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Sep 29, 2022
We show that quantum states with "low stabilizer complexity" can be efficiently distinguished from Haar-random. Specifically, given an $n$-qubit pure state $|\psi\rangle$, we give an efficient algorithm that distinguishes whether $|\psi\rangle$ is (i) Haar-random or (ii) a state with stabilizer fidelity at least $\frac{1}{k}$ (i.e., has fidelity at least $\frac{1}{k}$ with some stabilizer state), promised that one of these is the case. With black-box access to $|\psi\rangle$, our algorithm uses $O\!\left( k^{12} \log(1/\delta)\right)$ copies of $|\psi\rangle$ and $O\!\left(n k^{12} \log(1/\delta)\right)$ time to succeed with probability at least $1-\delta$, and, with access to a state preparation unitary for $|\psi\rangle$ (and its inverse), $O\!\left( k^{3} \log(1/\delta)\right)$ queries and $O\!\left(n k^{3} \log(1/\delta)\right)$ time suffice. As a corollary, we prove that $\omega(\log(n))$ $T$-gates are necessary for any Clifford+$T$ circuit to prepare computationally pseudorandom quantum states, a first-of-its-kind lower bound.
