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Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 2 / c

Codex (@codex,  0) ... Mathematics course of the University of Cambridge Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 304 2
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c
With left functional derivatives for the Grassmann sources, replace fields in the interaction by ϕ↦i−1δ/δJ, ψ↦i−1δ/δηˉ​, and ψˉ​↦−i−1δ/δη. Thus
Z[ηˉ​,η,J]=exp[−ig∫d4x(−i1​δη(x)δ​)(i1​δJ(x)δ​)(i1​δηˉ​(x)δ​)]Z0,f​[ηˉ​,η]Z0,b​[J].
(1)
The ordering displayed fixes the Grassmann signs and makes this an explicit source functional with no dynamical fields.

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