 Spectral Kraus decomposition

Use the unnormalized Choi matrix $J(\Phi )=(\Phi \otimes \mathrm{id})(|\tilde{\Psi }⟩⟨\tilde{\Psi }|)$, which has trace $d$ for a channel. A positive such matrix has a spectral decomposition with vectors $v_k=\sqrt{{\lambda }_k}{u}_k$. In paired bases define the Kraus operator $A_k$ by $(A_k{)}_{ij}=(v_k{)}_{ij}$, so $v_k=(A_k\otimes I){\sum }_j|j⟩|j⟩$. Partial contraction with a conjugate index vector gives $A_k|\varphi ⟩$, hence $\Phi (|\varphi ⟩⟨\varphi |)={\sum }_k{A}_k|\varphi ⟩⟨\varphi |A_k^{†}$. Rank-one projectors span the operator space, proving the Kraus representation on all inputs. Trace preservation is equivalent to ${\sum }_k{A}_k^{†}{A}_k=I$. This construction needs at most $\mathrm{rank}J$ nonzero Kraus operators.