Spacetime uncertainty makes quantum field theory finite

TL;DR

The paper proposes that considering spacetime points as quantum variables with fluctuations leads to a natural regularization of quantum field theory, making it finite and addressing issues like vacuum energy.

Contribution

It introduces a novel approach where spacetime points are quantum variables, resulting in finite quantum field theories without traditional renormalization.

Findings

Feynman diagrams become finite due to Gaussian averaging over spacetime fluctuations.

Vacuum zero-point energy density is finite but negative, challenging dark energy explanations.

The approach suggests a new perspective on quantum fields as functions of quantum coordinates.

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.