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Put $p=⟨\psi |M_1^2|\psi ⟩\ge 1-\epsilon$. Since $M_1$ is a positive contraction, the successful post-measurement state is the pure state represented by $|{\psi }^{\prime }⟩=M_1|\psi ⟩/\sqrt{p}$. Its squared quantum fidelity with the input isThe finite-dimensional spectral theorem and $0\le M_1\le I$ give $M_1^2\le M_1$, so $⟨M_1⟩\ge p$. Therefore the pure-state gentle measurement bound isThe positive-operator assumption matters: an arbitrary unitary measurement operator can have success probability one while rotating the state to an orthogonal vector.

Let $\Delta =\rho -\sigma ={\Delta }_{+}-{\Delta }_{-}$ be its positive-negative decomposition. Since $\mathrm{Tr}\Delta =0$, the positive and negative parts have equal trace, namely $D(\rho ,\sigma )$. For every positive contraction $M$,Here both products have nonnegative trace, and $M\le I$ bounds the first term. Equality is attained by the orthogonal projection $P_{+}$ onto the positive spectral subspace of $\Delta$, since $P_{+}{\Delta }_{+}={\Delta }_{+}$ and $P_{+}{\Delta }_{-}=0$. Thus the variational characterization of trace distance is

If $M$ is a positive contraction and $p=⟨\psi |M^2|\psi ⟩>0$, the normalized successful post-measurement state is $M|\psi ⟩/\sqrt{p}$. Since $M^2\le M$,A likely outcome therefore preserves a pure input well in quantum fidelity when its measurement operator is positive. Positivity excludes a hidden unitary rotation that could otherwise change the state even for an outcome of probability one.