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I am currently studying the renormalization group in quantum field theory and have gotten up to computing the $\beta$ function perturbatively . While I only have a basic understanding of it so far, the impression I get is that the $\beta$ function is more important as a qualitative tool than a quantitative one. What I mean by this is that the sign of the $\beta$ function is more meaningful than its magnitude (this is because the sign determines whether the theory becomes strongly coupled in the UV or in the IR, the latter of which gives rise to asymptotic freedom).

With this in mind, I wonder what importance the actual value or magnitude of the $\beta$ function has. Does it have any importance, if so, where?

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    $\begingroup$ Hundreds of HEP experiments describe their scattering events as functions of the QFT couplings as functions of energy, and connect those at varying energy, so the β function, sign and magnitude, serves to predict such experimental points, quite well. Where do you think these graphs in your introductory HEP and QFT texts come from? $\endgroup$ Commented Apr 4 at 9:26
  • $\begingroup$ For example, Fig 9.5 here. $\endgroup$ Commented Apr 4 at 10:53

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The sign of the β function is only more meaningful than its magnitude in crude qualitative arguments of the schematic asymptotic behavior of couplings—a speculative theorist's dream.

In real-life practice, QFT and its RG are mere tools (gimmicks!) for estimating (calculating) scattering amplitudes in particle physics (mostly), to check versus experiment, the ultimate arbiter of theories such as QCD, for example; or the weak interaction couplings, etc. There have been literally hundreds of such experiments, fitted nicely by such estimates. Your introductory HEP text explains all this.

For example, the detailed behavior of the strong coupling is fit to strong-produced (quark amp) jets here, enter image description here

Fig 2 of this popular articlet. The (negative) sign of the relevant β function merely tells you the theory estimate for the coupling squared is decreasing with energy, but the precise shape of the curve may only be computed by integrating the curve specified by the full derivative represented by β.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – CBBAM
    Commented Apr 4 at 17:30

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