Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 2

Take Schmidt decompositions of the two purifications across $A:R$. Their squared Schmidt coefficients and their $A$-side eigenspaces are fixed by the same reduced state ${\rho }_A$. The reference-side Schmidt vectors are two orthonormal families, so a unitary $U_R$ maps one family to the other, including arbitrary choices inside degenerate subspaces. Hence the unitary freedom of purification givesIf the reference supports have different dimensions, the corresponding statement uses an isometry.

Fororthonormality of the $|b_i⟩$ givesNormalization follows from ${\sum }_i{\lambda }_i=1$, so this is a purification of a density operator.

With $|\Omega {⟩}_{RA}=n^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_A$,The partial trace over $R$ isIts norm is $\mathrm{Tr}{\rho }_A=1$. The positive square root of an operator $\sqrt{{\rho }_A}$ is unique, proving the claim.

The states ${\rho }_A^i=|a_i⟩⟨a_i|$ are pure, so a purification isTracing out the orthonormal reference labels removes all cross terms and recovers the stated classical-quantum mixture. Its rank is the number of nonzero $p_{ij}$, so that is the minimum reference dimension for a particular state. The smallest dimension that can purify every state of the stated form is

Construct ensemble purificationsThey have the same reduced state exactly when ${\rho }_1={\rho }_2$. By the unitary freedom of purification, this holds exactly when $|{\Phi }_2⟩=(I\otimes U)|{\Phi }_1⟩$ for a unitary $U$. Comparing reference-basis coefficients gives the Hughston–Jozsa–Wootters theorem relationConversely, substituting this relation and using $U^{†}U=I$ immediately gives ${\rho }_2={\rho }_1$.

Yes. Uhlmann's theorem says that, for the fixed purification $|{\Psi }^{\rho }{⟩}_{AR}$,Choose a maximizing purification of ${\sigma }_A$ on the same reference space. In the unsquared fidelity convention,