<p>Quantum <i>k</i>-SAT (the problem of determining whether a <i>k</i>-local Hamiltonian is frustration-free) is known to be QMA<InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="220_2025_5377_Article_IEq1.gif" Format="GIF" Height="10" Rendition="HTML" Resolution="72" Type="Linedraw" Width="8" /> </InlineMediaObject> <EquationSource Format="TEX">\(_1\)</EquationSource> <EquationSource Format="MATHML"><math> <mmultiscripts> <mrow /> <mn>1</mn> <mrow /> </mmultiscripts> </math></EquationSource> </InlineEquation>-complete for <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="220_2025_5377_Article_IEq2.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\(k\ge 3\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>≥</mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation>, and hence likely hard for quantum computers to solve. Building on a classical result of Alon and Shapira, we show that quantum <i>k</i>-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 <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="220_2025_5377_Article_IEq3.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="27" /> </InlineMediaObject> <EquationSource Format="TEX">\(\epsilon n^k\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>ϵ</mi> <msup> <mi>n</mi> <mi>k</mi> </msup> </mrow> </math></EquationSource> </InlineEquation> constraints must be removed before a product state solution exists, for some fixed <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="220_2025_5377_Article_IEq4.gif" Format="GIF" Height="13" Rendition="HTML" Resolution="72" Type="Linedraw" Width="39" /> </InlineMediaObject> <EquationSource Format="TEX">\(\epsilon &gt;0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>ϵ</mi> <mo>&gt;</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>. The proof has two steps: we first show that for a satisfiable instance of quantum <i>k</i>-SAT, most subproblems on a constant number of qubits are satisfiable by a product state. We then show that for an instance of quantum <i>k</i>-SAT which is far from satisfiable by a product state, most subproblems are unsatisfiable by a product state. Given the promise, quantum <i>k</i>-SAT may therefore be solved by checking satisfiability by a product state on randomly chosen subsystems of constant size.</p>

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Testing Quantum Satisfiability

  • Ashley Montanaro,
  • Changpeng Shao,
  • Dominic Verdon

摘要

Quantum k-SAT (the problem of determining whether a k-local Hamiltonian is frustration-free) is known to be QMA \(_1\) 1 -complete for \(k\ge 3\) k 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 \(\epsilon n^k\) ϵ n k constraints must be removed before a product state solution exists, for some fixed \(\epsilon >0\) ϵ > 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.