We know the following definitions in calculating amplitude (or a cross section) in momentum space:(adsbygoogle = window.adsbygoogle || []).push({});

1, Ultraviolet divergences are due to the infinity of the integration measure;

2, Infrared divergences are due to the singularity of the integrand;

Now suppose we study a Feynman graph by calculating its value in momentum space, that is, we write it in terms of integration over internal loop momenta. Suppose it has a UV divergence but no IR divergence. Then we can rewrite it in coordinate space by a Fourier transformation of its external momenta. The result will contain an integration over internal vertices and should still be UV divergent.

My question is, can we assert that the resultant expression, has only a divergence due to the singularity of the integrand? This seems obvious because large momentum should correspond to small distance, so the type of the divergence should be like the IR divergence in momentum space. But is here a rigorous proof for ANY form of perturbative qft?

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# A IR divergences and UV divergences in perturbative QFT

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