Let $A=\{a_{ij}\}_{1\le i,j\le n}$ be an $n$ by $n$ normalized Gaussian random matrix with $E[a_{ij}]=0$ and $E[a_{ij}^2]=1/n$. Ordering its eigenvalues by $\lambda_1\le \lambda_2\le \cdots \lambda_n$ with corresponding eigenvectors $v_1,\dots, v_n \in \mathbb{R}^n$. Let $u_0$ be a vector on $\mathbb{R}^n$ uniformly distributed on the unit sphere. (We also know that $v_i$ is uniformly distributed on the unit sphere for $i=1,\dots, n$.)
Define $H_j(t)=u_t\cdot v_j$ for $j=1,\dots, n$ and time $t\ge 0$, solving the following ODE with initial value $H_j(0)$: $$ \frac{1}{2}H_j'(t)=\sum_{i=1}^n[(\lambda_i-\lambda_j)H_i^2(t)]H_j(t) $$
Assume that $$\sum_{i=1}^n H_i(t)=1, \mbox{ for } t\ge 0.$$
Note that $\sqrt{n}H_j(0)\to^d N(0,1)$, without loss of generality, assume that $H_j(0)>0$ for $j=1,\dots, n$.
Question: Fix $\epsilon>0$, define the hitting time $\tau_\epsilon=\inf_{t>0}\{H_1(t)\ge \epsilon\}$. I try to prove that $\tau_\epsilon\ge n^{2/3}$ with probability 1 as $n\to \infty$
Solving the above ODE $j=1,2,\dots, n$, we get $$
H_j(t):=\frac{1}{L}H_j(0)e^{-2\lambda_j t},
$$
where $L=\sqrt{\sum_{i=1}^n (H_i(0))^2 e^{-4\lambda_i t}}$.
Approach 1: I think the approach in the answer is still not rigorous enough. I still want to try to get the answer by the method of upper bound. I try to consider the difference between $H_1(t)-H_j(t)$, is it possible to get the result in this way?
\begin{align} H_1(t)-H_j(t)&=\frac{H_1(0)e^{-2\lambda_1 t}-H_j(0)e^{-2\lambda_j t}}{\sqrt{\sum_{i=1}^n H_i(0)^2 e^{-4\lambda_i t}}}\\ &=\frac{(H_1(0)-H_j(0)+H_j(0))e^{-2\lambda_1 t}-H_j(0)e^{-2\lambda_j t}}{\sqrt{\sum_{i=1}^n H_i(0)^2 e^{-4\lambda_i t}}}\\ &=\frac{(H_1(0)-H_j(0))e^{-2\lambda_1 t}+H_j(0)(e^{-2\lambda_1 t}-e^{-2\lambda_j t})}{\sqrt{\sum_{i=1}^n H_i(0)^2 e^{-4\lambda_i t}}}\\ &=\frac{(H_1(0)-H_j(0))e^{-2\lambda_1 t}+e^{-2\lambda_jt}H_j(0)(e^{2t(\lambda_j-\lambda_1)}-1)}{\sqrt{\sum_{i=1}^n H_i(0)^2 e^{-4\lambda_i t}}} \end{align}
Note that $\sqrt{n}(H_1(0)-H_j(0))$ are i.i.d. $N(0,1)$. Without loss of generality, assume that $H_1(0)>0$ and $H_1(0)-H_j(0)<0$, then $H_1(0)<H_j(0)$ for $j=2,\dots, N$.
Approach 2:
Since $$\sqrt{\sum_{i=1}^n H_i(0)^2 e^{-4\lambda_i t}}\ge \sqrt{H_1(0)^2 e^{-4\lambda_1 t}+H_2(0)^2 e^{-4\lambda_2 t}}$$, then we get as $t$ is large enough, $$ H_1(t)\le \left(1+(H_2(0)/H_1(0))^2 e^{-4t(\lambda_2-\lambda_1)}\right)^{-1/2}=1-\frac{1}{2}(H_2(0)/H_1(0))^2 e^{-4t(\lambda_2-\lambda_1)}+O(\delta^2), $$ where $\delta=e^{-4t(\lambda_2-\lambda_1)}$.
For $t\ge \tau_\epsilon$, we have $H_1(t)\ge \epsilon$. I am stuck on how to solve $$H_1(t)\ge \epsilon$$
Actually, I can show that $\tau_{\epsilon}^{(n-1)}:=\inf\{t\ge 0: H_{n-1}(t)\ge 0\}$ with order $O(n^{2/3})$.
Note that $$ H_{n-1}^2(t)=\frac{H_{n-1}^2(0)}{\sum_{i=1}^{n-2}H_i^2(0)e^{4t(\lambda_{n-1}-\lambda_i)}+H_{n-1}^2(0)+H_{n}^2(0)e^{4t(\lambda_{n-1}-\lambda_{n})}}. $$
By $e^{4t(\lambda_{n-1}-\lambda_{j})}\ge e^{4t(\lambda_{n-1}-\lambda_{n})}$ for $j=1,\dots, n-2$, we get $$ H_{n-1}^2(t)\le \frac{H_{n-1}^2(0)}{H_{n-1}^2(0)+(1-H_{n-1}^2(0))e^{4t(\lambda_{n-1}-\lambda_n)}}=\frac{1}{1+(H_{n-1}^{-2}(0)-1)e^{4t(\lambda_{n-1}-\lambda_n)}} $$ where using $\sum H_i^2(0)=1$.
For $t\ge \tau_{\epsilon}^{(n-1)}$, we have $H_{n-1}^2(t)\ge \epsilon^2$ $$ \tau_{\epsilon}^{(n-1)}\ge \frac{1}{4(\lambda_n-\lambda_{n-1})}\log\frac{H_{n-1}^{-2}(0)-1}{\epsilon^{-2}-1} $$ by the fact that $\lambda_n-\lambda_{n-1}=O(n^{-2/3})$.
Since $H_{n-1}(0)<\epsilon/2$ with high probability (with probability 1 as $n\to \infty$), then $$ \tau_{\epsilon}^{(n-1)}\ge n^{2/3} $$ with probability 1 as $n\to \infty$.
I do a simulation for $H_1(t)$ with same initial value $u_0$ with different $n$ from 200 to 2000: