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

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?

• This isn't my area of expertise, so this is more of a question. But I would guess the dielectric tensor for a uniaxial crystal sounds would have two degenerate eigenvalues (since only one direction is special by assumption). If there's a 2D degenerate subspace, that would give room for $D$ and $E$ to not be parallel. Commented Aug 3 at 23:08

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: $$\{\hat x, \hat y\}$$).

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

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: