I always see Snell's law
$$ n_1 \sin(\alpha_1) = n_2 \sin(\alpha_2) \tag{1} $$
for the electric field $E$ without any words mentioned of the electric displacement field $D$ which is also required for EM fields in matter. How does Snell's law look for the $D$ field and why is it never considered?
The derivation of Snell's law is indeed derived by assuming that the component of the E-field parallel to the interface between the two materials is continuous (i.e. the sum of the parallel components on either side of the interface are equal). So you get something like $$E_i \exp[i(k_i x\sin\theta_i - \omega_i t)]\cos\theta_i + E_r \exp[i(k_r x \sin\theta_r - \omega_r t)]\cos\theta_r = E_t \exp[i(k_t x \sin\theta_t - \omega_t]\cos\theta_t\ , $$ where the subscripts refer to incident, reflected and transmitted fields and $\theta_i$ is the angle of incidence.
You then argue that for this to be true for all values of $x$ and $t$, it must be the case that $$ \omega_i = \omega_r = \omega_t\ \ \ \ {\rm and} $$ $$ k_i \sin\theta_i = k_r \sin\theta_r = k_t \sin\theta_t\ . $$ Then, because you know that $\omega_i = k_i c/n_i$ etc. you get the law of reflection and Snell's law.
You are perfectly welcome to write down the equivalent expression for the components of the D-field normal to the interface, which must also be continuous (assuming there is no surface charge). In which case you have something like $$D_i \exp[i(k_i x\sin\theta_i - \omega_i t)]\sin\theta_i + D_r \exp[i(k_r x \sin\theta_r - \omega_r t)]\sin\theta_r = D_t \exp[i(k_t x \sin\theta_t - \omega_t]\sin\theta_t\ . $$ You can then go through exactly the same argument and get the same law of reflection and Snell's law.
Snell's law depends on the linearity assumption that $\bf D=\epsilon E$ (In Gaussian units, where $\epsilon_0=1$).
This is used at the start of derivations to eliminate $\bf D$.
