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2

Ok, I actually found the answer. As the question was up-voted, I'm going to write it down. I would argue as follows. Let's say that we want to define the cross section at an arbitrary renormalization scale. Then we put $$\sigma=\sigma(s,M,a_{s}(M))$$ as these are the only variables on which the cross section can explicitly depend (of course we are ...

0

The poor man's scaling is a perturbation method. The polynomial series that are produced by this method are not convergent (zero radius of convergence). This serie are called asymptotic serie. They give you good results until some small corrections ($\sim D_0 e^{-\frac{1}{\rho J}}$), when the series start to diverge. This is typical in Many-Body Physics and ...

1

Disclaimer: Renormalization is a huge subject with many facets, such as, e.g. overlapping divergences of subgraphs, regularization, renormalization group, etc. Here we will only elaborate on OP's quote from Ref. 1. Ref. 1 is considering a Feynman diagram ${\cal F}(q_1, \ldots, q_E)$ in momentum Fourier space, with external 4-momenta $(q_1, \ldots, q_E)$, ...

3

No, the coupling constants are not observable quantities. The only thing that we measure are correlation functions. When correlation functions are computed naively, they apparently depend on the cut-off as well as the coupling constants. The couplings must depend on the cut-off just in the right way for the dependence on the cut-off to cancel out.

2

Typos stops your evaluation of trilinears at[t] and ab[t] - you wrote (1/16Pi^2) as loop factor for these terms by mistake, and blow up at[0] and ab[0] seriously. There are also 2 other typos, one appears in the beta function of ab[t] \hbbis[t] should be hbbis[t]; the other one is systema should be system for the interpolation of mhu2 and mhd2. I feel, ...

1

I think the answer to the question is basically that not only the flow itself cannot be reversed, but more generally, and maybe more clearly, there is no flow which could take you in the reverse way, no matter what is the suggested path. Since there is a decreasing function characterizing any flow, then any RG flow violating this decreasing fashion is ...

1

You missed the fact that this operator $R$ is a linear operator, at least for sufficiently small $\vec{k}$. This is a linearization of the renormalization group transformation around the fixed point $\vec{K}^*$. Let $T$ be a renormalization group transformation. If you expand around the fixed point $\vec{K}^*$ you get: $$... 0 (This argument is for a one-dimensional system, but similar arguments can be given in higher dimensions. We work in units with c = \hbar = e = 1). Suppose we have some regulator procedure parameterized by a momentum cutoff \Lambda. Then, for distance L between two parallel plates, we can expand the regularized energy sum in powers of the cutoff as ... 1 Here's an attempt at clarification. We can expand the exact solution g'=f(g,\mu'/\mu) as a power series in g$$f(g,\mu'/\mu)=g+f_1(\mu'/\mu)\,g^2 + f_2(\mu'/\mu)\,g^3+\dots$$Now the idea is we can shift our renormalization scale from \mu to \mu''$$g'=f(g,\mu'/\mu)=f(g'',\mu'/\mu'').\quad (1) $g''$ itself can be expanded as a power series in ...

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