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I'm wonder what precisely is meant by the renormalizability of the standard model. I can imagine two possibilities:

  1. The renormalizability of all of the interaction described by the Lagrangian before spontaneous symmetry breaking (SSB) by the nonzero vacuum expectation value (VEV) of the Higgs field.

  2. The renormalizability of the Lagrangian obtain from the initial one after SSB, expressed in terms of suitable new fields (which has direct physical interpretation contrary to the fields appearing in initial Lagrangian).

It seems that in case (2) we obtain an effective (nonrenormalizable) theory only and this precisely was the reason to introduce the mechanism of generating mass by nonzero VEV of Higgs field. The original Lagrangian (case (1)) contains only power counting renormalizable vertices so if there are no anomalies then SM befor SSB is renormalizable. However, in physical prediction (actual computations being performed), as far as I know, Lagrangian after SSB is used. Does is require infinite number of counterterms (is it effective theory)?

Qmechanic
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user72829
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1 Answers1

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The Standard Model Lagrangian before and after spontaneous symmetry breaking (SSB) is renormalizable. To see that recall that the rule is (though it may not be immediately obvious as to why this rule holds) that a theory is renormalizable if all the terms in the Lagrangian are of dimension 4 or less. This is true by design for the Standard Model in which all other terms are omitted.

To see that this property of the SM is unaffected by SSB consider the part of the Lagrangian associated with the Higgs,

\begin{equation} {\cal L} = \mu ^2 \left| \phi \right| ^2 - \lambda \left| \phi \right| ^4 - \phi \psi _i \psi _j \end{equation} where, $ \psi $ are the set of SM fields which have Yukawas (I'm being a bit sloppy here about all keeping terms that are actually SU(2) invariant). SSB implies shifting the Higgs to its vacuum expectation value which is at some value $ v $: \begin{equation} \left( \begin{array}{c} \phi _1 + i \phi _2 \\ \phi _3 + i \phi _4 \end{array} \right) \rightarrow \left( \begin{array}{c} \phi _1 + i \phi _2 + v \\ \phi _3 + i \phi _4 \end{array} \right) \end{equation} This doesn't change the dimension of the Higgs field, since $ v $ is still of mass dimension $1$ and so each term containing the Higgs won't change dimensions after SSB. Every term will still be at most of mass dimension $4$. Therefore, whether the theory is renormalizable will hold equally well before or after SSB.

JeffDror
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