Quadratic-potential slow-roll solution

For $V(\varphi )=m^2{\varphi }^2$ in units where $3H^2\simeq V$ and for positive $\varphi$,and

For $V=m^2{\varphi }^2/2$ and unreduced Planck mass, the slow-roll e-fold count is $N=2\pi ({\varphi }_i^2-{\varphi }_f^2)/m_P^2$. A formal initial potential $V_i\sim m_P^4$ gives ${\varphi }_i^2\sim 2m_P^4/m^2$, hence the displayed large-field estimate and expansion factor $e^N$. The final-field contribution is order one in $N$. Without specifying $m/m_P$ there is no unique numerical expansion factor. The extrapolation to Planck density is a classical estimate, rather than a controlled treatment of quantum-gravity corrections.

For $V=m^2{\varphi }^2/2$ and unreduced Planck mass $m_P=G^{-1/2}$, the potential slow-roll parameter is ${ϵ}_V=m_P^2(V^{\prime }/V{)}^2/(16\pi )=m_P^2/(4\pi {\varphi }^2)$. Its conventional value one gives the displayed endpoint estimate. Exact accelerated expansion ends at ${ϵ}_H=1$, or ${\dot{\varphi }}^2=V$ for a canonical scalar alone. Substitution of slow-roll formulas into this exact condition gives a different order-one coefficient because slow roll is failing there; a precise endpoint requires integrating the full inflaton equation of motion and Friedmann equation. The large-field e-fold estimate is insensitive to this order-one endpoint uncertainty.

For a quadratic inflaton potential ending at the leading estimate ${ϵ}_V=1$, $\Phi /M_{\mathrm{Pl}}=\sqrt{4N+2}$. Cosmological horizon exit gives $d\log k/dN=-1+{ϵ}_H$. A narrow logarithmic interval therefore probes a field span $2\Delta \log k/(\sqrt{4N+2}(1-{ϵ}_H))$, to the accuracy of the slow-roll approximation. At sixty e-folds, five logarithmic wavenumber units probe about $0.64M_{\mathrm{Pl}}$ around a background field near $15.6M_{\mathrm{Pl}}$.

The quadratic potential has equal potential slow-roll parameters, ${ϵ}_V={\eta }_V=1/(2N+1)$. Consequently $n_s=1-4/(2N+1)$ and $r=16/(2N+1)$ at leading order. The relation $r=4(1-n_s)$ is independent of the inflaton mass, so a satisfactory scalar tilt does not by itself guarantee a satisfactory tensor prediction.

For a quadratic potential with the end set by ${ϵ}_V=1$, a mode leaving $N$ e-folds earlier has ${\varphi }_{\ast }^2=(4N+2)M_{\mathrm{pl}}^2$. Its curvature amplitude is $(m/M_{\mathrm{pl}}{)}^2(2N+1{)}^2/(24{\pi }^2)$. Thus the scalar amplitude fixes the inflaton mass while the tilt and tensor ratio depend only on the exit e-fold count at this order.