# What is the interpretation of a wave function of the Universe in Hawking's no boundary proposal?

In the path integral formalism we have an in state $\Psi_{in}[\phi]$ and and out state and we find the amplitude for going from one to the other:

$$\Delta[\Psi_{in},\Psi_{out}] = \int \Psi_{in}[\phi]\Psi_{out}[\phi]e^{i S[\phi]} D\phi$$

I am trying to understand quantum cosmology. Hawking describes his wave function of the universe as:

$$\Psi_{out}[h] = \int_{\partial g=h} e^{-S[g]} Dg$$

There is no "in" state since we assume the Universe is spatially closed and space-time is like a sphere with time starting at the "south pole".

But what's confusing me is that, unlike in the previous case where $\Psi_{out}[\phi]$ can take on many forms (for instance being highly peaked at various field configurations of our choosing), $\Psi[h]$ has only one form defined by the functional integral. Hence there can be no such thing as $\Delta[\Psi_{in},\Psi_{out}]$ in quantum cosmology since all states are the same and $\Psi_{in}=\Psi_{out}$. Is this related to the "problem of time?"

Instead I read that you can have $\Psi[h,h']$ which is a kind of amplitude for going from one geometry to another where you sum over 4D geometries bounded by both h and h'.

Is it true that in quantum cosmology $\Psi[h]$ can only have one form? It would be like a wave function only having one distribution!

Is there a simple way to show that the wave functional cannot change?

Edit

I think I understand a bit more. If $\Psi_{in}[g]=\delta[g(0)]$ and $\Psi_{out}[g]=\delta[g(1)-h]$ then $\Delta[\Psi_{in},\Psi_{out}] = \Phi[h]$ which is the wave functional of the Universe which satisfies the Wheeler-de-Witt equation. And so in general the amplitude for a particular distribution must be:

$$\Delta[\Psi_{out}] = \int \Psi_{out}[g] \Phi[g] Dg$$

I guess that makes sense. Perhaps I was mixing up states and wave functions.

PS

Is it true that there is only one quantum state of the Universe. In Hawking's first paper he says talks of the quantum state being the "ground state". Is this not the view now? Is there just a single quantum state?

$$\Psi[h] = \Delta[h,S_0]$$