Haag's Theorem: Importance & Implications in QFT

[Qubix]

So I'm currently studying QFT, and I got to the point where I realized that the S operator, initially assumed to be unitary, is not unitary anymore, since it is assumed to act between t0 = - infinity and tf = infinity. The author of the book I'm using says this is due to Haag's Theorem, so I gave that a search, and found a rather interesting bunch on information.

http://en.wikipedia.org/wiki/Haag's_theorem

It seems that the Interaction Picture does not exist in QFT. What is the importance of this theorem and what are its other implications?

 

[George Jones]

See also section 10.5, "How to stop worrying about Haag's theorem", in The Conceptual Framework of Quantum Field Theory by Anthony Duncan (Oxford, 2012).

Also, the interaction Hamiltonian doesn't exist as a genuine mathematical object. From Folland's "Quantum Field Theory for Mathematicians: A Tourist's Guide for Mathematicians" (a book that I highly recommend) page 123

"... the $H_I$'s that we shall need are too singular."

Too singular, because they products of distributions. From page 180

"A precise mathematical construction of the interacting fields that describe actual fundamental physical processes in 4-dimensional space-time is still lacking and may not be feasible without serious modifications to the theory. Similarly, we have no way to define the Hamiltonian $H$ in a mathematically rigorous way as a self-adjoint operator. It was presented as the sum of the free Hamiltonian $H_0$, which is well-defined and the interaction Hamiltonian $H_I$; but the latter was presented as the integral of a density consisting of products of fields, which are operator-valued distributions rather than functions."

 

[atyy]

Most physical QFT is not rigourous. You can roughly think of it as a many-body system on a lattice and QFT emerges at low energies. This sort of thinking works for QCD, but there is some problem for chiral gauge theories. Regardless the idea is that physical QFTs are only low energy effective theories, and may not exist at all energies. http://quantumfrontiers.com/2013/06/18/we-are-all-wilsonians-now/

In rigourous QFT, the aim is to construct a Lorentz invariant QFT that exists for all energies. Haag's theorem applies to such theories. I've found these comments about the theorem useful.

http://www.rivasseau.com/resources/book.pdf, p15: "The Gell-Mann Low formula ... is difficult to justify because the usual argument, based on the so called "interaction picture" is wrong, by a theorem of Haag ... Euclidean formulation ... theorems which allow us to go back from Euclidean to Minkowski space ..."

Also, https://webspace.utexas.edu/lupher/www/papers/KronzLupherPreprint.pdf , p7: "If representations are unitarily inequivalent, one is left to wonder whether they are at least empirically equivalent in some sense. If a reasonable notion of empirical equivalence cannot be found, then it will be necessary to introduce criteria for representation selection. To resolve this selection problem, Haag and Kastler (1964) introduced the notion of physical equivalence, which is related to Fell’s (1960) notion of weak equivalence, and a theorem proved by Fell, which they reformulate in terms of physical equivalence."

 

[dextercioby]

Please, also consider this source:

http://d-scholarship.pitt.edu/8260/1/D_Fraser_2006.pdf

 

[Avodyne]

In rigourous QFT, the aim is to construct a Lorentz invariant QFT that exists for all energies. Haag's theorem applies to such theories.

In 3+1 dimensions, most QFTs are believed not to exist at all (only asymptotically free theories are believed to exist). So any attempt at a rigorous construction of a non-asymptotically free theory in 3+1 dimensions should fail.

See also section 10.5, "How to stop worrying about Haag's theorem", in The Conceptual Framework of Quantum Field Theory by Anthony Duncan (Oxford, 2012).

Short version: QFT requires an ultraviolet regulator (such as a lattice), and Haag's theorem does not apply when the regulator is in place.

 

[atyy]

In 3+1 dimensions, most QFTs are believed not to exist at all (only asymptotically free theories are believed to exist). So any attempt at a rigorous construction of a non-asymptotically free theory in 3+1 dimensions should fail.

See also section 10.5, "How to stop worrying about Haag's theorem", in The Conceptual Framework of Quantum Field Theory by Anthony Duncan (Oxford, 2012).

Short version: QFT requires an ultraviolet regulator (such as a lattice), and Haag's theorem does not apply when the regulator is in place.

3+1? You must be a squalid state physicist who can't add. All HEP people say 4 :)

(I'm a biologist.)

 

[dextercioby]

For people seeing physics beyond the Standard Model, then 3+1 instead of 4 is a must.

 

[George Jones]

Stating the signature 3 + 1 is essential.

For people seeing physics beyond the Standard Model, then 3+1 instead of 4 is a must.

An for people working with models in 1 + 1 and 2 + 1.

 

[Qubix]

In 3+1 dimensions, most QFTs are believed not to exist at all (only asymptotically free theories are believed to exist). So any attempt at a rigorous construction of a non-asymptotically free theory in 3+1 dimensions should fail.

See also section 10.5, "How to stop worrying about Haag's theorem", in The Conceptual Framework of Quantum Field Theory by Anthony Duncan (Oxford, 2012).

Short version: QFT requires an ultraviolet regulator (such as a lattice), and Haag's theorem does not apply when the regulator is in place.

First of all, what is an "asymptotically free theory" ? Second, I don't know how you apply a lattice to QFT. From what wikipedia says, "Every lattice in ℝ^n can be generated from a basis for the vector space by forming all linear combinations with integer coefficients." So taking linear combinations of the eigenbasis of a Hilbert space, with integer coefficients gives you a lattice?

 

[Avodyne]

http://en.wikipedia.org/wiki/Asymptotic_freedom

http://en.wikipedia.org/wiki/Lattice_field_theory

 

[atyy]

One formal possibility in addition to asymptotic freedom is asymptotic safety. However, as Avodyne says, currently no 3+1 theory has been shown to be asymptotically safe without being asymptotically free. It is being researched if 3+1 gravity might be asymptotically safe, even though it is not asymptotically free.

Asymptotic freedom/safety are physics concepts and not rigourous. However, it is widely believed that it is not worth attempting a rigourous construction of a theory, unless it has been shown to be asymptotically free or safe. 3+1d Yang-Mills is asymptotically free, which is why its rigourous construction has been posed as a Clay Millenium problem.

 

[Demystifier]

See also section 10.5, "How to stop worrying about Haag's theorem", in The Conceptual Framework of Quantum Field Theory by Anthony Duncan (Oxford, 2012).

Short version: QFT requires an ultraviolet regulator (such as a lattice), and Haag's theorem does not apply when the regulator is in place.

That's exactly what I wanted to say.

 

[kith]

Could this lattice be associated with a fundamental discreteness of spacetime in the Planck regime or are these different concepts?

 

[Avodyne]

We have to distinguish between the real world and mathematical formulations of a particular theory. We've been discussing the latter here. In this context, the lattice is just a technical tool.

True discreteness at the Planck scale is an intriguing possibility. That would provide a real-world ultraviolet cutoff for the Standard Model. But then there are many other possibilities for an "ultraviolet completion" of the Standard Model + gravity that don't involve fundamental discreteness (e.g. zillions of string models).

 

[DarMM]

Short version: QFT requires an ultraviolet regulator (such as a lattice), and Haag's theorem does not apply when the regulator is in place.

QFTs do not require a regulator, although perhaps you mean this in a practical sense, i.e. to work with QFTs at our current level of mathematics requires a regulator at intermediate steps.

Also Haag's theorem is not affected by the ultraviolet cutoff, Haag's theorem applies even with an ultraviolet cutoff in place. You need an infrared cutoff to prevent Haag's theorem.

 

[Demystifier]

Also Haag's theorem is not affected by the ultraviolet cutoff, Haag's theorem applies even with an ultraviolet cutoff in place. You need an infrared cutoff to prevent Haag's theorem.

To prevent Haag's theorem, is the IR cutoff enough, or do we need both UV and IR cutoff?

 

[Demystifier]

QFTs do not require a regulator, although perhaps you mean this in a practical sense, i.e. to work with QFTs at our current level of mathematics requires a regulator at intermediate steps.

Can you be more specific how, at least in principle, one can work with QFT without a regulator?

 

[andrien]

BPHZ approach does not use a regulator.

 

[Demystifier]

BPHZ approach does not use a regulator.

OK, but BPHZ method subtracts an infinite quantity from an infinite quantity, which is not well defined at the mathematical level. The Haag's theorem, on the other hand, assumes that one works with quantities which are well defined mathematically. Physically, the BPHZ method postulates that the mathematically ill defined expression is actually equal to a finite coupling constant extracted from experiments. In this way, BPHZ is also a dirty trick which replaces an infinite quantity by a finite one, so in this sense it is not very different from a "regularization".

 

[atyy]

To prevent Haag's theorem, is the IR cutoff enough, or do we need both UV and IR cutoff?

Can you be more specific how, at least in principle, one can work with QFT without a regulator?

As I understand it, physical QFT does not require it to exist at all energies, a cut-off is fine, and QFT is just a low energy effective theory.

I think DarMM is saying that there are nonetheless some nonlinear relativistic QFTs that exist at all energies, and in infinite volume. In my understanding, the AdS/CFT proposal assumes that the CFT exists for all energies, which is why it is conjectured to provide a non-perturbative definition of some sector of string theory.

http://www.claymath.org/sites/default/files/yangmills.pdf (p8) mentions that relativistic QFTs that exist at all energies, and in infinite volume, have been constructed in fewer than 3+1 spacetime dimensions.

 

[George Jones]

Can you be more specific how, at least in principle, one can work with QFT without a regulator?

As I understand it, physical QFT does not require it to exist at all energies, a cut-off is fine, and QFT is just a low energy effective theory.

A cut-off is (one of the types of) regularization.

 

[atyy]

A cut-off is (one of the types of) regularization.

Yes, physically-relevant QFT, like the standard model, requires a cut-off. However, the article I linked to goes on to say that nonlinear relativistic QFTs in fewer than 4 spacetime dimensions in infinite volume at all energies have been rigourously constructed. Yang-Mills in 4D is asymptotically free, which is why it is a candidate for rigourous construction of a 4D QFT without a cut-off.

 

[Avodyne]

QFTs do not require a regulator

They don't require a regulator if they exist. But it is widely believed by experts that, in 3+1D, only asymptotically-free theories exist.

Also Haag's theorem is not affected by the ultraviolet cutoff, Haag's theorem applies even with an ultraviolet cutoff in place.

I disagree, and my opinion is backed up by Duncan's book.

 

[atyy]

I disagree, and my opinion is backed up by Duncan's book.

Duncan seems to say on his p369 that both the ultraviolet cutoff and the finite volume cut-off are needed to have the interaction picture. IIRC, sometime ago on these forums Dr Du said the same thing about condensed matter in the infinite volume limit. In condensed matter physically it can't really matter, since physically any material in the lab has a natural ultraviolet cutoff and finite volume.

 

[DarMM]

I disagree, and my opinion is backed up by Duncan's book.

Haag's theorem is a theorem proving the non-existence of the interaction picture provided the theory is translation invariant, i.e. if there is no infrared cutoff, it does not discuss the ultraviolet cutoff.

Basically Haag's theorem proves that free and interacting theories are unitarily inequivalent unless an infrared cutoff is in place.

There is no such general result for ultraviolet cutoffs because there are counterexamples. For example $\phi^4$ without an ultraviolet cutoff (but with an infrared cutoff) is unitarily equivalent to the free theory in 1+1 dimensions and so the interaction picture does exist.

For interaction picture to exist sometimes both an ultraviolet and infrared cutoff is needed, sometimes only an infrared cutoff. Haag's theorem is the statement that an infrared cutoff is always needed for the interaction picture to exist.

 

[DarMM]

To prevent Haag's theorem, is the IR cutoff enough, or do we need both UV and IR cutoff?

As above, Haag's theorem is the statement that you always need an IR cutoff. Sometimes, although it is rare, the interaction picture can exist without an ultraviolet cutoff.

 

[DarMM]

Can you be more specific how, at least in principle, one can work with QFT without a regulator?

A QFT doesn't require a regulator, in the sense that it can exist mathematically without one. For practical calculations however one almost always needs a cutoff, since we currently aren't capable of directly solving a QFT without help from a field theory, which is singular with respect to the interacting theory and so we see infinities.

However there is the Epstein-Glaser formalism which requires no cutoffs (Infrared or ultraviolet) nor subtracting infinite quantities, instead it uses the Hahn-Banach theorem.

 

[DarMM]

They don't require a regulator if they exist. But it is widely believed by experts that, in 3+1D, only asymptotically-free theories exist.

Yes true, so quantum field theories in general do not require a regulator. Only the ones without a continuum limit do.

 

[tom.stoer]

Examples for IR cutoffs are torus compactifications (which do not break translational invariance but require e.g. periodic boundary conditions)

Examples for theories w/o UV cutoffs are super-renormalizable theories in 1+1 dimension.

 

[Demystifier]

For interaction picture to exist sometimes both an ultraviolet and infrared cutoff is needed, sometimes only an infrared cutoff. Haag's theorem is the statement that an infrared cutoff is always needed for the interaction picture to exist.

As above, Haag's theorem is the statement that you always need an IR cutoff. Sometimes, although it is rare, the interaction picture can exist without an ultraviolet cutoff.

Thank you for the clarification.

 

[Demystifier]

However there is the Epstein-Glaser formalism which requires no cutoffs (Infrared or ultraviolet) nor subtracting infinite quantities, instead it uses the Hahn-Banach theorem.

Where can I learn more about that?

 

[DarMM]

since we currently aren't capable of directly solving a QFT without help from a field theory

Sorry this should read "without help from a free field theory". The exception being lattice formulations, which solve theories without reference to the free theory. However in that case we need a cutoff since we have no mathematical control over the continuum path integral.

 

[Qubix]

Haag's theorem is a theorem proving the non-existence of the interaction picture provided the theory is translation invariant, i.e. if there is no infrared cutoff, it does not discuss the ultraviolet cutoff.

Basically Haag's theorem proves that free and interacting theories are unitarily inequivalent unless an infrared cutoff is in place.

There is no such general result for ultraviolet cutoffs because there are counterexamples. For example $\phi^4$ without an ultraviolet cutoff (but with an infrared cutoff) is unitarily equivalent to the free theory in 1+1 dimensions and so the interaction picture does exist.

For interaction picture to exist sometimes both an ultraviolet and infrared cutoff is needed, sometimes only an infrared cutoff. Haag's theorem is the statement that an infrared cutoff is always needed for the interaction picture to exist.

Ok, so what is Infrared cutoff / UV cutoff, specifically, what is a cutoff ? An upper/lower bound placed on the energy / frequency spectra? If so, by an infinite volume cutoff, do you mean the volume that appears in all field equation solutions of QFT (Klein-Gordon, Dirac, Vector) is imposed as a finite (even though large) volume ?

 

[atyy]

A QFT doesn't require a regulator, in the sense that it can exist mathematically without one. For practical calculations however one almost always needs a cutoff, since we currently aren't capable of directly solving a QFT without help from a field theory, which is singular with respect to the interacting theory and so we see infinities.

However there is the Epstein-Glaser formalism which requires no cutoffs (Infrared or ultraviolet) nor subtracting infinite quantities, instead it uses the Hahn-Banach theorem.

http://www.rivasseau.com/resources/book.pdf (p15) mentions some of Epstein and Glaser's work, but it seems that it only provides "rigourous justification" in the sense of "formal power series". Is that enough to define a theory?

 

[Demystifier]

However there is the Epstein-Glaser formalism which requires no cutoffs (Infrared or ultraviolet) nor subtracting infinite quantities, instead it uses the Hahn-Banach theorem.

Where can I learn more about that?

In the meantime, I have learned something about Epstein-Glaser approach from the book:
G. Scharf, Finite Quantum Electrodynamics: The Causal Approach
which is a book completely devoted to the Epstein-Glaser theory and its developments.

In particular, at the top of page 180 the author writes:
"This is an ultraviolet "regularization" in the usual terminology. It should be stressed, however, that here this is a consequence of the causal distribution splitting and not an ad hoc recipe".

Therefore, I do not think it is correct to say that there is no regularization in the Epstein-Glaser approach. It's only that the regularization is mathematically better justified.

More technically, the Epstein-Glaser approach starts from the observation that time ordering of field operators introduces step functions theta(t-t'), which are ill defined at zero. Therefore, the theta functions are replaced by certain better defined regularized functions, which avoid problematic UV divergences.

For a very brief review of Epstein-Glaser approach see also
http://www.google.hr/url?sa=t&rct=j...pwW1s-CYzKoSs4Wmu3o8i1Q&bvm=bv.58187178,d.bGQ

By the way, Glaser was a Croat
http://en.wikipedia.org/wiki/Vladimir_Glaser
just as I am.

 

[DarMM]

In particular, at the top of page 180 the author writes:
"This is an ultraviolet "regularization" in the usual terminology. It should be stressed, however, that here this is a consequence of the causal distribution splitting and not an ad hoc recipe".

Sorry, I should have explained this better. There is still a "regularisation" of sorts, but no physical cutoff.

Therefore, I do not think it is correct to say that there is no regularization in the Epstein-Glaser approach. It's only that the regularization is mathematically better justified.

More technically, the Epstein-Glaser approach starts from the observation that time ordering of field operators introduces step functions theta(t-t'), which are ill defined at zero. Therefore, the theta functions are replaced by certain better defined regularized functions, which avoid problematic UV divergences.

That is the original form of the Epstein-Glaser approach, if you are interested the modern form of the Epstein-Glaser approach is as follows.

Let's say the propagator of a scalar field theory is $D(x-y)$, then the bubble second order diagram in $\phi^4$ theory in position space is:

$\int{D(x_1 - x_3)D(x_2 - x_3)D^{2}(x_3 - x_4)D(x_4 - x_5)D(x_4 - x_6)d^{4}x_{3}d^{4}x_{4}}$

Of course $S(x_1,x_2,x_3,x_4,x_5,x_6) = D(x_1 - x_3)D(x_2 - x_3)D^{2}(x_3 - x_4)D(x_4 - x_5)D(x_4 - x_6)$ is not a well-defined distribution on $\mathbb{R}^{24}$, in the sense that there are some test functions that when integrated against it have a divergent result.

So you restrict the space of test functions from $\mathcal{D}(\mathbb{R}^{24})$ to some subspace $\mathcal{A}$ on which $S(x_1,x_2,x_3,x_4,x_5,x_6)$ is a sensible linear functional, typically the space of test functions which vanish on the $x_i = x_j$ hyperplanes. This is the regularisation in the Epstein-Glaser approach, but hopefully you can see why it's not really a physical cutoff. I don't know what it would correspond to physically.

You then prove that there is essentially a unique distribution $\tilde{S}$ defined on all test functions in $\mathcal{D}(\mathbb{R}^{24})$ which:
1. Agrees with $\tilde{S}$ on $\mathcal{A}$.
2. Obeys relativity and causality.

Essentially $S$ has extensions thanks to the Hahn-Banach theorem to all of $\mathcal{D}(\mathbb{R}^{24})$, but only one (up to a constant) obeys relativity and causality.

Hence we have renormalization via Hahn-Banach + Relativity.

However really the only way to work with a QFT without any regulator at all, is to know in advance which Hilbert Space it is defined on. On that Hilbert space there will be no divergences. However we aren't really able to do this at the moment, except with Algebraic QFT, which removes Hilbert Spaces but at the cost of being so general you can't work with a specific QFT.

 

[vanhees71]

Another, much more physical way, is not to regularize at all, but take the Feynman rules to provide the integrands of divergent integrals and then directly subtract the divergences, choosing a renormalization scheme (and in case of anomalies choose which currents should stay conserved in the quantum version of the considered field theory). This is known as the BPHZ approach and works for UV divergences.

You still have to regularize the IR divergences somehow, which occur when massless particles are present (as IR and collinear divergences of the loop integrals) and then do the appropriate resummations of diagrams (Bloch-Nordsieck/Kinoshita-Lee-Nauenberg).

While the UV divergences are a "mathematical decease", i.e., multiplying distributions, leading to problems with the definition of the products, the IR divergences are a "physical decease". The Epstein-Glaser approach cures this in a mathematical way, which is somewhat complicated but very reassuring in the sense that it shows that the usual renormalization theory provides the unique solution for the problem, obeying all physical constraints on the S matrix (in a perturbative sense).

The IR divergences come into the game, because at the very beginning we evaluate unphysical S-matrix elements in using the wrong asymptotic free states. E.g., it doesn't make sense to look at elastic electron-electron scattering in the sense of just two free electrons in the initial and the final state, because any accelerated charge radiates and thus you always have soft photons around the electrons. So the correct asymptotic free states of interacting electrons are always states acompanied by an indefinite number of (soft) photons (coherent states).

This problem already occurs in non-relativistic quantum theory for the much simpler problem of Coulomb scattering. Also there the plain waves are not the correct asymptotic states but the socalled "Coulomb distorted waves". There it becomes clear that this problem is due to the long-range nature of the Coulomb potential, which vanishing at infinity only with 1/r.

 

[Demystifier]

This is the regularisation in the Epstein-Glaser approach, but hopefully you can see why it's not really a physical cutoff. I don't know what it would correspond to physically.

Essentially $S$ has extensions thanks to the Hahn-Banach theorem to all of $\mathcal{D}(\mathbb{R}^{24})$, but only one (up to a constant) obeys relativity and causality.

Just for an analogy: Dimensional regularisation is also not really a physical cutoff and I don't know what it would correspond to physically. It is based on analytic continuation, which is more-or-less unique due to the (Cauchy or whoever) theorem.

 

[atyy]

Does the EG method really avoid the UV problem?

For example, http://arxiv.org/abs/hep-th/0403246 says "In general the series does not converge (in any norm), but in a few cases it is at least Borel summable, cf. [45]."

Also http://arxiv.org/abs/0810.2173 says "One should note that no statements about the convergence of the full series eq. (3.2) can be made in general."

 

[DarMM]

Just for an analogy: Dimensional regularisation is also not really a physical cutoff and I don't know what it would correspond to physically. It is based on analytic continuation, which is more-or-less unique due to the (Cauchy or whoever) theorem.

That's true actually, I don't know what I was thinking! They also both share the weakness that there is no obvious non-perturbative version of either of them.

By the way, dimensional regularisation is essentially part of the theory of hyperfunctions, at least I find that a clearer way of understanding it. There's an old paper by Yasunori Fujii on this.

 

[DarMM]

Does the EG method really avoid the UV problem?

For example, http://arxiv.org/abs/hep-th/0403246 says "In general the series does not converge (in any norm), but in a few cases it is at least Borel summable, cf. [45]."

Also http://arxiv.org/abs/0810.2173 says "One should note that no statements about the convergence of the full series eq. (3.2) can be made in general."

That's not the UV problem. That's the problem of convergence of the perturbative series, which occurs even in quantum mechanics where there are no divergent integrals. Renormalisation methods like Epstein-Glaser, Dim Reg, Hard cutoffs, e.t.c. have nothing to say about this.

 

[atyy]

That's not the UV problem. That's the problem of convergence of the perturbative series, which occurs even in quantum mechanics where there are no divergent integrals. Renormalisation methods like Epstein-Glaser, Dim Reg, Hard cutoffs, e.t.c. have nothing to say about this.

I guess the more standard term might be the problem of "UV completeness". At the non-rigourous level, QED and Einstein gravity are usually thought not to be UV complete, and that there is truly a cut-off energy above which the theories don't exist, and some other more complete theory has to be used. Usually only asymptotically free or asymptotically safe theories are considered to have a chance of being UV complete. So either QED and Einstein gravity are asymptotically safe (since they are not asymptotically free), or they are not UV complete. Does the Epstein-Glaser method (or other points of view from rigourous QFT) challenge this heuristic thinking?

 

[DarMM]

I guess the more standard term might be the problem of "UV completeness". At the non-rigourous level, QED and Einstein gravity are usually thought not to be UV complete, and that there is truly a cut-off energy above which the theories don't exist, and some other more complete theory has to be used. Usually only asymptotically free or asymptotically safe theories are considered to have a chance of being UV complete. So either QED and Einstein gravity are asymptotically safe (since they are not asymptotically free), or they are not UV complete. Does the Epstein-Glaser method (or other points of view from rigourous QFT) challenge this heuristic thinking?

A few points:

1. Epstein-Glaser is not really rigorous QFT. It's a way of handling divergences in perturbation theory that's more rigorous than some others, although it isn't really more rigorous than the precise version of dimensional regularization. In perturbing a interacting quantum field theory, all of these methods assume there is something to perturb, i.e. that the theory exists, a non-rigorous statement. Rigorous QFT is really about proving the theories do exist.
Even if the theory did exist, you'd still have to prove its correlation functions and scattering cross-sections were smooth, in order for a perturbative series to exist. All perturbative methods like Epstein-Glaser assume the QFT exists and has smooth scattering cross-sections.

2. UV completeness is a non-perturbative phenomena, so perturbative methods like Epstein-Glaser can't say anything about it unfortunately.

3. UV completeness is a separate issue from convergence of the series. UV completeness is related to the UV divergences, but both issues are unconnected with series summation discussed in your links.

 

[atyy]

A few points:

1. Epstein-Glaser is not really rigorous QFT. It's a way of handling divergences in perturbation theory that's more rigorous than some others, although it isn't really more rigorous than the precise version of dimensional regularization. In perturbing a interacting quantum field theory, all of these methods assume there is something to perturb, i.e. that the theory exists, a non-rigorous statement. Rigorous QFT is really about proving the theories do exist.
Even if the theory did exist, you'd still have to prove its correlation functions and scattering cross-sections were smooth, in order for a perturbative series to exist. All perturbative methods like Epstein-Glaser assume the QFT exists and has smooth scattering cross-sections.

2. UV completeness is a non-perturbative phenomena, so perturbative methods like Epstein-Glaser can't say anything about it unfortunately.

3. UV completeness is a separate issue from convergence of the series. UV completeness is related to the UV divergences, but both issues are unconnected with series summation discussed in your links.

Is rigourous existence of a theory equivalent to "UV completeness"?

Also, is there any way for the perturbative series to make sense if the theory doesn't exist? For example, Scharf's book seems to develop the perturbation theory for QED, which at the heuristic level is usually thought not to exist (in the sense of not being UV complete), since it is not asymptotically free, and suspected not to be asymptotically safe. Heuristically, the perturbation series is thought to make sense as a low energy effective theory. Does this have a counterpart in more rigourous views of perturbation theory, or do they require that what they are perturbating does indeed exist?

 

[DarMM]

Is rigourous existence of a theory equivalent to "UV completeness"?

Yes, basically. One way you could see it is that a field theory exists rigorously, if you can prove that it is its own UV completion.

Also, is there any way for the perturbative series to make sense if the theory doesn't exist?

The perturbative method of dealing with a QFT constructs a formal series in the interaction strength. This construction is well-defined even if the supposed QFT doesn't exist. If it does exist then (under assumptions) the formal perturbative series is equal to the Taylor expansion of the QFT. If the quantum field theory doesn't exist, then it's just a meaningless formal construction.

You'd have to read Connes and others, but the basic picture is that the perturbative series is just something you can build from a Lagrangian, a formal construction (technically it's a map to a certain complete symmetric algebra, but that's not important). Whether that series actually means anything depends on whether the theory described by the Lagrangian actually exists or not. However you can always construct it, it is a mathematically well-defined operation, even if it is physically meaningless.

 

[atyy]

The perturbative method of dealing with a QFT constructs a formal series in the interaction strength. This construction is well-defined even if the supposed QFT doesn't exist. If it does exist then (under assumptions) the formal perturbative series is equal to the Taylor expansion of the QFT. If the quantum field theory doesn't exist, then it's just a meaningless formal construction.

You'd have to read Connes and others, but the basic picture is that the perturbative series is just something you can build from a Lagrangian, a formal construction (technically it's a map to a certain complete symmetric algebra, but that's not important). Whether that series actually means anything depends on whether the theory described by the Lagrangian actually exists or not. However you can always construct it, it is a mathematically well-defined operation, even if it is physically meaningless.

Hmmm, this does seem different from the Wilsonian heuristic, in which the theory need not be UV complete in order for the perturbation series to be meaningful as a low energy effective theory.

If the formal series requires the rigourous existence of the quantum field theory in order to be physically meaningful, and given that experiments indicate that the perturbative series in QED seems physically meaningful, does this then suggest that QED may be rigourously constructed, possibly through asymptotic safety?

 

[DarMM]

Hmmm, this does seem different from the Wilsonian heuristic, in which the theory need not be UV complete in order for the perturbation series to be meaningful as a low energy effective theory.

Well, if you make a field theory effective with an explicit cutoff, then the theory is mathematically well-defined and the perturbative series is equal to the Taylor series of the theory. So the perturbative series does make sense in the presence of a cutoff.

If the formal series requires the rigourous existence of the quantum field theory in order to be physically meaningful, and given that experiments indicate that the perturbative series in QED seems physically meaningful, does this then suggest that QED may be rigourously constructed, possibly through asymptotic safety?

No. QED on its own and QED + Weak Theory, both give the same diagrams at low energy, so these results are not really a test of just QED. QED + Weak + Strong, probably rigorously exists.

However I should say, there are some suggestions that QED may be constructable through asymptotic safety, but they're not very conclusive. I would suggest reading Montvay and Munster's book on lattice field theory, their chapter on gauge theories contains a lot of references about the continuum limit of QED.

 

[DarMM]

Actually atyy, please keep asking questions, the Wilsonian and rigorous point of views on QFT often seem to contradict each other, but that is simply because they come at it from completely different angles.

The easiest way I could explain it is that the Wilsonian viewpoint is concerned with the relations between various effective field theories. The main idea being the renormalization group flow, which maps an action at scale $\Lambda$ to the action at scale $\Lambda^{'}$ which has the same physics/produces the same expectation values.
In this framework, there is the critical point in the space of actions, the action which corresponds to physics at $\Lambda \rightarrow \infty$. The space of pure QED actions probably poses no critical point except the free theory. However if we extend the space of actions to include QED + Yang-Mills, then there are non-trivial critical points. We say that we have to "ultraviolet complete" QED, i.e. extend the space of actions.

The rigorous point of view disposes with the cutoff actions and how they relate to each other, the physics as $\Lambda\rightarrow\infty$ isn't built up from a renormalization group flow. Instead we attempt to define the theory directly at $\Lambda = \infty$ of in lattice terms $a = 0$.

Really there is currently no proof that anything exists in the $\Lambda\rightarrow\infty$ limit. Rigorous QFT attempts to show that there actually is something in that limit. That there really is a continuum theory.

 

[atyy]

DarMM, thanks a lot! The last two posts, and your comments throughout the thread have been very helpful. (I tried to clicking on "Thanks" for both your posts, but must have done something wrong, since I'm getting a message about "negative thanks".)

 

[A. Neumaier]

A QFT doesn't require a regulator, in the sense that it can exist mathematically without one. For practical calculations however one almost always needs a cutoff, since we currently aren't capable of directly solving a QFT without help from a field theory, which is singular with respect to the interacting theory and so we see infinities.

However there is the Epstein-Glaser formalism which requires no cutoffs (Infrared or ultraviolet) nor subtracting infinite quantities, instead it uses the Hahn-Banach theorem.

The Epstein-Glaser formalism has an intrinsic IR cutoff, since the x-dependent coupling function must have compact support. One can take the IR limit g(x) --> 1 only for selected results.