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For the transposition map $\Theta :M_d\to M_d$, $\Theta (X)=X^T$, the induced trace norm is one, while the diamond norm isThe first equality follows because transposition preserves every singular value and hence the trace norm.

For the lower bound, apply $\Theta \otimes {\mathrm{id}}_d$ to the normalized maximally entangled state projector $\rho =|{\Phi }_d⟩⟨{\Phi }_d|$. Its partial transpose is $F/d$, with $F$ the swap operator. All $d^2$ singular values of $F$ are one, so $\Vert F/d{\Vert }_1=d$ while $\Vert \rho {\Vert }_1=1$.

For the upper bound, partial transpose preserves the Frobenius norm $\Vert \cdot {\Vert }_2$. On matrices of size $d^2$,A $d$-dimensional ancilla suffices in the definition of the diamond norm, proving equality. ^[ref]

For $d>1$, transposition is not completely positive: $F/d$ has a negative eigenvalue on the antisymmetric subspace. Thus this is a mathematical example of norm amplification by an entangled ancilla; transposition itself is not a physically implementable quantum channel.

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.

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.