Decoding quantum information via the Petz recovery map
Salman Beigi, Nilanjana Datta, Felix Leditzky
TL;DR
This work advances the finite-blocklength theory of quantum information transmission by proving a second-order achievability bound for the quantum capacity of a channel using an explicit Petz recovery map as the decoder. The key insight is that the decoding via the Petz map, combined with a one-shot bound expressed through the information spectrum relative entropy, leads to a second-order term governed by the ε-quantum dispersion V_ε and the inverse normal quantile Φ^{-1}(ε). The method builds a random-code construction, analyzes average fidelity through the collision relative entropy, and introduces a weak monotonicity property under dephasing as a crucial technical ingredient. The results are illustrated on the 50-50 erasure channel, where the bound reveals a sharp transition: for ε > 1/2 the maximum transmissible qubits scale as √n, while for ε < 1/2 the one-shot capacity remains constant with n, underscoring the nuanced finite-blocklength behavior of quantum channels.
Abstract
We obtain a lower bound on the maximum number of qubits, $Q^{n, ε}(\mathcal{N})$, which can be transmitted over $n$ uses of a quantum channel $\mathcal{N}$, for a given non-zero error threshold $ε$. 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 $Q^{n, ε}(\mathcal{N})$ scales as $\sqrt{n}$.