Past exam of the laser science and quantum information processing course of the University of Oxford / 2019 / 1 / c

For a nonzero photon density, divide the laser rate equations to remove time:Integrating from the initial inversion and retaining a small seed density $n_i$ givesThe source's pulse approximation neglects $n_i$ compared with the macroscopic pulse density, givingA positive seed, or a spontaneous-emission source omitted from these equations, is nevertheless needed to start the pulse. Setting $n(0)=0$ exactly in the unforced equations gives the counterexample $n(t)=0$, $N^{\ast }(t)=N_i^{\ast }$ for all time, even above threshold. The displayed formula is therefore used in the negligible-seed limit, rather than as an assertion that a pulse starts from an identically zero field.

At the end of the pulse, the photon density returns to negligible size. Put $u=N_f^{\ast }/N_i^{\ast }$ and $r=N_i^{\ast }/N_{\mathrm{th}}^{\ast }$. The result of i givesUsing $x=\beta \eta =1-u$, the nonzero-pulse solution obeysFor $r>1$, select $0<x<1$, corresponding to depletion below threshold. The root $x=0$ is also an algebraic solution for every $r$ but represents no macroscopic pulse. A positive nonzero root exists only above threshold.

The physical branch can be plotted without a nonlinear solver by parameterizingIt begins at $(r,x)=(1,0)$ and increases monotonically to $x=1$ as $r\to \infty$. Expanding near threshold gives $r=1+x/2+O(x^2)$, so $x\simeq 2(r-1)$. For strong over-pumping the residual inversion fraction is exponentially small, $1-x\simeq e^{-r}$. Thus the curve saturates at the upper bound $\beta \eta =1$.

Extracting 95 percent of the maximum available inversion energy means $\eta =0.95/\beta$, or $x=\beta \eta =0.95$. Substitution in ii givesThe degeneracy factor cancels because the target is a fraction of the maximum energy $N_i^{\ast }{V}_g\hslash {\omega }_L/\beta$, rather than a fraction of $N_i^{\ast }{V}_g\hslash {\omega }_L$ without that factor.