Weak-coupling matching gives
$$
f(z)=z^2+\frac23z^3,
$$
because
$$
\frac12\frac{m^3}{g^2}
\left(z^2+\frac23z^3\right)
=\frac12m\Phi^2+\frac13g\Phi^3.
$$
Thus $W_{\rm eff}=W_{\rm tree}$. The holomorphic superpotential receives neither perturbative nor nonperturbative corrections compatible with the limits and symmetries. The Kähler potential and wavefunction normalization can still be renormalized, so canonically normalized couplings can run through anomalous dimensions.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 307 / 3 / iii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-307/3/iii/solution