Choi matrix

For normalized maximally entangled $|\varphi ⟩=d^{-1/2}{\sum }_j|jj⟩$, the Choi matrix of $T$ is $C_T=(T\otimes \mathrm{id})(|\varphi ⟩⟨\varphi |)$. Choi's theorem says $T$ is completely positive exactly when $C_T\ge 0$.

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.

A channel's Choi state is its normalized Choi matrix,It is a density operator and satisfies ${\mathrm{Tr}}_{\mathrm{out}}{C}_{\Lambda }=I_{\mathrm{in}}/d$. Some authors use an unnormalized Choi matrix of trace $d$, so reconstruction factors depend on the convention.

With normalized Choi state $C_{\Lambda }$ and the output factor first,Expand $C_{\Lambda }=d^{-1}{\sum }_{i,j}\Lambda (|i⟩⟨j|)\otimes |i⟩⟨j|$. The partial trace multiplies each coefficient by $\mathrm{Tr}(|i⟩⟨j|X^T)=X_{ij}$, reconstructing $\Lambda (X)$ by linearity.