Does QED Originate from Non-Relativistic Systems?

[A. Neumaier]

Try
https://www.amazon.com/dp/0521599172/?tag=pfamazon01-20

From the short paragraph summarizing the content of the book on the page you linked to:

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

This is the opposite of atyy's and your claim. They distinguish perfectly between approximations and the real thing, already in the advertisenment of their book.

 

[atyy]

1-loop Lorentz invariant QED is fully (and higher loop QED conceptually) constructed without any cutoff or regularization or lattices in Scharf's book on quantum electrodynamics. And his book on a true ghost story does the same for other gauge theories.

Scharf's construction is divergent, so it is not an example of Poincare invariant QED.

 

[A. Neumaier]

Scharf's construction is divergent,

There is no divergence in any of the Lorentz invariant 1-loop results provided by Scharf, and these give already much better agreement with experiment than lattice QED. 1-loop QED is the version of QED (though with a different derivation) for which Feynman, Tomonaga and Schwinger got the Nobel prize!

 

[atyy]

There is no divergence in any of the Lorentz invariant 1-loop results provided by Scharf, and these give already much better agreement with experiment than lattice QED. 1-loop QED is the version of QED (though with a different derivation) for which Feynman, Tomonaga and Schwinger got the Nobel prize!

Although each term is finite, the series is divergent.

 

[RockyMarciano]

Although each term is finite, the series is divergent.

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.

 

[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?