# 4 nonlinear optics problems: susceptibility, polarization

• snickersnee
In summary, the conversation covers various problems related to perturbation theory, nonlinear susceptibility, and crystal symmetry. The first problem involves using perturbation theory to derive the 3rd order nonlinear susceptibility. The second problem discusses the calculation of nonlinear and linear polarization in a nonlinear crystal with only one nonzero component for the second order nonlinear optical susceptibility tensor. The third problem asks for the calculation of the elements of a d_il matrix for a specific crystal with 3m symmetry. The fourth problem involves calculating the effective nonlinear coefficient for second harmonic generation in a nonlinear crystal with only one nonlinear coefficient. The conversation also touches on relevant equations and attempts at solutions for each problem.
snickersnee
[Wasn't sure if each problem needed a separate post. Please feel free to edit if needed.]
Also \~ and \^ are tilde and hat respectively.

1a. Homework Statement

Use perturbation theory to derive the 3rd order nonlinear susceptibility$\chi^{(3)}(3w;w,w,w)$ (problem gives potential energy, etc. but I already know what I have to do, I just need help calculating it)1b. Relevant equations

The equation I need to solve is this: $\ddot{\tilde{x}}^{(3)}+2\gamma \dot{\tilde{x}}^{(3)} + w_0^2 \tilde{x}^{(3)} +2a\~x^{(1)}\~x^{(2)}=0$

I need to solve it for $\~x^{(3)}$.

The first-order and second-order solutions were given in lecture. I plugged them into the equation, in the 4th term on the left hand side, getting some horrendous expression.

We are looking for solutions of the form (*) $\~x^{(3)}(t)=x^{(3)}(3w)e^{-3iwt}$ (is that the right form of the solution?) Since the differential equation has derivatives with respect to time, I guess I need to differentiate (*) twice, but is it only the exponential that depends on time? What about the x, and omega? Are those constants?

Also, E times its complex conjugate is |E|^2, right?

1c. The attempt at a solution

(explained in section 1b)
--------------------------------------2a. Homework Statement

Nonlinear crystal has an EM wave propagating in x direction, linear polarization along $1/\sqrt{2}(\^y+\^z)$ direction, frequency w, intensity 1MW/cm2. The second order non-linear optical susceptibility tensor for second harmonic generation has only one nonzero component, $\chi^{(2)}_{zzz}(2w,w,w)=10pm/V (10^{-11} m/V)$
- calculate amplitude and direction of nonlinear polarization at frequency 2w. (use Poynting vector to get E field, be careful with geometry and various factors of 2)
- calculate amplitude and direction of linear polarization P^(1) at frequency w2b. Relevant equations
k vector is in x direction.
Poynting vector is $\vec{S}=\vec{E} \times \vec{H}$. But I thought there was no magnetic field in optical materials at optical frequencies, so when taking the cross product wouldn't everything just go to 0? I wasn't given any magnetic field info. I guess the intensity is what I'd plug in for S.
I also know the formulas for second-order susceptibility and polarization: $\chi^{(2)}(2w)=-\frac{a(e/m)^2 E^2}{D(2w)D^2(w)},\ \~P^{(2)}=\epsilon_0 \chi^{(2)}\~E^2(t)$

2c. The attempt at a solution
given in section 2b--------------------------------------

3a. Homework Statement

Write all elements of d_il matrix (3x6) for lithium niobate (crystal symmetry 3m)
(values given: d_33, d_31=d_15, and d_22.)

3b. Relevant equations
(from table 1.5.1, Boyd)
Form of the 2nd order susceptibility tensor. Each element denoted by Cartesian indices
For 3m crystal class: xzx=yzy, xxz=yyz, zxx=zyy,zzz,yyy=-yxx=-xxy=-xyx (mirror plane perpendicular to x^)

To convert d_ijk to d_il:
http://snag.gy/kIrzU.jpg

3c. The attempt at a solution

The answer is supposed to be this:
http://snag.gy/WlbCk.jpg
But I don't know how to get that from the Cartesian indices given above. maybe someone could please do the first one as an example, (xzx=yzy) and then I could figure out the rest?
Also, when it says "zxx=zyy,zzz,yyy=-yxx=-xxy=-xyx" is that one big equation? --------------------------------------

4a. Homework Statement

Calculate the d_eff value for second harmonic generation in a nonlinear crystal with only one nonlinear coefficient d_33=10pm/V. Input beam at frequency w incident along x-axis, polarized along $1/\sqrt{2}(\^y+\^z)$ and:
- (case a) emission at frequency 2w along x-axis and polarized along ^z direction
- (case b) polarized along ^y direction
- what is the physical meaning of the value of d_eff in case b?

4b. Relevant equations

For SHG, $P(2w)=2\epsilon_0 d_{eff}E(w)^2$, and
http://snag.gy/ALUI9.jpg

4c. The attempt at a solution

We're told that d_il has only one nonzero component (d_33) so the right hand side of the matrix equation reduces to $2\epsilon_0 d_{33}E_z(w)^2$, right? But it seems that P and E are both unknown, so how can we solve for d_eff?

For problem 1, the 3rd order equation becomes $-27w^3x^{(3)}e^{-3iwt}-18\gamma iw^2 x^{(3)} e^{-3iwt}=(2a)(\frac{e}{m})(\frac{2E}{D(w)}e^{-iwt}+\frac{2E}{D(w)}e^{iwt})(-2a)(\frac{e}{m})^2 (\frac{2E^2}{D(0)D(w)D(-w)}+\frac{E^2}{D(2w)D^2(w)}e^{-i2wt}+\frac{E^2}{D(2w)D(w)D(-w)}e^{-i2wt}+\frac{E^2}{D(0)D(w)D(-w)})$.

What I'm supposed to get from that is: $x^{(3)}(3w;w,w,w)=\frac{-4a^2(\frac{e}{m})^3 E^3}{D(3w)D(w)}$
But I'm not sure how to get there. Specifically, how to cancel out all the exponentials and omegas, and create D(3w) and get rid of D(0) and D(w).

[edit: I should mention that $D(w_j)=w_0^2-w_j-2iw_j \gamma$

----------------

For problem 2, I got the direction as 1/sqrt(2) x^ by doing the cofactor thing, is it right?
$\begin{matrix} \^x & \^y & \^z\\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & 0 & 1 \end{matrix}$

Also, we need chi1 but the problem gives chi2 but not chi1. According to Miller's rule, we can get chi1 from chi2 but it seems to be only for sum-frequency generation. Is it valid for second harmonic generation as well? Would the constant be the same? I'll paste the lecture slide excerpt as soon as I can. The image hosting site is acting up.

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Actually for problem 2, according to the professor the problem provides enough information to
deduce the exact value of chi1 at a given frequency. Miller's rule shouldn't be needed. I'm still stumped though.

For problem 4, the equation is this:
$\chi_{eff}=\^e \cdot \underset{\chi}{\leftrightarrow} \cdot \^e_1 \^e_2 = \^z \cdot 10pm/V \cdot \frac{1}{\sqrt{2}}(\^y +\^z) \cdot \frac{1}{\sqrt{2}}(\^y+\^z)$
But how do you take a dot product of directions? Also, any hints on what the physical meaning might be?

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## What is the concept of susceptibility in nonlinear optics?

Susceptibility is a measure of a material's response to an applied electric field. In nonlinear optics, it is used to describe the nonlinear relationship between the electric field and the induced polarization of a material.

## How is polarization related to nonlinear optics?

Polarization is a measure of the alignment of electric dipoles within a material in the presence of an electric field. In nonlinear optics, the induced polarization is a key factor in the nonlinear response of a material to an applied electric field.

## What are the most common types of nonlinear optics problems?

The most common types of nonlinear optics problems include second-harmonic generation, frequency mixing, self-focusing, and optical parametric amplification.

## What is the difference between linear and nonlinear optics?

Linear optics deals with the linear relationship between the electric field and the induced polarization of a material, while nonlinear optics considers the nonlinear relationship between these two factors.

## What are some potential applications of nonlinear optics?

Nonlinear optics has a wide range of applications, including telecommunications, laser technology, optical computing, and medical imaging.

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