Does QED Originate from Non-Relativistic Systems?

[A. Neumaier]

Although each term is finite, the series is divergent.

This doesn't matter for prediction. The series is asymptotic, and the divergence is expected not to show up before loop order 100. The results at loop order 6 are already as accurate as the best experimental results.

Compare this with the poor accuracy obtained by noncovariant lattice theories. They give finite results at each lattice spacing, but none of the lattice spacings for which computations can be carried out gives results matching expoeriment, and it is not known whether (or in which sense) the limit exists in which the lattice spacing goes to zero. Thus lattice QED is according to your own criterion (convergence) in a worse state than the covariant theory.

 

[atyy]

That each term is finite is enogh to claim local Poincare invariance which is the actual mathematical claim of Poincare invariance in perturbative QED, the one that gives outstanding predictions.

This doesn't matter for prediction. The series is asymptotic, and the divergence is expected not to show up before loop order 100. The results at loop order 6 are already as accurate as the best experimental results.

Compare this with the poor accuracy obtained by noncovariant lattice theories. They give finite results at each lattice spacing, but none of the lattice spacings for which computations can be carried out gives results matching expoeriment, and it is not known whether (or in which sense) the limit exists in which the lattice spacing goes to zero. Thus lattice QED is according to your own criterion (convergence) in a worse state than the covariant theory.

If you only care about predictions, then QED is not a quantum theory!

What is the Hilbert space?

 

[Demystifier]

''Quantum fields exist in space and time, which can be approximated by a set of lattice points.''

Well, I don't remember where exactly I have seen the statement that lattice QFT can be viewed as a rigorous definition of QFT. But I know I have seen that, at more than one place.

 

[RockyMarciano]

If you only care about predictions, then QED is not a quantum theory!

What is the Hilbert space?

Quantum theory in its renormalized interacting quantum field form that gives accurate predictions is not depending on a Hilbert space in the usual concept of space of square integrable functions, but on more specific instances of Hilbert spaces like Hardy fixed point spaces.

 

[atyy]

Here is the state of the art:

At the physics level of rigour: we do not have a version of QED that extends to all energies, hence we do not have Poincare invariant QED. We do believe that QCD extends to all energies because of Asymptotic Freedom, hence we do have Poincare invariant QCD. However, gravity is not renornalizable, so we do not have a version of quantum gravity that is Poincare invariant. Certainly, these do not preclude future discoveries of Asymptotic Safety, or UV completions with more degrees of freedom introduced.

At the rigourous level, we do not have any 3+1 dimensional interacting Poincare invariant QFT.

 

[martinbn]

Just to be able to follow the subtle discussion, can some one explain what is meant by Poincare invariant QFT.

 

[RockyMarciano]

Here is the state of the art:

At the physics level of rigour: we do not have a version of QED that extends to all energies, hence we do not have Poincare invariant QED. We do believe that QCD extends to all energies because of Asymptotic Freedom, hence we do have Poincare invariant QCD. However, gravity is not renornalizable, so we do not have a version of quantum gravity that is Poincare invariant. Certainly, these do not preclude future discoveries of Asymptotic Safety, or UV completions with more degrees of freedom introduced.

At the rigourous level, we do not have any 3+1 dimensional interacting Poincare invariant QFT.

True, but this is not what is meant by the local, i.e. term by term Poincare invariance of QED, that is what many physicist hope for but hasn't been proved and is even subject of a million dollar prize.

 

[A. Neumaier]

What is the Hilbert space?

For practical purpuses (indeed, to make predictions with QED at finite times, e.g., to derive the macroscopic Maxwell equations) the Hilbert space is defined through the closed time path (CTP) formalism. Whether this has a rigorous formulation is an open problem.

If you only care about predictions, then QED is not a quantum theory!

Most of theoretical physics (including quantum mechanics) is not fully rigorous. Quantum mechanics is not a sub-discipline of mathematical physics; only the latter requires everything to be rigorously defined.

 

[vanhees71]

A Poincare invariant QFT is one leading to a Poincare invariant S matrix. There is no rigorous definition of a nonperturbative/exact interacting QED in 1+3 spacetime dimensions. We only have the formalism for loop-order-by-loop-order covariant perturbation theory, and for QED that's quite satisfactory in comparison to experiment (Lamb shift of the hydrogen atom, anomalous magnetic moment of the electron, etc.).

Further, it's clear that to define the perturbative expressions beyond tree level you have to either first regularize and then renormalize (the standard way since 1971 is to use dimensional regularization, because it's most convenient by satisfying the important symmetries like gauge and Poincare symmetries for a large class of models) or to just renormalize by using the BPHZ formalism which subtracts systematically all divergences of the loop integrals directly in the integrands. This is, however just a technical question. The final result are the S-matrix elements expressed in renormalized and finite quantities at any order of perturbation theory.

The lattice approach is a special way to regularize QFT, and it's used as a non-perturbative approach (mostly in QCD, not QED). Also there of course, the final results one is really interested in are the continuum extrapolations, as Arnold stressed several times.

Of course, there's also a non-relativistic limit of both QED and QCD, but I don't see, in how far this is related to this quite general discusssion.

 

[Demystifier]

Just to be able to follow the subtle discussion, can some one explain what is meant by Poincare invariant QFT.

I think you brilliant answer can be applied:
An argument against Bohmian mechanics?


Now seriously, it is QFT whose action is invariant under Poincare group, where Poincare group is
https://en.wikipedia.org/wiki/Poincaré_group
See also Sec. 1.5 of Ticciati, QFT for Mathematicians (I think you said you liked that book).

 

[atyy]

Most of theoretical physics (including quantum mechanics) is not fully rigorous. Quantum mechanics is not a sub-discipline of mathematical physics; only the latter requires everything to be rigorously defined.

Sure, but at the non-rigourous level it is widely agreed that QCD, but not QED, has been demonstrated to exist at all energies.

 

[A. Neumaier]

What is the Hilbert space?[/QUOTE]

lattice QFT can be viewed as a rigorous definition of QFT.

It can be viewed as a rigorous definition of approximations only. In the view of theoretical physics this may be rigorously enough, but renormalized perturbation theory (with selective resummation in case of strong coupling) is as rigorous as lattice theory.

 

[A. Neumaier]

At the physics level of rigour: we do not have a version of QED that extends to all energies, hence we do not have Poincare invariant QED. We do believe that QCD extends to all energies because of Asymptotic Freedom, hence we do have Poincare invariant QCD.

This is a strange view of the state of the art.

We have a covariant version of QED valid to any loop order, giving excellent computational results at all energies that will ever be experimentally accessible, far beyond the Planck level. We also have a Poincare invariant QCD, not because of a belief that only affects energies beyond all experimental accessibility but because of renormalization group improved perturbation theory (needed because of the strong interaction), though it is still in a far less good state than QED.

 

[A. Neumaier]

Sure, but at the non-rigourous level it is widely agreed that QCD, but not QED, has been demonstrated to exist at all energies.

It is just a matter of belief, not of demonstration. Whereas it is a well-known fact that QED is Poincare invariant in each finite loop approximation. and is demonstated to have a predictive power far beyond QCD. What happens at energies above the Planck scale is completely irrelevant for QED; it is a matter for the future unified theory.

Perturbative QED does not suffer from the Landau pole; only a nonperturbative version possibly does, if one attempts to construct the theory using lattices (!) or using a cutoff. The Landau pole invalidates a construction only if the (lattice or energy) cutoff has to move through the pole in order to provide a covariant limit.

On the other hand, while the causal, covariant construction of QED also has a Landau pole, Landau's argument that a Landau pole invalidates perturbation theory no longer applies to this version. In the causal construction there are no cutoffs that must be sent to zero or infinity and move through the pole. The pole is only in the choice of the renormalization point (or mass $M$). But in the exact theory, the theory is completely independent of this renormalization point; so it can be chosen at low energy without invalidating the construction!

The associated Callan-Symanzik equation shows how the observables depend on the chosen renormalization mass $M$, giving an identical theory - apart from truncation errors, which are of course small only if $M$ is of the order of the energies at which predictions are made. Thus for any range of $M$ for which the Callan-Symanzik equation is solvable, one gets the same theory. The Landau pole found (nonrigorously) in causal QED in the Callan-Symanzik equation only means that one cannot connect the theory defined by a superhigh renormalization point irrelevant for the physics of the universe to the theory defined by a renormalization point below the Landau pole. This is a harmless situation.

It is quite different from Landau's argument that the cutoff cannot be removed because on the way from a small cutoff energy to an infinite cutoff energy perturbation theory becomes invalid since the terms become infinite when the cutoff passes the pole. This means that perturbation theory fails close to this cutoff. Thus any approximate construction with cutoff featuring a Landau pole cannot be made to approximate the covariant QED to arbitrary precision. In particular, the Landau pole believed to exist in lattice QED is of this fatal kind and proves (at this level of rigor) that lattice QED can never approach the covariant QED, hence is a fake theory. Your insistence on the Landau pole defeats even your basic goal!

 

[A. Neumaier]

Just to be able to follow the subtle discussion, can some one explain what is meant by Poincare invariant QFT.

Poincare invariant = Lorentz invariant and translation invariant.

Translation invariance guarantees the conservation of energy and momentum, and Lorentz invariance is the minimal requirement for relativistic causality. (It is not quite sufficient since for causality one also needs hyperbolicity.)

 

[RockyMarciano]

Poincare invariant = Lorentz invariant and translation invariant.

Why be sloppy and leave room for confusing people when it is so easy being precise? In this context Poincare invariant refers to invariance under the proper orthochronous component of the Lorentz group and infinitesimal translations.

 

[A. Neumaier]

Why be sloppy and leave room for confusing people when it is so easy being precise? In this context Poincare invariant refers to invariance under the proper orthochronous component of the Lorentz group and infinitesimal translations.

This is not more precise but only more technical and confusing.

There are as many different versions of Poincare invariance as there are different Lorentz groups. Singling out only one of them is distracting. And translation invariance is commonly not assumed to be infinitesimal but global.

 

[RockyMarciano]

This is not more precise but only more technical and confusing.

There are as many different versions of Poincare invariance as there are different Lorentz groups. Singling out only one of them is distracting.

Not if it is the one that applies.

And translation invariance is commonly not assumed to be infinitesimal but global.

Mathematically is only justified locally, you have admitted that there is Poincare invariance only at every finite approximation order by order, this only allows to talk about infinitesimal translations, if you disagree prove it and there is a million dollars prize and probable a nobel for you.

 

[A. Neumaier]

that there is Poincare invariance only at every finite approximation order by order, this only allows to talk about infinitesimal translations

Nonsense. One can compute order by order all $N$-point functions, and these are globally Poincare invariant.

Not if it is the one that applies.

In case of QED, the QFT under discussion in this thread, the full Poincare group applies. It is 4 times as big as the one you mentioned.

 

[RockyMarciano]

Nonsense. One can compute order by order all $N$-point functions, and these are globally Poincare invariant.

In case of QED, the QFT under discussion in this thread, the full Poincare group applies. It is 4 times as big as the one you mentioned.

I guess you have never heard of spacetime anomalies, and clearly you don't care about microcausality. We are talking about the QFT that gives precise predictions, not about free QED, you keep mixing them.

 

[RockyMarciano]

I'll quote vanhees71 to show why you are wrong about global Poincare invariance

A Poincare invariant QFT is one leading to a Poincare invariant S matrix. There is no rigorous definition of a nonperturbative/exact interacting QED in 1+3 spacetime dimensions. We only have the formalism for loop-order-by-loop-order covariant perturbation theory, and for QED that's quite satisfactory in comparison to experiment .

 

[A. Neumaier]

I guess you have never heard of spacetime anomalies, and clearly you don't care about microcausality. We are talking about the QFT that gives precise predictions, not about free QED, you keep mixing them.

What you write is nonsense. Anomalies are absent in QED, the N-point functions of k-loop QED are invariant under the full Poincare group and satisfy the consequences of microcausality to order k, and QED has never been free.

 

[RockyMarciano]

Here's another description of what I'm saying http://physics.stackexchange.com/qu...state-of-a-poincare-symmetric-qft-vanish?rq=1
It should be enough to clarify Neumaier's confusion, he is taking for granted that the million dollar question mentioned in the link has already been affirmatively answered. That is just wishful thinking.

 

[dextercioby]

There's only one QED and that exists only perturbatively using formal (i.e. mathematically not rigorous) mathematics. QED is an interaction theory under the assumption that the interactions between electrons, muons, taons, photons and their antiparticles take place at finite times, but at + and - infinite time each particle (its field) is free, hence described by quantum fields satisfying the Wightman axioms (or their Segal-Haag-Ruelle algebraic counterparts).

This mathematically inaccurate QED which eveybody learns about at university level is full Poincare invariant, i.e. both spatial and time reflections and their products included, but the general assumptions about a mathematically sound QFT (spelled in the Wightman's axiomes) include only the identity-connected component of the Poincare group, known in the physics community as the restricted Poincare group. If full Poincare symmetry is required instead of the restricted one, then there would be no room for neutrinos and weak interactions in the Standard Model.

 

[dextercioby]

I guess you have never heard of spacetime anomalies, and clearly you don't care about microcausality. We are talking about the QFT that gives precise predictions, not about free QED, you keep mixing them.

The QFTs that give "precise predictions" are the sub-teories of the Standard Model, which makes sense (formally, i.e. not mathematically) as a whole essentially perturbatively and globally (at the level of asymptotic (free) relativistic fields) Poincare invariant. What is a spacetime anomaly?

 

[atyy]

It is just a matter of belief, not of demonstration. Whereas it is a well-known fact that QED is Poincare invariant in each finite loop approximation. and is demonstated to have a predictive power far beyond QCD. What happens at energies above the Planck scale is completely irrelevant for QED; it is a matter for the future unified theory.

Perturbative QED does not suffer from the Landau pole; only a nonperturbative version possibly does, if one attempts to construct the theory using lattices (!) or using a cutoff. The Landau pole invalidates a construction only if the (lattice or energy) cutoff has to move through the pole in order to provide a covariant limit.

On the other hand, while the causal, covariant construction of QED also has a Landau pole, Landau's argument that a Landau pole invalidates perturbation theory no longer applies to this version. In the causal construction there are no cutoffs that must be sent to zero or infinity and move through the pole. The pole is only in the choice of the renormalization point (or mass $M$). But in the exact theory, the theory is completely independent of this renormalization point; so it can be chosen at low energy without invalidating the construction!

The associated Callan-Symanzik equation shows how the observables depend on the chosen renormalization mass $M$, giving an identical theory - apart from truncation errors, which are of course small only if $M$ is of the order of the energies at which predictions are made. Thus for any range of $M$ for which the Callan-Symanzik equation is solvable, one gets the same theory. The Landau pole found (nonrigorously) in causal QED in the Callan-Symanzik equation only means that one cannot connect the theory defined by a superhigh renormalization point irrelevant for the physics of the universe to the theory defined by a renormalization point below the Landau pole. This is a harmless situation.

It is quite different from Landau's argument that the cutoff cannot be removed because on the way from a small cutoff energy to an infinite cutoff energy perturbation theory becomes invalid since the terms become infinite when the cutoff passes the pole. This means that perturbation theory fails close to this cutoff. Thus any approximate construction with cutoff featuring a Landau pole cannot be made to approximate the covariant QED to arbitrary precision. In particular, the Landau pole believed to exist in lattice QED is of this fatal kind and proves (at this level of rigor) that lattice QED can never approach the covariant QED, hence is a fake theory. Your insistence on the Landau pole defeats even your basic goal!

So if QED does not even exist at energies above the Planck scale, how can it be Poincare invariant?

There is no need for lattice QED to reproduce QED to arbitrary precision, since QED does not have arbitrary precision as it is only a low energy effective theory.

There is of course a "Landau pole" in lattice QED, because the point is it acknowledges we only need a low energy effective theory. The point of the lattice is to provide a conceptually secure starting point for Wilsonian renormalization.

What you are assuming is that Wilsonian renormalization starts from a UV complete Poincare invariant QFT, whose low energy effective theory is covariant perturbative QED. That is fine. However, since we don't know of any such theory at the moment, it is just as good to start from lattice QED if one wants to start from a well-defined quantum theory.

 

[Demystifier]

It can be viewed as a rigorous definition of approximations only.

All theories in physics are approximations. If we are condemned to work only with approximations, it helps when, at least, the approximation can be defined rigorously.

 

[vanhees71]

Why be sloppy and leave room for confusing people when it is so easy being precise? In this context Poincare invariant refers to invariance under the proper orthochronous component of the Lorentz group and infinitesimal translations.

This is important since indeed the standard model is invariant only under the continuous part of the Poincare group, which is a semidirect product of the proper orthochronous Lorentz group with the group of translations. The weak interaction breaks P, T, and when extended by the charge-conjugation transformation also CP. As local relativistic QFT, of course CPT is a symmetry, and so far there's indeed no empirical evidence that CPT is broken anywhere.

Concerning the rest of the debate, from a strict mathematical view there's no conclusive answer to the question whether QED in (1+3) spacetime dimensions exists. It's quite likely that it doesn't exist because of a quite probable Landau pole. Arnold is right in saying that QED exists at any order of perturbation theory in the sense that you can define the renormalized effective action (and thus the proper vertex functions) truncated at any finite loop order. However, already the propagators, i.e., the resummation of the Dyson equation to any order is plagued by a Landau pole, but in the case of QED at very high energies. In this sense QED is indeed only an effective theory valid for low enough energies. However this energy limit is so high that it doesn't play a practical role at all for today's particle physics.

Of course, from a theoretical point of view, it's well justified to seek for a better theory than the Standard Model. Unfortunately so far, there's no success in this endeavor, neither from the axiomatic-QFT side nor from the phenomenological side since at the moment all the "signals indicating physics beyond the standard model" are disproven at the $5 \sigma$ or higher significance level.

 

[atyy]

This is important since indeed the standard model is invariant only under the continuous part of the Poincare group, which is a semidirect product of the proper orthochronous Lorentz group with the group of translations. The weak interaction breaks P, T, and when extended by the charge-conjugation transformation also CP. As local relativistic QFT, of course CPT is a symmetry, and so far there's indeed no empirical evidence that CPT is broken anywhere.

Concerning the rest of the debate, from a strict mathematical view there's no conclusive answer to the question whether QED in (1+3) spacetime dimensions exists. It's quite likely that it doesn't exist because of a quite probable Landau pole. Arnold is right in saying that QED exists at any order of perturbation theory in the sense that you can define the renormalized effective action (and thus the proper vertex functions) truncated at any finite loop order. However, already the propagators, i.e., the resummation of the Dyson equation to any order is plagued by a Landau pole, but in the case of QED at very high energies. In this sense QED is indeed only an effective theory valid for low enough energies. However this energy limit is so high that it doesn't play a practical role at all for today's particle physics.

Of course, from a theoretical point of view, it's well justified to seek for a better theory than the Standard Model. Unfortunately so far, there's no success in this endeavor, neither from the axiomatic-QFT side nor from the phenomenological side since at the moment all the "signals indicating physics beyond the standard model" are disproven at the $5 \sigma$ or higher significance level.

But that is not really what is being discussed. The question being discussed is:

In the Wilsonian viewpoint, is lattice QED at fine but finite lattice spacing a conceptually adequate starting point for obtaining perturbative Poincare invariant QED as a low energy effective QFT?

 

[Demystifier]

But that is not really what is being discussed. The question being discussed is:

In the Wilsonian viewpoint, is lattice QED at fine but finite lattice spacing a conceptually adequate starting point for obtaining perturbative Poincare invariant QED as a low energy effective QFT?

Exactly! Or to put it in different words, the question is not whether lattice QED is more fundamental than Poincare invariant QED. The question is whether Poincare invariant QED at low energies can be derived from lattice QED in the low energy limit. Just because theory A can be derived from theory B does not imply that theory B is more fundamental than theory A.

Let me use an analogy. Fluid mechanics can be derived from a naive theory of atoms, in which atoms are hard balls. But it does not mean that such a naive theory of atoms is more fundamental than fluid mechanics.

U funny thing about QFT is that lattice QFT can be obtained as an approximation of continuous QFT, but also continuous QFT can be obtained as an approximation of lattice QFT. A string theorist would conjecture that this implies the existence of a mysterious M-theory, in which both QFT theories are special limits of the fundamental M-theory.

 

[A. Neumaier]

he is taking for granted that the million dollar question mentioned in the link has already been affirmatively answered.

No. The latter is about proving rigorously the existence of quantum Yang-Mills theory and is still an open problem. This problem has nothing to do with QED, which is the topic of the present discussion.

 

[A. Neumaier]

If full Poincare symmetry is required instead of the restricted one, then there would be no room for neutrinos and weak interactions in the Standard Model.

True but irrelevant for QED, which (in the version under discussion) by definition is only about electromagnetic fields, electrons, and positrons. We are not discussing the standard model.

 

[A. Neumaier]

The question is whether Poincare invariant QED at low energies can be derived from lattice QED in the low energy limit.

This was not the original question.

But even this question has a negative answer. Getting Poincare invariance requires the limit where the lattice spacing goes to zero, which is incurably problematic because of a Landau pole. The literature indicates that this limit gives a free theory only. Thus lattice QED is probably unrelated to true Poincare invariant QED as discussed in the textbooks. The whole question of triviality is discussed in detail on PhysicsOverflow.

 

[A. Neumaier]

if QED does not even exist at energies above the Planck scale, how can it be Poincare invariant?

Because the hypothesis of your question is invalid. Your argument from belief means nothing; it is a mere belief.

The question of the existence of QED is widely open, even the question of existence of $\Phi^4$ theory (see Section 8 of this paper, which discusses the state of the art from a rigorous point of view), for which the triviality arguments (i.e., the impossibility to construct it as a scaling limit from the corresponding lattice theory) are overwhelming.

The causal construction of QED avoids all these triviality arguments as it doesn't construct QED as a limit of theories with a cutoff but directly from a mathematical characterization by means of covariant axioms by Bogoliubov - not the Wightman axioms. Each loop order is valid at all energies. The only unsolved question is how to resum the series to make the construction nonperturbative. There is not the slightest argument indicating that this is impossible. And as mentioned, the constructive question is wide open.

 

[Demystifier]

But even this question has a negative answer. Getting Poincare invariance requires the limit where the lattice spacing goes to zero ...

Note that my question contains "at low energies" caveat. For physics at low energies, it is not relevant to know what happens in the limit of zero lattice spacing.

Frankly, I don't give a damn how any QFT without quantum gravity behaves at very small distances (Planck distance and less) because I believe that, at such small distances, QFT without quantum gravity has nothing to do with reality.