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For a subsystem $A$ of a multipartite system with density operator ${\rho }_{A{E}_1\cdots E_n}$, its reduced density matrix is the unique operatorsuch that $\mathrm{Tr}(M_A{\rho }_A)=\mathrm{Tr}[(M_A\otimes I_E){\rho }_{AE}]$ for every observable $M_A$ on $A$. In an orthonormal basis $\{|e_j⟩\}$ of the other subsystems, the partial trace iswhich is independent of the chosen basis.

Write $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. Before any interaction the system is in the pure state $|{\psi }_S⟩=\alpha |0⟩+\beta |1⟩$, soOne interaction leaves the environmental state $|+⟩$ unchanged on the system branch $|0⟩$, while on the branch $|1⟩$ it producesAfter interactions with $n$ distinct environment qubits, the joint state isand the decoherence factor is the product $(\cos ϵ{)}^n$. Taking the partial trace givesFor small $ϵ$, $(\cos ϵ{)}^n\simeq e^{-n{ϵ}^2/2}$: the populations remain fixed while the phase coherence decays. This is environmental decoherence, caused by entanglement with unobserved records rather than by a nonunitary evolution of the complete state.

A measurement in the computational basis cannot reveal the decay because its probabilities remain $|\alpha {|}^2$ and $|\beta {|}^2$. An interference measurement can. For example, measuring in the Hadamard basis givesChanging the measurement phase similarly accesses the imaginary part, so the shrinking interference visibility directly displays decoherence.

If the same environment qubit is reused, the conditional rotation accumulates coherently. After $n$ interactions the overlap is $\cos (nϵ)$ rather than $(\cos ϵ{)}^n$, and the off-diagonal entries oscillate. Coherence vanishes at some times but returns periodically: this finite environment exhibits quantum recoherence rather than effectively irreversible decay.

For common classical weights, the matrix logarithm on each positive-weight block is $\log (p_i{\rho }_i)=(\log p_i)I+\log {\rho }_i$ on its support. The two logarithms of $p_i$ cancel in the quantum relative entropy, giving the displayed identity. Zero-weight blocks are omitted, including if their unweighted relative entropy is infinite. If a positive-weight block violates support inclusion, both sides are infinite. Applying the data-processing inequality for quantum relative entropy to the partial trace over the flag gives joint convexity of quantum relative entropy.

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.