= Matrix product state as an unravelling of a completely positive map
Matrices $A^i$ acting on a finite-dimensional virtual space define a <completely positive map>
$$
\mathcal E(X)=\sum_iA^iX(A^i)^\dagger.
$$
The same matrices form the local tensor of a <matrix product state>. To interpret $\mathcal E$ as a <quantum channel>, require the additional <trace> <normalization> $\sum_i(A^i)^\dagger A^i=I$.
Under this condition,
$$
W|v\rangle=\sum_i|i\rangle\otimes A^i|v\rangle
$$
is an <isometry>, because $W^\dagger W=\sum_i(A^i)^\dagger A^i=I$. Keeping the physical label $i$ preserves a coherent record of each <Kraus operator>. Tracing it out gives $\operatorname{Tr}_{\rm phys}(WXW^\dagger)=\mathcal E(X)$. This is the finite-dimensional <Stinespring representation> of the <quantum channel>. <The Theory of Quantum Information by John Watrous/Basic notions of quantum information>[Section 2.2, Theorem 2.22 and Corollary 2.27, Kraus and isometric Stinespring channel conditions]{ref}
Starting from $|L\rangle$ and applying this <isometry> repeatedly produces amplitudes $A^{i_n}\cdots A^{i_1}|L\rangle$ for the physical chain and remaining virtual system. A final boundary contraction with $\langle R|$ gives the <matrix product state> amplitudes $\langle R|A^{i_n}\cdots A^{i_1}|L\rangle$. Tracing all physical labels instead gives $\mathcal E^n(|L\rangle\langle L|)$.
Without the stated <normalization>, the matrices still define a <completely positive map> and a <matrix product state> tensor. They do not automatically describe a deterministic <quantum channel>; <normalization> or conditioning must be specified.
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