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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-301/3/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 3 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

In phi cubed theory, each vertex has three legs. A connected four-point tree-level Feynman diagram therefore needs two cubic vertices and one internal propagator. There are exactly three ways to partition the labeled external legs between these vertices:

       s channel               t channel              u channel

 p \          / p'         p ---o--- p'            p ---o--- q'
    o--------o                  |                       |
 q /          \ q'         q ---o--- q'            q ---o--- p'

 internal: p+q             internal: p-p'          internal: p-q'

Solid lines denote the real scalar field propagator. The labels specify which external pair meets at each vertex, so the last two graphs are different even though their unlabeled shapes agree. This is the four-point tree amplitude in phi cubed theory:

three exchange diagrams, in the s, t, u channels; each is of order λ^{2} .

(1)

There is no four-point contact vertex in the given cubic interaction.

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