Laser Physics: Second Harmonic Generation

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
8 replies · 4K views
Niles
Messages
1,834
Reaction score
0
Hi

I was told that I is proportional to Iω2. It does not say so in my book. How is it one can easily see that it is the case?


Niles.
 
Physics news on Phys.org
Well, [tex]P_{2\omega}=\beta E_\omega^2[/tex] and [tex]I_{\2\omega}=P_{2\omega}^2/2 \epsilon_{2\omega} =\beta ^2 /\epsilon_{2\omega} E_\omega^4\propto I_\omega^2[/tex] as [tex]I_\omega=E^2_\omega/2 \epsilon_\omega[/tex].
 
Thanks, but how do we know that

[tex] I_{\2\omega}=P_{2\omega}^2/2 \epsilon_{2\omega}[/tex]

?
 
Hm, I have been maybe a bit too floppy. What I had in mind is [tex]I=\epsilon E^2/2=D^2/2 \epsilon[/tex]. So you have to express D in terms of P, i.e., [tex]P=(\epsilon -\epsilon_0)E=\chi E[/tex] to get [tex]I=\epsilon P^2/2 \chi^2[/tex].
 
The relation

[tex] P=\chi E [/tex]

only holds for linear materials. Isn't it wrong to use it to show what we are after?
 
Once you solve for the polarization at 2w (which is proportional to Iw), you then need to solve the driven wave equation with the nonlinear polarization as the source term. I can't tell you how it's done off the top of my head, but you can see it in Boyd's nonlinear optics for one(also Rick Trebino's FROG book, I think any book that goes into nonlinear optics will have this). I've never gone through it exactly, but I believe the slowly varying envelope approximation is made and other than that it's just some math which ends up showing that E(2w) is proportional to P(2w). And therefore I(2w) is proportional to I(w)^2.
 
Niles said:
The relation

[tex] P=\chi E [/tex]

only holds for linear materials. Isn't it wrong to use it to show what we are after?
Yes, but you can use this relation once the wave has left the zone where light of frequency omega is present.
 
johng23 said:
Once you solve for the polarization at 2w (which is proportional to Iw), you then need to solve the driven wave equation with the nonlinear polarization as the source term. I can't tell you how it's done off the top of my head, but you can see it in Boyd's nonlinear optics for one(also Rick Trebino's FROG book, I think any book that goes into nonlinear optics will have this). I've never gone through it exactly, but I believe the slowly varying envelope approximation is made and other than that it's just some math which ends up showing that E(2w) is proportional to P(2w). And therefore I(2w) is proportional to I(w)^2.

Thanks, I will have to find the Boyd book at my library. It looks good.


DrDu said:
Yes, but you can use this relation once the wave has left the zone where light of frequency omega is present.

I see. Is this equivalent of saying that the generated higher-harmonic wave behaves linearly in the material, because of low intensity, and also - possibly - because its frequency is too far off resonance?
 
You can expand P in a Taylor's series in terms of powers of E. Coefficients are linear and nonlinear susceptibility terms.

[tex]P=\chi_1 E+\chi_2 E^2+...[/tex]

The intensity of the second order term [tex]I_2=|P_2|^2[/tex] is proportional to [tex]I_1^2=|E|^4[/tex]