first of all that the perturbed $i-j$ equation of (\[dyn\]) for $i\not= j$ leads to the identity $\Psi=\Phi$. Thus, perturbing the $i-0$, $i-i$ and $0-0$ equations of (\[dyn\]) one gets, respectively, $$\begin{aligned}
\dot{\Phi}+H\Phi=\frac{1}{2}\dot{\bar\phi}\delta\phi-\frac{\bar f_{\chi\chi}}{2}\delta\chi,\end{aligned}$$ $$\begin{aligned}
2\left(  \ddot{\Phi}+4H\dot{\Phi}+(3H^2+2\dot{H})\Phi
\right)=-\dot{\bar\phi}(\Phi\dot{\bar\phi}-\delta\dot{\phi})-\bar V_{\phi}\delta\phi +\delta(f-\chi f_{\chi}+u^{\alpha}\chi_{,\alpha}f_{\chi\chi}),
$$ $$\begin{aligned}
2\left(3{ H}^2\Phi+3{ H}\dot{\Phi}-\frac{1}{a^2}\Delta \Phi\right)=\dot{\bar\phi}(\Phi\dot{\bar\phi}-\delta\dot{\phi})-\bar V_{\phi} \delta\phi
+\delta(f-\chi f_{\chi}-u^{\alpha}\chi_{,\alpha}f_{\chi\chi})+\Phi\dot{\bar f}_{\chi}-\partial_t(\bar{f}_{\chi\chi}\delta\chi).\end{aligned}$$ Adding equations $0-0$ and $i-i$ and using $i-0$, one gets $$\begin{aligned}
\ddot{\Phi}-\frac{1}{a^2}\Delta\Phi+H\dot{\Phi}+2\dot{H}\Phi = \ddot{\bar\phi}\delta\phi+
\frac{1}{2}\Phi\dot{\bar f}_{\chi}-\frac{1}{2}\partial_t(\bar{f}_{\chi\chi}\delta\chi),
$$ where we have introduced the notation $\Omega\equiv \frac{1}{1-\frac{3}{2}\bar{f}_{\chi\chi}}$.

We may write equation $i-0$ as: $$\begin{aligned}
\ddot{\bar\phi}\delta\phi=\frac{2\ddot{\bar\phi}}{\dot{\bar\phi}}\frac{\dot{\Phi}+H\Phi}{\Omega}-\frac{\ddot{\bar\phi}}{\dot{\bar\phi}}\bar f_{\chi\chi}\frac{\Delta\delta\phi}{a^2\dot{\bar\phi}},\end{aligned}$$ which leads to the following equation for the potential $\Phi$, $$\begin{aligned}
 \label{lqcperturbed}
{\ddot{\Phi}}-\frac{\Omega}{a^2}\Delta\Phi+\left(H-2\frac{\ddot{\bar\phi}}{\dot{\bar\phi}}-\frac{\dot{\Omega}}{\Omega}\right){\dot{\Phi}}+\left(2\left(\dot{H}-H\frac{\ddot{\bar\phi}}{\dot{\bar\phi}}\right)-H\frac{\dot{\Omega}}{\Omega} \right){\Phi}=
\frac{\Omega{\dot{\bar{\phi}}}^2}{2}\partial_t\left(\frac{\bar{f}_{\chi\chi}\Delta\delta\phi}{a^2{\dot{\bar{\phi}}}^3}   \right),
$$ which, for the case of the choice of $f$ given in (\[fLQC\]), differs from the corresponding equation of LQC in the [*deformed algebra approach*]{} [@grain; @caitelleau] only in the right hand side term, which in the last approach vanishes.

On the other hand, the equation for $\delta\phi$ is obtained from the linearization of the conservation equation (\[conser\]) $$\begin{aligned}
\label{linearization}
\delta\ddot{\phi}+3H\delta\dot{\phi}-\frac{1}{a^2}\Delta \delta\phi+V_{\phi\phi}\delta\phi-4\dot{\bar{\phi}}\dot{\Phi}+2V_{\phi}\Phi=
\frac{\Omega\dot{\bar{\phi}}}{a^2}\Delta\left( \frac{\bar{f}_{\chi\chi}\Delta\delta\phi}{a^2{\dot{\bar{\phi}}}^3} \right).\end{aligned}$$

From these equations, we will calculate the Mukhanov-Sasaki (M-S) equation for scalar perturbations in our approach. First of all, note that equation $i-0$ can be written as $$\begin{aligned}
\frac{d}{dt}\left(\frac{a\Phi}{H}  \right)=\frac{a\Omega{\dot{\bar{\phi}}}^2}{2H^2}
\left[\frac{H\delta\phi}{\dot{\bar{\phi}}}+\Phi+\frac{H}{a^2{\dot{\bar{\phi}}}^3}\bar{f}_{\chi\chi}\Delta\delta\phi  \right].\end{aligned}$$ On 