Vector Laplacian in electromagnetic wave theory I am a bit confused as to how we obtain regular wave equations from Maxwell's equations when the vector Laplacian is defined the way it is. We have the differential equation for waves in the form: $$ \nabla^{2} \mathbf{E}=\mu \epsilon \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}+\mu \sigma \frac{\partial \mathbf{E}}{\partial t} $$ and solutions of the form: $$\tilde{\mathrm{E}}(z, t)=\tilde{\mathrm{E}}_{0} e^{i(\bar{k} z-\omega t)}$$
 (this is from Maxwell's equations for a conductor).
When we expand the vector Laplacian we get: $$\nabla^{2} \mathbf{E}= \left(\nabla^{2} E_{x}, \nabla^{2} E_{y}, \nabla^{2} E_{z}\right)$$ but then we have the issue of having to expand these laplacians out to find for example in the x directions: $$\frac{\partial^{2} E_x}{\partial x^{2}}+\frac{\partial^{2} E_x}{\partial y^{2}}+\frac{\partial^{2} E_x}{\partial z^{2}}=\mu \epsilon \frac{\partial^{2} E_x}{\partial t^{2}}+\mu \sigma \frac{\partial E_x}{\partial t}$$
which is not the standard wave equation. 
Is this the real form and when we solve the wave equation we are assuming some components are zero or have I messed up somewhere/am not understanding it correctly?
 A: In general, a first order derivative in a wave equation represents a dissipative term (think of ODE's: friction for a damped harmonic oscillator or resistance in an RLC circuit).
This term is present in your wave equation because you are dealing with a conducting medium. The amplitude of your resulting EM wave will decay exponentially.
A: If you are dealing with E-field in vacuum or in a neutral medium then Gauss's law tells you that
$$\nabla \cdot {\bf E} = 0$$
For any wave solution of the form ${\bf E} = {\bf E_0} f({\bf k}\cdot {\bf r} - \omega t)$ then this mean that ${\bf E_0}\cdot {\bf k} = 0$.
Thus if your E-field is in the x-direction then $k$ must be perpendicular to the x-direction and ${\bf k} \cdot {\bf r} = k_y y + k_z z$ (in general).
That means your final equation has one term that is zero
$$\frac{\partial^2 E_x}{\partial x^2} = 0$$
Thus
$$\frac{\partial^2 E_x}{\partial y^2} + \frac{\partial^2 E_x}{\partial z^2} = \frac{\partial^{2} E_x}{\partial t^{2}}+\mu \sigma \frac{\partial E_x}{\partial t}$$
This is still a wave equation, that allows for the wave to be travelling in any direction perpendicular to the $x$-axis. The final term is just a damping term that makes the amplitude of the wave die away.
