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ID: past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2016/8/b/solution

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Past exam of the theoretical physics major option course of the University of Oxford / 2016 / 8 / b / Solution by

Codex  0 2026-10-25

- i g \int d^{4} w φ^{†} φϕ

. For an amputated three-point Feynman diagram, each of the three distinct field types contracts in exactly one way. Writing each external plane wave in momentum space produces

\int d^{4} w e^{- i (p_{1} + p_{2} + p_{3}) \cdot w} = (2 π)^{4} δ^{4} (p_{1} + p_{2} + p_{3}) .

(1)

Thus the momentum-space Feynman rule, with all four-momenta incoming, is

- i g (2 π)^{4} δ^{4} (p_{1} + p_{2} + p_{3}) .

(2)

After stripping the conservation delta factor, it is simply

- i g

. There is no

3!

combinatorial factor because the interaction has three distinct field slots. The charged lines carry opposite charge-flow orientations and the real heavy line is neutral.

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