= Solution
{c}
With no access to the discarded <qubit>, the two preparations are exactly the <reduced density matrices> in <2/c>, with equal prior <probabilities>. Their difference is
$$
\Delta=\rho_1-\rho_2=\frac16\begin{pmatrix}-1&1\\1&1\end{pmatrix},
\qquad \Delta^2=\frac1{18}I.
$$
Since $\operatorname{Tr}\Delta=0$, its <eigenvalues> are $\pm\sqrt2/6$. The <trace norm> and <trace distance> are consequently
$$
\|\Delta\|_{\rm tr}=\frac{\sqrt2}{3},
\qquad
\boxed{T(\rho_1,\rho_2)=\frac{\sqrt2}{6}.}
$$
The <Holevo–Helstrom theorem> gives the optimum
$$
\boxed{P_{\rm success}^{\rm max}=\frac12\left(1+\frac{\sqrt2}{6}\right)
=\frac12+\frac{\sqrt2}{12}\simeq0.617851.}
$$
To establish attainability, rather than only an upper bound, measure the <Hermitian operator>
$$
B=\frac{X-Z}{\sqrt2},\qquad
X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad
Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},
$$
where $X,Z$ are <Pauli matrices>. Since $\Delta=(\sqrt2/6)B$, report \b[first qubit for the positive outcome; second qubit for the negative outcome]. The corresponding <orthogonal projections> are $E_\pm=(I\pm B)/2$. Their success <probability> is
$$
\frac12\operatorname{Tr}(E_+\rho_1)+\frac12\operatorname{Tr}(E_-\rho_2)
=\frac12+\frac12\operatorname{Tr}(E_+\Delta)
=\frac12+\frac{\sqrt2}{12},
$$
which reaches the bound. Identical <eigenvalues> of the two <density matrices> do not make them indistinguishable: their eigenvectors differ, and this <quantum measurement> detects that difference.
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