Write the scalar components as $\phi_i,\widetilde\phi_i,x$. The <superpotential> gives
$$
W_x=\sum_i\widetilde\phi_i\phi_i,
\qquad W_{\phi_i}=\widetilde\phi_i(x-m_i),
\qquad W_{\widetilde\phi_i}=(x-m_i)\phi_i.
$$
Choosing the sign of the <Fayet–Iliopoulos term> so that positive $\zeta$ favors the positively charged fields, the full <Supersymmetric quantum electrodynamics> potential is
$$
\boxed{
V=\sum_i|x-m_i|^2\left(|\phi_i|^2+|\widetilde\phi_i|^2\right)
+\left|\sum_i\widetilde\phi_i\phi_i\right|^2
+\frac{g^2}{2}\left(\sum_i|\phi_i|^2-\sum_i|\widetilde\phi_i|^2-\zeta\right)^2}.
$$
Changing the FI sign convention exchanges $\phi$ with $\widetilde\phi$ in the descriptions below.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 307 / 2 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-307/2/solution