# [Physics] How to use the electro-optic tensor

optics

I would like to calculate the performance of an optical phase modulator, in which a varying electric field across a crystal modulates the effective refractive index of light passing through the crystal.

The effect depends on the crystal orientation, polarization (i.e. direction of the E-field of the optical beam), and the direction of the applied electric field. The dependence on these vectors is encoded into the electro-optic coefficients, which form the electro-optic tensor.

My question: How do I apply these coefficients to give the magnitude of the electro-optical effect (in radians of optical phase per applied volt of potential difference) when the light is polarized in direction $i$, propagating in direction $j$, with an applied electric field in direction $k$? In short, how do I use the electro-optical coefficients?

For concreteness, I am interested in lithium niobate. A datasheet gives the coefficients as:

The simple answer is that your LiNbO3 crystal will almost certainly be cut such that the coefficient of importance is $r_{33}^T$. This means that the electric field is applied along the $z$ axis, the polarization of your light field is along the $z$ axis, and the direction of propagation through the crystal is along the $y$ axis (or maybe it is the arbitrarily defined $x$ axis). The resulting index of refraction of the crystal with an applied electric field $E_z$ is then given by $$n_z\simeq n_{z0}-\frac{1}{2}n_{z0}^3\ r_{33}^T\ E_z,$$ where $n_{z0}$ is the nominal index of refraction of the crystal along the $z$ axis. The phase picked up when passing through the crystal is then given by $$\phi=\frac{\omega \ell}{c}n_z=\frac{\omega \ell n_{z0}^3 r_{33}^T E_z}{2c}=\frac{\pi\ell n_{z0}^3 r_{33}^T V}{\lambda d},$$ where $\ell$ is the length of the electrodes on the crystal, $d$ is the distance between the electrodes, and $V$ is the applied voltage. Also, we have dropped the static phase picked up by propagation through the crystal with no applied field.
The typical application of a phase modulator is to apply RF sidebands for use in sensing the degrees of freedom of the downstream optical apparatus, since the frequency of light is too high for direct sensing. In this case the voltage applied to the crystal is given by $V=V_m\sin(\Omega t)$ and the resulting light field coming out of the crystal is given by $$E=E_0 e^{-i(\omega t+\delta\sin(\Omega t))},$$ where $$\delta=\frac{\pi\ell n_{z0}^3 r_{33}^T}{\lambda d}.$$ If you expand the exponential using the Fourier-Bessel expansion $$E=E_0e^{-i\omega t}\ \sum_{n=-\infty}^\infty J_n(\delta)e^{-i\ n\ \Omega t} \simeq E_0 e^{-i\omega t}+E_0\frac{\delta}{2}e^{-i(\omega+\Omega)t}- E_0\frac{\delta}{2}e^{-i(\omega-\Omega)t}$$ where the final approximation drops all but the lowest terms and expands the Bessel functions to first order; then you find that the effect of RF phase modulation of a light beam to first order is to add two components to the light at frequencies $\omega+\Omega$ and $\omega-\Omega$. If you want to understand how this phase modulated light interacts with your optical system, you can simply ask how each of the three frequencies interacts independently with your system and sum them together in the end.