发表机构
The Hong Kong University of Science and Technology (Guangzhou); QudeLeap Research（香港科技大学（广州）; 量子跃迁研究）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文针对具有纯输出的量子比特到量子比特信道，消除了经典容量与纠缠代价的正则化，证明振幅阻尼信道的经典容量等于单次Holevo信息，纠缠代价为$h_2((1+\sqrt p)/2)$，并给出支撑准则与有限块亏缺结果。
AI 中文摘要
确定噪声量子信道的经典容量和纠缠代价通常需要对多次信道使用进行正则化处理。我们对于每个具有纯输出的量子比特到量子比特信道，消除了这两种正则化。对于每个此类信道，Holevo信息、信道形成纠缠以及并行纠缠代价与任何有限维伙伴信道的相应量是可加的。该类别包括所有Kraus秩至多为二的量子比特到量子比特信道。对于阻尼概率为$p$的振幅阻尼信道，无辅助经典容量等于已知的单次使用Holevo信息，由二元纯态系综通过集体解码实现，纠缠代价为$h_2((1+\sqrt p)/2)$ ebits每次使用。其共同机制是纠缠形成强超可加性的支撑准则：一个边缘在局部系统均正交于固定区分向量的扇区上没有支撑。我们利用三角块矩阵熵不等式和保持两个期望的分解，在任意有限维度下证明了该准则。对于振幅阻尼，我们还推导了精确的有限块Holevo亏缺，确定了$p<1$时唯一的最优平均输入，并构造了达到均匀形成界的二元Kraus表示。
英文摘要
Determining a noisy quantum channel's classical capacity and entanglement cost generally requires regularization over many channel uses. We remove both regularizations for every qubit-to-qubit channel admitting a pure output. For each such channel, Holevo information, channel entanglement of formation, and parallel entanglement cost are additive with those of any finite-dimensional partner channel. This class includes all qubit-to-qubit channels of Kraus rank at most two. For the amplitude damping channel with damping probability $p$, the unassisted classical capacity equals the known single-use Holevo information, attained by a binary pure-state ensemble with collective decoding, and the entanglement cost is $h_2((1+\sqrt p)/2)$ ebits per use. The common mechanism is a support criterion for strong superadditivity of entanglement of formation: one marginal has no support on the sector in which both local systems are orthogonal to fixed distinguished vectors. We prove this criterion in arbitrary finite dimensions using a triangular block-matrix entropy inequality and decompositions preserving two expectations. For amplitude damping, we also derive an exact finite-block Holevo deficit, identify the unique optimal average input for $p<1$, and construct a binary Kraus representation attaining the uniform formation bound.
Comments21+8 pages, 1 figure
