equation can be integrated formally as $$\begin{aligned}
\label{c30}
\underline{\vec{X}}_\omega(\vec{k},t)&=&\underline{\vec{Q}}_\omega(\vec{k},0)\frac{\sin
\omega t }{\omega}+ \underline{\vec{X}}_\omega(\vec{k},0)\cos
\omega t\nonumber\\
&+&\int _0^t d t' \frac{\sin\omega(t-t')}{\omega}\int d^3q\
\underline{f}^\dag( \omega , \vec{q} , \vec{k} )\cdot
\underline{\vec{E}}(\vec{q},t')\end{aligned}$$ Now, by substituting $\underline{\vec{X}}_\omega(\vec{k},t)$ from (\[c30\]) in the definition of electric polarization density given by (\[c13\]), we find the constitutive equation of the medium in reciprocal space relating the electric polarization to the electric field $$\label{c31}
\underline{\vec{P}}(\vec{k},t)=\underline{\vec{P}}_N(\vec{k},t)+\varepsilon_0\int_0^{|t|}
dt' \int d^3q\ \underline{\chi}^e(\vec{k},\vec{q},
|t|-t')\cdot\underline{\vec{E}}(\vec{q},\pm t')$$ where the upper (lower) sign corresponds to $t>0$ ($ t<0 $). Here the $\underline{\chi}^e$ is the spatial Fourier transform of the electric susceptibility tensor $$\label{c32}
\underline{\chi}^e(\vec{k},\vec{q}, t)=\left\{
\begin{array}{cc}
  \frac{1}{\varepsilon_0}\int_0^\infty d\omega \frac{\sin \omega t}{\omega}\int d^3p \ \underline{f}(\omega , \vec{k} , \vec{p}) \cdot
  \underline{f}^\dag(\omega , \vec{q} , \vec{p}) & t> 0 \\
  \\
  0 & t\leq 0
\end{array}\right.$$ and is guaranteed to possess solutions for the coupling tensor $
\underline{f} $ in terms of the temporal Fourier transform of $\underline{\chi}^e$ using a type of eigenvalue problem[@18]. It must be pointed out that for a given $\underline{\chi}_e$ the coupling tensor $ \underline{f} $ satisfying the relation (\[c32\]) is not unique. In fact if $ \underline{f} $ satisfy (\[c32\]), for a given susceptibility tensor ,then $$\label{c33}
\underline{f}'(\omega , \vec{k},\vec{q})=\int d^3p'\
\underline{f}(\omega , 