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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2016/paper-1/3/b/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2016 / Paper 1 / 3 / b / Solution by

Codex  0 2026-10-20

Time independence of

V

makes the Green function depend on

t, t^{'}

through

τ = t - t^{'}

. A spatially varying potential does not imply spatial translation invariance. Fix

r^{'}

and set

x = r - r^{'}

; write the Fourier coefficient more fully as

G_{r^{'}} (k; ω)

. The source position is an implicit parameter in the printed notation. Define

V_{r^{'}} (q) = \int d^{3} x V (r^{'} + x) e^{- i q \cdot x} = e^{i q \cdot r^{'}} V (q) .

The ordinary derivative terms transform into

(- ω^{2} + k^{2} + m^{2}) G_{r^{'}}

. The product

V \partial_{t} G

becomes a convolution, and

2 i e (- iω) = 2 e ω

. More explicitly, insert both inverse spatial Fourier transforms into

V G

: the integral over

x

enforces

q + k^{'} = k

. The transformed delta source is one, leaving

(- ω^{2} + k^{2} + m^{2}) G_{r^{'}} (k; ω) + 2 e ω \int \frac{d ^{3} k ^{'}}{( 2 π ) ^{3}} V_{r^{'}} (k - k^{'}) G_{r^{'}} (k^{'}; ω) = 1.

This is a source-position-dependent Fourier Green function. Dropping the explicit

r^{'}

subscripts recovers the requested notation with exactly the potential transform defined in the source. No assumption of a homogeneous potential is needed. The factor

(2 π)^{- 3}

and the source-position phase are necessary for the stated Fourier convention.

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