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.

  • 2
    $\begingroup$ Often said where? Which page? $\endgroup$
    – Qmechanic
    Mar 31, 2022 at 11:58
  • $\begingroup$ The Hamiltonian of TQFTs does not usually vanish identically. Your argument only shows that it vanishes on-shell, and indeed $H$ is almost always non-trivial, but proportional to constraints. Check the canonical quantization of Chern-Simons theories, for example. $\endgroup$ Mar 31, 2022 at 12:20
  • $\begingroup$ @Qmechanic see e.g first para of arxiv.org/abs/gr-qc/9506070 $\endgroup$ Mar 31, 2022 at 13:06
  • $\begingroup$ @AccidentalFourierTransform What do you mean by "on-shell". I've only previously seen that term used to describe classical trajectories, in which case it means those which make the action stationary. But in my question, $H$ acts on kets labelled by "geometry on a slice", and I don't know what it means for geometry on a single slice to be "on-shell". $\endgroup$ Mar 31, 2022 at 13:10
  • $\begingroup$ The canonical hamiltonian being zero is imposed as a constraint on the phase space. It's like the Gauss-law $\nabla\cdot\vec{E}=0$ in QED and QCD. Physical states should be gauge invariant, i.e $\nabla\cdot\vec{E}|\psi\rangle=0$. The Wheeler-de Witt equation you wrote is like the "Gauss-law" for diffeomorphism. $\endgroup$
    – Valac
    Jun 6, 2022 at 12:00

2 Answers 2


"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.

  • $\begingroup$ Technical follow-up question: just to be clear, in the case of GR, are the $\chi_\alpha$ the usual constraint operators $\mathcal{H},\mathcal{H}^i$, and $u_\alpha$ are the lapse and shift? And $\mathcal{H}$ is distinct from both $H$ and $H'$? $\endgroup$ Mar 31, 2022 at 16:21
  • $\begingroup$ Conceptual follow-up: I see the point of the spatial-diff constraint: physical wavefunctionals on a slice $\Sigma$ should be invariant under $diff(\Sigma)$. But I don't see the "point" of the Hamiltonian constraint. What's stopping us from using any old state from $\mathcal{H}_\Sigma$ (as defined in the question) as our initial state, and just turning the crank of the path integral to produce a final state? Isn't the unitary evolution produced by the gravitational path integral (at least formally) fully diff-invariant? $\endgroup$ Mar 31, 2022 at 16:30
  • $\begingroup$ @nodumbquestions 1. My answer is not specific to GR, the $u^\alpha$ are generic Lagrange multipliers enforcing the constraints $\chi_\alpha$. 2. You don't have a path integral to crank if you ignore the constraints! This is also a general fact in the quantization of constrained Hamiltonian systems: Since your actual (classical) states are equivalence classes of gauge orbits in a constraint surface, the naive space of states you would associated to your system (either classically or quantumly)`is too big and trying to integrate over that overcounts equivalent configurations. $\endgroup$
    – ACuriousMind
    Mar 31, 2022 at 16:35
  • $\begingroup$ 1. I appreciate your general answer, I just want to understand how it maps onto the specific case of GR. 2. Surely I do have a path integral. I've already quotiented by $Diff(\Sigma)$ in my definition of $\mathcal{H}_\Sigma$, so there's no overcounting. Use [ ] to denote equivalence classes under surface diffeos. Then we can calculate amplitudes $\langle [h_{ij}^{(2)}]|U_{1\to 2}|[h_{ij}^{(1)}]\rangle$ by path integrating over all geometries (mod diffeos) that restrict to the given geometries on the boundary. What's wrong with that? $\endgroup$ Mar 31, 2022 at 21:27
  • $\begingroup$ @nodumbquestions 1. Ah, I think in the ADM formalism the Lagrange multipliers are the lapse and shift, yes. 2. a) Whether or not "diffeomorphism invariance" corresponds to the gauge freedom in GR is somewhat thorny issue, see physics.stackexchange.com/q/346793/50583 and its linked questions. b) Generically in constrained quantization, restricting to the constraint surface is not enough to avoid overcounting, since the gauge orbits are orbits inside the constraint surface. Unfortunately I don't know enough QG to tell you how your TQFT model meshes with the canonical approach. $\endgroup$
    – ACuriousMind
    Mar 31, 2022 at 21:43

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.


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