Nontrivial mutually degradable channel pairs

Problem

Does there exist an integer \(d\geq2\) and a pair of distinct channels \(\mathcal M,\mathcal N:\mathcal L(A)\to\mathcal L(B)\), with \(A\simeq B\simeq\mathbb C^d\), that both have Choi rank exactly \(d\), are mutually degradable, and are each nondegradable? Let \(E\simeq\mathbb C^d\) and choose minimal Stinespring isometries \(V_{\mathcal M},V_{\mathcal N}:A\to B\otimes E\) defining the channels and their complements by

\begin{equation} \begin{aligned} \mathcal M(\rho)&=\operatorname{Tr}_E (V_{\mathcal M}\rho V_{\mathcal M}^{\dagger}), &\mathcal M^c(\rho)&=\operatorname{Tr}_B (V_{\mathcal M}\rho V_{\mathcal M}^{\dagger}),\\ \mathcal N(\rho)&=\operatorname{Tr}_E (V_{\mathcal N}\rho V_{\mathcal N}^{\dagger}), &\mathcal N^c(\rho)&=\operatorname{Tr}_B (V_{\mathcal N}\rho V_{\mathcal N}^{\dagger}). \end{aligned} \tag{1} \end{equation}

Equation (1) fixes representatives of the complementary channels; changing a minimal dilation only applies an output unitary to a complement.

For \(\lvert\Omega_d\rangle:=\sum_{j=1}^d\lvert j\rangle_{A'}\lvert j\rangle_A\), the required Choi-rank condition is

\begin{equation} J(\mathcal T):=(\operatorname{id}_{A'}\otimes\mathcal T) (\lvert\Omega_d\rangle\!\langle\Omega_d\rvert), \qquad \operatorname{rank}J(\mathcal M) =\operatorname{rank}J(\mathcal N)=d. \tag{2} \end{equation}

The equality in Eq. (2) makes the environment dimension in Eq. (1) minimal.

Mutual degradability requires channels \(\mathcal X,\mathcal Y:\mathcal L(B)\to\mathcal L(E)\) such that

\begin{equation} \mathcal X\circ\mathcal M=\mathcal N^c, \qquad \mathcal Y\circ\mathcal N=\mathcal M^c. \tag{3} \end{equation}

In addition to Eq. (3), neither channel may admit its own degrading map:

\begin{equation} \begin{aligned} &\nexists\ \mathcal D_{\mathcal M}:\mathcal L(B)\to\mathcal L(E) \quad\text{CPTP with}\quad \mathcal M^c=\mathcal D_{\mathcal M}\circ\mathcal M,\\ &\nexists\ \mathcal D_{\mathcal N}:\mathcal L(B)\to\mathcal L(E) \quad\text{CPTP with}\quad \mathcal N^c=\mathcal D_{\mathcal N}\circ\mathcal N. \end{aligned} \tag{4} \end{equation}

Equation (4), together with \(\mathcal M\neq\mathcal N\), excludes the identity-channel and self-pair constructions described below. It does not exclude complementary pairs.

Source

Ruskai posed mutual degradability in Problem 23, Eq. (32), and singled out two Choi-rank-\(d\) channels that are not individually degradable [Rus07]. The existence statement is answered by the explicit complementary qutrit pair recorded in item A of the catalog’s scientific-review issue [Rev26].

Progress

Reports do not certify correctness or automatically change the problem's status. Progress policy.

An explicit solution has \(d=3\), \(a=3/4\), and \(b=1/3\). Define the isometry

\begin{equation} \begin{aligned} V|0\rangle&=|00\rangle,\\ V|1\rangle&=\sqrt{1-a}|10\rangle+\sqrt a|01\rangle,\\ V|2\rangle&=\sqrt{1-b}|20\rangle+\sqrt b|02\rangle. \end{aligned} \tag{5} \end{equation}

Set \(\mathcal M=\operatorname{Tr}_E V(\cdot)V^\dagger\) and \(\mathcal N=\operatorname{Tr}_B V(\cdot)V^\dagger\) in Eq. (5), identifying both output spaces with \(\mathbb C^3\). Use the swapped isometry for \(\mathcal N\). Then \(\mathcal M^c=\mathcal N\) and \(\mathcal N^c=\mathcal M\), so \(\mathcal X=\mathcal Y=\operatorname{id}\) satisfies Eq. (3). The Kraus operators of \(\mathcal M\) are \(\operatorname{diag}(1,\sqrt{1-a},\sqrt{1-b})\), \(\sqrt a|0\rangle\langle1|\), and \(\sqrt b|0\rangle\langle2|\); those of \(\mathcal N\) replace \((a,b)\) by \((1-a,1-b)\). Both triples are linearly independent, proving Eq. (2), and the outputs on \(|1\rangle\langle1|\) differ. If degrading maps \(\mathcal D\mathcal M=\mathcal N\) and \(\mathcal D'\mathcal N=\mathcal M\) existed, their actions on the ground state and the relevant excited state would force

\begin{equation} \begin{aligned} \mathcal D(|1\rangle\langle1|) &=3|1\rangle\langle1|-2|0\rangle\langle0|,\\ \mathcal D'(|2\rangle\langle2|) &=2|2\rangle\langle2|-|0\rangle\langle0|. \end{aligned} \tag{6} \end{equation}

Neither output in Eq. (6) is positive, proving Eq. (4). This is the unpublished construction from the scientific-review issue [Rev26].
At unrestricted rank, Ruskai observed that the identity channel and an arbitrary channel form a mutually degradable pair. Also, any degradable channel paired with itself satisfies Eq. (3). The rank, distinctness, and nondegradability requirements exclude both constructions [Rus07].
Cubitt, Ruskai, and Smith proved that every qubit channel with two Kraus operators is either degradable or antidegradable. Consequently, any \(d=2\) solution satisfying Eq. (4) must consist of two antidegradable channels. Their classification neither constructs nor excludes such a pair satisfying Eq. (3) [CRS08].

Comment

The displayed existence question is solved by Eqs. (5) and (6). The resolving construction is an unpublished issue contribution, not a peer-reviewed result. Excluding complementary pairs would define a stronger question; no such exclusion appears in the archived statement or in Ruskai’s Problem 23.

References

[Rus07]: M. B. Ruskai, “Open Problems in Quantum Information Theory,” arXiv preprint arXiv:0708.1902 (2007).DOI arXiv
[CRS08]: T. S. Cubitt, M. B. Ruskai, and G. Smith, “The Structure of Degradable Quantum Channels,” Journal of Mathematical Physics 49, 102104 (2008).DOI arXiv
[Rev26]: QIQCOP Zoo, “Scientific review: confirm channel, resource and QEC statements in 10 catalog records,” GitHub issue #41, item A (2026). Unpublished construction. Issue #41.link

Nontrivial mutually degradable channel pairs

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