So before and after transformation, we are actually talking about different “adiabatic perturbations", which is in spite of the fact that mathematically the formulae in the two frames can be transformed to each other smoothly. Contrarily, in solo-degree-of-freedom (s.d.o.f.) system where the isocurvature perturbations do not appear, this difference will vanish and $\tilde{\cal R}$ and ${\cal R}$ will coincide, which gives the equivalence between curvature perturbations in two frames. Note that although we only show this point by taking a small example in this paper, in like manner, it can also be applicable for more complicated modified gravity models with m.d.o.f.. This conclusion is consistent with that in Ref. [@Gong:2011qe] obtained by using different methods.

According to (\[differenceR\]), the power spectrum of $\tilde{\cal R}$ is: \_&\~&||\^2 \
&&|[R]{}+|\^2 \
&&[P]{}\_[R]{}+\^2[P]{}\_[S]{}\
&&+2\_[[R]{}[S]{}]{} , where ${\cal P}_{\cal R}$, ${\cal P}_{\cal S}$ and ${\cal C}_{{\cal R}{\cal S}}$ can be related to their values at Hubble-crossing via relations (\[spectrum2\]).

From the expressions of $A$ and $B$ in Eq. (\[AandB\]) and making use of Eqs. (\[mus\]) and (\[Vs\]), it is straightforward to express $A$ and $B$ as: \[AandB\] A=-2\_[s]{}-b\_\^3 , B=--+b\_\^2-A\^2 , and one can see that all the terms in $A$ and $B$ are 