Spacetime uncertainty makes quantum field theory finite

Abstract

Since Einstein's equations $G_{ij} = 8\pi \, G \, T_{ij} \, / c^4 $ relate the metric $g_{ij}$ of spacetime to the energy-momentum tensor $T_{ij}$ which is a quantum field, the metric $g_{ij}$ must be a quantum field. And since the metric $g_{ij}(x)$ is the dot product $g_{ij}(x) = \partial_i p^\alpha(x) \, \partial_j p_\alpha(x)$ of the derivatives of the points $p(x)$ of spacetime, spacetime must be a quantum field. Its points have average values $\langle p(x) \rangle$ that obey general relativity and fluctuations $q(x) = p(x) - \langle p(x) \rangle$ that obey quantum mechanics. It is suggested that the fields of quantum field theory be regarded not as functions $\phi(x)$ of their classical coordinates $x$ but as functions $\phi(p(x))$ of their quantum coordinates $p(x)$. In empty flat spacetime where $p(x) = x + q(x)$ and $x = (t, \boldsymbol x)$, the Fourier exponentials $\exp(i k(x+q(x))$ averaged over normally distributed fluctuations $q(x)$ are gaussians $\exp(i kx -\ell^2 \boldsymbol k^2 - \ell^2 m^2/2)$. These gaussians make Feynman diagrams finite. The zero-point energy density of the vacuum also is finite -- but negative and too large to explain dark energy unless new bosons exist.