The <trace distance> and <quantum fidelity>, in the unsquared convention, are
$$
\boxed{D(\rho,\sigma)=\frac12\|\rho-\sigma\|_1},
\qquad
\boxed{F(\rho,\sigma)
=\operatorname{Tr}\sqrt{\sqrt\rho\,\sigma\sqrt\rho}}.
$$

Solution

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

For outcome probabilities $p_k=\operatorname{Tr}(\Lambda_k\rho)$ and $q_k=\operatorname{Tr}(\Lambda_k\sigma)$, the hint gives
$$
\frac12\sum_k|p_k-q_k|
\geq\frac12\sum_k(\sqrt{p_k}-\sqrt{q_k})^2
=1-\sum_k\sqrt{p_kq_k}.
$$
The left side is at most $\|\rho-\sigma\|_1/2$ by part d. Choose a POVM attaining the stated minimum characterization of <quantum fidelity>; then
$$
\boxed{\frac12\|\rho-\sigma\|_1\geq1-F(\rho,\sigma)}.
$$

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 323 / 4 / e / Solution

{title2=$F(\rho,\sigma)$}
{wiki=Fidelity_of_quantum_states}

In the unsquared convention, quantum fidelity is $F(\rho,\sigma)=\operatorname{Tr}\sqrt{\sqrt\rho\sigma\sqrt\rho}$. It equals the minimum classical Bhattacharyya coefficient over all POVMs.


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