Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 50 / 2

Work with the local Wilsonian effective action, retaining the elementary chiral superfield and using a nonzero infrared cutoff. Its superpotential is holomorphic in both the field and chiral spurions representing the couplings. A convenient pair of spurion symmetries assigns chargesBoth couplings and $\Phi$ transform nontrivially under each displayed symmetry. Each superpotential monomial has ordinary charge zero and R-charge two, so the integral $\int d^2\theta W$ is invariant. This R assignment differs from the conventional $(1,0,-1)$ assignment for $(\Phi ,m,g)$ by twice the ordinary charge, and describes the same two-dimensional charge space. These are formal spurion symmetries; fixing the couplings to numerical values need not leave them as actual symmetries of the fields alone.

For $m\ne 0$, $m{\Phi }^2$ has the required charges and mass dimension three, while $z=g\Phi /m$ has zero charge under both symmetries and is dimensionless. Thus the allowed holomorphic form isOne can see its generality directly. A holomorphic monomial ${\Phi }^u{m}^v{g}^w$ must obey $u-2v-3w=0$ and $3u-4v-7w=2$. Solving gives $u=w+2$ and $v=1-w$, hence precisely $m{\Phi }^2(g\Phi /m{)}^w$. A field-independent constant is irrelevant to this global supersymmetric action and can be dropped.

The symmetries alone leave the function arbitrary. To determine its perturbative form, use regularity as well as the two limits. A regular weak-coupling expansion at $g=0$ hasThe $g\to 0$ limit is a free massive theory, so $c_0=1/2$. Since the elementary field has not been integrated out, the Wilsonian superpotential must remain regular term by term as $m\to 0$ with the infrared cutoff fixed. Every term with $r\ge 2$ would introduce a pole in $m$ and is excluded. Negative powers of $g$ were already excluded by perturbative regularity at $g=0$. Consequently $F=c_0+c_{1}z$.

At $m=0$, the only surviving allowed interaction is proportional to $g{\Phi }^3$. Its coefficient is first order in $g$, and matching the tree three-field vertex gives $c_1=1/3$. No closed-loop three-field diagram can be made from a single cubic vertex; extra interaction vertices would produce higher coupling order. Equivalently free superspace propagators join chiral to antichiral fields, so a closed loop would require additional vertices or conjugate couplings, which the holomorphic charge constraints exclude for this term. This fixes the coefficient without assuming the desired conclusion. ThereforeThe regularity condition is essential: it concerns the local Wilsonian superpotential with the field retained, not a nonlocal one-particle-irreducible functional at zero infrared cutoff or a superpotential obtained by eliminating a massive field. Such different operations can generate inverse masses or infrared singularities without contradicting this proof.

The Kähler potential is a real D-term, not a holomorphic F-term, and is not protected by this argument. For example its quadratic part becomes $Z(g,\bar{g},m,\bar{m};\mu ){\Phi }^{†}\Phi$, with higher real operators also allowed. Terms involving $|g{|}^2$ are compatible with the symmetries and produce wave-function renormalization. After the canonical rescaling ${\Phi }_c=Z^{1/2}\Phi$, the parameters in the superpotential expressed using canonical fields are $m_c=m/Z$ and $g_c=g/Z^{3/2}$, so physical couplings can run despite the holomorphic non-renormalization statement.

Holomorphy is crucial. Without it, neutral factors such as $1+c|g{|}^2$ could multiply either superpotential monomial while respecting both spurion symmetries. Excluding conjugate couplings, together with regularity and tree matching, is what turns the charge argument into the non-renormalization theorem.