Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 46 / 1 / a

Decompose the original scalar field as $\Phi =\varphi +\hat{\varphi }$, where $\tilde{\varphi }(p)$ has support in $|p|\le {\Lambda }^{\prime }$ and $\tilde{\hat{\varphi }}(p)$ has support in ${\Lambda }^{\prime }<|p|\le \Lambda$. This is a momentum-shell decomposition of a scalar field. The two collections of integration variables are disjoint. Define the lower-scale Wilsonian effective action by integrating out the second collection:A field-independent normalization may be retained as a vacuum term or absorbed into the measure. Integrating this identity over the low modes recovers the original Euclidean path integral, so it preserves all observables depending only on those modes.

Put $\Delta S=S_{\Lambda }^{\mathrm{eff}}[\varphi +\hat{\varphi }]-S_{\Lambda }^{\mathrm{eff}}[\varphi ]$. The quadratic cross terms integrate to zero: in Fourier transform variables each pairs a low momentum with its negative, which cannot be a shell momentum. Expanding the interaction therefore givesFactoring $e^{-S_{\Lambda }^{\mathrm{eff}}[\varphi ]}$ out of the shell integral and taking minus its logarithm yieldsThis definition is a Wilsonian effective action, rather than a Legendre transform generating only one-particle-irreducible Feynman diagrams.