15 The above expression vanishes when µ = 0. It coincides with the quantum capacity of a 1 → N cloning channel when µ = 1 2 (Equation 30 in Ref. [33]): 0 0.2 0.4 0.6 0.8 1 1.2 0 0.5 1 0.2 0.4 0.6 0.8 1 1.2 1.4 Classical communication rate Quantum communication rate Entanglement consumption rate We now turn to the proof of the CE trade-off curve for the 1 → N cloning channel. Lemma 6 An ensemble of the same form as in Lemma 5 parametrizes all points on the CE trade-off curve for a 1 → N cloning channel. Proof. The proof of this lemma proceeds along simi- lar lines as the proof of Lemma 5, using the same states and ideas concerning the cloning channel. We highlight the major differences by giving the following chain of in- equalities that holds for all λ such that 0 ≤ λ < 1: I ( A 〉 BX )$_{µ}$$_{=1}$$_{/}$$_{2}$ = log$^{(}$ N + 1 N ) . FIG. 3: (Color online) The figure plots the CQE capacity re- gion for a 1 → 10 cloning channel. It features three distinct surfaces. The first is the flat vertical plane that arises from the bound R +2 Q ≤ log ( N +1 N $^{)}$+1, which is the entanglement- assisted classical capacity of a 1 → N cloning channel. The plane extends infinitely upward because we can always achieve these rate triples simply by wasting entanglement. The sec- ond surface is that below and to the left of the plane, formed by combining the CE trade-off curve with the inverse of super- dense coding, as described in Section III C. The final surface is that below and to the right of the plane, formed by combin- ing the CQ trade-off curve with the inverse of entanglement distribution, as described in Section III C. I ( AX ; B )$_{ρ}$ − λH ( A | X )$_{ρ}$ = (1 − λ ) H ( A | X )$_{ρ}$ + H ( B )$_{ρ}$ − H ( E | X )$_{ρ}$ = H ( B )$_{ρ}$ + (1 − λ ) H ( A | XJ )$_{σ}$ − H ( E | XJ )$_{σ}$ ≤ H ( B )$_{σ}$ + (1 − λ ) H ( A | XJ )$_{σ}$ − H ( E | XJ )$_{σ}$ = log ( N + 1) + (1 − λ ) H ( A | XJ )$_{σ}$ − H ( E | XJ )$_{σ}$ = log ( N + 1) + ∑ x p$_{X}$ ( x ) $^{[}$(1 − λ ) H ( A )$_{ψ}$$_{x}$ − H ( E )$_{ψ}$$_{x}$ ] ≤ log ( N + 1) + max$_{x}$ $^{[}$(1 − λ ) H ( A )$_{ψ}$$_{x}$ − H ( E )$_{ψ}$$_{x}$ ] = log ( N + 1) + (1 − λ ) H ( A )$_{ψ}$$_{∗}$ x − H ( E )$_{ψ}$$_{∗}$ $_{x}$. ( C$_{CQ}$ ( z,µ ) , Q$_{CQ}$ ( z,µ )) , ( C$_{CE}$ ( z,µ ) , E$_{CE}$ ( µ )) , linked to those of the cloning channel because the math- ematical structure of the output of the Unruh channel is an infinite-dimensional block-diagonal density matrix, with each block containing an occurence of a 1 → N cloning channel. In fact, maximizing the rate of trans- mission is equivalent to maximizing it in each block. This observation gives us the following theorem, and its proof appears in Appendix F. We do not provide justifications for the above chain of inequalities because it follows for similar reasons to the chain of inequalities in Lemma 5. Proof of CE trade-off in Theorem 5. The proof follows by noting that I ( AX ; B ) = H ( A | X ) + H ( B ) − H ( E | X ), H ( A | X ) = H$_{2}$ ( µ ), and that we have already computed H ( B ) and H ( E | X ) in the proof of the CQ trade-off in Theorem 5. Figure 2 plots the CQ and CE trade-off curves for a 1 → N cloning channel, for N = 1, 2, 3, 5, 8, 12, and 24. The figure demonstrates that both classically-enhanced quantum coding and entanglement-assisted classical cod- ing beat a time-sharing strategy for a cloning channel when N > 1. Theorem 6 All points on the CQ trade-off curve for an Unruh channel have the following form: and all points on the CE trade-off curve for an Unruh channel have the following form: Figure 3 plots the full CQE capacity region for a 1 → 10 cloning channel. The figure demonstrates that the classically-enhanced father protocol, combined with entanglement distribution and super-dense coding beats a time-sharing strategy because the first two surfaces described in Section III C are strictly concave for the cloning channels when N > 1. where We now compute the trade-off curves for the Unruh channel, defined in (13). Capacity questions are directly C. Unruh Channel C$_{CQ}$ ( z,µ ) ≡ 1 − ∞ ∑ l =2 p$_{l}$ ( z ) log ( l − 1) + ∞ ∑ l =2 p$_{l}$ ( z ) Δ$_{l}$$_{−}$$_{1}$ l − 1 ∑ i =0 λ ( l − 1) i ( µ ) log ( λ ( l − 1) i ( µ ) ) ,