$1/N\ll b_N\ll (\log N)/N$) this time does not correspond to the time needed to reach macroscopic equilibrium.

We could not prove such a result in the crossover regime $b_N\approx N^{-1}\log N$: In this case the time to reach macroscopic equilibrium and that to equilibrate local fluctuations are of the same order and the two phenomena are difficult to separate. In Section \[conjectos\] we provide a conjecture for the mixing time in this regime in the case of vanishing density. However the techniques developed here are not sufficient to obtain sharp results in this case.

Model and results
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Mixing time for the WASEP
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Given $N\in \mathbb N$, $k\in {\llbracket}1,N-1 {\rrbracket}$ (we use the notation ${\llbracket}a,b{\rrbracket}=[a,b]\cap {\mathbb{Z}}$) and $p\in(1/2,1]$, the Asymmetric Simple Exclusion Process on ${\llbracket}1, N  {\rrbracket}$ with $k$ particles and parameter $p$ is the random process on the state space $${\Omega}_{N,k}^0:=\Big\{ \xi\in\{0,1\}^{N} \ : \ \sum_{x=1}^N \xi(x)=k\Big\},$$ associated with the generator $$\label{defgen}
 {\mathcal{L}}_{N,k} f(\xi):= \sum_{y=1}^{N-1} \left( q{\mathbf{1}}_{\{\xi(y)< \xi(y+1)\}}+p{\mathbf{1}}_{\{\xi(y)> \xi(y+1)\}}\right)(f(\xi^y)-f(\xi)),$$ where $q=1-p$ and $$\label{flipz}
 \xi^y(x):=\begin{cases} \xi(y+1) \quad &\text{ if } x=y,\\
                         \xi(y)  \quad &\text{ if } x=y+1,\\
                         \xi(x) \quad& \text{ if } x\notin \{y,y+1\}.
           \end{cases}$$ In a more intuitive manner we can materialize the positions of $1$ by particles, 