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When one applies dimensional regularization in QED, in the end, one often gets an expression like $$\Gamma(n/2)\left(\frac{s}{\mu^{2}}\right)^{-n/2}$$, where $n$ is a small number, $\Gamma()$ is the gamma function, $\mu$ is a mass scale and $s$ is a constant related to the specific Feynman diagram.

On doing a small $n$ approximation to the above equation, one gets an expression like:

$$\frac{2}{n}-\gamma -\ln\left(\frac{s}{\mu^2}\right) $$

Here, in textbooks like 'Quantum field theory' (Mandl and Shaw, chapter 10: Regularization, equation 10.54), the first term is ignored because it diverges in the limit as $n$ goes to zero. $\mu$ is replaced by $m$- the mass of electron, maybe because $\mu$ is the mass scaling factor. Also, the second term $\gamma$, is dropped, maybe because it is a constant without any mass scaling factor $\mu$ in it, and hence is non-physical.

All this appears very ad hoc to me. Are these rules generally applicable while doing dimensional regularization? Is $\mu$ always replaced by $m$? Are all terms like $\gamma$ dropped in the end? Can all terms like $\frac{2}{n}$ that diverge always be safely ignored?

In other words, I need a general procedure to be followed in order to get a finite answer after applying the dimensional regularization. I know that in dimensional regularization, we analytically expand a function away from poles in the complex plane. But how do we get a finite answer in the process?

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    $\begingroup$ Oof. I find renormalized perturbation theory the best for understanding how to carefully implement dim reg and what happens to the $1/\epsilon$'s and the constants. Peskin and Schroeder only does an ok job of describing things. Sterman's book can be a bit hard to read because it's thorough, but that's also its strength. See especially Sterman's discussion of renormalization schemes starting on p 285 of his book. $\endgroup$
    – WAH
    Sep 14 '18 at 0:23
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    $\begingroup$ You do not "drop" terms -- you reabsorb them into counter-terms. They also allow you to replace $\mu$ for any other $\mu'$ you want, essentially by reabsorbing a term of the form $\log\frac{\mu}{\mu'}$. In a few words: if a book tells you that reanormalisation consists in dropping terms, get a better book. $\endgroup$ Sep 14 '18 at 2:13
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When it comes to renormalization, the procedure of reabsorbing certain quantities into counter-terms seems a bit ad hoc. Here are some general guidelines,

  1. The reabsorbed terms should be local corresponding to local counter-terms (with constraints on external momentum dependence, e.g, a mass counter-term should be momentum-independent). In OP's case, the reabsorbed terms should be independent of $s$.
  2. The infinite part should always be reabsorbed. In OP's case, the reabsorbed terms must include $\frac{2}{n}$.
  3. Whether or not to reabsorb the finite part (the $\gamma$ part in OP's case ) is just a matter of convenience (more on it below), which has no physical bearing. This is where the differences between $MS$ and $\bar{MS}$ schemes come from.

Some comments on the renormalization scale $\mu$,

  • One can argue for the existence of $\mu$ by just invoking that the $x$ in $ln(x)$ must be dimensionless. A standalone $ln(s)$ with $s$ of mass-square dimension does not make any sense, while $ln(\frac{s}{\mu^2})$ is acceptable.
  • The renormalization group equation as framed in most QFT book is in terms of $\mu$, which is a round about way of doing RG with $s$. For example, if we are concerned with a differential equation for $x(s)$ with initial condition $x(s)|_{s=\mu^2} = x_0$, the solution can be parameterized as $x(s, \mu^2, x_0)$. One can replace an original differential equation of $dx/ds$ with that of $dx/d\mu$.
  • As pointed out by @AccidentalFourierTransform, you can replace $\mu$ for any other $\mu'$ you want, essentially by reabsorbing an additional finite term of the form $ln(\frac{\mu^2}{\mu'^2})$. As illustrated in the previous bullet point, $\mu$ just sets the initial point of running in RG. One is allowed to start the running from a different initial point $\mu'$.
  • The miracle of RG enhancement can be readily achieved via simple resummation of geometric series of Feynman diagrams, as least in the context of perturbative QFT. Whether this kind of perturbative resummation can be dimmed as non-perturbative is a moot point.
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  • $\begingroup$ Thank you @AccidentalFourierTransform! Yes I am fully aware of that. The example is just a momentum-independent mass term, instead of a kinetic term. $\endgroup$
    – MadMax
    Sep 14 '18 at 21:32

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