For
$$
|\Psi\rangle_{AR}
=\sum_i\sqrt{\lambda_i}|a_i\rangle_A|b_i\rangle_R,
$$
orthonormality of the $|b_i\rangle$ gives
$$
\operatorname{Tr}_R|\Psi\rangle\langle\Psi|
=\sum_i\lambda_i|a_i\rangle\langle a_i|
=\rho_A.
$$
Normalization follows from $\sum_i\lambda_i=1$, so this is a <purification of a density operator>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 2 / a / ii / 1 / Solution

 Solution

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