On the non existence of two valued lattice homomorphisms of quantum logic

On the non-existence of two-valued lattice homomorphisms of quantum logic

Minoru Koga

@Misoca

2017/7/27

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1 Logic of a physical system: basic idea

2 Preliminaries: lattices, homomorphisms and filters

3 Main theorem

Part I

Logic of a physical system: basic idea

Logic of a physical system: basic idea

Reference

G. Birkhoff and J. von Neumann, The logic of quantum mechanics, Ann. Math., 37, (1936) 823-843.

Basic idea

experimental propositions: “observable A has a value in E ⊆ R” L: the set of experimental propositions

implication: order ≤

P1, P2: experimental propositions

P1≤ P2⇐⇒ P^def 1“ implies ”P2

ordered set (L , ≤): logic of a physical system

The logic of classical mechanics (Birkhoff, von Neumann (1936))

P: phase space (symplectic manifold)

f, g : P → R: observables (measurable functions)

experimental proposition: “observable f has a value in E” f⁻¹(E) = {p ∈ P | f (p) ∈ E} ⊆ P

⇒ (implication) : f⁻¹(E) ⊆ g⁻¹(F )

“f has a value in E”implies“g has a value in F ”

The logic of classical mechanics (Birkhoff, von Neumann (1936))

Logical connectives:

∧ (conjunction) : f⁻¹(E) ∩ g⁻¹(F )

∨ (disjunction) : f⁻¹(E) ∪ g⁻¹(F )

¬ (negation) : P \ f⁻¹(E) = f⁻¹_{(R \ E)}

The logic of quantum mechanics (Birkhoff, von Neumann (1936)) H: separable complex Hilbert space

L(H ): the set of all closed linear subspaces of H A, B: observables (self-adjoint operators) , A =^∫_RλdQA(λ), B =^∫_RλdQB(λ)

Axiom 1 (Born’s rule)

The probability that an observableA has a value in a Borel set E ⊆ R in a stateψ ∈ H is given by

∥QA(E)ψ∥²/∥ψ∥².

experimental proposition: “observable A has a value in E ⊆ R” MQ_A(E):={ψ ∈ H | QA(E)ψ = ψ}

=⇒ (implication) : “A has a value in E”implies“B has a value in F ” MQ_A(E)⊆ MQ_B(F )

The logic of quantum mechanics (Birkhoff, von Neumann (1936))

Logical connectives:

∧ (conjunction) : MQ_A(E)∩ MQ_B(F )

∨ (disjunction) : MQ_A(E)∨ MQ_B(F )= span(MQ_A(E)∪ MQ_B(F ))

¬ (negation) : M_Q^⊥

A^{(E)={ψ ∈ H | QA(E)ψ = 0}}

The logic of quantum mechanics: (L (H ), ⊆) (The lattice of all closed linear subspaces of H )

Part II

Preliminaries: lattices, homomorphisms and

filters

Posets

Definition 1 (posets)

Let L be a non-empty set. A binary relation ≤ on L is called a partial order if

≤ is reflexive, transitive and antisymmetric, i.e., reflexivity: a ≤ a for all a ∈ L ;

transitivity: a ≤ b and b ≤ c imply a ≤ c for all a, b, c ∈ L ; antisymmetry: a ≤ b and b ≤ a imply a = b for all a, b ∈ L .

A pair (L , ≤) consists of a non-empty set L and a partial order ≤ on L is called a partially ordered set (or poset for short).

A poset (L , ≤) is said to be bounded if it has both the maximum element 1 and minimum element 0 with respect to ≤, and denote it by (L , ≤, 0, 1).

Lattices

Definition 2 (lattices)

A poset (L , ≤) is a lattice if for any a, b ∈ L , there exist both the supremum (least upper bound) a ∨ b and the infimum (greatest lower bound) a ∧ b.

Definition 3 (distributivity)

A lattice (L , ≤) is said to be distributive if the following two conditions hold: a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) for all a, b, c ∈ L ; (1a) a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) for all a, b, c ∈ L . (1b)

Example 4

1 For a set X, (2^X, ⊆, ∅, X) is a bounded distributive lattice.

2 For a Hilbert space H with dim(H ) ≥ 2, (L (H ), ⊆, {0}, H ) is a bounded lattice but not distributive.

We call (L (H ), ⊆, {0}, H ) the quantum logic associated with H .

The quantum logic associated with a Hilbert space

Fact 5

Let(L (H ), ⊆, {0}, H ) be the quantum logic associated with a Hilbert space H . Then for anyM, N ∈ L (H ), we have

M ∨ N = span(M ∪ N ), M ∧ N = M ∩ N,

and the orthogonal complementM^⊥ofM is a complement of M , i.e., M ∨ M^⊥= H ,

M ∧ M^⊥={0}.

Lattice homomorphisms

Definition 6 (lattice homomorphisms)

For two bounded lattices (L1, ≤1, 0, 1), (L2, ≤2, 0, 1), a mapping ϕ : L1→ L2

is called a lattice homomorphism if ϕ satisfies the following conditions:

1 _{ϕ(0) = 0, ϕ(1) = 1;}

2 _{ϕ(a ∨1}b) = ϕ(a) ∨2ϕ(b), ϕ(a ∧1b) = ϕ(a) ∧2ϕ(b) for all a, b ∈ L1. In particular, if L2= 2 ={0, 1}, we say that ϕ is two-valued.

A lattice homomorphism ϕ is called a lattice isomorphism if ϕ is bijective; and we say that (L1, ≤1, 0, 1) is isomorphic to (L2, ≤2, 0, 1).

Fact 7

A lattice homomorphismϕ : L1→ L2 is order-preserving, i.e., for any a, b ∈ L1,

a ≤1b implies ϕ(a) ≤2ϕ(b).

Filters

Definition 8 (filters)

Let (L , ≤, 0, 1) be a bounded lattice. A subset F of L is called a (proper) filterin (L , ≤, 0, 1) if the following three conditions hold:

1 ₀̸∈ F and 1 ∈ F;

2 a ∧ b ∈ F for all a, b ∈ F ;

3 a ≤ b and a ∈ F imply b ∈ F for all a, b ∈ L . A filter F is called a prime filter if for all a, b ∈ L ,

a ∨ b ∈ F implies a ∈ F or b ∈ F .

Filters

Fact 9

LetF be a filter in a bounded lattice (L , ≤, 0, 1). Then for any a, b ∈ L , we have

a ∧ b ∈ F iff a ∈ F and b ∈ F . IfF is a prime filter, then we have

a ∨ b ∈ F iff a ∈ F or b ∈ F ;

moreover, if(L , ≤, 0, 1) is a Boolean algebra, then for any a ∈ L , we have a ∈ F or a^⊥∈ F,

wherea^⊥is the complement ofa in L .

Prime filter VS. two-valued lattice homomorphism

For a bounded lattice (L , ≤, 0, 1), if we consider L as a set of propositions, then two-valued lattice homomorphisms can be seen as valuation functions assigning values of truth and falsity for each proposition.

Lemma 10

Let(L , ≤, 0, 1) be a bounded lattice and F a subset of L . Then the following conditions are equivalent:

(a) F is a prime filter;

(b) F = ϕ⁻¹(1) for some two-valued lattice homomorphism ϕ : L → 2.

Representation theorem

Every bounded distributive lattice has “enough” prime filters in the following sense:

Fact 11 ([4, Theorem 0.7])

For every bounded distributive lattice(L , ≤, 0, 1), there exists a lattice of subsets of the set of all prime filters which is isomorphic to(L , ≤, 0, 1).

Proof.

Let P(L ) be the set of all prime filters in (L , ≤, 0, 1). A mapping ϕ : a 7→ P(a) := {F ∈ P(L ) | a ∈ F } .

gives a lattice isomorphism from (L , ≤, 0, 1) to ({P(a)}a∈L, ⊆, ∅, P(L )).

Question

Question 1

Are there any prime filters in L(H )?

Equivalently, are there any two-valued lattice homomorphisms in L(H )?

Part III

Main theorem

Main theorem

Theorem 12

Let H be a Hilbert space withdim(H )≥ 2. Then there exists no two-valued lattice homomorphismϕ : L (H ) → 2.

References

G. Birkhoff and J. von Neumann, The logic of quantum mechanics, Ann. Math., 37, (1936) 823-843.

G. W. Mackey, The Mathematical Foundations of Quantum Mechanics, W. A. Benjamin, Inc., (1963).

M. R´edei, Quantum Logic in Algebraic Approach, Kluwer Academic Publishers, (1998).

J. L. Bell, Set Theory: Boolean-Valued Models and Independence Proofs, 3rd. ed., Oxford University Press, Oxford Logic Guides, 47, (2011).