Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 65 / 1

Set $X=\rho -\sigma$. The trace distance is $D(\rho ,\sigma )=\frac{1}{2}\Vert X{\Vert }_1$, where $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}$. Since $X$ is a Hermitian operator, its spectral decomposition is $X={\sum }_j{x}_j|j⟩⟨j|$. Define its positive part of a Hermitian operator and negative part of a Hermitian operator byThey are positive semidefinite operators, satisfy $X=Q-R$ and $QR=0$, and obey $|X|=Q+R$. ThereforeThe states have equal trace, so $\mathrm{Tr}X=0$ and $\mathrm{Tr}Q=\mathrm{Tr}R=D(\rho ,\sigma )$. This is the spectral-parts formula for trace distance.

For every orthogonal projector $P$, positivity givesThese inequalities do not require $P$ to commute with $Q$ or $R$: for example, $\mathrm{Tr}(PR)=\mathrm{Tr}(R^{1/2}P{R}^{1/2})\ge 0$, and apply the same observation to $I-P$ and $Q$.

Let $P_{+}$ project onto the positive spectral subspace of $X$. Then $P_{+}Q=Q$ and $P_{+}R=0$, so $\mathrm{Tr}(P_{+}X)=\mathrm{Tr}Q=D(\rho ,\sigma )$. Hence the variational characterization of trace distance over projectors isOne may include or exclude the zero eigenspace without changing the optimum.

The allowed operators are exactly the positive contractions $0\le T\le I$, or binary measurement effects. As in (ii),because $\mathrm{Tr}(TR)\ge 0$ and $\mathrm{Tr}[(I-T)Q]\ge 0$. The positive-spectral orthogonal projector $P_{+}$ is among these effects and attains the bound. ThusAllowing general binary effects instead of projectors does not improve the optimal trace distance test.

Use the unsquared convention for quantum fidelity:Choose a purification of a density operator $|\Psi {⟩}_{RA}$ of $\rho$, and put ${\tau }_{RA}=({\mathrm{id}}_R\otimes \Lambda )(|\Psi ⟩⟨\Psi |)$. By definition,The monotonicity of quantum fidelity under partial trace follows from Uhlmann's theorem: maximizing purifications of two joint states are also candidate purifications of their marginals, with the discarded subsystem included in the purifying system. The optimization for the marginals therefore cannot give a smaller overlap. Here those marginals are $\rho$ and $\Lambda (\rho )$. ConsequentlyThis entanglement fidelity bound by state fidelity distinguishes preserving a reference entanglement from merely reproducing the input marginal.