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 known degenerate constructions.

Source

Ruskai posed mutual degradability explicitly and singled out the case of two Choi-rank-\(d\) channels that are not individually degradable [Rus07].

Progress

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 remaining task is either to construct a distinct pair satisfying Eqs. (2)–(4) for some \(d\), or to prove that mutual degradability at Choi rank \(d\) forces equality of the channels or individual degradability.

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

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