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What are the general conditions for a scalar field theory to be renormalizable? What additional conditions arise for spin-1/2 and spin-1(if any)?

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    $\begingroup$ Have you done any research on this? Most standard QFT texts introducing the notion of "renormalizability" should do power-counting shortly thereafter. $\endgroup$
    – ACuriousMind
    Dec 8, 2016 at 21:38
  • $\begingroup$ Possible duplicate of physics.stackexchange.com/q/88884 $\endgroup$
    – user108787
    Dec 8, 2016 at 22:41
  • $\begingroup$ I have done some research and I couldn't find any source which is not clustered with Jargon. If you can explain it in layman(Physics Major) terms, it will be helpful. $\endgroup$
    – Invariance
    Dec 8, 2016 at 23:37
  • $\begingroup$ As soon as I follow it, you will be the second to know:). I have read about it and to me anyway, I think it takes a while to get the idea , but once you have it , you wil have learnt a lot of physics on the way. There are two good non math books imo, Deep Down Things by Schumm and The Infinity Puzzle by Close, which do their best to give you an intuitive picture and one math QFT book, QFT Demystified by McMahon that cuts out text and gives you an overall view of the rudiments of QFT in general, using worked examples. $\endgroup$
    – user108787
    Dec 9, 2016 at 1:43

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One short definition is that if a theory is renormalizable, then the adjustment of a finite number of parameters (such as the bare electron charge and mass) allows us to calculate the results of all observable in finite terms.

My question, as part of this short answer to the OP, (and which I lack the knowledge to answer) is:

Based on the extract below: Are we satisfied today that Feymans reservations (as just one example) about renormalization are deemed out of date?

Or is the process still slightly in the realm of quantum interpretations, in that, without being blase or glib, we can say: it works, we would like to know more about it as much as we would like know more about the real nature of a photon or the "size" of an electron, but mainly we are happy that it works.

Renormalization

The renormalization group as developed along Wilson's breakthrough insights relates effective field theories at a given scale to such at contiguous scales. It thus describes how renormalizable theories emerge as the long distance low-energy effective field theory for any given high-energy theory. As a consequence, renormalizable theories are insensitive to the precise nature of the underlying high-energy short-distance phenomena (the macroscopic physics is dominated by only a few "relevant" observables). This is a blessing in practical terms, because it allows physicists to formulate low energy theories without detailed knowledge of high-energy phenomena. It is also a curse, because once a renormalizable theory such as the standard model is found to work, it provides very few clues to higher-energy processes.

I have just found a duplicate here Semantic Problem about Renormalization but I am not sure if it is at the level the OP is looking for (no offence, it's certaintly beyond me)

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  • $\begingroup$ What you described is called "nonperturbative renormalizability", and it is not what is usually meant by renormalizability in QFT. Usually we mean that given any loop order in advance, we can adjust parameters such that the predictions are finite with respect to the given order. This is called "perturbative renormalizability". +1 anyways $\endgroup$ Dec 9, 2016 at 0:34
  • $\begingroup$ @SolenodonParadoxus. Thank you very much for that, you told me before to ask you directly if needed but with these subjects I need to read up first. I can do, and understand (YES!! :) basic Moeller scattering QED calculations OK, but there is one last part of the chapter I might hit you with, in a month or so, just the elements of a 4 by 1 matrix that is not clicking as to its derivation. Thank you again $\endgroup$
    – user108787
    Dec 9, 2016 at 1:31
  • $\begingroup$ ofcourse, my contacts are in my SE profile. Please don't hit me in comments though :) $\endgroup$ Dec 9, 2016 at 15:07
  • $\begingroup$ @SolenodonParadoxus. No, I have never used the chat rooms, but I would imagine that is where the real action occurs. As you can see from my profile, I intended to take a break, but there is always just one more question I can't resist. Thanks very much. Regards $\endgroup$
    – user108787
    Dec 9, 2016 at 15:40

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