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From my understanding of renormalization, we fix infinities in QFT by writing the Lagrangian's parameters (the bare ones) as a sum of a renormalized part and a cutoff dependant part. The latter vanish when we compute measurable quantities. The renormalized part is fixed by experiments. This renormalized parameter is the "physical parameter". For example, when we refer to the value of the mass of the electron, the number we can find in experimental tables, we are talking about the value of the renormalized mass.

Am I right to assume that the renormalized mass $m_R$ is such that the bare mass $m$ of the standard model Lagrangian is $ m = m_R + \sum _i \delta m_i $ with {$\delta m_i $} the cutoff dependent part due to the different interactions? Meaning, if we had a new theory to the Standard Model, we can renormalized the mass in the analytical computations to get rid of infinities and of the cut-off and use the current $m_R$ that we find in the tables.

Do I get it right?

Qmechanic
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Samael
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  • what is the $i$ indexing in your counterterms? – QCD_IS_GOOD Oct 10 '22 at 02:18
  • It corresponds to the different interactions of the standard model. $\delta m_i $ is the mass counter terms of the interaction "i" with i $\in$ {QED, Weak or Strong}. – Samael Oct 10 '22 at 09:48
  • Can you break up the counter terms like that? In principle I would have thought you have diagrams that contain multiple types of interactions, so it’s not clear the counter terms are 1-to-1 with the types of interaction (i.e. you suggest delta m_QED is a series in alpha_EM, delta m_strong a series in alpha_QCD, but really the total counterterm is a series in both alpha_EM and alpha_QCD)? – QCD_IS_GOOD Oct 10 '22 at 14:08
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    I hope you can appreciate the difference between the renormalized mass $m_R$ and pole/physical mass $m_p$, unless you resort to a very specific on-shell renormalization scheme $-i{\Sigma(p^2)}|_{p^2=m_r^2}=0$. When we talk about, say, electron mass, it's pole mass $m_p$ we are referring to. To get pole mass $m_p$, you have to combine a lot of factors, including the renormalized mass $m_R$, wave function renormalization and higher order terms $O(p^2)$ in self-energy. More details here: https://physics.stackexchange.com/q/617334/ – MadMax Oct 10 '22 at 18:09

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