## Abstract

An error was made in calculating the polarization dependent second harmonic response of barium titanate nanoparticles. We have corrected the error and repeated the comparison with the experimental results.

© 2010 Optical Society of America

Full Article |

PDF Article

**Related Articles**
### Equations (1)

Equations on this page are rendered with MathJax. Learn more.

(1)
$$P\phantom{\rule{.2em}{0ex}}\left(2\mathrm{\omega}\right)={\chi}^{\phantom{\rule{.2em}{0ex}}\left(2\right)}\bullet {E}_{1}\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right)\bullet {E}_{1}\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right)=\left[\begin{array}{cccccc}{d}_{11}& {d}_{12}& {d}_{13}& {d}_{14}& {d}_{15}& {d}_{16}\\ {d}_{21}& {d}_{22}& {d}_{23}& {d}_{24}& {d}_{25}& {d}_{26}\\ {d}_{31}& {d}_{32}& {d}_{33}& {d}_{34}& {d}_{35}& {d}_{36}\end{array}\right]\phantom{\rule{.2em}{0ex}}\left[\begin{array}{c}{E}_{x}{\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right)}^{2}\\ {E}_{y}{\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right)}^{2}\\ {E}_{z}{\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right)}^{2}\\ 2{E}_{y}\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right){E}_{z}\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right)\\ 2{E}_{x}\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right){E}_{z}\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right)\\ 2{E}_{x}\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right){E}_{y}\phantom{\rule{.2em}{0ex}}\left(\mathrm{\omega}\right)\end{array}\right].$$