\lambda(n-1,n) = (1-\rho)(1+\varepsilon_n)\eta(n-1), \label{eq:rightboundary}\\
    \lambda(1,0)& = (1-\rho)\eta(1), \qquad   \qquad \quad\;\;\lambda(n,n-1) = \rho\big(1-\eta(n-1)\big).\label{eq:leftboundary}\end{aligned}$$

In and , the factor $1+\varepsilon_n$ breaks the symmetry of the jumps: there is a non-trivial drift towards the right. But since $\varepsilon_n \to 0$, we are in the weakly asymmetric setting. Moreover, note that the product $\eta(x)(1-\eta(x+1))$ in (for instance) corresponds to the *exclusion rule* explained above: for the jump $x\curvearrowleft x+1$ to have a non-zero intensity, there has to be a particle at site $x$, and no particle at site $x+1$.

An invariant measure for these dynamics is the Bernoulli product measure on $\{0,1\}^{\Lambda_n}$ of parameter $\rho$, which we denote by $\nu_\rho^n$, and whose marginals satisfy: $$\nu_\rho^n\big\{\eta(x)=1\big\}=1-\nu_\rho^n\big\{\eta(x)=0\big\}=\rho.$$ We start the Markov process $\eta_t$ under the invariant measure $\nu_\rho^n$ and we look at the evolution on the diffusive time scale $tn^2$, where $t{\geqslant}0$ is the macroscopic time. The *microscopic density fluctuation field* is then defined as $$\label{eq:fieldintro}
{{\mathcal Y}}_t^n(\cdot) = \frac{1}{\sqrt n}\sum_{x=1}^{n-1} \big(\eta_{tn^2}(x) - \rho \big) \delta_{x /n}(\cdot),$$ where $\delta_{x/n}$ is the Delta dirac distribution, and therefore $(\cdot)$ is meant to be a test function. We prove in Theorem \[theo:convergence\] two main results on the large $n$ limit of that field, assuming that the initial density is 