# Wheeler-deWitt equation in words?

Looking at the Wheeler-deWitt equation. My attempt to describe it in words is this:

"For each 3d-manifold given by metric tensor field $$\gamma$$, associate a complex number $$\Psi$$. The Wheeler-de-Witt equation says that the rate of change of the phase $$\Psi$$ compared to a small change in the manifold is proportional to the total curvature, $$R$$, of the manifold (plus a constant $$\Lambda$$)."

But then I guess it depends what one means by a "small change". If one can put a "small change in the metric" into more pedestrian terms. And I guess this is given by the super-metric $$G$$.

From this one would conclude the Universe "favours" smoother spaces as the wave function of highly curved spaces would tend to cancel as it would change a lot under a small perturbation.

Is this a good way to describe the equation in words? If not what is a better way?

$$\begin{eqnarray} \left[ -G_{ijkl} \frac{\delta^2}{\delta \gamma_{ij} \delta \gamma_{kl}} - ^{(3)}R(\gamma) \gamma^{\frac{1}{2}} + 2 \Lambda \gamma^{\frac{1}{2}} \right] \Psi[\gamma_{ij}] &=& 0\\ \frac{1}{2} \gamma^{-\frac{1}{2}} \left( \gamma_{ik} \gamma_{jl} + \gamma_{il} \gamma_{jk} - \gamma_{ij} \gamma_{kl} \right)&=& G_{ijkl} \end{eqnarray}$$

• Please use MathJax, not scans of math. – G. Smith Nov 13 '19 at 21:44
• the rate of change of the phase $\Psi$ Did you mean to write “phase of”? How do you conclude that the phase of $\Psi$ is being extracted? – G. Smith Nov 13 '19 at 21:52
• Well it's more or less a wave equation in the form $y''+a^2y=0$ which has a solution $y=e^{iax}$. So $a$ would correspond the the frequency of the wave or how fast the phase changes. In fact $a$ would be the square root of the curvature in this case. – zooby Nov 14 '19 at 5:22