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Every entanglement-breaking channel $\mathcal{E}$ has additive Holevo capacity with every finite-dimensional quantum channel $\mathcal{N}$. A measure-and-prepare channel factorization exposes a classical measurement outcome. The quantum mutual information balance identity and data processing for quantum mutual information then bound the joint output Holevo quantity by the sum of the individual capacities, using the conditional input ensemble after a local measurement for the other channel. Independent product ensembles attain the reverse inequality.

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.

Apply the measure-and-prepare channel locally to an arbitrary bipartite density operator ${\rho }_{AR}$. The resulting state iswhereTo see that this is the correct output, the partial trace is cyclic for operators acting only on the traced subsystem, so the displayed expression equals ${\mathrm{Tr}}_A[(E_a\otimes I){\rho }_{AR}]$. Put $q_a=\mathrm{Tr}{R}_a$. The POVM completeness relation implies ${\sum }_a{q}_a=1$, and terms with $q_a=0$ vanish. Thereforeis a separable quantum state. Thus the channel is an entanglement-breaking channel, for every reference-system dimension.

With $P_i=|y_i⟩⟨y_i|$, the channel has Kraus form $Y(\rho )={\sum }_i{P}_i\rho P_i$ and ${\sum }_i{P}_i^{†}{P}_i=I$, so it is completely positive and trace preserving. It measures in the $y$ basis and prepares the observed basis state, hence is a measure-and-prepare channel. If $p_i=⟨y_i|\rho |y_i⟩$, then $Y(\rho )={\sum }_i{p}_i{P}_i$ andMoreover $\log Y(\rho )={\sum }_i(\log p_i)P_i$ on its support, so

Choose a basis $|a⟩$ of $H$ and define Kraus operators from $H$ to ${\mathbb{C}}^m$ byThenandThe displayed map is therefore a completely positive trace-preserving measure-and-prepare 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.