I've been keeping my work as short and sweet as possible, so sorry if my work seems to gloss over at times.
I begin with a relationship I noticed that would hold in theory:
$G\cdot\rho = \Box^2 \phi$
The proceedings idea is very simple, we replace the appropriate terms in the Friedmann equation for Cosmic Evolution
$(\frac{\dot{R}}{R})^2\ \psi = \frac{8\pi G}{3} \rho\ \psi$
$\frac{8\pi}{3} \Box^2\ \psi= \mathbf{k}(\frac{\partial \psi}{\partial t} - \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} + \frac{\partial \psi}{\partial z})$
With c=1 in natural units with metric signature of (-,+,+,+). The other issue I looked at was an origin scale. It doesn't take much to prove that
$m\dot{R}^2\ \psi = \frac{8\piR}{3}\ \hbar c\ \Box^2\ \psi$
From it, we rearrange the mass and make use of
$\frac{R}{m} = \frac{G}{c^2}$
Giving
$\dot{R}^2\ \psi = \frac{8\pi G\hbar}{3c}\ \Box^2\ \psi$
And further divide through by c^2
$(\frac{\dot{R}}{c})^2\ \psi = \frac{8\pi G\hbar}{3c^3}\ \Box^2\ \psi$
The object
$\frac{G\hbar}{c^3} =\ell^2_P$
Is the Planck length squared. By a reparameterization
$\ell^2_P \rightarrow R_0$
Allows us to construct from rearrangement, now in c=1 units
$(\frac{\dot{R}}{R_0})^2\ \psi = \frac{8\pi}{3}\ \Box^2\ \psi$
Further, you can express the four operator in curved spacetime. For s simple one-dimensional case, it is well-known that it is given as
$(\frac{\dot{R}}{R_0})^2\ \psi = \frac{8\pi}{3}\ \Box^2\ \psi$
$=\mathbf{k} \frac{1}{g} (\frac{\partial}{\partial x}[\sqrt{-g} g \frac{\partial}{\partial x}])\psi$
Which would seem to say that the instinsic evolution of a universe, occurs such that the wave function is guided by its own geometry.
Conclusions
In this version, we avoid the nuances of a timeless theory purported by the Wheeler DeWitt Equation. Not a lot of work or supposition was made to keep time in there in the form of the d'Alembertian operator to which we rewrote it in the language of GR. Also to finish, it seems like a Planck origin (squared) is at least on the tables.
