The bare electron mass $m_0$, in QED, changes as $$m_0\to m=m_0+\delta m\Big(\frac{\Lambda}{E}\Big)$$ where high momentum modes from $E$ to $\Lambda$ has been integrated out.

What scale does the cut-off $\Lambda$ stand for in the theory of QED and why? Is it the top quark mass $\Lambda=m_{top}$, the GUT scale $\Lambda=M_{GUT}$ or the Planck scale $\Lambda=M_{Pl}$? I never understood which scale corresponds to the correct cut-off of a theory.


1 Answer 1


There is usually no unique cutoff scale $\Lambda$ in renormalization. The reason is that generally we don't know what the ultimate microscopic physics is.

So the rationale is to pick any scale $\Lambda$ to be much, much larger than any physical scale of interest (particle mass or energy $E$ of an experiment you're doing) and then adjust couplings as you lower $\Lambda$ - the usual RG flow story.

In some cases you do get information about the range of validity of a theory. Say you have a theory with some fields $\Phi_i$ which couple to a heavy particle $X$ of mass $M$, and you integrate out the heavy particle. Then you obtain an effective action for the $\Phi_i$ fields which is valid up to energies $E < M$. This is reflected by the fact that you generate dimensionful couplings of size $M$ to the appropriate power. A physically interesting example is the chiral Lagrangian for pion physics: $$L = \frac{f_\pi^2}{2} \text{Tr}(\partial_\mu \Sigma \partial_\mu \Sigma^\dagger) = \frac{1}{2} | \partial \vec{\pi}|^2 + \frac{1}{f_\pi^2} \left[\vec{\pi}^2(\partial \vec{\pi})^2 - (\vec{\pi} \partial \vec{\pi})^2 \right] + \ldots $$ All couplings scale like the pion decay constant $f_\pi$, and this action is useful to compute $\pi \pi$ scattering at low energies but breaks down below $\Lambda_\text{QCD}$. QED is however not of this form: it has a single, dimensionless coupling $\alpha$, which clearly doesn't carry any information about scales. Moreover, QED isn't a theory of quarks, gravity or weak interactions, so there's no way to tie $\Lambda$ to $m_\text{top}$, $M_\text{GUT}$ or $1/\ell_\text{pl}$.

  • $\begingroup$ If GUT is true QED even in the broadest sense (as the $U(1)$ sector of the Standard Model, I don't think people that often call that QED) will not longer work at $m_{GUT}$ $\endgroup$
    – OON
    Nov 1, 2017 at 18:20
  • $\begingroup$ If we don't introduce any new term in the Lagrangian and calculate the corrections to the parameters of the theory using the same Lagrangian, we must have to be sure that the theory is valid up to that scale. For example, we experimentally know that the Standard Model (SM) is valid up to a few TeV. Shouldn't that mean the cutoff of SM must be a few TeV in order to calculate the corrections to masses and couplings of the SM. $\endgroup$
    – SRS
    Nov 1, 2017 at 18:24
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    $\begingroup$ In principle you can make the cutoff $\Lambda$ as low as you want, keeping it only slightly higher than the scale you're probing: if you're doing an experiment at $E = 14$ TeV, then as a matter of principle you could take $\Lambda = 18$ TeV or whatever as long as you adjust the bare couplings $g_0(\Lambda)$ in the action according to the RG equations. In practice this is not what we do: we trade in the bare couplings for renormalized values (like $m_\text{higgs}$ = 125 GeV) and we send the cutoff to infinity, which is possible in renormalizable theories. $\endgroup$
    – user159249
    Nov 1, 2017 at 20:45

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