Let $G$ be a symmetric Gaussian random matrix with iid $E[G\_{ij}]=0$ and $E[G\_{ij}^2]=\frac{1}{n}$, and ordering its eigenvalues $\lambda\_1\le \lambda\_2\le \dots \le \lambda\_n$ corresponding eigenvectors $v\_1,\dots, v\_n$. Define $X\_t=\{X^i\_t\}$ is a vector on $R^n$ for time $t\ge 0$ and $X\_0$ is distributed uniformly on the unit sphere. Let $h(t)= X\_t\cdot v\_1$.


Let $T\_\epsilon:=inf\_{t>0}\{h(t)\ge \epsilon\}$. Assume that
$$
h(t)\ge h(0)e^{2(\lambda\_2-\lambda\_1)t}\ge h(0)e^{2\delta t}
$$
where $\delta=\min\{\lambda\_i-\lambda\_j\}$ is the smallest gap.


I try to find the upper bound of $T\_\epsilon$ with high probability $1-Ce^{-c n}$ (or something like that).



How to prove that upper bound of the hitting time holds with high probability?
I don't think this upper bound holds.


Notice that the probability distribution of $h(0)$ is a Gaussian of mean 0 and width $1/\sqrt n$, so the probability that $|h(0)|<\alpha$ is $\alpha\sqrt n$ for $0<\alpha<1/\sqrt n$.


Since $\lambda\_2-\lambda\_1\simeq n^{-2/3}$, the hitting time $T\_\epsilon$ for $h(0)=\alpha$ is
$$T\_\epsilon\lesssim n^{2/3}\log(\epsilon/\alpha).$$
Choose $\alpha<e^{-\sqrt n}$, then $T\_\epsilon$ can be larger than $n^{1/2}n^{2/3}>n^{2/3}\log n$ with probability $\sqrt{n}e^{-\sqrt n}$, so a larger probability than the $e^{-n}$ conjectured in the OP.


