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

**DrFaustus** · Nov6-09, 06:40 AM

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 $LaTeX Code: \\varphi(x)$ is a quantum field, then $LaTeX Code: \\int \\textrm{d} x \\varphi(x) f(x)$ is a well defined operator on some Hilbert space (with

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 $LaTeX Code: \\varphi^2(x)$ is ill defined and needs an actual proper defnition. (A mathematically correct definition for the square is $LaTeX Code: :\\varphi^2(x): = \\lim_{x \\to y}\\Big[ \\varphi(x) \\varphi(y) - \\langle \\varphi(x)\\varphi(y)\\rangle_0 \\Big]$ , 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 $LaTeX Code: \\varphi(x)\\varphi(y)$ and $LaTeX Code: \\langle\\varphi(x)\\varphi(y) \\rangle_0$ have the same "singularity structure". In other words, by subracting them you are removing the singularity in $LaTeX Code: \\varphi(x)\\varphi(y)$ 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** · Nov6-09, 07:01 AM

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** · Nov6-09, 08:42 AM

Originally Posted by 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?

...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** · Nov6-09, 08:45 AM

Originally Posted by DrFaustus

In other words, an object as simple as $LaTeX Code: \\varphi^2(x)$ 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** · Nov6-09, 06:36 PM

Originally Posted by strangerep

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

Originally Posted by meopemuk

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** · Nov6-09, 07:26 PM

Originally Posted by DrFaustus

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):
$LaTeX Code: <BR>[a_x, a^*_y] ~=~ \\delta_{xy}<BR>$
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:

LaTeX Code: <BR>[a_f, a^*_g] ~=~ (f,g)<BR>

(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
$LaTeX Code: \\phi_f ~:=~ \\int \\textrm{d}x \\; \\varphi(x) f(x)$
$LaTeX Code: \\varphi(x)$ is only a distribution, but we really want to construct
interactions by multiplying $LaTeX Code: \\varphi(x)$ terms. Hence the problem.

**meopemuk** · Nov6-09, 09:05 PM

Originally Posted by strangerep

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

I still don't see the problem.

For example, in QED interaction is built as a product of three $LaTeX Code: \\varphi(x)$ 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 $LaTeX Code: aaa, a^{\\dag}aa, a^{\\dag}a^{\\dag}a, a^{\\dag}a^{\\dag}a^{\\dag}$ . 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** · Nov7-09, 09:04 AM

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** · Nov7-09, 01:37 PM

Originally Posted by meopemuk

I still don't see the problem.

For example, in QED interaction is built as a product of three $LaTeX Code: \\varphi(x)$ 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 $LaTeX Code: aaa, a^{\\dag}aa, a^{\\dag}a^{\\dag}a, a^{\\dag}a^{\\dag}a^{\\dag}$ . 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 $LaTeX Code: \\psi, \\bar\\psi$ and $LaTeX Code: A_{\\mu}$ , 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 $LaTeX Code: \\phi^{4}$ 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** · Nov7-09, 02:18 PM

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

Originally Posted by DarMM

... 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.

Originally Posted by DarMM

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** · Nov7-09, 04:04 PM

Originally Posted by Bob_for_short

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.

Originally Posted by Bob_for_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?

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 $LaTeX Code: \\phi^{4}$ and $LaTeX Code: \\psi \\bar\\psi A_{\\mu}$ mathematically, if they are the experimentally correct interactions is a different question. However it is a question answered in the affirmative for $LaTeX Code: \\psi \\bar\\psi A_{\\mu}$ .

**meopemuk** · Nov7-09, 04:27 PM

Originally Posted by DarMM

... 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** · Nov7-09, 04:29 PM

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** · Nov7-09, 04:48 PM

Originally Posted by DarMM

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".

Originally Posted by DarMM

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 L_int = jA, it works for j_extA and jA_ext 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** · Nov7-09, 05:34 PM

Originally Posted by meopemuk

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 $LaTeX Code: B = \\int{\\psi \\bar\\psi A_{\\mu}(x)dx}$ and take any Fock state $LaTeX Code: \\lambda$ . Then create the state $LaTeX Code: B\\lambda = \\phi$ , you will find that $LaTeX Code: (\\phi,\\phi) = \\infty$ , $LaTeX Code: \\forall \\lambda \\neq 0$ . So

is undefined for all states.

**DarMM** · Nov7-09, 06:00 PM

Originally Posted by 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.

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" 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" David Brydges
"Quantum field theory in ninety minutes" 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 $LaTeX Code: \\phi^{4}$ 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.