1
$\begingroup$

What is the gravitational path integral (which roughly goes like $\int [dg]e^{iS_{\text{EH}}[g]}$) computing?

Usually, path integrals arise from transition amplitudes such as these: $\lim_{T\to\infty}\langle f|e^{2iHT}|i\rangle$ (the "$S$-matrix"). Is the gravitational path integral (GPI) computing a transition amplitude? But that's not possible because in quantum gravity the Hamiltonian is zero and hence there are no "transitions" -- see the problem of time in quantum gravity. So what, really, is the GPI computing?

This is similar to a previous question I asked.

$\endgroup$

1 Answer 1

1
$\begingroup$

I don't know if this answers directly your question, but based on your previous question, and what you said in this one, I thought it was worth it to write a few things. Tell me if this is completely off and I will delete this attempt to answer.

Having a zero Hamiltonian on-shell does not mean there are no transitions. For example, the free particle's Hamiltonian action is : \begin{equation} S=\int_0^1 dt \left( p_\mu\dot{x}{}^\mu-\frac{e}{2}(g^{\mu \nu}p_\mu p_\nu+m^2) \right) \tag{1} \end{equation} Where it is crystal clear that, on-shell, $\mathcal{H}=g_{\mu \nu}p_\mu p_\nu+m^2=0$. The ADM formulation of gravity, on its side, enjoys the Hamiltonian action: \begin{equation} S_G=\int_{\mathcal{M}} d^4x\left( \pi^{ij} \dot{\gamma}_{ij}-N\mathcal{H}_G-N_i \mathcal{P}^i \right) \tag{2} \end{equation} Where $\mathcal{H}_G$ is the Hamiltonian constraint and $\mathcal{P}^i$ is the diffeomorphism constraint.

Both $(1)$ and $(2)$ exhibit a null Hamiltonian, but we know that the free particle is evolving, so having this property does not imply there are no transitions.

One thing you can do is use the BFV formalism on the Hamiltonian path integral arising from $(2)$, as Qmechanic suggests in his answer to you previous question: \begin{equation} Z=\int\mathcal{D}\Phi_A \mathcal{D} \Phi^{\ast A}\,e^{i\int_{\mathcal{U}}d^4x (\Phi^{\ast A}\dot{\Phi}_A-\{Q,\psi\})} \end{equation} Where $\Phi$ are the fields, $\Phi^\ast$ the anti-fields, $Q$ the BRST charge of the theory, and $\psi$ a gauge fixing fermion you are free to choose as long as it is fermionic. Note that I used $\mathcal{U} \subset \mathcal{M}$, such that $\partial \mathcal{U}=f-i$, in your notations. You can follow https://journals.aps.org/prd/abstract/10.1103/PhysRevD.47.1420, "Microcanonical functional integral for the gravitational field" by Brown and York, and see that, upon ignoring the integral on the lapse function, you obtain indeed something of the form: \begin{equation} T_{i\rightarrow f}(N)={}_{\mathcal{P}}\langle f | e^{i\int_{\mathcal{U}}d^4x N\mathcal{H}_G} | i \rangle_{\mathcal{P}} \end{equation} Now, $N$ does not depend on the time parameter anymore. Moreover, the initial and final states satisfy the diffeomorphism constraint. So in fact, the problem seems to be that, contrarily to the free particle's Worldline action where reparametrizing the time parameter can be done without harm, you cannot repramatetrize the time. Indeed if you try to do so, you will not end up with one time interval, but with a field of time intervals: \begin{equation} I=\bigcup_{x\in i}[t_i(x),N(x)t_f(x)],\,\,\,\partial I \neq f-i \end{equation} Where $t_i(x)$ and $t_f(x)$ are the time position of the point $x$ in $i,f$. Of course, if you try to absorb $N$ in $t_f$, you end up with a final hypersurface $f$ which is not space-like anymore...


To conclude with your question: The gravitational path integral aims to describe a transition amplitude between two space-like hypersurfaces, but it is plagued with this reparametrization invariance and not by its null hamiltonian, and the fact that there is this very "problem of time" means that we actually don't quite know what to do. Of course, there are different approaches https://en.wikipedia.org/wiki/Problem_of_time#Proposed_solutions_to_the_problem_of_time, but there doesn't seem to be one solution that is accepted as consensus. There are transitions, but right now, no one can say "I have the right approach to quantum gravity, and it solves the problem of time". Remember that this is an entire field of research...

$\endgroup$
1
  • $\begingroup$ Interesting. I need time to think about what you've said. $\endgroup$
    – dennis
    Nov 5, 2023 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.