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> gives
$$
\boxed{|\Psi\rangle_{AR}
=(I_A\otimes U_R)|\Phi\rangle_{AR}}.
$$
If the reference supports have different dimensions, the corresponding statement uses an isometry.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-323/2/a/i/solution