\right)' d\bar{\tau}\cong 
B_2(k)z\int_{-\infty}^{\infty}\frac{1}{z^2}d\bar\tau,\end{aligned}$$ due, as we will show at the end of Section VI, to the small value of the pivot scale. One can also see it taking into account that $$\begin{aligned}
B_1(k)\sim \frac{k^{3/2}}{\rho_c}, \quad B_2(k)\int_{-\infty}^{\infty}\frac{1}{z^2}d\bar\tau\sim \frac{\sqrt{\rho_c}}{k^{3/2}}, \quad
k^2 \int_{-\infty}^{\tau}\frac{\bar{\phi}'}{z}\left( \frac{\bar{f}_{\chi\chi}\delta\sigma_k^{(0)}}{({\bar{\phi}')}^3 }  \right)' d\bar{\tau}\sim \frac{k^{7/2}}{\rho_c^2}.\end{aligned}$$ Then, since the pivot scale $k_*$ leaves the Hubble radius at time $-t_*$, when holonomy corrections could be disregarded, that is, when $-t_*\gg \frac{1}{\sqrt{\rho_c}}$, which is equivalent to $k_*=a(t_*)H(t_*)\ll \sqrt{\rho_c}$, one can see that the dominant term in (\[zzz\]) is $B_2(k)z\int_{-\infty}^{\infty}\frac{1}{z^2}d\bar\tau$.

Therefore, we have obtained approximately the same value as in the first iteration, which coincides with the result of LQC [@we; @haro13], meaning that in this iteration one will have $$\begin{aligned}
{\mathcal P}\equiv \frac{k^3}{2\pi^2}\left|\frac{v_k}{z}  \right|^2\cong\frac{\rho_c}{576}.\end{aligned}$$

Tensor perturbations
--------------------

The metric for tensor perturbations is [@mukhanovbook] $$\begin{aligned}
ds^2=-dt^2+a^2(\delta_{ij}-h_{ij})dx^idx^j,\end{aligned}$$ where $h_{ij}$ is a symmetric, traceless and transverse tensor ($h_i^i=\partial_ih^{ij}=0$).

Since the field $\phi$ does not affect the tensor perturbations, it is clear that in this theory this equation will coincide with the corresponding one in GR, that is, [@mukhanovbook] $$\begin{aligned}
\ddot{h}_i^j+3H\dot{h}_i^j-\frac{1}{a^2}\Delta {h}_i^j=0.\end{aligned}$$ Denoting $h_i^j$ by $h$ and introducing the variable $v_T=a h$, in Fourier space this equation becomes $$\begin{aligned}
v_{T, k}''+\left(k^2-\frac{a''}{a}  \right)v_{T,k}=0.\end{aligned}$$

Comparison with other models
