Interacting theory lives in a different Hilbert space [...]

**Bob_for_short** · Nov9-09, 09:14 AM

I would like you participants to express your feelings about the following:

In the Scharf’s - "Finite QED" and in all other books there are calculations of elastic processes like Compton scattering, Rutherford or Mott or Moeller (i.e., charge from charge) scattering and comparison of these results with experimental data. I speak of the first non-vanishing order of the perturbation theory, without loops. At first sight these results look good. But later on, much later on we discover that the probability of any elastic process is identically equal to zero. It is the inclusive cross sections and probabilities that are different from zero. And it is the inclusive quantities that correspond to the experimental situations.

So, in the first non-vanishing order the standard QED predicts events that never happen. And it does not predict the phenomenon that happen always (soft radiation). Don’t you consider this theory "feature" as a complete failure in the physics description? Isn’t it a too bad start for the perturbation theory?

Please answer these questions explicitly. I am waiting for your opinions.

To your information, in my pet theory the probabilities and cross sections of elastic processes, calculated in the first non-vanishing order, are just zero, as it should be. It is more correct, isn’t it? And the inclusive cross sections correspond to the Compton or Rutherford formulas with high accuracy, again, as it should be. It is also more correct, don’t you think so? I obtain correct physics in the first non-vanishing order, not in higher orders by the price of forced discarding self-action contributions, painful treatment of the infrared divergences and inventing renormalization ideology with its bare notions to cover this practice. I mention this just to show you the difference in quality of physics description.

**strangerep** · Nov9-09, 06:02 PM

Originally Posted by meopemuk

(1) We can simply take scattering amplitudes from high-order QED calculations and/or experiment and fit coefficient functions in (1) to these data.

(2) We can apply the so-called "unitary dressing transformation" to the renormalized Hamiltonian of QED to bring it to the desired form (1).

If we rely on existing high-order QED calculations to construct a "new" theory,
then this new theory is not predictive in its own right. How could we perform the next higher
order calculations if standard QED hasn't already done it?

If we rely heavily on experimental data to construct a "new" theory, then this new theory
is more phenomenological than standard QED. I don't see how it can be predictive in its
own right. How could it predict what the (future) experimental data would be when higher
accuracy becomes technically possible in the apparatus?

Either way guarantees that the S-matrix calculated with (1) is exactly the same as the S-matrix computed in QED or measured in experiments. The benefits of using interaction (1) are: (i) there is no need for renormalization, (ii) both free and interacting theories live in the same Fock space.

(i) Since the new theory is based on standard QED calculations, it implicitly relies on
the renormalization performed therein.

(ii) Since the new theory is not a non-perturbative theory, we cannot say anything
mathematically rigorous about such limits.

**Bob_for_short** · Nov9-09, 06:05 PM

And how about my questions? They are nearly "Yes or No" ones. Do you have your own opinion?

**strangerep** · Nov9-09, 06:20 PM

Originally Posted by Bob_for_short

And how about my questions? They are nearly "Yes or No" ones. Do you have your own opinion?

Bob, your questions are not "nearly yes or no", but require closer study of your paper.
Part of the problem is one of communication: I know that English is not your first
language, but you must understand that this makes it difficult at times for me to
understand properly what you really mean in your papers.

Also, badgering someone who is clearly willing to make the effort to study
your papers is not the way to win friends.

**Bob_for_short** · Nov9-09, 06:22 PM

No, without studying my papers, please answer the questions of post 65 concerning the standard QED. Forget for instance my papers.
You can answer privately, if you prefer.

**meopemuk** · Nov9-09, 07:07 PM

Hi strangerep,

Yes, I fully agree with your criticism. In its present form, the "dressed particle" approach is not developed to the point, where it can derive the Hamiltonian from some "first principles". The best we can do is to guess the Hamiltonian by relying on experiment or on high-order QED calculations. In this sense the theory is not predictive. However, it has the benefit of being consistent both physically and mathematically. On the other hand, the renormalized QFT is highly predictive, while being inconsistent. I think that both approaches have the right to exist. We'll see who will reach the goal (of being both consistent and predictive) first.

In spite of what I've said above, the "dressed particle" theory CAN make at least one important prediction. "Dressed" Hamiltonians have a characteristic property that interactions propagate instantaneously. In the last 16 years quite a few experiments have shown some signs of superluminal behavior. I am most impressed by the works done by A. Ranfagni et al. in Florence. Most theorists tend to dismiss these observations as some inconsequential curiosities. But I believe that they will be forced to change their attitudes soon. This will be the best argument in favor of the "dressed particle" theory.

Note that despite textbook claims, the traditional QFT cannot say anything about the speed of propagation of interactions. The "commutators of fields outside the light cone" have nothing to do with the time interval between the cause and the effect. In order to evaluate the speed of interactions, one must have a theory possessing a well-defined Hamiltonian and unitary time evolution. As we discussed earlier, the traditional QFT does not have these pieces. It can only calculate the S-matrix and energies of bound states, which do not reveal any time-dependent information.

Eugene.

**DarMM** · Nov10-09, 05:04 AM

Originally Posted by meopemuk

On the other hand, the renormalized QFT is highly predictive, while being inconsistent.

How is it inconsistent?

Originally Posted by meopemuk

Note that despite textbook claims, the traditional QFT cannot say anything about the speed of propagation of interactions. The "commutators of fields outside the light cone" have nothing to do with the time interval between the cause and the effect. In order to evaluate the speed of interactions, one must have a theory possessing a well-defined Hamiltonian and unitary time evolution. As we discussed earlier, the traditional QFT does not have these pieces. It can only calculate the S-matrix and energies of bound states, which do not reveal any time-dependent information.

The investigation of the propogation of effects in QFT was performed by Segal and Guenin in the 1960s for specific models. Also Haag's algebraic apporach allows it to be treated in general, where you can show that effects do propogate at the speed of light.

**meopemuk** · Nov10-09, 05:30 AM

Originally Posted by DarMM

How is it inconsistent?

It is undisputable that renormalized QFT (such as QED) can calculate the S-matrix (i.e., the result of time evolution between - and + infinite times) very accurately. I think we also agreed that this theory does not have a well-defined finite Hamiltonian. Without a Hamiltonian it is impossible to calculate the finite time evolution. That's the major inconsistency I am talking about.

Originally Posted by DarMM

The investigation of the propogation of effects in QFT was performed by Segal and Guenin in the 1960s for specific models. Also Haag's algebraic apporach allows it to be treated in general, where you can show that effects do propogate at the speed of light.

I would appreciate exact references. Though, I am rather sceptical, for the same reason: the lack of a well-defined finite Hamiltonian in renormalized QFT.

Perhaps you are talking about 2D models? I admit, I know almost nothing about them.

**DrFaustus** · Nov10-09, 11:13 AM

Bob_for_short -> What do you mena by later on we discover that the probability of an elastic process is zero? What do you mean that QED predicts an event that never happens?

No one ever claimed perturbation theory is perfect and the ultimate answer to all QFT problems. It has its limits, but renormalization is not one of them.

What has the Taylor power series expansion got to do with the perturbative expansion of, say, QED?

Also, it is very interesting how you ask people to answer your questions whereas you have not answered to my question: what kind of mathematical objects are your quantum fields?

strangerep & DarMM -> As I understand it, the renormalization procedure a la Epstein & Glaser (and hence the "infinite subtraction" one) is not more ad hoc then, say, solving

. Let me elaborate this a bit. I'm trying to obtain an answer from my (perturbation) theory, and to get that answer I must extend my product of distributions to coinciding points where it is in general ill defined. Of course, I'll have to satisfy some conditions (causality and locality), but in principle this is "just" another mathematical problem one has to solve. Much like trying to get an answer about some physical problem which would involve the solution of the above (simple) equation. So in this sense, it does not appear more ad hoc than every other problem in physics. Comments?

meopemuk -> The vanishing of the fields commutator outside the light cone is precisely a statement about the finite speed of the propagation of signals. Don't know the details about interacting theories, but if you consider the simple free massless scalar quantum field you can compute the commutator, which is just the difference of the advanced and the retarded propagator, and study its support properties. And what you find out is that it is only supported ON the light cone, i.e. that "light travels at the speed of light". (I know a scalar field does not describe photons correctly, but the idea is the same - a massless particle travels at the speed of light, which is finite.) For the massive case, the commutator will not be supported only on the light cone, but it has causal support, i.e. on and inside the light cone. Outside the light cone the commutator vanishes identically. And this is just a mathematical fact which is a consequence of the hyperbolic character of the underlying field equations.
(You can find a discussion on causality for the free massive scalar field in Chapter 2.4 of Peskins & Schroeder. It's not really in the "rigorous QFT" spirit, but it is meaningful nevertheless.)

**Bob_for_short** · Nov10-09, 11:38 AM

Originally Posted by DrFaustus

Bob_for_short -> What do you mena by later on we discover that the probability of an elastic process is zero? What do you mean that QED predicts an event that never happens?

I clearly write what I mean. First we calculate some elastic processes. On page 500 (I exaggerate, of course), when the IR divergence is treated, it is stated that the elastic cross section is identically equal to zero. So, instead of predicting zero for elastic processes the standard QED predicts some finite value (in the first non-vanishing order). That means a too bad start. If the exact value were 0.5 and the initial approximation would give 0.45, it would be OK. But the exact value is 0 and the initial calculation gives 1 (the probability of elastic process). It is too far from reality. Take any Taylor expansion around x₀=0 and calculate it at very big x, say x=10. You will obtain the same blinder. Why not to take a Taylor expansion around a closer point, say, at x₀ = 9.9. Then f(9.9) ≈ f(10) and the remaining corrections will be small.

Originally Posted by DrFaustus

What has the Taylor power series expansion got to do with the perturbative expansion of, say, QED?

The magnetic moment is a series in powers of alpha/2pi. I speak of its numerical precision.

Originally Posted by DrFaustus

...what kind of mathematical objects are your quantum fields?

My quantum fileds? Oh, they are terrible distributions. What saves me is their coming with natural regularization factors (charge form-factors).

**Avodyne** · Nov10-09, 01:14 PM

Well I thought I would check in with this thread after not doing so for a while.

Originally Posted by Bob_for_short

So, in the first non-vanishing order the standard QED predicts events that never happen. And it does not predict the phenomenon that happen always (soft radiation). Don’t you consider this theory "feature" as a complete failure in the physics description? Isn’t it a too bad start for the perturbation theory?

Not at all, it's exactly what should happen. The cross section for processes that include up to an energy $LaTeX Code: \\Delta$ in observed soft photons takes the form

$LaTeX Code: {d\\sigma\\over d\\Omega}\\biggr|_{\\rm soft} = {d\\sigma\\over d\\Omega}\\biggr|_{\\rm elastic}\\left({c_1\\Delta\\over E}\\right)^{\\!\\! c_2\\alpha}$

where

is the CM energy, $LaTeX Code: \\Delta$ is the maximum energy of the soft photons, $LaTeX Code: \\alpha$ is the fine-structure constant, and $LaTeX Code: c_{1,2}$ are numerical constants. This does indeed go to zero as $LaTeX Code: \\Delta$ goes to zero. $LaTeX Code: (d\\sigma/d\\Omega)_{\\rm elastic}$ starts with the tree-level term, which is $LaTeX Code: O(\\alpha^2)$ . So, expanding in powers of $LaTeX Code: \\alpha$ , one finds first the tree level term.

**Bob_for_short** · Nov10-09, 01:29 PM

Originally Posted by Avodyne

The cross section for processes that include up to an energy $LaTeX Code: \\Delta$ in observed soft photons takes the form

$LaTeX Code: {d\\sigma\\over d\\Omega}\\biggr|_{\\rm soft} = {d\\sigma\\over d\\Omega}\\biggr|_{\\rm elastic}\\left({c_1\\Delta\\over E}\\right)^{\\!\\! c_2\\alpha}$

where is the CM energy, $LaTeX Code: \\Delta$ is the maximum energy of the soft photons, $LaTeX Code: \\alpha$ is the fine-structure constant, and $LaTeX Code: c_{1,2}$ are numerical constants. This does indeed go to zero as $LaTeX Code: \\Delta$ goes to zero. $LaTeX Code: (d\\sigma/d\\Omega)_{\\rm elastic}$ starts with the tree-level term, which is $LaTeX Code: O(\\alpha^2)$ . So, expanding in powers of $LaTeX Code: \\alpha$ , one finds first the tree level term.

What you wrote is called the inclusive cross section, and I speak of elastic one. The elastic one (the exact and experimental) equals zero rather than > 0. The QED's one on the tree level is > 0 which is wrong. You have to work hard to cope with the IR divergence to finally obtain a physically reasonable result instead of obtaining it automatically, if your theory "catches" the physics right.

Let me put it in another way: whatever momentum q is transferred to the electron, no soft radiation appears on the tree level. Is it physical?

**meopemuk** · Nov10-09, 02:17 PM

Originally Posted by DrFaustus

The vanishing of the fields commutator outside the light cone is precisely a statement about the finite speed of the propagation of signals.

I agree that free particles move at the speed of light or slower. But this fact has no relationship to the issue of the speed of propagation of interactions or (what is basically the same) signals. In order to say something about the speed of interactions/signals you must have a well-defined interacting theory. In the simplest case you should consider two particles at a distance R and solve a time-dependent problem in which you perturb the position of the particle 1 at time t=0 and determine the time at which this perturbation reaches the particle 2. I don't think this kind of problem can be analyzed within renormalized QFT, due to the lack of a well-defined finite Hamiltonian. And I don't think that "vanishing of field commutators" is relevant to the solution.

**Vanadium 50** · Nov10-09, 02:47 PM

This thread has turned into a discussion of Bob For Short's theory. I remind everyone that personal theories can be discussed only in the IR section.

I'm locking this thread. If the OP feels that his question hasn't been adequately addressed before this thread was derailed, he should start another one.