Impact of incidence angle in nonlinearities

Question

I know this is a fairly simple question, but my mind is completely blanking.

Say if I have some incident field $\mathbf{E} = E_0 \hat{e} \exp(\mathbf{k_1}\cdot \mathbf{r} - \omega t)$ to a crystal. The subsequent polarisation due to second order nonlinearities is given by:

$\mathbf{P} = \epsilon_0 \chi^{(2)} \mathbf{E}^2$

To calculate this, one can express the polarisation $\hat{e} = e_x \hat{e_x} + e_y \hat{e_y} + e_z \hat{e_z}$, and then write $\mathbf{E} = \begin{pmatrix} E_0 e_x \\E_0 e_y \\ E_0 e_z \end{pmatrix}$ and substitute into the above expression (which gives you a vector with 6 entries).

How does this change if the incoming wave had the same polarisation, but was propagating in a direction $\mathbf{k_2}$ instead of $\mathbf{k_1}$? My instinct is that this should change in some way, but there isn't a way to account for propagation direction in the conventional expression. I was considering using Maxwell's equations, but I wasn't sure what exactly I'd have to solve.

Thanks in advance.

Answer 1

When considering monochromatic plane waves (that have an infinite extent), the induced polarization density indeed only depends on the electric-field components of the EM wave, and not its propagation direction. You see the same thing in linear optics as well: the refractive index seen by a propagating wave inside a birefringent medium doesn't depend on its propagation direction, but on its polarization. Waves propagating in the same direction but with different polarizations can experience different refractive indices, and waves with the same polarization but different propagation directions can experience the same refractive index.

However, if you write down the nonlinear wave equation:

$$ \nabla^2 \mathbf{E} = \epsilon_0 \mu_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} +\mu_0 \frac{\partial^2 \mathbf{P}}{\partial t^2} $$

the spatial derivatives of $\mathbf{E}$ still of course enter, which means that the propagating waves generated by the time varying polarization density $\mathbf{P}$ propagate in a direction that depends on the momentum of $\mathbf{E}$.