Does one need the O(epsilon)- and O(epsilon^2)-terms of one-loop amplitudes in an NNLO calculation ?

TL;DR

The paper analyzes how NNLO calculations incorporate one-loop amplitudes and proves that the potentially required $O(\varepsilon)$ and $O(\varepsilon^2)$ pieces of one-loop amplitudes cancel in the final result. It shows that, given a method to obtain the $O(\varepsilon^0)$ finite remainder of the two-loop amplitude (and the corresponding one-loop remainder), only tree-level amplitudes and finite remainder terms are needed. This reduces the problem to four-dimensional numerical evaluation of finite remainders, compatible with subtraction- and unitarity-based techniques. The result provides a practical framework for NNLO computations and guides future numerical implementations.

Abstract

This article discusses the occurences of one-loop amplitudes within a next-to-next-to-leading order calculation. In an NNLO calculation the one-loop amplitude enters squared and one would therefore naively expect that the O(epsilon)- and O(epsilon^2)-terms of the one-loop amplitudes are required. I show that the calculation of these terms can be avoided if a method is known, which computes the O(epsilon^0)-terms of the finite remainder function of the two-loop amplitude.