${\cal N} = 1$ SUSY Non-renormalization theorem

Question

In Ref. 1, on Page 53, the ${\cal N} = 1$ SUSY non-renormalization theorem is derived. One first specifies the symmetries of the general ${\cal N} = 1$ 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

$$H ~=~ Y h(\Phi) + (\alpha X + g(\Phi))W^{\alpha}W_{\alpha},\tag{1} $$

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?

References:

1. S. Krippendorf, F. Quevedo & O. Schlotterer, Cambridge Lectures on Supersymmetry and Extra Dimensions, arXiv:1011.1491; Subsection 4.2.2.

Answer 1

OP writes (v2):

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?

TL;DR: The point is that the spurious superfield/coupling constant $Y$ has $R$-symmetry$^1$ charge 2, so that the Wilsonian action (with $R$-symmetry charge 0) cannot generate (and hence contain) holomorphic $F$-terms with higher powers of $Y$, unlike any old coupling constant.

Sketch proof of perturbative non-renormalization theorem for holomorphic $F$-terms:

1. This is explained in more detail in Ref. 3. The original ${\cal N}=1$ SUSY action with $X=0,Y=1$ is not $R$-symmetric. The holomorphic $F$-sector is $$ W(\Phi) + f(\Phi) W^{\alpha}W_{\alpha}.\tag{A}$$

2. The extended ${\cal N}=1$ SUSY model with the spurious superfield $X$ and $Y$ have $R$-symmetry and Peccei-Quinn symmetry $$X\to X + \text{imaginary constant}.\tag{B}$$ The holomorphic $F$-sector is $$Y W(\Phi) + (X+f(\Phi)) W^{\alpha}W_{\alpha}.\tag{C}$$

3. The Wilsonian effective action is defined as $$\begin{align} \exp&\left\{-\frac{1}{\hbar}W_c[J^H,\phi_L] \right\}\cr ~:=~~~&\int \! {\cal D}\frac{\phi_H}{\sqrt{\hbar}}~\exp\left\{ \frac{1}{\hbar} \left(-S[\phi_L+\phi_H]+J^H_k \phi_H^k\right)\right\}\cr~\stackrel{\text{Gauss. int.}}{\sim} &{\rm Det}\left( (S_2)_{mn}\right)^{-1/2} \exp\left\{-\frac{1}{\hbar} S_{\neq 2}\left[\phi_L+ \hbar \frac{\delta}{\delta J^H}\right] \right\} \cr &\exp\left\{ \frac{1}{2\hbar} J^H_k (S_2^{-1})^{k\ell} J^H_{\ell} \right\}.\end{align}\tag{D}$$ The Wilsonian effective action consists only of connected Feynman diagrams. Only Feynman diagrams made entirely out of holomorphic $F$-vertices and $F$-propagators can generate $F$-terms in the Wilsonian effective action. The holomorphic $F$-sector of the Wilsonian effective action is protected by $R$-symmetry: $$H ~=~ Y h(\Phi) + (\alpha X+g(\Phi)) W^{\alpha}W_{\alpha}.\tag{E}$$

4. In the limit $Y\to 0$ and $X\to \infty$ all heavy-field Feynman diagrams linear in $Y$ vanish beyond tree-level, so that the superpotential $W(\Phi)=h(\Phi)$ is unrenormalized perturbatively.

5. Let $Y=0$ from now on. In more detail, a $X$-vertex is proportional to $X$, while the gauge propagator is proportional to $1/X$. Therefore the number $L$ of loops is related to the number of $X$s, $$\#(X)~=~V-I~=~1-L~\leq~1. \tag{F}$$ In the limit $X\to \infty$, we must also have $\#(X)\geq 0$. And $\#(X)=0$ corresponds to 1-loop, while $\#(X)=1$ corresponds to tree-level. We conclude that $\alpha=1$ and that $g(\Phi)-f(\Phi)$ contains only one-loop corrections. $\Box$

References:

1. S. Krippendorf, F. Quevedo & O. Schlotterer, Cambridge Lectures on Supersymmetry and Extra Dimensions, arXiv:1011.1491; Subsection 4.2.2.

2. N. Seiberg: arXiv:hep-ph/9309335 & arXiv:hep-ph/9408013.

3. S. Weinberg, Quantum Theory of Fields, Vol. 3; Section 27.6, p. 150-151.

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$^1$ Explicitly, $\theta$, $\overline{\theta}$ and $Y$ are the only fundamental objects that carry $R$-charge, which is $1$, $-1$ and $2$, respectively. The super-field-strength $W_{\alpha}$ only has $R$-charge $1$ because it consists of one ${\cal D}$ and two $\overline{\cal D}$'s. Similarly, the chiral measure $\int d^2\theta~=~\partial^2_{\theta}$ has $R$-charge $-2$. We stress that this $R$-symmetry is not the conventional $R$-symmetry.