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The nonlinear susceptibilities have an intrinsic permutation symmetry. This symmetry treats two frequency components that are equal differently than it treats two frequency components that are slightly different.

Typically, the susceptibility is found by assuming the electric field is a Fourier sum of plane waves:

$ E(t) = \sum\limits_{\omega = -\infty}^{\infty} \tilde{E}(\omega_n) e^{i\omega_n t} $

Implicitly, this makes an assumption:

$ E(\omega_n) = \frac{1}{T} \int\limits_{-T/2}^{T/2}dt E(t) e^{-i \omega_n t}$

Where $\omega_n = 2\pi n/T$. This converges to the Fourier integral in the limit that $T \rightarrow \infty$.

Unfortunately, the third order polarization is written as (from Boyd, pg. 13):

$ P^{(3)}(\omega) = 3 \chi^{(3)}(\omega; \omega, -\omega, \omega) |E(\omega)|^2E(\omega) + \sum\limits_{\omega_i,\omega_j\neq\omega}\left(6\chi^{(3)}(\omega; \omega, -\omega_i, \omega_i) |E_i(\omega_i)|^2E(\omega) + 6\chi^{(3)}(\omega; \omega, -\omega_j, \omega_j) |E_j(\omega_j)|^2E(\omega) \right)$

My issue lies with the "$6$" and "$3$" here, and how they interact with the sum, as $T\rightarrow\infty$.

If I have a pulse, my $E$ field is some distribution over frequencies, and I would find the total third order polarization by summing over all the frequencies present in my pulse. We assume the pulse is Gaussian for simplicity (though any pulse that is sufficiently smooth should do). We also note that the susceptibilities $\chi^{(3)}(\omega_i, \omega_j, \omega_k)$ are continuous in the $\omega$s (so to first order, the susceptibility doesn't change much when the frequencies change very little).

For small $T$, the first term ($\chi^{(3)}(\omega,\omega,-\omega)$) will dominate. The spacing between $\omega_n$ and $\omega_{n+1}$ is large enough that the electric field component $E_j(\omega_j)$ for $\omega_j \neq \omega$ is small enough to make the second two terms small.

However, as $T$ gets large, the frequency difference between $\omega_j$ and $\omega$ will get smaller, and for a sufficiently short pulse (compared to $T$), $E_j(\omega_j)$ and $E(\omega)$ will be roughly the same magnitude. In this case, the first term will be 1/2 as large as the second two terms.

The problem is that the polarization then becomes dependent on $T$, getting LARGER at $T$ gets larger, due to the fact that, in the infinite $T$ limit, the "3" has no effect on the sum/integral, but the "$6$" terms do, while in the small $T$ limit, the "3" term dominates.

How do I address this discrepancy? Both Boyd and Butcher and Cotter describe this intrinsic permutation symmetry, and assign different "permutation numbers" to $\chi^{(3)}(\omega, - \omega, \omega)$ and $\chi^{(3)}(\omega, -\omega_i, \omega_i)$.

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  • $\begingroup$ You assume your pulse is monochromatic, and then you switch to a gaussian one without changing much. Shouldn't you go back to the start and carry that continuum through? $\endgroup$ – Emilio Pisanty Jun 30 '16 at 22:08
  • $\begingroup$ I don't make any assumption that my pulse is monocromatic. The electric field vector is a Fourier sum of plane waves, not a single plane wave. $\endgroup$ – Andrew Spott Jun 30 '16 at 22:25
  • $\begingroup$ Yeah, but it's still a discrete spectrum, which can never be achieved with a finite pulse. If you want to do a gaussian pulse, I would take as spectrum a superposition of gaussians centered at the $\omega_n$ - but maybe that's just me. For sure, though, if you want a pulse so short that its bandwidth is bigger than the frequency resolution you want (i.e. $1/T$ determining the smallest relevant splitting between the $\omega_n$) then you should be putting in an explicit pulse with a corresponding continuous spectrum. $\endgroup$ – Emilio Pisanty Jun 30 '16 at 23:56
  • $\begingroup$ To be fair, I am not the one that is making this assumption, I'm just formalizing it (the first equation is frequently in the literature, but the second is, I assume, implied). It is in both Butcher and Cotter and Boyd nonlinear optics textbooks. Second, as I extend $T$, the Fourier series converges on a single pulse, getting there when I get to $T\rightarrow\infty$. At some point along that journey of extending $T$, I run into the problem I described. $\endgroup$ – Andrew Spott Jul 1 '16 at 0:16
  • $\begingroup$ Fair enough, we can talk about a pulse train, but I don't think this gets rid of the problem. The polarization at some frequency for a pulse train will depend more on the cross frequencies as the pulses get further apart (essentially $T$ gets longer), so an approximation with just the central frequency (the first term) should become LESS accurate as pulses get further apart (tending towards being a factor of two off), which doesn't make any sense. $\endgroup$ – Andrew Spott Jul 1 '16 at 1:14
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The confusion is why there the term $E(\omega_2)E^∗(\omega_2)E(\omega_1)$ has factor of $6$ and $E(\omega_1)E^∗(\omega_1)E(\omega_1)$ has factor of $3$

I understand this confusion as follows.

In first term there are two fields and in second term there is only one field. Hence if $E(\omega_1)=E(\omega_2)$ you are actually pumping double power in first case then in second (actually nonlinear processes goes with field and not with power hence talking about power is not so appropriate) and that is why you see this discrepancy.

As explained by the author, the factor 3 or 6 in the expression of $\chi_{123}$ comes from definition. They can equally well be 1 or 2 if you change the definition. He also explained that even if you move $\omega_2\rightarrow\omega_1$ then also the factor 6 do not change to 3 but at $\omega_2=\omega_1$ the factor of 6 changes to 3.

EDIT:

I will again try to explain from some physical arguments.

Usually the nonlinear effects come in picture when pulsed lasers interact with the atoms/matter.

Let us say there is a pulse with bandwidth $d\omega$ and Intensity $I$. The corresponding field is $E(\omega)$.

Let us think of a nonlinear crystal through which this pulse is passing and generating third harmonic. If you think this pulse as single entity then the factor in front of $\chi^3$ is 3.

Now if you divide this pulse in two parts one is centered at $\omega_1$ and other is at $\omega_2$ and bandwidth $\frac{d\omega}{2}$, then you think of the interaction between these two the factor in front of $\chi^3$ will be 6 but now the energy content in each pulse is halved and final result will be same.

I hope this will clarify your confusion.

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  • $\begingroup$ I understand where the factor comes from, but the problem arises when we consider $\chi^{(3)}(\omega; \omega, -\omega - \delta, \omega + \delta)$. This term is indistinguishable from the degenerate term, in the values of $\chi$ or the electric field components, but has a larger factor in front. Where is the problem with this? $\endgroup$ – Andrew Spott Jul 3 '16 at 16:36
  • $\begingroup$ Here is another way of thinking about it: what defines a second efield? Does a second frequency component of the same efield count? If it doesn't, why not? $\endgroup$ – Andrew Spott Jul 3 '16 at 16:53
  • $\begingroup$ Yes second frequency component of the same field counts for second field because the energy content in the second frequency component is different than the same frequency. Now question is how fine one can go. In my opinion there is an inherent spectral width in the nonlinear crystal comes from phase matching that defines the bandwidth of single frequency. This is my guess I am not sure about bandwidth explanation. $\endgroup$ – hsinghal Jul 4 '16 at 1:14
  • $\begingroup$ I'm thinking about the regime (single atom) where phase matching doesn't matter. I'm not sure what the limiting factor of resolution is either. $\endgroup$ – Andrew Spott Jul 4 '16 at 1:40

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