 # If we have two laser beams with central wavelengths lambda_1 and lambda_2 (one with negligible spectral width), interacting in a non-linear crystal, how would we calculate the spectral width of the difference-frequency-generated (DFG) output? I know it's a Lorentzian given by Delta lambda=(4 lambda_1^2 Delta lambda_2)/(4(lambda_2-lambda_1)^2-Delta lambda_2^2). But I don't know how to prove this. Kade Rosales 2022-08-06 Answered
If we have two laser beams with central wavelengths ${\lambda }_{1}$ and ${\lambda }_{2}$, (one with a spectral width of $\mathrm{\Delta }{\lambda }_{2}$ and one with negligible spectral width), interacting in a non-linear crystal, how would we calculate the spectral width of the difference-frequency-generated (DFG) output? I know it's a Lorentzian given by $\mathrm{\Delta }\lambda =\frac{4{\lambda }_{1}^{2}\mathrm{\Delta }{\lambda }_{2}}{4\left({\lambda }_{2}-{\lambda }_{1}{\right)}^{2}-\mathrm{\Delta }{\lambda }_{2}^{2}}$. But I don't know how to prove this.
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Difference frequency generation (DFG) is a stimulated parametric down-conversion process, which implies conservation of energy. It manifests in the relationship among the frequencies of the three beams (the pump, the seed and the idler or DFG beam):
${\omega }_{p}-{\omega }_{s}={\omega }_{i}.$
In terms of wavelengths it becomes
${\lambda }_{i}=\frac{{\lambda }_{2}{\lambda }_{1}}{{\lambda }_{2}-{\lambda }_{1}},$
where ${\lambda }_{1}$ is the pump wavelength and ${\lambda }_{2}$ is the seed wavelength. Assuming the pump has a negligible bandwidth, the bandwidth of the idler is determined by the bandwidth of the seed. To determine the idler bandwidth, we use
$\mathrm{\Delta }{\lambda }_{i}=|\frac{\mathrm{\partial }{\lambda }_{i}\left({\lambda }_{2}\right)}{\mathrm{\partial }{\lambda }_{2}}|\mathrm{\Delta }{\lambda }_{2}=|\frac{{\lambda }_{1}}{{\lambda }_{2}-{\lambda }_{1}}-\frac{{\lambda }_{2}{\lambda }_{1}}{\left({\lambda }_{2}-{\lambda }_{1}{\right)}^{2}}|\mathrm{\Delta }{\lambda }_{2}=\frac{{\lambda }_{1}^{2}\mathrm{\Delta }{\lambda }_{2}}{\left({\lambda }_{2}-{\lambda }_{1}{\right)}^{2}}.$
The resulting expression follows from that provided by the OP by setting $\mathrm{\Delta }{\lambda }_{2}^{2}=0$ since it would be small compared to the other term in the denominator. If this is not the case, the bandwidth would also depend on the shape of the spectrum.