PhyslibAlpha.ProbabilisticTheory.State.Convex

Given two states $\omega, \varphi \in \mathcal{S}[\mathbb{R}, E]$ and a probability $t \in [0, 1]$, this function defines a new state representing the probabilistic mixture of $\omega$ and $\varphi$. The resulting state is the convex combination $t\omega + (1-t)\varphi$, which maps an element $A \in E$ to the value $t \omega(A) + (1 - t) \varphi(A)$.

For any two states $\omega, \varphi \in \mathcal{S}[\mathbb{R}, E]$, any probability $t \in [0, 1]$, and any element $A \in E$, the evaluation of the probabilistic mixture state $\text{mix}(\omega, \varphi, t)$ at $A$ is given by the convex combination of the individual evaluations: $$\text{mix}(\omega, \varphi, t)(A) = t \omega(A) + (1 - t) \varphi(A).$$

The algebraic state space of a space $E$ is the set of linear maps in the algebraic dual space $E \to_{\ell} \mathbb{R}$ that are states. Specifically, it is the range of the function that maps each state $\omega \in \mathcal{S}[\mathbb{R}, E]$ (a unital positive linear map) to its underlying linear map.

The algebraic state space of a space $E$, which consists of the linear maps in the algebraic dual $E \to_{\ell} \mathbb{R}$ that are states (unital positive linear maps), is a convex set over the real numbers $\mathbb{R}$.

A state $\omega$ in the state space $\mathcal{S}[\mathbb{R}, E]$ is **pure** if its underlying linear map is an extreme point of the convex set of all states. Geometrically, this means $\omega$ cannot be expressed as a non-trivial convex combination of two distinct states in the state space.

A state $\omega$ in the state space $\mathcal{S}(E)$ is mixed if it is not a pure state. Formally, this means $\omega$ is not an extreme point of the convex set $\mathcal{S}(E)$.

Let $\mathcal{S}[\mathbb{R}, E]$ be the state space consisting of unital positive linear maps. For any states $\omega, \varphi, \psi \in \mathcal{S}[\mathbb{R}, E]$, the linear map $\omega$ lies in the open segment between $\varphi$ and $\psi$ if and only if there exists a probability $t$ in the unit interval $[0, 1]$ with $t \neq 0$ and $t \neq 1$ such that $\omega$ is the proper mixture $t\varphi + (1-t)\psi$.

Let $\mathcal{S}(E)$ be the space of states (unital positive linear maps). A state $\omega \in \mathcal{S}(E)$ is pure if and only if for every pair of states $\phi, \psi \in \mathcal{S}(E)$ and every $t$ in the unit interval $[0, 1]$ with $t \neq 0$ and $t \neq 1$, the mixture $t\phi + (1-t)\psi = \omega$ implies that $\phi = \omega$ and $\psi = \omega$.

Let $E$ be an ordered vector space and $\mathcal{S}$ denote the state space of unital positive linear maps on $E$. Let $X$ be a real vector space and $F: \mathcal{S} \to X$ be an injective map that preserves convex mixtures, such that for all states $\phi, \psi \in \mathcal{S}$ and all $t \in [0, 1]$, $F(t\phi + (1-t)\psi) = t F(\phi) + (1-t)F(\psi)$. Then a state $\omega \in \mathcal{S}$ is pure if and only if its image $F(\omega)$ is an extreme point of the image of the state space $F(\mathcal{S})$ in $X$.

A state $\omega$ is mixed if and only if it can be represented as a proper convex combination of states that are not equal to $\omega$. Formally, $\omega$ is mixed if and only if there exist states $\phi, \psi$ and a scalar $t \in (0, 1)$ such that $\omega = t\phi + (1-t)\psi$ and $(\phi \neq \omega \lor \psi \neq \omega)$.