For $p=(d+1)/(2d)$, the weighted difference of the two <Werner–Holevo channels> is
$$
\begin{aligned}
\bigl(pT_+-(1-p)T_-\bigr)(X)
&=\frac1{2d}\bigl(\operatorname{Tr}(X)I+X^T\bigr)
-\frac1{2d}\bigl(\operatorname{Tr}(X)I-X^T\bigr)\\
&=\frac1dX^T.
\end{aligned}
$$
Thus 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> give
$$
\boxed{\|pT_+-(1-p)T_-\|_\diamond
=\frac1d\|\Theta\|_\diamond=1},
\qquad
\boxed{\|pT_+-(1-p)T_-\|_1
=\frac1d\|\Theta\|_1=\frac1d}.
$$
Substitution into the optimal-error formulas shows that an entangled <quantum ancilla> permits perfect discrimination, whereas every ancilla-free strategy has
$$
\boxed{P_{\mathrm{err}}^{\mathrm{no\ anc}}=\frac12\left(1-\frac1d\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 323 / 1 / iii / Solution

{title2=$A^T$}
{wiki=Transpose}

The matrix transpose interchanges rows and columns: $(A^T)_{ij}=A_{ji}$. On a fixed matrix space, $A\mapsto A^T$ is the <transposition map>, a <linear map> that preserves the <trace norm> but is not completely positive in dimensions greater than one.

= Transposition map
{synonym}

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