QuantumInfo.Finite.Distance.Fidelity

Given two quantum states $\rho$ and $\sigma$ represented as density matrices in a $d$-dimensional Hilbert space, their fidelity is defined as $$F(\rho, \sigma) = \text{tr}\left( \sqrt{\sqrt{\rho} \sigma \sqrt{\rho}} \right)$$ where $\sqrt{\cdot}$ denotes the unique positive semi-definite square root of a matrix, and $\text{tr}$ is the trace operator. This quantity represents the quantum version of the Bhattacharyya coefficient between two states.

For any two quantum states $\rho$ and $\sigma$ in a $d$-dimensional Hilbert space, their fidelity $F(\rho, \sigma)$ is non-negative, i.e., $0 \le F(\rho, \sigma)$.

For any mixed states $\rho$ and $\sigma$ in a $d$-dimensional Hilbert space, the fidelity $F(\rho, \sigma)$ satisfies the inequality $F(\rho, \sigma) \le 1$.

The fidelity between two mixed states $\rho$ and $\sigma$ in a $d$-dimensional Hilbert space, denoted as $F(\rho, \sigma)$, defined as a probability value in the interval $[0, 1]$. This is implemented by bundling the real-valued fidelity function `MState.fidelity` with the proofs that $0 \le F(\rho, \sigma)$ and $F(\rho, \sigma) \le 1$.

For any quantum state $\rho$ represented as a density matrix in a $d$-dimensional Hilbert space, the fidelity of the state with itself is equal to 1, i.e., $F(\rho, \rho) = 1$.

Let $\rho$ and $\sigma$ be two quantum states represented as density matrices in a $d$-dimensional Hilbert space. The fidelity between these states, defined as $F(\rho, \sigma) = \text{tr}\left( \sqrt{\sqrt{\rho} \sigma \sqrt{\rho}} \right)$, is equal to $1$ if and only if the two states are identical, i.e., $F(\rho, \sigma) = 1 \iff \rho = \sigma$.

For any two mixed states $\rho$ and $\sigma$ in a Hilbert space of dimension $d$, the fidelity between them is symmetric, such that $F(\rho, \sigma) = F(\sigma, \rho)$.

For any two quantum states $\rho$ and $\sigma$ represented as matrix states in $\mathcal{M}_d$, and for any completely positive trace-preserving (CPTP) map $\Lambda: \mathcal{M}_d \to \mathcal{M}_{d_2}$, the fidelity $F$ satisfies the inequality $F(\Lambda(\rho), \Lambda(\sigma)) \geq F(\rho, \sigma)$.