How can one derive the conversion efficiency for sum frequency generation. In most standard textbooks, only SHG is used as an example.

$$\eta =\frac{P_\text{SHG}}{P_\text{pump}} \, .$$ In the case, where $\omega_{1}$ not equal to $\omega_{2}$. I imagine $P_\text{SHG}$ would split into two contributions for the two different components. $\omega_{1}=\omega_{2}$, $\omega_{1}+\omega_{2}=\omega_{3}$, where $\omega_{1} \neq \omega_{2}$.


1 Answer 1


Generally speaking, and making the naive assumption that phase-matching has a minimal effect on the produced intensity, the power emitted at a frequency $\omega_3=\omega_1+\omega_2$ in sum-frequency generation using pumps at frequencies $\omega_1$ and $\omega_2$ is given by $$ P_{\omega_3} = \eta(\omega_1,\omega_2)P_{\omega_1}P_{\omega_2}, $$ i.e. it is proportional to the product of the intensity of the two pumps; within that formalism, second-harmonic generation can be seen as the degenerate process with $\omega_1=\omega_2=\omega$, so that the produced intensity $$ P_{2\omega} = \eta(\omega,\omega)P_{\omega}^2 $$ is quadratic in the pump's intensity.

Note, however, that the efficiency is generally a function of the pump frequencies, and the equality $$ \eta(\omega,\omega)\stackrel{?}=\eta(\omega+\Delta,\omega-\Delta) $$ is never guaranteed. This might occasionally hold if you're very far away from any resonances, but typically the two efficiencies (essentially stand-ins for the nonlinear susceptibility) are never required to be equal.

And, of course, the assumption that phase-matching can be neglected is obviously wrong for any real-world case; once you include that, then the efficiency will be whatever the phase-matching dictates, which will come from the solution of a complex problem with no hard-and-fast rules for any aspect of the result.


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