Past exam of the quantum, atomic and molecular physics course of the University of Oxford / 2016 / 3 / d / Solution

Keep the given emission cross section in square centimetres so that the density remains per cubic centimetre. For the first transition, $r=5/3$ andThusThe separate populations are $N_2^{(0)}=1.667\times 10^{13}$ cm${}^{-3}$ and $N_1^{(0)}=1.00\times 10^{12}$ cm${}^{-3}$. Subtracting $5N_1/3$, rather than $N_1$ alone, reproduces the population inversion used above.

Here $rA{\tau }_1=0.10<1$, so upper-state pumping and sufficiently rapid lower-level removal produce positive small-signal optical gain. This coefficient is an intensity optical gain; a weak field amplitude would grow with half this exponent. A positive single-pass coefficient also does not by itself establish laser oscillation without a cavity-loss budget. No transition wavelength was supplied for this numerical example, so a numerical saturation intensity is not needed or invented to calculate ${\alpha }_0$.