QuantumInfo.Finite.Ensemble

A mixed-state ensemble is defined as a random variable $\text{var}$ taking values in the set of mixed states $\mathcal{M}(d)$ of a $d$-dimensional Hilbert space. Given a finite index set $\alpha$, an ensemble consists of a map $\text{var} : \alpha \to \text{MState } d$ that assigns a mixed state to each outcome, and a probability distribution $\text{distr}$ over $\alpha$ that provides the statistical weights for each state.

Given a finite-dimensional Hilbert space (represented by the type of its basis $d$) and a finite set of outcomes $\alpha$, a pure-state ensemble $E$ is a random variable mapping from $\alpha$ to the space of state vectors (kets) $\mathcal{H}_d$. It consists of a probability distribution $p: \alpha \to [0, 1]$ and a collection of pure states $|\psi_a\rangle \in \text{Ket}(d)$ for each $a \in \alpha$.

Given a mixed-state ensemble $E$ of type `MEnsemble d α`, where $d$ represents the Hilbert space dimension and $\alpha$ is a finite index set, the function `MEnsemble.states` maps the ensemble to a collection of mixed states $\rho_i \in \text{MState } d$ for each $i \in \alpha$. This is an alias for the random variable component of the underlying probability distribution.

Given a pure-state ensemble $\mathcal{E}$ over a finite set of outcomes $\alpha$ and a Hilbert space of dimension $d$, the function `PEnsemble.states` maps the ensemble to its corresponding collection of state vectors $|\psi_i\rangle$ for each $i \in \alpha$. Each state is represented as a unit vector (Ket) in the $d$-dimensional Hilbert space.

Let $\mathcal{E}_P = \{(p_i, |\psi_i\rangle)\}_{i \in \alpha}$ be a pure-state ensemble, where each $|\psi_i\rangle$ is a ket in a $d$-dimensional Hilbert space and $p_i$ is the probability associated with state $i$. The function maps this pure-state ensemble to a mixed-state ensemble $\mathcal{E}_M = \{(p_i, \rho_i)\}_{i \in \alpha}$, where each pure state $|\psi_i\rangle$ is converted into its corresponding density matrix representation $\rho_i = |\psi_i\rangle\langle\psi_i|$.

Let $d$ be a finite type (representing the dimension or Hilbert space) and $\alpha$ be a finite index set. A pure ensemble `PEnsemble d α` (a collection of pure states $|\psi_i\rangle$ with associated probabilities $p_i$) can be naturally coerced into a mixed ensemble `MEnsemble d α` (a collection of density matrices $\rho_i$ with associated probabilities $p_i$) by mapping each pure state to its corresponding projection operator $\rho_i = |\psi_i\rangle\langle\psi_i|$. This coercion is defined via the function `Ensemble.toMEnsemble`.

Let $d$ and $\alpha$ be finite types. For a physical ensemble (represented as `PEnsemble d α`) with a state mapping $ps: \alpha \to \text{Ket } d$ and a probability distribution $distr$, its conversion to a mixed ensemble $toMEnsemble \langle ps, distr \rangle$ is a mixed ensemble whose states are given by the pure state density matrices $\rho_a = |ps(a)\rangle\langle ps(a)|$ for each $a \in \alpha$, while retaining the same probability distribution $distr$.

Let $d$ be a finite index set and $\alpha$ be a finite set of indices for an ensemble. For a mixed-state ensemble $me$ of type `MEnsemble d α`, there exists a pure-state ensemble $pe$ of type `PEnsemble d α` such that its natural embedding into the space of mixed-state ensembles equals $me$ (i.e., $\uparrow pe = me$) if and only if there exists a collection of kets $|\psi_i\rangle$ for each $i \in \alpha$ such that every state $\rho_i$ in the ensemble $me$ is a pure state of the form $|\psi_i\rangle\langle\psi_i|$.

Given a quantum ensemble $e$ consisting of a finite set of mixed states $\{\rho_i\}_{i \in \alpha}$ and an associated probability distribution $\{p_i\}_{i \in \alpha}$ (where $\alpha$ is a finite index set), the function returns the resulting mixed state $\rho \in \text{MState}_d$ defined by the convex combination: \[ \rho = \sum_{i \in \alpha} p_i \rho_i \] where each $p_i$ is the probability weight of the state $\rho_i$.

Let $e$ be a mixed ensemble (`MEnsemble`) of dimension $d$ indexed by a finite set $\alpha$. The matrix representation $m(\text{mix } e)$ of the mixed state obtained from the ensemble $e$ is equal to the convex sum of the matrix representations of its constituent states, weighted by their probabilities. That is, $$m(\text{mix } e) = \sum_{i \in \alpha} p_i \cdot m(\rho_i)$$ where $p_i$ is the probability distribution (`e.distr i`) and $\rho_i$ are the states (`e.states i`) of the ensemble.

Given an equivalence (bijection) $\sigma: \alpha \simeq \beta$ between two indexing sets, there exists a corresponding equivalence between the types of mixed-state ensembles $MEnsemble(d, \alpha)$ and $MEnsemble(d, \beta)$. This map relates ensembles indexed by $\alpha$ to those indexed by $\beta$ by reindexing the probability distributions and states according to $\sigma$.

Given an equivalence (bijection) $\sigma: \alpha \simeq \beta$ between two index sets, we define an equivalence between the sets of pure-state ensembles $PEnsemble(d, \alpha)$ and $PEnsemble(d, \beta)$. This map re-indexes the states of an ensemble according to $\sigma$ while preserving the underlying physical content.

Let $d$ be a finite type and let $\alpha, \beta$ be finite types with decidable equality. For any bijection $\sigma: \alpha \simeq \beta$ and any mixed-state ensemble $e$ over $\alpha$ (of dimension $d$), the mixed state resulting from the mixture of $e$ is equal to the mixed state resulting from the mixture of the reindexed ensemble produced by $\sigma$. That is, $\text{mix}(\text{congrMEnsemble}(\sigma, e)) = \text{mix}(e)$.

Let $d$ be a finite index set and $\alpha, \beta$ be finite sets. For any pure-state ensemble $e$ indexed by $\alpha$ (of type `PEnsemble d α`) and any bijection $\sigma : \alpha \simeq \beta$, let $e'$ be the equivalent ensemble indexed by $\beta$ obtained via $\sigma$ (denoted `congrPEnsemble σ e`). Then the resulting mixed state (density matrix) $\rho$ obtained by mixing the ensemble $e'$ is equal to the mixed state obtained by mixing the original ensemble $e$. That is, $\text{mix}(\text{toMEnsemble}(\text{congrPEnsemble } \sigma \, e)) = \text{mix}(e)$.

Given a finite-dimensional Hilbert space with dimension type $d$ and an index set $\alpha$, let $e$ be a mixed-state ensemble $(\rho_i, p_i)_{i \in \alpha}$ where each $\rho_i$ is a mixed state ($\text{MState } d$) and $p_i$ is its probability weight. For a function $f: \text{MState } d \to T$ mapping mixed states to a type $T$ that is "mixable" (i.e., it supports convex combinations in a vector space $U$), the average is defined as the expectation value of $f$ over the ensemble $e$: $$\mathbb{E}_e[f] = \sum_{i \in \alpha} p_i f(\rho_i)$$ The result is an element of $T$.

Given a mixed ensemble $e$ (represented by `MEnsemble d α`) and a function $f: \text{MState } d \to \mathbb{R}_{\ge 0}$ mapping mixed states to non-negative real numbers, the average value of $f$ over the ensemble is defined as $$\langle f \rangle_e = \sum_{i \in \alpha} p_i f(\rho_i)$$ where $p_i$ are the probabilities and $\rho_i$ are the states associated with the ensemble $e$. This specific definition ensures that since $f$ maps to non-negative reals $\mathbb{R}_{\ge 0}$, the resulting average is also a non-negative real number $\mathbb{R}_{\ge 0}$.

Let $d$ and $\alpha$ be finite types. Given a function $f: \text{Ket } d \to T$ and a pure-state ensemble $e$ (represented as `PEnsemble d α`), the average of $f$ over $e$ is defined as the expectation value of $f$ acting on the states in $e$ with respect to the probability weights $w_i$ provided by its distribution $e.\text{distr}$. Formally, it is calculated as $\sum_{i \in \alpha} w_i \cdot f(\psi_i)$, where $\{w_i\}$ are the probabilities and $\{\psi_i\}$ are the pure states (Kets) in the ensemble. The result is an element of $T$, provided $T$ supports mixed operations via the `Mixable` instance.

Let $\mathcal{H}_d$ be a Hilbert space (represented by the type of its vectors `Ket d`) of finite dimension $d$. Given a pure ensemble $\mathcal{E} = \{ (p_i, |\psi_i\rangle) \}_{i \in \alpha}$ (represented by `PEnsemble d α`), where $p_i$ are probabilities and $|\psi_i\rangle$ are pure states, and a function $f: \text{Ket } d \to \mathbb{R}_{\ge 0}$ that maps states to non-negative real numbers, the function `pure_average_NNReal` computes the expected value: $$\mathbb{E}_{\mathcal{E}}[f] = \sum_{i \in \alpha} p_i f(|\psi_i\rangle)$$ The result is returned as a non-negative real number $\mathbb{R}_{\ge 0}$, utilizing the fact that the expectation of a non-negative function under a probability distribution is non-negative.

Let $d$ and $\alpha$ be finite types. Let $T$ be a type equipped with a mixing operation (specifically, a type $T$ that can be embedded into a real vector space $U$). Let $e$ be a pure-state ensemble in a $d$-dimensional Hilbert space with index set $\alpha$, denoted as $e \in \text{PEnsemble } d\ \alpha$. Let $\uparrow e$ denote the corresponding mixed-state ensemble (an element of $\text{MEnsemble } d\ \alpha$). For any function $f: \text{MState } d \to T$ mapping density matrices to $T$, the average of $f$ over the ensemble $\uparrow e$ is equal to the pure-state average of the composition $f \circ \text{pure}$ over $e$, where $\text{pure}(|\psi\rangle) = |\psi\rangle\langle\psi|$. That is, $$\text{average}(f, \uparrow e) = \text{pure\_average}(f \circ \text{pure}, e)$$

Let $e$ be a pure ensemble (an object of type `PEnsemble d α`) with state space $d$ and index set $\alpha$. Then the probability distribution associated with its conversion to a general mixed ensemble (denoted by `toMEnsemble e`) is equal to the probability distribution of the original pure ensemble $e$.

Let $e$ be a pure ensemble of quantum states in a $d$-dimensional Hilbert space, indexed by a finite set $\alpha$, where $e.distr(i)$ denotes the probability of state $|\psi_i\rangle = e.states(i)$. The mixed state $\rho$ formed by the ensemble, defined as the convex combination $\rho = \text{mix}(e) = \sum_{i \in \alpha} e.distr(i) |\psi_i\rangle\langle\psi_i|$, is equal to a specific pure state $|\psi\rangle\langle\psi|$ if and only if for every index $i$ with non-zero probability ($e.distr(i) \neq 0$), the corresponding pure state $|\psi_i\rangle\langle\psi_i|$ is equal to $|\psi\rangle\langle\psi|$.

Let $d$ and $\alpha$ be finite index sets. Let $e$ be a pure ensemble (represented as `PEnsemble d α`) consisting of kets $|\psi_i\rangle$ with associated probabilities $p_i$. Let $f: \text{Ket } d \to T$ be a function from kets to some type $T$, where $T$ is equipped with a mixing structure (allowing for weighted averages). Suppose $f$ is invariant under phase equivalence, i.e., for any kets $|\psi\rangle$ and $|\phi\rangle$, if $|\psi\rangle \sim |\phi\rangle$ (meaning $|\psi\rangle\langle\psi| = |\phi\rangle\langle\phi|$), then $f(|\psi\rangle) = f(|\phi\rangle)$. If the total mixed state of the ensemble $\rho = \sum_i p_i |\psi_i\rangle\langle\psi_i|$ is equal to a pure state $|\psi\rangle\langle\psi|$, then the average of $f$ over the ensemble $e$, denoted as $\sum_i p_i f(|\psi_i\rangle)$, is equal to $f(|\psi\rangle)$.

Let $\iota$ be a finite set. Let $(p_i)_{i \in \iota}$ and $(x_i)_{i \in \iota}$ be families of real numbers such that for all $i \in \iota$, $p_i \ge 0$, and the sum of the weights is $\sum_{i \in \iota} p_i = 1$. Suppose further that each $x_i$ satisfies $x_i \le 1$. Then the weighted sum $\sum_{i \in \iota} p_i x_i$ is equal to $1$ if and only if for every $i \in \iota$ where $p_i \neq 0$, it holds that $x_i = 1$.

Let $\rho$ be a quantum mixed state (density matrix) in a $d$-dimensional Hilbert space and $|\psi\rangle$ be a normalized ket vector. The expectation value of the pure state projector $|\psi\rangle\langle\psi|$ in the state $\rho$, defined as $\text{Tr}(\rho |\psi\rangle\langle\psi|)$, is equal to $1$ if and only if $\rho$ is the pure state $|\psi\rangle\langle\psi|$.

Let $e$ be a quantum ensemble over a finite set $\alpha$, where each state in the ensemble is denoted by $e.\text{states}(i)$ and occurs with probability $e.\text{distr}(i)$. The mixed state $\rho_{\text{mix}}$ associated with this ensemble is given by $\sum_i e.\text{distr}(i) \rho_i$. This mixed state is equal to a pure state $|\psi\rangle\langle\psi|$ (denoted as $\text{pure } \psi$) if and only if every state $\rho_i$ in the ensemble with a non-zero probability $e.\text{distr}(i) \neq 0$ is itself equal to the pure state $|\psi\rangle\langle\psi|$.

Let $e$ be a quantum ensemble of states in $d$ dimensions indexed by $\alpha$ (an `MEnsemble d α`), and let $f$ be a function mapping mixed states to some mixable type $T$ (such as an observable or property). If the mixture of the ensemble $e$ is a pure state $|\psi\rangle\langle\psi|$ (i.e., $\text{mix}(e) = \text{pure } \psi$), then the average value of $f$ over the ensemble is equal to the value of $f$ evaluated at that pure state: $\text{average}(f, e) = f(\text{pure } \psi)$.

Given a mixed state $\rho$ of dimension $d$ and an index $i$ in a finite set $\alpha$, the trivial mixed-state ensemble is defined as the ensemble where every state in the collection is $\rho$ (i.e., $e_j = \rho$ for all $j \in \alpha$), and the probability distribution is the constant distribution concentrated at $i$ (i.e., $p_i = 1$ and $p_j = 0$ for $j \neq i$).

Let $\rho$ be a quantum mixed state of dimension $d$ and $i$ be an index in a finite type $\alpha$. If we define a "trivial mixed-state ensemble" where all probability mass is concentrated on the single state $\rho$ at index $i$, then the result of mixing this ensemble (taking the weighted sum of its states) is equal to the original state $\rho$.

Let $f: \text{MState } d \to T$ be a function from mixed states to a type $T$ that supports mixing (an instance of `Mixable`). For any mixed state $\rho$ and any index $i$ in a type $\alpha$, the average of $f$ over the trivial ensemble consisting solely of the state $\rho$ (denoted as `trivial_mEnsemble` $\rho$ $i$) is equal to $f(\rho)$.

For a finite-dimensional Hilbert space (or index set) $d$ and a set of indices $\alpha$, if $d$ is non-empty and there exists a default element in $\alpha$, then there exists a default ensemble of mixed states $E \in \text{MEnsemble}(d, \alpha)$. This default ensemble is defined as the trivial ensemble where every index $i \in \alpha$ is mapped to the default mixed state (the maximally mixed state) of $d$.

Given a quantum state vector $|\psi\rangle$ (represented as a `Ket d`) and an index $i \in \alpha$, the trivial pure-state ensemble is defined as a probability distribution over states where every outcome is the state $|\psi\rangle$, and the probability distribution is the constant (Dirac) distribution centered at $i$. That is, the ensemble consists of copies of $|\psi\rangle$ such that the probability of the $i$-th element is 1 and all other probabilities are 0.

Let $d$ be a finite dimension and $\alpha$ be an index set. For any ket $|\psi\rangle$ of dimension $d$ and any index $i \in \alpha$, let $\mathcal{E}$ be the trivial pure-state ensemble centered on $|\psi\rangle$ at index $i$. The mixed state resulting from the ensemble $\mathcal{E}$ is equal to the pure state density matrix $|\psi\rangle\langle\psi|$.

Let $f: \text{Ket}(d) \to T$ be a function from a quantum state vector to a type $T$ that supports mixing. For any state vector $\psi \in \text{Ket}(d)$ and any index $i \in \alpha$, the average of $f$ over the trivial ensemble consisting solely of the state $\psi$ (denoted as `trivial_pEnsemble ψ i`) is equal to $f(\psi)$.

Given a finite-dimensional Hilbert space (represented by the type \( d \)) and a finite set of indices \( \alpha \), if the Hilbert space is non-empty (\( [ \text{Nonempty } d ] \)) and the index set has a default element (\( [ \text{Inhabited } \alpha ] \)), then there exists a default pure ensemble (\( \text{PEnsemble } d \alpha \)). This default ensemble is defined as the trivial pure ensemble consisting of the default state \( | \psi_{\text{default}} \rangle \) associated with the default index.

Given a mixed quantum state $\rho$ of finite dimension $d$, the spectral pure-state ensemble $\mathcal{E}_\rho$ is the pure-state ensemble $(p, \psi)$ where: 1. The probability distribution $p$ is the eigenvalue spectrum of the density matrix $\rho$. 2. For each index $i \in \{1, \dots, d\}$, the associated pure state $\psi_i$ is the $i$-th eigenvector from the orthonormal basis of the underlying Hermitian matrix of $\rho$. The ensemble consists of pairs $(\lambda_i, |\psi_i\rangle)$ such that the probabilities $\lambda_i$ are the eigenvalues of $\rho$ and $|\psi_i\rangle$ are its corresponding normalized eigenvectors.

Let $d$ be a finite index set and $\mathbb{k}$ be a real or complex field (an `RCLike` field). For any Hermitian matrix $A \in \text{Matrix}(d, d, \mathbb{k})$, the matrix can be decomposed as the sum $$A = \sum_{i \in d} \lambda_i (v_i v_i^*)$$ where $\lambda_i$ are the eigenvalues of $A$ (denoted by `hA.eigenvalues i`), $v_i$ are the corresponding orthonormal eigenvectors (the basis vectors of `hA.eigenvectorBasis i`), and $v_i v_i^*$ represents the outer product (the rank-one projection) of the $i$-th eigenvector with its adjoint.

Let $\rho$ be a quantum mixed state of finite dimension $d$. Let $\mathcal{E}$ be the spectral pure-state ensemble associated with $\rho$, which is constructed from the eigenvalues and eigenvectors of its density matrix. Then, the mixture of this ensemble—defined as the weighted sum of its component states—is equal to the original state $\rho$. Specifically, if $\rho$ has spectral decomposition $\rho = \sum_i \lambda_i | \psi_i \rangle \langle \psi_i |$, then $\text{mix}(\mathcal{E}) = \rho$.