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Lets assume that we have sourceless anisotropic medium with $\epsilon_1 , \epsilon_2,\epsilon_3 $ as the diagonal values. Also assume $\vec{k}=k_z\hat{z}$ and $e^{i \omega t} e^{-i \vec{k} \cdot \vec{r}}$ form.

We have $\vec{k} \cdot \vec{D} = 0 \implies \vec{k} \cdot \underline{\underline{\epsilon}} \vec{E} = 0 \implies k_z \epsilon_3 E_z = 0 $.

From the curl equations and the fact that from above $E_z = 0$ and $\vec{k} \cdot \vec{E} = 0$,then we have $\vec{E} k_z^2 = \omega \mu^2_0 \underline{\underline{\epsilon}} \vec{E}$ which implies $E_x k_z^2 = \omega^2 \mu_0 \epsilon_1 E_x$ and $E_y k_z^2 = \omega^2 \mu_0 \epsilon_2 E_y$

So is this saying that the wave can only propagate in two modes? One where $E_x=0, E_y \neq 0$ and one where $E_y=0, E_x \neq 0$? For if there were a nonzero $x$ and $y$ component then $k_z^2$ would equal two different values.

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You are right that the propagation constant differs for these two polarizations, but remember also that any linear combination of these two modes is also a perfectly valid solution. It's just that the components of the wave polarized along $x$ and $y$ directions propagate with different propagation constants (assuming they have the same frequency).

This situation, arising from the anisotropy of the medium, is called birefringence in optics. The difference of the propagation constants for the two polarizations cause their relative phase relationship to change as a function of position ($z$-coordinate), which results in a polarization state that depends on position. This property is exploited in quarter-wave plates, which can be used to create circularly polarized light from a linearly-polarized source.

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  • $\begingroup$ But if it is true that "For if there were a nonzero x and y component then $k^2_z$ would equal two different values" then how can there be a solution which is a linear combination of both? That lin. comb. has nonzero E_x and E_y. What is more confusing is there seems to be two ways to go about solving for field components. The first is to work with $\nabla$ and the second is to recast with $\vec{k}$. The first leads to the derivation what you are mentioning. And I can follow that derivation fine. The second method which I used here leads to this result which is seemingly not as general. $\endgroup$
    – ngc1300
    Commented Jul 7, 2020 at 22:37
  • $\begingroup$ The second method assumes that the solution is characterized by a single propagation constant $\vec{k}$. This certainly doesn't hold for all solutions to the wave equation, including linear combinations of the two solutions you have found. But the nice thing about it is that any solution to the wave combination can be expressed as a linear combination of solutions with a fixed $\vec{k}$, via a Fourier transform. In other words, these special solutions constitute a complete basis that spans the set of all (well-behaved) solutions. $\endgroup$
    – Puk
    Commented Jul 7, 2020 at 22:44
  • $\begingroup$ Ok, then why is that in the next example in my book, they do the same problem except letting $\vec{k} $ be in the xz-plane, but they derive the field components using the $\nabla$ equations? How do I know when to use what?! I am sick of physics\engineering books skipping over the details. Also, how do we know if assuming $e^{i \omega t}$ is misleading in certain problems (for example non-linear media), just like assuming $e^{-i k z}$ was in this one?? I want a more rigorous approach to solving wave equations. Do you know of any such book? $\endgroup$
    – ngc1300
    Commented Jul 8, 2020 at 21:59
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    $\begingroup$ The wave equation (the one with $\nabla$) is always valid, so there is nothing wrong with using it. The time variation described by $e^{i\omega t}$ means you are seeking a monochromatic (sinusoidal steady state) solution, which also constitute a complete basis spanning the whole solution space. These assumptions are usually clear at the outset or from the context. Not sure if they are as rigorous as you want, but you can try introductory books like Griffiths, or something like Collin's Foundations for Microwave Engineering for more solutions to the wave equation, mostly in waveguides. $\endgroup$
    – Puk
    Commented Jul 8, 2020 at 22:28
  • $\begingroup$ I understand that the $\nabla$ will always hold. Let me be more clear. For this problem, my book assumed $e^{i \omega t}$, that dx and dy operators are replaced by 0 (uniform plane wave), and they used the $\nabla$ form of ME's. But in the other example of propagation along the xz-plane, they assumed $e^{i \omega t}$ AND $e^{- i k \cdot r}$, they did NOT assume uniformity, and the used the k version of ME's. Why the different assumptions? $\endgroup$
    – ngc1300
    Commented Jul 10, 2020 at 16:06

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