respect to $\mu_{\epsilon}$ and the microscopic energy density $${\frak e}_x:=\frac12\left({\frak p}_x^2+\sum_{x'\in{{\mathbb Z}}}{\alpha}_{x-x'}{\frak q}_x{\frak q}_{x'}\right).$$ The assumption ensures that the macroscopic energy density of the chain is finite.

Kinetic scaling of the wave-function
------------------------------------

To observe macroscopic effects of the inter-particle scattering consider time of the order $t/{\epsilon}$. It is also convenient to introduce the wave function that, adjusted to the macroscopic time, is given by $$\label{011307}
\psi^{({\epsilon})}(t):=\tilde{{\omega}} \star {\frak q}\left(\frac{t}{{\epsilon}}\right)+i{\frak p}\left(\frac{t}{{\epsilon}}\right),$$ where $({\frak p}(t),{\frak q}(t))$ satisfies and $\left(\tilde {\omega}_x\right)_{x\in{{\mathbb Z}}}$ are the Fourier coefficients of the dispersion relation, see . We shall consider the Fourier transform of the wave function, given by $$\label{011307a}
\hat\psi^{({\epsilon})}(t,k)={\omega}(k)\hat {\frak q}^{({\epsilon})}\left(t,k\right)+i\hat{\frak p}^{({\epsilon})}\left(t,k\right).$$ Here $\hat {\frak q}^{({\epsilon})}\left(t\right)$, $\hat{\frak
  p}^{({\epsilon})}\left(t\right)$ are given by the Fourier series for ${\frak q}^{({\epsilon})}_x(t):={\frak q}_x(t/{\epsilon})$ and ${\frak
  p}^{({\epsilon})}_x(t):={\frak p}_x(t/{\epsilon})$, $x\in{{\mathbb Z}}$, respectively. They satisfy $$\begin{aligned}
 \label{basic:sde:2a}
&&
 d\hat\psi^{({\epsilon})}(t,k)=\left\{-\frac{i}{{\epsilon}} {\omega}(k)\hat\psi^{({\epsilon})}(t,k)
 -2i{\gamma}_0R(k)\hat{\frak p}^{({\epsilon})}\left(t,k\right)-\frac{i{\gamma}_1}{{\epsilon}}\int_{{{\mathbb T}}}\hat{\frak p}^{({\epsilon})}\left(t,k'\right)dk'\right\}dt
 \nonumber\\
 &&
 -2\sqrt{{\gamma}_0}\int_{{{\mathbb T}}}r(k,k')\hat{\frak p}^{({\epsilon})}\left(t,k-k'\right)B(dt,dk')+i\sqrt{\frac{2{\gamma}_1T}{{\epsilon}}}dw(t),\\
 &&
\hat\psi^{({\epsilon})}(0)= \hat\psi,\nonumber
 \end{aligned}$$ where $$\begin{aligned}
&
\hat{\frak p}^{({\epsilon})}\left(t,k\right):=\frac{1}{2i}\left[\hat\psi^{({\epsilon})}(t,k)-\left(\hat\psi^{({\epsilon})}\right)^\star(t,-k)\right],\\
&
r(k,k'):=4 {\frak s}(k) {\frak s}(k-k') {\frak s}(2k-k')
 \quad k,k'\in {{\mathbb T}},\\
&
R(k):=\int_{{{\mathbb T}}}r^2(k,k')dk'={\frak s}^2(2k)+2 {\frak s}^2(k)=\frac{\hat\theta(k)}{4}.
 \end{aligned}$$ Here $B(t,dk)=\sum_{x\in{{\mathbb Z}}}w_x(t)e_x(k)dk$ is a cylindrical Wiener process on $L^2({{\mathbb T}})$, i.e. $$\label{cnoise}
{\mathbb E}[B(dt,dk)B^\star(ds,dk')]=\delta(k-k')\delta(t-s)dt ds dkdk'.$$

Energy density - Wigner function
--------------------------------

One can easily 