have for every ${\varepsilon}>0$, every $\alpha \in [0,1]$ and every sequence $k_N$ such that $k_N/N \to \alpha$

$$\lim_{N\to \infty}\frac{{T_{\rm mix}}^{N,k_N}({\varepsilon})}{N}= \frac{(\sqrt{\alpha}+\sqrt{1-\alpha})^2}{p-q}\;.$$

The result implies in particular that at first order, the mixing time does not depend on ${\varepsilon}\in(0,1)$, meaning that on the appropriate time-scale, for large values of $N$ the distance to equilibrium drops abruptly from $1$ to $0$. This phenomenon is referred to as *cutoff* and was first observed in the context of card shuffling [@AldDia; @DiaSha]. It is known to occur for a large variety of Markov chains, see for instance [@LevPerWil]. In the context of the exclusion process, it has been proved in [@Lac16] that cutoff holds for the Symmetric Simple Exclusion Process (SSEP) which is obtained by setting $p=1/2$ in the generator .

When $p=1/2$, for any sequence $k_N$ that tends to infinity and satisfies $k_N\le N/2$ for all $N$, we have

$$\lim_{N\to \infty}\frac{{T_{\rm mix}}^{N,k_N}({\varepsilon})}{N^2\log k_N}= \frac{1}{\pi^2}.$$

While cutoff occurs in the two cases, it appears to be triggered by different mechanisms. When $p>1/2$, the mixing time is determined by the time needed for the particle density profile to reach its macroscopic equilibrium: After rescaling time and space by $N$, the evolution of the particle density 