0+1D dimensional quantum gravity (QG) action?

Question

I'm trying to understand how we might sum over the metric (or vielbein) in a QG theory. So I'm looking at a simple (perhaps too simple) 0+1D action.

Imagine a scalar field in 0+1D. The vielbein in 0+1D would be have one component $e(t)$. Also we know that $\det(e(t))=e(t)$ and $g^{00}(t)=e(t)^2$.

Therefor I imagine the Lagrangian would be:

$$L(t) = e(t)^{-1}\dot{\phi}^2(t) - m^2 e(t) \phi(t)^2.$$

There is no curvature term. Although perhaps we could include a cosmological constant $\Lambda e(t)$.

The amplitude to start at $\phi_0,e_0$ and end at $\phi_1,e_1$ would presumably be:

$$K(\phi_0,e_0;\phi_1,e_1) = \int_{\phi(0)=\phi_0,e(0)=e_0}^{\phi(1)=\phi_1,e(1)=e_1}\exp\left(i\int_0^1 L(t)dt \right ) D\phi De$$

Note, $K$ should not depend on a time as this is a QG theory.

Where we arbitrarily take the integral from $0$ to $1$ as the length of the interval is entirely governed by the metric $e(t)$ which gives the amount time is scaled by. Weirdly because of this the integration range shouldn't matter?!

Is this correct? I don't think it can be correct since the time metric $e_0$ and $e_1$ would not have any meaning on the bounds. If so can this be solved?

If it can be solved should it satisfy: $\int K(\phi_1,e_1;\phi_2,e_2)K(\phi_2,e_2;\phi_3,e_3)d\phi_2 de_2 = K(\phi_1,e_1;\phi_3,e_3)$ or some variation of this principle?

(In fact, should 0+1D GQ reduce to 0+1D QFT without gravity? Since gravity can't really exist in 0+1D?)

On second thoughts, perhaps this is a bad example as $e(t)$ might be gauged away by a redefinition of $t$.

Answer 1

• For what it's worth, Witten discusses 1D quantum gravity (QG) in Refs. 1-2 using a relativistic point particle action $$I= \int dt ~L,\qquad L~=~\frac{1}{2e}G_{IJ} \dot{X}^I\dot{X}^J - \frac{e(mc)^2}{2},$$ with a worldline einbein field $e=\sqrt{g_{tt}}$.

• He states that 1D QG describes QFT in a (possible curved) spacetime background, cf. OP's question.

• He argues furthermore that it does not describe QG in spacetime, since there is no state-operator correspondence in 1D.

• He goes on to state that 2D QG with conformal/Weyl symmetry describes QG in spacetime, since there is a state-operator correspondence in 2D.

References:

1. E. Witten, What Every Physicist Should Know About String Theory, Physics Today 68, 11, 38 (2015),

2. E. Witten, What Every Physicist Should Know About String Theory, 100 years of GR at strings 2015, YouTube video.

Answer 2

$\newcommand{\d}{\mathrm{d}}$Essentially what you are interested in, is the path integral of the Polyakov particle on a segment, say of length $\ell$. For it to be a theory of quantum gravity, the path integral measure over the einbein is constructed so that it sums over all $\ell$. For the path integral measures to be well-defined, I will work in Euclidean signature. The relevant quantity you want to compute is $$K[\phi_0;\phi_1]:=\int \frac{\mathrm{D}e}{\mathrm{vol}(\mathrm{diff})} \int_{\phi(0)=\phi_0}^{\phi(1)=\phi_1} \mathrm{D}\phi\ \exp(-S[e,\phi]),$$ with $$S[e,\phi]:= \int \d{\tau}\left(\frac{1}{2}e(\tau)^{-1} \dot{\phi}(\tau)^2 + \frac{e m^2}{2}\right).$$ Notice that the quantity you suggested, that depends on $e(0)$ and $e(1)$ is not a good quantity in a quantum gravity theory since we must sum over all possible intervals in the end.

Now, notice that there is diffeomorphism invariance. So, we have to divide by the volume of the diffeomorphism group, to make sure we do not overcount configurations. Moreover, the einbein can be written as pure gauge, up to the gauge-invariant modulus $$\ell := \int_0^1 \d\tau\; e(\tau).$$ I.e. any $e(\tau)$ can be written as $e(\tau) = \ell\;\partial_\tau \sigma$, where $\sigma:[0,1]\to[0,1]$ is a diffeomorphism. Then the measure of the einbein can be written as $$\mathrm{D}e = \d{\ell}\; \mathrm{D}\sigma.$$ To see that, you could follow a Fadeev-Popov procedure to gauge-fix the diffeos. Following through the Faddeev-Popov procedure and calculating the functional determinants that arise from the ghost and the $\phi$ path integrals we obtain $$K[\phi_0,\phi_1]=\int_0^\infty \frac{\d{\ell}}{\ell}\ \exp\left(-\frac{(\phi_1-\phi_0)^2}{2\ell} - \frac{\ell m^2}{2}\right).$$ A couple of comments here:

1. The $\frac{\d{\ell}}{\ell}$ is not the usual $\frac{\d{\beta}}{\beta}$ you might have seen in discussions of the Polyakov particle on a circle/1d quantum gravity on a circle. That $\frac{1}{\beta}$ comes from the zero-mode of the diffeomorphisms on the circle, which is absent on the line. Here the $\frac{1}{\ell}$ comes from the functional determinant of the $\phi$ path integral.

2. If we have a $D$-dimensional curved target space, as @QMechanic suggests the answer changes to $$K[\phi_0,\phi_1]=\int_0^\infty {\d{\ell}}\ {\ell^{-\frac{D+1}{2}}}\ \exp\left(-\frac{G_{IJ}(\phi_1-\phi_0)^I(\phi_1-\phi_0)^J}{2\ell} - \frac{\ell m^2}{2}\right).$$