When we renormalize to obtain the physical mass, the $\Lambda$ dependence of the physical mass is removed by introducing the counterterms in the Lagrangian. So whether we put $\Lambda\rightarrow\infty$ or not, it does not matter. Even we can take $\Lambda\rightarrow\infty$ at the end of the calculation. But can we really take $\Lambda\rightarrow\infty$ limit (even at the end of the calculation) if a quantum field theory is not complete at high energies? If yes, what would that mean? Certainly it cannot imply that since we can take $\Lambda\rightarrow\infty$, the QED is valid at all energies. Because we know it is not. I think that this point is blurred in all QFT texts.

Moreover, since the $\Lambda$ dependence of the physical mass is removed due to the counterterms, doesn't this imply the physical mass (of the electron, say) has a constant value, independent of the cut-off?

  • $\begingroup$ Just out of curiosity: Which QFT texts are you talking about? $\endgroup$
    – Siva
    Jun 25, 2015 at 23:24
  • $\begingroup$ @ Siva- For example, in Ryder he says $\Lambda\rightarrow\infty$ limit must be taken. Look at section 9.2. Same is there in Maggiore. $\endgroup$
    – SRS
    Jun 28, 2015 at 16:12
  • $\begingroup$ Try one of the more recent books (there's been a surge of QFT textbooks in the last few years) -- one of them might have a more modern viewpoint. Alternatively, here are some of the EFT reviews; pick one you like: inspirehep.net/search?ln=en&p=65027a%3AI-c-11 $\endgroup$
    – Siva
    Jun 28, 2015 at 19:11

1 Answer 1


It's not quite correct to take $\Lambda \rightarrow \infty$, even at the end of the calculation. That comes from ancient mistaken notions that the field theory under consideration needs to describe physics upto arbitrarily small distances.

The modern way to think about this is that you're making your theory agnostic of the value of $\Lambda$.

A theory is nothing but a way to turn (a finite number of) measurements into predictions.

You're recycling a measurement at scale $E_1$ to make a prediction at $E_2$. That means that all the modes at energies larger than both $E_1$ and $E_2$ are common to the two calculations (the actual value of the cutoff is moot), and you only need to account for the corrections from the modes between the two energies, to make your prediction.

When you "renormalize" the theory by adding a counterterm such that $m^2_{\textrm{phys}} = m^2_{\textrm{bare}} + \delta m^2$, you are essentially saying that your prediction for the particle mass includes the same combination of unknown terms (which might naively contain infinities) as any other prediction (eg: scattering cross-section) you might compute. So long as you measure the former, you can recycle the estimated value of $m^2_{\textrm{phys}}$ into your prediction for the latter.

EDIT: To elaborate, if you cannot parametrize a field theory with a finite number of measurements, then it's called "non-renormalizable". In the modern way of thinking about QFT (Effective field theories), higher dimensional operators are suppressed by powers of some high energy scale. So, if you decide on the desired (finite) precision, then you can truncate your effective action and drop the higher dimensional terms. This now means that you can characterize your theory with a finite number of measurements; the only caveat being that you've restricted your precision.

  • $\begingroup$ I love this: "A theory is nothing but a way to turn (a finite number of) measurements into predictions". $\endgroup$ Jun 26, 2015 at 15:27
  • $\begingroup$ It's just something I came up with to phrase my understanding, once things finally clicked. I also love it, because it allows me to capture the sentiment :-) $\endgroup$
    – Siva
    Jun 28, 2015 at 13:53

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