发表机构
University of Texas at Austin; Hon Hai (Foxconn) Research Institute; University of Waterloo（德克萨斯大学奥斯汀分校; 鸿海（富士康）研究院; 滑铁卢大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本研究证明存在全布尔函数在单向通信中，有纠缠时仅需对数级经典通信，无纠缠时需多项式级量子通信，实现指数级分离。
AI 中文摘要
量子通信复杂性中的一个长期悬而未决的问题是：某些任务在存在纠缠的情况下是否可以用少量通信完成，而在没有纠缠的情况下却需要多得多的量子通信。此前，这种分离已在关系问题以及同时消息传递模型中的部分函数上得到证明。但对于全布尔函数是否存在这样的分离，此前仍未得到解决。我们通过单向设置中的指数级分离解决了这个问题：我们构造了一族全布尔函数 $f_n\colon \{0,1\}^n \times \{0,1\}^n \to \{0,1\}$，在预先共享纠缠的情况下，该函数可以用 $O(\log n)$ 比特的单向经典通信计算，但在没有纠缠的情况下，需要 $\Omega(n^{1/3})$ 量子比特的单向量子通信。我们的函数是子群成员问题的一个特例，该问题最早由 Aaronson、Le Gall、Russell 和 Tani 在通信设置中研究。
英文摘要
A longstanding question in quantum communication complexity is whether some task can be accomplished with a small amount of communication in the presence of entanglement, yet require much more quantum communication in the absence of entanglement. Separations of this nature were previously known for relational problems and, in the simultaneous message passing model, for partial functions. But it has remained unresolved whether any such separation exists for a total Boolean function. We resolve this question with an exponential separation in the one-way setting: we exhibit a family of total Boolean functions $f_n\colon \{0,1\}^n \times \{0,1\}^n \to \{0,1\}$ that can be computed with $O(\log n)$ bits of one-way classical communication given prior entanglement, but that require $Ω(n^{1/3})$ qubits of one-way quantum communication without entanglement. Our function is a special case of the subgroup membership problem, first studied in the communication setting by Aaronson, Le Gall, Russell, and Tani.
