Concentration and anti-concentration of gap between largest and second largest value in Gaussian iid sample Let $n \ge 3$ be an integer and let $X=(X_1,\ldots,X_n)$ be random vector with iid coordinates from $N(0,1)$. For $1 \le k \le n$, let $X_{(k)}$ be the value of the $k$th largest coordinate of $X$.

Question. What are good  (anti-)concentration inequalities for $X_{(1)} - X_{(2)}$ ?

References welcome!
A crude concentration inequality
Note that $X_{(1)}  - X_{(2)} \le \Delta := \max_{i,j} X_i - X_j = X_{(1)} - X_{(n)}$. Moreover, $E[\Delta] \le 2\max_i X_i \le 2\sqrt{2\log n},$ and so using the result from this post, we have
$$
\begin{split}
P(X_{(1)} - X_{(2)} &\ge 2\sqrt{2 \log n} + t) \le P(\Delta \ge E[\Delta] + t)\\
&\le 2 P(|\max_i X_i - \mathbb E[\max_i X_i]| \ge t/2) \le 2e^{-t^2/8}.
\end{split}
$$
I wonder if my above somewhat naive bounds can be improved.
Edit: $P(X_{(1)} - X_{(2)} > t)$ when $n \ge 3$ and $t \ge 4 \sqrt{2 \log n}$
Inspired by the posted answers and the above Borell-TIS inequality, one may compute
$$
\begin{split}
P(X_{(1)} - X_{(2)} > t) &\le P(X_{(1)} > E X_{(1)} + t/2) + P(X_{(2)} < E X_{(1)} - t/2)\\
&\le e^{-t^2/8} + P(X_{(2)} < \sqrt{2 \log n} - t/2)\\
&= e^{-t^2/8} + P(X_{(2)} \le -t/4), \text{ if }t \ge 4\sqrt{2 \log n}\\
&= e^{-t^2/8} + (\Phi(-t/2))^n + n(1 - \Phi(-t/4))\cdot (\Phi(-t/4))^{n-1}\\
&=e^{-t^2/8} + (\Phi^c(t/4))^n + n\Phi(-t/4)\cdot (\Phi^c(t/4))^{n-1} \\
&\le e^{-t^2/8} + (\Phi(t/4)+n\Phi(-t/4))\cdot e^{-(n-1)t^2/16}\\
&\le e^{-t^2/8}+(1+n/2)e^{-(n-1)t^2/16} \le 2e^{-t^2/8}, \text{ if }n \ge 3.
\end{split}
$$
We deduce that

If $n \ge 3$ and $ t \ge 4\sqrt{2 \log n}$, then we have the concentration inequality
$$
P(X_{(1)} - X_{(2)} \ge t) \le 2e^{-Ct^2}.
$$

 A: Let us show that, after proper rescaling, $X_{(1)}-X_{(2)}$ has an asymptotically exponential distribution.
Let $Y_n:=X_{(1)}$ and $Y_{n-1}:=X_{(2)}$. By the known formula for the joint pdf of two order statistics, the joint pdf of $Y_{n-1}$ and $Y_n$ is given by
\begin{equation}
    f_{n-1,n}(y_{n-1},y_n)=n(n-1)F(y_{n-1})^{n-2}f(y_{n-1})f(y_n)\,1(y_{n-1}<y_n),
\end{equation}
where $F$ and $f$ denote, respectively, the cdf and pdf of $N(0,1)$.
Hence, for any fixed real $c>0$,
\begin{equation}
    x:=\frac c{\sqrt{2\ln n}},
\end{equation}
\begin{equation}
    l:=\ln n,
\end{equation}
and all large enough $n$
we have
\begin{equation}
    P(X_{(1)}-X_{(2)}>x)=P(V>x)=n(n-1)J,
\end{equation}
where
\begin{align*}
    J&:=\int_{-\infty}^\infty dw\,F(w)^{n-2}f(w)\int_x^\infty dv\, f(v+w) \\
&   =\int_{-\infty}^\infty dw\,F(w)^{n-2}f(w)G(x+w) \\ 
&   =\int_0^1 du\,h(u)=J_2+O(J_1+J_3),
\end{align*}
\begin{equation}
    h(u):=u^{n-2}G(x+Q(u)),\quad G:=1-F,\quad Q:=F^{-1}, 
\end{equation}
\begin{equation}
    J_1:=\int_0^{1-l^2/n}du\,u^{n-2}\le(1-l^2/n)^{n-2}\le\exp\Big\{-\frac{n-2}n\,l^2\Big\}
=o(1/n^2) 
\end{equation}
(as $n\to\infty$),
\begin{equation}
    J_3:=\int_{1-1/(nl)}^1du\,u^{n-2}G(Q(u))=\int_{1-1/(nl)}^1du\,u^{n-2}(1-u)
    \le\frac1{nl}\,\int_0^1 du\,u^{n-2}=o(1/n^2),
\end{equation}
and
\begin{equation}
    J_2:=\int_{1-l^2/n}^{1-1/(nl)} du\,u^{n-2}G(x+Q(u)). 
\end{equation}
For $u\in[1-l^2/n,1-1/(nl)]$ and $w:=Q(u)$, we have $u\uparrow1$ and hence $w\to\infty$; therefore and because $x\downarrow0$, again for $u\in[1-l^2/n,1-1/(nl)]$,
\begin{equation}
    G(x+Q(u))=G(x+w)\sim\frac{f(x+w)}{x+w}\sim\frac{f(w)}{w}\,e^{-xw}
    \sim G(w)\,e^{-xw}=(1-u)e^{-xw}, 
\end{equation}
and also
\begin{equation}
    |\ln(1-u)|\sim\ln n  
\end{equation}
and
\begin{equation}
    w=Q(u)\iff u=F(w)\iff 1-u=G(w)\implies w\sim\sqrt{2|\ln(1-u)|},
\end{equation}
since $G(w)=\exp\{-w^2/(2+o(1))\}$ as $w\to\infty$.
It follows that
\begin{align*}
    J_2&\sim\int_{1-l^2/n}^{1-1/(nl)} du\,u^{n-2}(1-u)e^{-xQ(u)} \\ 
    &=\int_{1-l^2/n}^{1-1/(nl)} du\,u^{n-2}(1-u)\exp\Big\{-\frac c{\sqrt{2\ln n}}\,\sqrt{2\ln n}\Big\} \\  
    &=\int_{1-l^2/n}^{1-1/(nl)} du\,u^{n-2}(1-u) e^{-c} \\  
    &=\int_0^1 du\,u^{n-2}(1-u) e^{-c}+O(J_1+J_3) \\  
    &=\frac{e^{-c}}{n(n-1)}+O(J_1+J_3)\sim\frac{e^{-c}}{n(n-1)}.  
\end{align*}
Collecting pieces, we conclude that for any fixed real $c>0$
\begin{equation}
    P\Big(X_{(1)}-X_{(2)}>\frac c{\sqrt{2\ln n}}\Big)\to e^{-c},
\end{equation}
so that indeed, after proper rescaling, $X_{(1)}-X_{(2)}$ has an asymptotically exponential distribution.
A: Let $\Phi(r)=P(X_1>r)$. Then for $s<t$ we have
$$P(X_{(1)}<t, \; X_{(2)}<s)=\Phi(s)^n+n(\Phi(t)-\Phi(s))\cdot \Phi(s)^{n-1}$$
and from this one can obtain the exact distribution of $X_{(1)}- X_{(2)}$.
For a simple but still suboptimal upper bound:
$$\Psi(2t):=P(X_{(1)}- X_{(2)}>2t )\le P(X_{(1)}>\sqrt{2 \log n}+t)+P(X_{(2)}<\sqrt{2 \log n}-t)$$
so
$$ \Psi(2t) \le 1-[\Phi(\sqrt{2 \log n}+t)]^n + [\Phi(\sqrt{2 \log n}-t)]^{n}+n [ 1-\Phi(\sqrt{2 \log n}-t)] [\Phi(\sqrt{2 \log n}-t)]^{n-1} .$$
Now one can plug in the tail bounds for $\Phi$, see e.g. https://www.johndcook.com/blog/norm-dist-bounds/
