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For the transposition map $\Theta :M_d\to M_d$, the induced trace norm is $\Vert \Theta {\Vert }_1=1$, while stabilization givesThus an entangled quantum ancilla can amplify the distinguishability witnessed by transposition by a factor of $d$.

Attach a quantum ancilla prepared with amplitudeon $|1⟩$; this is possible because $p>1/4$. Mark a state as good only when both the original success qubit and this ancilla equal one. The enlarged state's good probability is $p{c}^2=1/4$, so its good amplitude is $1/2=\sin (\pi /6)$. One amplitude amplification iteration rotates the angle from $\pi /6$ to $3\pi /6=\pi /2$. It therefore prepares $|{\psi }_1⟩$ exactly, after which the flag and ancillary qubits may be discarded. This is an instance of exact amplitude amplification.

Implement the assumed efficient classical algorithm for $x\mapsto {\theta }_x$ as a reversible circuit. On input $|x⟩|0^s⟩$, it computes an $O(m)$-bit binary expansion of the angle in a work register using $\mathrm{poly}(m)$ Toffoli gates and elementary reversible gates. Write the computed angle as a sum of binary-weighted angles. For each angle bit, apply the corresponding controlled single-qubit $R_y$ rotation to the target. These rotations have the same axis, so their angles add and produceFinally apply uncomputation to erase the work register. Each Toffoli gate and controlled rotation has a constant-size decomposition into one- and two-qubit gates when arbitrary one-qubit rotations are available. Ignoring the stipulated precision costs, the resulting circuit has size $\mathrm{poly}(m)$ and returns every quantum ancilla to zero.

For $p=(d+1)/(2d)$, the weighted difference of the two Werner–Holevo channels isThus the map is the transposition map divided by $d$. The diamond norm of the transposition map and invariance of the trace norm under matrix transpose giveSubstitution into the optimal-error formulas shows that an entangled quantum ancilla permits perfect discrimination, whereas every ancilla-free strategy has

For a fixed input, the Holevo–Helstrom theorem gives the optimal error for the two output states. Optimizing the input and quantum ancilla therefore givesThe stabilization in the diamond norm is exactly the optimization over ancillary systems. Without an ancilla the same argument instead giveswhere $\Vert \cdot {\Vert }_1$ is the induced trace norm.

In quantum channel discrimination, prepare a density operator ${\rho }_{HR}$ on the channel input $H$ and an optional quantum ancilla $R$. Under hypothesis $j\in \{1,2\}$ the output isUse a two-outcome quantum measurement $\{Q,I-Q\}$ and decide for $T_1$ on outcome $Q$. The conditional Type I error and Type II error areWith prior probabilities $p$ and $1-p$, symmetric Bayesian discrimination minimizes the average error $p\alpha +(1-p)\beta$ over the input and measurement.

Quantum channel discrimination feeds a common input into one of several candidate quantum channels and measures the output. An entangled quantum ancilla permits inputs ${\rho }_{HR}$ and compares $(T_j\otimes {\mathrm{id}}_R)({\rho }_{HR})$; optimal binary discrimination is governed by the diamond norm of the weighted channel difference.