For $K=|\phi|^2$,
$$
D_\phi W=m^2[1+\phi^*(\phi+\beta)],
$$
and the <supergravity F-term potential> is
$$
V(\phi)=m^4e^{|\phi|^2}
\left(
\left|1+\phi^*(\phi+\beta)\right|^2
-3|\phi+\beta|^2
\right).
$$
The potential is invariant under complex conjugation. Writing $\phi=x+iy$, its derivative with respect to $y$ vanishes at $y=0$, so the extremisation equations consistently admit a real solution. On that slice,
$$
V(x)=m^4e^{x^2}
\left[(1+x^2+\beta x)^2-3(x+\beta)^2\right].
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 307 / 2 / ii / Solution

In Planck units, chiral multiplets in four-dimensional $\mathcal N=1$ supergravity have
$$
V_F=e^K\left(K^{i\bar j}D_iW\,D_{\bar j}\overline W-3|W|^2\right),
\qquad
D_iW=\partial_iW+(\partial_iK)W.
$$


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 Incoming links: Supergravity F-term potential