Deriving the ∫dω1 integral in 2nd-order polarization

In summary, the Fourier transform is defined as and is related to the temporal version by which is also known as the Chirality Modulus.
  • #1
Ngineer
64
1
Homework Statement
Starting with the second order polarization in the time domain:
upload_2018-12-6_5-13-6.png
(1)

I am trying derive the frequency domain form:
upload_2018-12-6_5-5-58.png
(2)

Multiple sources give essentially the same formula with the same integral, I have obtained the particular ones in here from those lecture notes.

My issue is finding the origin of the ∫dω1 integral. After a day of attempts I still can't figure out how it comes into play.

The attempt at a solution
Attempt 1

I started by identifying the time-domain formula (equation 1) as a double convolution:
upload_2018-12-6_5-21-43.png


Which would map nicely to
upload_2018-12-6_5-27-26.png

But I did not get far as to deriving equation 2 from here.

Attempt 2
Another approach I have attempted is that recognizing the the desired frequency-domain form (equation 2) is very close to convolution with respect to ω=ω1+ω2:
upload_2018-12-6_5-52-57.png


However,
- Why would a convolution in the time domain map to a convolution in the frequency domain?
- The original formula (equation 2) does not have an integral with respect to ∫dω2.

Any help is incredibly appreciated.
Thank you!
 

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  • #2
How is your Fourier transform defined?

Assuming it is
##E\left(t\right)=\frac{1}{2\pi}\int d\omega \exp\left(-i\omega t\right) \tilde{E}\left(\omega\right)## Eq.(1)
##\tilde{E}\left(\omega\right)=\int dt \exp\left(i\omega t\right) E\left(t\right)##

Where ##\tilde{\dots}## denotes the frequency domain functions.

How about substituting Eq. (1) into your expressions, and then carrying out the integrals with respect to ##t'## and ##t''##? In the process of doing this you will need to explain what does ##\chi^{2}\left(\omega, \omega_1, \omega_2\right)## mean. How does it relate to temporal version?
 
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  • #3
Thank you for your response, the definition of the Fourier transform is

upload_2018-12-16_18-14-35.png

The derivation checks out if it is assumed to be separable with respect to t' and t''
 

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  • #4
Ngineer said:
The derivation checks out if it is assumed to be separable with respect to t' and t''

So does it settle your question then? If there are still problems can you show how far you got before getting stuck?
 
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  • #5
Yes. I have actually corresponded with the professor that posted the online lectures I linked in the original post, and he kindly answered in detail, so I marked this post as solved.

Thanks for your kind help.
 

Related to Deriving the ∫dω1 integral in 2nd-order polarization

1. What is the purpose of deriving the ∫dω1 integral in 2nd-order polarization?

The ∫dω1 integral in 2nd-order polarization is used to calculate the second-order nonlinear response of a material to an applied electric field. This is important in understanding the optical properties of materials and can be applied in various fields such as photonics, spectroscopy, and telecommunications.

2. How is the ∫dω1 integral derived in 2nd-order polarization?

The ∫dω1 integral is derived by applying the perturbation theory to the microscopic equations of motion for a system of interacting particles. This involves solving the time-dependent Schrödinger equation and using the perturbation expansion to obtain the second-order polarization term.

3. What is the significance of the ∫dω1 integral in 2nd-order polarization in quantum mechanics?

The ∫dω1 integral in 2nd-order polarization is a fundamental concept in quantum mechanics that describes the response of a system to an external electric field. It is used to calculate the transition probability between two states and is crucial in understanding the behavior of particles in an electric field.

4. Are there any assumptions made when deriving the ∫dω1 integral in 2nd-order polarization?

Yes, there are several assumptions made when deriving the ∫dω1 integral in 2nd-order polarization. These include the dipole approximation, which assumes that the electric field is approximately constant over the size of the particle, and the weak-field limit, which assumes that the electric field is not strong enough to cause significant changes in the system.

5. How is the ∫dω1 integral related to other integrals in 2nd-order polarization?

The ∫dω1 integral is just one term in the overall expression for the second-order polarization. It is related to other integrals, such as the ∫dω2 and ∫dω3 integrals, which correspond to different physical processes. Together, these integrals make up the full second-order polarization term, which gives a comprehensive understanding of the nonlinear response of a material.

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