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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)?

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  • $\begingroup$ If any model is said to be renormalizable, it entails all infinities may be absorbed by a finite number of counter-terms, expressing all quantities in terms of the renormalized, physical parameters. In addition, the Standard Model should indeed be viewed as an effective field theory. $\endgroup$
    – JamalS
    Jul 1, 2014 at 20:29
  • $\begingroup$ @JamalS So to make the prediction of SM finite one needs infinite number of counterterms? $\endgroup$
    – user72829
    Jul 1, 2014 at 20:33
  • $\begingroup$ No, if you read my post, I specifically said a finite number of counter-terms. $\endgroup$
    – JamalS
    Jul 1, 2014 at 20:58
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    $\begingroup$ By the way, have a look at: physics.stackexchange.com/q/4184 $\endgroup$
    – JamalS
    Jul 1, 2014 at 21:04
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    $\begingroup$ Visit einstein-schrodinger.com/Standard_Model.pdf for the full SM Lagrangian, with detailed explanations of each part and conventions. See also Prof. Wise's lectures on the SM available on the Perimeter Institute site. $\endgroup$
    – JamalS
    Jul 2, 2014 at 11:35

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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.

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