I Bare/renormalised parameters and predictions

[CAF123]

I came across the following statements about renormalisation in the context of Compton scattering $(\gamma q \rightarrow \gamma q$) and its QCD corrections:

1) 'The bare and renormalised couplings are equal at $\mathcal O(\alpha_s^0)$, thus the renormalisation of the strong coupling in a bare coefficient function of order $\alpha_s^j$ only contributes at order $\alpha_s^{j+1}$'

At tree level ($\mathcal O(\alpha_s^0)$) we have no loops so there is nothing to renormalise so I understand the first part of this sentence. But how does the latter part of the sentence come about?

2) 'With tree level considerations alone, there is nothing to predict because we don't really know what the couplings are and could take them at any scale. We have to go to $\mathcal O(\alpha_s)$ to get meaningful predictions out.'

What does 2) mean?

Thanks!

 

[vanhees71]

Where came these strange ideas from? I can't say what's meant by 1) at all. For that I'd need more context. For 2) it's obviously nonsense. At tree level Compton scattering is purely electromagnetic, and you don't need any value for $\alpha_s$ (the strong coupling constant), because it doesn't occur anywhere. Concerning the em. process you have two parameters, the quark mass and $\alpha$ (the electromagnetic coupling), and these have to be determined by experiment as all the fundamental parameters of the standard model. You'll get a value at the energy scale used to fit the measured cross section to the QED expression.

 

[nrqed]

I came across the following statements about renormalisation in the context of Compton scattering $(\gamma q \rightarrow \gamma q$) and its QCD corrections:

1) 'The bare and renormalised couplings are equal at $\mathcal O(\alpha_s^0)$, thus the renormalisation of the strong coupling in a bare coefficient function of order $\alpha_s^j$ only contributes at order $\alpha_s^{j+1}$'

At tree level ($\mathcal O(\alpha_s^0)$) we have no loops so there is nothing to renormalise so I understand the first part of this sentence. But how does the latter part of the sentence come about?

2) 'With tree level considerations alone, there is nothing to predict because we don't really know what the couplings are and could take them at any scale. We have to go to $\mathcal O(\alpha_s)$ to get meaningful predictions out.'

What does 2) mean?

Thanks!

I think that here we must keep in mind that they are focusing on pure QCD effects, and what they are more specifically discussing is the RG flow of the QCD parameters, in other words the dependence on $\mu$, the RG scale.

In 1), they mean the following: Let's say you compute a process that contains an explicit factor $\alpha_s^j$ and you are only working at that order in terms of precision. Then you can ignore any RG correction and just use the tree level values of the coupling constant, any correction due to the running of the parameters can be ignored since they will introduce corrections of higher order in the coupling constant.

For 2), what they are saying is rather unfortunately confusing. Of course one can make predictions at tree levels, one simply fixes all the free parameters of the theory with a sufficient number of data and then one can calculate any other process at tree level. I think they were really focusing on the RG flow here and what they meant was simply that on does not see any running of the parameters if one works strictly at tree level. Because of this, one does not need to fix the renormalization scale. I think that by "nothing to predict" they did not mean "no physical prediction can be done using the theory" but rather that "nothing can be said about the choice of the RG scale". In any case, this is my interpretation. Is this from a book? From a review paper?