Let $|\Phi\rangle$ be a normalized vector in $\mathbb{C}^d$ and let $|\psi\rangle$ be a random stabilizer state. I am trying to compute the quantity

$$\mathsf{Pr}\big[|\langle \Phi|\psi \rangle|^2 \geq \epsilon \big].$$

Note that if $|\psi\rangle$ is Haar random, then, by equation $2$ of this paper,

$$\mathsf{Pr}\big[|\langle \Phi|\psi \rangle|^2 \geq \epsilon \big] \leq \mathsf{exp}(-(2d-1) \epsilon).$$

Does a similar concentration bound hold for random stabilizer states too?

  • $\begingroup$ Well, the linked paper is very .. sketchy. Without having looked into the details, I expect that you need bounds on the generating function (i.e. all higher moments) to prove this using a Chernoff bound or similar. That implies that it doesn't work for random stabilizer states as the higher moments of the Clifford group are large (c.f. arxiv.org/abs/2212.06240). There, the best you can do is probably a Markov inequality with the third moment. I can try to work this out later this week ... $\endgroup$ Feb 28, 2023 at 7:41

1 Answer 1


No such bound holds for general $|\Phi\rangle$. The set $\mathcal{S}_n$ of $n$-qubit stabilizer states is finite, so $m=\min_{|\psi\rangle\in\mathcal{S}_n} |\langle\Phi|\psi\rangle|^2$ is well-defined and if $|\Phi\rangle$ is not a stabilizer state then $m>0$. But then for any $\epsilon\in[0,m]$ we have $\mathrm{Pr}\left[|\langle\Phi|\psi\rangle|^2\ge\epsilon\right]=1$ which rules out any general bound $\mathrm{Pr}\left[|\langle\Phi|\psi\rangle|^2\ge\epsilon\right]\le f(\epsilon)$ with $f(\epsilon)<1$ for $\epsilon>0$. In particular, no general bound of the form $\mathrm{Pr}\left[|\langle\Phi|\psi\rangle|^2\ge\epsilon\right]\le\exp(-a\epsilon)$ with $a>0$ is possible.


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