Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 2 / a / i

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- Solution i

Solution

Split the field into disjoint ranges of Fourier modes,

ϕ_{tot} = ϕ + ϕ^{+}, ϕ (p) = 0 (p^{2} > Λ^{2}), ϕ^{+} (p) = 0 unless Λ^{2} < p^{2} \leq Λ_{0}^{2} .

The Wilsonian effective action is defined by

e^{- S_{Λ}^{eff} [ϕ]} = \int_{Λ}^{Λ_{0}} D ϕ^{+} e^{- S_{Λ_{0}} [ϕ + ϕ^{+}]} .

Δ S [ϕ, ϕ^{+}] = S_{Λ_{0}} [ϕ + ϕ^{+}] - S_{Λ_{0}} [ϕ]

immediately gives

S_{Λ}^{eff} [ϕ] = S_{Λ_{0}} [ϕ] - lo g \int_{Λ}^{Λ_{0}} D ϕ^{+} e^{- Δ S [ϕ, ϕ^{+}]} .

Quadratic cross terms vanish because the momentum supports do not overlap. For

h_{0} \neq = 0

,

Δ S = S_{0, >} [ϕ^{+}] + \int d^{4} x {\frac{h _{0}}{3 !} [3 ϕ^{2} ϕ^{+} + 3 ϕ (ϕ^{+})^{2} + (ϕ^{+})^{3}] + \frac{g _{0}}{4 !} [4 ϕ^{3} ϕ^{+} + 6 ϕ^{2} (ϕ^{+})^{2} + 4 ϕ (ϕ^{+})^{3} + (ϕ^{+})^{4}]},

where

S_{0, >} [ϕ^{+}] = \frac{1}{2} \int d^{4} x [(\partial ϕ^{+})^{2} + m_{0}^{2} (ϕ^{+})^{2}] .

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