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Adding and subtracting the canonical scalar field density and pressure givesA real scalar with positive kinetic energy consequently needs $w\ge -1$. On the expanding scalar-dominated branch, divide $\dot{\varphi }$ by $H$ and use the Friedmann equation:Choose the positive rolling sign. Integration and elimination of $a$ using ${\rho }_{\varphi }={\rho }_{\varphi }^0(a/a_0{)}^{-3(1+w)}$ give the scalar-field reconstruction for a constant equation of state:Changing the rolling sign reverses the exponential slope. At $w=-1$, $\varphi$ is constant and $V={\rho }_{\varphi }^0$; elimination by an invertible $\varphi (a)$ then degenerates. The cosmological continuity equation gives $\ddot{\varphi }+3H\dot{\varphi }+V_{,\varphi }=0$ whenever $\dot{\varphi }\ne 0$, so the reconstruction satisfies the field equation as well as the Friedmann equation.

Compare the reconstructed exponent with $V_0{e}^{-\lambda \sqrt{8\pi G}\varphi }$. In this constant-$w$, scalar-dominated scaling solution,Strict ${\rho }_{\varphi }+P_{\varphi }>0$ requires $w>-1$, and acceleration requires $w<-1/3$. HenceThe accelerating branch has $V_0>0$. The zero-slope limit is a constant potential with $\rho +P=0$, so it does not meet the strict inequality. This result characterizes the exponential-potential power-law inflation solution; an arbitrary trajectory with the same potential need not have constant $w$ throughout its evolution.