# Fidelity concentration bound for random stabilizer states

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?

• 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 ... Feb 28, 2023 at 7:41

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.