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

Time independence of $V$ makes the Green function depend on $t,t^{\prime }$ through $\tau =t-t^{\prime }$. A spatially varying potential does not imply spatial translation invariance. Fix ${\mathbf{r}}^{\prime }$ and set $\mathbf{x}=\mathbf{r}-{\mathbf{r}}^{\prime }$; write the Fourier coefficient more fully as $G_{{\mathbf{r}}^{\prime }}(\mathbf{k};\omega )$. The source position is an implicit parameter in the printed notation. DefineThe ordinary derivative terms transform into $(-{\omega }^2+k^2+m^2)G_{{\mathbf{r}}^{\prime }}$. The product $V{\partial }_{t}G$ becomes a convolution, and $2ie(-i\omega )=2e\omega$. More explicitly, insert both inverse spatial Fourier transforms into $VG$: the integral over $\mathbf{x}$ enforces $\mathbf{q}+{\mathbf{k}}^{\prime }=\mathbf{k}$. The transformed delta source is one, leavingThis is a source-position-dependent Fourier Green function. Dropping the explicit ${\mathbf{r}}^{\prime }$ subscripts recovers the requested notation with exactly the potential transform defined in the source. No assumption of a homogeneous potential is needed. The factor $(2\pi {)}^{-3}$ and the source-position phase are necessary for the stated Fourier convention.