Is the conversion efficiency of second harmonic generation the same for sum frequency generation? 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 $\omega_{1} \neq \omega_{2}$.
 A: 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.
