Given the coupling costant $\alpha_s$ of QCD and it's RGE equation $\frac{d\ln \alpha_s(\mu^2)}{d\ln\mu^2}=\beta(\alpha_s)$, with $\beta(\alpha_s)=-\beta_0\alpha_s-\beta_1\alpha_s^2 -\beta_2\alpha_s^3+ \dots$ the perturbative expansion of the beta function. It can be proved that only the first two coefficents $\beta_0$ $\beta_1$ are independent by the renormalization scheme.
The positivity of the first beta coefficent is directly related to the property of asymptotic freedom, since $\alpha_s(\mu^2)$ decrease with $\mu^2$ only for $\beta_0>0$. So it is nice that this coefficent is independent by the renormalization scheme, given it's relation to a physical phenomenon.
My questions are the following
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Does exist a physical phenomenon directly related to some property of the second beta coefficent? Such as for the first one. A property that makes important the fact that the related coefficent is independent by the renormalization scheme.
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why it shouldn't bother me that the others coefficents are not independent by the renormalization scheme? They contributes to $\mu^2$ dependence of the coupling constant which is a property that can be measured experimentally.
The only solution I can see is that, order by order, the terms that are different in the different renormalization schemes cancel out. In such a way the full beta function is scheme independent, and so there are no problems concerning the evolution of $\alpha_(\mu^2)$. Is this correct?
Best Answer
When Wikipedia (and all other sources) quote values for the strong coupling constant, they don't just state the energy scale that the experiments use. They state the scheme as well. This is because experiments cannot measure the coupling constant. They measure a cross section at a given center of mass energy $\mu_1$. If we want to talk about the coupling constant, we have to do a loop calculation, choose a scheme in order for it to be finite, and then solve for the value of $g(\mu_1)$ which maximizes agreement with the experiment. This will inevitably be scheme dependent.
The reason why it's useful to quote $g(\mu_1)$ is that this only has to be done once. As soon as you choose a scheme and input the value of the coupling at one scale, renormalizable theories make an unambiguous prediction for cross sections at all other scales. This comes from evolving the coupling and plugging it back into the $\sigma(g)$ formula used in the first step. The one loop solution to the RG equation that you've probably seen before is \begin{equation} g(\mu_2) = \frac{g(\mu_1)}{1 + 9\log(\mu_2 / \mu_1)g(\mu_1)}. \end{equation} At a higher number of loops, the equation becomes longer but still accomplishes the same goal: predicting $g(\mu_2)$ in a given scheme from $g(\mu_1)$ in the same scheme.