PhyslibAlpha.ProbabilisticTheory.State.PureMeasurement

A preparation procedure $p$ with outcome space $X$ and observable space $E$ is defined to be **trivial** if, for every observable $A \in E$, the expectation value of $A$ in the state prepared given a specific outcome $x \in X$ is almost surely equal to the expectation value of $A$ in the average state $p.\text{state}$. Mathematically, $p$ is trivial if for every $A \in E$: $$(p.\text{prepare}(x))(A) = p.\text{state}(A)$$ for almost every $x$ with respect to the probability measure $p.\text{law}$ on $X$.

Let $p$ be a preparation procedure with outcome space $X$ and observable space $E$. Suppose that for every effect $e \in E$ (an element such that $0 \le e \le 1$), the expectation value of the state prepared given a specific outcome $x$, denoted $(p.\text{prepare}(x))(e)$, is almost surely equal to the expectation value of the average state $p.\text{state}(e)$ with respect to the probability measure $p.\text{law}$ on $X$. Then the preparation $p$ is trivial, meaning that for every observable $A \in E$, $(p.\text{prepare}(x))(A) = p.\text{state}(A)$ holds for almost every $x$.

Let $p$ be a preparation procedure with outcome space $X$ and observable space $E$. Let $\omega$ be the average state prepared by the procedure (denoted $p.\text{state}$). If $\omega$ is a pure state, then for any measurable event $s \subseteq X$ whose probability $\mathbb{P}(s)$ (given by $p.\text{law}$) is strictly between $0$ and $1$, the conditional state of the preparation given $s$ is equal to $\omega$.

Let $p$ be a preparation procedure with outcome space $X$ and observable space $E$, and let $\rho$ be the average state prepared by $p$. If $\rho$ is a pure state, then for any measurable subset $s \subseteq X$ and any observable $A \in E$, the integral of the expectation value of $A$ given the outcome $x$ (denoted by $(p.\text{prepare}(x))(A)$) over the set $s$ is equal to the probability of $s$ times the expectation value of $A$ in the state $\rho$: $$\int_s (p.\text{prepare}(x))(A) \, d\mu(x) = \mu(s) \rho(A)$$ where $\mu$ is the probability measure on $X$ associated with the preparation $p$.

Let $p$ be a preparation procedure with outcome space $X$ and observable space $E$. If the average state $\rho$ prepared by $p$ (denoted $p.\text{state}$) is a pure state, then the preparation $p$ is trivial. That is, for every observable $A \in E$, the expectation value of $A$ in the state prepared given a specific outcome $x \in X$ is almost surely equal to the expectation value of $A$ in the average state $\rho$: $$(p.\text{prepare}(x))(A) = \rho(A)$$ for $p.\text{law}$-almost every $x \in X$.

Let $\omega$ be a normal state on an observable space $E$. If every preparation procedure $p$ with a two-element outcome space (the type `Bool`) that averages to the state $\omega$ is trivial, then $\omega$ is a pure state. A preparation $p$ is trivial if, for every observable $A \in E$, the state prepared for a specific outcome $x$ satisfies $(p.\text{prepare}(x))(A) = \omega(A)$ for almost every $x$ with respect to the probability measure of the preparation.

Let $\omega$ be a normal state on an observable space $E$. The state $\omega$ is pure if and only if every preparation procedure $p$ with outcome space $X$ that yields $\omega$ as its average state (i.e., $p.\text{state} = \omega$) is trivial. A preparation $p$ is trivial if for every observable $A \in E$, the expectation value of $A$ given a specific outcome $x \in X$ is almost surely equal to the average expectation value: $$(p.\text{prepare}(x))(A) = \omega(A)$$ for $p.\text{law}$-almost every $x \in X$.

Given a preparation procedure $p$ with an outcome space $X$ and a law (probability measure) $\mu$, and a measurement $M$ with an outcome space $\Omega$, let $K: X \to \mathcal{P}(\Omega)$ be the outcome law kernel that assigns a probability measure on $\Omega$ for each roll $x \in X$. The preparation $p$ has **no side information** about the measurement $M$ if the random variables representing the roll (the projection onto $X$) and the measurement outcome (the projection onto $\Omega$) are independent with respect to the joint measure $\mu \otimes K$ on $X \times \Omega$.

Let $p$ be a preparation procedure with an outcome space $X$ (the "die roll") and a probability measure $\mu = p.\text{law}$ on $X$. Let $M$ be a measurement with an outcome space $\Omega$. The preparation $p$ has **no side information** about the measurement $M$ if and only if the joint measure on $X \times \Omega$ formed by the roll law and the outcome law (which depends on the roll), denoted $\mu \otimes p.\text{outcome } M$, is equal to the joint measure formed by the roll law and the Born law of the average state (which ignores the roll), denoted $\mu \otimes p.\text{forgetfulOutcome } M$.

Let $p$ be a preparation procedure with an outcome space $X$ and a probability measure $\mu$, and let $M$ be a measurement with an outcome space $\Omega$. Let $K: X \to \mathcal{P}(\Omega)$ be the outcome law kernel (where $K(x)$ is the distribution of measurement $M$ given the preparation roll $x$), and let $K_{\text{avg}}$ be the constant kernel representing the Born law of $M$ for the average state prepared by $p$. If $K(x) = K_{\text{avg}}$ for $\mu$-almost every $x \in X$, then the preparation $p$ has no side information about the measurement $M$, meaning the random variables representing the roll and the measurement outcome are independent.

Let $p$ be a preparation procedure with roll space $X$ and probability measure $\mu = p.\text{law}$, and let $M$ be a measurement with an outcome space $\Omega$. If $\Omega$ is countably generated, then $p$ has no side information about $M$ (meaning the roll and the measurement outcome are independent) if and only if the outcome law of $M$ given the roll $x$ is almost surely equal to the forgetful outcome law (the Born law of $M$ on the average prepared state $p.\text{state}$). Mathematically, this equivalence is expressed as: $$p.\text{NoSideInfo } M \iff p.\text{outcome } M = p.\text{forgetfulOutcome } M \text{ a.e. } [\mu]$$ where $p.\text{outcome } M$ is the Markov kernel mapping a roll $x$ to the measurement law of the state prepared by $x$.

Let $p$ be a preparation procedure with roll space $X$ and probability law $\mu$, and let $M$ be a measurement with a countably generated outcome space $\Omega$. The preparation $p$ has no side information about $M$ (meaning the random variables representing the preparation roll and the measurement outcome are independent) if and only if the Markov kernel representing the outcome law given the roll ($p.\text{outcome } M$) and the Markov kernel representing the outcome law ignoring the roll ($p.\text{forgetfulOutcome } M$) are almost surely equivalent under mutual post-processing with respect to $\mu$.

Let $p$ be a preparation procedure with roll space $X$, observable space $E$, and a probability measure $p.\text{law}$ describing the rolls. If $p$ is a **trivial preparation** (meaning the expectation values of any observable given a specific roll $x$ are almost surely equal to the expectation values of the average state $p.\text{state}$), then for any measurement $M$ with a countably generated outcome space $\Omega$, the **outcome law** of $M$ given the roll $x$ is almost surely equal to the **forgetful outcome law** (the Born law of $M$ on the average state $p.\text{state}$). Mathematically, the Markov kernels satisfy: $$p.\text{outcome } M = p.\text{forgetfulOutcome } M$$ almost everywhere with respect to the measure $p.\text{law}$.

Let $p$ be a preparation procedure with a space of "rolls" $X$ and a probability measure $\mu$ on $X$. Let $e$ be an effect in the observable space $E$. If $p$ has no side information about the binary (yes/no) measurement associated with $e$ (meaning the roll and the measurement outcome are independent), then for $\mu$-almost every roll $x \in X$, the probability of the effect $e$ in the state prepared given $x$ is equal to its probability in the average state $p.\text{state}$. That is, $$(p.\text{prepare } x)(e) = (p.\text{state})(e)$$ for $\mu$-almost all $x$.

Let $p$ be a preparation procedure with outcome space $X$ (representing the "roll") and observable space $E$. The preparation $p$ is **trivial** if and only if, for every effect $e \in E$ (an element such that $0 \le e \le 1$), $p$ carries **no side information** about the binary (yes/no) measurement associated with $e$. In this context, a preparation is trivial if the expectation value of any observable given a roll $x \in X$ is almost surely equal to the expectation value in the average state, and "no side information" means the outcome of the roll and the outcome of the measurement are independent random variables.

Let $p$ be a preparation procedure with an outcome space $X$ and a law $\mu$. Let the average state prepared by the procedure, $\rho = \int_X p.\text{prepare}(x) \, d\mu$, be a pure state (i.e., an extreme point of the state space). For any measurement $M$ with a countably generated outcome space $\Omega$, the preparation $p$ has no side information about $M$. That is, the random variable representing the "roll" $x \in X$ and the random variable representing the measurement outcome $\omega \in \Omega$ are independent.

A state $\omega \in \mathcal{S}[\mathbb{R}, E]$ has **no side information** if for every preparation procedure $p$ (with an outcome space $X$) such that its average state is $\omega$ (i.e., $p.\text{state} = \omega$), the outcome of the preparation (the "roll") and the outcome of any binary measurement corresponding to an effect $e \in [0, 1]_E$ are independent random variables.

Let $\omega$ be a normal state. Then $\omega$ is a pure state if and only if $\omega$ has no side information. That is, $\omega$ is an extreme point of the state space if and only if for every preparation procedure $p$ whose average state is $\omega$, the outcome of the preparation (the "roll") and the outcome of any binary measurement (corresponding to an effect $e$) are independent random variables.

Let $\omega$ be a normal state on an ordered vector space $E$. Then $\omega$ is a mixed state if and only if there exists a preparation procedure $p$ with a binary outcome space (i.e., a choice between two states) whose average state is $\omega$, and there exists an effect $e \in [0, 1]$ such that $p$ leaks side information regarding the binary measurement defined by $e$. In this context, side information exists when the "roll" (the internal choice made during preparation) and the outcome of the measurement are not independent random variables.