# Phase space suppression in loop integrals

Often (if not always) when calculating loop integrals in QFT one encounters extra 2 $\pi$'s that serve to suppress higher order corrections more so than the most naive guess would give. This happens to the extent that the fine structure constant is defined as $\frac{e^2}{4\pi}$ in anticipation of the fact that get this extra factor beyond the $e^2$ when we first start doing our perturbation theory after reading the Feynman rules from the Lagrangian. That is, people are so used to these extra factors we anticipate them and go ahead and define our perturbation theory in alpha instead of just $e^2$.

Is there a deep reason for this? Will this happen in all diagrams for scalar, vector or Dirac particles? This seems like something worth understanding since it will determine if a theory stays within the perturbative window for larger values of the coupling constant appearing in the Lagrangian than the most naive guess would imply.

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Isn't this just a matter of the units used...? The wikipedia page defines $\alpha=\frac{k_e e^2}{\hbar c}$ where $k_e$ is the Coulomb constant. – Vijay Murthy Apr 11 '12 at 16:32

It seems to me that these $\pi$'s come from the transition between real-space and $k$-space. Historically, the charge $q$ definition comes from the Coulomb potential (single boson exchange) in the real space, where it is naturally to set $V(k)=q^2/r$ (as CGS does). Now if you to the momentum space, you inevitable get these extra $\pi$'s between the coupling constant $\alpha$ and $q^2$.
@user793 I am not sure I really get what is your concern about. Anticipation or not, is not it just a matter of choosing what exactly to call $\alpha$? It's an aesthetic choice, beautiful in some contexts and ugly in other. Not sure there anything deeper than, e.g., $\int_{-\infty}^{+\infty} e^{-x^2} dx=\sqrt{\pi}$. – Slaviks Apr 11 '12 at 18:56