Transpose degradability beyond degradability
- Field
- Topics
Problem
Does there exist a finite-dimensional transpose-degradable quantum channel that is not degradable? Let \(V:A\to B\otimes E\) be an isometry defining a channel and a complementary channel by
Equation (1) fixes the output space \(B\) and environment space \(E\). For the transpose \(\mathsf T_E\) in a fixed basis of \(E\), transpose degradability means that a completely positive trace-preserving map \(\mathcal D:\mathcal L(B)\to\mathcal L(E)\) satisfies
Ordinary degradability instead requires a completely positive trace-preserving map \(\widetilde{\mathcal D}:\mathcal L(B)\to\mathcal L(E)\) satisfying
The question is whether Eq. (2) can hold while no map satisfying Eq. (3) exists.
Source
Singh and Datta explicitly ask whether transpose-degradable channels differ from ordinary degradable channels [SD22]. Brádler had posed the same strict-separation question using the earlier “ conjugate degradable” terminology [Bra15].
Progress
Writing \(J(\mathcal N)\) for the Choi operator of a channel \(\mathcal N\), Eq. (2) implies
\begin{equation} J(\Phi^c)^{T_E}\geq0, \qquad Q(\Phi)=Q^{(1)}(\Phi) :=\max_{\rho_A} \left[S(\Phi(\rho_A))-S(\Phi^c(\rho_A))\right]. \tag{4} \end{equation}Thus Eq. (4) gives a PPT complementary Choi operator and additive coherent information, but neither property supplies an ordinary degrading map [SD22].
If \(J(\Phi^c)\) is separable, then \(\Phi^c\) is entanglement breaking and \(\Phi\) is a Hadamard channel, hence degradable. Any strict example must therefore have a PPT-entangled complementary Choi operator. Moreover, transpose-degradable pcubed channels are always ordinarily degradable, so that structured family contains no strict example [Bra15], [SG16].
The universal-cloning family also provides no strict example. For all \(d\geq2\) and \(N,K\geq1\), the optimal symmetric cloner \(\mathcal C_{N\to N+K}^{(d)}\) and optimal pure-state transposition channel \(\mathcal T_{N\to K}^{(d)}\) obey
\begin{equation} \left(\mathcal C_{N\to N+K}^{(d)}\right)^c =\mathcal T_{N\to K}^{(d)}, \qquad \mathcal T_{N\to K}^{(d)}\ \text{is entanglement breaking}, \qquad \mathcal C_{N\to N+K}^{(d)}\ \text{is degradable}. \tag{5} \end{equation}Equation (5) eliminates the cloning channels that motivated conjugate degradability, but does not prove a general containment theorem [BGS+26].
Comment
No strict example and no equality theorem are known. Complementation turns Eq. (2) into transpose antidegradability and Eq. (3) into ordinary antidegradability. Consequently, the source document’s transpose-antidegradable separation question is exactly the same existence problem, not a distinct problem.
References
- [SD22]
- S. Singh and N. Datta, “ Detecting Positive Quantum Capacities of Quantum Channels,” npj Quantum Information 8, 50 (2022).DOIarXiv
- [Bra15]
- K. Brádler, “ The Pitfalls of Deciding Whether a Quantum Channel Is (Conjugate) Degradable and How to Avoid Them,” Open Systems & Information Dynamics 22, 1550026 (2015).DOIarXiv