timeless Wheeler-DeWitt equation, but this interpretation works for equation (42). What is important to show that whether this time is buried in the Wheeler-DeWitt equation or not. We will take up this point in the next section.

**[Solution of Schroedinger Wheeler-Dewitt Equation]{}**
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Let us now take the solution of (42) $$\Psi = N (t) e^{- {{\Omega (t)} \over 2} \phi^2},$$ with a choice of $V(\phi)={\lambda\over 2}(1-m^2\phi^2), \lambda$ being a constant. The time appearing in (42) is the physical time, at least in the region we are considering. Now substituting (44) in (42) we find $$i{d \over {dt}} \ln N = {\Omega \over {2a^3}} + {a^3 \lambda},$$ $$i{ {\partial\Omega}\over {\partial t}} = {{\Omega^2 + a^6 \lambda m^2} \over
a^3}\,.$$ Making an substitution $$\Omega = - i a^3 {{\dot{y}} \over y},$$ and using the conformal coordinate $dt=ad\eta$, equation (46) gives $$y'' + 2 {{a'} \over a} y' - \lambda m^2 a^2 y = 0\,.$$ Taking an inflationary background $a(\eta)=-{1\over
{\sqrt{\lambda}\eta}}$, the solution $y$ is given by $$y =C_1 \eta^{{3/2} \pm {\sqrt{{9/4} + m^2}}}\,.$$ For practical purposes $m^2<{9\over 4}\, and\,\sqrt{{9/4} + m^2} \simeq ({3/2} +{{m^2}/3}).$ Taking the negative sign in (49) and in conformal coordinate $$\Omega=-ia^2(\eta){y'\over y}\,,$$ we get $$\Omega = -i 