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Reading about supersymmetry I often read the phrase

because of the non-renormalization theorems the superpotential is not renormalized.

I would like someone to be more explicit on what is meant by this sentence. For concreteness sake, let's consider a theory with only a chiral superfield and with superpotential $$W(\Phi)=M\Phi^2+\Phi^3$$ so, what are the consequences of the non-renormalization theorems for the different vertices of this theory?

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The non-renormalization theorems mean that there is only wave-function renormaliziation for parameters in the superpotential.

The important consequence is that there are no quadratic divergences for a mass in the superpotential - there are only logarithmic correction. The $\mu^2\phi^2$ operator is safe from large radiative corrections from, for example, the Planck scale.

Because $\mu^2$ ultimately sets the scale of EWSB (along with a few other soft-breaking parameters), this is very important. It's closely related to how supersymmetry solves the hierarchy problem.

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  • $\begingroup$ I am a beginner at this so I would appreciate a lot if you were more explicit with what you mean by "wave-function renormalization" for the superpotential i wrote in the question. thanks! $\endgroup$ – Yossarian Jun 3 '15 at 16:14
  • $\begingroup$ It just means that only the fields are renormalized and not the couplings. $\endgroup$ – Surgical Commander Jun 4 '15 at 5:23

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