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

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.