Why is QCD not perturbative at low energies?

[Safinaz]

Hi all ,

I'm confused about the definition of a perturbative coupling for QFT that it should be less than 4 ## \pi ## , because the higher order corrections comes of order ## \lambda / ( 4 \pi ) ## ..

Now why QCD is not perturvative at low energy because the coupling constant approaches just 1 at this regine ?

 

[AndyW]

Hello,

I can understand your confusion regarding the definition of a perturbative coupling in quantum field theory (QFT). Let me start by explaining what a perturbative coupling is. In QFT, we use Feynman diagrams to represent interactions between particles. These interactions are described by a coupling constant, denoted by ##\lambda##. In perturbative QFT, we assume that the coupling constant is small, i.e. ##\lambda \ll 1##. This allows us to expand the calculations in powers of ##\lambda##, known as perturbation theory.

Now, coming to your question about why QCD is not perturbative at low energy when the coupling constant approaches 1. This is because QCD is a non-Abelian gauge theory, which means that the coupling constant is not a constant but rather a function of the energy scale. At high energies, the coupling constant is small and perturbative calculations are valid. However, at low energies, the coupling constant increases and perturbative calculations break down. This is known as asymptotic freedom.

In QCD, the coupling constant is given by ##\alpha_s = \frac{g^2}{4\pi}##, where ##g## is the strong coupling constant. As the energy scale decreases, the value of ##\alpha_s## increases and at low energies, it approaches 1. This is why QCD is not perturbative at low energies.

I hope this helps to clarify your confusion. Let me know if you have any further questions.