Renormalizable quantum field theories

[Bob_for_short]

Let me see if I can explain this.
Take H as the original unrenormalized Hamiltonian. When I introduce cutoffs I create H^{\Lambda}, the cutoff Hamiltonian. To this I add the counterterm part of the Hamiltonian with cutoffs, let's call it H^{\Lambda}_{c.t.}.
Now H^{\Lambda} and H^{\Lambda}_{c.t.} are both totally divergent, non-self-adjoint, unbounded operators, in fact they are so badly behaved that they're not even operators.

However the renormalized Hamiltonian which is:
H_{ren} = \lim_{\Lambda \rightarrow \infty} H^{\Lambda} + H^{\Lambda}_{c.t.}
is a well-defined self-adjoint and semi-bounded operator. Although the counterterms diverge, (there are no cancellations between different orders in perturbation theory), they exactly cancel the divergences coming from powering operator-valued distributions like \phi(x) or A_{\mu}(x), so H_{ren} is well-defined.<br /> <br /> Does that make sense?

<br /> <br /> No, it does not make sense, in my opinion. This &quot;counter-term&quot; ideology starts from a wrong idea that the original Hamiltonian contains non-observable masses and charges. In fact, it is the badly guessed interaction (self-action) that brings corrections (divergent or not) to the fundamental (known) constants. The interaction Hamiltonian should be changed from the very beginnig rather than perturbatively - namely one can reformulate the theory without self-action. Then the perturbation corrections are finite and small, and the theory becomes physical - its initial approximations are physical and it describes correctly the soft radiation and other effects quite naturally, without appealing to not physical counter-terms. The counter-terms detract, order by order, the contributions of the self-action term of the total Hamiltonian. There is a short-cut based on good physics rather than on good luck.<br /> <br /> Bob.

 

[Avodyne]

The transition frequency for harmonic and slightly anharmonic oscillators (g is small) is still \omega_{10}=\omega

No, it isn't, it's

\omega_{10}=\omega+{\textstyle{1\over8}}g\hbar^2\omega^{-2}+O(g^2).

It doesn't matter how small g is (unless it's exactly zero), you are still only able to measure \omega_{10}, which is not the same as \omega. And to find out experimentally how small g is, you have to do some other measurements. And these won't measure g directly either, but some other complicated function of g and \omega. Then, knowing the theory, you could compute g and \omega. But you can't measure them. And this is why it's OK, in field theory, for hamiltonian coefficients to be cutoff dependent.

 

[meopemuk]

What you mean is that divergences cancel when you calculate observable quantities (or certain generalizations, such as renormalized Green's functions, which are not directly observable but still come out finite). The renormalized hamiltonian can then be given an abstract definition such that it has finite matrix elements between certain classes of states. But if you try to write it down explicitly, it will still have cutoff-dependent coefficients in it.

Yes, this is my understanding too. The full Lagrangian/Hamiltonian of the renormalized theory L_0 + L_1 + L_2 is cutoff-dependent and divergent. However, S-matrix elements (scattering amplitudes) computed with this Hamiltonian are cutoff-independent, finite, and agree with experiment perfectly well. That's the great achievement of the renormalization theory.

If you care only about scattering theory (as in all QFT textbooks), then the ill-defined Hamiltonian does not pose a serious problem. However, if you want to study the time evolution of states and observables, then the cutoff-dependence and divergences in the Hamiltonian is a show-stopper.

 

[Bob_for_short]

No, it isn't, it's

\omega_{10}=\omega+{\textstyle{1\over8}}g\hbar^2\omega^{-2}+O(g^2).

It doesn't matter how small g is (unless it's exactly zero), you are still only able to measure \omega_{10}, which is not the same as \omega. And to find out experimentally how small g is, you have to do some other measurements. And these won't measure g directly either, but some other complicated function of g and \omega. Then, knowing the theory, you could compute g and \omega. But you can't measure them. And this is why it's OK, in field theory, for hamiltonian coefficients to be cutoff dependent.

Even when g is not small, the anharmonic oscillator has a discrete spectrum which is directly measurable. You have chosen a bad example to show "complexity" of the QFT problems.

The Hamiltonian coefficients - masses and charges are not cutoff dependent if one uses a better Hamiltonian - without self-action. These fundamental constants are observable and constant. What is the energy dependent is not the "running" charge but the exact scattering amplitude.

Bob.

 

[Bob_for_short]

Yes, this is my understanding too. The full Lagrangian/Hamiltonian of the renormalized theory L_0 + L_1 + L_2 is cutoff-dependent and divergent. However, S-matrix elements (scattering amplitudes) computed with this Hamiltonian are cutoff-independent, finite, and agree with experiment perfectly well. That's the great achievement of the renormalization theory.

Eugene, you give too much flattery to the renormalized S-matrix. It is known/shown that the elastic processes are impossible - the S-matrix elements are identically equal to zero! It is the inclusive cross sections that are different from zero. So even after the renormalizations you have to work hard with the IR problem.

Bob.

 

[meopemuk]

Eugene, you give too much flattery to the renormalized S-matrix. It is known/shown that the elastic processes are impossible - the S-matrix elements are identically equal to zero! It is the inclusive cross sections that are different from zero. So even after the renormalizations you have to work hard with the IR problem.

Bob.

Yes, I know that. There are ultraviolet divergences and there are infrared divergences. But let us solve one problem at a time. Otherwise the whole thing becomes too confusing. I am interested to solve the ultraviolet problem first. Temporarily, we can assign a small non-zero mass to the photon, and thus avoid the annoying issues with "soft" photons and exclusive/inclusive cross sections.

 

[Bob_for_short]

Yes, I know that. There are ultraviolet divergences and there are infrared divergences. But let us solve one problem at a time. Otherwise the whole thing becomes too confusing. I am interested to solve the ultraviolet problem first. Temporarily, we can assign a small non-zero mass to the photon, and thus avoid the annoying issues with "soft" photons and exclusive/inclusive cross sections.

It turns out the the both problems can be solved at once: writing a physical Hamiltonian without self-action but with a potential interaction of coupled charges (electroniums) eliminates both mathematical problems and makes the theory physical: the soft radiation in present automatically as excitations of the internal (or relative) degrees of freedom of compound systems - electroniums. The inclusive consideration is now obligatory, otherwise the elastic S-matrix contains only zero-valued elements. I would love to work out the Novel QED with you. You have everything for that. Just take may ideas and solutions and make calculations. If anything seems to you strange or non-comprehensible - I will stand by. You always can count on me. I have outlined how the Lamb shift and the anomalous magnetic moment can be calculated (straightforwardly, without artificial cut-offs). We can start tomorrow, if you like.

Vladimir Kalitvianski.

 

[ytuab]

It turns out the the both problems can be solved at once: writing a physical Hamiltonian without self-action but with a potential interaction of coupled charges (electroniums) eliminates both mathematical problems and makes the theory physical:ted (straightforwardly, without artificial cut-offs). We can start tomorrow, if you like.
Vladimir Kalitvianski.

Sorry. I have not read your paper completely.

Is your theory relativistic or non-relativistic?
Does your theory keep Lorentz invariant?

 

[Bob_for_short]

Sorry. I have not read your paper completely.
Is your theory relativistic or non-relativistic?
Does your theory keep Lorentz invariant?

Yes, relativistic and Lorentz invariant. If you like to discuss details, I invite you to the Independent Research forum where I posted a thread on this subject.

Bob.

 

[DarMM]

I see your point, but it doesn't make sense to me.

If we return to the Weinberg's Lagrangian, then what you call "the original non-renormalized Hamiltonian" is equivalent to L_0 and L_1 in (11.1.7) and (11.1.8). These are finite operators, and their definition does not involve loop integrals. So, they are independent on the cutoff.

Okay here is the first difficulty. In Weinberg L_{0} in (11.1.7) is a well defined self-adjoint operator. L_{1} in (11.1.8) is not. It isn't self-adjoint and it's not semi-bounded. Hence L_{0} + L_{1} is not a well defined operator. I can sketch a proof if you wish, it involves some functional analysis.

The cutoff-dependent divergent part is L_2. So, the sum L_0 + L_1 + L_2 is still cutoff-dependent and divergent. I don't see where the cancelation comes from.

What am I missing here?

Let me describe the procedure.
Okay, so L_{0} is a well behaved operator so let's leave it alone. L_{1} is not well defined, so let's make it well defined by intorducing a cutoff. Taking the cutoff to infinity would also mean L_{2} is undefined so let's leave it cutoff dependant.
Now let's add the cutoff L^{\Lambda}_{1} and L^{\Lambda}_{2} to obtain:
L^{\Lambda}_{int} = L^{\Lambda}_{1} + L^{\Lambda}_{2}.

The \Lambda \rightarrow \infty limit of L^{\Lambda}_{int} is a well-defined self adjoint operator with no cutoff dependence. We can then add it to L_{0} (although by "adding" it you have to deal with certain technical difficulties but I'm being lose), to obtain a well-defined cutoff independant theory with well-defined finite time evolution.

 

[DarMM]

What you mean is that divergences cancel when you calculate observable quantities

I hope my previous post makes things clearer, but what I'm saying is that you can remove all cutoff dependence from the Hamiltonian and have it finite after you have performed renormalization.

But if you try to write it down explicitly, it will still have cutoff-dependent coefficients in it.

No it doesn't. It has no cutoff dependence once you take the limit and it is finite in that limit.

 

[Bob_for_short]

Dear DarMM,

What would you prefer: to get rid of UV and IR infinities perturbatively or to do with a physically correct theory without physical and mathematical difficulties?

Bob.

 

[ytuab]

Yes, relativistic and Lorentz invariant. If you like to discuss details, I invite you to the Independent Research forum where I posted a thread on this subject.

Bob.

I have now read " Atom as a Dressed Nucleus".

"Positive charge cloud" theory is very interesting, and I think your idea is true.

But I'm sorry if I make mistakes.

I think your paper is not relavent to the divergent problems of QFT ( is only relevant to 1/r?).
Your paper is nonrelativistic and does not keep Lorentz invariant?
It is almost impossible to solve the all divergent problems of QFT keeping Lorentz invariant.

How do you think about it?

 

[meopemuk]

L_{1} is not well defined, so let's make it well defined by intorducing a cutoff.

I don't understand why you are saying that L_1 is not well defined and cutoff-dependent. First note that L_1 is interaction "density", so in order to obtain the interaction operator, L_1 must be integrated on d^3x. So, in total there are 4 integrations in this expression: one on d^3x and 3 momentum integrals (which come from definitions of quantum fields). Quantum fields also supply exponential factors like exp(ipx). Their integration on d^3x results in one momentum delta function (which simply expresses the fact that our interaction conserves the total momentum). As a result we obtain a sum of terms. Each term has 3 momentum integrals whose integrands contain 1 momentum delta function and a product of 3 creation/annihilation operators. This is exactly the standard form of any interaction operator as shown in Weinberg's eq. (4.4.1). There is no cutoff in this equation.

 

[DarMM]

I don't understand why you are saying that L_1 is not well defined

I'm saying it because it's not. It's not self-adjoint or semi-bounded, there are proofs of this fact in a few books. The basic problem is that \bar\psi\gamma^{\mu}\psi is an operator valued distribution, but in more than two dimensions...

First note that L_1 is interaction "density", so in order to obtain the interaction operator, L_1 must be integrated on d^3x. So, in total there are 4 integrations in this expression: one on d^3x and 3 momentum integrals (which come from definitions of quantum fields). Quantum fields also supply exponential factors like exp(ipx). Their integration on d^3x results in one momentum delta function (which simply expresses the fact that our interaction conserves the total momentum). As a result we obtain a sum of terms. Each term has 3 momentum integrals whose integrands contain 1 momentum delta function and a product of 3 creation/annihilation operators. This is exactly the standard form of any interaction operator as shown in Weinberg's eq. (4.4.1).

...after performing integration it's not a well-defined operator. This was proven by Wightman in the 1960s. I can provide you with some references. So \int{\bar\psi\gamma^{\mu}\psi}A_{\mu} is not a well defined operator, this is rigorous mathematical fact.

and cutoff-dependent.

It's not cutoff-dependent, I'm just doing that so I can add it to the counterterm part without getting into the rigours of distribution theory. I'm introducing a cutoff in order to turn it into an operator which is well-defined and then adding it to the cutoff counterterm part obtaining another operator which is well-defined when the cutoff is removed.

There is no cutoff in this equation.

Of course, but the problem is that equation doesn't give you an operator unless it is cutoff.
Think about it, if the integral of L_{1} was an operator why would you even need renormalization? If it was well-defined there would be no loop divergences.

 

[meopemuk]

...if the integral of L_{1} was an operator why would you even need renormalization? If it was well-defined there would be no loop divergences.

In my opinion, L_1 is a well defined operator. In any case, whatever subtle irregularities were found in it by Wightman, they are no match for the explicitly divergent constants deltam, Z_2 and Z_3 in L_2. So, I don't see how any cancelation of divergences is possible in L_1 + L_2.

My understanding of the origin of loop divergences is different from yours. In my view the major problem is that L_1 (when expressed in terms of creation/annihilation operators) contains "trilinear" terms, like

L_1 = a*ac + a*c*a + ...

where a is annihilation operator for electrons and c is annihilation operator for photons. When you calculate the 2nd order S-operator with interaction L_1 you need to take the product of two copies of L_1

S = (a*ac + a*c*a + ...)(a*ac + a*c*a + ...)

After normal ordering you may notice that there is a non-zero term of the type

(loop integral) a*a

This term describes some kind of "scattering of the electron on itself", i.e., self-interaction. One effect produced by this term is that the electron mass in the interacting theory is different from the electron mass in the free theory. That's why the electron mass renormalization is needed. In QED things are even worse: the momentum dependence of the trilinear interactions is such that the loop integral is divergent, so the electron mass correction is infinite.

As long as you have trilinear interaction terms in your Hamiltonian you'll always have renormalization problems. In the "dressed particle" approach these trilinear interaction terms are called "bad". The idea of this approach is to change the Hamiltonian so that these bad terms are not present. All interactions must be written in terms of "good" operators only. An example of such a "good" operator is

a*a*aa

You may notice that if you take a product of two such operators and normal-order this product, you'll never get terms of the type a*a. So, there can be no "corrections" to the electron mass. No renormalization is needed.

The question is: can we get rid of the "bad" terms in the Hamiltonian and still obtain the same accurate S-matrix as we know it from renormalized QED? The answer is "yes", and the "dressed particle" approach shows how this can be done.

 

[Bob_for_short]

I have now read " Atom as a Dressed Nucleus".
"Positive charge cloud" theory is very interesting, and I think your idea is true.
But I'm sorry if I make mistakes.
I think your paper is not relavent to the divergent problems of QFT ( is only relevant to 1/r?).
Your paper is nonrelativistic and does not keep Lorentz invariant?
It is almost impossible to solve the all divergent problems of QFT keeping Lorentz invariant.
How do you think about it?

Hamiltonian formulation of QED looks like non Lorentz invariant but it is in fact. It's been proven many times. I use the Hamiltonian formulation in the gauge invariant (Dirac's) variables (known also as a Coulomb gauge). I build the relativistic Hamiltonian basing on the physics of the quantum mechanical charge smearing outlined in the first part of the article. The resulting Hamiltonian (see also "Reformulation instead of Renormalizations", formula (60)) is relativistic but free from the self-action. This, new formulation, is free from non-physical entities and describes the right physics analogous to the atomic scattering description. Preliminary non relativistic estimations show that it is right. I have not presented the detailed relativistic calculations but it is clear that they only bring some numerical corrections to the right physics obtained already in the non-relativistic approximation.

Bob.

 

[DarMM]

In my opinion, L_1 is a well defined operator.

It's not though and this isn't some subtlety, it is the origin of the divergences in quantum field theories. Operating on any vector in the Hilbert space twice with L_{1} (which is second order in perturbation theory) maps to a vector outside Fock space, hence the divergences.

In any case, whatever subtle irregularities were found in it by Wightman, they are no match for the explicitly divergent constants deltam, Z_2 and Z_3 in L_2.

Yes, in fact they are. You can prove that they are matches for these counterterms, are you contesting the proofs? If you are I suggest we start with the quartic scalar case.

So, I don't see how any cancelation of divergences is possible in L_1 + L_2.

Maybe so, but that doesn't change the fact that these cancellations indeed occur. Take a look at Glimm's "Boson fields with the \Phi^{4}interaction in three dimensions", to see it at work in the case of quartic scalar theory in three dimensions.

My understanding of the origin of loop divergences is different from yours. In my view the major problem is that L_1 (when expressed in terms of creation/annihilation operators) contains "trilinear" terms, like

L_1 = a*ac + a*c*a + ...

where a is annihilation operator for electrons and c is annihilation operator for photons. When you calculate the 2nd order S-operator with interaction L_1 you need to take the product of two copies of L_1

S = (a*ac + a*c*a + ...)(a*ac + a*c*a + ...)

After normal ordering you may notice that there is a non-zero term of the type

(loop integral) a*a

This term describes some kind of "scattering of the electron on itself", i.e., self-interaction. One effect produced by this term is that the electron mass in the interacting theory is different from the electron mass in the free theory. That's why the electron mass renormalization is needed. In QED things are even worse: the momentum dependence of the trilinear interactions is such that the loop integral is divergent, so the electron mass correction is infinite.

As long as you have trilinear interaction terms in your Hamiltonian you'll always have renormalization problems. In the "dressed particle" approach these trilinear interaction terms are called "bad". The idea of this approach is to change the Hamiltonian so that these bad terms are not present. All interactions must be written in terms of "good" operators only. An example of such a "good" operator is

a*a*aa

You may notice that if you take a product of two such operators and normal-order this product, you'll never get terms of the type a*a. So, there can be no "corrections" to the electron mass. No renormalization is needed.

The question is: can we get rid of the "bad" terms in the Hamiltonian and still obtain the same accurate S-matrix as we know it from renormalized QED? The answer is "yes", and the "dressed particle" approach shows how this can be done.

Remember that by Haag's theorem, interacting theories do not live in Fock space and hence cannot in general be written using creation and annhilation operators.
Anyway if you look at any of the literature on rigorous field theory you will see that the origin of ultraviolet divergences has been known for a long time as being due to two facts:
1. The producting of operator valued distributions resulting in ill-defined powers which will not not give rise to operators when integrated.
2. The non-Fock representations needed for interacting theories.

These are the origins of the ultraviolet difficulties. They are the reason L_{1} is divergent. In fact L_{1} is a tremendously divergent object, it even has what physicists would call nonperutrbative infrared divergences. To say it's well-defined up to some subtleties is totally provably wrong.

See the paper of Glimm above for an example of how badly divergent L_{1} is even in the case of a scalar field theory.

 

[meopemuk]

There can be no cancelation between L_1 and L_2 for the simple reason that L_1 is first order in the coupling constant (1st power in e) and L_2 is composed of 2nd, 3rd, 4th etc. order terms.

Operating on any vector in the Hilbert space twice with L_{1} (which is second order in perturbation theory) maps to a vector outside Fock space, hence the divergences.

This agrees with what I was saying: the (2nd order) product L_1 * L_1 is divergent (due to loop integrals). This divergence is compensated by the divergent 2nd order term in L_2. However, as I said above, there is no cancelation of divergences in the sum L_1 + L_2.

 

[DarMM]

There can be no cancelation between L_1 and L_2 for the simple reason that L_1 is first order in the coupling constant (1st power in e) and L_2 is composed of 2nd, 3rd, 4th etc. order terms.

This stuff is strange and perhaps I've been explaining badly. Think of what L_{1} is meant to be a function of, it's a function of the interacting fields \psi and A_{\mu} which are themselves functions of e. So L_{1} does contain terms of higher order in e through its dependence on the interacting fields.

Does this make sense?

However I must say that the best way of thinking of this probably isn't through the use of perturbative additive renormalizations, but with nonperturbative multiplicative renormalizations where you can this whole thing become a problem related to the theory of distributions.
At the end of the day however it has been proven that cancellations do take place between L_{1} and L_{2}.

 

[ytuab]

Hamiltonian formulation of QED looks like non Lorentz invariant but it is in fact. It's been proven many times. I use the Hamiltonian formulation in the gauge invariant (Dirac's) variables (known also as a Coulomb gauge). I build the relativistic Hamiltonian basing on the physics of the quantum mechanical charge smearing outlined in the first part of the article. The resulting Hamiltonian (see also "Reformulation instead of Renormalizations", formula (60)) is relativistic but free from the self-action. This, new formulation, is free from non-physical entities and describes the right physics analogous to the atomic scattering description. Preliminary non relativistic estimations show that it is right. I have not presented the detailed relativistic calculations but it is clear that they only bring some numerical corrections to the right physics obtained already in the non-relativistic approximation.

Bob.

I think you misunderstand "the relativistic and Lorentz invariant" of divergent problems.

I have not yet read your paper "Reformulation instead of Renormalizations",
But as far as I read your first paper, there is nothing about solving the divergent problems
keeping Lorentz invariant.

The divergence is caused by the infinit loop (the action of infinit photon, particles and antiparticles) and divergent 4-momentum integral (which keep Lorentz invariant).
It is not caused only by 1/r as you say.

I do not like the idea of "bare mass or bare charge ".
I think the idea of QFT has reached the limit.

 

[meopemuk]

This stuff is strange and perhaps I've been explaining badly. Think of what L_{1} is meant to be a function of, it's a function of the interacting fields \psi and A_{\mu} which are themselves functions of e. So L_{1} does contain terms of higher order in e through its dependence on the interacting fields.

Does this make sense?

Not yet. I see that our disagreement about QFT is even deeper than I thought. I always thought that quantum fields present in Weinberg's L_0 + L_1 + L_2 are *free* quantum fields. So, L_1 has only 1st order contributions. Apparently, you disagree with that.

 

[Bob_for_short]

I think you misunderstand "the relativistic and Lorentz invariant" of divergent problems.

I have not yet read your paper "Reformulation instead of Renormalizations",
But as far as I read your first paper, there is nothing about solving the divergent problems
keeping Lorentz invariant.

As I said previously, the relativistic theory of interacting particles or fields can be cast in the Hamiltonian form, so it is a multi-particle quantum mechanics. Such a form is covariant, it's been proven.

The divergence is caused by the infinit loop (the action of infinit photon, particles and antiparticles) and divergent 4-momentum integral (which keep Lorentz invariant).

Do you understand what you are writing? The divergences are caused by divergences. It is a tautology. There is no physical mechanism behind such statements.

It is not caused only by 1/r as you say.

Yes, it does. I give an example in "Atom...". If your potential in the integral is, roughly speaking,

1/(r+a) (i.e., "cut-off" or finite at r=0),

but you try to use a perturbation theory like that:

1/(r+a)=1/r - a/r² + a²/r³ -...,

your integral will diverge at small r.

As I said previously, I not only use a better initial approximation for interacting fields, but also remove the self-action. So my Hamiltonian is different. It is well defined physically and mathematically, contrary to the standard QED Hamiltonian.

Bob_for_short.

 

[ytuab]

As I said previously, the relativistic theory of interacting particles or fields can be cast in the Hamiltonian form, so it is a multi-particle quantum mechanics. Such a form is covariant, it's been proven.
Bob_for_short.

I'm sorry to displease you.
But I'm still convinced that your paper is nonrelativistic and doesn't keep Lorentz invariant.

Because If you solve the divergent problems (infinit bare charge and mass ...) under the Lorentz invariant condition,
your paper will be immediately accepted by the top journal such as "Nature" or "Science".
So?

The relativistic particle is a point particle.
If you use " (natural) cut off ", the part of integral becomes noncontinuous and the upper and lower limit of momentum will appear. So this state doesn't keep Lorentz invariant.

I don't believe a point particle, so I don't believe QFT (and QM).

 

[Bob_for_short]

I'm sorry to displease you.
But I'm still convinced that your paper is nonrelativistic and doesn't keep Lorentz invariant.

It is a very superficial impression. I had no objections from experienced researchers.

Because If you solve the divergent problems (infinit bare charge and mass ...) under the Lorentz invariant condition, your paper will be immediately accepted by the top journal such as "Nature" or "Science".
So?

Not immediately. They all require the complete relativistic calculations, not only formulation.

The relativistic particle is a point particle.

As I showed in "Atom...", the point-like particle is the inclusive rather than "elastic" picture of scattering in QM.

If you use " (natural) cut off ", the part of integral becomes noncontinuous and the upper and lower limit of momentum will appear. So this state doesn't keep Lorentz invariant.

Yes, look at the atomic form-factors: they contain characteristic "cloud" sizes a₀ or (m_e/M_A)a₀, for example. There is nothing wrong with it. On the contrary, it is natural unlike artificial cut-offs in the standard QFT.

I don't believe a point particle, so I don't believe QFT (and QM).

And I am trying to build a trustful and working theory.

Bob_for_short.

 

[DarMM]

Not yet. I see that our disagreement about QFT is even deeper than I thought. I always thought that quantum fields present in Weinberg's L_0 + L_1 + L_2 are *free* quantum fields. So, L_1 has only 1st order contributions. Apparently, you disagree with that.

No not really and to be honest we don't really have deep disagreements, I've just been vague at times for brevity. Let me try to explain in full. Some of the issues come from me talking in general rather than sticking to Weinberg. I'm wasn't clear about the non-Fock nature of the problem.
If you'll allow me, I will take the case of \phi^{4} in three dimensions since it is somewhat easier to deal with.

The first thing I should say is that the cancellations that I'm talking about probably can't be understood best as perturbative cancellations, but rather as direct operator cancellations or operator identities.

Anyway let's take the Hamiltonian of \phi^{4}_{3}. The interacting part, is \int{\lambda\phi^{4}}, this is the analogue of L_{1} in Weinberg. Immediately you can prove this isn't a well defined operator on Fock space. In my previous post I tried to demonstrate how badly behaved the operator is by showing that it can't act twice on a vector. Let me say the real problem, it's not self adjoint. This is true even for the L_{1} term in QED. An even bigger problem is that when added to the free part of the Hamiltonian it causes the total Hamiltonian to be unbounded, meaning there is no positivity of energy.
Now I know it seems strange, but it is a proven fact that these L_{1} terms are just as divergent or badly behaved as the counterterms. Even if you can't "see it" and I accept that it may be difficult, it is a fact that they are highly divergent.

In \int{\lambda\phi^{4}} physicists usually get around this with mass renormalization. We add a term to the Hamiltonian \delta m^{2}\int{\phi^{2}}. Now \delta m^{2} contains terms up to order \lambda^{2}. I'm claiming that this results in a well-defined Hamiltonian. However you rightly ask how can this be possible if \int{\lambda\phi^{4}} is only first order in \lambda and \delta m^{2} goes up to second order?

The truth is, it can't in Fock Space, which is the crux of Haag's theorem. If you move to the correct representation of the canonical commutation relations, or in physicist's speak "the interacting Hilbert space", then the cancellations are possible. See the paper by Glimm for details.

So yes, since Weinberg remains in Fock space then these cancellations cannot occur. However we know that interacting theories can't live in Fock space.

If one wants to stick to Fock space then you'll be presented with an odd situation, you'll have order by order cancellations for the S-matrix, but you'll have a poorly defined Hamiltonian. Not just because of the counterterms, but also because of L_{1}.

I also just want to mention that renormalization basically turns out to be Wick ordering in a non-Fock space.

Is that better?

 

[ytuab]

Yes, look at the atomic form-factors: they contain characteristic "cloud" sizes a₀ or (m_e/M_A)a₀, for example. There is nothing wrong with it. On the contrary, it is natural unlike artificial cut-offs in the standard QFT.
.

I think you probably forgot that "the integral part " of the Hamiltonian(QED) should be Lorentz invariant. this part must be continuous and has no upper and lower limit of momentum.
In your papar, this is not commented anywhere. It is strange, I think.

If we only get the part of the hamiltonian Lorentz invariant, it is insufficient.

And I think both the natural and artificial cut-off doesn't keep Lorentz invariant.

 

[Bob_for_short]

I think you probably forgot that "the integral part " of the Hamiltonian(QED) should be Lorentz invariant. this part must be continuous and has no upper and lower limit of momentum. In your papar, this is not commented anywhere. It is strange, I think. If we only get the part of the hamiltonian Lorentz invariant, it is insufficient.

There are no limits in the Fourier integral itself. It is a form-factor that "cuts-of" certain parts in integration.

I forgot nothing. You just do not believe to that I wrote. There are well known things that go without saying. The proof that I wrote about is valid for all Hamiltonian terms including the four-fermion Coulomb term.

And I think both the natural and artificial cut-off doesn't keep Lorentz invariant.

As I showed in "Atom...", the elastic cross section can be measured. It contains the positive charge cloud size. Very very roughly, there is a dimensionless ratio of this size and the impact parameter in the elastic cross section.

The same is valid for inelastic cross sections.

But if you add up all cross sections, these dependencies smooth out and you obtain the Rutherford cross section, as if the target charge were point-like. That it what is observed in inclusive experiments. The inclusive picture is illusory, not real. That is why starting from assigning 1/r to a charge leads to bad mathematical expressions.

Bob.

 

[meopemuk]

the crux of Haag's theorem... However we know that interacting theories can't live in Fock space.

I think I know what Haag's theorem is, and in my (perhaps ill-informed) opinion this theorem does not present a significant obstacle for developing QFT in the Fock space. This theorem basically says that "interacting field" cannot have a manifestly covariant Lorentz transformation law. Some people say that this violates the relativistic invariance and, therefore, is unacceptable. However, I would like to disagree.

A quantum theory is relativistically invariant if its ten basic generators (total energy, total momentum, total angular momentum, and boost) satisfy Poincare commutation relations. These commutators have been proven for interacting QFT. For example, in the case of QED the detailed proof is given in Appendix B of

S. Weinberg, "Photons and Gravitons in S-matrix theory: Derivation of charge conservation and equality of gravitational and inertial mass", Phys. Rev. 135 (1964), B1049.

So, relativistic non-invariance is out of question. I think that the absence of the manifestly covariant transformation law of the "interacting field" is not a big problem. Actually, one can perform QFT calculations without even mentioning "interacting field" at all. It is quite sufficient to have a Hamiltonian and obtain the S-operator from it by usual Rules of Quantum Mechanics.

My claim remains that the Hamiltonian L_0 + L_1 is well-defined. However S-matrix divergences appear when products like L_1 * L_1 are calculated. In the renormalization theory these divergences get canceled by the addition of (divergent) counterterms L_2 in the Hamiltonian. So, the full Hamiltonian

H = L_0 + L_1 + L_2

is cutoff-dependent and divergent in the limit of removed cutoff. This divergence is not a big deal in regular QFT, where we are interested only in the S-matrix. However, if one day we decide to study the time evolution of states and observables in QFT, we may hit a difficult problem due to the absence of a well-defined Hamiltonian. Fortunately, this day seems to be quite far away, because experimental information about the time evolution of colliding particles is virtually non-existent.

I think that our disagreement reflects two different philosophies about dealing with interacting QFT. In your approach (which is widely accepted), you seek solution by leaving the Fock space. In my approach (which is less known) I stay in the Fock space and try to change the original Hamiltonian by "dressing". It may well happen that both philosophies are correct (or that both are wrong).

 

[Bob_for_short]

I think I know what Haag's theorem is, and in my ... opinion this theorem does not present a significant obstacle for developing QFT in the Fock space.

I agree with you here. In fact, there may be different QFTs with different interaction Hamiltonians. In my Novel QED I stay within Fock spaces without problem.

I think that our disagreement reflects two different philosophies about dealing with interacting QFT. In your approach (which is widely accepted), you seek solution by leaving the Fock space. In my approach (which is less known) I stay in the Fock space and try to change the original Hamiltonian by "dressing". It may well happen that both philosophies are correct (or that both are wrong).

Both of you perform perturbative renormalizations of the standard QED (i.e., with self-action) however it is named. Eugene's approach keeps the fundamental constants intact and discards perturbative corrections to them. This is a typical renormalization prescription. Of course, it is also a perturbative dressing. What I propose is a non perturbative dressing and a physical interaction without wrong self-action and without wrong renormalizations.

Bob_for_short.