Let

K_{M} = - \partial^{2} + M^{2}

and let

Δ_{M} = K_{M}^{- 1}

be its Euclidean quantum field theory propagator. With the source-sign convention suited to the expression in the question, define

Z_{0} [J] = N_{0} \int D χ exp [- S_{2} [χ] - \int d^{d} x J (x) χ (x)] = exp [\frac{1}{2} \int d^{d} x d^{d} y J (x) Δ_{M} (x - y) J (y)],

where

N_{0}

makes

Z_{0} [0] = 1

. Since inserting

χ (x)

is equivalent to acting with

- δ / δ J (x)

χ

gives

W [ϕ] = S_{1} [ϕ] - lo g {exp [- S_{3} (ϕ, - \frac{δ}{δ J})] Z_{0} [J]}_{J = 0} .

Solved by gpt-5.6-sol high.

ii

Solution

The interaction is linear in the Gaussian field

χ

, so completing the functional square makes its order-

λ^{2}

contribution exact:

W [ϕ] = S_{1} [ϕ] - \frac{λ ^{2}}{2} \int d^{d} x d^{d} y ϕ^{2} (x) Δ_{M} (x - y) ϕ^{2} (y) .

Δ_{M} (p) = (p^{2} + M^{2})^{- 1}

. If every external momentum satisfies

p^{2} ≪ M^{2}

Δ_{M} (p) = \frac{1}{M ^{2}} - \frac{p ^{2}}{M ^{4}} + O (M^{- 6} p^{4})

starts with the local effective interaction

- \frac{λ ^{2}}{2 M ^{2}} \int d^{d} x ϕ^{4} (x),

which is the field-theory version of the zero-dimensional result.

Solved by gpt-5.6-sol high.

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