Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 60 / 1

Use the unsquared convention for quantum fidelity. For density operators on a common finite-dimensional Hilbert space,Here $\Vert A{\Vert }_1=\mathrm{Tr}\sqrt{A^{†}A}$ is the trace norm, and all square roots are the positive operator square roots. The two displayed expressions agree because $\sqrt{\rho }\sqrt{\sigma }$ and its adjoint have the same singular values. This convention has $0\le F\le 1$; some literature squares this quantity, but that convention is not used here.

For normalized pure states, the rank-one operator $\sqrt{\rho }\sqrt{\sigma }=⟨\phi |\psi ⟩|\phi ⟩⟨\psi |$ has a single nonzero singular value. Therefore

Uhlmann's theorem identifies quantum fidelity with the largest absolute overlap of purifications of a density operator. Choose a common reference Hilbert space $R$ of dimension at least that of the original system. Thenwhere $|u⟩,|v⟩\in {\mathcal{H}}_A\otimes {\mathcal{H}}_R$ have reduced density operators ${\rho }_A,{\sigma }_A$. One purification may be fixed in advance: the maximum is over the other, with the freedom to apply a unitary on the reference system. This is the unitary freedom of purification. Enlarging the reference by unused dimensions does not change the maximum.

Take maximizing purifications $|u⟩,|v⟩$ of ${\rho }_{AB},{\sigma }_{AB}$ on $ABR$. Their overlap has magnitude $F({\rho }_{AB},{\sigma }_{AB})$. Regard these same vectors as purifications of ${\rho }_A,{\sigma }_A$ with reference system $BR$. They are candidates in the larger optimization, so Uhlmann's theorem givesThis is monotonicity of quantum fidelity under partial trace: discarding a system cannot make two states more distinguishable according to their quantum fidelity.

Embed the given purifications into a common reference Hilbert space $R$, enlarging it if necessary, and add a flag register $X$ with orthonormal basis $\{|x⟩\}$. A flagged purification of a quantum ensemble isOrthogonality of the flags gives $⟨\Psi |\Psi ⟩={\sum }_x{p}_x=1$. Moreover,Thus it is a purification of a density operator. The extra flag is essential: simply superposing the purifications without orthogonal labels would generally leave unwanted cross terms.

For each $x$, choose purifications $|u_x⟩$ of ${\rho }_x$ and $|v_x⟩$ of ${\sigma }_x$ in a common reference space. By Uhlmann's theorem, the second can be chosen, including its overall phase, so thatConstruct the two flagged purifications of a quantum ensembleTheir reduced states are the respective mixtures, and orthogonality of the flags gives $⟨U|V⟩={\sum }_x{p}_{x}F({\rho }_x,{\sigma }_x)$. A particular purification overlap cannot exceed the maximizing overlap in Uhlmann's theorem. HenceThis proves joint concavity of quantum fidelity. Choosing each overlap nonnegative prevents cancellation of different phases; the same probability weights in the two mixtures yield $p_x$ rather than distinct square-root weights.