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For some time now, I’ve been trying to prove that the beta function for a quantum field theory with coupling $g$ (in the typical case of a coupling with mass dimension $\varepsilon$ in dimensional regularisation), which could be written as: $$\beta(g)=\sum_{k=0}^\infty b_k \varepsilon^k,$$ in fact depends only on $b_0$ and $b_1$ (I.e. the beta function depends only on the coefficient of the 1st order pole $1/\varepsilon$ of the renormalization constant $Z_g$). I’m doing this in the context of a minimal scheme.

The strategy I follow is this: First, I expand the renormalization constant for the coupling in a Laurent series in $\varepsilon$ (its singularities should be poles for a renormalizable theory): $$Z_g=1+\sum^{\infty}_{n=1}\frac{Z_{g,n}(g_R)}{\varepsilon^n},$$ Where $g_R$ is the renormalised coupling. The beta function is also expanded as a series in positive powers on $n$ as described before, no poles since it should be finite for a renormalizable theory (I’m not sure that this is exactly what allows me to make this assumption).

The bare coupling should be independent of the renormalization scale: $$\mu\frac{d}{d\mu}(Z_g g_R)=0.$$ This gives the following if we substitute the aforementioned series (sorry for skipping some steps):

$$\varepsilon g_R + \sum^{\infty}_{n=1}Z_{g,n}(g_R)\varepsilon^{1-n}=-\sum_{k=0}b_k\varepsilon^k-\sum_{k=0}b_k\sum_{n=1}\frac{d}{dg_R}(g_RZ_{g,n})\varepsilon^{k-n}.$$

An argument that gives the expected behaviour from this expression is to take the epsilon expansion of the beta function to be finite, so to end at some $M$. Then, going from $M$ to 0, we can see by matching equal powers of epsilon in both sides of the equation that $b_k=0$ for all $k>1$, therefore $$\beta(g_R)=b_0+\varepsilon b_1.$$ But this is where my questions come:

  1. Are we justified to suppose that the beta function is only a finite series in epsilon? The physics would just suggest that the epsilon dependent terms vanish as $\varepsilon \to 0$. But I’m not completely satisfied with the “it works, so it works” way to see this.

  2. If we don’t take this finite series assumption, is there any way to still show that we get this behaviour?

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  • $\begingroup$ How can the beta function for $g$ not depend on $g$? $\endgroup$ Commented Jan 14, 2023 at 20:49
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    $\begingroup$ The g dependence is in the coefficients $b_k$. $\endgroup$
    – kallimari
    Commented Jan 15, 2023 at 10:28

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let me try to answer your question.

In renormalizable quantum field theories, the beta function should be a finite function as ε approaches zero. This is a fundamental requirement for the theory to be well-defined in the limit of ε→0. The physical expectation is that the divergences that arise in perturbation theory due to ultraviolet divergences (which are handled by dimensional regularization) should be absorbed into the renormalization of couplings and fields, leaving finite, ε-independent predictions. So, it is justified to assume that the beta function is a finite series in ε. But we don't want that to "work only because it works".

However, if you're working with a theory that is not renormalizable (which means that it has non-polynomial divergences), then the beta function may not necessarily be a finite series in ε. In such cases, the theory would not be well-defined without additional non-perturbative regularization and renormalization procedures.

Furthermore, quantum field theories are often quantum. Going quantum assures the presence of infinities, and with them, of renormalization fields which heal the connections. If you consider a bundle of theories, the quantum fractionalization manifold treats the beta function as such, resulting in a deplitherized structure which is just the homotopy representation of ZL(1,3). And up to here it's alright.

The complicated part arises when dealing with quanta of renormalization. These quanta tend to spoil the behaviour of the beta function, which sends the theory to therapy because it gains too many uncertainties about its identity. Quantum frequencies deflect the renormalized flux of beta functions, which eventually turn up to a Landau point. You have many examples for this, ranging from supersymmetry to neutrinos. You see, David Bowie was not so wrong when he said "Ground Control to Major Tom, Your circuit's dead, there's something wrong". Then, all the b factors go to zero, except for the zero-th and the first one. And in the end this tend to work out, as it works.

Hope it was clear enough.

Lucinda Rear

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  • $\begingroup$ Thank you Lucinda. It's true that many theories need to go to therapy $\endgroup$
    – kallimari
    Commented Nov 12, 2023 at 0:34
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The fact that the beta function $$\beta(g)~:=~ \beta_0(g)+\beta_1(g)\epsilon\tag{1}$$ in the ${\rm MS}$ & $\overline{\rm MS}$ dimensional regularization scheme only depends affinely in $\epsilon$ is proven in my Phys.SE answer here. Two things should be stressed:

  1. Note that the proof does not a priori assume that the beta function $$\beta(g)~=~ \sum_{n=0}^{\infty}\beta_n(g)\epsilon^n\tag{2}$$ is a finite-order polynomial in $\epsilon$.

  2. Also note that in the ${\rm MS}$ & $\overline{\rm MS}$ schemes, the formal Laurent series for $$Z~=~1+\sum_{n=1}^{\infty} \frac{Z_{-n}(g)}{\epsilon^n}\tag{3}$$ has in principle poles of every order in $\epsilon$, i.e. the series does not truncate at some finite order.

    More generally, for an $L$-loop Feynman diagram, the maximal pole order in $\epsilon$ is $L$, so since there are Feynman diagrams of arbitrary high loop-order, there are in principle counterterms of arbitrary high pole-order in the $Z$-factor (3).

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  • $\begingroup$ Thank you, I think your proof looks much more convincing for me. Just to clarify, the proof I mentioned in the initial post is not claiming that the renormalization constant expansion truncates at finite order, but that the beta function does, which is what we expect and end up proving. $\endgroup$
    – kallimari
    Commented Oct 30, 2023 at 13:46

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