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

Let $\tau =t-t^{\prime }$ and $\Omega =\sqrt{k^2+m^2-{\lambda }^2}>0$. First take $\lambda =e\gamma >0$. Since $|e^{-i\omega \tau }|=e^{\tau \mathrm{Im}\omega }$, close the contour below for $\tau >0$, and above for $\tau <0$. On the chosen semicircle the exponential is bounded and the denominator grows as $R^2$, so the arc integral tends to zero. The upper contour encloses no pole, giving $G=0$ for $t<t^{\prime }$. The lower contour is clockwise, hence the real-axis integral divided by $2\pi$ is $-i$ times the sum of residues. Since $D^{\prime }({\omega }_{+})=-2\Omega$ and $D^{\prime }({\omega }_{-})=2\Omega$, that sum isThusThis is the causal Green function of a damped oscillator. Its value is continuous through $\tau =0$, and its derivative jumps from zero to one; integrating the differential equation through zero verifies the delta-source normalization and the residue sign. The limits are $H(\tau )\tau e^{-\lambda \tau }$ at critical damping and $H(\tau )e^{-\lambda \tau }\sinh (\kappa \tau )/\kappa$ in the overdamped regime.

For completeness, if the unspecified charge has $e<0$, the same real-axis integral instead encloses upper poles for $\tau <0$ and gives $-H(-\tau )e^{-\lambda \tau }\sin (\Omega \tau )/\Omega$, an advanced Green function. A causal growing response would require an inverse-Laplace contour above those poles, not the stated real-axis Fourier integral. At $e=0$ the real-axis integral needs a pole prescription; shifting both poles below it gives the retarded zero-damping limit. These distinctions are necessary because $\gamma >0$ alone does not fix the sign of damping.