Past exam of the theoretical physics I course of the University of Cambridge / 2014 / Paper 1 / 5 / c / Solution

The delta function in the artificial constant potential collapses the convolution. Put $b=e\gamma$ and $K^2=k^2+m^2$. ThenInverse Fourier transform gives the stated real-frequency integral with exponential $e^{-i\omega (t-t^{\prime })}$. For $K^2>b^2$, the poles have opposite real parts and common imaginary part $b$. For $0<K^2<b^2$, both are imaginary, at $i[b\pm \sqrt{b^2-K^2}]$. They remain in the upper half-plane if $b>0$, or the lower half-plane if $b<0$. At equality they merge into a second-order pole at $ib$. For $K^2=0$ one pole reaches the real axis, requiring a prescription.