Interacting theory lives in a different Hilbert space [ ]

In summary, the conversation revolves around the use of Hilbert spaces in quantum mechanics, particularly in the context of interacting and non-interacting systems. The question of whether the same Hilbert space can be used for both types of systems is discussed, with some arguing that the same logical propositions can be asked in both cases while others point out the importance of considering the full spectrum of operators. The concept of unitary transformations is also brought up, along with the role of detraction (or subtraction) in QFT. Ultimately, the differences between the relativistic and non-relativistic cases are considered, with the conclusion that the use of Hilbert spaces in QFT may be influenced by the nature of infinite degrees of freedom and unbounded
  • #36
DrFaustus said:
In other words, an object as simple as [tex]\varphi^2(x)[/tex] 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.
 
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  • #37
strangerep said:
I'm pretty sure Klauder's stuff with quartic interaction qualifies as a QFT.

meopemuk said:
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.
 
  • #38
DrFaustus said:
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):
[tex]
[a_x, a^*_y] ~=~ \delta_{xy}
[/tex]
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:
[tex]
[a_f, a^*_g] ~=~ (f,g)
[/tex]
(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
[tex]\phi_f ~:=~ \int \textrm{d}x \; \varphi(x) f(x)[/tex]
[itex]\varphi(x)[/itex] is only a distribution, but we really want to construct
interactions by multiplying [itex]\varphi(x)[/itex] terms. Hence the problem.
 
  • #39
strangerep said:
... but we really want to construct
interactions by multiplying [itex]\varphi(x)[/itex] terms. Hence the problem.

I still don't see the problem.

For example, in QED interaction is built as a product of three [itex]\varphi(x)[/itex] 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 [itex] aaa, a^{\dag}aa, a^{\dag}a^{\dag}a, a^{\dag}a^{\dag}a^{\dag} [/itex]. There are no bad singularities. The only problem is that these terms act non-trivially on the vacuum and 1-particle states.
 
  • #40
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?
 
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  • #41
meopemuk said:
I still don't see the problem.

For example, in QED interaction is built as a product of three [itex]\varphi(x)[/itex] 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 [itex] aaa, a^{\dag}aa, a^{\dag}a^{\dag}a, a^{\dag}a^{\dag}a^{\dag} [/itex]. 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 [tex]\psi, \bar\psi[/tex] and [tex]A_{\mu}[/tex], 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 [tex]\phi^{4}[/tex] 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 gauge 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.
 
  • #42
- Excuse me! I am lost! Please, tell me, where I am?
- You are in a car, Sir.

DarMM said:
... 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.

DarMM said:
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?
 
  • #43
Bob_for_short said:
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.

Bob_for_short said:
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 [tex]\phi^{4}[/tex] and [tex]\psi \bar\psi A_{\mu}[/tex] mathematically, if they are the experimentally correct interactions is a different question. However it is a question answered in the affirmative for [tex]\psi \bar\psi A_{\mu}[/tex].
 
  • #44
DarMM said:
... 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.
 
  • #45
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.
 
  • #46
DarMM said:
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".
DarMM said:
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!
 
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  • #47
meopemuk said:
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 [tex]B = \int{\psi \bar\psi A_{\mu}(x)dx}[/tex] and take any Fock state [tex]\lambda[/tex]. Then create the state [tex]B\lambda = \phi[/tex], you will find that [tex](\phi,\phi) = \infty[/tex], [tex]\forall \lambda \neq 0[/tex]. So [tex]B[/tex] is undefined for all states.
 
  • #48
Fredrik said:
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 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:
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183550013" David Brydges
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 [tex]\phi^{4}[/tex] 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.
 
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  • #49
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.)
 
  • #50
DarMM said:
...What I'm talking about is the following:
Take the interaction Langrangian operator [tex]B = \int{\psi \bar\psi A_{\mu}(x)dx}[/tex] and take any Fock state [tex]\lambda[/tex]. Then create the state [tex]B\lambda = \phi[/tex], you will find that [tex](\phi,\phi) = \infty[/tex], [tex]\forall \lambda \neq 0[/tex]. So [tex]B[/tex] 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?
 
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  • #51
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 [tex]P(\phi)_{2}[/tex] 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.
 
  • #52
DrFaustus said:
... 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.
DrFaustus said:
...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.
 
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  • #53
DarMM said:
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 [tex]B = \int{\psi \bar\psi A_{\mu}(x)dx}[/tex] and take any Fock state [tex]\lambda[/tex]. Then create the state [tex]B\lambda = \phi[/tex], you will find that [tex](\phi,\phi) = \infty[/tex], [tex]\forall \lambda \neq 0[/tex]. So [tex]B[/tex] 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 [itex]B[/itex] is the interaction operator of QED and [itex]\lambda = a^{\dag}|0 \rangle[/itex] 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 [itex]S_0 = 1[/itex]. 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

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

(which is basically the same as your formula [itex](\phi,\phi) [/itex]). After normal ordering of the product [itex]BB [/itex] 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 [itex]S_2 = 0[/itex] cannot be satisfied.

In the standard QED this problem if fixed as follows. It is admitted that the interaction [itex]B[/itex] is not correct. The "correct" interaction is obtained by adding certain "renormalization counterterms" [itex]C[/itex] to the original Hamiltonian. The new "renormalized" interaction is then [itex]B+C[/itex]. 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 [itex]C_2[/itex] such as to cancel exacly the infinite contribution (1)

[tex] \langle 0|a C_2 a^{dag}|0\rangle= -\langle 0|a BB a^{dag} |0\rangle [/tex]

to make sure that the "renormalized" S-matrix satisfies the physical condition [itex]S_2 = 0[/itex]. 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 [itex]B+C[/itex] has lost any physical meaning, because the counterterms [itex]C[/itex] 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.
 
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  • #54
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 https://www.amazon.com/dp/079230540X/?tag=pfamazon01-20 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?
 
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  • #55
I didn't follow this discussion from start but I just want to throw in a comment here.

DrFaustus said:
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
 
  • #56
Fredrik said:
Have any of you read https://www.amazon.com/dp/079230540X/?tag=pfamazon01-20 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.

Fredrik said:
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 into 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 [tex]\phi^{4}_{2}[/tex], 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.
 
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  • #57
meopemuk said:
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.

meopemuk said:
The renormalization works great if our only goal is to calculate scattering amplitudes and cross-sections. However, the full "renormalized" interaction Hamiltonian [itex]B+C[/itex] has lost any physical meaning, because the counterterms [itex]C[/itex] 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.

meopemuk said:
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 [tex]\phi^{4}_{3}[/tex] (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.
 
  • #58
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.
 
  • #59
Bob_for_short said:
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.
 
  • #60
DrFaustus said:
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.
 
  • #61
DrFaustus said:
...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?
 
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  • #62
DrFaustus said:
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?

Bob_for_short said:
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.
 
  • #63
DrFaustus said:
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

[tex] V = a^{\dag}a^{\dag}aa + a^{\dag}a^{\dag}aaa + a^{\dag}a^{\dag}a^{\dag}aa + ... [/tex]...(1)

The characteristic feature of this interaction is that usual QFT terms [itex] aaa + a^{\dag}aa + a^{\dag}a^{\dag}a + a^{\dag}a^{\dag}a^{\dag} [/itex] 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.
 
  • #64
strangerep said:
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 [tex]S[/tex].
What you do is restrict [tex]S[/tex] 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 [tex]S[/tex]. There are many such possible modifications, but only one is consistent with causality and locality. You make this modification and you get [tex]S'[/tex]. 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.
 
  • #65
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.
 
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  • #66
meopemuk said:
(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.
 
  • #67
And how about my questions? They are nearly "Yes or No" ones. Do you have your own opinion?
 
  • #68
Bob_for_short said:
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.
 
  • #69
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.
 
  • #70
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.
 
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