Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 3

Complete the square in the heavy variable:The shifted Gaussian integral is independent of $\varphi$. For $M>0$, choosing $N=M/\sqrt{2\pi }$ removes it and gives

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The induced four-$\varphi$ interaction is the tree-level diagram with two $\lambda {\varphi }^2\chi$ vertices joined by one internal $\chi$ propagator. The zero-dimensional propagator is $M^{-2}$, and the second-order expansion of the exponential supplies the factor $1/2$. The resulting effective-action coefficient is therefore $-{\lambda }^2/(2M^2)$, exactly the value of $\tilde{\lambda }$ found by the Gaussian integral.

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Let $K_M=-{\partial }^2+M^2$ and let ${\Delta }_M=K_M^{-1}$ be its Euclidean quantum field theory propagator. With the source-sign convention suited to the expression in the question, definewhere $N_0$ makes $Z_0[0]=1$. Since inserting $\chi (x)$ is equivalent to acting with $-\delta /\delta J(x)$, the integral over $\chi$ gives

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The interaction is linear in the Gaussian field $\chi$, so completing the functional square makes its order-${\lambda }^2$ contribution exact:In momentum space, ${\Delta }_M(p)=(p^2+M^2{)}^{-1}$. If every external momentum satisfies $p^2\ll M^2$, the derivative expansionstarts with the local effective interactionwhich is the field-theory version of the zero-dimensional result.

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