Quantum Gravity – Why Isn’t the Wheeler-De Witt Equation Automatically Satisfied if Quantum Gravity is a TQFT?

constrained-dynamicsgeneral-relativityquantum-gravitytopological-field-theory

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.

Best Answer

"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.

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