In the process of renormalization, regularization is usually cited as indispensable in taming infinities encountered in quantum field theory. Is explicit regularization really necessary?

Let's take for example the fermion propagator $$ G = \frac{i}{\not{p}-m_0 - \Sigma(\not{p})+i\epsilon} = \frac{i}{(1-b(p^2))\not{p}-(m_0 + a(p^2)) + i\epsilon}, $$ where self energy is expressed as $$ \Sigma(p) = a(p^2) + b(p^2)\not{p}. $$ The propagator has a pole at $$ (1-b(p^2))^2p^2-(m_0 + a(p^2))^2 = 0 $$ that is $$ p = m_p = \frac{m_0 + a(m_p^2)}{1-b(m_p^2)}, $$ where $m_0$ is bare mass (infinite) and $m_p$ is the physical mass (finite).

One may rearrange the above Fermion propagator via introducing modified self energy $\hat{\Sigma}(\not{p})$ so that $$ G = \frac{iZ}{\not{p}-m_p - \hat{\Sigma}(\not{p})+i\epsilon}, $$ where $\hat{\Sigma}(\not{p})$ is defined as $$ Z^{-1}\hat{\Sigma}(\not{p}) = [a(p^2)-a(m_p^2)] + [b(p^2)-b(m_p^2)]\not{p}, $$ and $$ Z = \frac{1}{1-b(m_p^2)}. $$

Note that the difference $a(p^2)-a(m_p^2)$ is finite, even though $a(p^2)$ and $a(m_p^2)$ are individually infinite. If we follow the regime of sticking with finite differences (i.e. $a(p^2)-a(m_p^2)$) and measurable quantities (i.e. $m_p$) only, then the explicit regularization schemes (such as the widely used dimensional regularization) are not needed at all.

Of course, one can follow the same procedure at a different energy scale (renormalization scale $\mu$), rather than at the physical mass scale ($m_{p}$).

An added note on the difference between two infinite quantities. Take the following example, $$ \int_{0}^r \frac{1}{x}dx - \int_{0}^{r_0} \frac{1}{x}dx = \int_{r_0}^r \frac{1}{x}dx = \ln(\frac{r}{r_0}). $$ Hard core mathematicians will be leery of the first step and demand some form of regularization. Do physicists, while not fazed by the lack of mathematical rigor with things like path integral, really need a formal explicit regularization to arrive at the final result?

One may call the above procedure implicit regularization. Similar idea has already been picked up by some researchers (see Jackiw's approach, approach of the Australian school, and approach of the Brazilian school) though in a different fashion as framed here. The merit of implicit regularization is that it circumvents various pitfalls besieging explicit regularization, e.g. violation of gauge invariance in cutoff regularization or the $\gamma^5$ issue in dimensional regularization.

So my question is:

  1. In light of the vice of explicit regularization mentioned above (moreover, given the intricacies and pitfalls, explicit regularization is often perplexing rather than elucidating when renormalization is taught in text books), shall we skip it in the process of renormalization?
  2. How do the different implicit regularization schemes (Jackiw, the Australian school, and the Brazilian school) stack against each other?
  • 2
    $\begingroup$ Physicists may not be concerned with explicit rigor, but they aren't sloppy. You might say that they're worried about the content rather than the notation. $\endgroup$
    – user1504
    Commented Jun 21, 2018 at 14:01
  • 2
    $\begingroup$ The implicit cancellation of $\ln 0 - \ln 0$ is regularisation, it's just you've done it informally. Both are divergent quantities but you've "held" them at the unresolved form $\ln 0$ and cancelled them. $\endgroup$
    – JamalS
    Commented Jun 21, 2018 at 14:18
  • 1
    $\begingroup$ Implicit regularisation works fine for scalar and spinor theories, but it fails for gauge theories. I invite you to compute the photon self-energy without introducing an explicit (and gauge-invariant) regulator, and check whether the Ward identity is satisfied. Spoiler alert: it is not. $\endgroup$ Commented Jun 21, 2018 at 15:16
  • $\begingroup$ As a matter of fact, one may imagine a theory where the perturbative corrections to the fundamental constants are finite. The real problem is not in their values. The real problem is in their existence - they are not necessary at all, they are harmful, so they must be subtracted. And this still does not guarantee the agreement with experiment. $\endgroup$ Commented Jun 21, 2018 at 15:37
  • $\begingroup$ @AccidentalFourierTransform Please do a search for key word "implicit regularization". Someone else has already picked up similar idea, though in a different fashion as framed in my question. Spoiler alert: Ward identity is satisfied. $\endgroup$
    – MadMax
    Commented Jun 21, 2018 at 15:44

2 Answers 2


I will address your comment on regularisation at the end. Note that in performing,

$$\int_0^r \frac{\mathrm dx}{x} - \int^{r_0}_0 \frac{\mathrm dx}{x} = \ln \frac{r}{r_0}$$

you are implicitly cancelling the divergent terms $\ln 0 - \ln0$. Indeed, one does not need to formally introduce a regularisation scheme, you can do it in this handwavy way.

However, one can arguably say you have sort of regularised things, since you keep $\ln0$ in its unevaluated form, precisely so that you can cancel it with the other logarithm. The key point to take from this is there is some need for regularisation, in the broader sense, of being able to tame infinities into a form that allows for algebraic reasoning to cancel them.

  • $\begingroup$ Please see the added comments at the end of the question on implicit vs explicit regularization. $\endgroup$
    – MadMax
    Commented Jun 21, 2018 at 15:06

If you don't have some kind of regularization scheme, then there's a problem with saying that the difference of two infinite values is finite. Such a difference is not defined. You need to take some kind of approach to deal with this. You can define it, for example, as some kind of limit. But that is then an explicit regularization scheme.

Another option is to introduce some characteristic to your theory that makes everything finite. Then you calculate the values you are interested in. If the results don't depend on the characteristic, you can set it back to "zero". One that I have rather fond feelings for (for shameless non-physics reasons) is this.


The basic scheme is, introduce non-locality, and notice that the integrals are now finite. Work out the values. Notice that the results can be arranged not to depend on the non-locality order by order. Then at the end, turn off the non-locality.

  • $\begingroup$ With regard to the mathematical rigor of the difference between two infinite values, please see the added note to the question. $\endgroup$
    – MadMax
    Commented Jun 21, 2018 at 3:50

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