 Solution

The delta function in the specified convolution kernel collapses the ${\mathbf{k}}^{\prime }$ integral to $-i\gamma G$. Set $\lambda =e\gamma$ and ${\omega }_0^2=k^2+m^2$. ThenThere is a change of assumptions here: the specified kernel is the transform of $V=-i\gamma$, which is not real. These later parts define a formal analytic continuation to an effective damped equation, $[{\partial }_{\tau }^2+2\lambda {\partial }_{\tau }+{\omega }_0^2]G=\delta (\tau )$; they cannot simultaneously describe the earlier real external potential. The following pole analysis uses that continued operator.

Completing the square gives $D(\omega )=-(\omega +i\lambda {)}^2+({\omega }_0^2-{\lambda }^2)$, soFor $e>0$, hence $\lambda >0$, the underdamped regime has two poles at $\pm \Omega -i\lambda$, with $\Omega =\sqrt{k^2+m^2-{\lambda }^2}>0$. Both lie below the real axis. In the overdamped regime they lie on the imaginary axis at $-i(\lambda \mp \kappa )$, where $\kappa =\sqrt{{\lambda }^2-k^2-m^2}<\lambda$ for $k^2+m^2>0$; both are again below the real axis. At equality they coalesce into a double pole at $-i\lambda$. If $m<|\lambda |$, the crossover occurs at $k_c=\sqrt{{\lambda }^2-m^2}$; increasing $k$ moves the two imaginary poles together and then apart horizontally. If $m>|\lambda |$, every $k$ is underdamped; if equal, only $k=0$ is critical.

The sign of $e$ is not specified: for $e<0$ all nonzero-frequency poles are instead in the upper half-plane, describing antidamping; for $e=0$ they lie on the real axis and a causal prescription is required. When $m=k=0$, one pole is at zero even for nonzero $\lambda$ and a real-axis prescription is again needed. The figure uses illustrative $\lambda =1$, not supplied numerical data.