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P. Dirac was worried with the infinities and their discarding in QED. He wanted us to reformulate the theory in order to eliminate infinities and renormalizations from the very beginning. Is there anybody who works professionally on this subject? I mean, who earns his salary for that. I know several amateurs but no one professional.

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locked by mbq Apr 14 '11 at 7:51

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Well, this is a controversial question but it managed to live peacefully -- thus I reopen it to prevent escalation. @Close-voters, please use downvotes to criticize questions. –  mbq Apr 13 '11 at 10:14
    
@mbq: many thanks! –  Vladimir Kalitvianski Apr 13 '11 at 10:18
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@mbq This is peaceful only when the moderator is fast enough to delete all offensive comments. It also had 5 votes to close and 13 down votes, which must be some sort of a record. It is not formulated as an answerable question and Vladimir is dismissive of any attempt to explain to him the very well known physics involved. If you want "controversy" and discussion of fringe theories, suit yourself. I will not be willing to be associated with this kind of stuff, and i predict the same applies to any self-respecting professional. –  user566 Apr 13 '11 at 15:45

4 Answers 4

Kenneth Wilson. He won a Nobel prize for it.

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Maybe you could expand this into something less snappy and more informative? :-) –  Sklivvz Apr 12 '11 at 20:54
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I think the upvotes to this answer serve a very bad example. (even though they're probably motivated by a notorious history of the OP) –  yayu Apr 12 '11 at 21:59
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Just to clarify, there are two users with the same first name. I kind of like this answer, snappy and all, but it wasn't mine. –  user566 Apr 12 '11 at 22:23
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K. Wilson justifies renormalizations and does not do any reformulation. His lattice is similar to the old idea of W. Heisenberg about the "fundamental length" in the frame of the standard theory. It is not what I am speaking of. –  Vladimir Kalitvianski Apr 13 '11 at 10:52

Anyone working on non-perturbative QFT really. (temporary) infinities arise from our mathematical approach.

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Anyone? Are you sure? Someone, perhaps... –  Marek Apr 11 '11 at 16:51
    
I do not ask why there are infinities in the standard QFT. I think the standard QFT has no physical exact (non-perturbative) solutions. One needs reformulated QFT to have something reasonable exactly and perturbatively. –  Vladimir Kalitvianski Apr 11 '11 at 16:55
    
@Vladimir: There are known exact, non-perturbative solutions to QFT's in D < 4. Many of the known problems with renormalization really are problems with perturbation theory, anyway. Beyond that, if someone can make something mathematically rigorous and precise that makes unambiguous physical predictions then what is your problem? –  Jerry Schirmer Apr 12 '11 at 22:49
    
Jerry, I do not want discussions here. You closed my question. –  Vladimir Kalitvianski Apr 12 '11 at 22:58
    
yes, but totally no-perturbative QFT will be hard to construct and solve, so I think of partially taking into account some (essential) part of interaction and considering the rest with the perturbation theory. In other words, I think a better (more physical) initial approximation can resolve the problem of infinities in the perturbative series. The closer the initial approximation to the exact solution, the smaller perturbative corrections. –  Vladimir Kalitvianski Apr 13 '11 at 15:39

What about the Twistor uprising of Arkani Hamed discussed in Lumos' blog?

Nima explains that the goal is to think different - to eliminate the word "Feynman" from the QFT calculations as completely as possible. ;-) In particular, the new description and calculational methods proudly make locality in the ordinary spacetime obscure while they succeed in making many other, more exotic properties of the N=4 gauge theory manifest

I do not know what twistors do to renormalization, quite out of my depth. But the claim in the lecture is that they simplify QCD calculations, which is one of your worries.

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I think eliminating Feynman is different from eliminating infinities. As well, if something is made manifest within the same theory, it is not a reformulation. –  Vladimir Kalitvianski Apr 11 '11 at 19:09
    
@Vladimir did you listen to the lecture? It is a completely different way of calculating. –  anna v Apr 11 '11 at 19:13
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I started listening but quit. No variable change can change physics implemented in the original equations. –  Vladimir Kalitvianski Apr 11 '11 at 19:25
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@Vladimir Strange answer. You are asking for a change of physics? I thought that your problem was use of the renormalization technique, reformulation of QFT. Mathematically infinities may disappear with changes of variables after all ( appear too). –  anna v Apr 12 '11 at 4:39
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Yes, a reformulation means a better physics implemented in the equations. "Success of renormalizations" signifies that this physics is close, not so far from us. You are right, with a bad variable change one can worsen convergence of the perturbative series but I believe that the original QFT equations are based on a wrong physics. This wrongness is removed with "repairing" the original solutions. There must be a direct way (a shortcut) to the final (good) results. I wonder if there is anybody who proceeds from this understanding in his activity? –  Vladimir Kalitvianski Apr 12 '11 at 10:50

I think you need to look for the following book, Finite Quantum Electrodynamics: this is not something "fringe" nor some "crackpot" off-shoot.

The name of the game is Causal Perturbation Theory, and was pioneered by Epstein, Glaser: "The role of locality in perturbation theory".

As far as i understand your question (in the context of your comments, etc), this answer should suffice. However, maybe it'd be a good idea to take a look at what's known as Local Quantum Physics: i think this can give you a broader view.

PS: Another route that seems worth pursuing is that of Vertex Operator Algebra (VOAs) (sometimes known just as 'Borcherds Algebra'). Note that these "infinities problems" appear in QFT because, ultimately, we're multiplying distributions (in the sense of generalized functions — in fact, operator-valued distributions) at the same point, which is not an allowed operation mathematically speaking. Thus, the way out is to do what people have suggested above and follow Ken Wilson's approach of "point-splitting": this is just the physicist's version of the idea behind VOAs. The bottom line is that every QFT is valid only on some energy scale, so when you cross such line, you're bound to get meaningless results (think of it as the sites of your lattice — defined by this energy scale — merging: completely distinct points gradually become the same). In this fashion, you can take the following path: define your QFT via a certain VOA that explicitly sets your energy scale in its Operator product expansion — which is nothing but the 'multiplication rule' of the VOA you have at hands. In principle, this is mathematically rigorous, although very difficult. In any case, this also answers your question.

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Hi Daniel, I have both editions of the Scharf's book. It is the same as the others - the same equations, the same S-matrix but a different starting point. You know, QED equations with the initial data contain everything in a causal way. You have just to make calculations. Namely these calculations contain infinities. So the conclusion should be - we have a bad theory. Scharf's trick consists in saying that the delta-function arguments in the product $[\delta (x-x_1)]^2$ do not coincide. It is funny. I did not find any difference between subtraction and Scharf's way to put the integral to zero. –  Vladimir Kalitvianski Apr 12 '11 at 13:21
    
I agree that any theory is incomplete because it does not include possibly undiscovered yet heavy particles or excitations. But it does not mean that any theory should be singular at short distances. These "points" are quite distant! –  Vladimir Kalitvianski Apr 12 '11 at 13:25
    
Dear Daniel, I am not interested in a "mathematically rigorous" way of working with the standard QFT. I am interested in a physical reformulation. –  Vladimir Kalitvianski Apr 12 '11 at 13:39
    
Regarding your first comment above, i explicitly addressed it in my "PS" — in fact, this was the raison d'être for my editing my original answer in order to add the "PS". So, i simply don't understand what you meant. Further, "these points" are not "quite distant" as you seem to think: they are $1/\Lambda$ from each other, where $\Lambda$ is the energy scale of your problem: the more energy, the less apart they are. –  Daniel Apr 13 '11 at 1:10
    
I have spoken of the following "points" or statements: 1) Any theory is incomplete with respect to the nature because it does not include highly excited states properly (we do no know them yet); 2) Any theory should be "singular at short distances" because of such incompleteness. These two statements ("points") are not the same! I believe we can advance phenomenological constructions (theories) free of infinities from the very beginning and flexible enough to incorporate possible higher excitations. Constructing such theories is a physical reformulation rather than VOA of the standard theory. –  Vladimir Kalitvianski Apr 13 '11 at 10:14

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