Rigorous Quantum Field Theory.

In summary, strangerep and DarMM are discussing rigorous issues in quantum field theory. Strangerep says that the Epstein-Glaser approach is not more ad hoc then solving LaTeX Code: a x^2 + b x +c =0. DarMM says that the upshot is that at the end you've constructed the perturbative expansion for S(g) (the S-matrix in finite volume) in a completely rigorous way. Modern work on the Epstein-Glaser approach tries to take the limit g \rightarrow 1, to go to infinite volume, although it has proven extremely difficult. They agree that renormalized QFT (such as QED) can calculate the S-matrix (i.
  • #211
DrFaustus I believe you had posted a question, but it seems to have disappeared and I can't remember what it was.

In addition I've finally gathered my notes and references, so I'll soon be doing a series of three posts climbing the ladder of [tex]\phi^{4}_{2}[/tex], [tex]\phi^{4}_{3}[/tex], [tex]\phi^{4}_{4}[/tex].
 
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  • #212
DarMM -> I did post something which disappeared indeed. In short, I understand your answer (on a finite volume local Moeller operators act on Fock space, but in the infinite volume case they do not), but I do not *see* the reason. Is there an easy way to understand why am I kicked out of Fock space?

Second, I was asking you whether you are aware of the work of Schlegelmilch. In his PhD thesis he claimed he constructed local Moeller operators. It seems, however, that there's a mistake in his argument and its not an easy to correct it (if at all possible). You can find his work on the AQFT thesis archive on the DESY webpage (workgroup of Fredenhagen). Maybe you'll want to have a look. I browsed it quickly so cannot really comment on it.

It's good to know about your notes... might I ask you where and who did you learn all this stuff from? Personal study? Lectures?

sheenashirley -> Welcome to Physics Forums :) It's funny that you probably chose the toughest subject (imo) to join, but if you know *rigorous* aspects of QFT, it will be great to have someone else to discuss it with.
 
  • #213
DrFaustus said:
DarMM -> I did post something which disappeared indeed. In short, I understand your answer (on a finite volume local Moeller operators act on Fock space, but in the infinite volume case they do not), but I do not *see* the reason. Is there an easy way to understand why am I kicked out of Fock space?
Good question. First of all, let me say that this is the result Haag's theorem directly deals with. In essence the theorem says that an interaction being present over all of spacetime necessarily leads to a different Hilbert space. Of course there are several models were you must move to a different Hilbert space even in the finite volume case, for example [tex]\phi^{4}_{3}[/tex].

The easiest way to see it is to look at the interaction term [tex]\int_{\mathbb{R}}{:\phi^{4}(x):dx}[/tex]. The whole inequivalent reps of the canonical commutation relations issue means that I several choices for what [tex]\phi(x)[/tex] could be. Each choice will act on a different Hilbert space.

Now for [tex]\int_{\mathbb{R}}{:\phi^{4}(x):dx}[/tex] to be defined we require that:
1. [tex]\phi(x)[/tex] to the fourth power be well defined
and
2. That this fourth power when integrated over all of space and combined with [tex]H_{0}[/tex] leads to a well-define operator.

In a finite volume we would only need
2'. That this That this fourth power when integrated over a given region and combined with [tex]H_{0}[/tex] leads to a well-define operator.

If you use the Fock choice for [tex]\phi(x)[/tex] you can satisfy 1 and 2'. However to satisfy 1 and 2 you need a different choice.

Haag basically proved that an interaction term involving an integral over all of space cannot be defined in the Fock representation. You here many versions of Haag's theorem, but I think that is the definitive version.

If you want a somewhat physical picture of why, the integral over all of space means the interaction is everywhere so things cannot "tend" to being free/Fock.

If you feel anything is unanswered just ask.

Second, I was asking you whether you are aware of the work of Schlegelmilch. In his PhD thesis he claimed he constructed local Moeller operators. It seems, however, that there's a mistake in his argument and its not an easy to correct it (if at all possible). You can find his work on the AQFT thesis archive on the DESY webpage (workgroup of Fredenhagen). Maybe you'll want to have a look. I browsed it quickly so cannot really comment on it.
Ah, yes. He constructs the local Scattering operator [tex]S(g)[/tex] of Bogoliubov, Stuckelberg, Epstein and Glaser. I'm more familiar with this object from the work of Epstein and Glaser than Bogoliubov and Stuckelberg. What he constructs is in a [tex]\mathcal{P}(\phi)_{2}[/tex] theory with a finite volume cutoff, so the different Hilbert space issues are avoided.

It's good to know about your notes... might I ask you where and who did you learn all this stuff from? Personal study? Lectures?
Well, when I was learning QFT two questions began to play through my mind constantly:
What is an interacting QFT nonperturbatively and what is renormalization?

I eventually realized that I needed to understand a bit more about what the mathematical objects in quantum field theory were in order to even phrase this question properly. Eventually I found out that an entire discipline was dedicated to this. So I did my Master Thesis on constructive field theory as a way to force myself to learn the subject. So basically it was all personal study.
 
  • #214
DarMM said:
In essence [Haag's] theorem says that an interaction being present over all of spacetime necessarily leads to a different Hilbert space.

I know a different formulation of the theorem, which says that interacting quantum field (i.e., the field whose time evolution is governed by the full interacting Hamiltonian) cannot obey usual Lorentz transformation laws. See, e.g.,

M.I. Shirokov, "Dressing" and Haag's theorem, http://www.arxiv.org/abs/math-ph/0703021

How the two formulations are related to each other?

Eugene
 
  • #215
meopemuk said:
I know a different formulation of the theorem, which says that interacting quantum field (i.e., the field whose time evolution is governed by the full interacting Hamiltonian) cannot obey usual Lorentz transformation laws. See, e.g.,

M.I. Shirokov, "Dressing" and Haag's theorem, http://www.arxiv.org/abs/math-ph/0703021

How the two formulations are related to each other?

Eugene
In truth the above paper does not contain Haag's theorem. The full statement of Haag's theorem is that given certain assumptions about quantum field theory (a subset of the Wightman axioms), then a translation and rotation invariant interacting theory lives in a different Hilbert space.

The above paper says that in a theory in which the interacting field transforms in a way different to what the Wightman axioms suggest, then this would not necessarily happen.

So Haag's theorem says under a set of assumptions:
Interacting field theory => Different Hilbert space.
or its contraposition
Same Hilbert Space => Non-Interacting theory.

The paper above simply says losing one of the assumptions means the theorem may not hold. In particular it gives some arguments as to why the assumption does not hold in QED. However:
(a) Even with this assumption gone the theorem may still hold.
(b) There is no proof that the assumption doesn't hold.
(c) In all theories constructed so far the assumptions do hold, even QED in lower dimensions.
 
  • #216
DarMM said:
Same Hilbert Space => Non-Interacting theory.

I can easily provide a counterexample by building an interacting theory in the Fock space. For example, I can treat each N-particle sector separately and define there usual quantum-mechanical N-particle interacting Hamiltonians. I can even make this interaction relativistically invariant. See, for example

B. Bakamjian and L. H. Thomas, "Relativistic particle dynamics. II", Phys. Rev., 92 (1953),1300.

I agree that in my example the full Hamiltonian (in the entire Fock space) cannot be represented as a product of fields, and Wightman axioms may not be satisfied. But why should I care? What is the reason to build interactions from field products? Why should I trust Wightman axioms? I am my opinion, they are just artificial restrictions.

Eugene.
 
  • #217
meopemuk said:
I can easily provide a counterexample
No you can't. You can provide a counterexample to the direct statement
Same Hilbert Space => Non-Interacting theory
but nobody is interested in that statement in particular. Rather we are discussing Haag's theorem which says this statement follows from the Wightman axioms and you cannot give a counterexample to that.
It would be like giving a counterexample to a theorem about Hilbert spaces by producing a general Banach space.

meopemuk said:
I agree that in my example the full Hamiltonian (in the entire Fock space) cannot be represented as a product of fields, and Wightman axioms may not be satisfied. But why should I care? What is the reason to build interactions from field products? Why should I trust Wightman axioms? I am my opinion, they are just artificial restrictions.
Well:
(a) A lot of modern particle physics is built on the theory of quantum fields, so it would seem sensible to take the idea seriously.
(b) Every single field theory known to exist nonperturbatively satisfies the Wightman axioms, does this not suggest there may be something to them?
 
  • #218
DarMM said:
(b) Every single field theory known to exist nonperturbatively satisfies the Wightman axioms, does this not suggest there may be something to them?

What about QED? If I understand Haag's proof, then interacting fields in QED should not transform by Lorentz formulas wrt boosts. This looks like a violation of one of Wightman axioms.
 
  • #219
DarMM said:
(a) A lot of modern particle physics is built on the theory of quantum fields, so it would seem sensible to take the idea seriously.

If the field-based approach leads to such absurd results like Haag's "no-interaction" theorem and ultraviolet divergences, then it makes sense to try something different.

Eugene.
 
  • #220
meopemuk said:
What about QED? If I understand Haag's proof, then interacting fields in QED should not transform by Lorentz formulas wrt boosts. This looks like a violation of one of Wightman axioms.
No, as I have said that is not what Haag's theorem says. Haag's theorem says given a subset of the Wightman axioms then an interacting field theory lives in a different Hilbert space. It says nothing at all about boosts.

The paper quoted above says that perhaps, if the interacting fields do not transform by Lorentz formulas in regards to boosts, then Haag's theorem can be avoided. However it offers no proof of this assertion.

There is no direct link between what Haag's theorem says and boosts.

(In fact what you are saying doesn't make any sense. How could Haag's theorem imply the violation of a property it assumes.)

meopemuk said:
If the field-based approach leads to such absurd results like Haag's "no-interaction" theorem
Once again that is not what Haag's theorem says. It says that an interacting theory must live in a different Hilbert space. It does not forbid interactions. (It couldn't since interacting Wightman field theories are known)

So let me be clear:
1. Haag's theorem does not say that interacting fields do not transform according to Lorentz formulas.
2. Haag's theorem does not say there are no interactions.
3. Haag's theorem does say interacting theories must live in a separate Hilbert space.

You should understand what Haag's theorem says before calling it absurd.

meopemuk said:
and ultraviolet divergences
The origin of the ultraviolet divergences is understood. It is known that if the theory was written in the correct Hilbert space there would be no ultraviolet divergences.
 
  • #221
DarMM said:
There is no direct link between what Haag's theorem says and boosts.

I have a feeling that we are talking about two different Haag's theorems. The theorem I am talking about can be found, for example, in

O. W. Greenberg "Haag's theorem and clothed operators" Phys. Rev. 115 (1959), 706.

Greenberg said:
We denote by Haag's theorem the statement that any quantum field theory which has the following four properties: I- relativistic transformation properties, II - unique, normalizable, invariant, vacuum state [tex] \Psi_0 [/tex] an no negative-energy states or states of spacelike momenta, III - canonical commutation relations at equal times, and IV - being related to the free-field theory at a given time by a unitary transformation, is completely equivalent to the free-field theory.

The boost transformation condition for interaction field (see eq. (1)) plays an important role in this theorem. My point was that if this condition is relaxed (I don't see any reason why this condition should be imposed in the first place) then the theorem does not hold, which means that non-trivial interaction theories become possible.

The theorem in this form does not mention "different Hilbert spaces". I guess this was a later invention.

I also don't think that Greenberg misrepresented the statement of the theorem, because in the Acknowledgement he mentions discussions with Schweber, Haag, and Wightman who are supposed to be experts on this.

Eugene.
 
  • #222
meopemuk said:
I also don't think that Greenberg misrepresented the statement of the theorem, because in the Acknowledgement he mentions discussions with Schweber, Haag, and Wightman who are supposed to be experts on this.

Eugene.
He isn't misrepresenting it. In a previous post you said that Haag's theorem states that the interacting field should not transform by Lorentz formulas with regards to boosts. This is not what it says. What Greenberg is saying is correct.

The boost transformation condition for interaction field (see eq. (1)) plays an important role in this theorem. My point was that if this condition is relaxed (I don't see any reason why this condition should be imposed in the first place) then the theorem does not hold, which means that non-trivial interaction theories become possible.
The condition was initially imposed because it was what most people expected to be true. Since then it has been shown to be true non-perturbatively in every field theory we have constructed. Hence I would say it is somewhat justified, certainly more than the alternative.

Also relaxing the theorem does not suddenly allow nontrivial interactions, since they are already allowed. Haag's theorem does not say nontrivial interactions aren't allowed, it says they're unitarily inequivalent or in other words they live in a separate Hilbert space.

Relaxing the assumption doesn't allow non-trivial interactions, because they are already permitted. It may allow them to exist in the same Hilbert space, however this has never been shown by anyone.

I would like to repeat that Haag's theorem does not state there are no interactions.

The theorem in this form does not mention "different Hilbert spaces". I guess this was a later invention.
It's in the paper, he just says unitary inequivalence, not different Hilbert spaces. It certainly wasn't a latter invention. Friedrichs had already found other Hilbert spaces in 1953.
 
  • #223
DarMM said:
In a previous post you said that Haag's theorem states that the interacting field should not transform by Lorentz formulas with regards to boosts. This is not what it says. What Greenberg is saying is correct.

I see two logically equivalent ways to formulate the Haag-Greenberg theorem (I simplify the theorem by omitting other conditions which seem reasonable, like unique vacuum.):

1) If "interacting" quantum field transforms as shown in eq. (1), then there can be no interaction.

2) If the theory has interaction, then "interacting" quantum field cannot transform as in eq. (1).

In my previous post I used the second version. Yes, in both cases I assume just one Hilbert space of states. I cannot understand the physical meaning of "different Hilbert spaces". In my opinion, hydrogen atom (an interacting system) lives in the same Hilbert space as separated electron and proton (a non-interacting system).

Eugene.
 
  • #224
meopemuk said:
I cannot understand the physical meaning of
"different Hilbert spaces".
There doesn't need to be a "physical meaning" for all the different
Hilbert spaces. Only the one that carries a representation of the full
interacting dynamics is physically significant.

In my opinion, hydrogen atom (an interacting system) lives in the same
Hilbert space as separated electron and proton (a non-interacting system).
Widely separated (physical) electron and proton do not obey the dynamics
of the free Hamiltonian, but rather what remains of the full Hamiltonian
in the asymptotic limit. Hence there is no contradiction with these two
(physical) systems living in the same Hilbert space.

But that's different from saying that the "free" and "interacting" theories
live in the same Hilbert space, (which is what Haag's theorem forbids).
 
  • #225
strangerep,

the main reason I don't like the idea of "different Hilbert spaces" is that this idea goes against the basic structure of quantum theory as I understand it.

In QM we first need to identify the studied physical system. Then we build its Hilbert space of states according to the particle content. If the system is one elementary particle then the Hilbert space is the one corresponding to an irreducible unitary representation of the Poincare group. For a 2-particle system (e.g., hydrogen atom) the Hilbert space is built as a tensor product of two 1-particle spaces. For a system with variable number of particles we build the Fock space, which is a direct sum of N-particle tensor products. These constructions also provide basic particle observables (momentum, spin, position, etc.) in the Hilbert space. They are independent on interactions.

The interaction is defined independent on the above Hilbert space structure. This is achieved by definiing an unitary representation of the Poincare group in the Hilbert space. Generators of this representation are identified with operators of (total) momentum, angular momentum, energy, and boost. Different representations (non-interacting and interacting) can be built in the same Hilbert space. The presence/absence of interactions does not have any effect on the Hilbert space itself.

This logic is implied in the Weinberg's textbook.

Your approach, in which the interaction induces changes of the Hilbert space of states, is completely different. I haven't seen any logical and consistent exposition of these ideas.

Eugene.
 
  • #226
meopemuk said:
For a 2-particle system (e.g., hydrogen atom) the Hilbert space is built as a
tensor product of two 1-particle spaces.
I'm pretty sure the dynamical algebra for a physical hydrogen atom does not arise
just from a product of algebras applicable to a free electron and a free proton.

Different representations (non-interacting and interacting) can be built in the same Hilbert space. The presence/absence of interactions does not have any effect on the Hilbert space itself.
If you mean non-perturbatively, then these statements are only wishful thinking
at the present time.

This logic is implied in the Weinberg's textbook.
Weinberg does not discuss the subtleties associated with unitarily inequivalent
representations, afaict.

Your approach, in which the interaction induces changes of the Hilbert space of states, is completely different. I haven't seen any logical and consistent exposition of these ideas.
Well, it is not "my" approach as such. I merely repeat stuff that is given in
various textbooks. E.g., Umezawa, Barton, and even those QED IR papers
which I mentioned a while back.

But whichever approach one favors, there is always the issue of mathematical rigor,
which is supposed to be the subject of this thread. For my part, it was illuminating
when DarMM mentioned that paper by Glimme which (iiuc) shows that dressing
can be performed successfully to all orders in the theory considered. (I.e., convergence
of the procedure is established, again iiuc since I still find that paper rather difficult.)
Hence I'd like to learn more about attempts to tackle rigorous convergence in more
realistic theories, without being diverted off on tangents.
 
  • #227
meopemuk said:
For a 2-particle system (e.g., hydrogen atom) the Hilbert space is built as a
tensor product of two 1-particle spaces.

strangerep said:
I'm pretty sure the dynamical algebra for a physical hydrogen atom does not arise
just from a product of algebras applicable to a free electron and a free proton.

Why not? If we neglect particle creation/annihilation (e.g., absorption and emission of photons by hydrogen) then the relevant Hilbert space is the tensor product of two 1-particle spaces. In this tensor product 1-particle operators (position, spin, momentum, etc.) for both the electron and the proton can be defined naturally.


meopemuk said:
Different representations (non-interacting and interacting) can be built in the same Hilbert space. The presence/absence of interactions does not have any effect on the Hilbert space itself.

strangerep said:
If you mean non-perturbatively, then these statements are only wishful thinking
at the present time.


In the above 2-particle space I can define an interacting representation of the Poincare group, which accommodates the Coulomb interaction between the two charges plus relativistic corrections.

This is a pretty good approximation to the real hydrogen atom. Some improvements can be made by treating the system in the Fock space, where more states (with photons and electron-positron pairs) are involved, and additional terms are added to the Hamiltonian, which describe the photon absorption and emission.

I agree that presently there is no systematic way to construct all corrections (radiative and relativistic) in the Hamiltonian to get a 100% accurate description of hydrogen or any other system. So, you may say that this approach is heuristic. However, I think it is important that this approach keeps the usual quantum-mechanical separation between constructing the Hilbert space of states and defining the interaction (=unitary representation of the Poincare group).


strangerep said:
But whichever approach one favors, there is always the issue of mathematical rigor,
which is supposed to be the subject of this thread.

I think that before discussing the mathematical rigor of QFT one should first define which set of axioms one is using. I see two radically different approaches here. One of them is the particle-based approach as I roughly described above. It basically treats QFT as usual quantum theory of systems with variable number of particles. I found the most complete description of this approach in Weinberg's textbook. So, I will call it "Weinberg's approach".

Another approach is based on Wightman axioms already mentioned in this thread. So, I will call it "Wightman's approach". It's main idea is that QFT is a quantum theory of some physical objects called "fields". One can imediately ask a lot of questions about it. What are fields? If they are operators, then what is the Hilbert space where they are acting? What is the meaning of arguments x of the fields? Are these positions measured in real experiments? Are these eigenvalues of some position operator? What is this operator? What is the reason for postulating the field transformation law?


[tex] \Psi_i(x) \to \sum_j D_{ij}(\Lambda^{-1})\Psi_j (\Lambda x + a) [/tex]

How this transformation law is related to the assumed relativistic invariance of the theory (i.e., the presence of an unitary representatrion of the Poincare group)?

I can agree that the problem of "different Hilbert spaces" is relevant in the Wightman's approach. However, I think this is not the most serious of its problems. In my opinion, despite being "heuristic", the Weinberg's approach has a more solid theoretical foundation.

Eugene.
 
  • #228
Climbing the Ladder - Part 1, Post 1.

Okay, so now I begin climbing the ladder. First I'll begin with the two dimensional quartic scalar theory. Or as it is commonly labeled [tex]\phi^{4}_{2}[/tex].

The reason I begin with this is exactly the same reason it was treated first historically:
(1) It's easier.
(2) It has only one real divergence, a divergence which crops up in all other field theories. It is the infinite volume divergence demanded by Haag's theorem. The idea is to understand that divergence so well that it becomes trivial in other theories.

Before I begin, let's recall some things about the free field [tex]\phi_{0}(x)[/tex] and its Hamiltonian [tex]H_{0}[/tex] in two dimensions.

1. [tex]\phi_{0}(x)[/tex] obeys the canonical commutations and as such provides a representation of them.
2. The Hilbert space which supports this representation is called Fock space.
3. The Hamiltonian [tex]H_{0}[/tex] is a self-adjoint operator on Fock space.
4. When smeared with functions [tex]\int{f(x)\phi_{0}(x)d^{2}x}[/tex], the free field gives an unbounded operator on Fock space. That is, the free field is an unbounded operator valued distribution or a OVD for short.
5. The Hamiltonian involves smearing the field in space only, since it only involves a spatial integral. This is not important for now, but will become important in four dimensions so I just mention it here.

Now we begin by creating the [tex]\phi^{2}_{4}[/tex] theory. We add the following term to the Hamiltonian:
[tex]H_{I} = \int_{\mathbb{R}}{\phi^{4}dx}[/tex]

So the Hamiltonian of the new theory is meant to be:
[tex]H_{0}(\phi) + H_{I}(\phi)[/tex]

However unlike QM, we have several representations for the canonical commutation relations, so we have several choices for what [tex]\phi[/tex] should be. The representation we understand the best is the Fock rep and that rep already worked for the free theory, so let's cross our fingers and hope it works.

We then have:
[tex]H(\phi_{0}) = H_{0}(\phi_{0}) + H_{I}(\phi_{0})[/tex]

Unfortunately, when we check this Hamiltonian it is only defined on the null-vector, so we cannot exponentiate it to obtain the time evolution operator. Hence the theory as it stands does not exist.

So we take things in steps and see if we can set things right.

First of all where might the problem be?
Firstly [tex]H_{I}(\phi_{0})[/tex] involves an integral over all of space, necessary for the theory to be translationally invariant. Haag's theorem tells us this will be a problem, so let's restrict the integration to a region [tex]\Lambda[/tex]

In the [tex]H_{I}(\phi_{0})[/tex] term we have:
[tex]H_{I}^{\Lambda}(\phi_{0}) = \int_{\Lambda}{\phi_{0}^{4}dx}[/tex]

The issue here is [tex]\phi_{0}^{4}[/tex]. The free field [tex]\phi_{0}[/tex] is an OVD, like anything distributional its fourth power isn't clearly defined. Can we find a meaning for the fourth power? By a meaning for the fourth power I mean:
1. Is well-defined as an OVD.
2. When integrated solely in space it leads to a well-defined operator.

The first condition can be satisfied quite easily. Wick has discovered the correct definition of powers of the free-field with the normal ordering prescription. Hence instead of [tex]\phi_{0}^{4}[/tex], we use [tex]:\phi_{0}^{4}:[/tex]. This results in a well-defined OVD.
Unique to two dimensions is the result that Wick ordering also solves the second condition. When integrated [tex]:\phi_{0}^{4}:[/tex] gives a densely defined operator.
[tex]H_{I}^{\Lambda}(\phi_{0}) = \int_{\Lambda}{:\phi_{0}^{4}:dx}[/tex]

So now that [tex]H_{I}^{\Lambda}(\phi_{0})[/tex] is well defined, we add it to [tex]H_{0}(\phi_{0})[/tex] and hope the full Hamiltonian is self-adjoint.
Glimm and Jaffe showed in 1968 [1] that this is indeed the case. [tex]H_{\Lambda}(\phi_{0})[/tex] is a self-adjoint operator on Fock space.

However another problem arises is that [tex]:\phi_{0}^{4}:[/tex] is no longer positive definite because of subtractions used to define it. This means that [tex]H_{\Lambda}(\phi_{0})[/tex] might not be bounded below and energy might not be positive. Luckily however, this does not occur.
Nelson in 1966 [2] showed that the full Hamiltonian is semi-bounded, Glimm in 1968 [3] also proved this. Glimm's proof is the one that is applicable to the problem as I have set it up.
 
  • #229
Climbing the Ladder - Part 1, Post 2.

So [tex]H_{\Lambda}(\phi_{0})[/tex] is self-adjoint and semi-bounded. The only thing left is to take the infinite volume limit [tex]\Lambda \rightarrow \infty[/tex]. To do this we need some detailed knowledge of the spectrum of [tex]H_{\Lambda}(\phi_{0})[/tex]. This was done by Glimm and Jaffe in 1970 [4]. There they found that [tex]H_{\Lambda}(\phi_{0})[/tex] has a lowest eigenvalue [tex]\Omega(\Lambda)[/tex], with energy [tex]E(\Lambda)[/tex]. They develop some impressive lore about this finite volume theory, but I'll leave it aside in this post.

Latter in 1970 [5] they performed the last step. Using their knowledge of [tex]\Omega(\Lambda)[/tex] and [tex]E(\Lambda)[/tex] as functions of the volume, as well as the lore mentioned above they were able to construct the dynamical algebra of the field, which I'll call [tex]\mathcal{A}[/tex].

The problem with taking the [tex]\Lambda \rightarrow \infty[/tex] limit was that [tex]\Omega(\Lambda)[/tex] converged to zero in Fock space, even though no expectation values seemed to go wrong.

So they rewrote the theory in a Hilbert space independent way. [tex]\mathcal{A}[/tex] was used and [tex]\Omega(\Lambda)[/tex] became a state* [tex]\rho_{\Lambda}[/tex] on [tex]\mathcal{A}[/tex] and [tex]H_{\Lambda}(\phi_{0})[/tex] when exponentiated gave an automorphism [tex]\alpha_{\Lambda}[/tex] on [tex]\mathcal{A}[/tex].

In this language, removed from Hilbert spaces, they took the limit [tex]\Lambda \rightarrow \infty[/tex]. The limit existed, there was limiting state [tex]\rho_{infty}[/tex] on the algebra and a limiting automorphism [tex]\alpha_{infty}[/tex].

Now a return to the Hilbert space language (which you can accomplish with the GNS theorem) showed that one had arrived in a new Hilbert space, unitarily inequivalent to Fock space. The Hamiltonian [tex]H[/tex] was a self-adjoint semi-bounded operator on this space and a function of a new field [tex]\phi[/tex], which is a different rep of the canonical commutation relations from [tex]\phi_{0}[/tex]. On this space everything is well-defined: finite time evolution, S-matrix, Lorentz invariance, e.t.c.
Even the field equations and the OPE, [6].

Hence there is a nonperturbatively defined [tex]\phi^{4}_{2}[/tex] theory obeying everything we expect.

[1] Glimm J. and Jaffe A., "A [tex]\lambda(\phi^{4})_{2}[/tex] quantum field theory without cutoffs. I." Phys. Rev. 176, p.1945-1951.
[2] Nelson, E. "A quartic interaction in two dimensions", In: Mathematics of Elementary Particle, Eds.: Goodman, R. and Segal, I. Cambridge: MIT Press
[3] Glimm J., "Boson Fields with non-linear self-interaction in two dimensions", Comm. Math. Phys. 8, p.12-25
[4] Glimm J. and Jaffe A., "A [tex]\lambda(\phi^{4})_{2}[/tex] quantum field theory without cutoffs. II. The field operators and the approximate vacuum" Ann. Math., 91, p.362-401.
[5] Glimm J. and Jaffe A., "A [tex]\lambda(\phi^{4})_{2}[/tex] quantum field theory without cutoffs. III. The physical vacuum" Acta. Math., 125, p.204-267.
[6] Schrader R., "Local Operator Products and field euqations in [tex]\mathcal{P}(\phi)_{2}[/tex] theories", Fort. Physik., 22, p.611-631.
 
  • #230
meopemuk said:
I see two logically equivalent ways to formulate the Haag-Greenberg theorem (I simplify the theorem by omitting other conditions which seem reasonable, like unique vacuum.):

1) If "interacting" quantum field transforms as shown in eq. (1), then there can be no interaction.

2) If the theory has interaction, then "interacting" quantum field cannot transform as in eq. (1).
Neither of these are Haag's theorem. I have never seen either of these statements proved. There is no point in continuing to call either statement "Haag's theorem".
 
  • #231
meopemuk said:
This logic is implied in the Weinberg's textbook.

Your approach, in which the interaction induces changes of the Hilbert space of states, is completely different. I haven't seen any logical and consistent exposition of these ideas.

Eugene.

meopemuk said:
I can agree that the problem of "different Hilbert spaces" is relevant in the Wightman's approach. However, I think this is not the most serious of its problems. In my opinion, despite being "heuristic", the Weinberg's approach has a more solid theoretical foundation.

Eugene.
Yet, Weinberg himself references Streater and Wightman and Glimm and Jaffe, as well as Frohlich, Haag and others when referring the reader to a more rigorous exposition. That is, Weinberg himself is referring to the two approaches as one and the same. This is because they are one and the same.

Wightman's approach to QFT is not built around the idea
meopemuk said:
that QFT is a quantum theory of some physical objects called "fields"

Rather it is built around the idea that all physically correct theories of relativistic particle interactions seem to use operators called field operators. Hence let's try to understand what those objects are mathematically. All of the first five chapters of Weinberg hold in the Wightman framework, simply because the Wightman framework is the exact same except rigorous. The only difference is that Weinberg uses the creation and annihilation operators and never worries about which representation of them to use. This never effects the earlier five chapters because he is doing things abstractly and it doesn't effect the latter chapters because he does everything perturbatively. All that happens in constructive field theory is that you trying to do things nonperturbatively so you need to worry about this representation problem. That's the only difference between Weinberg and constructive field theory.

Finally how can you say that there is no logical and consistent exposition of the Wightman framework when it is mathematically rigorous? Maybe you could say it's a lot of work for very little or it obscures the physics or something, but the one one thing you can't say is that it's not logical and consistent. That's the aim of the entire field, a logical and consistent mathematical framework for QFT.
 
  • #232
DarMM said:
Finally how can you say that there is no logical and consistent exposition of the Wightman framework when it is mathematically rigorous?

I am not arguing about the mathematical rigor of the Wightman's framework. My questions are about its physical relevance. The physical content of a theory is captured by its axioms. In order to understand axioms one needs to understand definitions of all ingredients there. In the case of a physical theory it is important to specify how these ingredients are related to what is measured in experiments.

You haven't answered my specific questions about Wightman's approach, so I'll repeat them here.

1. What is the physical meaning of the electron-positron field? Can we measure it in experiments? Note that in ordinary quantum mechanics there is no analog for the quantum field. QM operates with such notions as Hilbert space, state vectors, and operators of observables. Note also that in Weinberg's approach quantum fields do no play a central role. The fields are introduced simply as formal mathematical objects. They are useful, because by building interaction terms in the Hamiltonian as products of fields one can easily guarantee the relativistic invariance and cluster separability.

2. What is the meaning of quantum field arguments "x"? As I understand, they are supposed to be related to real space positions. However, we know that position measurements are always associated with quantum uncertainties. This is the reason why in quantum mechanics we have the operator of position, whose eigenvalues are labeled by "x". What is the position operator associated with quantum field arguments "x"? In Weinberg's approach, where quantum fields are formal entities, there is no need to worry about the physical meaning of "x". They are absolutely formal parameters. They are actually dummy integration variables, because all physically relevant results (e.g., scattering amplitudes) are expressed as integrals on "x".

If "rigorous QFT" is just a mathematical exercise, then you don't need to answer my questions. However, if it pretends to be a physically relevant theory then I would like to see connections to experimentally measured stuff at each step along the way.

Eugene.
 
  • #233
meopemuk said:
1. What is the physical meaning of the electron-positron field? Can we measure it in experiments? Note that in ordinary quantum mechanics there is no analog for the quantum field. QM operates with such notions as Hilbert space, state vectors, and operators of observables. Note also that in Weinberg's approach quantum fields do no play a central role. The fields are introduced simply as formal mathematical objects. They are useful, because by building interaction terms in the Hamiltonian as products of fields one can easily guarantee the relativistic invariance and cluster separability.
The meaning is the same as in any other view of QFT. That is Wightman field theory is just a rigorous version of standard field theory, so this particular issue doesn't really apply in specific to Wightman field theory. I can give you an answer, but it would have nothing in particular to do with rigorous field theory.

Since the electron positron field is not Hermitian it is not observable. Only bilinears of fermion fields can be measurable.

Haag has said that the reason for fields is simply that they are the most convenient way to have a local Hamiltonian which describes a theory with charges. For example the electron-positron field has a definite electric charge so it is a useful object in creating a local Hamiltonian with charges.

meopemuk said:
2. What is the meaning of quantum field arguments "x"? As I understand, they are supposed to be related to real space positions. However, we know that position measurements are always associated with quantum uncertainties. This is the reason why in quantum mechanics we have the operator of position, whose eigenvalues are labeled by "x". What is the position operator associated with quantum field arguments "x"? In Weinberg's approach, where quantum fields are formal entities, there is no need to worry about the physical meaning of "x". They are absolutely formal parameters. They are actually dummy integration variables, because all physically relevant results (e.g., scattering amplitudes) are expressed as integrals on "x".
The meaning is the same as standard QFT. It is a parameter on which the fields depend, which gives meaning to their field equations.

meopemuk said:
If "rigorous QFT" is just a mathematical exercise, then you don't need to answer my questions. However, if it pretends to be a physically relevant theory then I would like to see connections to experimentally measured stuff at each step along the way.

Eugene.
Rigorous QFT is exactly what it says. You seem to be talking about and constantly questioning it as if it was some other theory or another version of QFT.
Let me repeat again, Rigorous QFT is exactly what it says it is. A rigorous version of quantum field theory.

For example imagine two classes in Quantum Mechanics were the lecturer as reached a point of discussing the Harmonic Oscillator. At this point they:
1. Solve the Schrodinger equation using standard methods, get the energy eigenvalues and show how to use creation and annihilation operators.
or
2. Show the Harmonic Oscillator Hamiltonian is self-adjoint, semi-bounded and obtain the spectrum using methods from Reed and Simons. Then using Stone's theorem to obtain the finite time evolution operator as a Unitary operator.

Now imagine asking if (2.) is physically relevant and connected to experiment. (2.) is discussing the same thing as (1.) and achieves the same aims. (2.) is just a rigorous version of (1.), it will have the same physical relevance as (1.), so there is no point in asking this question.

You seem to have issues with standard QFT and QM anyway, as seen with your disagreements over how scattering theory works and what an observable is. You are not going to find you answers in rigorous QFT, because it is just a rigorous approach to usual QFT and not a new theory as you mistakingly seem to think.
 
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  • #234
Just a quick remark before going to bed...

meopemuk -> You seem to have issues with *fields*. Let me stress out once again why people are so stubborn to use fields. It's for a very *physical* reason, and that is *relativistic causality*. More specifically, if the fields obey hyperbolic partial differential equations, then perturbations to a field configuration will propagate with *finite speed*. (And particles, which still lack a proper definition in QFT as far as I know, are heuristically interpreted as field perturbations.) This in turn means that if I manage to construct a *quantum* field obeying hyperbolic field equations, then two field operators (that is, fields smeared with test functions) that are supported at spacelike separations will commute, i.e. we can measure them at the same with arbitrary precision.

To put it simply, fields offer a very natural way to implement *relativistic causality* (no signal can propagate faster than the speed of light) and the *quantum principle* that two commuting operators can simultaneously be measured with arbitrary precision. Needless to say, all the fields in the Standard Model satisfy hyperbolic field equations...

As for the Wightman axioms, they put so few restrictions on what a quantum field should be, that it is actually really surprising that an incredible amount of work has to be done in order to give examples of interacting fields. I'll soon post them and explain their meaning. Some are technical, but others are *very* physical.

DarMM -> Have more than one question, but will limit myself to a quick one for now. From your post it is clear that the infrared problem is the crucial one in 2D. How does such a construction come across in the algebraic framework where the IR and UV problems are disentangled? And, perhaps even more importantly, why would such an algebraic construction not be feasible in higher dimensions?
 
  • #235
meopemuk said:
If "rigorous QFT" is just a mathematical exercise, then you don't need to answer my questions. However, if it pretends to be a physically relevant theory then I would like to see connections to experimentally measured stuff at each step along the way.
This looks like a very strange requirement. I would say that a theory of physics is defined by a set of axioms that tells us how to associate probabilities with possible results of experiments. So I don't think we need "connections to experimentally measured stuff at each step along the way". Why isn't it sufficient that the end result is a theory that tells us how to calculate probabilities of possibilities?
 
  • #236
Fredrik said:
This looks like a very strange requirement. I would say that a theory of physics is defined by a set of axioms that tells us how to associate probabilities with possible results of experiments. So I don't think we need "connections to experimentally measured stuff at each step along the way". Why isn't it sufficient that the end result is a theory that tells us how to calculate probabilities of possibilities?

It would be OK to have some abstract axioms if they allowed us to calculate all possible physical results in a consistent manner. Then there will be no urgency in understanding the physical meaning of these axioms. However the problem is that modern QFT is far from being a successful theory like that. There are numerous problems and inconsistencies in QFT (ultraviolet divergences, the lack of finite time evolution operator, to name a few). Even proponents of "rigorous QFT" would agree that their rigorous approach works only in toy model theories.


I think in order to move forward we need to understand exactly what we are doing in QFT. It would be nice to revisit (Wightman's) axioms to see what is their physical meaning (if any). For example, one axiom postulates how quantum fields transform with respect to inertial frame changes (see my post #227). DarMM has agreed with me that quantum fields are not directly observable objects/properties. This means that the mentioned transformation law cannot be verified in experiments even in principle. So, this transformation law is simply an unjustified assumption. There is a good chance that this assumption is just wrong. Then no matter how "rigorous" our math is, we'll not get anything useful from a wrong axiom.

Eugene.
 
  • #237
meopemuk said:
It would be OK to have some abstract axioms if they allowed us to calculate all possible physical results in a consistent manner. Then there will be no urgency in understanding the physical meaning of these axioms.
Are you saying that a successful rigorous QED in 3+1 dimensions wouldn't associate probabilities with possible results of experiments in a consistent manner? Probably not, but if that's not what you're saying, I really don't know what your argument is. It sounds like you're just saying that there's still work to be done in rigorous QFT, and that we shouldn't be doing that work because it hasn't been done already.

meopemuk said:
However the problem is that modern QFT is far from being a successful theory like that. There are numerous problems and inconsistencies in QFT (ultraviolet divergences, the lack of finite time evolution operator, to name a few).
You're describing the problems with non-rigorous QFT. Isn't this precisely what rigorous QFT is trying to do something about?

People were probably saying the same thing about the Dirac delta in 1930. Do you also think that the "inconsistencies" of the delta "function" made it pointless to develop distribution theory?

meopemuk said:
Even proponents of "rigorous QFT" would agree that their rigorous approach works only in toy model theories.
So? To me that sounds like a good reason to continue with this, and not at all like a reason to give up. I know how to prove that the group of transition functions between inertial coordinate systems in 1+1-dimensional SR is either the Galilei group or isomorphic to the Poincaré group, given a few reasonable assumptions about the properties of those functions. But I haven't been able to do it in 3+1 dimensions. Does the fact that I've only been able to prove it for a "toy model" mean that the whole idea is flawed? (It certainly doesn't).

meopemuk said:
So, this transformation law is simply an unjustified assumption.
The time when we could make progress by only trying out assumptions that had already been verified by experiments (like the invariance of the speed of light) is long gone. We have no choice but to make "unjustified" assumptions and see what theories we end up with. And the specific assumption you mention, isn't that a formula that shows up in all the non-rigorous QFTs that make absurdly accurate predictions about results of experiments? I'm having a hard time imagining a better justification than that.

meopemuk said:
There is a good chance that this assumption is just wrong. Then no matter how "rigorous" our math is, we'll not get anything useful from a wrong axiom.
Technically all axioms in all theories are wrong, but I guess you mean that this one could be so wrong that the theory it produces will make predictions that are clearly inconsistent with the results of experiments. That's a possibility, but there's no way to know unless we actually find the theory first so that we can see what it's predictions are.
 
  • #238


DarMM said:
In the [tex]H_{I}(\phi_{0})[/tex] term we have:
[tex]H_{I}^{\Lambda}(\phi_{0}) = \int_{\Lambda}{\phi_{0}^{4}dx}[/tex]

The issue here is [tex]\phi_{0}^{4}[/tex]. The free field [tex]\phi_{0}[/tex] is an OVD, like anything distributional its fourth power isn't clearly defined. Can we find a meaning for the fourth power? By a meaning for the fourth power I mean:
1. Is well-defined as an OVD.
2. When integrated solely in space it leads to a well-defined operator.

The first condition can be satisfied quite easily. Wick has discovered the correct definition of powers of the free-field with the normal ordering prescription. Hence instead of [tex]\phi_{0}^{4}[/tex], we use [tex]:\phi_{0}^{4}:[/tex]. This results in a well-defined OVD.

I don't understand how normal-ordering [itex]:\phi_{0}^{4}:[/itex] gives a well-defined OVD,
since there's still quartic products of creation operators therein. What am I missing?
 
  • #239


strangerep said:
I don't understand how normal-ordering [itex]:\phi_{0}^{4}:[/itex] gives a well-defined OVD,
since there's still quartic products of creation operators therein. What am I missing?
Two ways of seeing it:
1. If you integrate [itex]:\phi_{0}^{4}:[/itex] against a test function it always results in a densely-defined operator. This is in constrast to [itex]\phi_{0}^{4}[/itex], which after smearing does not give a densely-defined operator
2. From a Feynman graph point of view, any graphs associated with [itex]:\phi_{0}^{4}:[/itex] do not contain tadpole loops. Tadpole loops are the only ultraviolet divergent loops in 2D, so it is ultraviolet finite.

However when I say a well defined OVD, I mean (1.). (2.) is just for perturbative intuition.
 
  • #240
Fredrik said:
Are you saying that a successful rigorous QED in 3+1 dimensions wouldn't associate probabilities with possible results of experiments in a consistent manner? Probably not, but if that's not what you're saying, I really don't know what your argument is. It sounds like you're just saying that there's still work to be done in rigorous QFT, and that we shouldn't be doing that work because it hasn't been done already.

You're describing the problems with non-rigorous QFT. Isn't this precisely what rigorous QFT is trying to do something about?

I believe that people doing "rigorous QFT" are trying to solve these problems (e.g., time evolution and renormalization). I wish them well. However, in my personal (uneducated) opinion, they chose a wrong (formalistic) approach. I think one can also try an alternative approach which pays more attention to the physical meaning of theoretical constructions.



Fredrik said:
And the specific assumption you mention, isn't that a formula that shows up in all the non-rigorous QFTs that make absurdly accurate predictions about results of experiments? I'm having a hard time imagining a better justification than that.

Yes, the formula for field transformations is a necessary ingredient of all quantum field theories. However, note that this forrmula applies to non-interacting fields only. Actually, according to Weinberg, non-interacting fields are specifically defined in such a way that this "Lorentz" transformation law is valid. The reason given by Weinberg is that if we build interactions as products as thus defined fields, then the theory becomes Poincare-invariant and cluster separable automatically.

Wightman's axioms go beyond that and postulate that the same transformation law should be valid for interacting fields as well. As far as I know, there is no justification for this requirement. Moreover, Haag's theorem (in the formulation given by Greenberg) says that if interacting fields transform like that (plus some other conditions, which I find reasonable and therefore omit) then the theory must be equivalent to the non-interacting one.

I have a strong feeling that if one succeeds in constructing the interacting field operators in QED (which is an absurdly accurate theory, as you say) one would find that the "Lorentz" transformation law does not apply to them. Unfortunately, as far as I know, nobody was able to construct interacting fields in QED in any reasonable approximation and study their inertial transformations. However, this kind of study has been performed in a simple model example.

H. Kita, "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties", Progr. Theor. Phys., 35 (1966), 934.

Fredrik said:
Technically all axioms in all theories are wrong, but I guess you mean that this one could be so wrong that the theory it produces will make predictions that are clearly inconsistent with the results of experiments. That's a possibility, but there's no way to know unless we actually find the theory first so that we can see what it's predictions are.

The answer is given in Haag's theorem: If the "Lorentz" transformation condition is postulated for interacting fields, then there can be no interaction. So a theory having this postulate is simply inconsistent. (DarMM will tell you that there CAN be interaction, but in a different Hilbert space. That's something I can't comprehend.)

Eugene.
 
  • #241
meopemuk said:
I believe that people doing "rigorous QFT" are trying to solve these problems (e.g., time evolution and renormalization). I wish them well. However, in my personal (uneducated) opinion, they chose a wrong (formalistic) approach. I think one can also try an alternative approach which pays more attention to the physical meaning of theoretical constructions.
Yes, but remember that the rigorous approach is the only which has accomplished this goal nonperturbatively in any model. And they have done so in several models in two and three dimensions. They haven't accomplished four dimensions yet, but they have a better track record than approaches which have accomplished nothing nonperturbatively.

Wightman's axioms go beyond that and postulate that the same transformation law should be valid for interacting fields as well. As far as I know, there is no justification for this requirement.
No justification is a bit of a stretch. Let me list the theories where it is known to be true:
1. All pure scalar theories in 2D
2. All pure scalar theories in 3D
3. All Yukawa theories in 2D
4. All Yukawa theories in 3D
5. Yang-Mills in 2D
6. The Abelian Higgs-Model in 2D and 3D
7. The Gross-Neveu model in 2D and 3D
8. The Thirring model
and finally
9. All scalar theories in 4D.

The caveat on (9.) is that the only purely scalar theory which exists in 4D is probably the trivial one. However any field theory which exists has been proven to have this transformation property.
This list is basically every single theory we have constructed and understood nonperturbatively. So for every theory we have nonperturbative knowledge of, the transformation law holds.

The list of theories which exist nonperturbatively and don't obey the transformation law is an empty list. Hence I would say the assumption is justified, or at least far more justified than its negation.

The answer is given in Haag's theorem: If the "Lorentz" transformation condition is postulated for interacting fields, then there can be no interaction. So a theory having this postulate is simply inconsistent.
I don't know how many times I can repeat this, that is not what Haag's theorem says. Not even Shirokov, in the paper you quoted, mentions this. To transcribe what Haag's theorem says, again, into language you might understand:

Haag's theorem says that if the theory lives in the same Hilbert space as the free theory and obeys relativistic transformations and is translationally invariant, then it is free.

That is it says:
(Same Hilbert space) + (Normal transformation law) + (Translationally invariance) => Non-interacting

It does not say:
(Normal transformation law) => Non-interacting.

DarMM will tell you that there CAN be interaction, but in a different Hilbert space. That's something I can't comprehend.
It doesn't matter if you can't comprehend it or that I'm saying it. It is true and has been known to be true since 1969. I have even left references to papers which prove it in this thread, including in my two-part post above. It's perfectly fine if you can't imagine it, but it is true.
 
  • #242
DarMM said:
That is it says:
(Same Hilbert space) + (Normal transformation law) + (Translationally invariance) => Non-interacting

Agreed.

Eugene.
 
  • #243


DarMM said:
If you integrate [itex]:\phi_{0}^{4}:[/itex] against a test function it always results in a densely-defined operator. This is in contrast to [itex]\phi_{0}^{4}[/itex], which after smearing does not give a densely-defined operator.

Could you please give me a specific reference where these statements are derived rigorously?
(Or are they easy to derive but I'm still missing something?)

-------
[Edit: I sense a note of frustration in your post #241, so I just like to say two things:

a) THANK YOU for going to the effort in those earlier posts, and THANK YOU in advance
for (hopefully) future episodes of the climbing-the-ladder saga.

b) I do want to understand these things rigorously, including how one goes about
proving convergence since (among other things) acquiring such functional-analytic
skill is clearly valuable in any other non-Wightman approach that one might wish to
investigate.
-------
 
Last edited:
  • #244


strangerep said:
Could you please give me a specific reference where these statements are derived rigorously?
(Or are they easy to derive but I'm still missing something?)
Oh, they're certainly not easy to derive. The fact that Wick products give densely defined operators was first proved by Jaffe in 1966 [1]. However I personally fined a later derivation by Segal in 1967 to be much clearer [2]. Segal has a very erudite way of writing, which you will either love or find very difficult to read.

I should also say the theorem is much harder to prove in the Hamiltonian approach that I'm discussing. In the Functional-Integral (Path-Integral) approach it's just a matter of evaluating a single Feynman diagram. See Glimm and Jaffe's book Section 8.5, Proposition 8.5.1.

[1] Jaffe, A. : Wick polynomials at a fixed time. J. Math. Phys. 7, 1250 — 1255

[2] Segal, I. "Notes toward the construction of nonlinear relativistic quantum
fields, I. The Hamiltonian in two space-time dimensions as the generator
of a C*-automorphism group." Proc. Natl. Acad. Sci. U. S. 57, p.1178—1183

[Edit: I sense a note of frustration in your post #241, so I just like to say two things:

a) THANK YOU for going to the effort in those earlier posts, and THANK YOU in advance
for (hopefully) future episodes of the climbing-the-ladder saga.

b) I do want to understand these things rigorously, including how one goes about
proving convergence since (among other things) acquiring such functional-analytic
skill is clearly valuable in any other non-Wightman approach that one might wish to
investigate.
You're more than welcome, I will be glad to continue the series of posts.
 
  • #245
DrFaustus said:
DarMM -> Have more than one question, but will limit myself to a quick one for now. From your post it is clear that the infrared problem is the crucial one in 2D. How does such a construction come across in the algebraic framework where the IR and UV problems are disentangled? And, perhaps even more importantly, why would such an algebraic construction not be feasible in higher dimensions?
In a purely algebraic approach, this whole construction is quite easy to carry out. The C*-algebra of observables for the finite and infinite volume theories are exactly the same. The only difference is the representation of the algebra.

Let's say the representation of the algebra which gives you the finite volume theory is [tex]\rho_{\Lambda}[/tex]. All [tex]\rho_{\Lambda}[/tex] are unitarily equivalent, only the infinite volume theory [tex]\rho_{\infty}[/tex] is unitarily inequivalent. Also, something which allows making estimates and bound easier, the [tex]\rho_{\Lambda}[/tex] are all unitarily equivalent to the Fock/Free rep [tex]\pi[/tex].

So the entire construction of the theory is "merely" a matter of passing from one rep to another.

In higher dimensions though things are not so easy. As I will explain in detail, in three dimensions due to ultraviolet divergences one must renormalize. In the algebraic approach this shows up in the fact that the ultraviolet cutoff theory and the theory with no UV cutoff have the same C*-algebra, but different reps. Put another way, even though the algebra is again unchanged, the finite volume reps [tex]\rho_{\Lambda}[/tex] are not unitarily equivalent to the Fock/Free rep [tex]\pi[/tex]

Also unlike the 2D case [tex]\rho_{\Lambda}[/tex] and [tex]\rho_{\Lambda'}[/tex] for [tex]\Lambda \neq \Lambda'[/tex] are unitarily inequivalent.

Let me sum up. In the Algebraic approach, ultraviolet divergences associated with mass and vacuum renormalization show up as changes in representations as you take some limit.

In the 2D case there is only ever one change in rep. If you take the UV limit, the rep stays the same. When you then take the infinite volume limit the rep change only shows up in the limit.

In the 3D case there is a change of rep in the UV limit. Then there is a change of rep for every single value of [tex]\Lambda[/tex] in the ultraviolet limit.

In the 4D case things become incredibly difficult, unlike all previous cases the algebra itself changes as you take the UV limit. It's not just a rep change. It's difficult enough to control the reps, but controlling the algebra is something truly difficult. The change in the algebra itself is associated with coupling constant renormalization.

(If anybody is curious, Field Strength renormalization is associated with something you can't really see in the Algebraic approach. I'll explain it when I do my post on the 4D field.)
 

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