TransWikia.com

Intrinsic permutation in nonlinear susceptibilities and pulses

Physics Asked by Andrew Spott on March 7, 2021

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) = sumlimits_{omega = -infty}^{infty} tilde{E}(omega_n) e^{iomega_n t} $

Implicitly, this makes an assumption:

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

Where $omega_n = 2pi n/T$. 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) + sumlimits_{omega_i,omega_jneqomega}left(6chi^{(3)}(omega; omega, -omega_i, omega_i) |E_i(omega_i)|^2E(omega) + 6chi^{(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 $Trightarrowinfty$.

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 $chi^{(3)}(omega, -omega_i, omega_i)$.

One Answer

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 $omega_2rightarrowomega_1$ 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 $domega$ 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 $frac{domega}{2}$, 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.

Answered by hsinghal on March 7, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP