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

The trace distance and quantum fidelity, in the unsquared convention, are

Let $\Delta =\rho -\sigma ={\Delta }_{+}-{\Delta }_{-}$ be the positive-negative decomposition of a Hermitian operator. Since $\mathrm{Tr}\Delta =0$,For $0\le P\le I$,Equality is attained by the projector onto the positive eigenspace of $\Delta$. This proves the variational characterization of trace distance

Distinct Bell states are orthogonal. For orthogonal pure states, the trace distance is one and the fidelity is the absolute overlap, zero. Hence

The difference $\tau ={\rho }_1-{\rho }_2$ determinesso it suffices for trace distance.

It does not determine fidelity. For $0<t<1$, both pairsandhave the same difference $\mathrm{diag}(t,-t)$. Their fidelities are respectively $\sqrt{1-t^2}$ and $\sqrt{1-t}$, which are generally unequal.

For the binary POVM $\{E,I-E\}$ and equal priors,Maximizing with the variational characterization of trace distance gives the equal-prior Holevo–Helstrom theorem

For priors $p_1,p_2$,If $\Delta =p_1{\rho }_1-p_2{\rho }_2$, the maximum is $p_2+\mathrm{Tr}{\Delta }_{+}$. Sinceone obtains