. The full polytope characterizes a world where almost anything goes; the world of the bare non-signaling polytope and the associated PBA is less permissive, to be sure, but is still a loose-jointed sort of place. Quantum systems understood lattice-theoretically are an intriguing combination of non-classical freedom and systematic constraint. This paper makes a case for the usefulness of proposition systems in quantum foundations. Nonetheless, we certainly don't see the attention we've paid to such lattices as in competition with information-theoretic approaches. The beautiful results from studies of quantum information stand on their own. Our belief is that the two approaches complement one

19 another. In particular, we believe, lattices may tell us something about the structure of the events in terms of which quantum informational exchanges play out. They reinforce the idea that there is something properly called quantum information, and that what quantum information reflects is the structure of the underlying space of possibilities.

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