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In QFT Vol2 written by Weinberg (Chap 16-17), or very much similarly in Adel Bilal's notes (Chap 7), a powerful way of proving renormalizability is presented: Analyze the symmetries of the quantum effective action (QEA), and since QEA generates all 1-particle irreducible diagrams, the symmetry of it tells us what kind of counterterms are needed. If the symmetry is strong enough to limit the form of counterterms such that they are already contained in the original Lagrangian, the renormalizability is proved.

But it seems that, for the terms in their intermediate derivations to make sense at all, they must be understood as already regularized. However, it is not obvious that there exists a regularization scheme that respects all the utilized symmetries (In the QCD case they are Lorentz invariance, global gauge invariance, antighost translation invariance, ghost number conservation). Is this a missing step in their proof?

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  • $\begingroup$ I think the regulator is unimportant, as long as at every step the limit that removes it from the theory is well-defined and in the end that limit reproduces the symmetry you broke introducing it. $\endgroup$
    – Neuneck
    Mar 22, 2014 at 12:56
  • $\begingroup$ What are you referring to when you say "terms in their intermediate derivations"? Could you please give some specific equation numbers? Thanks $\endgroup$
    – Heterotic
    Mar 22, 2014 at 16:29
  • $\begingroup$ @Heterotic, I mean couterterms like $S_{\infty}$ or $\Gamma_{N,\infty}$. Now after thinking abut the proof harder, I'm more confused that I'm not even sure my original question makes sense. But at least it is weird to me that in this proof the choice of regulator seems immaterial, in conflict with my memory that people often claim "dimensional regularization is essential to the renormalization of QCD, though not essential to QED." $\endgroup$
    – Jia Yiyang
    Mar 23, 2014 at 3:27
  • $\begingroup$ There is nothing wrong with the statement: symmetries of the effective 1PI action constrains the possibles counter-terms. If the effective 1PI action have symmetries, there is no quantum anomaly for that symmetries. Quantum anomalies arises when is impossible to regularize and add suitable counter terms in order to maintain the symmetry. For non-abelian gauge theories is very unpractical to maintain the gauge symmetry in an arbitrary regularization by adding suitable counter terms. $\endgroup$
    – Nogueira
    Oct 7, 2017 at 20:19

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Not an expert on this, but I would think about from a top-down approach rather than the other way round: So we start from a "fundamental" theory (FT), valid at all energy scales (high energy included) and we then derive the low energy effective field theory (LEFT), which is sufficient to describe physics at the low energies. There are two logical possibilities:

i)The LEFT is "of the same type" as the FT, ie the terms in the lagrangian for both of them are of the same type, or

ii)The LEFT is of "different type" than the FT, ie the lagrangian of the LEFT contains more terms that are needed to describe the effects of the high energy effects that have been integrated out.

If for a given low energy theory all the counter-terms are constrained by the symmetries to be of the same type as the ones we already have, then clearly we are in case i) and the theory is renormalizable. I do not see why the existence of a regulator would affect the argument. Maybe in one regularization scheme you can prove that the counter-terms are of the same type and in another one you cannot, so you should choose the scheme that works best if you are trying to prove renormalizability. However, this choice should not change the fact that the theory is renormalizable or not. It just makes it easy/hard to give a proof for it. Furthermore, I don't think it tells you anything about which scheme to use to actually make the calculations easiest.

Finally, regarding the statement "dimensional regularization is essential to the renormalization of QCD", I would interpret it as "dimensional regularization is essential to QCD, because it respects the gauge symmetry" and not as "dimensional regularization is essential to prove the renormalization of QCD".

Hope this helps!

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    $\begingroup$ Thanks for the reply. But I don't think this is what I want. In fact I don't have a clear idea exactly what kind of answer I want. Sorry for that. $\endgroup$
    – Jia Yiyang
    Mar 24, 2014 at 2:07

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