Hamiltonian constraint corresponding to (6) now reads $$H=-{1 \over {2Ma}} {P_a}^2 + {1 \over {2 a^3}}{P_{\phi}}^2 - {M \over 2} ka 
+ a^3 V(\phi) = 0\,.$$ In equation (8), $M =  {{3 \pi}\over {2G}}$, where $G$ is Newton’s constant, $k  =  0, \pm 1 $ for flat, closed and open models, and $P_a = - M a {\dot{a}},\; P_{\phi} = a^3 {\dot{\phi}}\,.$ Identifying $P_i = {{\partial S} \over {\partial q_i}},\; q_i=a,\; \phi,$ the Einstein-Hamilton-Jacobi equation is $${-{ 1 \over {2M}} }{({{\partial S} \over {\partial a}})^2} + {1 \over {2a^2}}
{({{\partial S} \over {\partial \phi}})^2} -{1 \over 2}Mka^2 + a^4 V(\phi) = 0\,.$$ Now to obtain the prescription like (5), we define a time operator consistent with Liouville’s equation as $$\begin{aligned}
{\partial\over {\partial t}}&=& \sum_{i}({{{\partial H}\over {\partial P_i}}
{\partial\over {\partial q_i}}-{{\partial H}\over {\partial q_i}}
{\partial\over {\partial P_i}}})\nonumber \\
&=&
{{\partial H}\over {\partial P_a}} {\partial\over {\partial a}}
+{{\partial H}\over {\partial P_\phi}} {\partial\over {\partial \phi}}
-{{\partial H}\over {\partial a}} {\partial\over {\partial P_a}}
-{{\partial H}\over {\partial \phi}} {\partial\over {\partial P_\phi}}\,.\end{aligned}$$ that would satisfy (8) or (9). From (10) we get $$\begin{aligned}
{{\partial a}\over {\partial t}}&=& {{\partial H}\over {\partial P_a}}=
-{P_a\over Ma}\\
{{\partial \phi}\over {\partial t}}&=&{ 1\over a^3} P_\phi\\
{{\partial P_a}\over{\partial t}}&=& -[{{P_a}^2\over {2Ma^2}}
-{3{P_\phi}^2\over 