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

An open-boundary <matrix product state> with physical dimension $q$ and bond dimension $\chi$ is
$$
|\Psi_N\rangle
=\sum_{i_1,\ldots,i_N=1}^q
\langle \ell|A^{i_1}A^{i_2}\cdots A^{i_N}|r\rangle
|i_1i_2\cdots i_N\rangle,
$$
where the $A^i$ are $\chi\times\chi$ matrices and $|\ell\rangle,|r\rangle$ are boundary vectors. For periodic boundary conditions, replace the boundary contraction by the <matrix trace> $\operatorname{Tr}(A^{i_1}\cdots A^{i_N})$.

The same matrices define the <matrix product state transfer map>
$$
\mathcal E(X)=\sum_iA^iX(A^i)^\dagger.
$$
When $\sum_i(A^i)^\dagger A^i=I$, the <Stinespring dilation>
$$
V|\psi\rangle=\sum_i|i\rangle\otimes A^i|\psi\rangle
$$
is an <isometry>. Repeatedly applying $V$ stores each Kraus label in a fresh physical register:
$$
V_N\cdots V_1|r\rangle
=\sum_{i_1,\ldots,i_N}|i_1\cdots i_N\rangle
\otimes A^{i_N}\cdots A^{i_1}|r\rangle.
$$
Contracting the remaining virtual system with $\langle\ell|$ gives an MPS, while tracing over all recorded labels gives repeated application of the <completely positive map> $\mathcal E$. The MPS is therefore a coherent <matrix product state as an unravelling of a completely positive map>[unravelling] of the channel. Equivalently, retaining the Kraus-label registers realizes a <purification of a density operator>[purification] of its output. A non-normalized MPS tensor gives the same construction with a general completely positive map; an appropriate canonical gauge normalizes the transfer map on its support.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 343 / 1 / b / Solution

A purification of a density operator $\rho_A$ is a pure state $|\Psi\rangle_{AE}$ on a larger system such that $\operatorname{Tr}_E|\Psi\rangle\langle\Psi|=\rho_A$. A <spectral decomposition> $\rho_A=\sum_jp_j|j\rangle\langle j|$ gives the canonical construction $|\Psi\rangle=\sum_j\sqrt{p_j}|j\rangle_A|j\rangle_E$.

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 Incoming links: Purification of a density operator