m^2 \sqrt{\lambda} {a^3 \over 3}.$$ Since $m^2\sqrt{\lambda}$ is small and $a$ is large and ${d\over {dt}}=\sqrt{\lambda}a{d\over {da}}$, we get from (45) $$N = N_o \exp { \left[ - {{i a^3 \sqrt{\lambda}} \over 3} \right]},$$ the integration constant $N_o$ has to be determined through a suitable initial condition.We now write down the solution of the SWD equation (42) as $$\begin{aligned}
\Psi &=& N_0\exp{\left[\,i\frac{ m^2\sqrt{\lambda} a^3}{6}\phi^2
-i\frac {a^3\sqrt{\lambda}}{3}\right]} \nonumber \\
&\simeq& N_0\exp{\left[\,-i\frac
{a^3[\lambda(1-m^2\phi^2)]^{1/2}}{3}\,\right]}\nonumber\\
     &\simeq& N_0\exp{\left[\,-i~\frac{(a^2 2V)^{3/2}}{3.2V}\,\right]}\nonumber
     \\
     &\simeq& N_0\exp{\left[\,-i~\frac{(a^2 2V-1)^{3/2}}{3.2V}\,\right]}\end{aligned}$$ In obtaining (53) we have made the approximation $\sqrt{\lambda}(1-{1\over 2}m^2\phi^2)
\simeq (1-m^2\phi^2)^{1\over 2}\sqrt{\lambda}=\sqrt{2V(\phi)}$ and $2a^2V\simeq 2a^2V-1$ for $2a^2V>1$, which is quite valid in classically allowed region. This is what we meant earlier by saying ‘the region of interest’ while neglecting $S_1$ compared to $S_o$. Obviously the wavefunction (53) is not normalizable because of absence of real part of $\Omega$ in equation (51). The general procedure is to consider higher mode solutions of the scalar field. We take however the complex trajectory approach as in [@bis:prd]. It should be pointed out that though we started with a Gaussian form, the wavefunction very closely resembles the WKB form that we guessed earlier (see (27)). To compare our result with that of the Wheeler-DeWitt equation, 