# How to derive gravitational path integral from the Hamiltonian operator formalism?

How does one derive the gravitational path integral $$\int [dg]\exp(iS_{\text{EH}}/\hbar)$$ from the Hamiltonian operator formalism?

The connection between the Hamiltonian operator formalism and the path integral is standard lore for QM and most QFTs but I've never seen it applied to gravity.

• @Qmechanic I'm not questioning that the EH Lagrangian density $\mathcal L=\pi^{nm}\dot h_{nm}-\mathcal H$, where $\mathcal H$ is the ADM Hamiltonian density. My question is about the path integral itself. How do we justify the integrations over the $g_{0\mu}$ components of the metric when they don't appear in the Hamiltonian formalism. It probably has to do with enforcing constraints, but this requires a 'rigorous' study. Commented Mar 14, 2023 at 15:57
• @Qmechanic Also, the Legendre transform is a classical construction, and doesn't immediately justify the path integral which is a quantum mechanical object. In particular, what is the gravitational path integral computing in the Hamiltonian formalism? Some matrix element? Which one? Commented Mar 14, 2023 at 16:32
• So you're asking about the equivalence between the operator formalism and the path integral formalism. Commented Mar 15, 2023 at 16:27
• @Qmechanic Yes exactly. Commented Mar 15, 2023 at 17:50

1. At the classical level, the equivalence goes as follows:

• The Lagrangian Einstein-Hilbert (EH) action with variables $$g_{\mu\nu}$$

• is replaced with the first-order Palatini action with variables $$(g_{\mu\nu},\Gamma^{\lambda}_{\mu\nu})$$,

• which is then reduced to the Hamiltonian Arnowitt-Deser-Misner (ADM) action with variables $$(g_{ij},\pi^{ij}, N, N_i)$$ and 3+1 first-class constraints, cf. e.g. Ref. 1.

• If one integrate out the momenta $$\pi^{ij}$$, the ADM action reduces to the EH action.

2. To define a formal perturbative (although non-renormalizable) quantum gravity, it is most systematic to use the BRST language.

• In the Lagrangian path integral formulation, one can use the Batalin-Vilkovisky (BV) quantization scheme.

• Be aware that the Lagrangian BV method may fail to correctly identify gauge-invariant measure factors in the path integral, as compared to Hamiltonian methods.
• In the Hamiltonian phase space path integral formulation and in the operator formulation, one can use the Batalin-Fradkin-Vilkovisky (BFV) quantization scheme. In both the path integral and operator formalism, one works systematically with

• an extended phase space of original variables, gauge-generating & gauge-fixing Lagrange multipliers, ghosts, antighosts, and their corresponding momentum variables. The equivalence between the operators/Hilbert space and the path integral and is most easily seen in this extended setting.

• Gauge-fixing. BRST-invariant observables do not depend on gauge-fixing.

References:

In the operator formalism the goal is to solve for the physical states, i.e. states which satisfy $$$$\mathscr H_\mu(x)|\Psi\rangle=0 \tag{1}$$$$ where $$\mathscr H_\mu(x)$$ are the four constraints in gravity. Let us assume that $$\mathscr H_\mu(x)$$ is self-adjoint. Then a simple solution to (1) is (see the 'master constraint') $$$$|\Psi\rangle=\lim_{t\to\infty}e^{-t\Delta}|\chi\rangle \tag{2}$$$$ where $$\Delta=\int \mathscr{H}_\mu(x)\mathscr{H}_\mu(x)dx$$ and $$|\chi\rangle$$ is an arbitrary state.
Interpreting $$t$$ as the time coordinate, the operator $$e^{-t\Delta}$$ can be written as a path integral: $$$$\lim_{t\to\infty}e^{-t\Delta}=\int[dg_{ij}][d\pi^{ij}]\exp\int dtdx\left(i\pi^{ij}\dot g_{ij}-\mathscr{H}_\mu\mathscr{H}_\mu\right)$$$$ (where the space and time arguments have been suppressed). Introducing 'Lagrange multipliers' $$N^\mu(x)$$ (aka the lapse and shift functions), this can be equivalently written as $$$$\lim_{t\to\infty}e^{-t\Delta}=\int[dg_{ij}][d\pi^{ij}][dN^\mu]\exp\int dtdx\left(i\pi^{ij}\dot g_{ij}-N^\mu N^\mu/4-iN^\mu\mathscr{H}_\mu\right)$$$$ Now you integrate $$\pi^{ij}$$ out and you get $$$$\lim_{t\to\infty}e^{-t\Delta}=\int[dg_{ij}][dN^\mu]|\det J|\exp\int dtdx\left(i\mathcal L_{EH}-N^\mu N^\mu/4\right) \tag{3}$$$$ where $$\mathcal L_{EH}$$ is the Einstein-Hilbert Lagrangian and $$|\det J|$$ is the field dependant determinant you get from integrating $$\pi^{ij}$$ out.
The $$N^\mu N^\mu$$ term can be interpreted as a gauge fixing term (probably arising from the gauge $$N^\mu=c^\mu$$ which is then integrated w.r.t. $$c^\mu$$ against a Gaussian weight like $$e^{-c^{\mu}c^{\mu}/4}$$).