IR divergences and UV divergences in perturbative QFT

[Primroses]

We know the following definitions in calculating amplitude (or a cross section) in momentum space:
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?

 

[AndyW]

I can confirm that the statement made in the forum post is correct. In perturbative quantum field theory, we often encounter UV divergences and IR divergences in our calculations. These divergences arise due to the nature of the theory and the way we perform calculations in momentum space.

UV divergences are associated with the infinity of the integration measure in momentum space. This means that as we integrate over higher and higher momenta, the integrand becomes larger and larger, resulting in an infinite value for the integral. On the other hand, IR divergences are caused by the singularity of the integrand at small momenta. This singularity arises due to the nature of the particles involved in the interaction and can lead to an infinite value for the integral when integrated over all momenta.

Now, let's consider the scenario described in the forum post. We have a Feynman graph that has a UV divergence but no IR divergence when calculated in momentum space. This means that the integral over the internal loop momenta diverges at high momenta, but remains finite at small momenta. However, when we perform a Fourier transformation to rewrite the expression in coordinate space, we will still have an integration over internal vertices. Since the UV divergence is related to the infinity of the integration measure, it will still exist in the coordinate space expression.

But, as the forum post suggests, we can assert that the resultant expression in coordinate space will only have a divergence due to the singularity of the integrand. This is because, in coordinate space, the large momenta correspond to small distances. So, the type of divergence we encounter will be similar to an IR divergence in momentum space. This can be understood intuitively, but there is also a rigorous proof for this statement in perturbative quantum field theory.

In perturbative quantum field theory, we use a technique called renormalization to deal with UV divergences. This involves introducing counterterms to cancel out the divergences in the calculations. In coordinate space, the counterterms are related to the singularity of the integrand, and thus, the resultant expression will only have a divergence due to the singularity of the integrand.

In conclusion, as a scientist, I can confirm that the resultant expression in coordinate space will only have a divergence due to the singularity of the integrand when the original expression has a UV divergence but no IR divergence in momentum space. This