# A Is renormalization the ideal solution?

#### A. Neumaier

There are many works in dressed/clothed particle theory. They usually pay tribute to the seminal paper

O. W. Greenberg, S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim. 8 (1958), 378.

You can use Google Scholar to find all citations of this work. Currently this search shows 113 results.

Eugene.
Why don't you say that the best exposition is in your [meomepuk = Eugene Stefanovich] 3 volume treatise on quantum electrodynamics ? I think this is a valid statement. Your treatise shows both the potential and the limitations of the dressed particle approach.

It uses perturbatively constructible (but rigorously ill-defined) ''unitary'' transformations to renormalize standard perturbative QED. It misses, like any purely perturbative treatment of QFT, the infrared aspects of the theory. The faults show up in your version by predicting small superluminal effects.

4. This Hamiltonian is relativistically covariant: there exists a corresponding interacting boost operator such that all commutators of the Poincare Lie algebra are satisfied.
No. You didn't prove the existence of the boost operator in your version of QED; you only give a perturbative construction for it, without showing its convergence. The superluminal effects you inherit are proof of the lack of true Lorentz invariance.

The devil is in the details, and getting the details right (as Glimm and Jaffe did in 2D) requires abandoning the Fock structure.

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#### vanhees71

Gold Member
No. His book is called quantum field theory, and particles don't play a big role. Moreover, he formulated only perturbative scattering theory, which doesn't get rid of all infinities since the perturbative series diverges.
Well Weinberg in his books takes explicitly the particle picture as fundamental. It's not so simple though. He does not discuss the IR problem from the "infraparticle picture" either but uses the conventional "ladder summation" approach to deal with the "soft-photon problem". It's also not a book about axiomatic QFT but a physics book with all the sloppiness used by physicists to use perturbative QFT to describe particles, but indeed he presents the "particle point of view" right from the very beginning.

#### A. Neumaier

Well Weinberg in his books takes explicitly the particle picture as fundamental. It's not so simple though. He does not discuss the IR problem from the "infraparticle picture" either but uses the conventional "ladder summation" approach to deal with the "soft-photon problem". It's also not a book about axiomatic QFT but a physics book with all the sloppiness used by physicists to use perturbative QFT to describe particles, but indeed he presents the "particle point of view" right from the very beginning.
He presents this view, but as an asymptotic one, not as a fundamental one. This is obvious from his preface, where particles appear only in two sentences - as ingredients for the unitary representations of the Poincare group, and as a collection of free particles in the far past.

Fundamental are the quantum fields, but they need a motivation, which is given through the picture of free particles. See also his paper ''What is quantum field theory, and what did we think it is?'', where he says on p.2,
Steven Weinberg said:
In its mature form, the idea of quantum field theory is that quantum fields are the basic ingredients of the universe, and particles are just bundles of energy and momentum of the fields.

#### Geonaut

I don't take Feynman as the ultimate authority.
I love Feynman, but the man didn't believe in brushing his teeth or even washing his hands after using the bathroom... so I'm with you there, otherwise, I wouldn't be very hygienic. Actually, I don't think even the physics community should be seen as the ultimate authority. Humans are smart, but we're not all knowing. Which is why I'm still not 100% sure about renormalization, although it seems like I'll come to accept it.

@A. Neumaier so to be clear, it sounds like you're saying that we should abandon the idea of a Fock space, and that this means that particles don't exist, right? If particles are "just bundles of energy and momentum of the fields" then are you claiming that bundles of energy and momentum don't exist? Perhaps the better question is: How does constructing a QFT in a non-Fock space change our interpretation of things?

naively quantized gravity is non-renormalizable.
Wow, are you saying that there is a way to quantize gravity that gives us a renormalizable theory? How have I never heard of that before. What's this theory called, and does it have any problems within it?

#### meopemuk

Why don't you say that the best exposition is in your [meomepuk = Eugene Stefanovich] 3 volume treatise on quantum electrodynamics ? I think this is a valid statement. Your treatise shows both the potential and the limitations of the dressed particle approach.
Thank you for the generous introduction of my humble book.

It uses perturbatively constructible (but rigorously ill-defined) ''unitary'' transformations to renormalize standard perturbative QED. It misses, like any purely perturbative treatment of QFT, the infrared aspects of the theory. The faults show up in your version by predicting small superluminal effects.
I deliberately avoided infrared aspects. Otherwise I would have to write a 4th volume of the book. The handling of IR divergences in QED is well understood, so I do not see any problem with applying the same ideas within the dressed particle approach. For experimentally measured things like the anomalous magnetic moment or the Lamb shift, IR divergences do not show up or cancel out. I've explicitly demonstrated this cancellation in the book.

The superluminal effects are not bugs, but unavoidable features of the dressed particles approach. In this approach we eliminated fields as carriers of interactions. Then the energy-momentum conservation implies that the energy-momentum exchange between particles (=interaction) occurs without retardation. This is not an approximation, but a solid prediction of the theory. It appears that this prediction was confirmed in the experiment of G. Pizzella et al. If you think that instantaneous Coulomb potentials violate relativity or causality, then please read this paper, which proves otherwise:

E. V. Stefanovich, "Does Pizzella's experiment violate causality?", J. Phys. Conf. Series, 845 (2017), 012016.
PDF file

No. You didn't prove the existence of the boost operator in your version of QED; you only give a perturbative construction for it, without showing its convergence. The superluminal effects you inherit are proof of the lack of true Lorentz invariance.
In Appendix E.2 of volume 2, I copied Weinberg's proof that 10 Poincare generators of QED $(\boldsymbol{P}_0, \boldsymbol{J}_0, \boldsymbol{K}_0 + \boldsymbol{Z}, H_0+V)$ satisfy the required commutation relations. This proof is non-perturbative. Ten generators of the dressed theory are obtained from QED generators by means of a unitary "dressing" transformation. This means that all commutators are preserved and the dressed theory is rigorously relativistic.

In practice, this unitary transformation can be performed only at low perturbation orders, so -- you are right -- some violations of the Poincare invariance are to be expected. This is not different from the Poincare-non-invariance of QED in finite orders. However, as I explained above, predictions of superluminal effects would remain in the dressed theory even non-perturbatively.

Eugene.

#### meopemuk

Renormalization in all cases studied so far in constructive detail seems to render Hamiltonians finite, so there is no reason to view it as problematic.
Do I understand it correctly that this field-based theory has not advanced enough to handle the Hamiltonian of the hydrogen atom?

Eugene.

#### haushofer

I feel another Insight being born here :P

#### DarMM

Gold Member
Do I understand it correctly that this field-based theory has not advanced enough to handle the Hamiltonian of the hydrogen atom?

Eugene.
The non-relativistic Hamiltonian is well defined and self-adjoint. In QED there isn't really a Hamiltonian for hydrogen, just the QED Hamiltonian.

#### DarMM

Gold Member
Wow, are you saying that there is a way to quantize gravity that gives us a renormalizable theory? How have I never heard of that before. What's this theory called, and does it have any problems within it?
It's possible that gravity is asymptotically safe, so that gravity quantized with extra terms may be well defined. However it is currently unknown if that is true.

#### meopemuk

The non-relativistic Hamiltonian is well defined and self-adjoint. In QED there isn't really a Hamiltonian for hydrogen, just the QED Hamiltonian.
I am OK with that. In the dressed theory there is also a full Hamiltonian $H= H_0+V$ in the entire Fock space. This Hamiltonian is obtained by a unitary transformation from the "QED Hamiltonian". $H$ is a normally-ordered polynomial in particle creation and annihilation operators, like in eq. (4.4.1) of Weinberg's vol. 1. This polynomial is Poincare invariant, cluster separable and has huge (infinite) number of different terms. However, if we are interested only in the hydrogen atom (which I define as a bound state of two particles -- an electron and a proton), we don't need the full interaction $V$, we can choose only few terms that can be relevant to this physical system.

Let me introduce the following creation/annihilation operators: $a^{\dagger}, a$ for electrons, $d^{\dagger}, d$ for protons, $c^{\dagger}, c$ for photons. Then in a pretty decent approximation, the part of the polynomial $V$ that is relevant to the description of the hydrogen atom is

$V = d^{\dagger}a^{\dagger}da + d^{\dagger}a^{\dagger}dac + d^{\dagger}a^{\dagger}c^{\dagger}da + \ldots$

For brevity, I left only operator symbols and omitted coefficient functions, integral signs and other paraphernalia. If necessary, I can supply the missing info. These three terms have clear physical meanings:

$d^{\dagger}a^{\dagger}da$ is the direct electron-proton interaction potential. Its leading part is the usual Coulomb potential, plus there are relativistic (contact, spin-orbit, spin-spin) corrections, plus -- starting from the 4th perturbation order -- there are corrections responsible for the Lamb shift. If we diagonalize $H_0 + d^{\dagger}a^{\dagger}da$ in the electron+proton sector of the Fock space, we will get a decent energy spectrum of hydrogen.

$d^{\dagger}a^{\dagger}dac$ is interaction by which the atom can absorb a free photon and jump to a higher energy level

$d^{\dagger}a^{\dagger}c^{\dagger}da$ is interaction by which photons are emitted from excited atomic levels. So, the energy widths of the excited levels can be properly described.

In brief, the dressed approach provides a clear physical picture of the hydrogen atom with all important interactions. The relevant part of the Hamiltonian is resembling the familiar non-relativistic hydrogen Hamiltonian. However, there are also relativistic and radiative corrections, which ensure that the whole theory is Poincare invariant and cluster separable.

Now, you are saying that the "proper" field-based QED does not use the ideas of particles, their creation/annihilation operators and their Fock space, when interacting systems (such as the hydrogen atom) are involved. On the other hand, you claim that this theory possesses a finite well-defined cutoff-independent Hamiltonian. I am really interested in learning how this Hamiltonian looks like and how one can extract from it useful info, like the energy spectrum, lifetimes, wave functions, etc?

Eugene.

#### DarMM

Gold Member
Ten years ago I went through all of this with you, giving you the relevant papers, links etc that mathematically prove these facts and how Fock space and creation and annihilation operators will not work and explaining every point in detail. I'm not repeating it.

#### A. Neumaier

@A. Neumaier so to be clear, it sounds like you're saying that we should abandon the idea of a Fock space
No. A Fock space is alright for approximations, but not for the limit of these approximations.

#### A. Neumaier

I deliberately avoided infrared aspects.
You had to, because you only work perturbatively. As a result, your 2-point function for the electron has a pole at the electron mass while the correct QED 2-point function has there a branch point (as found by a renormalization group analysis). This conclusively proves that your dressed particle QED is not equivalent to QED proper.
In Appendix E.2 of volume 2, I copied Weinberg's proof that 10 Poincare generators of QED $(\boldsymbol{P}_0, \boldsymbol{J}_0, \boldsymbol{K}_0 + \boldsymbol{Z}, H_0+V)$ satisfy the required commutation relations. This proof is non-perturbative. Ten generators of the dressed theory are obtained from QED generators by means of a unitary "dressing" transformation. This means that all commutators are preserved and the dressed theory is rigorously relativistic.
Weinberg has no dressing transform, hence his derivation is alright. But your dressing transforms are only perturbatively defined, and do not have a limit when the order goes to infinity. Thus they are no true unitary transformations in the limit relevant for true QED. This is indeed impossible by Haag's theorem.
In practice, this unitary transformation can be performed only at low perturbation orders, so -- you are right -- some violations of the Poincare invariance are to be expected.
This is not different from the Poincare-non-invariance of QED in finite orders.
It is very different from standard renormalized QED, which is Poincare invariant at every loop order.

#### meopemuk

...you only work perturbatively. As a result, your 2-point function for the electron has a pole at the electron mass while the correct QED 2-point function has there a branch point (as found by a renormalization group analysis). This conclusively proves that your dressed particle QED is not equivalent to QED proper.
The 2-point function is a field concept, but the dressed particle theory is about particles, not fields. It deals with particle positions, momenta, spins, wave functions, etc. But n-point functions is a foreign concept for the dressed particle theory.

As for the non-equivalence to QED, I can argue that one can make the dressed particle theory as close to QED as one wishes. For example one quick and dirty way to get the dressed Hamiltonian is simply to fit its interactions to the S-matrix calculated at the specific loop order in QED.

But your dressing transforms are only perturbatively defined, and do not have a limit when the order goes to infinity. Thus they are no true unitary transformations in the limit relevant for true QED.
You claimed that the renormalized QED has a finite cutoff-independent Hamiltonian. However this is not what textbooks say. You can find the Lagrangian of QED in eqs. (11.1.6) - (11.1.9) of Weinberg's vol. 1 and in eq. (10.38) of Peskin & Schroeder. (Obviously, the QED Hamiltonian $H_{QED}$ has a similar structure.) These expressions have counterterms that depend on divergent (=cutoff-dependent) constants $Z_1, Z_2, Z_3, \delta m$. The counterterms are divergent, but (in Weinberg's words) "these terms just suffice to cancel the ultraviolet divergences that arise from loop integrals." That's the whole idea of renormalization, as far as I understand it.

Now, the dressing approach says: I would like to make a unitary transformation of the QED Hamiltonian

$H_d = e^{i\Phi} H_{QED} e^{-i \Phi}$

where the operator $\Phi$ is chosen to satisfy 3 conditions:

1. It is Hermitian in each perturbation order. This means that the dressing transformation $e^{i\Phi}$ is unitary in each perturbation order, thus your objection about the non-unitarity is not valid.
2. It is sufficiently smooth. Then one can prove that the S-matrix computed with $H_d$ is exactly the same as the S-matrix computed with $H_{QED}$
3. It cancels order-by-order certain "bad" terms in $H_{QED}$. The "bad" terms are those which act non-trivially on the vacuum and 1-particle states. They are the ones responsible for self-interactions and ultimately for the divergences in the traditional QED.

At the end of this procedure one obtains the dressed Hamiltonian $H_d$ without self-interactions. This Hamiltonian makes exactly the same predictions as $H_{QED}$ about scattering and bound state energies. Not surprisingly, $H_d$ resembles the familiar Hamiltonian of the non-relativistic theory. For example, the leading interaction term is the Coulomb potential between charged particles. As a little bonus, $H_d$ is finite, i.e., all the divergent renormalization constants cancel out in all orders. More precisely, they get absorbed into the transformation operator $\Phi$, which becomes divergent now, but we shouldn't be bothered, because this operator has no physical meaning.

Now we can use $H_d$ to make all kinds of electrodynamics calculations without ever thinking about cutoffs, counterterms and renormalization. All integrals are ultraviolet-convergent.

Eugene.

#### A. Neumaier

The 2-point function is a field concept, but the dressed particle theory is about particles, not fields. It deals with particle positions, momenta, spins, wave functions, etc. But n-point functions is a foreign concept for the dressed particle theory.
The 2-point function is the integral kernel of the inner product in the renormalized 1-particle Hilbert space (which exists since a single electron does not scatter), hence can be compared with the inner product in the dressed particle theory. They do not agree.
As for the non-equivalence to QED, I can argue that one can make the dressed particle theory as close to QED as one wishes. For example one quick and dirty way to get the dressed Hamiltonian is simply to fit its interactions to the S-matrix calculated at the specific loop order in QED.
You claimed that the renormalized QED has a finite cutoff-independent Hamiltonian.
This is assuming that QED exists as a Wightman field theory. (You need to assume as well that QED exists since your symmetry arguments depend on it....)
However this is not what textbooks say. You can find the Lagrangian of QED in eqs. (11.1.6) - (11.1.9) of Weinberg's vol. 1 and in eq. (10.38) of Peskin & Schroeder. (Obviously, the QED Hamiltonian $H_{QED}$ has a similar structure.)
This is only the approximate, cutoff-dependent unrenormalized Hamiltonian in Fock space describing a noncovariant approximation of QED; without cutoff the Hamiltonian and the other generators satisfy formally the Poincare commutation relations, but mathematically these expressions are meaningless. The limiting Hamiltonian of true QED is an operator on the renormalized Hilbert space, and cannot given in terms of Fock space but in terms of the Wightman reconstruction theorem, as the infinitesimal generator of the translations on the constructed states.
Now, the dressing approach says: I would like to make a unitary transformation of the QED Hamiltonian
$H_d = e^{i\Phi} H_{QED} e^{-i \Phi}$
which is a cutoff-dependent approximation to QED only. No Poincare invariance there; the generators don't satisfy exact Poincare relations.
where the operator $\Phi$ is chosen to satisfy 3 conditions:

1. It is Hermitian in each perturbation order. This means that the dressing transformation $e^{i\Phi}$ is unitary in each perturbation order, thus your objection about the non-unitarity is not valid.
2. It is sufficiently smooth. Then one can prove that the S-matrix computed with $H_d$ is exactly the same as the S-matrix computed with $H_{QED}$
3. It cancels order-by-order certain "bad" terms in $H_{QED}$. The "bad" terms are those which act non-trivially on the vacuum and 1-particle states. They are the ones responsible for self-interactions and ultimately for the divergences in the traditional QED.

At the end of this procedure one obtains the dressed Hamiltonian $H_d$ without self-interactions.
Well, you proceed formally without any consideration of limits, thus you only prove formal relations, not true ones. Note that in general, limits are not interchangeable, which is one of the reasons why subtle issues like the existence of QED are technically so difficult.
This Hamiltonian makes exactly the same predictions as $H_{QED}$ about scattering and bound state energies.
No. It makes exactly the same noncovariant predictions as the noncovariant approximations used at the start of the procedure. You don't even try to analyze the limit. Indeed, the limit does not exist, by Haag's theorem.

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#### bhobba

Mentor
Abdus Salam was writing before the more rigorous understanding of QFTs we possess today. Renormalization does result in well-defined Hamiltonians that is a mathematically established fact.
Indeed. Interestingly as was shown by Hawking Zeta Function renormalization is equivalent, and in some cases maybe better than dimensional regularization. What is more, no counter-terms are required and it's basis is a mathematically rigorous and a very well understood area of math called analytic continuation:

Evidently something similar was tried by Schwinger but he couldn't make it work.

Thanks
Bill

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#### A. Neumaier

There are many works in dressed/clothed particle theory. They usually pay tribute to the seminal paper

O. W. Greenberg, S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim. 8 (1958), 378.

You can use Google Scholar to find all citations of this work. Currently this search shows 113 results.
Why don't you say that the best exposition is in your [meomepuk = Eugene Stefanovich] 3 volume treatise on quantum electrodynamics ? I think this is a valid statement. Your treatise shows both the potential and the limitations of the dressed particle approach.
Thank you for the generous introduction of my humble book.
The interesting part about your books is the extent to which the dressed particle approach can be made to work, as a (though somewhat pedestrian) application of what nowadays is usually called similarity renormalization. (The linked Google scholar search for this turns up over 1400 results.)

The problematic part of your books is that they lack a self-critical attitude and do not discuss the (to experts obvious) limitations of the approach, but sell them as nonexistent or even progress:
The superluminal effects are not bugs, but unavoidable features of the dressed particles approach.
4. This Hamiltonian is relativistically covariant: there exists a corresponding interacting boost operator such that all commutators of the Poincare Lie algebra are satisfied.
At the end of this procedure one obtains the dressed Hamiltonian $H_d$ without self-interactions. This Hamiltonian makes exactly the same predictions as $H_{QED}$ about scattering and bound state energies.
No. It makes exactly the same noncovariant predictions as the noncovariant approximations used at the start of the procedure. You don't even try to analyze the limit. Indeed, the limit does not exist, by Haag's theorem.
And, most sadly, you seem to be de facto uninterested to learn what is outside your horizon:
Now, you are saying that the "proper" field-based QED does not use the ideas of particles, their creation/annihilation operators and their Fock space, when interacting systems (such as the hydrogen atom) are involved. On the other hand, you claim that this theory possesses a finite well-defined cutoff-independent Hamiltonian. I am really interested in learning how this Hamiltonian looks like and how one can extract from it useful info, like the energy spectrum, lifetimes, wave functions, etc?
Ten years ago I went through all of this with you, giving you the relevant papers, links etc that mathematically prove these facts and how Fock space and creation and annihilation operators will not work and explaining every point in detail. I'm not repeating it.

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#### Geonaut

@bhobba that's interesting, but I lost hope once I read the passage "This form of re-calculate a series has been widely used in Physics, one of the most famous examples being the calculation of the force due to “Casimir Effect” using the Zeta regularized value zeta (-3)=-1/120, or as another example, the value zeta(-1)=-1/12 for the divergent sum 1+2+3+4+5+6+7+…. That appears in theoretical physics."

I've seen people use the zeta function to claim that the sum of all positive integers is equal to -1/12, but it's my understanding that this is actually incorrect. It seems this paper does something similar at first glance, and if this paper uses that same "form of re-calculate a series" then it should contain the same error. With that in mind I stopped reading it, but if this isn't the case then please let me know.

#### A. Neumaier

I've seen people use the zeta function to claim that the sum of all positive integers is equal to -1/12, but it's my understanding that this is actually incorrect.
This caricature is of course incorrect, but the sum of $kx^k$ for all positive integers defines an analytic function with the value of -1/12 at x=1 after removal of the singularity, and its Taylor expansion around zero at x=1 agrees with the series you mention. Something similar happens in all cases where the series of interest is a special case of a power series expansion of an analytic function, and then one can use the method to find the intended sum, though the literal sum diverges. (Note that the singular terms in individual infinite contributions to Feynman diagrams cancel by design of any renormalization procedure, so dropping them in each term of the sum is justified!)

#### Geonaut

@A. Neumaier I wish I was as educated as you, I'm not sure what you mean by "intended sum" and "literal sum" (Edit: Oh, OK). More importantly, are you saying that you think that paper provides a legitimate technique?

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#### Geonaut

@A. Neumaier regarding everything else, I only have a Bachelor's in physics at the moment, but I understood enough to find your arguments to be very persuasive. I admire your willingness to go to the lengths that you have here to contribute to sites like this one for the sake of spreading the truth.

The fact that I didn't understand everything makes me want to learn more. You have me asking myself the question: What are all of the ways that we can test the limitations of a quantum theory? I know of many, but clearly, I'm not aware of all of them. Which is very worrisome, not knowing all of these methods can determine whether or not someone spends years confused about an important topic. Unfortunately, it might be impossible to adequately discuss such a broad question on Physics Forums, but maybe it would be appropriate to make a new post regarding that question to avoid going too far off topic. Do you think this question is too broad for Physics Forums?

#### Geonaut

Fyi, if you don't ignore me at some point then I will never shut up. I have a divergent number of questions about physics (I also have subtle jokes), and a desire to avoid making myself seem uneducated in public in order to find the answers to them is nothing in comparison to my desire to find those answers.

#### bhobba

Mentor
@bhobba I've seen people use the zeta function to claim that the sum of all positive integers is equal to -1/12, but it's my understanding that this is actually incorrect. It seems this paper does something similar at first glance, and if this paper uses that same "form of re-calculate a series" then it should contain the same error. With that in mind I stopped reading it, but if this isn't the case then please let me know./
Dr Neuaimer explained it correctly. If its true or not is often argued about. Its simply how you define infinite sums. If you replace the terms as the special case of a series where the terms involve a variable, and interpret the variable as a complex number, you can often sum it for some values of that variable - lets call it x. Then you can use analytic continuation to define the sum for a much larger range of values of x. Then you substitute the values you started with and viola - you get an answer. There are a number of ways of doing it - my favorite is Borel Summation because explaining it is so simple (it will not work on the Zeta Function though). There other ways to sum it - the general one I like best is Ramanujan Summation - despite Hardy waning against it - he wrote a textbook on summing divergent series.

As an example lets look at Borel Summation. ∑an = ∑an*n!/n! = ∑an*∫(t^n)*e^-t/n!
= ∑∫(an*t^n)*e^-t/n! where we have used n! = ∫(t^n)*e^-t.

The above is rigorously correct but sometimes in physics and applied math we get sloppy and reverse integrals and sums without showing you can do it. This gives ∫∑(an*t^n)*e^-t/n! and is called the Borel Sum. It turns out for series that converge normally you can reverse the sum and integral. So we have a more general summation method allowing the summation of things like 1-1+1-1......... A deeper analysis than I will give shows its really a form of analytic continuation. The great mathematician, Borel, when an unknown young man, discovered that his summation method gave the 'right' answer for many classical divergent series. He decided to make a pilgrimage to Stockholm to see Mittag-Leffler, who was the recognized lord of complex analysis. Mittag-Leffler listened politely to what Borel had to say and then, placing his hand upon the complete works by Weierstrass, his teacher, he said in Latin, 'The Master forbids it'. So even experts get fooled by this. Mittag-Leffler eventually came around and cam up with his own method, Mittag-Leffler summation.

We are very lucky having Dr Neumaier and others like him post here - you will find you will look back after a while and be surprised at what you have learned.

Thanks
Bill

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#### Geonaut

The fact that we're talking about debatable math is crazy to me. Math is supposed to be the one thing that is black and white, but I guess we've reached the realm of theoretical math where that's no longer the case.

What is more, no counter-terms are required
This is very interesting, honestly, I expected someone to shoot this down. Now I'm extremely curious about this, and I'm reading the paper very closely. If this is legit then what happens to renormalization conditions? I've never been a fan of their presence in QFT, and if this means that we no longer need them then that would be interesting. That's currently what's going through my mind... I'm going to have to think about this some more. Thank you for your input Bill.