We obtain a lower bound on the maximum number of qubits, $Q^{n, \epsilon}(\mathcal{N})$, which can be transmitted over $n$ uses of a quantum channel $\mathcal{N}$, for a given non-zero error threshold $\epsilon$. To obtain our result, we first derive a bound on the one-shot entanglement transmission capacity of the channel, and then compute its asymptotic expansion up to the second order. In our method to prove this achievability bound, the decoding map, used by the receiver on the output of the channel, is chosen to be the \emph{Petz recovery map} (also known as the \emph{transpose channel}). Our result, in particular, shows that this choice of the decoder can be used to establish the coherent information as an achievable rate for quantum information transmission. Applying our achievability bound to the 50-50 erasure channel (which has zero quantum capacity), we find that there is a sharp error threshold above which $Q^{n, \epsilon}(\mathcal{N})$ scales as $\sqrt{n}$.

翻译：我们针对给定非零误差阈值$\epsilon$，获得了量子信道$\mathcal{N}$在$n次使用中可传输的最大量子比特数$Q^{n, \epsilon}(\mathcal{N})$的下界。为得到该结果，我们首先推导了信道单次纠缠传输容量的界限，进而计算其二阶渐进展开式。在证明该可达性界限的方法中，接收端对信道输出采用的解码映射被选为\emph{Petz恢复映射}（亦称\emph{转置信道}）。该结果特别表明，该解码器的选择可用于确立相干信息作为量子信息传输的可达速率。将我们的可达性界限应用于50-50擦除信道（其量子容量为零），发现存在一个临界误差阈值，当超过该阈值时$Q^{n, \epsilon}(\mathcal{N})$以$\sqrt{n}$的速率增长。