Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 48 / 2 / f / Solution

For canonical $K$, extract the ${\theta }^2{\bar{\theta }}^2$ component of ${\Phi }_i^{†}{\Phi }_i$. The part depending on the auxiliary fields is $F_i^{\ast }{F}_i$. To extract the F-term of the superpotential, expand around the complex scalar fields:All higher terms contain at least three identical-type Grassmann variables and vanish. In the convention $(\theta {\psi }_i)(\theta {\psi }_j)=-\frac{1}{2}{\theta }^2{\psi }_i{\psi }_j$, the highest component is $W_i{F}_i-\frac{1}{2}{W}_{ij}{\psi }_i{\psi }_j$. ThereforeHere $W_i=\partial W/\partial {\varphi }_i$ and $W_{ij}={\partial }_i{\partial }_{j}W$. The Yukawa interaction is displayed to distinguish it from the terms actually containing $F$. With signature $(+---)$, the other canonical terms are ${\partial }_{\mu }{\varphi }_i^{\ast }{\partial }^{\mu }{\varphi }_i+i{\bar{\psi }}_i{\bar{\sigma }}^{\mu }{\partial }_{\mu }{\psi }_i$, up to a total derivative.