发表机构
National Tsing Hua University; National Center for Excellence in Quantum Information Science and Engineering, National Tsing Hua University（国立清华大学; 国立清华大学量子信息科学与工程卓越中心）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
通过乘法对手方法，我们证明了量子近似计数两权重判定版本的下界，并给出基于单查询进展的推导，同时将唯一OR归约到该问题。
AI 中文摘要
我们研究量子近似计数的两权重判定版本：给定对$x\in\{0,1\}^N$的预言机访问，以成功概率$1/2+\zeta$区分$|x|=M$与$|x|=M+\Delta$。利用乘法对手方法，我们证明$\Omega\left(\max\left\{\zeta\sqrt{(N-M)(M+\Delta)}/\Delta,\sqrt{\zeta N/\Delta}\right\}\right)$。相同的参数依赖也由Podder、Yao和Ye对两层对称函数的多项式方法刻画得出。我们的贡献在于一种乘法对手推导，它追踪了单个预言机查询所产生的进展。对于第一项，在必要时对输入取补后，我们假设$M+\Delta\le N-M$。我们使用Ambainis、Spalek和de Wolf的本征空间方法中的Hamming层子空间，并组合其相邻层酉映射以关联两个不相邻的承诺层。在固定查询坐标后，分析块对角化为四维子空间。对单查询进展比的精确计算给出了第一个下界。同样的估计也蕴含$\left\\|(I-\widehat{\Pi}_{\mathrm{bad}})\lvert\Psi^T\rangle\right\\|^2=O\left(T^2\Delta^2/((N-M)(M+\Delta))\right)$，其中$\lvert\Psi^T\rangle$是对手论证中使用的相干输入叠加态。对于第二项，我们直接使用三本征值乘法对手证明：在$n$比特上以成功概率$1/2+\zeta$求解唯一OR需要$\Omega(\sqrt{\zeta n})$次查询，然后将唯一OR归约到两权重计数问题。
英文摘要
We study the two-weight decision version of quantum approximate counting: given oracle access to $x\in\{0,1\}^N$, distinguish $|x|=M$ from $|x|=M+Δ$ with success probability $1/2+ζ$. Using the multiplicative adversary method, we prove $Ω\left(\max\left\{ζ\sqrt{(N-M)(M+Δ)}/Δ,\sqrt{ζN/Δ}\right\}\right)$. The same parameter dependence follows from the polynomial-method characterization of the two-layer symmetric function by Podder, Yao, and Ye. Our contribution is a multiplicative-adversary derivation that tracks the progress produced by individual oracle queries. For the first term, after complementing the input if necessary, we assume $M+Δ\le N-M$. We use the Hamming-layer subspaces from the eigenspace method of Ambainis, Spalek, and de Wolf and compose their adjacent-layer unitary maps to relate the two nonadjacent promise layers. After fixing the queried coordinate, the analysis block-diagonalizes into four-dimensional subspaces. An exact calculation of the one-query progress ratio gives the first lower bound. The same estimate also implies $\left\|(I-\widehatΠ_{\mathrm{bad}})\lvertΨ^T\rangle\right\|^2=O\left(T^2Δ^2/((N-M)(M+Δ))\right)$ for the coherent input superposition used in the adversary argument. For the second term, we prove directly using a three-eigenvalue multiplicative adversary that unique OR on $n$ bits with success probability $1/2+ζ$ requires $Ω(\sqrt{ζn})$ queries, and then reduce unique OR to the two-weight counting problem.
Comments18 pages
