Some related questions on Renormalization:

  1. Why is renormalization even necessary? My understanding is that the supposed problem is that the sums of certain amplitudes end up being infinite. But actual observations are made at the level of probabilities, not amplitudes. Once we actually make an observation, all the different possible results of a particular observable "collapse" into a single observed result that occurs with a certain probability and that is always finite. So if we always get a single finite result, why do we care that there are infinities in the equations at the level of amplitudes? The only reason I can think of is that "purifying" the equations of these infinite amplitudes makes it possible for us to make realistic computations and predictions. Is that the practical reason for renormalization? If so, can anyone give an actual example of such a computation/prediction that requires renormalization?

  2. The standard description of what charge renormalization represents physically is that the supposedly infinite "bare charge" of, say, an electron gets screened by an infinite number of virtual positrons, leaving a small finite net observed negative charge. OK, this makes a certain amount of sense (with one qualification I'll get to). However, there doesn't seem to be an equivalently reasonable physical picture for mass renormalization. If the mechanism is similar to that of charge screening, then presumably an infinite positive mass would have to be "screened" by an infinite negative mass, leaving a small positive remainder (or the bare mass would have to be negative and screened by a positive mass). Some sources on renormalization imply that this is what happens; other sources do not. If this is actually what's supposed to happen, what are the specifics of this mechanism, and what possible sense could it make to talk about a "negative mass" given that we never observe any such thing? If some type of "mass screening" is not what's going on here, is there any other easily understood physical explanation?

  3. Even if we can come up with good physical pictures for both charge and mass renormalization, the business of "cancellation of infinities" seems mathematically nonsensical. In ordinary mathematics, you can't do ordinary arithmetic on infinities. If I add 2 infinities of the same order, I don't get something twice as infinite, I get one infinity of the same order: I + I = I. And if I subtract one infinity from another I don't get a finite result, I just get the same infinity. So is there a mathematically reasonable way of describing what happens during this "cancellation of infinities" or are we just presuming that the ordinary mathematical rules don't apply to the physics? And what possible warrant would we have for such a presumption?

  • $\begingroup$ Just a small remark on the "mathematics of infinities": if you sum two infinite cardinal numbers, you get the bigger of the two; however if you sum two ordinals the result depends on the order of summation. And if you sum two infinite surreal numbers, then you almost do the same as with finite numbers. $\endgroup$
    – yuggib
    Jun 23, 2015 at 17:00
  • $\begingroup$ Excellent question, to point number 3; yes, it looks like , to me, we do throw away any ordinarary math laws and I have never met a physicist who is happy with the normalisation process, but they are happy it does work, in that predictions are very precisely confirmed once you play by the new "rules". But we currently have perform math contortions, saying for example, that the electron is a dimensionalless point. Good luck with your question. $\endgroup$
    – user81619
    Jun 23, 2015 at 17:12
  • $\begingroup$ More on $\infty$ "math" via Mathematics: math.stackexchange.com/q/36289, math.stackexchange.com/q/37327 and math.stackexchange.com/q/60766 (probably more) $\endgroup$
    – Kyle Kanos
    Jun 23, 2015 at 17:31
  • $\begingroup$ More on subtraction physics: bayes.wustl.edu/etj/articles/prob.in.qm.pdf, part "Classical subtraction physics" on page 15. $\endgroup$ Jun 23, 2015 at 20:41

2 Answers 2


Actually there are no infinities in physics. These "infinities" come from integrating over loop momenta when the momentum goes to infinity. But we don't know what actually happens to particles when they have infinite momentum. Maybe they turn into strings, maybe loop quantum gravity and black holes become relevant, maybe there's infinitely small unicorns guiding the particles... There is no way to tell without building even bigger accelerators.

The point of renormalization is that all this does not matter for the physics we do. If you introduce a cut-off at energies higher than what accelerators could possibly measure, you find that --- in a renormalizable theory --- all dependence on the cut-off goes away if the cut-off is simply largish. So physicists like to imagine that the cut-off is infinity, because that makes things easier. In reality the cutoff if finite, but not yet known, and in the Standard Model it is sufficiently high as to make to difference when we put it to infinity.

A beautiful example is QED. It has been argued that, whatever the bare charge be (up to infinity), the physical charge will go to zero when taking the cut-off to infinity. The solution is, of course, that beyond the electroweak unification scale QED is just a part of the electroweak theory. QED has a physical cut-off at the electroweak unification scale. If you do low-energy physics, you can ignore that, just do the usual renormalization procedure and set the cut-off equal to infinity at the end (because the actual cut-off is sufficiently infinity as far as low-energy physics are concerned). But you cannot go all philosophical without taking into account that QED is part of a larger whole.

The Standard Model will, some day, also be part of a larger whole, and maybe some day that larger whole will be able to stand on itself. But we don't know what it is yet, so for now don't worry about it.


1.why is renormalization even necessary?

It is because, experiments force us to do that. You suppose to get finite values i.e, for charge or for mass of a particle.

2.if we always get a single finite result, why do we care that there are infinities in the equations at the level of amplitudes?

Actually we don't. Your integrals in the loop corrections, would be infinite and the way you get rid of them is counter-terms. You cancel the infinite part of the loop corrections with counter-terms and then finite parts still remains. In QED. after regularization and cancellation of infinities, you will get something like $M=iC\lambda^2 log(\frac{\mu^2}{s})$. M amplitude, C, numerical numbers, s, center of mass energy and $\mu^2$ is what we put cut off for energy scale.But do not forget, this is what we got after regularization and cancel the infinite parts with the counter-terms. If one wants to calculate the amplitude, need to know cut off scale. For an experimentalist, $\mu$ makes perfect sense that any given theory has a finite domain of validity. But what is the precise value? If someone wants to change $\mu$ what happens? M is suppose to be finite and should not depend on value $\mu$. On the other hand, we also do not know the value of $\lambda$. What is $\lambda$? It is basically a coupling constant, here not a physical parameter. If someone wants to change $\mu$, one just shift the $\lambda$ in such a way that M does not change!So $\lambda$ is secretly a function of $\mu$. This gives rise to RGE. But this is not the topic here. We have still no information about cut off scale. How can we get rid of it?

Here $\lambda$, as I said in the first paragraph, not a physical value. And the expression above, depends on an arbitrary scale $\mu$.In observables, one need to define $\lambda$ as a physical parameter(observable). I mean, If we renormalize the theory at some scale i.e, $s_0$ then the cut off scale vanishes and the lagrangian comes out as, $$ L-L_0=iC\lambda^2 log(\frac{s_0}{s}) $$ Then you can be sure that when the scattering amplitude is expressed in terms of physical coupling constant $\lambda_p$, the cut off disappear completely! And all the values remain physical and observable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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