What is the intuition behind $\vec{D}$ and $\vec{E}$ not being parallel for the extraordinary beam?

Question

In anisotropic dielectrics, the electric field $\vec{E}$ and the displacement field $\vec{D}$ are no longer necessarily orthogonal. The mathematical reasons are the connection between the two fields:

$$ \vec{D} = \underline{\epsilon} \vec{E} \tag{1} $$

where $\underline{\epsilon}$ is tensor- (matrix) valued and therefore the two fields may point (in principle) in arbitrary directions. Let's consider an uniaxial crystal with one optical axis. If the wave vector $\vec{k}$ is not aligned parallel to it, it is possible to let the plane wave propagate as an extraordinary wave. The refractive index this wave sees depends on the angle between $\vec{k}$ and the optic axis.

In the derivation, one usually solves an equation and obtains the indicatrix equations. This basically boils down to a diagonalization.

Intuitively, I would expect that if $\vec{D}$ coincides with the extraordinary eigenvector of said diagonalization, that

$$ \vec{D} = \epsilon \vec{E} \tag{2} $$

where now $\epsilon$ is a scalar and therefore $ \vec{D} \parallel \vec{E} $. But that is obviously not correct.

My question is therefore, what is the correct intuition here? Why are the two fields in the same plane with $\vec{k}$ but not parallel for extraordinary waves? I sometimes see the construction with $\vec{E}$ being orthogonal to the surface of the indicatrix, but this does not really help me because it does not really explain why $\vec{E}$ should point that way in the first place. Is there a way to see this more clearly?

Answer 1

So suppose

$$\epsilon_{ij} = \epsilon_0{\rm diag(n_o^2, n_o^2, n_e^2)}$$

that is, a unary axis aligned with $\pm \hat z$ (or more precisely: $\{\hat x, \hat y\}$).

You're asking about a wave with direction defined by:

$$\vec k \cdot \hat z = 0$$

and polarization:

$$ \vec E = |E|\hat z $$

Then the displacement field is:

$$ \vec D \propto \epsilon_{ij}E_j = \epsilon_{zz}E_z = \epsilon_0 n_e^2 \hat z$$

which is parallel to $\hat z$. That doesn't make $\epsilon_{ij}$ a scalar.

But this is a very special case of the extraordinary wave.

The general extraordinary wave (with $\vec k$ in the x-z plane, there's no loss of generality in ignoring $y$) is:

$$ \vec k = k(\cos\theta \hat x + \sin\theta \hat z) $$

with $\theta \ne 0$. Then:

$$ \vec D \propto -\sin\theta \hat x + \cos\theta \hat z $$

so that

$$ \vec k \cdot \vec D = 0 $$

and the electric field:

$$ \epsilon^{-1}_{ij}D_j = E_i $$

is not parallel to $\vec D$

See:

http://www.sophphx.caltech.edu/Adv_Lab/References/Birefringence.pdf