PhyslibAlpha.AlgebraicFramework.JordanOrderUnit.Conditioning

Given a Jordan projection $p$ in a Jordan order unit space $E$ and a state $\omega \in \mathcal{S}(\mathbb{R}, E)$, if $\omega(p) > 0$ and the quadratic representation $U_p$ is a positive map, then the conditioned state is the unital positive linear map defined by $$x \mapsto \frac{\omega(U_p x)}{\omega(p)}$$ where $U_p$ denotes the quadratic representation of the projection $p$.

Let $E$ be a Jordan order unit space, $p \in E$ be a Jordan projection, and $\omega$ be a state on $E$ such that $\omega(p) > 0$. Suppose the quadratic representation $U_p$ is a positive map (i.e., it maps the positive cone of $E$ to itself). For any element $x \in E$, the state $\omega$ conditioned on the projection $p$ is evaluated as: $$\text{condition}(\omega, p)(x) = \frac{\omega(U_p x)}{\omega(p)}$$ where $U_p$ denotes the quadratic representation of the projection $p$.

Let $E$ be a Jordan order unit space and $p \in E$ be a Jordan projection. Let $\omega$ be a state on $E$ such that $\omega(p) > 0$, and let $U_p$ be the quadratic representation of $p$. Suppose $U_p$ is a positive map. Then the conditioned state $\omega_{|p}$ (defined by $x \mapsto \frac{\omega(U_p x)}{\omega(p)}$) is normalized, i.e., $\omega_{|p}(1) = 1$.

Let $E$ be a Jordan order unit space, $p \in E$ be a Jordan projection, and $\omega$ be a state on $E$. Suppose that $\omega(p) > 0$ and the quadratic representation $U_p$ is a positive map (i.e., $x \geq 0 \implies U_p x \geq 0$). Then the conditioned state $\omega|_p$, defined by the mapping $x \mapsto \frac{\omega(U_p x)}{\omega(p)}$, is a positive functional. That is, for any $x \in E$, if $x \geq 0$, then $\omega|_p(x) \geq 0$.

Let $E$ be a Jordan order unit space and $p \in E$ be a Jordan projection. Let $\omega \in \mathcal{S}(\mathbb{R}, E)$ be a state such that $\omega(p) > 0$ and the quadratic representation $U_p$ is a positive map. Then the state $\omega$ conditioned on $p$, defined by the mapping $x \mapsto \frac{\omega(U_p x)}{\omega(p)}$, assigns the value $1$ to the projection $p$. That is, $$\omega|_p(p) = 1$$ where $\omega|_p$ denotes the conditioned state.

Let $E$ be a Jordan order unit space. Let $p \in E$ be a Jordan projection and $\omega$ be a state on $E$ such that $\omega(p) > 0$. Suppose the quadratic representation $U_p$ is a positive map. If $q \in E$ is Jordan-orthogonal to $p$ (that is, $p \cdot q = 0$), then the conditioned state $\omega|_p$ evaluated at $q$ is zero, i.e., $\omega|_p(q) = 0$. Here, the conditioned state $\omega|_p$ is defined as the functional $x \mapsto \frac{\omega(U_p x)}{\omega(p)}$.

Let $E$ be a Jordan order unit space and $p \in E$ be a Jordan projection. Let $\omega$ be a state on $E$ such that $\omega(p) > 0$. Suppose the quadratic representation $U_p$ of $p$ is a positive map (i.e., $x \ge 0 \implies U_p x \ge 0$). Then the state conditioned on $p$, defined by the map $x \mapsto \frac{\omega(U_p x)}{\omega(p)}$, assigns the value $0$ to the algebraic complement $1 - p$.

Given a Jordan projection $p$ in a Jordan order unit space $E$ and a state $\omega \in \mathcal{S}(\mathbb{R}, E)$ such that $\omega(p) > 0$, if $E$ satisfies the condition `IsQuadraticallyPositive` (which ensures that the quadratic representation $U_p$ of any projection is a positive map), then the conditioned state $\omega|_p \in \mathcal{S}(\mathbb{R}, E)$ is defined as the unital positive linear map: $$x \mapsto \frac{\omega(U_p x)}{\omega(p)}$$ where $U_p$ denotes the quadratic representation of the projection $p$.

Let $E$ be a Jordan order unit space and let $p \in E$ be a Jordan projection. Assume $E$ is quadratically positive. For any state $\omega$ on $E$ such that $\omega(p) > 0$ and any element $x \in E$, the state conditioned on $p$ (denoted by `conditionOfQuadraticPositive`) evaluates to: $$\omega|_p(x) = \frac{\omega(U_p x)}{\omega(p)}$$ where $U_p$ is the quadratic representation of the projection $p$.