克利福德层级中的深孔
Deep Holes in the Clifford Hierarchy
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中文总结 AI 辅助
该研究确定单量子比特克利福德层级拓扑闭包的覆盖半径为arccos√(5/6)，发现192个深孔轨道，将覆盖问题转化为SO(3)上ℓ^∞范数的极小极大问题并精确求解，等价于单量子比特全层级克利福德保真度最小值为5/6。
中文摘要 AI 辅助
我们确定了单量子比特克利福德层级在SU(2)≅S³中的拓扑闭包的覆盖半径。该闭包是18个大圆（即克利福德-泡利圆）的并集，我们证明其覆盖半径为arccos√(5/6)。这些极值点（我们称之为“深孔”）在克利福德门的左乘和右乘下形成一个大小为192的单一轨道，且以闭式形式描述。等价地，单量子比特幺正操作的全层级克利福德保真度的最小值为5/6。证明基于与R⁴中18个平面对应的两种结构：它们的中心化秩2投影算子构成不可约SO(4)模Sym₀(4)的正交基，而单位四元数的投影轮廓恰好是其在二重覆盖SU(2)→SO(3)下的像。这些将覆盖问题简化为SO(3)上ℓ^∞范数的极小极大问题，我们对其进行了精确求解并分类了等号情形。
英文摘要
We determine the covering radius of the topological closure of the single-qubit Clifford hierarchy in $\SU(2)\cong S^3$. This closure is a union of $18$ great circles --- the Clifford--Pauli circles --- and we prove that its covering radius is $\arccos\sqrt{5/6}$. The extremal points, which we call \emph{deep holes}, form a single orbit of size $192$ under left and right multiplication by Clifford gates, and are described in closed form. Equivalently, the minimum over one-qubit unitaries of the all-level Clifford fidelity is $5/6$. The proof rests on two structures attached to the configuration of $18$ planes in $\R^4$: their centered rank-two projectors form an orthonormal basis of the irreducible $\SO(4)$-module $\Sym_0(4)$, and the projection profile of a unit quaternion is exactly its image under the double cover $\SU(2)\to\SO(3)$. These reduce the covering problem to a minimax statement for the $\ell^\infty$-norm on $\SO(3)$ which we solve exactly, classifying its equality cases.
