QuantumInfo.Finite.Distance.TraceDistance

Given two quantum mixed states $\rho$ and $\sigma$ of dimension $d$, the trace distance is defined as $$D(\rho, \sigma) = \frac{1}{2} \| \rho - \sigma \|_{\text{tr}}$$ where $\| \cdot \|_{\text{tr}}$ denotes the trace norm (the sum of the singular values) of the difference between the matrix representations of the two states.

For any quantum mixed states $\rho$ and $\sigma$ of dimension $d$, the trace distance $D(\rho, \sigma)$ is non-negative, i.e., $0 \leq D(\rho, \sigma)$.

Let $\rho$ and $\sigma$ be quantum mixed states of finite dimension $d$. The trace distance between them, denoted as $\text{TrDistance}(\rho, \sigma)$, is less than or equal to $1$.

Given two mixed states $\rho$ and $\sigma$ in a $d$-dimensional Hilbert space, the trace distance $D(\rho, \sigma)$ is defined as a probability value in the interval $[0, 1]$. It is constructed using the real-valued trace distance $\text{TrDistance}(\rho, \sigma)$, constrained by the proofs that $0 \le \text{TrDistance}(\rho, \sigma)$ and $\text{TrDistance}(\rho, \sigma) \le 1$.

For any two quantum mixed states $\rho$ and $\sigma$ of finite dimension $d$, the trace distance between them is symmetric, meaning $D(\rho, \sigma) = D(\sigma, \rho)$, where $D$ denotes the `TrDistance` function.