PhyslibAlpha.ProbabilisticTheory.State.Discrimination

For two states $\omega_0, \omega_1 \in \mathcal{S}(\mathbb{R}, E)$, a prior probability $p \in [0, 1]$, and an effect $e \in E$, the **success probability** is defined as the real number: $$p \omega_0(e) + (1 - p) \omega_1(e^\perp)$$ where $e^\perp$ (or $1 - e$) denotes the complement of the effect $e$. This represents the probability of correctly guessing the state when the decision rule is to guess $\omega_0$ if the effect $e$ occurs ("clicks") and $\omega_1$ otherwise.

For two states $\omega_0, \omega_1 \in \mathcal{S}(\mathbb{R}, E)$, a prior probability $p \in [0, 1]$, and an effect $e \in E$ (where $0 \le e \le 1$), the **advantage** of the test $e$ is defined as the real number: $$p \omega_0(e) - (1 - p) \omega_1(e)$$ This value represents how much the measurement $e$ improves the success probability of identifying the prepared state compared to the baseline strategy of always guessing $\omega_1$.

Let $\omega_0, \omega_1 \in \mathcal{S}(\mathbb{R}, E)$ be two states of a system and $p \in [0, 1]$ be the prior probability that the system is in state $\omega_0$. For any effect $e$, the success probability of discriminating between the two states is equal to the baseline probability $1 - p$ (the probability of success achieved by always guessing $\omega_1$) plus the advantage provided by the effect $e$: $$\text{successProb}(\omega_0, \omega_1, p, e) = (1 - p) + \text{advantage}(\omega_0, \omega_1, p, e)$$

Let $E$ be an ordered vector space with a unit element $1$. For any two states $\omega_0, \omega_1 \in \mathcal{S}(\mathbb{R}, E)$ (unital positive linear maps), a prior probability $p \in [0, 1]$, and an effect $e \in E$ (where $0 \le e \le 1$), the **advantage** of the test $e$ in discriminating between $\omega_0$ and $\omega_1$ is bounded above by the prior probability $p$: $$p \omega_0(e) - (1 - p) \omega_1(e) \le p$$

Let $E$ be an ordered space with a unit element, and let $\omega_0, \omega_1 \in \mathcal{S}(\mathbb{R}, E)$ be two states (represented as unital positive linear maps). For a fixed prior probability $p \in [0, 1]$, the advantage of a binary test using an effect $e$ (where $0 \le e \le 1$) is the additional success probability gained over the baseline of always guessing state $\omega_1$. The set of all possible values for this advantage, obtained by varying the effect $e$, is bounded above in $\mathbb{R}$.

Let $E$ be an ordered space with a unit element $1$, and let $\omega_0, \omega_1 \in \mathcal{S}(\mathbb{R}, E)$ be two states (unital positive linear maps). For a given prior probability $p \in [0, 1]$, the success probability of a binary test using an effect $e$ ($0 \le e \le 1$) is defined as $p \omega_0(e) + (1 - p) \omega_1(1 - e)$. The set of all such success probabilities, obtained by ranging over all possible effects $e$, is bounded from above.

For two states $\omega_0, \omega_1 \in \mathcal{S}(\mathbb{R}, E)$ (represented as unital positive linear maps) and a prior probability $p \in [0, 1]$, the **optimal success probability** is the supremum of the success probabilities over all possible effects $e$ in the space $E$. It is expressed as: $$\sup_{e \in \text{Effect}(E)} \left( p \omega_0(e) + (1 - p) \omega_1(1 - e) \right)$$ where an effect $e$ (satisfying $0 \le e \le 1$) represents a binary measurement outcome that triggers a guess of $\omega_0$, and $1 - e$ represents the outcome for guessing $\omega_1$. This value represents the maximum probability of correctly identifying the state given the prior $p$, also known as the Helstrom bound.

Let $E$ be an ordered space with a unit element. For any two states $\omega_0, \omega_1 \in \mathcal{S}(\mathbb{R}, E)$ (represented as unital positive linear maps) and a prior probability $p \in [0, 1]$, the optimal success probability for state discrimination is equal to the baseline probability $1 - p$ plus the supremum of the advantage over all possible effects $e$: $$\text{optimalSuccessProb}(\omega_0, \omega_1, p) = (1 - p) + \sup_{e \in \text{Effect}(E)} \text{advantage}(\omega_0, \omega_1, p, e)$$ where an effect $e$ is an element of $E$ such that $0 \le e \le 1$. This expression provides a form of the Helstrom bound, characterizing the maximum probability of correctly identifying the state.

For any two states $\omega_0$ and $\omega_1$ (represented as unital positive linear maps) and an effect $e$ with complement $1 - e$, the difference between the states evaluated at the complement effect is the negation of the difference between the states evaluated at $e$: $$\omega_0(1 - e) - \omega_1(1 - e) = -(\omega_0(e) - \omega_1(e))$$

Let $E$ be a partially ordered additive commutative group with a unit element $1$. Let $\omega_0$ and $\omega_1$ be states, defined as unital positive linear maps from $E$ to $\mathbb{R}$. An effect is an element $e \in E$ such that $0 \le e \le 1$. Then the set of absolute differences $$\{ |\omega_0(e) - \omega_1(e)| : e \text{ is an effect} \}$$ is bounded above.

For any two states $\omega_0, \omega_1 \in \mathcal{S}[\mathbb{R}, E]$ (unital positive linear maps from $E$ to $\mathbb{R}$), the set of real values $\{\omega_0(e) - \omega_1(e) \mid 0 \le e \le 1\}$ is bounded above, where $e \in E$ is an effect.

Let $E$ be an ordered additive commutative group with a unit element $1$. Let $\omega_0$ and $\omega_1$ be states on $E$, defined as unital positive linear maps from $E$ to $\mathbb{R}$ (so $\omega_i(1) = 1$ and $\omega_i(x) \ge 0$ for $x \ge 0$). An element $e \in E$ is called an effect if $0 \le e \le 1$. Then the supremum of the difference $\omega_0(e) - \omega_1(e)$ over all effects $e$ is equal to the supremum of the absolute difference $|\omega_0(e) - \omega_1(e)|$ over all effects $e$: $$\sup_{e \in [0, 1]} (\omega_0(e) - \omega_1(e)) = \sup_{e \in [0, 1]} |\omega_0(e) - \omega_1(e)|.$$ This equality holds because for any effect $e$, its complement $1 - e$ is also an effect, and $\omega_0(1 - e) - \omega_1(1 - e) = -(\omega_0(e) - \omega_1(e))$, effectively flipping the sign of the difference.

Let $E$ be an Archimedean order-unit space. For any two states $\omega_0, \omega_1$ on $E$ (defined as unital positive linear maps from $E$ to $\mathbb{R}$), the distance $\text{dist}(\omega_0, \omega_1)$ is equal to the supremum of the absolute difference between their evaluations over all unit-ball effects: $$\text{dist}(\omega_0, \omega_1) = \sup_{e \in \text{Effect}(E)} |\omega_0(e') - \omega_1(e')|$$ where an effect $e$ is an element of $E$ satisfying $0 \le e \le 1$, and $e'$ is its corresponding element in the unit ball of the order-unit norm (via the mapping `Effect.equivBall`).

Let $E$ be an Archimedean order-unit space. For any two states $\omega_0, \omega_1 \in \mathcal{S}[\mathbb{R}, E]$ (defined as unital positive linear maps from $E$ to $\mathbb{R}$), the distance $\text{dist}(\omega_0, \omega_1)$ is equal to twice the supremum of the difference between their evaluations on any effect $e \in E$ ($0 \le e \le 1$): $$\text{dist}(\omega_0, \omega_1) = 2 \sup_{0 \le e \le 1} (\omega_0(e) - \omega_1(e))$$ The state distance is defined as the operator-norm distance, which measures the maximum possible difference between the predictions of the two states.

Let $\omega_0$ and $\omega_1$ be two states on an Archimedean order-unit space $E$. If the states are prepared with equal prior probabilities $p_0 = p_1 = 1/2$, then the optimal success probability of discriminating between them is given by $$\text{optimalSuccessProb}(\omega_0, \omega_1, 1/2, 1/2) = \frac{1}{2} + \frac{\text{dist}(\omega_0, \omega_1)}{4}$$ where $\text{dist}(\omega_0, \omega_1)$ is the distance between the states defined as the supremum of $|\omega_0(A) - \omega_1(A)|$ over all observables $A \in E$ with $\|A\|_1 \le 1$.