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  $N = 1$ SUSY Non-renormalization theorem

+ 4 like - 0 dislike

In reference [1], on Page 53, the $N = 1$ SUSY non-renormalization theorem is derived. One first specifies the symmetries of the general $N=1$ SUSY action in the superspace formalism, and then imposes holomorphicity of the Wilsonian effective action and finally takes limits of spurious fields. In this context my problem is as follows. After imposing consistency conditions due to the symmetries on the action, one arrives at the following form for the holomorphic term in the action

$$H = Y h(\Phi) + (\alpha X + g(\Phi))W^{\alpha}W_{\alpha},$$ 

where $X$ and $Y$ are the spurious fields. The contribution of this term to the action can be found after integrating with respect to the spacetime and $\theta$ coordinates. Immediately afterwards on Page 54, the following statement is made: 

"In the limit $Y \to 0$, there is an equality $h(Φ) = W(Φ)$ at tree level, so $W(Φ)$ is not renormalized."

I don't understand the reasoning here. Since the spurion field $Y$ plays the role of coupling here, then shouldn't the same argument hold for any QFT potential in general which respects symmetries, which is obviously not true? What exactly is the role of holomorphicity in the above statement which makes it work?

  [1]: https://arxiv.org/abs/1011.1491

asked Dec 28, 2018 in Theoretical Physics by Joyshaitan (85 points) [ no revision ]
recategorized Dec 28, 2018 by Joyshaitan

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