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

[strangerep]

In article #34 of a recent thread about Haag's theorem, i.e.,
http://www.physicsforums.com/showthread.php?t=334424&page=3
a point of view was mentioned which I'd like to discuss further.
Here's the context:


[...] The interacting theory lives in a different Hilbert space [...]

I often see this statement, but I am not sure about its validity. In my
opinion, this statement goes against basic postulates of quantum theory. Let
me explain why I think the Hilbert space used to describe a physical system
should be independent on whether the system is interacting or not.

Let us first ask why we use Hilbert spaces to describe physical systems (their
states and observables) in QM? The answer is given by "quantum logic". This
theory tells us that subspaces in the Hilbert space are representatives of
"yes-no experiments" or logical "propositions" or experimental "questions".
Meets, joins, and orthogonal complements of subspaces represent usual logical
operations OR, AND, and NOT. It seems reasonable to assume that the same
questions can be asked about interacting and non-interacting system. The
logical relationships between these questions should not depend on the
interaction as well. Therefore, the same Hilbert space (= logical
propositional system) should be applied to both interacting and
non-interacting system, if their particle content is the same. [...]


I think I see a flaw in the argument above.

Suppose I want to know the accelerations of the two particles (or
maybe just their relative acceleration wrt each other). In the
case of 2 free particles I can certainly ask the question, but
the answer is always 0. But for two interacting particles, the
answer is nonzero in general.

Expressing this in the language of quantum logic and yes-no
experiments, the question "is the acceleration 0" always yields "yes"
in the free case, but can yield "no" in the interacting case.
Similarly, the question "is the acceleration nonzero" always yields
"no" in the free case but might yield "yes" in the interacting case.

Denote the 2-free-particle Hilbert space as H_0 and the
2-interacting-particle Hilbert space as H.

Proceeding by contradiction, let's assume that H_0 and H are unitarily
equivalent, i.e., that any basis of H_0 also spans H. Assume as well that the
dynamical variable known as "acceleration" corresponds to (densely
defined) self-adjoint operators in H_0 and H, and that the two
operators are equivalent to each other up to a unitary transformation.

Every state in H_0 is a trivial eigenstate of the acceleration
operator, with eigenvalue 0. Expressed differently, the acceleration
operator annihilates every state in H_0. However, we expect to find
states in H corresponding to nonzero accelerations, i.e., states which
the acceleration operator does not annihilate.

This implies that the basis states of H_0 cannot span H, contradicting
the initial assumption. Conclusion: H_0 and H are not unitarily
equivalent.

So it's not enough that the same logical propositions (questions)
can be asked in both spaces. The spectrum of the corresponding
operator must also be considered, and whether both spaces
accommodate the full spectrum.

Or am I missing something?

[meopemuk]

Hi strangerep,

I must admit that my argument was not correct. Indeed, operators of observables (such as acceleration) have different properties and different commutation relations in the interacting and non-interacting cases. However, still I don't see the reason why these operators cannot coexist in the same Hilbert space. In ordinary non-relativistic quantum mechanics we use the same Hilbert space for both interacting and non-interacting system. What is fundamentally different in the relativistic case and in QFT?

[Bob_for_short]

However, still I don't see the reason why these operators cannot coexist in the same Hilbert space. In ordinary non-relativistic quantum mechanics we use the same Hilbert space for both interacting and non-interacting system.

That is a good point. Indeed, let us consider two Hamiltonians: one for a couple of free particles and another - with a repulsive potential. Are these Hamiltonians related with a "unitary" transformation? No. (Or, maybe, yes? Consider one-particle case). Yet, there is no problem with the perturbation theory (scattering) as well as with expansion of the exact solution of interacting case in series of free solutions. We have the same Hilbert space but two different basises to span this space.

In many cases it is so. I can also refer to the simplest 1D Sturm-Liouville problem considered in my publications where the perturbation theory may even give divergent matrix elements whereas the exact solutions are finite, physical, and they co-exist in the same Hilbert space.

[Bob_for_short]

What is fundamentally different in the relativistic case and in QFT?

In QFT there is an additional subtraction prescription. There is a principal difference before and after subtraction. Subtraction (discarding some corrections) changes the solution. The latter corresponds now to another Hamiltonian - a Hamiltonian without self-action. After that the rest is similar to a non-relativistic case.

[strangerep]

Indeed, operators of observables (such as acceleration) have different properties and different commutation relations in the interacting and non-interacting cases.
Hi meopemuk.

Thanks for your answer.

However, still I don't see the reason why these operators cannot coexist in the same Hilbert space.
As I mentioned in my original post, it's not so much about whether such an operator can
be represented in the Hilbert space(s), but whether the Hilbert spaces in question
account for the same (subset of) the full spectrum of the operator.

For finite degrees of freedom, one is saved by well-known theorems about isomorphisms
between Hilbert spaces, the Stone-von Neumann theorem, and all that. For infinite
degrees of freedom, continuous spectra, etc, these theorems fail.

In ordinary non-relativistic quantum mechanics we use the same Hilbert space for both interacting and non-interacting system. What is fundamentally different in the relativistic case and in QFT?

Afaict, it's all about how infinite degrees of freedom leads to unbounded operators.

[strangerep]

[...] let us consider two Hamiltonians: one for a couple of free particles and another - with a repulsive potential.
That sounds like a scenario with finite degrees of freedom. (?)

Are these Hamiltonians related with a "unitary" transformation? No. (Or, maybe, yes? Consider one-particle case).
If the answer was "yes", both Hamiltonians would have the same spectrum, so that
seems not correct.

In QFT there is an additional detraction prescription. There is a principal
difference before and after detraction. [...]
I'm not familiar with the word "detraction" in this context.
Did you mean "renormalization"?

[meopemuk]

Afaict, it's all about how infinite degrees of freedom leads to unbounded operators.

I don't see any big problem with unbounded operators. Physically, most relevant operators of observables are bound to be unbounded (pardon the pun). For example, the operator of position is unbounded, because the universe is infinite. The operator of momentum is unbounded, because there is no limit on the value of momentum.

If some mathematical formalism (functional analysis, theory of Hilbert spaces, etc.) has a problem with unbounded operators, then I consider this a problem of math rather than physics. Perhaps our math is build on unrealistic axioms (out of convenience). Perhaps, we need to be more creative and consider more general structures such as non-separable Hilbert spaces or non-standard Hilbert spaces, or something like that. In this particular case (unbounded operators) I don't think that math troubles indicate some new physics. I believe that physical considerations must take the lead, and math must follow.

Another point is that in the "dressed particle" approach to QFT (which is formulated in terms of particles rather than fields) the number of relevant degrees of freedom is always finite, because any realistic physical system is made of a finite number of particles.

[Bob_for_short]

I'm not familiar with the word "detraction" in this context.
Did you mean "renormalization"?
Yes, they call it "renormalization" although it is subtraction or discarding, to tell the truth .
I thought you had read my argumentations in "RiR".

P.S. I meant "subtraction", of course. My poor English...

[DrFaustus]

strangerep & meopemuk-> One crucial flaw of that argument is a purely logical one. While it is true that we can ask the same questions in both an interacting and a free theory, this does not imply at all that the underlying mathematical structure must be the same. Indeed, I can ask the same questions , say about particle scattering, in a classical, non quantum theory. Of course, physically the answers will be incorrect, but I can still ask those questions and obtain answers. But also, to an interacting theory I can ask questions a free theory simply cannot answer. For example, considering an out of equilibrium gas of particles, I could ask "How long will it take for the system to reach thermal equilibrium?". Clearly, if the particles cannot interact, an out of equilibrium system will never thermalize.

meopemuk -> strangerep nailed the answer: it's about the number of degrees of freedom. In ordinary QM you always have a finite nr of degrees of freedom and the Stone - Von Neumann theorem tells you, more or less, that all the representations of the canonical commutation relations are unitarily equivalent. That's why no one bothers talking about a different Hilbert space when dealing with QM other than the usual L^2(\mathbb{R}^n) . It's not so easy in QFT. If the nr of degrees of freedom is infinite, as it is when you deal with fields, the theorem no longer applies and it is in fact a physically relevant question as to what the appropriate Hilbert space is. And this holds for "simple" free theories. So the fundamental difference between QM and QFT is the number of degrees of freedom, not "additional subtraction procedures".

The problem with unbounded operators is precisely that they can have infinite expectation value in certain states. Sure, a purely mathematical problem. But if I'm to extract physics from some maths, then my maths better yield unique and finite answers or I'm in trouble. And it's not difficult to see how manipulating infinities can give you "paradoxes". That's the problem with unbounded operators: their maths is not clear yet we need them for physics.

And to say "consider a Hamiltonian with two particles" you must really specify if you're talking about QM or QFT. Actually, you don't need to: QFT is a multiparticle theory from the very beginning. So you're talking about QM, where things are "mathematically easy", so to say.

Finally, will now give you two more physical argument as to why the free field theory and the interacting one live in different Hilbert spaces. (Am still kinda struggling with the mathematical arguments myself :) )

Let's first clear what the Hilbert space of a (vacuum) free theory is. It's the usual Fock space, and the vectors of this space are multiparticle states. It is important however to observe now that these particles do not interact. So in the usual Fock space there is no space for interaction! Think about a self interacting particle propagating through space. The self-interaction will sure modify the one particle state of free theory and this state is in no sense a linear combination of multiparticle states of the same Fock state as all the "virtual particles" are also interacting whereas the Fock space ones are not.

Second, consider a system in thermal equilibrium at some finite temperature T. Such a state cannot be obtained in any way from the multiparticle states of the vacuum Fock space. Thermal states are not excitations of the vacuum state. Think about the quark-gluon plasma in the early Universe. The particle density in such states is non zero whereas the vacuum multiparticle states are zero density states. So even for free theories, the vacuum Hilbert space and a finite temperature Hilbert space will be different.

Yep, things are complicated in QFT - it's why you get $1m to prove the mass gap conjecture for Yang-Mills theories.

bob_for_short -> You insist on saying there are "problems" with renormalization because it involves subtraction of infinities. Read the book by Bogoliubov to see QFT without subtractions of infinities but with renormalization. Or read Scharf's book on "Finite QED" - as the title suggests, there is not a single infinity in that book. Or check out Epstein and Glaser papers about causal perturbation theory, where again renormalization is done without any subtraction of infinities. Renormalization is a (fairly) well understood issue in QFT and it's related to the mathematical nature of the various objects encountered in QFT, i.e. distributions.

[Bob_for_short]

...Read the book by Bogoliubov to see QFT without subtractions of infinities but with renormalization.
I read it in 1979.
Renormalization is a (fairly) well understood issue in QFT and it's related to the mathematical nature of the various objects encountered in QFT, i.e. distributions.
And I am a re-distribution master.

[meopemuk]

Let's first clear what the Hilbert space of a (vacuum) free theory is. It's the usual Fock space, and the vectors of this space are multiparticle states. It is important however to observe now that these particles do not interact. So in the usual Fock space there is no space for interaction! Think about a self interacting particle propagating through space. The self-interaction will sure modify the one particle state of free theory and this state is in no sense a linear combination of multiparticle states of the same Fock state as all the "virtual particles" are also interacting whereas the Fock space ones are not.


Hi DrFaustus,

I agree that in the traditional formulation of QFT particles do self-interact; "physical" particles of interacting theory are different from "bare" particles of free theory. I can also agree that this self-interaction makes definition of the interacting Hilbert (Fock) space problematic.

However, the more I think about the idea of "self-interacting particles" the less attractive it looks. The idea of "self-interacting QFT vacuum" is even less attractive.

Fortunately, there exists an alternative formulation of QFT, which I find more acceptable philosophically. In this formulation interactions do not modify 0-particle and 1-particle states of the theory. There is no distinction between "bare" and "physical" states. The interacting theory can happily occupy the Fock space of the free theory. This is called the "dressed particle" approach, which stems from the old work

O. W. Greenberg, S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim., 8 (1958), 378.

There are only few researchers who take this approach seriously, but I hope this number will grow. The "dressed particle" approach can yield exactly the same renormalized S-matrix as traditional QFT. One side benefit is that calculations of the "dressed particle" S-matrix do not involve divergences.

[Bob_for_short]

...Think about a self interacting particle propagating through space. ...
Drunk, you mean?
...Think about the quark-gluon plasma in the early Universe. ...
I don't remember anything from that time.

[Fredrik]

DarMM wrote a couple of interesting posts about the Hilbert space of the interacting theory here (http://www.physicsforums.com/showthread.php?p=2160934). See #50 and #54. If I understand him correctly, the usual Fock space of non-interacting states is equivalent to (the completion of?) the vector space of square-integrable distributions, and the only thing that changes about that space when we introduce interactions is the measure we use to define the integrals.

I don't really understand any of this myself, so I don't have much to contribute here, but I hope to learn this stuff some day.

[Fredrik]

So in the usual Fock space there is no space for interaction!
I think this is a mistake. Any time evolution operator that doesn't leave every n-particle subspace invariant would by definition be an interaction, and you can certainly define such operators.

[Bob_for_short]

Let me summarize: Interacting theory lives just in a different Hilbert space, so its Hamiltonian is alive and kicking.

[strangerep]

If some mathematical formalism (functional analysis, theory of Hilbert spaces, etc.) has a problem with unbounded operators, then I consider this a problem of math rather than physics. Perhaps our math is build on unrealistic axioms (out of convenience).
Yes.

Perhaps, we need to be more creative and consider more general structures such as non-separable Hilbert spaces or non-standard Hilbert spaces, or something like that.
IIUC, that's why people use rigged Hilbert space -- which is larger than ordinary
Hilbert space. In the space of tempered distributions (of which ordinary Hilbert space
is a subspace), one can define a distribution-valued inner product like

\langle p | q \rangle ~=~ \delta(p-q)

and also obtain a usable generalized spectral theorem.

Other people prefer the C*-algebra approach which can also be regarded
as "larger" than the usual Hilbert space methods (but it would take ages
for me to make that statement precise. :-)

Maybe someone will get that $1M prize one day when they
figure out which maths is right.

[strangerep]

Yes, they call it "renormalization" although it is subtraction or discarding, to tell the truth .
I thought you had read my argumentations in "RiR".
Yes, but it was some time ago and I didn't remember
the word "detraction". Since you intended "subtraction",
I now understand what you meant. :-)

[strangerep]

Another point is that in the "dressed particle" approach to QFT (which is formulated in terms of particles rather than fields) the number of relevant degrees of freedom is always finite, because any realistic physical system is made of a finite number of particles.
I think that point could only be valid if the full (non-perturbative) Hamiltonian of the
dressed theory turned out to be a finite polynomial in the a/c operators. But if one
needs ever-longer more complicated terms at each perturbation order, one ends up
with an inf-dim theory, afaict.

[Bob_for_short]

I think that point could only be valid if the full (non-perturbative) Hamiltonian of the dressed theory turned out to be a finite polynomial in the a/c operators. But if one needs ever-longer more complicated terms at each perturbation order, one ends up with an inf-dim theory, afaict.
An "inf-dim theory" is not a problem per se if the series can be summed up into a certain function, as in my electronium potential. The a/c operators are in the denominator and in such a combination (Q) that is integrable in the perturbation theory: V(r,Q) = 1/|r + εeQ|.

[strangerep]

Any time evolution operator that doesn't leave every n-particle subspace invariant would by definition be an interaction, and you can certainly define such operators.
Examples?

You've got to make sure that the full Hamiltonian still participates correctly
in a representation of the Poincare algebra, and that the Hamiltonian is
densely well-defined on the multiparticle Fock space.

[strangerep]

An "inf-dim theory" is not a problem if the series can be summed up into a certain function, as in my electronium potential. The a/c operators are in the denominator and in such a combination (Q) that is integrable in the perturbation theory.
I wait patiently for your forthcoming paper giving a fully relativistic,
fully multiparticle treatment. :-)

[Bob_for_short]

Better tell me how to get a grant or at least a part-time research position to carry out this "a fully relativistic, fully multiparticle treatment".

[meopemuk]

I think that point could only be valid if the full (non-perturbative) Hamiltonian of the
dressed theory turned out to be a finite polynomial in the a/c operators. But if one
needs ever-longer more complicated terms at each perturbation order, one ends up
with an inf-dim theory, afaict.

I agree about that. There is no fully rigorous non-perturbative formulation of the "dressed particle" theory in a closed form. But this seems to be a common drawback of many modern theories. Attempts to put them on a solid mathematical footing do not bring much progress. A good case in point is axiomatic QFT, which hasn't produced any tangible results, AFAICT. In my opinion, if we want to move forward, we should often replace mathematical rigor with physical intuition.

[DrFaustus]

meopemuk -> I'll have to defend rigorous (axiomatic, algebraic) QFT a bit. It's not true that it hasn't produced any tangible result. Interacting models have been rigorously constructed in both 2D and 3D. It's 4D that is the problem. In fact, it is essentially the only big open problem that the rigorous approach has yet to overcome. (Tightly related to the $1m prize.) But if models in 2D and 3D don't seem tangible, one tangible result of rigorous QFT is the proper justification of scattering theory in the vacuum as we know it today. In particular, the use of the Gell-Mann and Low formula to compute the amplitudes.

As for the "dressed particle" approach, I don't know it and won't comment on it. But I want to point out why is it that people use fields in the first place. (Trust me if I say that people would be more than happy to throw them out of the window if a valid alternative would be found.) Fields, when governed by hyperbolic PDEs, have the mathematical property that perturbations to the system will propagate at finite speed. And it is essentially this the main reason to use fields - relativistic causality. And this is also the meaning of the commutation relations at the quantum level - measurements at space like separations cannot influence each other. And QFT is a theory of fields that on some occasions admits a particle interpretation. Which is precisely when we use it for scattering. But it is a theory of fields.

Bob_for_short -> Recipe to get a grant or research position.

1. Re-read Bogloiubov's book to find, say in the conclusions on pag. 644 A method has thus been developed for the actual construction of the operator functions S_n free from divergences. It has turned out that this method is equivalent to the usual subtraction procedure. In this way the latter, which until now had the nature purely of a set of prescriptions, has been given a rigorous mathematical foundation within the framework of perturbation theory. And, on pag.647, last paragraph of the book Moreover, independently of the nature of possible changes in the mathematical methods and in the fundamental assumptions of local field theory, because of the singular nature of the local commutation and causal functions the principal quantities in terms of which the theory will have to be formulated will have to be generalized functions i.e., some modification of the subtraction formalism associated with the multiplication of generalized functions will be an unavoidable attribute of any future local theory. The question of the existence of divergences in a possible future non local theory of course remains open for the time being.
2. Stop wasting time on the internet ranting about renormalization and, instead, write a couple of interesting papers and publish them.

3. Find 3 academics that could understand and evaluate your papers and ask them for reference letters.

4. Enter the battlefield for the not many academic jobs.

Fredrik -> That claim was not meant to be perfectly rigorous, but more an a posteriori heuristic interpretation. But you should ask DarMM about more details - he is indeed the man to talk to when it comes to rigorous QFT. I have barely scratched the surface of it and not really delved into the details.

[Bob_for_short]

...If the answer was "yes", both Hamiltonians would have the same spectrum, so that seems not correct.
See Appendix 4 in http://arxiv.org/abs/0906.3504 where the same spectrum is obtained for two different Hamiltonians. Of course, the latter are related with a unitary ("dressing", if you like) transformation.
Bob_for_short -> Recipe to get a grant or research position.
1. Re-read Bogloiubov's book to find, say in the conclusions on pag. 644 And, on pag.647, last paragraph of the book...
What to do if I went in some respects farther than Bogoliubov? What if I have my own successful experience?
2. Stop wasting time on the internet ranting about renormalization and, instead, write a couple of interesting papers and publish them.
Done. By the way, I like mixing with clever guys.
3. Find 3 academics that could understand and evaluate your papers and ask them for reference letters.
Done. By the way, don't you want to read my papers? No? Anybody else?
4. Enter the battlefield for the not many academic jobs.
You know what they say? First you fulfil all calculations at your own account, and then we will see. It's a world upside down.

[meopemuk]

As for the "dressed particle" approach, I don't know it and won't comment on it. But I want to point out why is it that people use fields in the first place. (Trust me if I say that people would be more than happy to throw them out of the window if a valid alternative would be found.) Fields, when governed by hyperbolic PDEs, have the mathematical property that perturbations to the system will propagate at finite speed. And it is essentially this the main reason to use fields - relativistic causality. And this is also the meaning of the commutation relations at the quantum level - measurements at space like separations cannot influence each other. And QFT is a theory of fields that on some occasions admits a particle interpretation. Which is precisely when we use it for scattering. But it is a theory of fields.


I would like to call two witnesses who are going to support my view that the role of quantum fields is more formal than fundamental.

One of them is S. Weinberg, who explained the reasons to introduce quantum fields in his vol. 1. He introduces fields as a formal device to satisfy physical requirements of relativistic invariance and cluster separability. It appears that if one builds the interaction Hamiltonian as a polynomial in quantum fields, then these two conditions are satisfied. However, nobody's proved that quantum fields is the only possibility. There are examples of acceptable field-less models in the literature. In Weinberg's logic quantum fields appear as formal mathematical constructs rather than real physical entities. He even says that field commutators have nothing to do with measurability and causality. Also he doesn't see any physical relevance of the Dirac equation.

Just yesterday I "met" my other witness

N. D. Mermin, "What's bad about this habit", Physics Today, May (2009), 8.

http://www.ehu.es/aitor/irakas/mes/Reference/mermin.pdf

In particular, Mermin writes:

"But what is the ontological status of those quantum fields
that quantum field theory describes? Does reality consist of a
four-dimensional spacetime at every point of which there is a
collection of operators on an infinite-dimensional Hilbert space?
... But I hope you will agree that YOU are not a continuous
field of operators on an infinite-dimensional Hilbert space. Nor,
for that matter, is the page you are reading or the chair you are
sitting in. Quantum fields are useful mathematical tools. They
enable us to calculate things."

[Bob_for_short]

...Quantum fields are useful mathematical tools. They
enable us to calculate things."

OK, we will use fields.

[strangerep]

... (various complaints about the cruel real world) ....I had previously skimmed Bob's paper:

V. Kalitvianski, "On Perturbation Theory for the
Sturm-Liouville Problem with variable coefficients and its applications."
(Originally from around 1991-1994 in Russian.)
Available as arXiv: 0906.3504

and I figured I should read it a bit more carefully before continuing this
thread. In this post I'll summarize what I think Bob's main point is,
and then (in my next post) relate/compare it to another more recent paper:

J.R.Klauder, "Rethinking Renormalization",
Available as arXiv:0904.2869 [hep-th]

(J.R.Klauder is a renowned mathematical physicist, perhaps
best known for his vast body of work on coherent states.)

First to Bob's paper, which I'll present in a different sequence
from the paper itself (for reasons which I'll explain later)...

Consider a Schrodinger-like equation

\left( \frac{d^2}{dz^2} + \lambda - U(z) \right) \Phi(z) ~=~ 0
~~~~~~(A)

where \lambda is an eigenvalue (to be found), and
U(z) is a potential defined in terms of another function
r(z). We take a specific U of the form:

U(z) ~=~ \frac{(r^{1/4})''}{r^{1/4}}
~=~ -\frac{3}{16r^2}\left(\frac{dr}{dz}\right)^2
~+~ \frac{1}{4r} \frac{d^2r}{dz^2} ~.

(The primes denote derivatives wrt z, as usual.)

Now consider the case where r(z) is a step function (hence U
contains the square of a delta function) and is rather nasty. We wish to
find all the eigenvalues (ie values of \lambda by perturbation
about a "free" theory where U is 0.

Denote the eigenvalues and eigenfunctions of the free theory with a "(0)"
superscript, e.g., \Phi^{(0)}_n and \lambda^{(0)}_n
denote the n'th eigenfunction and eigenvalue, respectively, for the
free theory. Similarly, denote the eigenquantities of the perturbed
theory using a "PT" superscript.

A standard iterative procedure (assuming r changes slowly wrt z) yields
the perturbation series:

\Phi^{PT}_n ~=~ \Phi^{(0)}_n ~+~ \sum_{m\ne n}
\frac{U_{mn}}{\lambda^{(0)}_n - \lambda^{(0)}_m} \;
\Phi^{(0)}_m ~+~ \dots


\lambda^{PT}_n ~=~ \Phi^{(0)}_n ~+~ U_{nn} ~+~ \sum_{m\ne n}
\frac{U_{nm}U_{mn}}{\lambda^{(0)}_n - \lambda^{(0)}_m}
~+~ \dots

where

U_{mn} ~=~ \left(\Phi^{(0)}_m , U \Phi^{(0)}_m\right)
~=~ \int dz \Phi^{(0)}_m U \Phi^{(0)}_m

which are divergent because of the delta fn squared in U. In other words,
U is not an everywhere-defined operator on the Hilbert space made from
the free eigenfunctions \Phi^{(0)}_n. Letting the step in
r(z) occur at z_1, being a jump from r_1
to r_2, further evaluation of the above gives

U_{mn} ~=~ -\frac{1}{2} \Phi^{(0)}_n(z_1) \,{\Phi'}^{(0)}_n(z_1) \,
ln(r_2/r_1)
~-~ \frac{1}{16}\int dz \, \left[(ln r)' \right]^2
\left(\Phi^{(0)}_n\right)^2

As we decrease the step size, i.e., let r_2 \to r_1, the
first term approaches 0, but the second remains infinite no matter how
close r_2 gets to r_1.

Thus we have an example of what's known as a "discontinuous perturbation"
problem: the full solution does not approach the free solution continuously
in any sense. Rather, there's a discontinuous jump. This is another way of
seeing how/why the two theories "live in different Hilbert spaces". :-)

In his paper, Bob does not start from the Schrodinger-like equation (A) where
I began. Rather, he starts from a Sturm-Liouville problem:

\left( \frac{d^2}{dx^2} + \lambda r(x) \right) \psi(x) ~=~ 0
~~~~~~(B)

(plus some straightforward boundary conditions), and he notes that after
a change of variables x \to z equivalent to:

\frac{dz}{dx} ~=~ \sqrt{r(x)} ~~;~~~~
\Phi(z) ~=~ r^{1/4}(x(z)) \; \psi(z(z)) ~~,

one gets the Schrodinger-like equation (A) I started with above.
(Work it out if you don't believe me. :-)

BTW, the relevance of all this "Sturm-Liouville" stuff is that a great
many applications in physics can be cast in Sturm-Liouville form, and
there's a large body of theory one can call on to analyze this stuff,
including a functional-analytic Hilbert space setting. See the Wiki entry
http://en.wikipedia.org/wiki/Sturm-liouville
for a bit more background. Hence, understanding a problem in the
more general S-L context has wider potential applicability than one
might naively think.

In Bob's paper, he then applies a different change of variable to
the Sturm-Liouville equation (B) above, similar to the previous one
but now the new dependent variable is

\phi(z) ~=~ \psi(x(z))

yielding the equation:

\left( \frac{d^2}{dz^2} + \lambda - V(z) \right) \phi(z) ~=~ 0
~~~~~~(C)

where now the potential operator is

V(z) ~=~ -\left(ln \sqrt{r(z)}\right)' \; \frac{d}{dz} ~.

The derivative of the logarithm turns out to be proportional to a delta
function (ie not involving a delta function squared), and the matrix elements
V_{nm} are finite. For the r(z) step-function case
considered above, they turn out to be

V_{nm} ~=~ -\frac{1}{2} \, {\Phi}^{(0)}_n(z_1) \, {\Phi}^{(0)}_n(z_1)
\, ln(r_2/r1) ~.

and the perturbation procedure for iterating the eigenfunctions and eigenvalues
now remains finite.

Bob's paper goes on to analyze the eigenvalue behaviour in more detail,
and for some other cases, but the above is sufficient for my purposes
in this thread. I'll continue the story (relating to Klauder's work) in
the next post, so please don't post any comments yet until I've
completed it.

[strangerep]

(Continuation of previous posting...)

The paper of J.R. Klauder that I want to discuss in this post is:

J.R.Klauder, "Rethinking Renormalization",
Available as arXiv:0904.2869 [hep-th]

It summarizes, and builds on a lot of earlier work. Some of his earlier
ideas can be found in textbook form as

J.R.Klauder, "Beyond Conventional Quantization",
Cambridge University Press, 2000 & 2005, ISBN 0-521-25884-7.

In "Rethinking Renormalization", Klauder considers a 1D example with
classical Hamiltonian

H_\lambda(p,q) ~=~ \frac{1}{2}\left(p^2 + q^2 \right)
~+~ \frac{\lambda}{|q|^\alpha}

where \lambda is considered as a coupling constant. Clearly
the potential is singular at the origin q=0, so the particle can never reach
the origin for any \lambda > 0. The classical solutions to
the free problem (with \lambda=0) are just those of the
harmonic oscillator, proportional to cos(t-a). However, the
solutions of the "interacting" theory with infinitesimal (but nonzero)
\lambda are "rectified" waveforms, ie involving
\pm|cos(t-a)|. Klauder calls this the "pseudofree" theory,
since it clearly doesn't coincide with the free theory. One finds that
solutions to the full theory are indeed continuously connected to those
of the pseudofree theory. Klauder notes on p5 that:

If one were contemplating a perturbation series representation of the
interacting solution, that power series should not be about the free
theory (to which the interacting solutions are not continuously
connected!) but rather about the pseudofree theory.

Klauder then goes to analyze what happens when one quantizes the theory
for various values of \alpha. He takes a regularized form
of the potential:

V_\epsilon ~=~ \frac{\lambda}{(|q| + \epsilon)^\alpha}

and finds that for all \alpha < 2 one can construct a finite
series of counterterms with diminishing divergences, reminiscent of
super-renormalizable QFTs. For \alpha = 2, he finds that an
infinite series of counterterms is needed, all of equal divergence, like
a strictly renormalizable QFT. But for \alpha > 2 the
term-by-term divergences increase, reminiscent of nonrenormalizable QFTs.

In different words, for \alpha \le 2 it is possible to modify
the theory (using counterterms) such that it becomes continuously connected
to the free theory. But for \alpha > 2, Klauder finds that:

For \alpha > 2, ..., there is no modification of the theory that
can be made to prevent the theory from passing to a pseudofree theory as the
parameter \lambda \to 0. In other words, for
\alpha > 2 the interacting quantum theory passes to a pseudofree
theory with a set of eigenfunctions and eigenvalues that are generally
different from those that characterize the free theory.

Hopefully, the attentive reader can now see the parallels I'm trying to
draw between Bob's work, and the ideas of J.R.Klauder.

Even Klauder acknowledges that taking such experience, and hoping to apply
it to QFT, is still a "leap of faith" (or as someone else recently said of
Bob's ideas, "wishful thinking"). However, he then takes his ideas a little
further, attempting to characterize the singular nature of the interaction
that leads "to either a continuous connection with the original free theory
or instead leads to a continuous connection with a pseudofree theory". He
calls the latter a "hard-core interaction", and develops the idea in terms
of a functional integral approach. The distinction involves projecting
out all those paths in the free harmonic oscillator that reach or cross
the origin, and does indeed have its seeds in the classical theory in which
the feature of discontinuous perturbation can already be seen.

Klauder develops these ideas a lot further, however. He uses the example
of so-called "ultralocal" scalar QFT, which has been rigorously solved
nonperturbatively. IIUC, he shows that it is necessarily continuously
connected to a pseudofree theory distinct from the free theory. He remarks
on p23 that

.... the solution obtained from a rigorous viewpoint is identical to the one
obtained by the supremely simple prescription of choosing a suitable
pseudofree model ....

We shall see that this breathtakingly elementary procedure, coupled with
a judicious choice of further details of the pseudofree model, will provide
a divergence-free formulation of additional examples of nonrenormalizable
models, formulations that would be difficult to arrive at by any other means.


Hopefully, the similarities and difference between Bob's and Klauder's
hopes and dreams should now be clearer to the attentive reader. :-)

However, Klauder takes it even further. He then looks at more
relativistic models, specifically covariant quartic self-interacting
scalar fields and finds a pseudofree model suitable for that case. This
is treated more fully in an earlier paper:

J.R. Klauder. “Taming Nonrenormalizability”,
Available as: arXiv:0811.3386.

which I haven't yet studied properly.

In the conclusion, Klauder says

For covariant scalar nonrenormalizable quantum field models, we have shown
that the choice of a nonconventional counterterm, but one that is still non-
classical, leads to a formulation for which a perturbation analysis of both
the mass term and the nonlinear interaction term, expanded about the
appropriate pseudofree model, are term-by-term finite.


- - - - - - - - - - -
Finally, I should mention that my intent in writing these lengthy posts was
to introduce into this thread the distinction between renormalizable theories
(which many people find boring, regarding them as tractable), and
nonrenormalizable theories (which are much more difficult). The question still
remains open whether renormalizability is an essential guide in finding
physically-relevant theories, or just the mathematically easier ones. No doubt
this would become considerably more interesting if the Higgs is not found,
and a nonrenormalizable Higgs-less electroweak theory must be confronted.

[meopemuk]

Hi strangerep,

in my opinion, your examples have little relevance to the problem of renormalization in QFT. Your examples are related to ordinary quantum mechanics, where the number of particles is conserved (actually, this number is 1). But renormalization in QFT is required exactly because its interaction does not conserve the number of particles. More specifically, a QFT model requires renormalization if two conditions are satisfied:

1. The interaction operator can change the number of particles (the Hamiltonian does not commute with particle number operators).

2. The interaction operator contains terms, which act non-trivially on the vacuum and 1-particle states. Examples of such terms are (in the creation-annihilation operator notation) aaa, a^{\dag}aa, a^{\dag}a^{\dag}a, a^{\dag}a^{\dag}a^{\dag} .

As long as you have a theory with conditions 1. and 2. you are guaranteed to have renormalization difficulties. Even if the interaction is well-behaving and loop integrals are convergent, the vacuum will be "polarized" and particles will "self-interact". Hamiltonians with interactions 2. are simply physically unacceptable. This fact is tacitly admitted in the standard QFT by adding renormalization counterterms to such Hamiltonians. (By the way, I don't see much difference between renormalizable and non-renormalizable QFT's. The former ones have finite number of types of counterterms. The latter ones have infinite number of types. Not a spectacular difference.)

Condition 1. above is physically justifiable, because we know that in the real world the number of particles changes all the time. However, condition 2. does not follow from any physical observation. This condition is bad and evil. So, if you want a physically acceptable theory without renormalization, make sure that your Hamiltonian satisfies condition 1. and does not contain terms of the type 2.

[strangerep]

in my opinion, your examples have little relevance to the problem of renormalization in QFT.

Well, they are not "my" examples. :-)

Your examples are related to ordinary quantum mechanics, where the number of particles is conserved (actually, this number is 1).
I'm pretty sure Klauder's stuff with quartic interaction qualifies as a QFT.

[meopemuk]

I'm pretty sure Klauder's stuff with quartic interaction qualifies as a QFT.

Then we use different terminologies. In my opinion, the characteristic feature of QFT is the presence of interactions changing the number of particles. So, QFT requires the Fock space, in which the number of particles is allowed to vary from 0 to infinity. (Even the presence of "quantum fields" is not essential, as I tried to explain in one of my previous posts.).

If the number of particles does not change, then we have ordinary quantum mechanics, which can be formulated in an N-particle Hilbert space. The examples of Bob and Klauber belong to the latter category.

[DrFaustus]

When I quoted Bogoliubov in one of my previous posts I was really trying to emphasize the following point. There is one simple question everyone discussing QFT should answer and which is crucial to the whole renormalizability issue.

What is the mathematical nature of a quantum field? In other words, what kind of mathematical object is a quantum field?

You might try to say that it's an operator, but that is not correct. It cannot bean operator because with operators you simply cannot satisfy the canonical commutation relations. Up to now, no one has found a better way of describing quantum fields mathematically than using the concept of "operator valued distribution". This means that if \varphi(x) is a quantum field, then \int \textrm{d} x \varphi(x) f(x) is a well defined operator on some Hilbert space (with f(x) a compactly supported function). And that's the "generalized function" that Bogoliubov talks about (see my previous post). Simply stated, quantum field = operator valued distribution. And the emphasis here is on "distribution".

Now, no matter how much you dislike renormalization, the problem with distributions is that pointwise multiplication is in general ill defined. (Convince yourself by multiplying two Dirac deltas by approximating them with Gaussians and taking the limit - you get infinity.) And this is the true origin of divergences in QFT - pointwise multiplication of distributions. In other words, an object as simple as \varphi^2(x) is ill defined and needs an actual proper defnition. (A mathematically correct definition for the square is :\varphi^2(x): = \lim_{x \to y}\Big[ \varphi(x) \varphi(y) - \langle \varphi(x)\varphi(y)\rangle_0 \Big] , where the colon denotes normal ordering and the expectation value is taken in the vacuum.) Such an object is a smooth function of x and the reason for this is that \varphi(x)\varphi(y) and \langle\varphi(x)\varphi(y) \rangle_0 have the same "singularity structure". In other words, by subracting them you are removing the singularity in \varphi(x)\varphi(y) and afterwards you can take the limit. But note that this is already a first example of renormalization! In this sense you could talk about renormalizing a squared Dirac delta. (As an aside, strangerep and Bob_for_short, if in that paper appears a squared delta, then the equation is mathematically ill defined and all the conclusions that one might draw are dubious to say the least. It's similar to dividing by zero - you simply cannot do it without some modification, like considering a limit.) And obviously, when you start multiplying Feynman diagrams, which are distributions themselves, things cannot be any better and in fact things actually get worse. For Christ's sake, they wouldn't be giving you a million if properly defining a QFT would be that easy.

Bottom line, as long as quantum fields are to be regarded as operator valued distributions, and they have to be some sort of nontrivial mathematical object if we want them to satisfy canonical commutation relations, then renormalization is unavoidable. That's the meaning of Bogoliubov's words.

Now, please find a way around that and I'll be happy to listen. But first answer to my question above. Because that'll be the starting point.

[Bob_for_short]

Thank you, Strangerep, for mentioning my paper. It shows that I did manage to reformulate and find a finite perturbative solution of that particular problem. I would also add that the the precision of my solution is very good and quite sufficient for practical goals.


Dear Dr. Faustus,

Nobody argues about the nature of fields in QFTs. Nobody argues that they come in product in the perturbation theory. You yourself say that taking a finite (Gaussian or so) distribution makes the product well defined. What I found is a natural mechanism of keeping the product well defined due to quantum mechanical smearing. There is no need to "remove it". It follows from a new initial approximation and a new interaction Hamiltonian. It is natural rather than artificial. No corrections to the fundamental constants appear and the matrix elements are finite and small. The perturbation series thus are different from what you imply as inevitable.

[Bob_for_short]

When I quoted Bogoliubov in one of my previous posts I was really trying to emphasize the following point. There is one simple question everyone discussing QFT should answer and which is crucial to the whole renormalizability issue.

What is the mathematical nature of a quantum field? In other words, what kind of mathematical object is a quantum field?

...Up to now, no one has found a better way of describing quantum fields mathematically than using the concept of "operator valued distribution".

What we encounter in QFTs is not just a distribution product but a T-ordered product, i.e., a propagator, i.e., a solution of equation with a point-like source which is reduced to the Coulomb field in a non-relativistic case, i.e., a banal particle interaction. Make this interaction naturally smeared and no ill-definiteness arise. Roughly speaking, interaction of charge with the quantized EMF, taken into account in the first turn, brings this smearing in a natural way. The interaction reminder for the perturbation theory is then not so dangerous.

[Bob_for_short]

In other words, an object as simple as \varphi^2(x) is ill defined and needs an actual proper defnition.

When I encountered a delta-function product δ2(z-z1) in my integral, I had fun while thinking that this product should "actually" be determined with experimental data.

[strangerep]

I'm pretty sure Klauder's stuff with quartic interaction qualifies as a QFT.

Then we use different terminologies. In my opinion, the characteristic feature of QFT is the presence of interactions changing the number of particles.


I meant all the rest of Klauder's work, not merely the small subset I mentioned
in my earlier post.

[strangerep]

You might try to say that [a quantum field] is an operator, but that is not correct. It cannot be an operator because with operators you simply cannot satisfy the canonical commutation relations.
OK, so let's review that in a bit more detail before continuing...

In a unitary rep of the usual CCRs, at least one of the P,Q operators must be
unbounded. (The proof can be found in Reed & Simon.) Therefore at least one of
the operators cannot be defined on the whole Hilbert space.

In a standard field theory, we want an uncountably infinite collection of such
operators, parameterized by points of Minkowski space. If one tried to write (naively):

[a_x, a^*_y] ~=~ \delta_{xy}

then the "Kronecker delta" operator on the rhs is not trace class, right?
But is that the only difficulty? Or is it more problematic that such a
generalized "Kronecker delta" with continuous-valued indices (ie not a
Dirac delta distribution) is only nonzero on a set of Lebesgue measure zero?
That causes problems in spectral representation theorems, right?
Anything else?

Generalizing the rhs to a Dirac delta distribution still has the problem
that such things only make sense when integrated against a test function
from (eg) the Schwarz space.

Since test functions are square-integrable, one then adopts a different
form for the CCRs:

[a_f, a^*_g] ~=~ (f,g)

(where f,g are test functions). I.e., one parameterizes the set of operators
using functions from Schwarz space instead. In this way, one constructs
an infinite set of bona fide operators on a Hilbert space, but then must
confront the fact that in the definition
\phi_f ~:=~ \int \textrm{d}x \; \varphi(x) f(x)
\varphi(x) is only a distribution, but we really want to construct
interactions by multiplying \varphi(x) terms. Hence the problem.

[meopemuk]

... but we really want to construct
interactions by multiplying \varphi(x) terms. Hence the problem.

I still don't see the problem.

For example, in QED interaction is built as a product of three \varphi(x) terms. I can represent each term as a linear combination of creation and annihilation operators, perform the product and obtain a collection of quite regular terms like aaa, a^{\dag}aa, a^{\dag}a^{\dag}a, a^{\dag}a^{\dag}a^{\dag} . There are no bad singularities. The only problem is that these terms act non-trivially on the vacuum and 1-particle states.

[Bob_for_short]

I see we speak completely different languages. Eugene is concentrated on a+ and a, Dr. Faustus and Dr. Strangerep operate in terms of fields φ, and I talk about different interacting species (electroniums) involved in a new rather than fixed-once-and-forever Hamiltonian with self-action.

I agree, in a narrow sense of "different Hilbert spaces" your reasoning may be sufficient but what we finally seek is how to get a working theory with physically meaningful degrees of freedom, don't we?

[DarMM]

I still don't see the problem.

For example, in QED interaction is built as a product of three \varphi(x) terms. I can represent each term as a linear combination of creation and annihilation operators, perform the product and obtain a collection of quite regular terms like aaa, a^{\dag}aa, a^{\dag}a^{\dag}a, a^{\dag}a^{\dag}a^{\dag} . There are no bad singularities. The only problem is that these terms act non-trivially on the vacuum and 1-particle states.
The interaction density in QED is supposed to be an operator valued distribution (OVD). Upon integration to obtain the actual interaction it is supposed to be an operator.
The problem is that the interaction density is not an OVD. It's produced by multiplying three well defined OVDs, namely \psi, \bar\psi and A_{\mu}, however as pointed out by DrFaustus, the product of OVDs cannot be defined in general and the resulting integrated object is not an operator and is indeed quite singular. To show this for yourself test it by making it act on a few states and check the norm of the resulting state, you will find it is infinite.

Renormalization is the method of defining powers of OVDs.

Now for \phi^{4} in two and three dimensions Glimm and Jaffe (with simpler proofs being provided later by others), showed that a part of of this method involves choosing the correct representation of the canonical commutation relations, which means choosing the correct Hilbert space. A Hilbert space which must be different from Fock space.
In fact you can show that in Fock space the most you can "soften" a product of OVDs (in the hopes of making it well defined) is by using Wick ordering on it. If Wick ordering doesn't work then it cannot be defined and you must leave Fock space.

All other QFTs which have been shown to nonperturbativley exist (which includes guage theories and Yukawa models in two and three dimensions) do in fact live in another Hilbert space which is not Fock space. The work done so far on four dimensional theories by Balaban, Magnen, Seneor and Rivasseau shows that this is also the case in four dimensions.

[Bob_for_short]

- Excuse me! I am lost! Please, tell me, where I am?
- You are in a car, Sir.

... however as pointed out by DrFaustus, the product of OVDs cannot be defined in general and the resulting integrated object is not an operator and is indeed quite singular. To show this for yourself test it by making it act on a few states and check the norm of the resulting state, you will find it is infinite.

Infinite corrections, to be short.

Renormalization is the method of defining powers of OVDs.
From experimental data? Are you serious?

Where from do those "interactions" come like φ4 and jA ? Don't we put in Lagrangians some trial stuff ourselves and then forget about "trial character" of our activity?

[DarMM]

Infinite corrections, to be short.
No, the norm of a state created by the interaction Lagrangian is infinite. I'm just talking about operators and states, it has nothing to do with infinite corrections.


Renormalization is the method of defining powers of OVDs.
From experimental data? Are you serious?

Where from do those "interactions" come like φ4 and jA ? Don't we put in Lagrangians some trial stuff ourselves and then forget about "trial character" of our activity?
I'm not sure what you are talking about. I'm just saying the renormalization is a way of defining powers of operator valued distributions. Just like the Gram-Schmidt procedure is a way of producing an orthonormal basis. Nothing to do with experiment, it's just mathematical statement.
I'm talking about how you define \phi^{4} and \psi \bar\psi A_{\mu} mathematically, if they are the experimentally correct interactions is a different question. However it is a question answered in the affirmative for \psi \bar\psi A_{\mu}.

[meopemuk]

... as pointed out by DrFaustus, the product of OVDs cannot be defined in general and the resulting integrated object is not an operator and is indeed quite singular. To show this for yourself test it by making it act on a few states and check the norm of the resulting state, you will find it is infinite.


I've calculated such products in the case of QED interaction. The resulting expressions in terms of creation and annihilation operators are long and boring, but they are not singular. At least, they are no more singular than the usual Coulomb potential. You can check the details in Appendix L of http://arxiv.org/abs/physics/0504062

Eugene.

[Fredrik]

DarMM, I'm pretty impressed by the knowledge you seem to have about these things. I would appreciate if you could give us a few tips about the best books and articles that one would have to study to get closer to your level of awesomeness.

[Bob_for_short]

No, the norm of a state created by the interaction Lagrangian is infinite. I'm just talking about operators and states, it has nothing to do with infinite corrections.
But in physics we do not seek the state norms! Especially those from interaction Lagrangians. We calculate corrections.

You can say it is the "norm" which is infinite, or the "momentum integral" at hight momenta, or a "distribution product" is infinite, whatever. Such statements or observations lead to nothing. "You are in a car, Sir". It is not an answer. It is a restatement of the same useless statement. The right answer is: "You made a mistake at this and that places. That is why you have problems".
I'm not sure what you are talking about. I'm just saying the renormalization is a way of defining powers of operator valued distributions. Just like the Gram-Schmidt procedure is a way of producing an orthonormal basis. Nothing to do with experiment, it's just mathematical statement.
Then why to do the renormalizations? We calculated the norm, it is infinite, the job is done, everybody is happy, let us go home. No? You want to compare something with experiment, don't you? Isn't it the true motivation for renormalizations - patching a failed theory?

Concerning Lint = jA, it works for jextA and jAext where the subscript "ext" means an external, known function. Extrapolating this form to the self-consistent case has failed and the conceptual and mathematical difficulties testify it. Now the theory is something that starts from non-physical things, patched with non-physical counter-terms, and free (!?) constants are used to fit the experiment, as if a theory were a fitting curve. What a shame!

[DarMM]

I've calculated such products in the case of QED interaction. The resulting expressions in terms of creation and annihilation operators are long and boring, but they are not singular. At least, they are no more singular than the usual Coulomb potential. You can check the details in Appendix L of http://arxiv.org/abs/physics/0504062

Eugene.
Hey,
It's great to see such expressions documented somewhere. However I'm not referring to such expressions, because you can't see from the expressions themselves how singular the objects are. What I'm talking about is the following:
Take the interaction Langrangian operator B = \int{\psi \bar\psi A_{\mu}(x)dx} and take any Fock state \lambda. Then create the state B\lambda = \phi, you will find that (\phi,\phi) = \infty, \forall \lambda \neq 0. So B is undefined for all states.

[DarMM]

DarMM, I'm pretty impressed by the knowledge you seem to have about these things. I would appreciate if you could give us a few tips about the best books and articles that one would have to study to get closer to your level of awesomeness.
Hey Fredrik,
Absolutely, let me give you some "must reads" first. Then I'll have a think about more specific articles.

PCT, Spin and Statistics and all that by Streater and Wightman, it's basically a good summary of the general properties of QFTs and also describes what exactly fields are mathematically.
To get a handle on the fact that QFT uses different reps try starting off with either Strocchi's Spontaneous Symmetry Breaking book, publisher Springer Berlin/Heidelberg. A free alternative which is a fun read is Stephen Summer's paper "Yet more ado about nothing, the remarkable relativistic vacuum state" (http://arxiv.org/abs/0802.1854) Particularly check out note 5 on page 4 and the references there in.
Also check out Stephen Summer's webpage.

Some good general overview articles are:
"What is a quantum field theory" (http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183550013) David Brydges
"Quantum field theory in ninety minutes" (http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183553963) Paul Federbush
They basically run through what rigorous field theory is trying to do and what kind of mathematical problem constructive field theory is. Brydges says its the study of very general diffusion processes and showing such processes exist. Federbush says its a problem in measure theory.

Now most people recommend Glimm and Jaffe's "Quantum Physics, A Functional Integral point of view", but I would actually say not to read it until much later for two reasons:
(1)One third of the book is basically one giant proof, which is that \phi^{4} exists in two dimensions. This is only necessary reading if you want to see how these things are proved.
(2)It doesn't provide an overview of rigorous field theory like most people think it does. Rather it provides an overview of rigorous Statistical Mechanics and how techniques from Statistical Mechanics, combined with measure theory can be used to analytically control a quantum field theory enough to prove its existence and some of its properties rigorously. A big mistake of mine was trying to understand this book too early.

That's just a start, I'll have a think and come back with a more extensive list.

[DrFaustus]

Finally the heavy artillery has arrived: DarMM. (Am I allowed to so blatantly admire the knowledge of a user on here?) Anyway, it's good to see you're here to help us, me included, understand this stuff better :)

Bob_for_short ->Isn't it the true motivation for renormalizations - patching a failed theory? This is not true. In lower dimensions, 2 and 3, QFT is all but a failed theory. A number of models has been constructed rigorously and non perturbatively. Moreover, for these models the usual perturbative expansion has been shown to be an asymptotic expansion to the non-perturbative solution. (Hence no problems with the non convergence of the perturbative series.) And of course, it requires renormalization. And when a QFT will be constructed rigorously in 4D, then you will be able to say the same. Clearly, given the impressive experimental accuracy, I dare you to say it's a failed theory.

Renormalization is a well understood procedure. I know it has the feeling of "ad hoc" subtractions. And it might indeed have been like that in the early days of QED. But that is not so anymore. And as I said some time ago, it need not be done at all by infinite subtractions - you might do it by "extending the product of distributions to coinciding points", and if done in this way there are no infinities arising. (See book by Scharf - "Finite QED". As the title suggests, there is not a single infinity.)

[Bob_for_short]

...What I'm talking about is the following:
Take the interaction Langrangian operator B = \int{\psi \bar\psi A_{\mu}(x)dx} and take any Fock state \lambda. Then create the state B\lambda = \phi, you will find that (\phi,\phi) = \infty, \forall \lambda \neq 0. So B is undefined for all states.

What the use of your expression if it never appears in calculations? Take any other expression, make a right statement about it and it will have the same relation to calculations as the previous one, namely no relation at all. We speak of practical things.

Yes, corrections are divergent too. Now, who is guilty? Fields? Interactions at short distances? State norms? Or physicist who failed to construct something reasonable and trying to justify obvious patches?

[DarMM]

Apologies Fredrik, I forgot the best book of all! If you can get your hand on it, try Barry Simon's gem of a book "The P(\phi)_{2} Euclidean Quantum Field Theory".
It's full of lovely details about quantum field theories and their relations to stochastic processes and probability theory in general.

[Bob_for_short]

... And when a QFT will be constructed rigorously in 4D, then you will be able to say the same.
I am afraid that 2D and 3D cases are just traps rather than a right direction. Otherwise the 4D case would have been resolved by now.
Clearly, given the impressive experimental accuracy, I dare you to say it's a failed theory.
Look, take any reasonable function and expand it in a power series. Then calculate it with the small parameter of about 0.001 (≈ α/2π) to the fourth order. You will obtain even better accuracy than in QED.

Moreover, the perturbative series in QED, considered as correct, serve to determine the value of the small parameter (alpha) from some experiments. It is not really correct. In my practice I encountered a case where seemingly correct series turned out to be wrong (due to the phenomenon of perturbative-spectral non-commutativity). So using it for finding the small parameter is erroneous. We are still unsure about correctness of the renormalized solutions.
...See book by Scharf - "Finite QED". As the title suggests, there is not a single infinity.
Do you have this book at hand? I have a question on its results.

[meopemuk]

Hey,
It's great to see such expressions documented somewhere. However I'm not referring to such expressions, because you can't see from the expressions themselves how singular the objects are. What I'm talking about is the following:
Take the interaction Langrangian operator B = \int{\psi \bar\psi A_{\mu}(x)dx} and take any Fock state \lambda. Then create the state B\lambda = \phi, you will find that (\phi,\phi) = \infty, \forall \lambda \neq 0. So B is undefined for all states.

I basically agree with you, but I would like to choose a slightly different wording to formulate the problem. Let me consider an example in which B is the interaction operator of QED and \lambda = a^{\dag}|0 \rangle is an 1-electron state. Suppose that I want to calculate the scattering matrix element in which both the initial and the final states have just one electron. I chose this example, because from physical intuition it is easy to guess what the result should be. A single electron cannot experience any scattering. It should propagate freely, and the S-operator should be just the identity operator, which is what we get already in the 0-th order S_0 = 1. This means, that all higher order contributions must be zero. However, this is not what we obtain in QED.

The second order S-matrix element is obtained (roughly) as

S_2 = \langle 0|a BB a^{dag} |0\rangle ........................(1)

(which is basically the same as your formula (\phi,\phi) ). After normal ordering of the product BB one can easily show that the result is infinite. In the language of Feynman diagrams this expression is represented by the famous electron self-energy diagram, where the infinity comes from the loop integral. Of course, we can avoid the infinity by cutting artificially the loop integral at high momentum. But, still the physical requirement S_2 = 0 cannot be satisfied.

In the standard QED this problem if fixed as follows. It is admitted that the interaction B is not correct. The "correct" interaction is obtained by adding certain "renormalization counterterms" C to the original Hamiltonian. The new "renormalized" interaction is then B+C. If the counterterms are carefully chosen, then in each perturbation order S-matrix elements become finite and physically acceptable. For example, in the 2nd order we choose the counterterm C_2 such as to cancel exacly the infinite contribution (1)

\langle 0|a C_2 a^{dag}|0\rangle= -\langle 0|a BB a^{dag} |0\rangle

to make sure that the "renormalized" S-matrix satisfies the physical condition S_2 = 0. This is my understanding of how the renormalization program works in QED and in any other quantum field theory.

The renormalization works great if our only goal is to calculate scattering amplitudes and cross-sections. However, the full "renormalized" interaction Hamiltonian B+C has lost any physical meaning, because the counterterms C involve explicitly infinite constants, such as mass and charge "corrections". It would be impossible to use this Hamiltonian in any practical calculations of, e.g., time evolution of states and observables.

Do you agree with this assessment of renormalization difficulties in QFT? If yes, then we can talk about how these difficulties can be avoided. If no, then what is your proposal?

Eugene.

[Fredrik]

Thanks for the tips DarMM. Keep 'em coming if you think of anything else to add. I'm going to read the review articles now and save the rest for later. I just started with Summers. I have never heard of the books by Strocchi and Simon. Streater & Wightman has been on my bookshelf for a while, but I have only studied a small part of it carefully so far.

Have any of you read General principles of quantum field theory (http://www.amazon.com/General-Principles-Quantum-Mathematical-Mathematics/dp/079230540X/ref=sr_1_1?ie=UTF8&s=books&qid=1257642911&sr=8-1) by Bogolubov, Logunov, Todorov & Oksak? I've been curious about that one, but not curious enough to pay what the sellers are asking for. (You have to see it to believe it, so I suggest you click on the link).

Do any of you know the differences between the two editions of Glimm & Jaffe?

[Fra]

I didn't follow this discussion from start but I just want to throw in a comment here.

Clearly, given the impressive experimental accuracy, I dare you to say it's a failed theory.
...
Renormalization is a well understood procedure. I know it has the feeling of "ad hoc" subtractions. And it might indeed have been like that in the early days of QED. But that is not so anymore.

While I do not share Bob_for_short's way of reasoning, I share and appreciate his courage to keep looking for something better.

I have a feeling som of the discussion here, regards to what extent renormalization is mathematically well defined. But as has often been the case in history, a mathematically reasonably well defined model, can still be ambigous from the physical point of view.

I do not understand how people can have such confidence in the current framework given that we have a large set of unsolved issues. Before I am ready to setttle, I would like to see how this framework solves

gravity successfully included in the framework, hierarchy problem, BH information paradoxes, cosmological constant, unification

Until then, I find it speculative to not consider the option that our current framework of QFT/renormalizations will need revision.

What the specifics will be is more difficult. But some of the comment to Bob seems to convey and idea there is nothing wrong with the current renorm/qft framework whatsoever. On that point I disagree.

I might question Bob's specific suggestions, but I do not question his quest.

/Fredrik

[DarMM]

Have any of you read General principles of quantum field theory (http://www.amazon.com/General-Principles-Quantum-Mathematical-Mathematics/dp/079230540X/ref=sr_1_1?ie=UTF8&s=books&qid=1257642911&sr=8-1) by Bogolubov, Logunov, Todorov & Oksak? I've been curious about that one, but not curious enough to pay what the sellers are asking for. (You have to see it to believe it, so I suggest you click on the link).
Well those prices are ridiculous!
I have read the book, my opinion would be that it was well a head of its time, you'll see here for the first time the renormalization group, the problems with multiplying distributions as the source of the UV divergences, ideas on how to construct theories nonperturbatively, e.t.c.
However usually you can find most of its ideas distilled into a better form after a few years of improvement in other books and articles. You see Bogolubov, Wightman and Symanzik basically started trying to get a handle on what field theories actually were back in the 1950s and tried to prove things nonperturbatively. However since then their discoveries have been refined and placed in a broader context, so I would see it as a book mainly for historical interest or to get "inspiration from a master", rather than a book to learn from.

Do any of you know the differences between the two editions of Glimm & Jaffe?
Yes, the second edition includes extended comments on three things:
(1)More on rigorous gauge theories, since a few gauge theories had been shown to exist between the two editions.
(2)More on theories in dimensions other than two. This was because the proofs of the existence of most three-dimensional theories were horribly long when the first edition came out, so they didn't really go in to them.
(3)Different methods of proving the things from the first edition and more development of Hilbert space theory.

To be honest the first edition is as good as the second, since the main draw is that the book contains the best complete proof of the existence of an interacting quantum field theory \phi^{4}_{2}, which is in both editions.
Besides that, the best other thing in the book is the first six chapters, which offer an overview of rigorous statistical mechanics (chapters 4,5) and the axioms of quantum field theory (chapter 6).
I think if you want some good general reading, as opposed to wading through a huge proof, the main draw of Glimm and Jaffe is chapters 3 and 6. These chapters will give you a good idea what the path integral is supposed to be mathematically.

In fact if I were to recommend Glimm and Jaffe's book to a beginner, I would say read chapter 3 and 6 to see what the path integral actually is as a mathematical object.

[DarMM]

This is my understanding of how the renormalization program works in QED and in any other quantum field theory.
I would view your understanding as correct.

The renormalization works great if our only goal is to calculate scattering amplitudes and cross-sections. However, the full "renormalized" interaction Hamiltonian B+C has lost any physical meaning, because the counterterms C involve explicitly infinite constants, such as mass and charge "corrections". It would be impossible to use this Hamiltonian in any practical calculations of, e.g., time evolution of states and observables.
Yes, exactly. In fact it is hard to show, but it can be shown that such a Hamiltonian can not generate unitary time evolution.

Do you agree with this assessment of renormalization difficulties in QFT? If yes, then we can talk about how these difficulties can be avoided. If no, then what is your proposal?

Yes, I agree with this assessment. Allow me to take a quantum field theory with these problems which we actually fully understand mathematically, namely \phi^{4}_{3} (subscript is the dimension of spacetime).
Exactly the same thing happens there, the S-matrix converges as an operator on Fock space so scattering is fine. However finite time evolution is not. James Glimm proved (following ideas by Rudolf Haag) in 1968 that the renormalized Hamiltonian converges as an operator not on Fock space, but on another Hilbert space, which I will call Glimm space. It is only on this space that the Hamiltonian with counterterms is self-adjoint (proven later) and bounded below and can generate unitary time evolution.
This why you need another Hilbert space:
(1)To get a finite S-matrix the Hamiltonian needs counterterms.
(2)The counterterms mean the Hamiltonian doesn't work as an operator on Fock space.
(3)Move to another Hilber space, the S-matrix will remain finite and the Hamiltonian will be well-defined.

In fact you can show that if you just pick Glimm space from the start and never go near Fock space, then everything is finite and well-defined from the beginning. Things only go wrong because we pick the wrong representation of the canonical commutation relations to begin with.

[DrFaustus]

Bob_for_short -> I am afraid that 2D and 3D cases are just traps rather than a right direction. Otherwise the 4D case would have been resolved by now. This is just your own speculation and guessing. It is no argument at all. Still, if nothing else, that 2D and 3D QFT exist shows that renormalization is a non-issue in the sense that it is present even in mathematically rigorously defined theories. Look, take any reasonable function and expand it in a power series. Then calculate it with the small parameter of about 0.001 (≈ α/2π) to the fourth order. You will obtain even better accuracy than in QED. Please, show me how to do that. Maybe we could share a Nobel prize.

I have Scharf's book. Go on and ask.

Fra -> QFT is, first and foremost, a mathematical framework. Sure, physicists use it to compute real life quantities, but it is a mathematical framework that starts from some (physically motivated) axioms and goes on from there. That one can confidently use it to compute physical quantities is precisely because it has been under scrutiny for the past 80 or so years and it's been proved times and again that "everything fits". And it is within this framework that renormalization is understood. No one claims it is the final framework to do fundamental physics and the problems you point out are amongst the reasons to go beyond it. But that does not imply a thing about the mathematical structure of QFT. On the other hand, Bob_for_short is attacking renormalization within QFT. And all I've been trying to say is that as long as QFT is formulated in terms of operator valued distributions (which quantum fields are), then some sort of renormalization is unavoidable. It is an intrinsic feature of the theory. And the usual physical interpretation that considers it an "evil" property that one should try to eliminate apparently does nothing but confuse people. The bottom line here is that renormalization is a purely mathematical necessity arising out of multiplication of distributions. You want to get rid of renormalization? Find a way to express the principles of quantum physics and the finite speed of propagation of signals that does not use any kind of distribution.

[DarMM]

I am afraid that 2D and 3D cases are just traps rather than a right direction. Otherwise the 4D case would have been resolved by now.

In the sense we are talking about it is!
Balaban, as well as independantly Magnen, Seneor and Rivasseau, have shown that the ultraviolet limit of pure SU(2) Yang-Mills theory in four dimensions is well-defined with the renormalizations. What has yet to be rigorously studied is the infrared limit (infinite volume) and the mass-gap, but this isn't yet understood even on a physical non-rigorous level, hence the use of lattice gauge theory, e.t.c.

However renormalization has been shown to rigorously work in a four-dimensional gauge theory and give a finite ultraviolet limit.

[DarMM]

And all I've been trying to say is that as long as QFT is formulated in terms of operator valued distributions (which quantum fields are), then some sort of renormalization is unavoidable.
Indeed, in fact there is a theorem which proves that this is the case. See the paper of Glimm and Jaffe "Infinite Renormalization of the Hamiltonian Is Necessary" Journal of Mathematical Physics, Vol. 10, p.2213-2214.

[Bob_for_short]

...Please, show me how to do that. Maybe we could share a Nobel prize.
I am quite sure that you can expand functions in Taylor series. Take a textbook on calculus, choose a function and make such a calculation. Then you will see that the QED precision is not the best one obtained in the fourth order.

About Scharf's book. Has the author calculated something like Compton cross section in the first non-vanishing order? Or Rutherford-like (Mott or so) cross section? I speak of QED cross sections in the first non-vanishing order, without divergent loops. I want to know if he calculates some elastic cross sections at the beginning?

[strangerep]

See book by Scharf - "Finite QED". As the title suggests, there is not a single infinity.
I only have the 1st edition (1989), and it's been years since I read it, but I got the impression
that the Epstein-Glaser-Scharf approach works by choosing smoothing functions successively
at each order of perturbation. This seemed a bit ad hoc to me. Or am I missing something?


About Scharf's book. Has the author calculated something like Compton cross section in the first non-vanishing order? Or Rutherford-like (Mott or so) cross section? I speak of QED cross sections in the first non-vanishing order, without divergent loops. I want to know if he calculates some elastic cross sections at the beginning?
The 1st edition treats Compton and Moeller scattering up to at least the first loop order
where vacuum polarization, self-energy, etc, corrections arise.

Looking on Amazon, the 2nd edition is considerably expanded (to almost twice the size).
It has a modified title: "Finite Quantum Electrodynamics: The Causal Approach".

Product Description (from Amazon):

In this textbook for graduate students in physics the author carefully analyses the role of causality in Q.E.D. This new approach avoids ultraviolet divergences, so that the detailed calculations of scattering processes and proofs can be carried out in a mathematically rigorous manner. Significant themes such as renormalizability, gauge invariance, unitarity, renormalization group, interacting fields and axial anomalies are discussed. The extension of the methods to non-abelian gauge theories is briefly described. The book differs considerably from its first edition: Chap. 3 on Causal Perturbation Theory was completely rewritten and Chap. 4 on Properties of the S-Matrix and Chap. 5 on Other Electromagnetic Couplings are new.

[meopemuk]

And all I've been trying to say is that as long as QFT is formulated in terms of operator valued distributions (which quantum fields are), then some sort of renormalization is unavoidable.

I can agree with this statement. But I don't see a good reason why physical interactions should be always constructed as products of quantum fields. Perhaps, we can remove this artificial restriction and obtain a good theory (we should not call it *quantum field* theory then), in which there is no need for renormalization.

Examples of such a theory are not difficult to construct. See

O. W. Greenberg and S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim., 8 (1958), 378.

For example, we can choose the interaction operator in the normally-ordered form

V = a^{\dag}a^{\dag}aa + a^{\dag}a^{\dag}aaa + a^{\dag}a^{\dag}a^{\dag}aa + ... ..........(1)

The characteristic feature of this interaction is that usual QFT terms aaa + a^{\dag}aa + a^{\dag}a^{\dag}a + a^{\dag}a^{\dag}a^{\dag} are absent. These terms act non-trivially on the vacuum and 1-particle states. They are considered "bad" and forbidden. The "good" terms present in our interaction (1) have at least two annihilation operators and at least two creation operators.

One can easily see, that there is no need for mass renormalization with interaction (1). Free vacuum and free 1-particle states are eigenstates of the full interacting Hamiltonian with unchanged (free) eigenvalues. One can show that the charge renormalization can be avoided too. One can also make sure that all loop integrals are convergent.

The next question is how to make sure that the S-matrix computed with interaction (1) agrees with experiment (e.g., on scattering of charged particles and photons). There are basically two ways to do that:

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

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

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

Eugene.

[DarMM]

I only have the 1st edition (1989), and it's been years since I read it, but I got the impression
that the Epstein-Glaser-Scharf approach works by choosing smoothing functions successively
at each order of perturbation. This seemed a bit ad hoc to me. Or am I missing something?

The basic point of the Epstein-Glaser approach is to take some expression from perturbation, which is not a well defined distribution since it contains products of other distributions and try to turn it into a well defined product. Let me call this object from perturbation theory S.
What you do is restrict S to a smaller domain of test functions it is well-defined on and then try to extend it to all test functions to produce a well-defined distribution. This requires making a small modification to S. There are many such possible modifications, but only one is consistent with causality and locality. You make this modification and you get S'. You now have a well-defined, local, causal object.
However this is still ad hoc as you had to actual change it by hand, however you can then show that these modifications need not be done by hand, but can be implemented by the Feynman rules themselves provided the coeffecients of the Lagrangian contain extra distributional terms. These terms agree with the counterterms of standard field theory.

Which demonstrates that renormalizing using causality, locality and distribution theory is the same as renormalizing by using the condition that a few physical numbers be finite.

[Bob_for_short]

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

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

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

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

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

[strangerep]

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

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

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

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


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

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

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

[Bob_for_short]

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

[strangerep]

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

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

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

[Bob_for_short]

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

[meopemuk]

Hi strangerep,

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

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

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

Eugene.

[DarMM]

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

Note that despite textbook claims, the traditional QFT cannot say anything about the speed of propagation of interactions. The "commutators of fields outside the light cone" have nothing to do with the time interval between the cause and the effect. In order to evaluate the speed of interactions, one must have a theory possessing a well-defined Hamiltonian and unitary time evolution. As we discussed earlier, the traditional QFT does not have these pieces. It can only calculate the S-matrix and energies of bound states, which do not reveal any time-dependent information.
The investigation of the propogation of effects in QFT was performed by Segal and Guenin in the 1960s for specific models. Also Haag's algebraic apporach allows it to be treated in general, where you can show that effects do propogate at the speed of light.

[meopemuk]

How is it inconsistent?

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




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

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

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

[DrFaustus]

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

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

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

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

strangerep & DarMM -> As I understand it, the renormalization procedure a la Epstein & Glaser (and hence the "infinite subtraction" one) is not more ad hoc then, say, solving a x^2 + b x +c =0. Let me elaborate this a bit. I'm trying to obtain an answer from my (perturbation) theory, and to get that answer I must extend my product of distributions to coinciding points where it is in general ill defined. Of course, I'll have to satisfy some conditions (causality and locality), but in principle this is "just" another mathematical problem one has to solve. Much like trying to get an answer about some physical problem which would involve the solution of the above (simple) equation. So in this sense, it does not appear more ad hoc than every other problem in physics. Comments?

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

[Bob_for_short]

Bob_for_short -> What do you mena by later on we discover that the probability of an elastic process is zero? What do you mean that QED predicts an event that never happens?
I clearly write what I mean. First we calculate some elastic processes. On page 500 (I exaggerate, of course), when the IR divergence is treated, it is stated that the elastic cross section is identically equal to zero. So, instead of predicting zero for elastic processes the standard QED predicts some finite value (in the first non-vanishing order). That means a too bad start. If the exact value were 0.5 and the initial approximation would give 0.45, it would be OK. But the exact value is 0 and the initial calculation gives 1 (the probability of elastic process). It is too far from reality. Take any Taylor expansion around x0=0 and calculate it at very big x, say x=10. You will obtain the same blinder. Why not to take a Taylor expansion around a closer point, say, at x0 = 9.9. Then f(9.9) ≈ f(10) and the remaining corrections will be small.
What has the Taylor power series expansion got to do with the perturbative expansion of, say, QED?
The magnetic moment is a series in powers of alpha/2pi. I speak of its numerical precision.
...what kind of mathematical objects are your quantum fields?

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

[Avodyne]

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

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

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

[tex]{d\sigma\over d\Omega}\biggr|_{\rm soft} = {d\sigma\over d\Omega}\biggr|_{\rm elastic}\left({c_1\Delta\over E}\right)^{\!\! c_2\alpha}[/itex]

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

[Bob_for_short]

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

[tex]{d\sigma\over d\Omega}\biggr|_{\rm soft} = {d\sigma\over d\Omega}\biggr|_{\rm elastic}\left({c_1\Delta\over E}\right)^{\!\! c_2\alpha}[/itex]

where E is the CM energy, \Delta is the maximum energy of the soft photons, \alpha is the fine-structure constant, and c_{1,2} are numerical constants. This does indeed go to zero as \Delta goes to zero. (d\sigma/d\Omega)_{\rm elastic} starts with the tree-level term, which is O(\alpha^2). So, expanding in powers of \alpha, one finds first the tree level term.
What you wrote is called the inclusive cross section, and I speak of elastic one. The elastic one (the exact and experimental) equals zero rather than > 0. The QED's one on the tree level is > 0 which is wrong. You have to work hard to cope with the IR divergence to finally obtain a physically reasonable result instead of obtaining it automatically, if your theory "catches" the physics right.

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

[meopemuk]

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

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

[Vanadium 50]

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

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