Nonlinear Optics: Understanding NLO Chromophores and Polarization

[DanSandberg]

Hi All - I am trying to immerse myself in NLO and purchased Robert W. Boyd's Third Edition on Nonlinear Optics. I'm already struggling just 3 pages into the book.

We are looking at the polarization of a material in a NLO chromophore, so:

P(t)=$$\epsilon$$ [X(1)E(t)+X(2)E²(t)+X(3)E³(t)...]

where P is the polarization at time t, X is the NLO susceptibility for the corresponding ordered response, and E(t) is the strength of the applied electric field.

The text goes on to say "One might expect that the lowest-order correction term X(2)E²(t) to be comparable to the linear response, X(1)E(t), when the amplitude of the applied field, E(t), is equal to the characteristic atomic electric field E(atomic)=e/4$$\pi$$$$\epsilon$$₀a₀²"

Then they derive a whole bunch of stuff, which I follow, but my question is why do we expect the second-order polarization to equal the first-order polarization when the applied field equals E(atomic)?

Why is that

if E(applied)=E(atomic) then
X(1)E(t)=X(2)E²(t)

 

[Dr Lots-o'watts]

I think it's simply because in general, NLO effects happen when the E field is relatively large. But large compared to what? Large compared to the E field of the atom. And "naturally", we might expect the first NLO effects to manifest themselves are those related to the next order parameter X(2). (and as the field gets stronger, the next parameters come into play accordingly in order X(3), X(4), etc.) The text may go on to say that this is not necessarily the case.

But I don't have the book with me, so I'm not sure about the actual context of the phrase, and could be totally wrong.

 

[DanSandberg]

Thank you Dr. Lots-O'Watts. The book didn't really mention that this assumption was weak but as long as I'm not missing something fundamental I can accept it and move on. Thanks again. -Dan