硬币、量子测量和图灵势垒
Coins, Quantum Measurements, and Turing's Barrier
摘要：
量子计算是否有希望挑战图灵障碍，即解决一个不可判定的问题，计算一个不可计算的函数？根据Feynman '82的论证，答案是{\it negative}。本文重新打开的情况下：我们将讨论解决几个简单的问题，这表明{\it量子计算是{\it理论上}能够计算不可计算的功能}。本文构造了一个数学量子"器件"（具有灵敏度$ε$）来解决停机问题。"设备"在$T$时间单位内对随机选择的测试向量进行工作。如果"设备"产生一个点击，然后程序停止。如果它不产生点击，则程序不停止，或者测试矢量已从{\it不可区分矢量集} ${\IF}_{ε, T}$中选择。最后一种情况是不危险的，因为我们的主要结果证明：{\it} ${\IF}_{ε, T}$ {\it的维纳测度在} $T$ {\it趋于无穷大}时建设性地趋于零。这个"装置"，在适当计算的时间$T$中工作，将以预先建立的精度确定任意程序是否停止。{\it构建"停机机器"在数学上是可能的。}
英文摘要：
Is there any hope for quantum computing to challenge the Turing barrier, i.e. to solve an undecidable problem, to compute an uncomputable function? According to Feynman's '82 argument, the answer is {\it negative}. This paper re-opens the case: we will discuss solutions to a few simple problems which suggest that {\it quantum computing is {\it theoretically} capable of computing uncomputable functions}. In this paper a mathematical quantum "device" (with sensitivity $ε$) is constructed to solve the Halting Problem. The "device" works on a randomly chosen test-vector for $T$ units of time. If the "device" produces a click, then the program halts. If it does not produce a click, then either the program does not halt or the test-vector has been chosen from an {\it undistinguishable set of vectors} ${\IF}_{ε, T}$. The last case is not dangerous as our main result proves: {\it the Wiener measure of} ${\IF}_{ε, T}$ {\it constructively tends to zero when} $T$ {\it tends to infinity}. The "device", working in time $T$, appropriately computed, will determine with a pre-established precision whether an arbitrary program halts or not. {\it Building the "halting machine" is mathematically possible.}
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
