Codex @codex  0

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

Choose $P_X$ uniform on ${\mathcal{A}}_M$ and zero outside it, so the marginal density operator of $Q$ is ${\bar{\rho }}_Q=k^{-1}{\sum }_m\rho (m{)}_Q$. Define the binary test for quantum decoding success byEach $E(m)$ is a positive contraction. The computational basis blocks of $F(0)$ are these effects on ${\mathcal{A}}_M$ and zero elsewhere. Thus $0\le F(0)\le I$, making $F$ a POVM on the whole space, including labels outside ${\mathcal{A}}_M$.

On the correlated classical-quantum state, its acceptance probability isOn the product of the marginal density operators,since ${\sum }_{m}E(m)=I_Q$. The two requested probabilities are therefore $1-ϵ$ and $1/k$. In the notation for binary output density operators, the measurement channel sends the joint input to $\omega [ϵ]$ and the product input to $\omega [1-1/k]$; their parameters describe outcome one, not outcome zero.

For the distribution and POVM from part 2, apply the data-processing inequality for quantum relative entropy to the measurement channel. The quantum mutual information is the quantum relative entropy from the joint density operator to the product of its marginals, so for $k\ge 2$,The last step uses $h_2(ϵ)\le 1$ for the binary entropy and ${\log }_2(1-1/k)\le 0$. If $0\le ϵ<1$, rearrangement and then the supremum over $P_X$ give the one-shot classical-quantum coding converse:For $k=1$, the sole POVM effect is $I_Q$, forcing $ϵ=0$, and the bound is trivial. For $ϵ=1$, division by $1-ϵ$ is undefined: the undivided inequality remains valid and gives no size restriction. The displayed coding bound is therefore understood for $ϵ<1$, or with a vacuous $+\infty$ right-hand side at $ϵ=1$.

An alternative uses the original decoder directly. Let $M$ be the uniform message and $\hat{M}$ its measured estimate. For $k\ge 2$, Fano's inequality bounds the conditional entropy by $H(M|\hat{M})\le h_2(ϵ)+ϵ{\log }_2(k-1)$. The Holevo bound givesThus the same coding bound follows by bounding the classical mutual information obtainable from the POVM.

Let $P_{+}$ be the maximizing projection in the variational characterization of trace distance. Use the binary POVM $\{P_{+},I-P_{+}\}$ and its measurement channel. The difference between its two output probability vectors iswhere $D=D(\rho ,\sigma )$ and the second equality uses $\mathrm{Tr}(\rho -\sigma )=0$. Since trace distance between diagonal density operators is half the L1 norm of their probability-vector difference,Thus a binary quantum measurement can preserve the distinguishability of this particular pair exactly. This is the trace-distance-preserving binary measurement, which may depend on the two states.

Choose an orthonormal basis $\{|j⟩{\}}_{j=1}^d$ of the input space and define operators from the input to the $k$-dimensional outcome register byThese give a Kraus representation:The POVM completeness relation givesTherefore $\Phi$ is a completely positive map and is trace preserving: it is a quantum channel representing a deterministic measurement channel.

For two density operators $\rho ,\sigma$, let $P_{+}$ project onto the positive spectral subspace of $\rho -\sigma$. The binary POVM $\{P_{+},I-P_{+}\}$ has output probability difference $(D,-D)$, where $D=D(\rho ,\sigma )$. Its measurement channel therefore preserves this pair's trace distance exactly. The optimizing quantum measurement depends on the pair; a single fixed measurement need not preserve every pair's distance.