A major open problem in understanding shallow quantum circuits (QAC$^0$) is whether they can compute Parity. We show that this question is solely about the Fourier spectrum of QAC$^0$: any QAC$^0$ circuit with non-negligible high-level Fourier mass suffices to exactly compute PARITY in QAC$^0$. Thus, proving a quantum analog of the seminal LMN theorem for AC$^0$ is necessary to bound the quantum circuit complexity of PARITY. In the other direction, LMN does not fully capture the limitations of AC$^0$. For example, despite MAJORITY having $99\%$ of its weight on low-degree Fourier coefficients, no AC$^0$ circuit can non-trivially correlate with it. In contrast, we provide a QAC$^0$ circuit that achieves $(1-o(1))$ correlation with MAJORITY, establishing the first average-case decision separation between AC$^0$ and QAC$^0$. This suggests a uniquely quantum phenomenon: unlike in the classical setting, Fourier concentration may largely characterize the power of QAC$^0$. PARITY is also known to be equivalent in QAC$^0$ to inherently quantum tasks such as preparing GHZ states to high fidelity. We extend this equivalence to a broad class of state-synthesis tasks. We demonstrate that existing metrics such as trace distance, fidelity, and mutual information are insufficient to capture these states and introduce a new measure, felinity. We prove that preparing any state with non-negligible felinity, or derived states such as poly(n)-weight Dicke states, implies PARITY $\in$ QAC$^0$.

翻译：理解浅层量子电路（QAC$^0$）的一个主要开放问题是它们能否计算奇偶性。我们表明，这个问题仅与QAC$^0$的傅里叶谱有关：任何具有不可忽略的高阶傅里叶质量的QAC$^0$电路都足以在QAC$^0$中精确计算PARITY。因此，证明AC$^0$经典LMN定理的量子类比是限制PARITY量子电路复杂度的必要条件。另一方面，LMN并不完全捕捉AC$^0$的局限性。例如，尽管MAJORITY有$99\%$的权重集中在低阶傅里叶系数上，但没有任何AC$^0$电路能与其产生非平凡关联。相比之下，我们提供了一个与MAJORITY达到$(1-o(1))$关联的QAC$^0$电路，首次建立了AC$^0$与QAC$^0$之间的平均情形决策分离。这表明一个独特的量子现象：与经典情形不同，傅里叶集中可能在很大程度上刻画QAC$^0$的能力。已知PARITY在QAC$^0$中等价于制备高保真度GHZ态等固有量子任务。我们将这种等价性推广到一大类态合成任务。我们证明，现有度量如迹距离、保真度和互信息不足以刻画这些态，并引入一个新的度量——"猫性"（felinity）。我们证明，制备任何具有不可忽略猫性态的态，或导出态（如poly(n)权重的Dicke态），都意味着PARITY $\in$ QAC$^0$。