Strict inclusion of degradable channels in the less-noisy class
- Field
- Topics
Equivalent question: Regularized less-noisy channels beyond degradability. Counted once in question totals.
Problem
Do there exist finite-dimensional quantum channels that are less noisy in Watanabe’s regularized sense but are not degradable? Let \(V_{A\to BE}\) be a Stinespring isometry defining a channel and a complementary channel by
The channel in Eq. (1) is degradable if there is a completely positive trace-preserving map \(\mathcal D_{B\to E}\) such that
Writing \(P\) for the unassisted private classical capacity, Watanabe calls \(\mathcal N\) less noisy when
Equivalently, Eq. (3) requires that, for every \(n\ge1\) and every classical–quantum ensemble on \(UA^{n}\), the receiver and environment outputs obey
Thus the question asks whether the inclusion implied by Eqs. (2)–(4) is strict.
Source
Watanabe proved the inclusion of degradable channels in the regularized less-noisy class and explicitly left open whether it is strict [Wat12].
Progress
Watanabe proved that every degradable channel satisfies Eq. (3), but left open whether this inclusion is strict [Wat12].
Belzig, Gao, Smith, and Wu constructed nondegradable channels satisfying the one-copy version of Eq. (4). Their result separates degradability from the level-1 less-noisy condition, while explicitly leaving the all-blocklength condition in Eq. (4) unresolved [BGSW25].
Zhu and Wang subsequently considered the qutrit channel
\begin{equation} \Lambda(X) =\frac12X+\frac14\bigl(\operatorname{Tr}(X)I-X^{\mathsf T}\bigr) \tag{5} \end{equation}and proved in Theorem 1.1 that
\begin{equation} P(\Lambda)=Q(\Lambda)=0, \qquad \Lambda\ \text{is not antidegradable}. \tag{6} \end{equation}Setting \(\mathcal N=\Lambda^c\), Eq. (6) gives \(P(\mathcal N^c)=0\), so \(\mathcal N\) is Watanabe-less-noisy. If \(\mathcal N\) were degradable, then \(\Lambda=\mathcal D\circ\Lambda^c\) for some channel \(\mathcal D\), contrary to the non-antidegradability statement in Eq. (6). Hence \(\Lambda^c\) is less noisy but nondegradable [ZW26].
Comment
The answer is affirmative: the complement of the channel in Eq. (5) proves that degradable channels form a proper subset of Watanabe-less-noisy channels. The resolving result [ZW26] is, as of August 2026, a recent preprint; the solved status records its theorem rather than peer-review history. This is the same archived question as “Regularized less-noisy channels beyond degradability.” Both permanent identities are retained; the resolving theorem was checked in version 2 on 6 September 2026.