# Nonlinear optics: second order polarization calculation

1. Mar 10, 2014

### dnp33

This is a problem from Boyd Nonlinear Optics chptr 1 problem 2.

1. The problem statement, all variables and given/known data

Numerical estimate of nonlinear optical quantities. A laser beam of frequency ω carrying 1 W of power is focused to a spot size of 30μm diameter in a crystal having a refractive index of n =2 and a second order susceptibility of $\chi^{(2)}=4\times 10^{-11} m/V$. Calculate numerically the amplitude P(2ω) of the component of the nonlinear polarization oscillating at frequency 2ω.

2. Relevant equations

$$P(2\omega)=\epsilon_0^{(2)}E^2$$
$$I=\frac{cn\epsilon_0}{2}E_0^2$$

3. The attempt at a solution

I solved for $E_0$ assuming a uniform distribution across the spot, with I=P/A and got
$$E_0=\sqrt{\frac{2P}{Acn\epsilon_0}}$$
and put that into the equation I gave for $P(2\omega)$. The value I got was $1.89\times 10^{-11}$, which is almost exactly 4 times the value given in the text of $4.7\times 10^{-11}$

I feel like it's possible that the value given in the text accidentally uses the diameter of the spot to calculate the area, which would give them a factor of 1/4 that I don't have, but I also thought that maybe my problem lies in my assumption that the spot is uniform. Maybe I actually need to integrate numerically assuming a gaussian beam profile-which would kind of make sense considering the problem title (numerical estimate of nonlinear optical quantities).

2. Mar 11, 2014

### TSny

Hello, dnp33.

Here's a possibility. Note that below (1.2.14b), Boyd (3rd edition) gives the expression for intensity as $I = \frac{1}{2}n_0\epsilon_0c\varepsilon^2$ where $\varepsilon$ is the field amplitude as defined by (1.2.12). But this definition of field amplitude differs by a factor of 2 from the field amplitude $E$ defined by (1.2.1). Note $\varepsilon = 2E$.

Very confusing. Would have been nice if Boyd had given an explicit numerical example to help sort out the notation.

3. Sep 24, 2015