Trace distance ($D(\rho ,\sigma )$)

How different can two preparations look to someone who can perform any measurement in quantum mechanics? The trace distance answers this question for density operators $\rho ,\sigma$ on the same finite-dimensional vector space:Here $|A|$ is the absolute value of an operator, and $\Vert A{\Vert }_1$ 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 $1/2$ givesThe 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 measurement in quantum mechanics.

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 measurement in quantum mechanics. For one copy of either known preparation, each chosen with prior probability $1/2$, the Holevo–Helstrom theorem givesThus 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. ^[ref]

For a worked example with commuting density operators, take a qubit withTheir 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 $D\le ϵ$, every event in every measurement in quantum mechanics has a probability gap at most $ϵ$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.

For qubit density operators, the trace distance has a particularly simple geometry. Write their Bloch vectors as $\mathbf{r},\mathbf{s}$ in the Bloch ball:where $I$ is the identity operator and $\mathbf{\sigma }=({\sigma }_x,{\sigma }_y,{\sigma }_z)$ denotes the Pauli matrices. This Bloch vector representation includes both pure states on the Bloch sphere and mixed states in the interior of the Bloch ball. ^[ref]

The Pauli matrix multiplication law gives $[(\mathbf{r}-\mathbf{s})\cdot \mathbf{\sigma }{]}^2=|\mathbf{r}-\mathbf{s}{|}^{2}I$. Hence the two eigenvalues of $\Delta$ are $\pm |\mathbf{r}-\mathbf{s}|/2$, and their absolute sum givesThe trace distance is half the straight-line Euclidean distance between the Bloch vectors, not the angular or surface distance along the Bloch sphere. Opposite Bloch vectors have trace distance one; the centre $I/2$ is at trace distance $1/2$ from every pure state.

For the pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, the Bloch vectors are $\mathbf{r}=(0,0,1)$ and $\mathbf{s}=(1,0,0)$. Their straight-line separation is $\sqrt{2}$, soThese Bloch vectors are perpendicular, but the pure states are not orthogonal: $|⟨0|+⟩|=1/\sqrt{2}$. The Helstrom measurement for the zero and plus states measures along the difference direction $\mathbf{n}=(\mathbf{r}-\mathbf{s})/\sqrt{2}=(-1,0,1)/\sqrt{2}$, rather than along either input Bloch vector.

By the variational characterization of trace distance, maximize $\mathrm{Tr}[P\mathcal{E}(\rho -\sigma )]$ over $0\le P\le I$. The adjoint of a quantum channel is positive and unital, so $0\le {\mathcal{E}}^{\ast }(P)\le I$. Moving the channel onto $P$ therefore restricts, rather than enlarges, the effects available in the input variational problem. This proves the displayed bound for every pair of density operators.

For the pure qubit $\alpha |0⟩+\beta |1⟩$, subtracting the output of dephasing channel leaves a Hermitian matrix with zero diagonal and off-diagonal modulus $|\alpha \beta ||1-c|$. Its two eigenvalues are that number and its negative, giving the displayed trace distance. For real $c=1-2p$, this is $2p|\alpha \beta |$.

For two normalized pure states, the difference of their projectors has two eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, giving that trace distance. Write a mixed density operator as $\rho ={\sum }_j{q}_j|u_j⟩⟨u_j|$. Convexity of the trace norm and weighted Cauchy-Schwarz inequality give $D(\rho ,|v⟩⟨v|)\le {\sum }_j{q}_j\sqrt{1-|⟨u_j|v⟩{|}^2}\le \sqrt{1-⟨v|\rho |v⟩}$. Thus a small probability of failing the pure-state projector test forces trace-distance proximity, including for correlated multipartite outputs.

For the Hermitian difference $X=\rho -\sigma$, define its positive part of a Hermitian operator $Q=X_{+}$ and negative part of a Hermitian operator $R=X_{-}$. Their supports are orthogonal and $|X|=Q+R$. Equal traces give $\mathrm{Tr}Q=\mathrm{Tr}R$, so the trace distance is both $\frac{1}{2}(\mathrm{Tr}Q+\mathrm{Tr}R)$ and $\mathrm{Tr}Q$. This explains why the positive spectral projector attains the variational characterization of trace distance.

For a density operator $\rho$ and pure state $P=|\psi ⟩⟨\psi |$, the diagonal absolute-sum bound for the trace norm in a basis containing $|\psi ⟩$ gives $D(\rho ,P)\ge 1-⟨\psi |\rho |\psi ⟩=1-F(\rho ,P{)}^2$. The quantum fidelity is unsquared. Block-diagonal states with respect to the target and its complement attain equality.

The variational characterization of trace distance turns a matrix distance into the best possible binary experiment. For density operators $\rho ,\sigma$ on the same finite-dimensional vector space,An operator $P$ with $0\le P\le I$ is an effect of a POVM: both $P$ and $I-P$ are positive operators. The Born rule assigns the first outcome probabilities $\mathrm{Tr}(P\rho )$ and $\mathrm{Tr}(P\sigma )$. Their signed gap is $\mathrm{Tr}(P\Delta )$, where $\Delta =\rho -\sigma$. Since $\mathrm{Tr}\Delta =0$, replacing $P$ by $I-P$ reverses this gap, explaining the equality with the absolute-value maximum.

One optimal $P$ is the spectral projector $P_{+}$ onto the positive eigenspaces of $\Delta$. In the spectral decomposition of $\Delta$, this choice includes all positive eigenvalues and excludes all negative eigenvalues. Their positive sum equals the trace distance by the spectral-parts formula for trace distance. Any zero eigenspace can be assigned to either outcome without changing the gap. This is the effect selected by the equal-prior Holevo–Helstrom theorem. ^[ref]

The same optimum has a classical description. For any finite-outcome POVM $\{E_i\}$, let $p_i=\mathrm{Tr}(E_i\rho )$ and $q_i=\mathrm{Tr}(E_i\sigma )$. ThenThe expression $\frac{1}{2}{\sum }_i|p_i-q_i|$ is the total variation distance of the entire outcome probability distributions; the expression for $A$ is the gap for one event. Grouping outcomes in $A$ gives the single effect $P_A={\sum }_{i\in A}{E}_i$. Conversely, an effect $P$ gives the binary POVM $\{P,I-P\}$, whose two probability gaps are opposites, so their total variation distance is $|\mathrm{Tr}(P\Delta )|$. Thus the optimum over all POVMs is already attained by one binary projective measurement. ^[ref]

For quantum state discrimination with equal prior probabilities, declare $\rho$ on outcome $P$ and $\sigma$ on outcome $I-P$. Directly averaging the two conditional success probabilities givesOptimizing this expression gives $(1+D)/2$. For the diagonal example in trace distance, $P=|0⟩⟨0|$ gives a gap $0.8-0.3=0.5$. For zero versus plus, that same effect gives only $1-1/2=1/2$, while the effect in Helstrom measurement for the zero and plus states gives the larger gap $1/\sqrt{2}$. The trace-distance-preserving binary measurement depends on the pair of preparations; a fixed measurement basis does not generally attain the trace distance for every pair.

For two density operators $\rho ,\sigma$, let $P_{+}$ project onto the positive spectral subspace of $\rho -\sigma$. The binary POVM $\{P_{+},I-P_{+}\}$ has output probability difference $(D,-D)$, where $D=D(\rho ,\sigma )$. Its measurement channel therefore preserves this pair's trace distance exactly. The optimizing quantum measurement depends on the pair; a single fixed measurement need not preserve every pair's distance.