We explore the power of the unbounded Fan-Out gate and the Global Tunable gates generated by Ising-type Hamiltonians in constructing constant-depth quantum circuits, with particular attention to quantum memory devices. We propose two types of constant-depth constructions for implementing Uniformly Controlled Gates. These gates include the Fan-In gates defined by $|x\rangle|b\rangle\mapsto |x\rangle|b\oplus f(x)\rangle$ for $x\in\{0,1\}^n$ and $b\in\{0,1\}$, where $f$ is a Boolean function. The first of our constructions is based on computing the one-hot encoding of the control register $|x\rangle$, while the second is based on Boolean analysis and exploits different representations of $f$ such as its Fourier expansion. Via these constructions, we obtain constant-depth circuits for the quantum counterparts of read-only and read-write memory devices – Quantum Random Access Memory (QRAM) and Quantum Random Access Gate (QRAG) – of memory size $n$. The implementation based on one-hot encoding requires either $O(n\log^{(d)}{n}\log^{(d+1)}{n})$ ancillae and $O(n\log^{(d)}{n})$ Fan-Out gates or $O(n\log^{(d)}{n})$ ancillae and $16d-10$ Global Tunable gates, where $d$ is any positive integer and $\log^{(d)}{n} = \log\cdots \log{n}$ is the $d$-times iterated logarithm. On the other hand, the implementation based on Boolean analysis requires $8d-6$ Global Tunable gates at the expense of $O(n^{1/(1-2^{-d})})$ ancillae.

中文翻译：

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使用多量子比特门的布尔函数和量子存储设备的恒定深度电路


我们探讨了 Ising 型哈密顿量生成的无界扇出门和全局可调门在构建恒定深度量子电路方面的强大功能，特别关注量子存储设备。我们提出了两种类型的恒定深度结构来实现 Uniformly Controlled Gates。这些门包括由 $|x\rangle|b\rangle\mapsto |x\rangle|b\oplus f（x）\rangle$ 定义的扇入门，用于 $x\in\{0,1\}^n$ 和 $b\in\{0,1\}$，其中 $f$ 是一个布尔函数。我们的第一个构造是基于计算控制寄存器 $|x\rangle$ 的独热编码，而第二个是基于布尔分析，并利用 $f$ 的不同表示形式，例如它的傅里叶展开。通过这些结构，我们获得了内存大小为 $n$ 的只读和读写存储设备——量子随机存取存储器 （QRAM） 和量子随机存取门 （QRAG）的量子对应物的恒定深度电路。基于独热编码的实现需要 $O（n\log^{（d）}{n}\log^{（d+1）}{n}）$ 辅助门和 $O（n\log^{（d）}{n}）$ 扇出门或 $O（n\log^{（d）}{n}）$ 辅助门和 $16d-10$ 全局可调门，其中 $d$ 是任何正整数，$\log^{（d）}{n} = \log\cdots \log{n}$ 是 $d$ 次迭代对数。另一方面，基于布尔分析的实现需要 $8d-6$ 个全局可调门，代价是 $O（n^{1/（1-2^{-d}）}）$ 个辅助门。