where $ S_0= S(a,V_0) $ is now identified as zeroth order action.

Restricting ourselves to k=+1 universe and using the (13) and the constraint equation (9) we get $$a'^2 + aa'' =4a^2 {V_0\over M} -1,$$ where the prime now denotes differentiation with respect to $t$. Denoting its solution by $ a_0(t) $ we get $$\begin{aligned}
a_0(t)&=& \beta{\sqrt{M\over 2V_0}} sin{(i\sqrt{2V_0\over M}t)},\nonumber\\
    &=&{ \frac {\beta} {\nu}} \sin{(i~\nu~t)},\end{aligned}$$ where $\nu=\sqrt{2V_0/M}$. This solution satisfies the boundary conditions $$a(0)=0,~~ a'(0)=\beta~ i=constant,~~ a''(0)=0$$

Let us construct the zeroth order action $ S_0 $ for the solution (19). Using (11) and (19) we find $$P_a^0= {\frac {\partial S_0} {\partial a}}= -M\beta a_0 
(2a^2\frac {V_0} {M}-1)^{1/2}.$$ We note that for $2a^2\frac {V_0}{M}>1$, $P_a^0$ is real and is identified as classically allowed region. For $2a^2\frac {V_0}{M}<1,~P_a^0$ is imaginary and the corresponding region is identified as classically unallowed region. Thus we have the turning points at $a_0=0 $ and at $a_T=\sqrt{\frac {M}{2V_0}}$. We put the final boundary condition as $a_0 =0$ at $t=0,$ and $a_0=a_f$ at $t=t_f$. Integrating (21) we get $$S_0(a_f, V_0)= \frac {-M^2\beta} {6V_0} (2a^2\frac {V_0} {M}-1)^{3/2} + i~\frac {M^2} {6V_0}$$ We have used $ (-1)^{3/2} = \beta~~i $. The two possible values of the action (due to 