# What is meant by the phrase "the superpotential is not renormalized"?

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Reading about supersymmetry i often read the phrase "because of the non-renormalization theorems the superpotential is nor 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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Consider the typical WZ model with global symmetry $U(1)\times U(1)_R$. Usually we write the superpotential $W$ as

$$W = \frac{M}{2} \Phi^2 + \frac{\lambda}{3} \Phi^3$$

but you have simply done some rescaling, which is ok. I am writting it like this to agree with the usual convention. Then, we require that the, so to be, renormalized superpotential is still a holomorphic function and transforms correctly under the global transformations I quote above. For these requirements to hold the superpotential takes the form

$$W = \frac{M}{2} \Phi^2 \,\, f \left( \frac{\lambda \Phi^3}{3} \right)$$

Now what happens in the weak coupling limit of the theory? The quantum superpotential should be approaching the classical one, so Taylor expand the function $f$ as

$$W = \frac{M}{2} \Phi^2 \,\, \left( 1 + \frac{2}{3}\frac{\lambda \Phi}{m} + \text{higher corrections} \right) = \frac{M}{2} \Phi^2 \,\, f \left( \frac{\lambda \Phi^3}{3} \right)+ \text{higher corrections}$$ Finally, the superpotential should be well behaved as $M \to 0$ constraining W to not having negative powers of $M$. It is obvious then that no higher terms are generated at quantum level and thus no counterterms to absorb any divergences of $M$ and $\lambda$. This is also the reason for the phrase "holomorphicity puts constraints in renormalization". Most of these stuff can be found in Terning's book "Modern Supersymmetry: Dynamics and Duality"

answered Jun 3, 2015 by (3,625 points)

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