PhyslibAlpha.ProbabilisticTheory.Classical.Compatibility

Let $E$ be an ordered additive commutative group with a unit element $1$. If $E$ satisfies the Riesz decomposition property, then for any two effects $e, f \in E$ (where an effect is an element $x$ such that $0 \le x \le 1$), there exists an element $g \in E$ that is a binary joint effect of $e$ and $f$.

Let $F$ be an ordered space that satisfies the Riesz decomposition property. For any two effects $e$ and $f$ in $F$ (that is, elements such that $0 \le e \le 1$ and $0 \le f \le 1$), the binary measurements (also known as "yes/no" measurements) associated with $e$ and $f$ are jointly measurable.

Let $E$ be an order-unit space. Let $x$ be a nonnegative element of the bidual of $E$ (i.e., $x \ge 0$). For any two positive functionals $\phi$ and $\psi$ on $E$, if $\phi \le \psi$, then $x(\phi) \le x(\psi)$.

Let $E$ be an order-unit space, and let $P$ and $N$ be positive linear functionals on $E$. Suppose that for every effect $a$ in $E$ (i.e., any element $a \in E$ satisfying $0 \le a \le 1$), the difference $P(a) - N(a)$ is at most $m$ for some real number $m$. Then, for any effect $x$ in the bidual space $\text{Bidual}(E)$ (i.e., any $x$ satisfying $0 \le x \le 1$), the difference between the values of $x$ applied to $P$ and $N$ is also at most $m$: $$x(P) - x(N) \le m.$$

Let $E$ be an ordered vector space and $E \to_p \mathbb{R}$ denote the set of positive linear functionals on $E$. Given a finite collection of positive linear functionals $\{\phi_i\}_{i \in I}$ and real weights $w_i \in \mathbb{R}$, there exist two positive linear functionals $P, N \in E \to_p \mathbb{R}$ such that for all $a \in E$: $$P(a) - N(a) = \sum_{i \in I} w_i \phi_i(a),$$ and for every element $x$ in the bidual space $E^{**}$, the action of $x$ on these functionals satisfies: $$x(P) - x(N) = \sum_{i \in I} w_i x(\phi_i).$$

Given a finite set of positive linear functionals $S$ on an ordered vector space $E$, the set of effect values is the set of all functions $v: S \to \mathbb{R}$ such that there exists an element $a \in E$ satisfying $0 \le a \le 1$ (an "effect") where $v(\phi) = \phi(a)$ for every $\phi \in S$.

Let $E$ be an ordered vector space and $S$ be a finite set of positive linear functionals on $E$. The set of effect values $\mathcal{V}_S$, defined as the set of all functions $v: S \to \mathbb{R}$ such that there exists an element $a \in E$ with $0 \le a \le 1$ satisfying $v(\phi) = \phi(a)$ for all $\phi \in S$, is a convex set.

Let $E$ be an order-unit space and $\text{Bidual}(E)$ be its bidual space. Let $x \in \text{Bidual}(E)$ be an effect, meaning $0 \leq x \leq 1$. For any finite set $S$ of positive linear functionals on $E$, the vector of values $(x(\phi))_{\phi \in S}$ belongs to the closure of the set of values $(\phi(a))_{\phi \in S}$ attainable by effects $a \in E$ (where $0 \leq a \leq 1$).

Let $E$ be an order-unit space and $E^{**}$ be its bidual space. Let $x$ be an effect in the bidual space, such that $0 \leq x \leq 1$. For any finite set $S$ of positive functionals $\phi : E \to \mathbb{R}$ and any $\varepsilon > 0$, there exists an effect $a$ in the original space $E$ (where $0 \leq a \leq 1$) such that for all $\phi \in S$, the evaluation of $a$ by $\phi$ is within $\varepsilon$ of the value of the bidual element $x$ at $\phi$: $$|\phi(a) - x(\phi)| < \varepsilon.$$

Let $E$ be an order-unit space and let $E^*_+$ denote the space of positive functionals $E \to_{p} \mathbb{R}$. For two elements $x, y$ in the bidual $E^{**}$ and a pair $i = (S, n)$ consisting of a finite set of positive functionals $S \subset E^*_+$ and a natural number $n$, the set of **joint candidates** is the collection of additive and positively homogeneous functions $z: E^*_+ \to \mathbb{R}$ such that: 1. For all $\phi \in E^*_+$, the value $z(\phi)$ is bounded as $-\phi(1) \leq z(\phi) \leq 2\phi(1)$. 2. For all $\phi \in S$, $z$ satisfies the conditions for being a joint effect of $x$ and $y$ up to an error of $\epsilon = \frac{1}{n+1}$: - $z(\phi) \geq -\epsilon$ - $z(\phi) \leq x(\phi) + \epsilon$ - $z(\phi) \leq y(\phi) + \epsilon$ - $z(\phi) \geq x(\phi) + y(\phi) - \phi(1) - \epsilon$

Let $E$ be an order-unit space and $E^*_+$ be the space of positive functionals on $E$. For any two elements $x, y$ in the bidual space $E^{**}$ and a pair $i = (S, n)$ consisting of a finite set of positive functionals $S \subset E^*_+$ and a natural number $n$, the set of **joint candidates** is defined as the collection of additive and positively homogeneous functions $z: E^*_+ \to \mathbb{R}$ satisfying: 1. $-\phi(1) \leq z(\phi) \leq 2\phi(1)$ for all $\phi \in E^*_+$. 2. For all $\phi \in S$, the following inequalities hold with $\epsilon = \frac{1}{n+1}$: - $z(\phi) \geq -\epsilon$ - $z(\phi) \leq x(\phi) + \epsilon$ - $z(\phi) \leq y(\phi) + \epsilon$ - $z(\phi) \geq x(\phi) + y(\phi) - \phi(1) - \epsilon$ The theorem states that this set of joint candidates is closed in the topology of pointwise convergence.

Let $E$ be an order-unit space with order unit $1$, and let $E^*_+$ be the cone of positive functionals on $E$. Given elements $x, y$ in the bidual $E^{**}$, a finite set of positive functionals $S \subset E^*_+$, and a natural number $n$, let $\epsilon = \frac{1}{n+1}$ represent the error tolerance. Suppose there exist elements $a, b, c \in E$ and a real number $\delta \in \mathbb{R}$ such that: 1. $\delta \sum_{\phi \in S} \phi(1) \leq \frac{\epsilon}{4}$ and $\delta \leq 1$. 2. For all $\phi \in S$, the values of $a$ and $b$ are close to $x$ and $y$ respectively: $|\phi(a) - x(\phi)| < \frac{\epsilon}{4}$ and $|\phi(b) - y(\phi)| < \frac{\epsilon}{4}$. 3. $a \leq 1$, and $c$ serves as an approximate joint effect for $a$ and $b$ in the sense that $-\delta \cdot 1 \leq c$, $c \leq a + \delta \cdot 1$, $c \leq b + \delta \cdot 1$, and $a + b - 1 \leq c + \delta \cdot 1$. Then the evaluation functional induced by $c$ (the map $\phi \mapsto \phi(c)$) belongs to the set of joint candidates for $x$ and $y$ associated with the pair $(S, n)$.

Let $E$ be an order-unit space. Suppose that for any two effects $e, f \in E$ and any $\delta > 0$, there exists an effect $g \in E$ that is a $\delta$-approximate binary joint effect of $e$ and $f$. Then, for any two effects $x, y$ in the bidual space $E^{**}$ and any pair $i = (S, n)$ consisting of a finite set of positive functionals $S \subset E^*_+$ and a natural number $n$, the set of joint candidates for $x$ and $y$ associated with $i$ is non-empty.

Let $E$ be an order-unit space and $E^*_+$ be the space of positive functionals on $E$. For any two elements $x, y$ in the bidual $E^{**}$ and indices $i, j \in \text{Finset}(E^*_+) \times \mathbb{N}$, if $i \leq j$ (which implies $i.1 \subseteq j.1$ and $i.2 \leq j.2$), then the set of joint candidates for $j$ is a subset of the set of joint candidates for $i$: $$\text{jointCandidates}(x, y, j) \subseteq \text{jointCandidates}(x, y, i)$$

Suppose that in an order-unit space $E$, every pair of effects $e, f \in [0, 1]_E$ is compatible, in the sense that for any $\delta > 0$, there exists an approximate binary joint effect $g \in E$. Let $x$ and $y$ be effects in the bidual space $E^{**}$ (i.e., $0 \le x, y \le 1$). Then the intersection of all sets of joint candidates $\text{jointCandidates}(x, y, i)$ over all indices $i = (S, n)$ (where $S$ is a finite set of positive functionals on $E$ and $n \in \mathbb{N}$) is non-empty: $$\bigcap_{i} \text{jointCandidates}(x, y, i) \neq \emptyset$$

Let $E$ be an order-unit space and $E^{**}$ be its bidual space. Let $x$ and $y$ be effects in $E^{**}$, such that $0 \le x, y \le 1$. If $z: E^*_+ \to \mathbb{R}$ is a function belonging to the intersection of all sets of joint candidates $\text{jointCandidates}(x, y, i)$ over all indices $i \in \text{Finset}(E^*_+) \times \mathbb{N}$, then there exists an element $z' \in E^{**}$ which is a binary joint effect of $x$ and $y$.

Let $E$ be an order-unit space. Suppose that for every pair of effects $e, f \in [0, 1]_E$ and every $\delta > 0$, there exists an approximate binary joint effect $g \in E$ for $e$ and $f$. Then, for any two effects $x, y \in [0, 1]_{E^{**}}$ in the bidual space of $E$, there exists an element $z \in E^{**}$ that is a binary joint effect for $x$ and $y$. Here, an effect is an element $v$ such that $0 \le v \le 1$.

For an element $x$ in the bidual space $\text{Bidual}(E)$ of an order-unit space $E$, the gauge of $x$ is the infimum of all non-negative real numbers $t$ such that $x$ is bounded above by $t$ times the order unit $1$. That is, $$\text{gauge}(x) = \inf \{ t \in \mathbb{R} \mid 0 \le t \text{ and } x \le t \cdot 1 \}$$ where $1$ is the canonical order unit in the bidual space.

Given an order-unit space $E$, let $\text{Bidual}(E)$ be its bidual space and $1$ be the order unit in $\text{Bidual}(E)$. For any element $x \in \text{Bidual}(E)$, the set of non-negative real numbers $t$ such that $x \le t \cdot 1$ is non-empty. That is, there exists some $t \in \mathbb{R}$ with $t \ge 0$ satisfying $x \le t \cdot 1$.

For any element $x$ in the bidual space $\text{Bidual}(E)$ of an order-unit space $E$, the gauge of $x$ is non-negative, i.e., $\text{gauge}(x) \ge 0$.

Let $E$ be an order-unit space and $\text{Bidual}(E)$ be its bidual space. For any element $x \in \text{Bidual}(E)$ and any non-negative real number $t \ge 0$, if $x$ is bounded above by $t$ times the canonical order unit $1$ (i.e., $x \le t \cdot 1$), then the gauge of $x$ satisfies $\text{gauge}(x) \le t$.

Let $E$ be an order-unit space and $\text{Bidual}(E)$ be its bidual space. For any element $x \in \text{Bidual}(E)$, it holds that $x \le \text{gauge}(x) \cdot 1$, where $\text{gauge}(x)$ is the gauge of $x$ (the infimum of all $t \ge 0$ such that $x \le t \cdot 1$) and $1$ is the canonical order unit in the bidual space.

In the bidual space $\text{Bidual}(E)$ of an order-unit space $E$, the gauge of the zero element is $0$. The gauge of an element $x \in \text{Bidual}(E)$ is defined as the infimum of all non-negative real numbers $t$ such that $x$ is bounded above by $t \cdot 1$, where $1$ is the canonical order unit in the bidual space.

Let $E$ be an order-unit space and let $\text{Bidual}(E)$ be its bidual space. For any element $x \in \text{Bidual}(E)$ such that $x \ge 0$ and its gauge is strictly positive ($\text{gauge}(x) > 0$), the scaled element $(\text{gauge}(x))^{-1} \cdot x$ is an effect in $\text{Bidual}(E)$. That is, $0 \le (\text{gauge}(x))^{-1} \cdot x \le 1$, where $1$ is the unit element in the bidual space.

Let $E$ be an order-unit space and $E^{**}$ its bidual space. Let $r$ and $w$ be non-negative elements in $E^{**}$ that are orthogonal, meaning they have no common non-zero lower bound (i.e., for any $c \in E^{**}$, if $0 \le c \le r$ and $0 \le c \le w$, then $c = 0$). For any positive linear functional $\phi$ on $E$, the following inequality holds: $$\text{gauge}(r) \cdot w(\phi) \leq \text{gauge}(w) \cdot (\text{gauge}(r) \cdot \phi(1) - r(\phi))$$ where $\text{gauge}(x) = \inf \{ t \in \mathbb{R} \mid 0 \le t \text{ and } x \le t \cdot 1 \}$ is the gauge of $x$ in the bidual space. This inequality implies that the normalized elements are "spread apart," effectively satisfying $\frac{r}{\text{gauge}(r)} + \frac{w}{\text{gauge}(w)} \le 1$.

Given elements $v_1, v_2, u$ in the bidual space $E^{**}$, the set of decompositions consists of all pairs of elements $(p_1, p_2) \in E^{**} \times E^{**}$ such that $0 \le p_1 \le v_1$, $0 \le p_2 \le v_2$, and their sum satisfies $p_1 + p_2 \le u$.

Let $E^{**}$ be the bidual space of an order-unit space $E$. For elements $v_1, v_2, u \in E^{**}$, let $\mathcal{D}(v_1, v_2, u)$ be the set of decompositions defined as the set of pairs $(p_1, p_2) \in E^{**} \times E^{**}$ such that $0 \le p_1 \le v_1$, $0 \le p_2 \le v_2$, and $p_1 + p_2 \le u$. If $C \subseteq \mathcal{D}(v_1, v_2, u)$ is a non-empty chain with respect to the product order, then there exists an upper bound $ub \in \mathcal{D}(v_1, v_2, u)$ such that for every $z \in C$, $z \le ub$.

Let $E^{**}$ be the bidual space of an ordered vector space $E$. For elements $v_1, v_2, u \in E^{**}$, let $(p_1, p_2)$ be a maximal element (with respect to the product order) in the set of pairs $(q_1, q_2)$ satisfying $0 \le q_1 \le v_1$, $0 \le q_2 \le v_2$, and $q_1 + q_2 \le u$. Then, the remainder $u - p_1 - p_2$ and the remaining space below $v_1$ (given by $v_1 - p_1$) have no common positive lower bound; that is, if $0 \le c \le u - p_1 - p_2$ and $c \le v_1 - p_1$, then $c = 0$.

Let $E^{**}$ be the bidual of an order-unit space $E$. For elements $v_1, v_2, u \in E^{**}$, let $(p_1, p_2)$ be a maximal element (with respect to the product order) of the set of pairs $(q_1, q_2) \in E^{**} \times E^{**}$ such that $0 \le q_1 \le v_1, 0 \le q_2 \le v_2$, and $q_1 + q_2 \le u$. Then, for any $c \in E^{**}$, if $0 \le c \le u - p_1 - p_2$ and $c \le v_2 - p_2$, it must be that $c = 0$.

Let $E$ be an order-unit space and $E^{**}$ be its bidual space. If every pair of effects in $E^{**}$ (elements $e \in E^{**}$ such that $0 \le e \le 1$) is compatible, then $E^{**}$ satisfies the Riesz decomposition property. That is, for any elements $u, v_1, v_2 \in E^{**}$ such that $0 \le u \le v_1 + v_2$, there exist $p_1, p_2 \in E^{**}$ such that $u = p_1 + p_2$ with $0 \le p_1 \le v_1$ and $0 \le p_2 \le v_2$.

The canonical embedding of an order-unit space $E$ into its bidual $E^{**}$ as a positive linear map. It maps an element $x \in E$ to the functional in $E^{**}$ that evaluates any linear functional $\varphi \in E^*$ at $x$. This map is positive, meaning it maps the positive cone of $E$ to the positive cone of $E^{**}$.

Let $E$ be an order-unit space. If the bidual space $E^{**}$ of $E$ satisfies the Riesz decomposition property, then the system $E$ is classical. This means that the dual cone $E^*_+$ of positive linear functionals on $E$ forms a lattice under the natural ordering, such that for any two positive functionals $\phi, \psi \in E^*_+$, there exists a least upper bound $\phi \vee \psi$.

Let $E$ be an order-unit space. If for every pair of effects $e, f \in E$ (where an effect is an element $x$ satisfying $0 \le x \le 1$) and for every $\delta > 0$, there exists an approximate binary joint effect $g \in E$ for $e$ and $f$ with error $\delta$, then the space $E$ is classical. Here, a space is defined as classical if its dual cone $E^*_+$ of positive linear functionals forms a lattice.

Let $F$ be an Archimedean order-unit space. If for every pair of effects $e, f \in F$ (elements satisfying $0 \le e, f \le 1$), their associated binary (yes/no) measurements are jointly measurable, then $F$ is classical. Classicality means that the dual cone $F^*_+$ of positive linear functionals on $F$ forms a lattice, which is equivalent to the state space of the system being a Choquet simplex.

Let $F$ be an Archimedean order-unit space. If $F$ is classical (meaning its dual cone $F^*_+$ of positive linear functionals forms a lattice), then for any two effects $e, f \in F$ (elements such that $0 \le e \le 1$ and $0 \le f \le 1$), there exists an element $g \in F$ that is a binary joint effect for $e$ and $f$.

**Kuramochi's Theorem.** Let $F$ be a complete Archimedean order-unit space. Then $F$ is classical if and only if every pair of binary measurements is jointly measurable. Specifically, the following conditions are equivalent: 1. $F$ is **classical**, meaning the dual cone $F^*_+$ of positive linear functionals on $F$ forms a lattice (which is equivalent to the state space of $F$ being a Choquet simplex). 2. For any two effects $e, f \in F$ (where an effect is an element $x \in F$ such that $0 \le x \le 1$), their corresponding binary (yes/no) measurements are jointly measurable.