 Maser microwave energy budget

Multiplying the upper-level and photon equations by $h{\nu }_0$ makes the cancellation of internal transfer visible:Incoming excitation becomes unused excitation carried away, internal dissipation, useful output, or a change in stored energy. Translational kinetic energy, the source's other internal states and preparation losses lie outside this two-level budget. The output cannot exceed the incoming excitation power $h{\nu }_{0}F$.

Choose $G=10^{-4}{\mathrm{s}}^{-1}$ per unit inversion, $\gamma =10^4{\mathrm{s}}^{-1}$, ${\kappa }_{\mathrm{int}}={\kappa }_{\mathrm{out}}=10^6{\mathrm{s}}^{-1}$, and $F=4\times 10^{14}{\mathrm{s}}^{-1}$. These are stipulated teaching parameters. Then $N_0=4\times 10^{10}$, $F_{\mathrm{th}}=2\times 10^{14}{\mathrm{s}}^{-1}$, $N_{\mathrm{ss}}=2\times 10^{10}$ and $n_{\mathrm{ss}}=5\times 10^7$. Using the rounded ammonia inversion transition frequency, the loaded quality factor is approximately $7.50\times 10^4$. The incoming excitation power is $6.327\mathrm{nW}$: $4.745\mathrm{nW}$ leaves unused, $0.791\mathrm{nW}$ becomes internal dissipation and $0.791\mathrm{nW}$ is useful output, with rounding in these displayed values.

There is an instructive extraction optimum within this model. With the same $F$, $G$ and $\gamma$, put $\rho ={\kappa }_{\mathrm{out}}/{\kappa }_{\mathrm{int}}$ and $p=FG/(\gamma {\kappa }_{\mathrm{int}})=4$. For $0\le \rho <p-1$,Differentiation gives ${\rho }_{\mathrm{opt}}=\sqrt{p}-1=1$. At zero coupling no power is extracted; at $\rho =3$ the device reaches onset and the nonzero branch disappears. The optimum balances extraction against the inversion required to compensate that extraction. Equal residence rates and loss of molecular phase memory are simplifying assumptions; the calculated efficiency is not a universal bound for beam masers.