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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.

A measure-and-prepare channel measures a POVM $\{M_x\}$ and outputs a prescribed state ${\sigma }_x$: $T(\rho )={\sum }_x\mathrm{Tr}(M_x\rho ){\sigma }_x$. Such channels are exactly the entanglement-breaking channels.

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.

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.

For a bipartite input,Each second factor is positive and has trace equal to the probability of measurement outcome $k$. Dividing nonzero factors by their traces therefore writes the output as a convex combination of product states. Hence every measure-and-prepare channel is 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 ${\sum }_j\Vert a_j{\Vert }^2|b_j⟩⟨b_j|=I$.