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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.

Complex conjugation sends a complex number $z=a+ib$ to its complex conjugate $\overline{z}=a-ib$. It is an involution and a field automorphism of $\mathbb{C}$, with fixed field $\mathbb{R}$. Applied componentwise to a complex vector, it produces the conjugated vector appearing in a Choi state for a rank-one Kraus operator.

Use the unnormalized Choi matrixIt is positive because ${\Phi }_A$ is a completely positive map. Take its spectral decomposition, absorbing square roots of nonzero eigenvalues into the vectors:Expand $|v_k⟩={\sum }_{i,j}{v}_{k,ij}|i{⟩}_A|j{⟩}_B$ and define the Kraus operator $A_k$ by $(A_k{)}_{ij}=v_{k,ij}$. Equivalently,The preceding relative state identity then gives $(I\otimes ⟨{\varphi }_B^{\ast }|)v_k=A_k{\varphi }_A$. Inserting the spectral decomposition into the stated reconstruction identity yieldsThe conjugated index vector is required here; it is present in the PDF but missing in the converted TeX version of this identity.

Rank-one projectors span all operators by polarization. Explicitly, putting $P(w)=|w⟩⟨w|$ givesThus linearity extends the equality to every $X\in \mathcal{B}({\mathcal{H}}_A)$:Finally trace preservation and cyclicity of the trace giveFor pure-projector inputs this says the Hermitian operator ${\sum }_k{A}_k^{†}{A}_k-I$ has zero expectation on every vector, so it vanishes. ThereforeThis is the spectral Kraus decomposition. Using a normalized Choi state $J_{\Phi }/d$ instead would require the extra factor $\sqrt{d}$ when reshaping its eigenvectors into Kraus operators.

Fix an orthonormal basis $\{|i⟩{\}}_{i=1}^d$ of the input Hilbert space, where $d=\dim \mathcal{H}$, and a reference copy $R\simeq \mathcal{H}$. Use the normalized maximally entangled vector $|{\Omega }_d⟩=d^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_{\mathcal{H}}$. The normalized Choi–Jamiołkowski state isComplete positivity makes $J\ge 0$, and trace preservation gives ${\mathrm{Tr}}_{\mathcal{K}}J=I_R/d$, hence $\mathrm{Tr}J=1$. It is therefore a genuine density operator. In the unnormalized Choi matrix convention the factor $1/d$ is omitted and the trace is $d$; specifying normalization distinguishes a Choi state from that matrix convention. The reference basis also fixes the transpose in the Choi reconstruction formula, $\Lambda (X)=d{\mathrm{Tr}}_R[(X^T\otimes I)J]$.

The useful object is the Choi state of the entanglement-breaking channel. Put $d_A=\dim {\mathcal{H}}_A$, introduce a reference $R$ with the same computational basis as $A$, and use the normalized maximally entangled stateThe entanglement-breaking channel makes $C_{RB}$ a separable quantum state. Its partial trace is ${\mathrm{Tr}}_B{C}_{RB}=I_R/d_A$, because $\mathcal{E}$ is a quantum channel. Write a separable positive operator decompositionDiscard any term with $\mathrm{Tr}{\beta }_y=0$: a positive operator with zero trace is zero. DefineEach ${\tau }_y$ is a density operator. Each $E_y$ is a positive operator, and taking the matrix transpose of the partial trace identity givesThus the $E_y$ form a POVM. In this reference-first convention, the Choi reconstruction formula isThe factor $d_A$ is essential because we used a normalized Choi state. A measurement channel storing the outcome $y$, followed by the prescribed quantum state preparation of ${\tau }_y$, has precisely this action on every linear operator $L$. The desired factorization is therefore

Use the normalized Choi state $C_{\Lambda }=(\Lambda \otimes \mathrm{id})(|{\Phi }_d⟩⟨{\Phi }_d|)$, with $|{\Phi }_d⟩=d^{-1/2}{\sum }_i|ii⟩$. If $\Lambda$ is an entanglement-breaking channel, its action on this particular bipartite input makes $C_{\Lambda }$ a separable quantum state.

Conversely, suppose $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, with local density operators ${\sigma }_a,{\tau }_a$. The Choi reconstruction formula givesBecause $\Lambda$ is trace preserving, ${\mathrm{Tr}}_A{C}_{\Lambda }=I/d$, so ${\sum }_a{E}_a=I$. Thus the $E_a$ form a POVM and the channel is a measure-and-prepare channel.

For any bipartite input ${\rho }_{AR}$, define the positive, possibly unnormalized reference operatorsIts output is ${\sum }_a{\sigma }_a\otimes R_a$. Since ${\sum }_a\mathrm{Tr}{R}_a=1$, normalizing each nonzero $R_a$ expresses this as a convex combination of product states. Hence every output is separable, proving the separable Choi-state criterion for entanglement breaking.

Expanding the normalized maximally entangled state in the definition of the Choi state givesHere the bar denotes componentwise complex conjugation in the basis defining the Choi state. Every term is a positive tensor-product operator, and the whole operator has trace one because $\tilde{\Lambda }$ is assumed to be a quantum channel. It is therefore a separable quantum state, proving entanglement breaking by the separable Choi-state criterion for entanglement breaking.

If the vectors are unit vectors, the displayed $p_j$ are already the product-state weights. If they are not normalized, absorb their squared norms into the weights and normalize the nonzero vectors. Trace preservation requiresthe probability-distribution condition alone would not guarantee this. Equivalently, the Kraus operators are $K_j=\sqrt{d{p}_j}|a_j⟩⟨b_j|$, exhibiting the rank-one Kraus representation of an entanglement-breaking channel.

An entanglement-breaking channel admits a Kraus representation with rank-one Kraus operators $K_j=|a_j⟩⟨b_j|$. Its Choi state is a sum of positive tensor-product operators, hence separable. Conversely, spectral decompositions of each prepared state and each POVM effect in a measure-and-prepare channel produce rank-one Kraus operators. Trace preservation imposes ${\sum }_j\Vert a_j{\Vert }^2|b_j⟩⟨b_j|=I$.

A quantum channel is entanglement breaking exactly when its Choi state is a separable quantum state. If $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, the Choi reconstruction formula gives a measure-and-prepare channel with POVM $E_a=d{p}_a{\tau }_a^T$. Conversely, applying an entanglement-breaking channel to a maximally entangled state produces a separable Choi state. Horodecki, Shor and Ruskai's entanglement-breaking channel paper develops the equivalences.