测试量子可满足性

IF 2.6 1区 物理与天体物理 Q1 PHYSICS, MATHEMATICAL
Ashley Montanaro, Changpeng Shao, Dominic Verdon
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引用次数: 0

摘要

量子k-SAT(确定k-局部哈密顿量是否无挫折的问题)已知是QMA \(_1\) -对于\(k\ge 3\)是完整的,因此量子计算机可能很难解决。基于Alon和Shapira的经典结果,我们证明了量子k-SAT可以在随机多项式时间内解决,因为“属性测试”承诺实例要么是可满足的(任何状态),要么是远不能满足的产品状态;所谓“产品状态远不能满足”,我们的意思是,对于某些固定的\(\epsilon >0\),在产品状态解决方案存在之前,必须移除\(\epsilon n^k\)约束。证明有两个步骤:我们首先证明了对于量子k-SAT的可满足实例,在常数量子位上的大多数子问题都可以被乘积态满足。然后,我们证明了对于量子k-SAT的一个实例,它远远不能被积态满足,大多数子问题是不能被积态满足的。鉴于这一承诺,量子k-SAT可以通过在随机选择的恒定大小的子系统上通过乘积状态检查可满足性来解决。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Testing Quantum Satisfiability

Quantum k-SAT (the problem of determining whether a k-local Hamiltonian is frustration-free) is known to be QMA\(_1\)-complete for \(k\ge 3\), and hence likely hard for quantum computers to solve. Building on a classical result of Alon and Shapira, we show that quantum k-SAT can be solved in randomised polynomial time given the ‘property testing’ promise that the instance is either satisfiable (by any state) or far from satisfiable by a product state; by ‘far from satisfiable by a product state’ we mean that \(\epsilon n^k\) constraints must be removed before a product state solution exists, for some fixed \(\epsilon >0\). The proof has two steps: we first show that for a satisfiable instance of quantum k-SAT, most subproblems on a constant number of qubits are satisfiable by a product state. We then show that for an instance of quantum k-SAT which is far from satisfiable by a product state, most subproblems are unsatisfiable by a product state. Given the promise, quantum k-SAT may therefore be solved by checking satisfiability by a product state on randomly chosen subsystems of constant size.

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来源期刊
Communications in Mathematical Physics
Communications in Mathematical Physics 物理-物理:数学物理
CiteScore
4.70
自引率
8.30%
发文量
226
审稿时长
3-6 weeks
期刊介绍: The mission of Communications in Mathematical Physics is to offer a high forum for works which are motivated by the vision and the challenges of modern physics and which at the same time meet the highest mathematical standards.
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