Past exam of the laser science and quantum information processing course of the University of Oxford / 2022 / 1 / c / Solution

Put $b=\Delta {\omega }_b>0$ and $s=t-t_m$. At the peak time, $e^{-ip\Delta \omega t_m}=1$, so the exact comb sum becomesFor many occupied normal modes, $b\gg \Delta \omega$, replace the sum by an integral with density $1/\Delta \omega$. Setting $x=\omega -{\omega }_0$ and evaluating the Fourier transform of a Gaussian givesThus the isolated-optical pulse approximation hasThe optical pulse has a Gaussian function envelope and no chirp. Its carrier angular frequency is ${\omega }_0$, and its light intensity full width at half maximum is $2\sqrt{2\ln 2}/b$. Successive envelope maxima have different carrier-envelope phases, consistent with b.

Approximation in the printed assertion. A discrete optical frequency comb gives a train, not one exact isolated Gaussian pulse. Applying the Poisson summation formula to the ideal sum over all integers instead givesNear $t_m$, the $j=m$ term dominates when $bT\gg 1$. The other terms are exponentially small but not identically zero. A single Gaussian function used globally would decay between $0$ and $T$, whereas the exact light intensity has equal maxima there; this supplies a concrete reason the isolated form cannot be an exact global identity. The extension to negative frequencies is the usual narrow-positive-band analytic-field approximation when ${\omega }_0\gg b$.