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

Let $\tau =t-t^{\prime }$ and $\Omega =\sqrt{K^2-b^2}>0$. Because $|e^{-i(u+iv)\tau }|=e^{v\tau }$, close below for $\tau >0$ and above for $\tau <0$. The lower closure is clockwise. The two residues sum to $i{e}^{b\tau }\sin (\Omega \tau )/\Omega$. Hence, for nonzero $b$,In particular, the printed sign with $e\gamma >0$ gives zero for $t>t^{\prime }$ and an advanced Green function for $t<t^{\prime }$. Reversing that sign puts the poles below and gives a retarded Green function. Both solutions have the unit derivative jump required by $({\partial }_{\tau }^2-2b{\partial }_{\tau }+K^2)G=\delta (\tau )$. If $b=0$, the bare integral has real-axis poles and requires an additional contour prescription; a unique retarded or advanced answer cannot be inferred without one.