 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-45/2/e/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

v

gives

W^{'} (v + φ) = W^{'} (v) + W^{''} (v) φ + \frac{g}{2} φ^{2} = M_{v} φ + \frac{g}{2} φ^{2}, M_{v} = m + gv .

Therefore the fluctuation scalar potential is

V (φ) = ∣ M_{v} ∣^{2} ∣ φ ∣^{2} + \frac{1}{2} (M_{v}^{*} g φ^{*} φ^{2} + M_{v} g^{*} φ (φ^{*})^{2}) + \frac{∣ g ∣ ^{2}}{4} ∣ φ ∣^{4} .

For the nonzero vacuum

w = - 2 M / g

in the shifted coordinate,

M_{v} = M + g w = - M

. Thus it is also

∣ - M φ + g φ^{2} /2 ∣^{2}

. The cubic interaction changes sign compared with expansion at the other root, whereas the quadratic mass term is the same.

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