Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 301 / 4

Substitution of $\Phi =\phi +\lambda {\phi }^2$ givesand

With all momenta incoming, the Feynman rules areandEach vertex includes its momentum-conserving delta function. The combinatorial factors follow by assigning the identical fields to the differentiated and undifferentiated slots in every possible way.

At order ${\lambda }^2$, four connected tree-level Feynman diagrams contribute: one four-point contact diagram and three diagrams with two cubic vertices joined by a scalar propagator in the $s$, $t$, and $u$ channels.

On shell, the contact vertex is $4i{\lambda }^2{m}^2$. At a cubic vertex with two external on-shell legs and internal momentum $k$, the rule reduces to $2i\lambda (k^2-m^2)$. The propagator cancels one such factor, so the three exchange diagrams sum toFor equal-mass two-to-two scattering, $s+t+u=4m^2$, and the exchange sum is $-4i{\lambda }^2{m}^2$. It cancels the contact diagram exactly. Thus the connected on-shell amplitude vanishes, as required by the equivalence theorem for field redefinitions: an invertible local field redefinition cannot turn a free theory into a physically interacting one.