Past exam of the theoretical physics II course of the University of Cambridge / 2016 / Paper 1 / 4 / c / Solution

For this mode, abbreviateThe electric pulse is $\mathbf{E}(t)=-{\mathbf{A}}_0\delta (t)$. In temporal gauge the mode equation has a finite step in its frequency squared, with no delta multiplying the mode. Thus both ${\psi }_{\mathbf{k}}$ and its first time derivative are continuous at zero.

Writing the positive-frequency incident amplitude as $a_{\mathbf{k}}^{\mathrm{in}}$, the matched solution isIts initial value is $a_{\mathbf{k}}^{\mathrm{in}}/\sqrt{2\omega }$ and its initial derivative is $-i\omega a_{\mathbf{k}}^{\mathrm{in}}/\sqrt{2\omega }$, as required. In the normalized outgoing-frequency basis it isThe negative-frequency contribution appears even though only positive frequency was incident. These are temporal-mode matching coefficients, so their conserved difference is hyperbolic rather than the sum-one rule of ordinary spatial reflection and transmission probabilities.