= Trace distance
{title2=$D(\rho,\sigma)$}
{wiki}
How different can two preparations look to someone who can perform any <quantum measurement>? The <trace distance> answers this question for <density operators> $\rho,\sigma$ on the same <finite-dimensional vector space>:
$$
D(\rho,\sigma)=\frac12\lVert\rho-\sigma\rVert_1
=\frac12\operatorname{Tr}|\rho-\sigma|,\qquad
|A|=\sqrt{A^\dagger A}.
$$
Here $|A|$ is the <absolute value of an operator>, and $\lVert A\rVert_1$ is its <trace norm>. Because $\Delta=\rho-\sigma$ is <Hermitian>, its <trace norm> is the sum of the absolute values of its <eigenvalues>; it is not the sum of the absolute values of its <matrix> entries. The factor $1/2$ gives
$$
0\leq D(\rho,\sigma)\leq1.
$$
The <trace distance> is a <metric>: it is symmetric, vanishes exactly when $\rho=\sigma$, and obeys the <triangle inequality>. Identical preparations have <trace distance> zero; preparations with orthogonal occupied <eigenspaces> have <trace distance> one and can be identified perfectly in a single <quantum measurement>.
Operationally, the <trace distance> is the largest <total variation distance> between the outcome <probability distributions> obtainable by a <POVM>. Equivalently, the <variational characterization of trace distance> says it is the largest possible <probability> gap for one event in a binary <quantum measurement>. For one copy of either known preparation, each chosen with prior <probability> $1/2$, the <Holevo–Helstrom theorem> gives
$$
P_{\mathrm{success}}^{\mathrm{opt}}=\frac{1+D(\rho,\sigma)}2.
$$
Thus <trace distance> zero leaves only guessing, while <trace distance> one permits perfect <quantum state discrimination>. The optimum ranges over measurements; different measurement <bases> can expose different amounts of distinguishability. https://cs.uwaterloo.ca/~watrous/TQI/TQI.3.pdf[Watrous, The Theory of Quantum Information, §3.1, Theorem 3.4 and Proposition 3.5]{ref}
For a worked example with commuting <density operators>, take a <qubit> with
$$
\rho=\begin{pmatrix}0.8&0\\0&0.2\end{pmatrix},\qquad
\sigma=\begin{pmatrix}0.3&0\\0&0.7\end{pmatrix}.
$$
Their difference has <eigenvalues> $0.5,-0.5$, so $D=(|0.5|+|-0.5|)/2=0.5$. Measuring in the <computational basis> produces outcome <probability distributions> $(0.8,0.2)$ and $(0.3,0.7)$, whose <total variation distance> is also $0.5$. Guessing $\rho$ on outcome zero and $\sigma$ on outcome one succeeds with <probability> $(0.8+0.7)/2=0.75$. A <projective measurement> in the <Hadamard basis> instead produces $(0.5,0.5)$ for both preparations and gives no discrimination advantage.
The <trace-distance contraction under quantum channels> ensures that a common <quantum channel>, including the <partial trace>, cannot increase the <trace distance>. This makes <trace distance> a useful error measure: if $D\leq\epsilon$, every event in every <quantum measurement> has a <probability> gap at most $\epsilon$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.
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