Quantum logics with given centres and variable state spaces Mirko Navara 1, Pavel Ptak 2 Abstract We ask which logics with a given centre allow for en

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1 Quantum logics with given centres and variable state spaces Mirko Navara 1, Pavel Ptak 2 Abstract We ask which logics with a given centre allow for enlargements with an arbitrary state space. We show in this note that these are precisely those logics the centre of which possesses a two-valued state and the state space of which is nonempty. This extends the results of [1] and [9] and supplements the results of [3] and [10]. We also comment on some related questions. AMS Subj. Class: 81B10, 06C15, 46E27 Key words: quantum logic (=-orthocomplete partially ordered set), state (=-additive probability measure), centre, Boolean sublogic. Introduction and basic notions In the logico-algebraic approach to the foundation of quantum mechanics we usually identify the \quantum logic" of an experiment with a -orthocomplete orthomodular poset and the \state" of an experiment with a -additive probability measure (see [5], [14], [16], etc.). The set of all \absolutely compatible" events of an experiment is then identied with the centre of the corresponding -orthocomplete orthomodular poset. A natural question arises whether this identication allows for independence of the state space and the centre (it should be noted that, for instance, in the von Neumann algebra formalism this is not the case, see [6]). We want to show in this note that it is \almost" so. Under a very mild condition on the centre we are able to guarantee a full independence. Moreover, we also ensure an arbitrary degree of noncompatibility in our constructions (i.e., in proving that there is a quantum logic with a given centre and an arbitrary state space, we ensure that such a logic may be as \non-classical" as needed). The constructions are rather technical in places. Routine details are therefore sometimes omitted. Nevertheless, all essential ideas are at least sketched. A reader interested exclusively in orthomodular lattices can simply replace the word \logic" with \orthomodular lattice" { the results remain valid without changes. Let us recall some basic notions as we shall use them in the sequel. By a quantum logic (or simply by a logic) we mean a partially ordered set L = (L; ) with a least element, 0, and a greatest element, 1, and with a complementation operation 0 such that the following conditions are satised: (i) a = a 00, and if a b, then b 0 a 0 (a; b 2 L), (ii) if a i 2 L (i 2 N) and a i a 0 j for any i; j 2 N (i 6= j), then W i2n a i exists in L, (iii) if a b, then a _ (b ^ a 0 ) = b (a; b 2 L). Let us reserve the symbol L for logics. Typical examples of logics are Boolean -algebras or lattices of projections in a von Neumann algebra. We do not assume that quantum logics are necessarily lattices. Let us dene the notion of sublogic of a logic. Let K and L be logics. Then K is said to be a sublogic of L if there is an injective mapping e: K! L such that (i) e(0) = 0, (ii) e(a 0 ) = e(a) 0 (a 2 K), (iii) e?w i2n a W i = i2n e(a i) (a i 2 K for any i 2 N) provided a i a 0 j (i 6= j), (iv) a $ b (in K) if and only if e(a) $ e(b) (in L) (recall that the symbol a $ b means that a is compatible with b, i.e., a = (a ^ b) _ (a ^ b 0 ) and b = (a ^ b) _ (b ^ a 0 )). 1 Center for Machine Perception, Faculty of Electrical Engineering, Czech Technical University, Karlovo nam. 13, Praha, Czech Republic, navara@cmp.felk.cvut.cz 2 Department of Mathematics, Faculty of Electrical Engineering, Czech Technical University, Technicka 2, Praha, Czech Republic, ptak@math.feld.cvut.cz

2 If K is a sublogic of L, then K is a logic in its own right with the operations inherited from L. Thus, in this case we can understand K as a subset of L. If K is a Boolean -algebra, we call it a Boolean sublogic of L. The centre of L, C(L), is dened to be the intersection of all maximal Boolean sublogics of L. Of course, C(L) is again a Boolean sublogic of L. Moreover, a 2 C(L) if and only if a $ b for any b 2 L. Let L be a logic. By a state on L we mean a probability measure on L. Thus, a mapping s: L! [0; 1] is a state on L if (i) s(1) = 1, (ii) s?w i2n a P i = i2n s(a i) whenever a i 2 L (i 2 N) and a i a 0 j (i 6= j). Let us denote by S (L) (resp. S 2 (L)) the set of all states (resp. the set of all two-valued states) on L. The set S (L) naturally carries an ane and a topological structure inherited from the space [0; 1] L. The set S (L) endowed with this ane and topological structure is called the state space of L. It is easily seen that S (L) is convex (obviously, S (L) is almost never compact). There is a characterization of state spaces of logics among convex sets. It was proved in [12] that the state spaces are exactly s-semi-exposed faces in compact convex subsets of locally convex Hausdor topological spaces (or, equivalently, in a product of real lines). (Let us only recall the denition of an s-semi-exposed face. Consider the ane hull, A(K), of a compact convex set K, and its second dual, A(K). An element f 2 A(K) is said to be an s-functional if f is the weak limit of an isotone sequence of elements in [0; e] \ A(K) (e is the unit function on K). A face F of K is said to be s-exposed if there exists an s-functional f such that F = f?1 (1) \ K, and F is said to be s-semi-exposed if F is an intersection of s-exposed faces of K.) We are ready to state our result. It is convenient to employ the following notion (naturally, the sign \=" means the existence of an ane homeomorphism when applied for state spaces, the existence of a Boolean -isomorphism when applied for Boolean -algebras). Denition. Let L be a logic and let B be a Boolean -algebra. Then L is called state-space-exible with B xed for the centre if the following condition is satised: If we are given an s-semi-exposed face F, then there is a logic K such that (i) L is a sublogic of K, (ii) C(K) = B, (iii) S (K) = F. THEOREM. Let L be a logic and let B be a Boolean -algebra. Then L is state-space-exible with B xed for the centre if and only if S (L) 6= ; and S 2 (B) 6= ;. Proof. Let us rst show that the condition is necessary. Obviously, S (L) 6= ;, because if it is not the case then L cannot be embedded into any logic which possesses states. Let us show that S 2 (B) 6= ;. We have to verify that if L is state-space-exible with B xed for the centre then there is a two-valued state on B. Take a singleton for F. Thus, set F = f0g. Obviously, F is an s-semi-exposed face in the interval [0; 1]. Suppose that K is such a logic that allows for an embedding e: L! K and, moreover, suppose that C(K) = B and S (K) = F. If c 2 C(K), then evidently K = [0; c] K [0; c 0 ] K. (Here the symbols [0; c] K ; [0; c 0 ] K denote the corresponding intervals in K.) Consider now the state spaces S ([0; c] K ) and S ([0; c 0 ] K ). If S ([0; c] K ) 6= ; and, also, S ([0; c 0 ] K ) 6= ;, then we can easily construct two dierent states on K. This is absurd. Thus, one of these state spaces must be empty. Let us say that S ([0; c] K ) = ;. Then S ([0; c 0 ] K ) is a singleton. This means that for the restriction, ~s, of the state s, s 2 S (K), to C(K), we have ~s(c) = 0 and ~s(c 0 ) = 1. This can be applied to every element of C(K), obtaining a two-valued state on C(K). Since C(K) = B, we have S 2 (B) 6= ; which we wanted to show. Let us show that the condition is sucient. The proof follows to certain extent the technique of [1] and [9], we therefore omit technical details. Let us divide the proof into a few parts. A. Suppose that F is a singleton and B = f0; 1g. Then we claim that L can be embedded into a logic K such that C(K) = f0; 1g and S (K) = F. In order to obtain the proof of A, let us make the following series of constructions. a) Fix a state on L, some s 2 S (L). Choose an M -base, M, in L. (Let us recall that an M -base is such a collection M; M L, that the following conditions are satised: (i) a 2 M; a b ) b 2 M; (ii) a =2 M ) a 0 2 M. An M -base exists in every logic and each M -base contains no orthogonal pair { see [7] and [8].) We will now enlarge the logic L in such a way that all states on the larger 2

3 logic will be uniquely determined by their values attained on elements of L. Since the values of a state on L are already determined by the values on M, we will obtain an enlargement with a singleton state space. b) Suppose that r 2 [0; 1]. Then there is a logic, L r, such that S (L r ) is a singleton and, moreover, there is an atom a r 2 L r such that the only state s r 2 S (L r ) satises the condition s r (a r ) = r. This construction is demonstrated e. g., in [13]. c) Suppose that a 2 M 0, where M is the chosen M -base and M 0 = fx j x = y 0 for some y 2 Mg. We will now use the following construction manoeuvre which will indicate the way how we shall enlarge the logic L and, also, it will make the value of any state of the enlargement equal to s(a) (recall that s 2 S (L) is the xed state). Put r = s(a) and construct the logic L r (containing an atom a r ) with the properties listed in the above part b). We will utilize the construction called in [11] a replacement of an atom a r in L r with a logic I a = [0; a] L. This construction proves the following proposition. There is a logic L a with the properties (i) there is an isomorphism h a of L r and a sublogic of L a, (ii) there is an isomorphism i a of I a and [0; h a (a r )] L a, (iii) the logic L a is generated by h a (L r ) [ i a (I a ). The construction of such a logic is presented in detail in [11]. (All elements of the set-theoretical dierence L a? h a (L r ) can be expressed as orthogonal suprema h a (b) _ i a (c), where b 2 [0; a 0 r], c 2 I a.) Finally, we identify the corresponding elements of the isomorphic sublogics [0; a] L [ [a 0 ; 1] L and [0; h(a r )] L a [[h(a r ) 0 ; 1] L a. Each state on L a is uniquely determined by its values on [0; h(a r )] L a (= [0; a] L ) and attains the value r on a r (= a). d) Suppose that we have constructed the logic L a for any a 2 M 0. Thus, we now have a family P = flg [ fl a j a 2 M 0 g: We will perform the pasting of the whole family P, reaching the logic K we need. This can be done (as one veries in [11] and [9]). Let us only indicate the basic ideas of the construction. Observe rst that the following assertions hold true for all R; S 2 P, R 6= S: (i) if R; S 2 P, then R \ S is a sublogic of both R and S so that the orderings (resp. the orthocomplementations) of R and S coincide on R \ S, (ii) if x 2 R \ S, then either [0; x] R = [0; x] S R \ S or [x; 1] R = [x; 1] S R \ S (this follows from the fact that M is an M -base), (iii) R \ S is closed under the formation of countable orthogonal suprema. (iv) R \ S L, Put K = S P and dene the partial ordering and orthocomplementation so that a b in K (resp., a = b 0 in K) if and only if a b (resp., a = b 0 ) in some R 2 P. It is not dicult to verify that K endowed with this partial ordering and orthocomplementation is a logic. Obviously, S (K) is a singleton (S (K) contains exactly the (unique) extension of s over K!). This completes the proof of the proposition A. B. Thus, for the given logic L we have constructed a logic K such that L is a sublogic of K and the state space S (K) is a singleton. Suppose now that we are given an arbitrary s-semi-exposed face F. By [12], there is a non-boolean logic T such that S (T ) = F. Take the horizontal sum of T and K. Denote this logic by V. Then, again, S (V ) = F and C(V ) = f0; 1g. Moreover, L is a sublogic of V. We will now show that there is a logic, Q, such that S (Q) = F, C(Q) = B and such that V is a sublogic of Q. This will complete the proof. Take rst a nite stateless logic, Z (see [4]). Thus, S (Z) = ; and Z is nite. Let us denote by Y the horizontal sum of V and Z. Then S (Y ) = ; and V is a sublogic of Y. Consider now the Boolean -algebra B. By the Loomis-Sikorski theorem [15], there is a Boolean -algebra of subsets of a set S so that B is a -epimorphic image of. In other words, there is a Boolean -epimorphism h:! B. By our assumption, there is a two-valued state on B. Let us denote it by t. Then the composition mapping h t:! f0; 1g is a two-valued state on. Obviously, the set F = fa 2 j t(h(a)) = 1g forms an ultralter on which has the countable intersection property. We will use this ultralter in our construction that follows. Fix a state u on V. Let us consider the set of functions f : S! Y which have the following properties: 3

4 (i) there is an F f 2 F such that fjf f is a constant function attaining a value in V, (ii) for all z 2 Y? V = Z? f0; 1g, the set U z = f?1 (z) belongs to? F, (iii) the composition function f u: (S? S z2z U z)! [0; 1] is measurable with respect to Borel subsets of [0; 1] and the -algebra e, where e is the trace of on the set S? Sz2Z U z. It is not dicult to check that the set of all functions f : S! Y with the above properties (i), (ii) and (iii) is a logic (the most technical part is the verication of -orthocompleteness which follows from the fact that measurable functions are closed under the formation of countable sums). Let us denote this logic by W. Since f 2 C(W ) if and only if f(s) f0; 1g, we see that C(W ) =. Moreover, S (W ) = S (V ), since Z is stateless and therefore, for any state w 2 S (W ), w(f) depends only on the value of f on a set F f from the ultralter F. Since on F f the function f attains a constant value and this value belongs to V, we have S (W ) = S (V ). Let us nally factorize the logic W with respect to the -ideal of all functions from W which attain nonzero values only on the sets which belong to the kernel, h?1 (0), of the Loomis{Sikorski -epimorphism h:! B. We obtain a logic, Q, such that C(Q) = B. Moreover, S (Q) = S (V ) since no set belonging to the ultralter F can be in the kernel of h. The proof is complete. By a slight modication of our construction, we can prove the following result of a separate purely algebraic interest. Proposition. Let L be a logic and let B be a Boolean -algebra. Let S (L) 6= ;. Then L can be embedded in a logic K such that C(K) = B. (The result remains valid if we write \-orthocomplete orthomodular lattice" instead of \logic".) We conjecture that the above result is valid for all logics L and all Boolean -algebras, i. e., we conjecture that the assumption S (L) 6= ; is not necessary (observe that in the nite additive setup of the problem it is so the bounded Boolean power does the job, see e.g. [2] and [13]). Let us formulate the lattice version (resp. the complete lattice version) of this conjecture as an open question. Open question: Suppose that L is a -orthocomplete orthomodular lattice and suppose that B is a Boolean -algebra. Is there a -orthocomplete orthomodular lattice K such that L is a sub-orthocomplete orthomodular lattice of K and C(K) = B? Is there a positive answer to this question if we replace \-orthocomplete" with \orthocomplete"? Acknowledgement. The authors acknowledge the support by the project no. VS96049 of the Czech Ministry of Education and the grant no. 201/96/0117 of the Grant Agency of the Czech Republic. References [1] Binder, J. (1986). On the interplay of the centre and the state space in quantum logics, Reports on Mathematical Physics, 24, 337{341. [2] Bruns, G., Greechie, R.J., Harding, J., Roddy, M. (1990). Completions of orthomodular lattices, Order, 7, 67{76. [3] Foulis, D., Ptak, P. (1995). On absolutely compatible elements and hidden variables in quantum logics, Ricerche di Matematica, 44, 19{29. [4] Greechie, R.J. (1971). Orthomodular lattices admitting no states, J. Comb. Theory (A), 10, 119{132. [5] Gudder, S. (1979). Stochastic Methods in Quantum Mechanics, North Holland, New York. [6] Kadison, R.V. (1965). Transformations of states in operator theory and dynamics, Topology, 3, Suppl. 2, 177{198. [7] Katrnoska, F. (1982). On the representation of orthocomplete posets, Comment. Math. Univ. Carolin., 23, 489{498. 4

5 [8] Marlow, A.R. (1978). Quantum theory and Hilbert space, J. Math. Phys., 19, 1{15. [9] Navara, M., Ptak, P. (1988). Enlargements of logics (-orthocomplete case), Proc. Conf. Topology and Measure, V, Greifswald, 109{115. [10] Navara, M., Ptak, P., Rogalewicz, V. (1988). Enlargements of quantum logics, Pacic J. Math., 135, 361{369. [11] Navara, M., Rogalewicz, V. (1991). The pasting constructions for orthomodular posets, Math. Nachrichten, 154, 157{168. [12] Navara, M., Ruttimann, G. (1991). A characterization of -state spaces of orthomodular lattices, Expositiones Math., 9, 275{284. [13] Ptak, P. (1987). Exotic logics, Colloquium Math., 54, 1{7. [14] Ptak, P., Pulmannova, S. (1991). Orthomodular Structures as Quantum Logics, Kluwer, Dordrecht. [15] Sikorski, R. (1969). Boolean Algebras, Springer-Verlag, Berlin. [16] Varadarajan, V.S. (1968). Geometry of Quantum Theory I. Van Nostrand, Princeton, New Jersey. 5