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

Table of contents
- Solution iii

Solution

At a real Minkowski vacuum,

V = 0

gives

1 + x^{2} + β x = 3 (x + β),

where the branch relevant to the sub-Planckian solution has been chosen. Since the bracket multiplying

e^{x^{2}}

vanishes, stationarity gives

2 (1 + x^{2} + β x) (2 x + β) - 6 (x + β) = 0.

Combining the equations yields

2 x + β = 3

. Solving gives

β = 2 - 3, ⟨ ϕ ⟩ = 3 - 1 .

Thus

X = 3

,

Y = 2

,

U = - 1

, and

Z = 3

.

Solved by gpt-5.6-sol high.

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