If quantum gravity is a TQFT, why isn't the Wheeler-De Witt equation satisfied automatically?

Question

It is often said that QG is a topological QFT: given a bordism between $D$-manifolds $\Sigma_1$ and $\Sigma_2$, QG assigns a unitary between the Hilbert spaces associated with $\Sigma_1$ and $\Sigma_2$. For simplicity, here we won't sum over topologies, even though this is widely believed to be the right thing to do in general.

Concretely, the Hilbert space $\mathcal{H}_\Sigma$ associated to a $D$-manifold $\Sigma$ is the space of complex functionals $\Psi:Riem(\Sigma)/Diff(\Sigma)\to \mathbb{C}$. Note that these states are diff-invariant by construction, and so no spatial-diff constraint must be imposed.

Now, it's my understanding that in a TQFT, the Hamiltonian $H$ must vanish identically. To see this, fix the background topology $\mathbb{R}\times\Sigma$, with $t\in\mathbb{R}$ interpreted as time. Then all the bordisms are the identity bordism, so the evolution from $t_1\times \Sigma$ to $t_2 \times \Sigma$ is just the identity. But the generator of time translations is $H$, so we must have $H=0$.

On the other hand, in canonical quantum gravity, the Hamiltonian constraint of classical GR is quantised and leads to the Wheeler-de Witt equation $H|\Psi\rangle=0$. This equation is supposed to be a nontrivial constraint, selecting a "physical" sector of $H_\Sigma$. Much effort has gone into constructing explicit solutions for $|\Psi\rangle$. It surely can't be the case that $H$ vanishes identically on $\mathcal{H}_\Sigma$, otherwise no one would talk about "solving" the WdW equation.

What have I misunderstood? In answering, please feel free to assume $\Sigma$ is closed, compact and without boundary, if this simplifies things.

Answer 1

"The Hamiltonian is zero" is not really an interesting statement for reparametrization-invariant theories - the Hamiltonian is generically zero for such theories, see this answer of mine.

The crucial point is that a Hamiltonian theory is more than its "naive" Hamiltonian $H(p,q)$. The Hamiltonian theories that correspond to Lagrangian theories with gauge freedoms are typically constrained (see also this answer of mine), and the action of such a constrained theory looks like $$ S = \int (\dot{q}^ip_i - u^\alpha\chi_\alpha - H)\mathrm{d}t$$ where the $\chi_\alpha$ are the constraints that must be fulfilled as $\chi_\alpha(p(t),q(t)) = 0$ on-shell classically and as $\chi_\alpha \lvert \psi(t)\rangle = 0$ quantumly. Your argument shows $H=0$, but that's just a very elaborate way of showing the general fact that reprametrization-invariant systems have zero Hamiltonians. It does not remove at all the requirement that the constraints of the system need to be implemented. The confusion with the "Hamiltonian constraint" of the Wheeler-deWitt equation is that the people who talk about the Wheeler-deWitt equation consider as "the Hamiltonian" the extended Hamiltonian $$ H' = u^\alpha \chi_\alpha - H$$ so that $H'\lvert \psi\rangle = 0$ requires fulfillment of the constraints.

Answer 2

I think I've sorted out my confusion, which was very basic, so I'll answer my own question for posterity.

Let $Z$ be the functor in question from Cob(D+1)$\to$Hilb. The key point is that I was wrong in assuming the $Z([0,1]\times\Sigma) = \text{id}_{Z(\Sigma)}$. Instead, all we can say is that $Z([0,1]\times\Sigma)^2 = Z\left(([0,1]\times \Sigma) \sqcup ([0,1]\times\Sigma\right)) = Z(([0,1]\times \Sigma)$. That is, $Z([0,1]\times \Sigma)$ is a projector, which we'll call $P$.

Given some more complicated bordism $B$, say from $\Sigma \to \Sigma$ for simplicity, we can always produce a homeomorphic bordism by gluing $[0,1]\times\Sigma$ at the start and end. Hence $P Z(B) P = Z(B)$. In other words, the evolution maps corresponding to different bordisms all restrict to the image of $P$. The orthogonal complement of the image of $P$ is always projected to zero. For this reason, the image of $P$ is called the "physical Hilbert space" $\mathcal{H}_{phys}$.

In the TQFT literature, people tend to just restrict to the physical Hilbert space from the very start, sometimes without saying so explicitly. Therefore one might see $Z([0,1]\times \Sigma) = \text{id}_{Z(\Sigma)}$, or words to that effect. This was the original source of my confusion.

In quantum gravity, $P$ projects onto the space of solutions to the WdW equation. But now we're not just dealing with some abstract Hilbert space -- instead, states come with physical interpretations. So it is now an meaningful question to ask what kinds of states lie in $\mathcal{H}_{phys}$. That is, it is interesting to look for explicit solutions to the WdW equation, written in terms of pre-existing geometric variables. So, as opposed to the TQFT literature, the quantum gravity literature doesn't just restrict to $\mathcal{H}_{phys}$ from the very start; instead, working out exactly what $\mathcal{H}_{phys}$ looks like is an important challenge of its own.

Finally, note that if we don't restrict to $H_{phys}$, then any attempt at a Hamiltonian description of the dynamics will fail, since the evolution is not unitary (in fact, it's not even invertible!). So if one prefers to take the Hamiltonian formulation as a starting point, one will find that the constraint must be satisfied for the theory to make any sense at all. In my opinion, the situation is clearer from the TQFT perspective, where nonphysical states are just those that are projected out by the dynamics.