I am trying to follow a derivation for the reflectance and transmittance of a plane wave from an anisotropic material ($\varepsilon_a,\varepsilon_b,\varepsilon_c$). The derivation uses a $4 \times 4$ tranfser matrix method. Isotropic homogeneous materials occupy $z\leq 0$ ($n_1$) and $z \geq L$ ($n_2$), and the anisotropic material occupies $0 < z < L$, extending to infinity in the $x$ and $y$ directions. The book states that the field phasor in the material is best represented as
$$\textbf{E}(\textbf{r}) = \textbf{e}(z) \exp[i q (x \cos \psi + y \sin \psi)]$$
where $\textbf{e}(z) = e_x(z) \hat{\textbf{x}} + e_y(z) \hat{\textbf{y}} + e_z(z) \hat{\textbf{z}}$, $\textbf{h}(z) = h_x(z) \hat{\textbf{x}} + h_y(z) \hat{\textbf{y}} + h_z(z) \hat{\textbf{z}}$, $q$ is the complex-valued wavenumber, and $\psi$ is the angle from the $+x$-axis to the $+y$-axis.
The fields for the regions $0 \leq z$ and $z \geq L$ are written as:
$$ \textbf{e}_{inc}(z) = (a_s \hat{\textbf{s}} + a_p \hat{\textbf{p}}_{inc}) \exp[i k_0 n_1 ( x \cos \psi + y \sin \psi) + k_0 n_1 z \cos \theta_{inc}]$$
$$ \textbf{e}_{ref}(z) = (r_s \hat{\textbf{s}} + r_p \hat{\textbf{p}}_{ref}) \exp[i k_0 n_1 ( x \cos \psi + y \sin \psi) - k_0 n_1 z \cos \theta_{inc}]$$
$$ \textbf{e}_{tr}(z) = (t_s \hat{\textbf{s}} + t_p \hat{\textbf{p}}_{tr}) \exp[i k_0 n_2 ( x \cos \psi + y \sin \psi) + k_0 n_2 (z-L) \cos \theta_{tr}]$$
where
$$ \hat{\textbf{s}} = - \hat{\textbf{x}} \sin \psi + \hat{\textbf{y}} \cos \psi$$
$$ \hat{\textbf{p}}_{inc} = -(\hat{\textbf{x}} \cos \psi + \hat{\textbf{y}} \sin \psi) \cos \theta_{inc} + \hat{\textbf{z}} \sin \theta_{inc}$$
$$ \hat{\textbf{p}}_{ref} = (\hat{\textbf{x}} \cos \psi + \hat{\textbf{y}} \sin \psi) \cos \theta_{inc} + \hat{\textbf{z}} \sin \theta_{inc}$$
$$ \hat{\textbf{p}}_{tr} = -(\hat{\textbf{x}} \cos \psi + \hat{\textbf{y}} \sin \psi) \cos \theta_{tr} + \hat{\textbf{z}} \sin \theta_{tr}$$
$$ \sin \theta_{tr} = \frac{n_1}{n_2} \sin \theta_{inc}$$
Here are my questions:
The problem is solving for the $z$-dependence of the electric field. But why only have $\textbf{e}(z)$ and not $\textbf{e}(x,y,z)$? If we can assume an $\exp[i q (x \cos \psi + y \sin \psi)]$ form for $x$ and $y$, why not for $z$? Generalizing, could we have assumed the form for any two pairs of variables and then solved the problem for the third variable?
If we have a plane wave entering a complex material, how do we know that the wave propagating inside the material will also be a plane wave (which is what has been assumed for the $x$ and $y$ direction)?
Given that the material is anisotropic, how would we know that the transmitted wave will obey Snell's law? In addition, as given, $\psi$ is the same inside and outside of the anisotropic material. But, how do we know that $\psi$ won't change inside the anisotropic material?
The closest I have gotten to obtaining the first equation is by making use of the continuity of the tangential component at the first interface. So we have the following:
$$ \hat{\textbf{z}} \times (\textbf{E}_1 - \textbf{E}) = 0$$
where $\textbf{E} = \textbf{e}_{inc} + \textbf{e}_{ref}$ and $\textbf{E}$ is the field inside the anisotropic material. Then we can solve for the $x$ and $y$ component of $\textbf{E}$ at $z = 0$ to obtain
$$ \textbf{e}_{x} = (-a_s \sin \psi - r_s \sin \psi - a_p \cos \theta_{inc} \cos \psi + r_p \cos \theta_{inc} \cos \psi) \exp[i k_0 n_1 ( x \cos \psi + y \sin \psi)]$$
$$ \textbf{e}_{y} = (a_s \cos \psi + r_s \cos \psi - a_p \cos \theta_{inc} \sin \psi + r_p \cos \theta_{inc} \sin \psi) \exp[i k_0 n_1 ( x \cos \psi + y \sin \psi)]$$
But now there is all this junk at the beginning. Not to mention these contain $k_0 n_1$ while the former has $q$ and this doesn't address the $\textbf{e}_{x}$ component.