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

A vacuum preserves supersymmetry exactly whenA nonzero expectation value alone therefore does not decide the issue. On the real slice this condition is $1+{\varphi }^2+\beta \varphi =0$, which has real solutions only when $\beta \ge 2$. The sub-Planckian Minkowski solution found below has $\beta =2-\sqrt{3}<2$, so $D_{\varphi }W\ne 0$ there and supersymmetry is spontaneously broken.

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For $K=|\varphi {|}^2$,and the supergravity F-term potential isThe potential is invariant under complex conjugation. Writing $\varphi =x+iy$, its derivative with respect to $y$ vanishes at $y=0$, so the extremisation equations consistently admit a real solution. On that slice,

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At a real Minkowski vacuum, $V=0$ giveswhere the branch relevant to the sub-Planckian solution has been chosen. Since the bracket multiplying $e^{x^2}$ vanishes, stationarity givesCombining the equations yields $2x+\beta =\sqrt{3}$. Solving givesThus $X=3$, $Y=2$, $U=-1$, and $Z=3$.

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At $x_0=\sqrt{3}-1$ and $\beta =2-\sqrt{3}$,The stationary point is therefore a local minimum along the real direction. In fact $V(x)$ is positive away from this point on the real axis, tends to $+\infty$ as $x\to \pm \infty$, decreases to the zero-valued minimum at $x_0\approx 0.732$, and then rises again. Its sketch is a single well tangent to the $V=0$ axis at $x_0$.

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