\pi$. This means that by dimensional analysis, there are three possible candidates for the CTs. On the other hand, we have only one kind  of divergent term present in the dimensionally regulated expression for the $\langle\zeta_{\vec k}(\eta)\zeta_{\vec k'}(\eta)\rangle$ when $\eta = 0$ since there is just a monomial of $k^3$ in the front (see Eq. (\[G\_loop\_asym\]) as compared to Eq. (\[Pi\_usual\_QFT\])). In this case, we cannot determine the coefficients of the operators in the CT Lagrangian in any unique way.

When $\tilde{\cal G}_{\rm 1-loop}$ is worked out for $\eta$ which is non-zero but still such that $-k\eta \ll 1$, then we get an expression of the form (save for some numerical factors) $$\begin{aligned}
 \tilde{\cal G}_{\rm 1-loop} &\sim& k^3 ~\tilde{\cal G}_{\rm tree}^2 ~\frac{c_3^2 M^8}{\epsilon^2 M_{\rm Pl}^4 H^4}  \times \\ \nonumber 
 && \left( \frac{F_{-1}}{\epsilon} + \frac{F_{-1}}{2} \log \left( \frac{H(\eta_k)}{\mu} \right) + F_0 + \cdots \right) \times \\ \nonumber
 && \left( \sum_{i=0}^{n} c_i (- k \eta )^{i} \right) \; ,\end{aligned}$$ i.e., we get an extra term multiplied to the $\eta = 0$ result which is a polynomial, not in the external momentum $k$ but in $-k\eta$. Thus, in this case, even for a dimensionless integral (see the discussion after Eq.(\[eq:loop\_SZ\_gen\])) we get an 