Integrating out a field means performing its path integral while retaining the other fields as backgrounds. The result is generally a nonlocal <quantum effective action>; for a field much heavier than all external momenta, a derivative expansion turns it into local effective interactions.

Integrating out a field

{title2=1PI Feynman diagram}
{wiki=One-particle-irreducible}

A one-particle-irreducible Feynman diagram is a connected <Feynman diagram> that remains connected after any one internal line is cut. The vertices of the <quantum effective action> generate precisely these diagrams.

One-particle-irreducible Feynman diagram

For a bubble carrying momentum $p$, a <Feynman parameter> and a shift of the loop momentum give, after Wick rotation and <cutoff regularization>,
$$
B(p^2)=\int_0^1dx\int_{|\ell_E|<\Lambda}
\frac{d^4\ell_E}{(2\pi)^4}
\frac1{[\ell_E^2+F(p^2)]^2}
=\frac1{16\pi^2}\int_0^1dx
\left[\log\frac{\Lambda^2}{F(p^2)}-1\right]
+O(\Lambda^{-2}),
$$
where $F(p^2)=m^2-x(1-x)p^2-i0$. The <quantum effective action> therefore contains
$$
\Gamma_4(s,t,u)=\lambda+\delta_\lambda
-\frac{\lambda^2}{2}\{B(s)+B(t)+B(u)\}+O(\lambda^3).
$$
The <renormalization condition> $\Gamma_4(M^2,M^2,M^2)=\lambda$ fixes
$$
\delta_\lambda=\frac{3\lambda^2}{2}B(M^2)+O(\lambda^3).
$$
Substitution cancels both the cutoff and the scheme-dependent constant and leaves
$$
\Gamma_4(s,t,u)=\lambda-\frac{\lambda^2}{32\pi^2}
\int_0^1dx\log\left[
\frac{F(M^2)^3}{F(s)F(t)F(u)}
\right]+O(\lambda^3),
$$
as required.

Solved by gpt-5.6-sol high.

Solution

Paper 304 / 1 / b / ii / Solution

{title2=$\Gamma$}
{wiki=Effective_action}

The quantum effective action is the <Legendre transform> of the generating functional of connected correlation functions. Its functional derivatives are the <one-particle-irreducible Feynman diagram>[one-particle-irreducible vertex functions].

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