Embedding Quantum Universes into Classical Ones 1

Cristian S. Calude,2     Peter H. Hertling,3     Karl Svozil4

Abstract

Do the partial order and ortholattice operations of a quantum logic correspond to the logical implication and connectives of classical logic? Re-phrased, how far might a classical understanding of quantum mechanics be, in principle, possible? A celebrated result by Kochen and Specker answers the above question in the negative. However, this answer is just one among different possible ones, not all negative. It is our aim to discuss the above question in terms of mappings of quantum worlds into classical ones, more specifically, in terms of embeddings of quantum logics into classical logics; depending upon the type of restrictions imposed on embeddings the question may get negative or positive answers.

1  Introduction

Quantum mechanics is a very successful theory which appears to predict novel ``counterintuitive'' phenomena (see Wheeler [50], Greenberger, Horne and Zeilinger [12]) even almost a century after its development, cf. Schrödinger [42], Jammer [19,20]. Yet, it can be safely stated that quantum theory is not understood (Feynman [10]). Indeed, it appears that progress is fostered by abandoning long-held beliefs and concepts rather than by attempts to derive it from some classical basis, cf. Greenberger and YaSin [13], Herzog, Kwiat, Weinfurter and Zeilinger [18] and Bennett [4].

But just how far might a classical understanding of quantum mechanics be, in principle, possible? We shall attempt an answer to this question in terms of mappings of quantum worlds into classical ones, more specifically, in terms of embeddings of quantum logics into classical logics.

One physical motivation for this approach is a result proven for the first time by Kochen and Specker [26] (cf. also Specker [43], Zierler and Schlessinger [52] and John Bell [2]; see reviews by Mermin [32], Svozil and Tkadlec [48], and a forthcoming monograph by Svozil [46]) stating the impossibility to ``complete'' quantum physics by introducing noncontextual hidden parameter models. Such a possible ``completion'' had been suggested, though in not very concrete terms, by Einstein, Podolsky and Rosen (EPR) [9]. These authors speculated that ``elements of physical reality'' exist irrespective of whether they are actually observed. Moreover, EPR conjectured, the quantum formalism can be ``completed'' or ``embedded'' into a larger theoretical framework which would reproduce the quantum theoretical results but would otherwise be classical and deterministic from an algebraic and logical point of view.

A proper formalization of the term ``element of physical reality'' suggested by EPR can be given in terms of two-valued states or valuations, which can take on only one of the two values 0 and 1, and which are interpretable as the classical logical truth assignments false and true, respectively. Kochen and Specker's results [26] state that for quantum systems representable by Hilbert spaces of dimension higher than two, there does not exist any such valuation s: L {0,1} defined on the set of closed linear subspaces of the space L (these subspaces are interpretable as quantum mechanical propositions) preserving the lattice operations and the orthocomplement, even if one restricts the attention to lattice operations carried out among commuting (orthogonal) elements. As a consequence, the set of truth assignments on quantum logics is not separating and not unital. That is, there exist different quantum propositions which cannot be distinguished by any classical truth assignment.

The Kochen and Specker result, as it is commonly argued, e.g. by Peres [35] and Mermin [32], is directed against the noncontextual hidden parameter program envisaged by EPR. Indeed, if one takes into account the entire Hilbert logic (of dimension larger than two) and if one considers all states thereon, any truth value assignment to quantum propositions prior to the actual measurement yields a contradiction. This can be proven by finitistic means, that is, with a finite number of one-dimensional closed linear subspaces (generating an infinite set whose intersection with the unit sphere is dense; cf. Havlicek and Svozil [17]). But, the Kochen-Specker argument continues, it is always possible to prove the existence of separable valuations or truth assignments for classical propositional systems identifiable with Boolean algebras. Hence, there does not exist any injective morphism from a quantum logic into some Boolean algebra.

Since the previous reviews of the Kochen-Specker theorem by Peres [34,35], Redhead [38], Clifton [6], Mermin [32], Svozil and Tkadlec [48], concentrated on the nonexistence of classical noncontextual elements of physical reality, we are going to discuss here some options and aspects of embeddings in greater detail. Particular emphasis will be given to embeddings of quantum universes into classical ones which do not necessarily preserve (binary lattice) operations identifiable with the logical or and and operations. Stated pointedly, if one is willing to abandon the preservation of quite commonly used logical functions, then it is possible to give a classical meaning to quantum physical statements, thus giving raise to an ``understanding'' of quantum mechanics.

Quantum logic, according to Birkhoff [5], Mackey [28], Jauch [21], Kalmbach [23], Pulmannová [37], identifies logical entities with Hilbert space entities. In particular, elementary propositions p,q, are associated with closed linear subspaces of a Hilbert space through the origin (zero vector); the implication relation is associated with the set theoretical subset relation , and the logical or , and , and not operations are associated with the set theoretic intersection , with the linear span of subspaces and the orthogonal subspace ^, respectively. The trivial logical statement 1 which is always true is identified with the entire Hilbert space H, and its complement with the zero-dimensional subspace (zero vector). Two propositions p and q are orthogonal if and only if p q. Two propositions p,q are co-measurable (commuting) if and only if there exist mutually orthogonal propositions a,b,c such that p = ab and q = ac. Clearly, orthogonality implies co-measurability, since if p and q are orthogonal, we may identify a, b, c with 0,p,q, respectively. The negation of p q is denoted by p \not q.

2  Varieties of embeddings

One of the questions already raised in Specker's almost forgotten first article [43]5 concerned an embedding of a quantum logical structure L of propositions into a classical universe represented by a Boolean algebra B. Thereby, it is taken as a matter of principle that such an embedding should preserve as much logico-algebraic structure as possible. An embedding of this kind can be formalized as a mapping j:L B with the following properties.6 Let p,q L.

(i)
Injectivity: two different quantum logical propositions are mapped into two different propositions of the Boolean algebra, i.e., if p q, then j(p) j(q).

(ii)
Preservation of the order relation: if p q, then j(p) j(q).

(iii)
Preservation of ortholattice operations, i.e. preservation of the

(ortho-)complement:
j(p) = j(p),
or operation:
j(pq) = j(p)j(q),
and operation:
j(pq) = j(p)j(q).

As it turns out, we cannot have an embedding from the quantum universe to the classical universe satisfying all three requirements (i)-(iii). In particular, a head-on approach requiring (iii) is doomed to failure, since the nonpreservation of ortholattice operations among nonco-measurable propositions is quite evident, given the nondistributive structure of quantum logics.

2.1  Injective lattice morphisms

Here we shall review the rather evident fact that there does not exist an injective lattice morphism from any nondistributive lattice into a Boolean algebra. We illustrate this obvious fact with an example that we need to refer to later on in this paper; the propositional structure encountered in the quantum mechanics of spin state measurements of a spin one-half particle along two different directions (mod p), that is, the modular, orthocomplemented lattice MO2 drawn in Figure 1 (where p- = (p+) and q- = (q+)).

0.80mm
Picture Omitted

Figure 1: Hasse diagram of the ``Chinese lantern'' form of MO2.

Clearly, MO2 is a nondistributive lattice, since for instance,

p- (q- q+) = p- 1 = p-,
whereas
(p- q-)(p- q+) = 00 = 0.
Hence,
p- (q- q+) (p- q-)(p- q+).
In fact, MO2 is the smallest orthocomplemented nondistributive lattice.

The requirement (iii) that the embedding j preserves all ortholattice operations (even for nonco-measurable and nonorthogonal propositions) would mean that

j(p-) (j(q-) j(q+)) (j(p-)j(q-))(j(p-) j(q+)). That is, the argument implies that the distributive law is not satisfied in the range of j. But since the range of j is a subset of a Boolean algebra and for any Boolean algebra the distributive law is satisfied, this yields a contradiction.

Could we still hope for a reasonable kind of embedding of a quantum universe into a classical one by weakening our requirements, most notably (iii)? In the next three sections we are going to give different answers to this question. In the first section we restrict the set of propositions among which we wish to preserve the three operations complement , or , and and . We will see that the Kochen-Specker result gives a very strong negative answer even when the restriction is considerable. In the second section we analyze what happens if we try to preserve not all operations but just the complement. Here we will obtain a positive answer. In the third section we discuss a different embedding which preserves the order relation but no ortholattice operation.

2.2  Injective order morphisms preserving ortholattice operations among orthogonal propositions

Let us follow Zierler and Schlessinger [52] and Kochen and Specker [26] and weaken (iii) by requiring that the ortholattice operations need only to be preserved among orthogonal propositions. As shown by Kochen and Specker [26], this is equivalent to the requirement of separability by the set of valuations or two-valued probability measures or truth assignments on L. As a matter of fact, Kochen and Specker [26] proved nonseparability, but also much more-the nonexistence of valuations on Hilbert lattices associated with Hilbert spaces of dimension at least three. For related arguments and conjectures, based upon a theorem by Gleason [11], see Zierler and Schlessinger [52] and John Bell [2].

Rather than rephrasing the Kochen and Specker argument [26] concerning nonexistence of valuations in three-dimensional Hilbert logics in its original form or in terms of fewer subspaces (cf. Peres [35], Mermin [32]), or of Greechie diagrams, which represent orthogonality very nicely (cf. Svozil and Tkadlec [48], Svozil [46]), we shall give two geometric arguments which are derived from proof methods for Gleason's theorem (see Piron [36], Cooke, Keane, and Moran [7], and Kalmbach [24]).

Let L be the lattice of closed linear subspaces of the three-dimensional real Hilbert space \Bbb R3. A two-valued probability measure or valuation on L is a map v:L{0,1} which maps the zero-dimensional subspace containing only the origin (0,0,0) to 0, the full space \Bbb R3 to 1, and which is additive on orthogonal subspaces. This means that for two orthogonal subspaces s1, s2 L the sum of the values v(s1) and v(s2) is equal to the value of the linear span of s1 and s2. Hence, if s1, s2, s3 L are a tripod of pairwise orthogonal one-dimensional subspaces, then

v(s1) + v(s2) + v(s3) = v(\Bbb R3) = 1.
The valuation v must map one of these subspaces to 1 and the other two to 0. We will show that there is no such map. In fact, we show that there is no map v which is defined on all one-dimensional subspaces of \Bbb R3 and maps exactly one subspace out of each tripod of pairwise orthogonal one-dimensional subspaces to 1 and the other two to 0.

In the following two geometric proofs we often identify a given one-dimensional subspace of \Bbb R3 with one of its two intersection points with the unit sphere

S2 = {x \Bbb R3  |  ||x|| = 1} .
In the statements ``a point (on the unit sphere) has value 0 (or value 1)'' or that ``two points (on the unit sphere) are orthogonal'' we always mean the corresponding one-dimensional subspaces. Note also that the intersection of a two-dimensional subspace with the unit sphere is a great circle.

To start the first proof, let us assume that a function v satisfying the above condition exists. Let us consider an arbitrary tripod of orthogonal points and let us fix the point with value 1. By a rotation we can assume that it is the north pole with the coordinates (0,0,1). Then, by the condition above, all points on the equator {(x,y,z) S2 |  z = 0} must have value 0 since they are orthogonal to the north pole.

Let q = (qx,qy,qz) be a point in the northern hemisphere, but not equal to the north pole, that is 0 < qz < 1. Let C(q) be the unique great circle which contains q and the points (qy,-qx,0)/[(qx2+qy2)] in the equator, which are orthogonal to q. Obviously, q is the northern-most point on C(q). To see this, rotate the sphere around the z-axis so that q comes to lie in the {y = 0}-plane; see Figure 2. Then the two points in the equator orthogonal to q are just the points (0,1,0), and C(q) is the intersection of the plane through q and (0,1,0) with the unit sphere, hence

C(q) = {p \Bbb R3  |  ($ a,b \Bbb R)   a2 + b2 = 1  and p = aq + b (0,1,0) } .
This shows that q has the largest z-coordinate among all points in C(q).

Figure

Figure 2: The great circle C(q).

Assume that q has value 0. We claim that then all points on C(q) must have value 0. Indeed, since q has value 0 and the orthogonal point (qy,-qx,0)/[(qx2+qy2)] on the equator also has value 0, the one-dimensional subspace orthogonal to both of them must have value 1. But this subspace is orthogonal to all points on C(q). Hence all points on C(q) must have value 0.

Now we can apply the same argument to any point [q\tilde] on C(q) (by the last consideration [q\tilde] must have value 0) and derive that all points on C([q\tilde]) have value 0. The great circle C(q) divides the northern hemisphere into two regions, one containing the north pole, the other consisting of the points below C(q) or ``lying between C(q) and the equator'', see Figure 2. The circles C([q\tilde]) with [q\tilde] C(q) certainly cover the region between C(q) and the equator.7 Hence any point in this region must have value 0.

But the circles C([q\tilde]) cover also a part of the other region. In fact, we can iterate this process. We say that a point p in the northern hemisphere can be reached from a point q in the northern hemisphere, if there is a finite sequence of points q = q0, q1, , qn-1, qn = p in the northern hemisphere such that qi C(qi-1) for i = 1,,n. Our analysis above shows that if q has value 0 and p can be reached from q, then also p has value 0.

The following geometric lemma due to Piron [36] (see also Cooke, Keane, and Moran [7] or Kalmbach [24]) is a consequence of the fact that the curve C(q) is tangent to the horizontal plane through the point q:

If q and p are points in the northern hemisphere with pz < qz, then p can be reached from q.

This result will be proved in Appendix A. We conclude that, if a point q in the northern hemisphere has value 0, then every point p in the northern hemisphere with pz < qz must have value 0 as well.

Consider the tripod (1,0,0), (0,[1/( 2)],[1/( 2)]),(0,-[1/( 2)],[1/( 2)]). Since (1,0,0) (on the equator) has value 0, one of the two other points has value 0 and one has value 1. By the geometric lemma and our above considerations this implies that all points p in the northern hemisphere with pz < [ 1/(2)] must have value 0 and all points p with pz > [1/(2)] must have value 1. But now we can choose any point p with [1/(2)] < pz < 1 as our new north pole and deduce that the valuation must have the same form with respect to this pole. This is clearly impossible. Hence, we have proved our assertion that there is no mapping on the set of all one-dimensional subspaces of \Bbb R3 which maps one space out of each tripod of pairwise orthogonal one-dimensional subspaces to 1 and the other two to 0.

In the following we give a second topological and geometric proof for this fact. In this proof we shall not use the geometric lemma above.

Fix an arbitrary point on the unit sphere with value 0. The great circle consisting of points orthogonal to this point splits into two disjoint sets, the set of points with value 1, and the set of points orthogonal to these points. They have value 0. If one of these two sets were open, then the other had to be open as well. But this is impossible since the circle is connected and cannot be the union of two disjoint open sets. Hence the circle must contain a point p with value 1 and a sequence of points q(n), n = 1,2, with value 0 converging to p. By a rotation we can assume that p is the north pole and the circle lies in the {y = 0}-plane. Furthermore we can assume that all points qn have the same sign in the x-coordinate. Otherwise, choose an infinite subsequence of the sequence q(n) with this property. In fact, by a rotation we can assume that all points q(n) have positive x-coordinate (i.e. all points q(n), n = 1,2, lie as the point q in Figure 2 and approach the north pole as n tends to infinity). All points on the equator have value 0. By the first step in the proof of the geometric lemma in the appendix, all points in the northern hemisphere which lie between C(q(n)) (the great circle through q(n) and (0,1,0)) and the equator can be reached from q(n). Hence, as we have seen in the first proof, v(q(n)) = 0 implies that all these points must have value 0. Since q(n) approaches the north pole, the union of the regions between C(q(n)) and the equator is equal to the open right half {q S2 |  qz > 0, qx > 0} of the northern hemisphere. Hence all points in this set have value 0. Let q be a point in the left half {q S2 |  qz > 0, qx < 0} of the northern hemisphere. It forms a tripod together with the point (qy,-qx,0)/[(qx2+qy2)] in the equator and the point

(-qx,-qy,[(qx2+qy2)/( qz)]) / ||(-qx,-qy,[(qx2+qy2)/( qz)])|| in the right half. Since these two points have value 0, the point q must have value 1. Hence all points in the left half of the northern hemisphere must have value 1. But this leads to a contradiction because there are tripods with two points in the left half, for example the tripod (-1/2,[1/( 2)],1/2), (-1/2,-[1/( 2)],1/2), ([1/( 2)],0,[1/( 2)]). This ends the second proof for the fact that there is no two-valued probability measure on the lattice of subspaces of the three-dimensional Euclidean space which preserves the ortholattice operations at least for orthogonal elements.

2.3  Injective morphisms preserving order as well as or and and operations

We have seen that we cannot hope to preserve the ortholattice operations, not even when we restrict ourselves to operations among orthogonal propositions.

An even stronger weakening of condition (iii) would be to require preservation of ortholattice operations merely among the center C, i.e., among those propositions which are co-measurable (commuting) with all other propositions. It is not difficult to prove that in the case of complete Hilbert lattices (and not mere subalgebras thereof), the center consists of just the least lower and the greatest upper bound C = {0,1} and thus is isomorphic to the two-element Boolean algebra 2 = {0,1}. As it turns out, the requirement is trivially fulfilled and its implications are quite trivial as well.

Another weakening of (iii) is to restrict oneself to particular physical states and study the embeddability of quantum logics under these constraints; see Bell, Clifton [1].

In the following sections we analyze a completely different option: Is it possible to embed quantum logic into a Boolean algebra when one does not demand preservation of all ortholattice operations?

One method of embedding an arbitrary partially ordered set into a concrete orthomodular lattice which in turn can be embedded into a Boolean algebra has been used by Kalmbach [22] and extended by Harding [16] and Mayet and Navara [31]. In these Kalmbach embeddings, as they may be called, the meets and joins are preserved but not the complement.

The Kalmbach embedding of some bounded lattice L into a concrete orthomodular lattice K(L) may be thought of as the pasting of Boolean algebras corresponding to all maximal chains of L [15].

First, let us consider linear chains

0 = a0a1 a2 1 = am. Such chains generate Boolean algebras 2m-1 in the following way: from the first nonzero element a1 on to the greatest element 1, form An = an(an-1), where (an-1) is the complement of an-1 relative to 1; i.e., (an-1) = 1-an-1. An is then an atom of the Boolean algebra generated by the bounded chain

0 = a0a1 a2 1.

Take, for example, a three-element chain 0 = a0 {a} a1{a,b} 1 = a2 as depicted in Figure 3a). In this case,

A1
=
a1(a0) = a11 {a}{a,b} = {a},
A2
=
a2(a1) = 1(a1) {a,b}{b} = {b}.
This construction results in a four-element Boolean Kalmbach lattice K(L) = 22 with the two atoms {a} and {b} given in Figure 3b).

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Picture Omitted

Figure 3: Examples of Kalmbach embeddings.

Take, as a second example, a four-element chain 0 = a0 {a} a1 {a,b} {a,b,c} 1 = a3 as depicted in Figure 3c). In this case,

A1
=
a1(a0) = a11 {a}{a,b,c} = {a},
A2
=
a2(a1) {a,b}{b,c} = {b},
A3
=
a3(a2) = 1(a2) {a,b,c}{c} = {c}.
This construction results in an eight-element Boolean Kalmbach lattice K(L) = 23 with the three atoms {a}, {b} and {c} depicted in Figure 3d).

To apply Kalmbach's construction to any bounded lattice, all Boolean algebras generated by the maximal chains of the lattice are pasted together. An element common to two or more maximal chains must be common to the blocks they generate.

Take, as a third example, the Boolean lattice 22 drawn in Figure 3e). 22 contains two linear chains of length three which are pasted together horizontally at their smallest and biggest elements. The resulting Kalmbach lattice K(22) = MO2 is of the ``Chinese lantern'' type, see Figure 3f).

Take, as a fourth example, the pentagon drawn in Figure 3g). It contains two linear chains: one is of length three, the other is of length 4. The resulting Boolean algebras 22 and 23 are again horizontally pasted together at their extremities 0,1. The resulting Kalmbach lattice is given in Figure 3h).

In the fifth example drawn in Figure 3i), the lattice has two maximal chains which share a common element. This element is common to the two Boolean algebras, hence central in K(L). The construction of the five atoms proceeds as follows:

A1
=
{a}{a,b,c,d} = {a},
A2
=
{a,b,c}{b,c,d} = {b,c},
A3
=
B3 = {a,b,c,d}{d} = {d},
B1
=
{b}{a,b,c,d} = {b},
B2
=
{a,b,c}{a,c,d} = {a,c},
where the two sets of atoms {A1,A2,A3 = B3} and {B1,B2,B3 = A3} span two Boolean algebras 23 pasted together at the extremities and at A3 = B3 and A3 = B3. The resulting lattice is 2×MO2 = L12 depicted in Figure 3j).

2.4  Injective morphisms preserving order and complementation

In the following, we shall show that any orthoposet can be embedded into a Boolean algebra where in this case by an embedding we understand an injective mapping preserving the order relation and the orthocomplementation.

A slightly stronger version of this fact using more topological notions has already been shown by Katrnoka [25]. Zierler and Schlessinger constructed embeddings with more properties for orthomodular orthoposets [52] and mentioned another slightly stronger version of the result above without explicit proof [52].

For completeness sake we give the precise definition of an orthoposet. An orthoposet (or orthocomplemented poset) (L, ,0,1,) is a set L which is endowed with a partial ordering , (i.e. a subset of L ×L satisfying (1) p p, (2) if p q and q r, then p r, (3) if p q and q p, then p = q, for all p,q,r L). Furthermore, L contains distinguished elements 0 and 1 satisfying 0 p and p 1, for all p L. Finally, L is endowed with a function (orthocomplementation) from L to L satisfying the conditions (1) p = p, (2) if p q, then q p, (3) the least upper bound of p and p exists and is 1, for all p,q L. Note that these conditions imply 0 = 1, 1 = 0, and that the greatest lower bound of p and p exists and is 0, for all p L.

For example, an arbitrary sublattice of the lattice of all closed linear subspaces of a Hilbert space is an orthoposet, if it contains the subspace {0} and the full Hilbert space and is closed under the orthogonal complement operation. Namely, the subspace {0} is the 0 in the orthoposet, the full Hilbert space is the 1, the set-theoretic inclusion is the ordering , and the orthogonal complement operation is the orthocomplementation .

In the rest of this section we always assume that L is an arbitrary orthoposet. We shall construct a Boolean algebra B and an injective mapping j:L B which preserves the order relation and the orthocomplementation. The construction goes essentially along the same lines as the construction of Zierler and Schlessinger [52] and Katrnoka [25] and is similar to the proof of the Stone representation theorem for Boolean algebras, cf. Stone [45]. It is interesting to note that for a finite orthoposet the constructed Boolean algebra will be finite as well.

We call a nonempty subset K of L an ideal if for all p,q L:

  1. if p K, then p\not K,
  2. if p q and q K, then p K .

Clearly, if K is an ideal, then 0 K. An ideal I is maximal provided that if K is an ideal and I K, then K = I.

Let be the set of all maximal ideals in L, and let B be the power set of considered as a Boolean algebra, i.e. B is the Boolean algebra which consists of all subsets of . The order relation in B is the set-theoretic inclusion, the ortholattice operations complement, or, and and are given by the set-theoretic complement, union, and intersection, and the elements 0 and 1 of the Boolean algebra are just the empty set and the full set . Consider the map

j:L B
which maps each element p L to the set
j(p) = { I  |  p \not I }
of all maximal ideals which do not contain p. We claim that the map j

  1. is injective,
  2. preserves the order relation,
  3. preserves complementation.

This provides an embedding of quantum logic into classical logic which preserves the implication relation and the negation.8

The rest of this section consists of the proof of the three claims above. Let us start with claim (ii). Assume that p,q L satisfy p q. We have to show the inclusion

j(p) j(q)  .
Take a maximal ideal I j(p). Then p \not I. If q were contained in I, then by condition 2. in the definition of an ideal also p had to be contained in I. Hence q \not I, thus proving that I j(q).

Before we come to claims (iii) and (i) we give another characterization of maximal ideals. We start with the following assertion which will also be needed later:

If I is an ideal and r L with r \not I and r\not I,
then also the set J = I {s L \mid s r} is an ideal.
(1)
Here is the proof: It is clear that J satisfies condition 2. in the definition of an ideal. To show that it satisfies condition 1. assume to the contrary that there exists s J and s J, for some s L. Then one of the following conditions must be true; (I) s,s J, (II) s r and s r, (III) s I and s r, (IV) s r, s I. The first case is impossible since I is an ideal. The second case is ruled out by the fact that r 1 (namely, r = 1 would imply r = 0 which would contradict our assumption r\not I). The third case is impossible since s r implies r s which, combined with s I would imply r I, contrary to our assumption. Finally the fourth case is nothing but a reformulation of the third case with s and s interchanged. Thus we have proved that J is an ideal and have proved the assertion (1).

Next, we prove the following new characterization of maximal ideals:

An ideal I is a maximal ideal iff r \not I implies r I.
(2)
To prove this first assume that for all r L, if r\not I, then r I and suppose I is a proper subset of an ideal K. Then there exists p K such that p\not I. By our hypothesis (for all r L, r\not I implies r I), we have p I. Thus both p K and p K. This contradicts the fact that K is an ideal.

Conversely, suppose that I is a maximal ideal in L and suppose, to the contrary, that for some r L,

r\not I and r\not I .
(3)
Of course r 1, since 1 = 0 and 0 I. Let
J = I(r)
(4)
where (r) = {s L  | s r} is the principal ideal of r (note that (r) is indeed an ideal). Then, under assumption (3), using (1) above, we have that J is an ideal which properly contains I. This contradicts the maximality of I and ends the proof of the assertion (2).

For claim (iii) we have to show the relation:

j(p) = \j(p) ,
for all p L. This can be restated as
I j(p)   iff   I \not j(p)
for all I . But this means p\not I   iff   p I, which follows directly from condition 1. in the definition of an ideal and from assertion (2).

We proceed to claim (i), which states that j is injective, i.e., if p q, then j(p) j(q). But p q is equivalent to p\not or q\not p. Furthermore, if we can show that there is a maximal ideal I such that q I and p\not I then it follows easily that j(p) j(q). Indeed, p\not I means I j(p) and q I means I\not j(q). It is therefore enough to prove that:

If p\not q, then there exists a maximal ideal I such that q I and p\not I.

To prove this we note that since p\not q, we have p 0. Let

pq = {K L \mid K is an ideal and p\not K and q K}.
We have to show that among the elements of p,q there is a maximal ideal. Therefore we will use Zorn's Lemma. In order to apply it to p,q we have to show that p,q is not empty and that every chain in p,q has an upper bound.

The set p,q is not empty since (q) p,q. Now we are going to show that every chain in p,q has an upper bound. This means that, given a subset (chain) C of p,q with the property

for all   J,K C   one has   J K  or K J  ,
we have to show that there is an element (upper bound) U p,q with K U for all K C. The union
UC =

K C 
K
of all ideals K C is the required upper bound! It is clear that all K C are subsets of UC. We have to show that UC is an element of p,q. Since p \not K for all K C we also have p \not UC. Similarly, since q K for some (even all) K C, we have q UC. We still have to show that UC is an ideal. Given two propositions r,s with r s and s UC we conclude that s must be contained in one of the ideals K C. Hence also r K UC. Now assume r UC. Is it possible that the complement r belongs to UC? The answer is negative, since otherwise r J and r K, for some ideals J,K C. But since C is a chain we have J K or K J, hence r, r K in the first case and r, r J in the second case. Both cases contradict the fact that J and K are ideals. Hence, UC is an ideal and thus an element of p,q. We have proved that p,q is not empty and that each chain in p,q has an upper bound in p,q.

Consequently, we can apply Zorn's Lemma to p,q and obtain a maximal element I in the ordered set p,q. Thus

p\not I and q I.
(5)
It remains to show that I is a maximal ideal in L. Thus suppose, to the contrary, that I is not a maximal ideal in L.

By (2) there exists r L such that both r\not I and r\not I. Furthermore, since p 0, then either p\not r or p\not r. Without loss of generality suppose

p\not r.
(6)
It follows, by (1), and since r\not I and r\not I, that I(r) is an ideal properly containing I. But since, by Conditions (5) and (6), q I and p\not r, we have

p\not I(r) and q I(r).

Thus I(r) pq and, since r\not I, we deduce that I(r) properly contains I, contradicting the fact that I is a maximal element in Ipq. This ends the proof of claim (i), the claim that the map j is injective.

We have shown:

Any orthoposet can be embedded into a Boolean algebra where the embedding preserves the order relation and the complementation.

2.5  Injective order preserving morphisms

In this section we analyze a different embedding suggested by Malhas [29,30].

We consider an orthocomplemented lattice (L, , 0, 1, ), i.e. a lattice (L, , 0, 1) with 0 x 1 for all x L, with orthocomplementation, that is with a mapping : L L satisfying the following three properties: a) x = x, b) if x y, then y x, c) x·x = 0 and y y = 1. Here x·y = glb (x,y) and x y = lub(x,y) .

Furthermore, we will assume that L is atomic9 and satisfies the following additional property:

for all   x,y L, x y    iff for every atom   a L, a x   implies   a y.
(7)

Every atomic Boolean algebra and the lattice of closed subspaces of a separable Hilbert space satisfy the above conditions.

Consider next a set U10 and let W(U) be the smallest set of words over the alphabet U {, } which contains U and is closed under negation (if A W(U), then A W(U)) and implication (if

A, B W(U), then A B W(U)).11 The elements of U are called simple propositions and the elements of W(U) are called (compound) propositions.

A valuation is a mapping

t: W(U) 2
such that t(A) t(A) and t(A B) = 0 iff t(A) = 1 and t(B) = 0. Clearly, every assignment s: U 2 can be extended to a unique valuation ts.

A tautology is a proposition A which is true under every possible valuation, i.e., t(A) = 1, for every valuation t. A set K W(U) is consistent if there is a valuation making true every proposition in K. Let A W(U) and K W(U). We say that A derives from K, and write K\models A, in case t(A) = 1 for each valuation t which makes true every proposition in K (that is, t(B) = 1, for all B K). We define the set of consequences of K by

Con(K) = {A W(U) \mid K\models A}.
Finally, a set K is a theory if K is a fixed-point of the operator Con:
Con(K) = K.
It is easy to see that Con is in fact a finitary closure operator, i.e., it satisfies the following four properties:

The first three properties can be proved easily. A topological proof for the fourth property can be found in Appendix B.

The main example of a theory can be obtained by taking a set X of valuations and constructing the set of all propositions true under all valuations in X:

Th(X) = { A W(U)\mid t(A) = 1,    for all   t X}.

In fact, every theory is of the above form, that is, for every theory K there exists a set of valuations X (depending upon K) such that K = Th(X). Indeed, take

XK = { t: W(U) 2 \mid t    valuationwith   t(A) = 1,   for all   A K},
and notice that

Th(XK)
=
{B W(U)\mid t(B) = 1,   for all   t XK}
=
{ B W(U)\mid t(B) = 1,   for every valuation with  t(A) = 1,
for all   A K}
=
Con(K) = K.

In other words, theories are those sets of propositions which are true under a certain set of valuations (interpretations).

Let now T be a theory. Two elements p,q U are T-equivalent, written p T q, in case p q T. The relation T is an equivalence relation. The equivalence class of p is [p]T = {q U\mid p T q} and the factor set is denoted by U T; for brevity, we will sometimes write [p] instead of [p]T. The factor set comes with a natural partial order:

[p] [q]    if  p q T.
Note that in general, (U T, ) is not a Boolean algebra.12

In a similar way we can define the T-equivalence of two propositions:

A T B    if   A B T.
Denote by [[A]]T (shortly, [[A]]) the equivalence class of A and note that for every p U,
[p] = [[p]] U.
The resulting Boolean algebra W(U) T is the Lindenbaum algebra of T.

Fix now an atomic orthocomplemented lattice (L, , 0, 1, ) satisfying (7). Let U be a set of cardinality greater or equal to L and fix a surjective mapping f: U L. For every atom a L, let sa: U 2 be the assignment defined by sa (p) = 1 iff a f(p). Take

X = {tsa \mid a    is an atom of  L}13  and T = Th(X) .

Malhas [29,30] has proven that the lattice (U T, ) is orthocomplemented, and, in fact, isomorphic to L. Here is the argument. Note first that there exist two elements 0, 1 in U such that f(0) = 0,   f(1) = 1. Clearly, 0\not T, but 1 T. Indeed, for every atom a, a f(1) = 1, so sa(1) = 1, a.s.o.

Secondly, for every p,q U,

p q T    iff   f(p) f(q).

If p q\not T, then there exists an atom a L such that tsa(p q) = 0, so sa(p) = tsa(p) = 1,   sa(q) = tsa(q) = 0, which-according to the definition of sa-mean a f(p), but a\not f(q). If f(p) f(q), then a f(q), a contradiction. Conversely, if f(p) \not f(q), then by (7) there exists an atom a such that a f(p) and a\not f(q). So,

sa(p) = tsa(p) = 1,  sa(q) = tsa(q) = 0, i.e., (p q)\not T.

As immediate consequences we deduce the validity of the following three relations: for all p,q U,

Two simple propositions p,q U are conjugate in case f(p) = f(q).14 Define now the operation *: UT UT as follows: [p]* = [q] in case q is a conjugate of p. It is not difficult to see that the operation * is well-defined and actually is an orthocomplementation. It follows that (UT, T, *) is an orthocomplemented lattice.

To finish the argument we will show that this lattice is isomorphic with L. The isomorphism is given by the mapping y: UT L defined by the formula y([p]) = f(p). This is a well-defined function (because f(p) = f(q) iff [p] = [q]), which is bijective ( y([p]) = y([q]) implies f(p) = f(q), and surjective because f is onto). If [p] [q], then f(p) f(q), i.e. y([p]) y([q]). Finally, if q is a conjugate of p, then

y([p]*) = y([q]) = f(q) = f(p) = y([p]).

In particular, there exists a theory whose induced orthoposet is isomorphic to the lattice of all closed subspaces of a separable Hilbert space. How does this relate to the Kochen-Specker theorem? The natural embedding

G: U T W(U) T,  G([p]) = [[p]]
is order preserving and one-to-one, but in general it does not preserve orthocomplementation, i.e. in general G([p]*) G([p]). We always have G([p]*) G([p]), but sometimes G([p])\not G([p]*). The reason is that for every pair of conjugate simple propositions p,q one has (p q) T, but the converse is not true.

By combining the inverse y-1 of the isomorphism y with G we obtain an embedding j of L into the Boolean Lindenbaum algebra W(U) T. Thus, the above construction of Malhas gives us another method how to embed any quantum logic into a Boolean logic in case we require that only the order is preserved.15

Next we shall give a simple example of a Malhas type embedding j: MO2 24. Consider again the finite quantum logic MO2 represented in Figure 1. Let us choose

U = {A,B,C,D,E,F,G,H}.
Since U contains more elements than MO2, we can map U surjectively onto MO2; e.g.,
f(A)
=
0,
f(B)
=
p-,
f(C)
=
p-,
f(D)
=
p+,
f(E)
=
q-,
f(F)
=
q+,
f(G)
=
1,
f(H)
=
1.

For every atom a MO2, let us introduce the truth assignment sa:U 2 = {0,1} as defined above (i.e. sa(r) = 1 iff a f(r)) and thus a valuation on W(U) separating it from the rest of the atoms of MO2. That is, for instance, associate with p- MO2 the function sp- as follows:

sp-(A) = sp-(D) = sp-(E) = sp-(F) = 0,
sp-(B) = sp-(C) = sp-(G) = sp-(H) = 1.
The truth assignments associated with all the atoms are listed in Table 1.

ABCDEFGH
sp- 0 11 0 0 01 1
sp+ 0 00 1 0 01 1
sq- 0 00 0 1 01 1
sq+ 0 00 0 0 11 1

Table 1: Truth assignments on U corresponding to atoms p-,p+,q-,q+ MO2.

The theory T we are thus dealing with is determined by the union of all the truth assignments; i.e.,

X = {tsp-,tsp+,tsq-,tsq+}  and T = Th(X) .
The way it was constructed, U splits into six equivalence classes with respect to the theory T; i.e.,
U T = {[A],[B],[D],[E],[F],[G]}.
Since [p] [q] if and only if (p q) T, we obtain a partial order on U T induced by T which isomorphically reflects the original quantum logic MO2. The Boolean Lindenbaum algebra W(U) T = 24 is obtained by forming all the compound propositions of U and imposing a partial order with respect to T. It is represented in Figure 4.

0.85mm
Picture Omitted

Figure 4: Hasse diagram of an embedding of the quantum logic MO2 represented by Figure 1. Concentric circles indicate the embedding.

The embedding is given by

j(0)
=
[[A]],
j( p-)
=
[[B]],
j( p+)
=
[[D]],
j(q-)
=
[[E]],
j(q+)
=
[[F]],
j(1)
=
[[G]] .
It is order-preserving but does not preserve operations such as the complement. Although, in this particular example, f(B) = (f(D)) implies (B D) T, the converse is not true in general. For example, there is no s X for which s(B) = s(E) = 1. Thus, (B E) T, but f(B) (f(E)).

One needs not be afraid of order-preserving embeddings which are no lattice morphisms, after all. Even automaton logics (see Svozil [47], Schaller and Svozil [39,40,41], and Dvureenskij, Pulmannová and Svozil [8]) can be embedded in this way. Take again the lattice MO2 depicted in Figure 1. A partition (automaton) logic realization is, for instance,

{{{1},{2,3}},{{2},{1,3}}},
with
{1}
p-,
{2,3}
p+,
{2}
q-,
{1,3}
q+,
respectively. If we take {1},{2} and {3} as atoms, then the Boolean algebra 23 generated by all subsets of {1,2,3} with the set theoretic inclusion as order relation suggests itself as a candidate for an embedding. The embedding is quite trivially given by
j(p) = p 23.
The particular example considered above is represented in Figure 5.

a) 0.60mm
Picture Omitted
        b) 1.30mm
Picture Omitted

Figure 5: Hasse diagram of an embedding of MO2 drawn in a) into 23 drawn in b). Again, concentric circles indicate points of 23 included in MO2.

It is not difficult to check that the embedding satisfies the requirements (i) and (ii), that is, it is injective and order preserving.

It is important to realize at that point that, although different automaton partition logical structures may be isomorphic from a logical point of view (one-to-one translatable elements, order relations and operations), they may be very different with respect to their embeddability. Indeed, any two distinct partition logics correspond to two distinct embeddings.

It should also be pointed out that in the case of an automaton partition logic and for all finite subalgebras of the Hilbert lattice of two-dimensional Hilbert space, it is always possible to find an embedding corresponding to a logically equivalent partition logic which is a lattice morphism for co-measurable elements (modified requirement (iii)). This is due to the fact that partition logics and MOn have a separating set of valuations. In the MO2 case, this is, for instance

{{{1,2},{3,4}},{{1,3},{2,4}}},
with
{1,2}
p-,
{3,4}
p+,
{1,3}
q-,
{2,4}
q+,
respectively. This embedding is based upon the set of all valuations listed in Table 2. These are exactly the mappings from MO2 to 2 preserving the order relation and the complementation. They correspond to the maximal ideals considered in Section 2.3. In this special case the embedding is just the embedding obtained by applying the construction of Section 2.3, which had been suggested by Zierler and Schlessinger [52].

p-p+q-q+
s1 1 01 0
s2 1 00 1
s3 0 11 0
s4 0 10 1

Table 2: The four valuations s1,s2,s3,s4 on MO2 take on the values listed in the rows.

The embedding is drawn in Figure 6.

0.85mm
Picture Omitted

Figure 6: Hasse diagram of an embedding of the partition logic {{{1,2}, {3,4}}, {{1,3}, {2,4}}} into 24 preserving ortholattice operations among co-measurable propositions. Concentric circles indicate the embedding.

3  Surjective extensions?

The original proposal put forward by EPR [9] in the last paragraph of their paper was some form of completion of quantum mechanics. Clearly, the first type of candidate for such a completion is the sort of embedding reviewed above. The physical intuition behind an embedding is that the ``actual physics'' is a classical one, but because of some yet unknown reason, some of this ``hidden arena'' becomes observable while others remain hidden.

Nevertheless, there exists at least one other alternative to complete quantum mechanics. This is best described by a surjective map f:B L of a classical Boolean algebra onto a quantum logic, such that |B| |L|.

Plato's cave metaphor applies to both approaches, in that observations are mere shadows of some more fundamental entities.

4  Summary

We have reviewed several options for a classical ``understanding'' of quantum mechanics. Particular emphasis has been given to techniques for embedding quantum universes into classical ones. The term ``embedding'' is formalized here as usual. That is, an embedding is a mapping of the entire set of quantum observables into a (bigger) set of classical observables such that different quantum observables correspond to different classical ones (injectivity).

The term ``observables'' here is used for quantum propositions, some of which (the complementary ones) might not be co-measurable, see Gudder [14]. It might therefore be more appropriate to conceive these ``observables'' as ``potential observables.'' After a particular measurement has been chosen, some of these observables are actually determined and others (the complementary ones) become ``counterfactuals'' by quantum mechanical means; cf. Schrödinger's catalogue of expectation values [42]. For classical observables, there is no distinction between ``observables'' and ``counterfactuals,'' because everything can be measured precisely, at least in principle.

We should mention also a caveat. The relationship between the states of a quantum universe and the states of a classical universe into which the former one is embedded is beyond the scope of this paper.

As might have been suspected, it turns out that, in order to be able to perform the mapping from the quantum universe into the classical one consistently, important structural elements of the quantum universe have to be sacrificed:

Since per definition, the quantum propositional calculus is nondistributive (nonboolean), a straightforward embedding which preserves all the logical operations among observables, irrespective of whether or not they are co-measurable, is impossible. This is due to the quantum mechanical feature of complementarity.
One may restrict the preservation of the logical operations to be valid only among mutually orthogonal propositions. In this case it turns out that again a consistent embedding is impossible, since no consistent meaning can be given to the classical existence of ``counterfactuals.'' This is due to the quantum mechanical feature of contextuality. That is, quantum observables may appear different, depending on the way by which they were measured (and inferred).
In a further step, one may abandon preservation of lattice operations such as not and the binary and and or operations altogether. One may merely require the preservation of the implicational structure (order relation). It turns out that, with these provisos, it is indeed possible to map quantum universes into classical ones. Stated differently, definite values can be associated with elements of physical reality, irrespective of whether they have been measured or not. In this sense, that is, in terms of more ``comprehensive'' classical universes (the hidden parameter models), quantum mechanics can be ``understood.''

At the moment we can neither say if the nonpreservation of the binary lattice operations (interpreted as and and or) is a too high price for value definiteness, nor can we speculate of whether or not the entire program of embedding quantum universes into classical theories is a progressive or a degenerative case (compare Lakatosch [27]).

Appendix A: Proof of the geometric lemma

In this appendix we are going to prove the geometric lemma due to Piron [36] which was formulated in Section 2.2. First let us restate it. Consider a point q in the northern hemisphere of the unit sphere S2 = {p \Bbb R3  |  ||p|| = 1}. By C(q) we denote the unique great circle which contains q and the points (qy,-qx,0)/[(qx2+qy2)] in the equator, which are orthogonal to q, compare Figure 2. We say that a point p in the northern hemisphere can be reached from a point q in the northern hemisphere, if there is a finite sequence of points q = q0, q1, , qn-1, qn = p in the northern hemisphere such that qi C(qi-1) for i = 1,,n. The lemma states:

If q and p are points in the northern hemisphere with pz < qz, then p can be reached from q.

For the proof we follow Cooke, Keane, and Moran [7] and Kalmbach [24]). We consider the tangent plane H = {p \Bbb R3 |  pz = 1} of the unit sphere in the north pole and the projection h from the northern hemisphere onto this plane which maps each point q in the northern hemisphere to the intersection h(q) of the line through the origin and q with the plane H. This map h is a bijection. The north pole (0,0,1) is mapped to itself. For each q in the northern hemisphere (not equal to the north pole) the image h(C(q)) of the great circle C(q) is the line in H which goes through h(q) and is orthogonal to the line through the north pole and through h(q). Note that C(q) is the intersection of a plane with S2, and h(C(q)) is the intersection of the same plane with H; see Figure 7.


Picture Omitted
Figure 7: The plane H viewed from above.

The line h(C(q)) divides H into two half planes. The half plane not containing the north pole is the image of the region in the northern hemisphere between C(q) and the equator. Furthermore note that qz > pz for two points in the northern hemisphere if and only if h(p) is further away from the north pole than h(q). We proceed in two steps.

Step 1. First, we show that, if p and q are points in the northern hemisphere and p lies in the region between C(q) and the equator, then p can be reached from q. In fact, we show that there is a point [q\tilde] on C(q) such that p lies on C([q\tilde]). Therefore we consider the images of q and p in the plane H; see Figure 8. The point h(p) lies in the half plane bounded by h(C(q)) not containing the north pole.


Picture Omitted
Figure 8: The point p can be reached from q.

Among all points h(q) on the line h(C(q)) we set [q\tilde] to be one of the two points such that the line trough the north pole and h(q) and the line through h(q) and h(p) are orthogonal. Then this last line is the image of C([q\tilde]), and C([q\tilde]) contains the point p. Hence p can be reached from q. Our first claim is proved.

Step 2. Fix a point q in the northern hemisphere. Starting from q we can wander around the northern hemisphere along great circles of the form C(p) for points p in the following way: for n 5 we define a sequence q0, q1, ,qn by setting q0 = q and by choosing qi+1 to be that point on the great circle C(qi) such that the angle between h(qi+1) and h(qi) is 2p/n. The image in H of this configuration is a shell where h(qn) is the point furthest away from the north pole; see Figure 9.


Picture Omitted
Figure 9: The shell in the plane H for n = 16.

First, we claim that any point p on the unit sphere with pz < qnz can be reached from q. Indeed, such a point corresponds to a point h(p) which is further away from the north pole than h(qn). There is an index i such that h(p) lies in the half plane bounded by h(C(qi)) and not containing the north pole, hence such that p lies in the region between C(qi) and the equator. Then, as we have already seen, p can be reached from qi and hence also from q. Secondly, we claim that qn approaches q as n tends to infinity. This is equivalent to showing that the distance of h(qn) from (0,0,1) approaches the distance of h(q) from (0,0,1). Let di denote the distance of h(qi) from (0,0,1) for i = 0,,n. Then di / di+1 = cos(2p/n), see Figure 9. Hence dn = d0 ·(cos(2p/n))-n. That dn approaches d0 as n tends to infinity follows immediately from the fact that (cos(2p/n))n approaches 1 as n tends to infinity. For completeness sake16 we prove it by proving the equivalent statement that log((cos(2p/n))n) tends to 0 as n tends to infinity. Namely, for small x we know the formulae cos(x) = 1-x2/2 + O(x4) and log(1+x) = x+O(x2). Hence, for large n,

log((cos(2p/n))n)
=
n ·log(1-2 p2
n2
+ O(n-4))
=
n ·( - 2 p2
n2
+ O(n-4))
=
- 2 p2
n
+ O(n-3)  .
This ends the proof of the geometric lemma.

Appendix B: Proof of a property of the set of consequences of a theory

In Section 2.5 we introduced the set Con(K) of consequences of a set K of propositions over a set U of simple propositions and the logical connectives negation    and implication . We mentioned four properties of the operator Con. In this appendix we prove the fourth property:

Con(K) =

{X K, X   finite} 
Con(X)  .

The inclusion Con(K) {X K, X   finite} Con(X) follows directly from the second property of Con, i.e., from the monotonicity: if X K, then Con(X) Con (K). For the other inclusion we assume that a proposition A Con(K) is given. We have to show that there exists a finite subset X K such that A Con(X).

In order to do this we consider the set V(W(U)) of all valuations. This set can be identified with the power set of U and viewed as a topological space with the product topology of |U| copies of the discrete topological space {0,1}. By Tychonoff's Theorem (see Munkres [33]) V(W(U)) is a compact topological space. For an arbitrary proposition B and valuation t the set {t V(W(U)) \mid t(B) = 0} of valuations t with t(B) = 0 is a compact and open subset of valuations because the value t(B) depends only on the finitely many simple propositions occurring in B.

Note that our assumption A Con(K) is equivalent to the inclusion

{t V(W(U)) \mid t(A) = 0}

B K 
{t V(W(U)) \mid t(B) = 0}.
Since the set on the left-hand side is compact, there exists a finite subcover of the open cover on the right-hand side, i.e. there exists a finite set X K with
{t V(W(U)) \mid t(A) = 0}

B X 
{t V(W(U)) \mid t(B) = 0}.
This is equivalent to A Con(X) and was to be shown.

Acknowledgement

The authors thank the anonymous referees for their extremely helpful suggestions and comments leading to a better form of the paper.

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Footnotes:

1 This paper has been completed during the visits of the first author at the University of Technology Vienna (1997) and of the third author at the University of Auckland (1997). The first author has been partially supported by AURC A18/XXXXX/62090/F3414056, 1996. The second author was supported by DFG Research Grant No. HE 2489/2-1.

2 Computer Science Department, The University of Auckland, Private Bag 92019, Auckland, New Zealand, e-mail: cristian@cs.auckland.ac.nz.

3 Computer Science Department, The University of Auckland, Private Bag 92019, Auckland, New Zealand, e-mail: hertling@cs.auckland.ac.nz.

4 Institut für Theoretische Physik, University of Technology Vienna, Wiedner Hauptstraß e 8-10/136, A-1040 Vienna, Austria, e-mail: svozil@tph.tuwien.ac.at.

5 In German.

6 Specker had a modified notion of embedding in mind; see below.

7 This will be shown formally in the proof of the geometric lemma below.

8 Note that for a finite orthoposet L the Boolean algebra B is finite as well. Indeed, if L is finite, then it has only finitely many subsets, especially only finitely many maximal ideals. Hence I is finite, and thus also its power set B is finite.

9 For every x L\{0}, there is an atom a L such that a x. An atom is an element a L with the property that if 0 y a, then y = 0 or y = a.

10 Not containing the logical symbols ,, .

11 Define in a natural way A B = A B, A B = (A B),

A B = (A B) (B A).

12 For instance, in case T = Con({p}), for some p U. If U has at least three elements, then (U T, ) does not have a minimum.

13 Recall that ts is the unique valuation extending s.

14 Of course, this relation is symmetrical.

15 In Section 2.4 we saw that it is possible to embed quantum logic into a Boolean logic preserving the order and the complement.

16 Actually, this is an exercise in elementary analysis.


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