# What is the relation between renormalization in physics and divergent series in mathematics?

The theory of Divergent Series was developed by Hardy and other mathematicians in the first half of the past century, giving rigorous methods of summation to get unique and consistent results from divergent series. Or so.

In physics, it is said that the pertubative expansion for the calculation of QFT scattering amplitudes is a divergent series and that its summation is solved via the renormalisation group.

Is there some explicit connection between the mathematical theory and the physics formalism?

Perhaps the question can have different answers for two different layers: the "renormalized series", which can be still divergent, and the structure of counter-terms doing the summation at a given order. If so, please make clear what layer are you addressing in the answer. Thanks!

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What the renormalization group resums, in most applications, are only those terms in the perturbation series that involve large logarithms. This is not a solution to the problem Dyson pointed out that implies that the series as a whole is only asymptotic, not convergent. Physical attempts to construct means of summing asymptotic series are a different story from the renormalization group, and one I can't tell you much about. (There's an old argument by 't Hooft that perturbation series in QCD are not Borel-summable, for instance, which might be the first thing one would have attempted.) – Matt Reece Jan 16 '11 at 22:00
Yep, the upper layer, the whole series, is said to be asymptotic, I forgot to mention Dyson. Modernly, I do not know what is really proved, and if it is proved for the bare or the renormalized series, so please consider it as part of the question. As for the contributions for a fixed power in the expansion, it is not even argued to be related to an asymptotic series, is it? Still, it has structure (Connes Kreimer etc)... – arivero Jan 16 '11 at 22:32
Lubosh wrote: "No, Vladimir, you can't throw the quantum corrections altogether..." But I do not propose to discard all quantum corrections altogether! My question is: Can we consider renormalizations as discarding unnecessary corrections to the initially physical fundamental constants? In other words, doesn't discarding give the same result? Isn't it the same as adjusting (fitting) the constants? There is no linearization in it. – Vladimir Kalitvianski Jan 17 '11 at 17:30
@Vladimir: Converted your question to a comment; hope you don't mind. – Noldorin Jan 17 '11 at 17:31
this topic has always given me headaches to understand, and i am glad someone could articulate it into a question – lurscher Jul 10 '11 at 18:31

You are conflating three conceptually different categories of "regularizations" of seemingly divergent series (and integrals).

The type of resummations that Hardy would talk about are similar to the zeta-function regularization - the example that is most familiar to the physicists. For example, $$S=\sum_{n=1}^\infty n= -\frac{1}{12}$$ is the most famous sum. Note that this result is unique; it is a well-defined number. In particular, that allows one to calculate the critical dimension of bosonic string theory from $(D-2)S/2+1=0$ and the result is $D=26$. Fundamentally speaking, there is no real divergence in the sum. The "divergent pieces" may be subtracted "completely".

However, in the usual cases of renormalization - of a loop diagram - in quantum field theory, there are divergences. Renormalization removes the "infinite part" of these terms. A finite term is left but the magnitude of the term is not uniquely determined, like it was in the case of the sum of positive integers. Instead, every type of a divergence in a loop diagram produces one parameter - analogous to the coupling constant - that has to be adjusted. Because the finite results can be "anything", this is clearly something else than the zeta-regularization and, more generally, Hardy's procedures whose very goal was to produce unique, well-defined results for seemingly divergent expressions. Infinitesimally speaking, the Renormalization Group only mixes the lower-order contributions (by the number of loops) into a higher-order contribution.

So these are two different things that one should distinguish.

There is another category of problems that is different from both categories above: the summation of the perturbative expansions to all orders. It can be demonstrated that in almost all fields theories - and perturbative string theories as well - the perturbative expansions diverge. For a small coupling, one can sum them up to the smallest term, before the factorial-like coefficient begins to increase the terms again, despite the $g^{2L}$ suppression. The smallest term is of the same order as the leading non-perturbative contributions.

At the very end, if the theory can be non-perturbatively well-defined - and both QCD-like theories and string theory can, at least in principle - the full function as a function of the coupling constant $g$ exists. But it just can't be fully obtained from the perturbative expansion. The Renormalization Group won't really help you because it only mixes the perturbative terms of another order to a perturbative diagram you want to calculate. If you don't know the non-perturbative physics, the equations of the Renormalization Group won't fill the gap because they will keep you in the perturbative realm.

So I have sketched three different things: in the Hardy/zeta problems, the answer to the divergent series was unique; in the particular $L$-loop diagrams in QFT, it wasn't unique but the infinite part was subtracted and the finite part was obtained by a comparison with the experiments; and in the perturbative expansion resummed to all orders, the sum actually didn't converge and indeed, it didn't know about all the information about the full result for a finite $g$.

The last statement may have some subtleties; at least for some theories, the non-perturbative physics is fully determined by the perturbative physics. But I think it is not quite general and we have counterexamples - e.g. for AdS/CFT with orthogonal groups and different discrete values of $B$ etc. So it means that the perturbative expansion doesn't uniquely determine the theory non-perturbatively.

Because the three examples differ at the level of "what can be calculated" and "what cannot", they are different.

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I was only expecting to conflat two concepts, the renormalization of a diagram and the sum to all orders, and in fact I suggested, in the question body, that people should indicate which of the concepts (I told layers) was being addressed in each answer. Regretly my mention of Hardy brought the third concept (the first in your answer) into play too, it seems. – arivero Jan 16 '11 at 22:46
Lubosh wrote: "Renormalization removes the "infinite part" of these terms. A finite term is left but the magnitude of the term is not uniquely determined... Instead, every type of a divergence in a loop diagram produces one parameter - analogous to the coupling constant - that has to be adjusted." My question to Lubosh is: Can we consider renormalizations as discarding unnecessary corrections to the initially physical fundamental constants? In other words, doesn't discarding give the same result? – Vladimir Kalitvianski Jan 17 '11 at 4:22
No, Vladimir, you can't throw the quantum corrections altogether because the resulting theory wouldn't be unitary: it wouldn't conserve probabilities and they would no longer sum to one. For example, if one computes the scattering S-matrix, it just ends up being the time-ordered exponential of the integral of the Hamiltonian (or Lagrangian) - an exponential is what solves similar equations - which is a nonlinear function of the Hamiltonian, allowing any number of vertices. $\exp(iH)$ for a Hermitean $H$ is unitary; a linearization of it wouldn't be unitary. – Luboš Motl Jan 17 '11 at 9:12
Lubosh wrote: "No, Vladimir, you can't throw the quantum corrections altogether..." But I do not propose to discard all quantum corrections altogether! My question is: Can we consider renormalizations as discarding unnecessary corrections to the initially physical fundamental constants? In other words, doesn't discarding give the same result? Isn't it the same as adjusting (fitting) the constants? There is no linearization in it. – Vladimir Kalitvianski Jan 19 '11 at 22:54

There are overlaps between QFT renormalization problems and mathematical approaches to divergent series summations (see for instance the section 'physics' in the Wikipedia article $1 + 2 + 3 + 4 + \dots$ ) However, most of the infinities encountered in modern physics are 'harder' than those that can be attacked by existing divergent series summation methods.

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Here, is the method to deal with DIVERGENT INTEGRALS (single and multiple) in quantum physics by using the method of DIVERGENT SERIES and Zeta regularization

http://vixra.org/pdf/1009.0047v4.pdf

http://vixra.org/pdf/1009.0047v4.pdf

for multiple integrals we can apply the method on each variable, so we can get FINITE RESULTS using zeta regularization for multi-loop integrals.

The logarithmic divergence is handled with the FUNCTIONAL DETERMINANT or infinite product over all the integers (1+x)(2+x)........ using Hurwitz Zeta function.

so we have managed to get only finite results in QFT :) using Zeta regularization.

not only for DIVERGENT SUMS but also for DIVERGENT INTEGRALS we can obtain finite corrections using the Zeta regularization algorithm

my paper in the subject

http://vixra.org/abs/1009.0047

this can be used to regularize DIVERGENT INTEGRALS in quantum mechanics, and MULTIPLE integrals by doing a term by term integration on each variable, you are free to check it out and give your opinion.

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i think DIMENSIONAL REGULARIZATION is hidding the power of zeta function because physicist prefer dimensional regularization over zeta function regularization for divergent integrals – Jose Javier Garcia Feb 1 '14 at 20:34
can someone help me to publish my paper on divergent series :( i need help :'( so te relation between integral and divergent series will not be forgotten – Jose Javier Garcia Nov 9 '15 at 18:55