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For a homogeneous canonical inflaton, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires positive potential energy to dominate both kinetic energy and any competing material components. Then $P_{\varphi }\simeq -{\rho }_{\varphi }$ and the expansion is nearly vacuum-like. Exact scalar-driven acceleration requires ${\rho }_{\varphi }+3P_{\varphi }<0$, equivalently ${\dot{\varphi }}^2<V$; potential domination is a stronger condition.

The slow-roll approximation additionally neglects $\ddot{\varphi }$ compared with $3H\dot{\varphi }$ in the inflaton equation of motion. Define the unreduced Planck mass $m_{\mathrm{pl}}=G^{-1/2}$; it differs from the reduced Planck mass by $\sqrt{8\pi }$. Sufficient potential-flatness conditions aretogether with approach to the slow-roll attractor. The reduced equations are $H^2\simeq 8\pi V/(3m_{\mathrm{pl}}^2)$ and $3H\dot{\varphi }\simeq -V^{\prime }$.

For $V=m^2{\varphi }^2/2$, take $m>0$ and the positive-field rolling branch. ThenThe field velocity is constant to this order. Choosing the start at $t=0$ gives the quadratic-potential slow-roll solutionThe PDF puts $\pi$ inside this square root; the converted TeX incorrectly moves it outside. Next integrate $d\log a/d\varphi =H/\dot{\varphi }=-4\pi \varphi /m_{\mathrm{pl}}^2$:Thus the number of e-folds is the logarithmic expansion, not the expansion factor itself:Here ${ϵ}_V={\eta }_V=m_{\mathrm{pl}}^2/(4\pi {\varphi }^2)$. The usual quadratic-inflation endpoint estimate sets ${ϵ}_V\simeq 1$, givingThis is an endpoint estimate, not an exact result from the full equations. Exact acceleration ends at the Hubble slow-roll parameter ${ϵ}_H=1$, equivalently ${\dot{\varphi }}^2=V$. Inserting the slow-roll velocity into that last condition instead gives $\varphi \simeq m_{\mathrm{pl}}/\sqrt{6\pi }$; the different order-one coefficient illustrates the failing approximation at the endpoint. A precise value requires the full scalar and Friedmann evolution. The large initial-field contribution to $N$ is unaffected by that order-one uncertainty.

Finally $V({\varphi }_i)\sim m_{\mathrm{pl}}^4$ gives ${\varphi }_i^2\sim 2m_{\mathrm{pl}}^4/m^2$. The quadratic-inflation expansion from Planck density is thereforeFor $m\ll m_{\mathrm{pl}}$ this is enormous and the initial field is in the slow-roll regime. No unique numerical factor follows without specifying $m/m_{\mathrm{pl}}$. Extrapolating the classical model from Planck density is only a formal estimate; it does not establish control of quantum-gravity corrections at that starting scale.

At horizon exit $k=aH$, $d\log k/dt=H(1-{ϵ}_H)$, where ${ϵ}_H=-\dot{H}/H^2$ is the Hubble slow-roll parameter. To leading slow-roll order,The slow-roll curvature power spectrum is proportional to $V^3/V^{\prime 2}$. Therefore the scalar spectral index in potential slow-roll parameters isUsing the potential slow-roll parameter and the second potential slow-roll parameter givesThe neglected horizon-exit derivative correction contributes at higher slow-roll order.

For $t>0$, $\dot{a}=n{t}^{n-1}>0$ and $\ddot{a}=n(n-1)t^{n-2}$. Consequently inflation occurs exactly when $\boxed{n>1}$. The case $n=1$ is coasting, and $0<n<1$ is decelerating. The Hubble slow-roll parameter is ${ϵ}_H=-\dot{H}/H^2=1/n$, giving the same acceleration criterion ${ϵ}_H<1$.

Use the comoving curvature perturbation mode normalization. Take $H,ϵ,c_s$ to be effectively constant at leading slow-roll approximation order, with $H,ϵ,c_s>0$, and use the Bunch-Davies vacuum selected by the early-time $i\epsilon$ contour. The regulator $\epsilon$ is distinct from the Hubble slow-roll parameter $ϵ$.

Let $C=ϵ+1-c_s^2$ and use the Fourier transform convention $\zeta (\mathbf{x})=\int d^{3}k\zeta (\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}/(2\pi {)}^3$. Conversion to conformal time givesThe last sign follows from substituting $a=-1/(H\tau )$, which is positive for $\tau <0$; dropping the explicit minus sign in that substitution would reverse the final primordial bispectrum.

At the order needed by the in-in formalism, the unequal-time Wick contraction isIt uses the Gaussian quantum vacuum two-point function with the displayed operator ordering; an arbitrary classical unequal-time covariance cannot replace it. For the connected three-point function, each external field contracts with one of the three fields at the vertex. There are $3!=6$ such Wick contractions, all equal. The spatial integral conserves total momentum and leaves one factor $(2\pi {)}^3{\delta }^{(3)}({\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3)$. The other nine pairings are disconnected tadpole terms supported on a zero external momentum; they are excluded from the connected primordial bispectrum, equivalently by subtracting the one-point function or normal ordering the vertex.

The free Gaussian odd correlator is zero. Applying the stated first-order in-in formalism therefore givesBy definition, the full connected correlator is $(2\pi {)}^3{\delta }^{(3)}({\sum }_r{\mathbf{k}}_r)B$. The assumptions are a Gaussian Bunch-Davies vacuum, the given leading cubic vertex, tree order, the late-time limit $\tau \to 0^{-}$, effectively constant background parameters, and connected nonzero external modes. No symmetry factor $1/3!$ is inserted into the given Hamiltonian.

At cosmological horizon exit, $d\log k/dN=1-ϵ\simeq 1$. Differentiating the slow-roll curvature power spectrum and using the Hubble slow-roll parameters givesThe supplied relations implyConsequently the scalar spectral index is

For each mode, the right side ofis evaluated at cosmological horizon exit, $k=aH$. Sincefirst order in the Hubble slow-roll parameters gives