PhyslibAlpha.ProbabilisticTheory.Classical.NamiokaPhelps

For an ordered vector space $E$, a sequence of four elements $A = (A_0, A_1, A_2, A_3)$ in $E$ satisfies the facet sum property if there exist four non-negative elements $X_0, X_1, X_2, X_3 \in E$ (where $X_j \ge 0$ for each $j \in \{0, 1, 2, 3\}$) such that the components of $A$ are given by: - $A_0 = X_2 + X_3$ - $A_1 = X_0 + X_3$ - $A_2 = X_1 + X_2$ - $A_3 = X_0 + X_1$ This definition characterizes vertex values that arise as facet sums of four non-negative observables, typically used in the context of the Namioka–Phelps square test for classicality.

Let $E$ be an ordered vector space. For any four elements $X_0, X_1, X_2, X_3 \in E$, the vertex values of the composite observable $z = \sum_{j=0}^3 X_j \otimes \text{facet}_j$ in $E \otimes_{\mathbb{R}} \text{Square}$ are given by the 4-tuple: $$\text{vals}(z) = (X_2 + X_3, X_0 + X_3, X_1 + X_2, X_0 + X_1)$$ where $\text{facet}_j$ are the four facet observables of the square.

Let $E$ be an ordered real vector space and $\text{Square}$ be the space of observables on a square state space. If a composite observable $z \in E \otimes_{\mathbb{R}} \text{Square}$ belongs to the minimal tensor cone, then its vertex values $\text{vals}(z) = (A_0, A_1, A_2, A_3)$ are facet sums. Specifically, there exist four non-negative elements $X_0, X_1, X_2, X_3 \in E$ such that: - $A_0 = X_2 + X_3$ - $A_1 = X_0 + X_3$ - $A_2 = X_1 + X_2$ - $A_3 = X_0 + X_1$

Let $E$ be an ordered real vector space and $\text{Square}$ be the space of square observables (functions $v: \{0, 1, 2, 3\} \to \mathbb{R}$ satisfying $v_0 + v_3 = v_1 + v_2$). For any composite observable $z \in E \otimes_{\mathbb{R}} \text{Square}$, if the vertex values $\text{vals}(z) = (A_0, A_1, A_2, A_3) \in E^4$ are facet sums—meaning there exist four non-negative elements $X_0, X_1, X_2, X_3 \in E$ such that: - $A_0 = X_2 + X_3$ - $A_1 = X_0 + X_3$ - $A_2 = X_1 + X_2$ - $A_3 = X_0 + X_1$ then $z$ belongs to the minimal tensor cone of $E \otimes_{\mathbb{R}} \text{Square}$.

Let $E$ be an ordered real vector space with an order unit $1$. If the maximal tensor cone of $E \otimes_{\mathbb{R}} \text{Square}$ is contained in the Archimedean closure of the minimal tensor cone, then every two effects $e, f \in [0, 1] \subseteq E$ have an approximate binary joint effect. Specifically, for any $\delta > 0$, there exists an element $g \in E$ that serves as a binary joint effect of $e$ and $f$ within an error margin of $\delta$. Here, $\text{Square}$ denotes the space of observables on a square state space satisfying the diagonal relation $v_0 + v_3 = v_1 + v_2$ for vertex values $v_i$.

If an Archimedean order-unit space $E$ is nuclear—meaning that for every other Archimedean system, the minimal and maximal tensor cones of their composition coincide—then $E$ is a classical system. In this case, the space of positive functionals on $E$ forms a lattice, which corresponds to the state space being a Choquet simplex.

Let $E$ be an ordered real vector space. If the minimal tensor cone of the composite system $E \otimes \text{Square}$ is equal to its maximal tensor cone, then $E$ possesses the Riesz decomposition property. Here, $\text{Square}$ denotes the space of observables on a square state space, consisting of functions $v: \{0, 1, 2, 3\} \to \mathbb{R}$ that satisfy the diagonal relation $v_0 + v_3 = v_1 + v_2$.

Let $E$ be an ordered real vector space. If $E$ satisfies the Riesz decomposition property, then the maximal tensor cone of the tensor product $E \otimes_{\mathbb{R}} \text{Square}$ is contained in the minimal tensor cone of $E \otimes_{\mathbb{R}} \text{Square}$, where $\text{Square}$ is the space of observables on a square state space.

Let $E$ be an ordered real vector space. The minimal tensor cone of the tensor product $E \otimes \text{Square}$ is equal to the maximal tensor cone of $E \otimes \text{Square}$ if and only if $E$ satisfies the Riesz decomposition property. Here, $\text{Square}$ is the space of observables on a square state space, consisting of vectors $v \in \mathbb{R}^4$ that satisfy the diagonal relation $v_0 + v_3 = v_1 + v_2$.

Let $E$ be an ordered real vector space and $\text{Square}$ be the space of observables on a square state space (the space of functions on four vertices satisfying $v_0 + v_3 = v_1 + v_2$). If the minimal tensor cone of $E \otimes \text{Square}$ is equal to the maximal tensor cone, then $E$ is a classical system. This implies that the state space of $E$ is a Choquet simplex and the space of positive functionals on $E$ forms a lattice.

**Namioka–Phelps square test.** Let $E$ be a complete Archimedean order-unit space. The system $E$ is classical (meaning its state space is a Choquet simplex) if and only if the minimal and maximal tensor cones of $E$ composed with the square state space observables ($\text{Square}$) are equal: $$\text{minTensorCone}(E, \text{Square}) = \text{maxTensorCone}(E, \text{Square})$$ where $\text{Square}$ is the space of observables on a square, defined as the subspace of $\mathbb{R}^4$ satisfying $v_0 + v_3 = v_1 + v_2$.

Let $E$ be a complete Archimedean order-unit space. Every two yes/no measurements in $E$ (represented by effects $e, f \in E$ such that $0 \le e \le 1$ and $0 \le f \le 1$) are jointly measurable if and only if the minimal tensor cone of $E \otimes \text{Square}$ is equal to the maximal tensor cone.