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Generally it would be expected that a renormalizable/physical quantum field theory (QFT) would be regularization independent. For this I would first fix my regularization scheme and then compute stuff.

I'm interested to know whether there's a general way to convert results obtained from one regularization scheme to another, particularly those quantum corrections that affect the beta functions. More specifically, say I want to compare the coefficients of the $\ln{\Lambda}$ terms for computations done at the upper critical dimension $d = d_c$ of the QFT, with the coefficient of $\epsilon^{-1}$ terms obtained via dimensional regularization at $d = d_c - \epsilon$. In general, I would expect that the coefficients of $\ln{\Lambda}$ terms at $d = d_c$ and the $\epsilon^{-1}$ terms at $d < d_c$ would be different. But, is there a simple relationship between these numbers, such as a multiplicative factor?

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  • $\begingroup$ Were you able to find a nice answer to this question? $\endgroup$ May 24, 2019 at 10:18
  • $\begingroup$ No, I don't know a general way to answer this. But, it seems, at one-loop order these coefficients are usually easy to relate, and most of the time they are identical with proper redefinitions. $\endgroup$
    – vik
    Jun 2, 2019 at 17:07

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Eqs (3) and (4) of the arXiv paper state relations connecting the quadratic and logarithmic divergences in a 4-momentum cutoff scheme with those calculated using dimensional regularization. They arrive at this result by matching the one and two point Passarino-Veltman functions. $$ 4 \pi \mu^2 \left(\frac{1}{\epsilon-1}+1\right) = \Lambda^2,$$ and $$\frac{1}{\epsilon}- \gamma_{E} +\log(4 \pi \mu^2) +1 = \log \Lambda^2.$$

Here, $\mu$ is the mass-scale of dimensional regularization and $\gamma_E$ is the Euler-Macheroni constant.

Also see PRD paper and/or arXiv paper

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  • $\begingroup$ Is there a conceptual way to understand this result? As in, why do these approaches both work? This is a literal answer to the question, but I'm guessing the people who have this question generally haven't heard of the Passarino-Veltman functions. $\endgroup$ May 24, 2019 at 10:07
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    $\begingroup$ Yeah, I just got acquainted with Passarino-Veltman function. Thanks for the references @lepto. $\endgroup$
    – vik
    Jun 2, 2019 at 16:59

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