Conditional lower bound on approximate stabilizer rank of magic states

Question

I am currently reading about the approximate stabilizer rank and properties of the same.

I will quote the definitions from this paper.

The stabilizer rank of a quantum state $|\psi\rangle$ is the minimal $r$ such that \begin{equation} |{\psi}\rangle = \sum_{j=1}^{r} c_j |φ_{j}\rangle. \end{equation} for $c_j \in \mathbb{C}$ and stabilizer states $|φ_j\rangle$.

The $\delta$-approximate stabilizer rank of $|ψ\rangle$, denoted $\chi_{\delta}(ψ)$, is defined as the minimum of $\chi(\psi)$ over all states $|φ\rangle$ such that \begin{equation} |||ψ\rangle − |φ\rangle||_{2} \leq \delta. \end{equation}

The paper I quoted gives a loose unconditional lower bound on approximate stabilizer rank. But, is there a known stronger conditional lower bound for the approximate stabiliser rank of an $n$-fold tensor product of any single qubit magic state?

There is such a conditional lower bound for the exact stabilizer rank, given by

Much stronger hardness assumptions than $\text{P}=\text{NP}$, such as the exponential time hypothesis, imply that $\chi(H^{\otimes n}) = 2^{Ω(n)}.$ [MT19, HNS20]

$H$ is a single qubit magic state, as defined in the paper. But I could not find anything for approximate stabilizer rank.

Answer 1

In a recent work Lovitz and Steffan (theorem 3.5) showed that for any non-stabilizer $n$-qubit state, there is a constant $\delta>0$, such that for every $n\geq 2$, $$\chi_\delta(\psi^{\otimes n})\geq \frac{\sqrt{n}}{2\log_2 n}.$$