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Is an old-fashioned renormalization prescription unconvincing?

+ 2 like - 0 dislike
705 views

As you know, before Wilsonian POV on QFT, there was an "old-fashioned" renormalization prescription, which was justified/substantiated with different truthful reasonings.

When I was young, the following reasoning looked quite convincing, if not perfect, to me. Let us consider, for example, scattering a low-frequency EM wave from a free electron. QED must give the usual non-relativistic Thomson cross section for this scattering process, and indeed, QED gives it in the first approximation. However, in higher orders the charge acquires perturbative corrections, so that it is the initial charge plus all perturbative corrections who serve now as a physical charge in QED. (Let's not discuss their cut-off dependence.) Therefore, there is no problem at all since we just must define the physical charge as this sum. S. Weinberg, arguing with Dirac, writes in his "Dreams" (page 116), that it is just a matter of definition of physical constants.

We can still find this way of presenting things in many textbooks today.

Isn't it sufficiently convincing for QED? Do you see any loophole in this "old-fashion" reasoning?

asked Aug 17, 2014 in Theoretical Physics by Vladimir Kalitvianski (22 points) [ revision history ]
edited Aug 17, 2014 by Vladimir Kalitvianski
The old point of view is not wrong, it is comfortably embeded in and generalized by Wilson's modern EFT picture. Depending on the applications, it is still useful to apply it. This is probably why old renormalization comes first in many QFT textbooks and EFTs are treated as an advanced topic which comes later. Your example is probably such a case.

Thanks, Dilaton. So you do not see any loophole in this reasoning. Does anybody else?

2 Answers

+ 2 like - 0 dislike

The old-fashioned reasoning is partially defective as it defines the renormalized quantities by an ill-defined expression. It works with the usual hand-waving, but I don't like the latter.

I recommend the causal approach, which gives the same results as other approaches but works without any UV cutoff $\Lambda$ and always works with physical constants only. This is much cleaner conceptually. It still has an IR cutoff, but this causes far less problems in calculations. Also, it has a good axiomatic basis that is justified from ordinary QM.

answered Aug 18, 2014 by Arnold Neumaier (12,385 points) [ no revision ]
Most voted comments show all comments

Thank you, Arnold. To me, the ill-defined expressions are not a problem; that's why I proposed not to discuss the cutoff dependence. So let's concentrate on the expressions like $\mathcal{M}$ and on my question.

@VladimirKalitvianski I don't understand why you want to avoid talking about the cut-off (in)dependence: your question is if the classical interpretation of renormalisation is not convincing - and it would probably so if it were cut-off dependent (which it is not, obviously).

@dimension10: It is not an issue since after defining the physical constant, the cutoff disappears.

I thought I had answered your questions at the end: To me something that is ill-defined is not convincing at all and has glaring holes. For a mathematician, it may convey intuition and suggest a conjecture, but not more. Causal perturbation theory makes it fully precise, obtaining the same renormalized results without compromise. 

Thanks, Arnold, for clarifying your position. But if we define a QFT on a lattice, then nothing is ill-defined, isn't it?

Most recent comments show all comments

OK, thanks, Arnold. So it is a correct way of proceeding.

@VladimirKalitvianski That's exactly what I said, and when it is independent of the cut-off, it means it's also independent of the way you formulate it philosophically, so basically - convincing enough.

+ 1 like - 0 dislike

(I'm not sure if I understand your question correctly - how could we find renormalisation, something that works mathematically unconvincing?)

It's usually shown in most introductory textbooks on Quantum Field Theory, that the scattering amplitude, let's call it \(\mathcal{M}\), can be written without the cut-off \(\Lambda\). C.f. Zee p. 148-150. On the last page of that selection, the matrix element is written without any reference to \(\Lambda\) and other unmeasurable quantitites:

\[\mathcal{M}=-i\lambda_P+iC\lambda_P^2\left(\ln\left(\frac{s_0}{s}\right)+\ln\left(\frac{t_0}{t}\right)+\ln\left(\frac{u_0}{u}\right)\right)+O\left(\lambda_P^3\right)\]

I don't see why one should find something unconvincing when the physics is in terms of measurable quantities, and is furthermore experimentally observed. The modern EFT picture is more comfortable, and as has been said in the comments, some sort of a "generalisation" of the old picture.

answered Aug 18, 2014 by dimension10 (1,950 points) [ no revision ]

Thank you, dimension10. I know the Zee's argument, but it concerns a specific model, maybe a bit far from a realistic case, so I did not type it in.

@VladimirKalitvianski Yes, it concerns a specific example, but it is always true that the scattering amplitudes are independent of the cut-off.

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