[Paper Review] A map for simultaneous measurements for a quantum logic

Oľga Nánásiová|ArXiv.org|Jun 2, 2003

Quantum Information and Cryptography16 references3 citations

TL;DR

This paper introduces a simultaneous measurement map (s-map) on orthomodular lattices to generalize joint distributions for non-compatible quantum observables. By linking s-maps to Rényi-style conditional states, it enables conditional expectations even when observables are incompatible, revealing asymmetric independence—key for modeling non-classical causality in quantum logic systems.

ABSTRACT

In this paper we will study a function of simultaneus measurements for quantum events (s-map) which will be compared with the conditional states on an orthomodular lattice as a basic structure for quantum logic.

Motivation & Objective

• To define a function for simultaneous measurements (s-map) on orthomodular lattices (OMLs), the foundational structure of quantum logic.
• To establish a correspondence between s-maps and conditional states derived from Rényi's approach to conditioning in non-classical probability.
• To extend the concept of conditional expectation to non-compatible observables in quantum logic by defining $ E_f(x|B) $ for Boolean sub-algebras $ B $.
• To investigate asymmetric independence in quantum systems, where $ a \perp_f b $ does not imply $ b \perp_f a $, contrasting classical symmetry.
• To construct joint distributions for non-compatible observables using the s-map, enabling probabilistic analysis beyond standard quantum mechanics.

Proposed method

• Define an s-map $ p: L \times L \to [0,1] $ such that $ p(x,y) $ represents a joint probability for events $ x $ and $ y $, even when they are non-compatible.
• Use Rényi’s conditional probability framework to define a conditional state $ f: L \times L_c \to [0,1] $, where $ L_c $ is a $ \sigma $-conditional system.
• Establish equivalence: a conditional state $ f $ induces an s-map via $ p(x,y) = f(x, y) \cdot f(y,y) $, and vice versa.
• Define a conditional expectation $ z = E_f(x|B) $ for an observable $ x $ and Boolean sub-algebra $ B \subset L $, satisfying $ f(x,b) = f(z,b) $ for all $ b \in B $.
• Construct joint distribution $ p_{x,y}(E,F) = p(x(E), y(F)) $ for observables $ x, y $, using the s-map to define a distribution function $ F_{x,y}(r,s) $.
• Demonstrate existence of $ E_f(x|B) $ via algebraic consistency checks, showing $ f(z,1) = f(x,1) $ and matching moments across $ b \in B $.

Experimental results

Research questions

• RQ1 How can joint distributions be defined for non-compatible quantum observables in an orthomodular lattice framework?
• RQ2 What is the relationship between conditional states (Rényi-style) and the s-map for simultaneous measurements?
• RQ3 Can conditional expectations be defined for non-compatible observables, and if so, how do they differ from classical conditional expectations?
• RQ4 Does the independence relation induced by Rényi’s conditional states preserve symmetry in quantum logic, and what are the implications for causality?
• RQ5 Under what conditions does a conditional expectation $ E_f(x|B) $ exist, and how can it be constructed algebraically?

Key findings

• An s-map $ p: L \times L \to [0,1] $ exists and is equivalent to a conditional state $ f $, with $ p(x,y) = f(x,y) \cdot f(y,y) $, enabling joint probability assignment even for non-compatible events.
• The conditional state $ f $ induces an asymmetric independence: $ a \perp_f b $ does not imply $ b \perp_f a $, violating classical symmetry and reflecting non-causal quantum dependencies.
• For any observable $ x $ and Boolean sub-algebra $ B \subset L $, a conditional expectation $ z = E_f(x|B) $ exists such that $ f(x,b) = f(z,b) $ for all $ b \in B $, even when $ x $ and $ z $ are not compatible.
• In Example 3.1, $ z = E_f(x|B_b) $ is constructed explicitly with $ f(x,1) = f(z,1) = 1.6 $, and $ f(z,b) = f(z,b^{ot}) = 1.6 $, showing consistency across $ b \in B_b $.
• The joint distribution $ p_{x,y} $ is defined via $ p_{x,y}(E,F) = p(x(E), y(F)) $, and in the example, $ p_{x,y}(1,1) = 0.7 $, $ p_{x,y}(1,2) = 0.3 $, while $ p_{y,x}(1,1) = 0.4 $, $ p_{y,x}(1,2) = 0.6 $, showing asymmetry.
• For $ y $ with $ R(y) = B_b $, $ w = E_f(y|B_a) $ is constructed with $ f(y,1) = 1.7 = 0.4w_1 + 0.6w_2 $, and $ w_1 = 1.8 $, $ w_2 = 49/30 \approx 1.633 $, showing non-trivial conditional expectation.