Timelessness of the universal wavefunction In the Wheeler-DeWitt equation (universal wavefunction) the Hamiltonian constraint is zero. Representing that there's no time in such a scenario.
My question is - what term in the equation, or lack thereof, is causing the evolution of the equation to stall, thus representing a timeless universe?
 A: The canonical quantisation of general relativity stems from the replacement of the Poisson bracket by a commutation relation such that
$$\begin{align}
 \left\lbrace A(p,q),B(p,q)\right\rbrace \rightarrow -\frac{i}{\hbar}\left[\hat{A}(\hat{p},\hat{q}),\hat{B}(\hat{p},\hat{q})\right] \;.
\end{align}$$
When you choose a particular representation satisfying this, the dynamical evolution of the wave function $\psi$ is given by the Schrödinger equation
$$\begin{align}
 i\hbar \frac{\partial\psi}{\partial t} = \left(\hat{H}_T-\lambda_m\hat{\phi}_m(\hat{p},\hat{q}) \right)\psi \;,
\end{align}$$
where $\hat{H}_T$ is the total Hamiltonian of your theory, and each $\lambda_m\hat{\phi}_m$ is a constraint such that $\hat{\phi}_m(\hat{p},\hat{q})\psi =0$. These constraints (rigorously, primary constraints) appear when a momentum $p_i:=\partial \mathcal{L}/\partial \dot{q}^i$, with $\mathcal{L}$ the lagrangian of the theory and a dot means time derivative, is such that
$$\begin{align}
  det\left(\frac{\partial p_i}{\partial \dot{q}^{j}}\right) =0 \;.
\end{align}$$
If we consider general relativity, for simplicity, we can express the Hamiltonian in terms of the curvature using the ADM decomposition (I skip the details here). The associated metric is
$$\begin{align}
  ds^2 = -(N^2-N^{i}N_i) dt^2 + 2N_i dx^{i}dt + h_{ij} dx^{i} dx^{j} \;,
\end{align}$$
where, a priori, the dynamical variables are the lapse function $N$, the shift function $N_i$ and the spatial metric $h_{ij}$. However, we deduce from the GR lagrangian $\mathcal{L}=N h^{1/2}(K_{ij}K^{ij}-K^2+^3R)$, that the canonical momenta of $N$ and $N_i$ are zero! Therefore, the only true dynamical quantity is $h_{ij}$. If we note the momenta $P$, $P_i$ and $\Pi$, respectively, the canonical Hamiltonian density obtained by Legendre transform of the above Lagrangian is
$$\begin{align}
 \mathcal{H}_C = P\dot{N} + P_i \dot{N}^{i} + \Pi^{ij}\dot{h}_{ij} - \mathcal{L} \;,
\end{align}$$
and the total Hamiltonian density $\mathcal{H}_T$ is
$$\begin{align}
 \mathcal{H}_T = \mathcal{H}_C + \lambda P + \lambda^{i} P_i \;,
\end{align}$$
where $\lambda$ and $\lambda_i$ are Lagrange multipliers.
Now, expressing $\mathcal{H}_C$ in terms of $\Pi_{ij}$ and $h_{ij}$, we can show that $\mathcal{H}_C=N\mathcal{H}_0 + N_i \mathcal{H}^{i}$, with $\mathcal{H}_0=G_{ijkl}\Pi^{ij}\Pi^{kl}-h^{1/2} {\,}^3 R$ and $\mathcal{H}^{i}=-2\Pi^{ij}_{\,\,;j}$, where I use the DeWitt metric $G_{ijkl}$. But we now have that the time dependence of P is
$$\begin{align}
 \dot{P}&=-\frac{\partial \mathcal{H}_T}{\partial N}\\
  & = \left\lbrace P,H_T \right\rbrace \\
  & = \int d^3 y \left\lbrace P(x^{i},t),N(y^{i},t)  \right\rbrace \mathcal{H}_0(y^{i}) \\
  & = -\mathcal{H}_0 (x^{i}) \\
  & \simeq 0\;.
\end{align}$$
I used the definition of a Poisson bracket in the third line, and I reached the conclusion $\dot{P}=0$ since $P=0$. We conclude that the quantity $\hat{H}_C=\hat{H}_T-\lambda_m\hat{\phi}_m(\hat{p},\hat{q})$ is zero, hence
$$\begin{align}
  \frac{\partial\psi}{\partial t} = 0\;.
\end{align}$$
This is where the problem of time comes from. Note that in general relativity, this is expected since GR is invariant by time-reparametrisation.
If you are wondering how time can be restored in quantum cosmology, the most common way is to introduce an external scalar field, which can play the role of radiation or dark matter, for instance. Using the FRLW metric (which homogeneous and isotropic for simplicity)
$$\begin{align}
  ds^2=-dt^2+a^2(t)d\vec{x}\;,
\end{align}$$
we can identify $N=1$, $N_i=0$ and $h_{ij}=a^2(t)$, this will roughly add a matter contribution $H_M=Na^{-3w} \Pi_T$ to the Hamiltonian, where $a$ is the scale factor. Note that the momentum enters linearly, not quadratically, and this will play the role of time. The nature of the field depends on the equation of state $w$ you choose (remember that the energy density is roughly related to pressure through $\rho=wp$, then $w=0$ is for pressureless matter and $w=1/3$ is for radiation).
As time must appear in some form in the Hamiltonian, the matter we choose as time must eventually give a scalar. I suppose we could choose an additional vector field, but it will be contracted with another vector (think of the term $\lambda^{i}P_i$ in $\mathcal{H}_T$).
When we add the gravitational and matter parts, we obtain the total Hamiltonian
$$\begin{align}
 -\frac{\Pi_a^2}{24a}+\frac{\Pi_T}{a^{3w}} =0\;.
\end{align}$$
After quantising, we obtain the Wheeler-DeWitt equation (I choose a convenient ordering to maintain covariance)
$$\begin{align}
  \frac{1}{24}\left[a^{\frac{3w-1}{2}}\frac{\partial}{\partial a}\left(a^{\frac{3w-1}{2}}\frac{\partial}{\partial a}\right) \psi(a,T)\right]=i\frac{\partial}{\partial T} \psi(a,T) \;.
\end{align}$$
So we understand that the momentum $\Pi_T$ associated to the matter field $T$ gives the evolution of the equation.
