The difference $\tau=\rho_1-\rho_2$ determines
$$
\boxed{D(\rho_1,\rho_2)=\frac12\|\tau\|_1},
$$
so it suffices for trace distance.

It does not determine fidelity. For $0<t<1$, both pairs
$$
\rho_1=\begin{pmatrix}(1+t)/2&0\\0&(1-t)/2\end{pmatrix},
\quad
\rho_2=\begin{pmatrix}(1-t)/2&0\\0&(1+t)/2\end{pmatrix}
$$
and
$$
\rho_1'=\begin{pmatrix}t&0\\0&1-t\end{pmatrix},
\quad
\rho_2'=\begin{pmatrix}0&0\\0&1\end{pmatrix}
$$
have the same difference $\operatorname{diag}(t,-t)$. Their fidelities are respectively $\sqrt{1-t^2}$ and $\sqrt{1-t}$, which are generally unequal.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 4 / a / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-323/4/a/iv/solution