It is known that for degradable channels $$\mathcal{N}$$ and $$\mathcal{M}$$, the single-letter quantum capacity is aditive (Potential Capacities of Quantum Channels), i.e. $$$$Q^{(1)}(\mathcal{N}\otimes\mathcal{M}) = Q^{(1)}(\mathcal{N}) + Q^{(1)}(\mathcal{M}).$$$$

I am unsure if this kind of results can be extrapolated to the setting of anti-degradable channels, that is, is it true that for $$\mathcal{N}$$ and $$\mathcal{M}$$ anti-degradable the quantum capacity equals zero (additive)? i.e. $$$$Q^{(1)}(\mathcal{N}\otimes\mathcal{M}) = Q^{(1)}(\mathcal{N}) + Q^{(1)}(\mathcal{M})=0?$$$$ Moreover, would the following be also true: let $$\mathcal{N}$$ be a degradable channel and $$\mathcal{M}$$ an anti-degradable channel, then, $$$$Q^{(1)}(\mathcal{N}\otimes\mathcal{M}) = Q^{(1)}(\mathcal{N}) + Q^{(1)}(\mathcal{M})=Q^{(1)}(\mathcal{N})?$$$$

I have been looking in the literature for results of this type and have not found anything. Also, if this are not true in general, is it possible that they are true for particular channel, for example, if we have two amplitude damping channels, one with the damping parameter in the degradable region and the other with the damping parameter in the antidegradable region.

• You might add that, for degradable channels, the single-letter capacity is in fact equal to quantum capacity. In several superactivation results, one of the two 0-capacity channels is antidegradable, but you may already know this. It seems unlikely to me that superactivation could occur with two antidegradable channels, but I have no reference for such a claim. Jun 14 at 18:44