1
$\begingroup$

For a $\phi^{3}$ quantum field theory, the interaction term is $\displaystyle{\frac{g}{3!}\phi^{3}}$, where $g$ is the coupling constant.

The mass dimension of the coupling constant $g$ is $1$ in 4D, which means that $\displaystyle{\frac{g}{E}}$ is dimensionless.

Therefore, $\displaystyle{\frac{g}{3!}\phi^{3}}$ is a small pertubation at high energies $E \gg g$, but a large perturbation at low energies $E \ll g$.

Terms with this behavior are called relevant because they’re most relevant at low energies.


However, I do not understand why the interaction term is called relevant if we cannot use perturbation theory at low energies (where the term $\displaystyle{\frac{g}{3!}\phi^{3}}$ is a large pertubation). Is it because quantum field theory is only applicable in the relativistic limit, where $E \gg g$ and the perturbation is small?

$\endgroup$
3
$\begingroup$

"Relevant" doesn't mean "tractable," it means "important." It means that the interaction is strong, so you can't neglect it. So a perturbative treatment will indeed usually break down. As explained in section 8.5 of these notes, "super-renormalizable [another word for relevant] interactions aren't super at all." But sometimes things are okay if the energy scale below which the theory becomes strongly coupled is smaller than the mass, because then the particles can never reach those low energies anyway.

$\endgroup$
  • 1
    $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Sep 21 '16 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.