For a monochromatic plane wave in a lih material v = 1/(εμ)^˝ = c/n.  The index of refraction is n = c(εμ)^˝.  For most dielectric materials μ ~ μ₀, and n = (ε/ε₀)^˝.  For a linear, isotropic, homogeneous material we have P = ε₀χ_eE, and ε = ε₀(1 + χ_e).
The permittivity ε is determined by how easily a material can be polarized.

Anisotropic materials:

In a non-magnetic material let us define H = B/μ₀ and D = ε₀E + P.
In a non-magnetic material with polarization P, Maxwell’s equations can then be written in macroscopic form as

(1)  ∇∙D = ρ_f,
(2)  ∇×E = -∂B/∂t,
(3)  ∇·B = 0, 
(4)  ∇×H = j_f + ∂D/∂t.
(SI units)

Let us assume that the material contains no free charges, ρ_f = 0,  j_f = 0.  Then
∇∙D = 0,  ∇×E = -μ₀∂H/∂t,  ∇∙H = 0,  ∇×H = ∂D/∂t.

If we are looking for plane wave solutions that satisfy these equations, i.e.
E(r,t) = E₀exp(i(k·r-ωt)),   D(r,t) = D₀exp(i(k·r-ωt)),   H(r,t) = H₀exp(i(k·r-ωt)),
then we need
k∙D = 0,  ik×E = iμ₀ωH,  k∙H = 0,  ik×H = -iωD.

What is the relationship between D and E?

In a lossless macroscopic medium, Maxwell's equations (k∙D = 0) require that for any direction of propagation e_k of a sinusoidal plane electromagnetic wave, D is perpendicular to e_k.  Here e_k is the unit vector pointing in the direction of the wave vector k.  We also have from Maxwell's equations 

D is always perpendicular to k, i.e. D is perpendicular to the normal to the wave front.  This is not necessarily true for E.  D and E are not necessarily parallel vectors.  Maxwell's equations in macroscopic form do not, in general, require, that for plane waves propagating through the medium E is perpendicular to k, as is the case for plane waves propagating through free space or a lih material.

Consider a linear, homogeneous, nonmagnetic, anisotropic medium.  For such a medium the magnitude of the polarization vector P is proportional to the magnitude of E, but P and E are not necessarily parallel vectors.

Example:

Assume that in a material the electron moves in a potential that has a local minimum and that near this minimum can be modeled as an anisotropic harmonic oscillator potential.  In a simple model in two dimensions, we may picture the electron in a square box, connected by strong springs to the right and left walls and by weak springs to the top and bottom walls.

Let a force of magnitude F_h = F' pointing towards the right produce a displacement of magnitude d and a dipole moment of magnitude -q_ed.
Let a force of magnitude F_v = F' pointing upwards produce a displacement of magnitude 2d and a dipole moment of magnitude -2q_ed.

If we choose the coordinate system shown in the figure below and apply a force F = F_h + F_v in the x-direction, then this force produces a displacement Δx = (d, 2d)·(1/√2, 1/√2) = 3d/√2 in the x-direction and a displacement Δy = (d, 2d)·(-1/√2, 1/√2) = d/√2 in the y-direction.  The dipole moment and the polarization therefore have a x- and a y-component.  If the force F is due to an electric field E, then P and E are not parallel vectors.

In a linear, homogeneous, nonmagnetic, anisotropic medium we have for the Cartesian components of P,

P_i = ε₀ Σ_jχ_ijE_j, with  i, j = 1, 2, 3  = x, y, z.
In vector form we write P = χ ∙ E, where χ denotes the susceptibility tensor.

We have
D_i = ε₀E_i + P_i = ε₀ Σ_j(δ_ij + χ_ij)E_j = Σ_jε_ijE_j,  or  D = ε ∙ E,
where ε denotes the dielectric tensor.

.

D and E are not necessarily parallel vectors.

For a lossless (non-absorbing) medium that has no optical activity, the elements of ε are real and ε is a real, symmetric tensor.  For such a tensor we can always find a Cartesian coordinate system in terms of which its matrix is diagonal.  In that coordinate system ε has real elements and is of the form

.

The quantities ε_x, ε_y, and ε_z are called the principal values of ε.  They are all non-negative and in general are functions of the frequency ω.  The directions of the eigenvectors of the matrix ε are the principal axes.

Now consider a plane wave with wave vector k propagating through this medium. 

For a given direction of propagation e_k, there are in general two values for the refractive index, n₁ and n₂.  We can solve for the corresponding components of E up to a multiplicative constant.  (See mathematical details.) 

Results:

Let the solutions E₁ and E₂ correspond to n₁ and n₂, respectively.
If n₁² is not equal to n₂², then D₁ and D₂ are perpendicular to each other.
If n₁² is equal to n₂², then the directions of D₁ and D₂ are arbitrary as long as they are perpendicular to e_k and it is convenient to choose them perpendicular to each other.

If the parallel components of both E₁ and E₂ are not zero, then E₁·E₂ =  E_1|| ·E_2|| ≠ 0.  (Parallel refers to the direction of k.) 

The parallel components of E₁ and E₂ are only zero if the wave propagates along one of the principal axes of the crystal.

AI Study Tip:

Example prompt:  'Explain the physical meaning of the electric susceptibility tensor χ in anisotropic materials.  How does it differ from the scalar χ in isotropic materials, and what does it imply about the direction of the induced polarization vector P relative to the electric field vector E?'

Problem:

Consider a uniaxial crystal  with the diagonal dielectric tensor 

,

(a)  Show that a circularly polarized wave with the field E = E₀j + iE₀k at x = 0, propagating in the crystal along the x-axis, becomes periodically linearly polarized.  A plate of the appropriate thickness so as to transform circularly polarized light into linearly polarized light is called a quarter-wave plate.
(b)  Find the thickness d of a quarter wave plate in terms of n_o, n_e, and the free-space wavelength l.

Solution:

• Reasoning:
  In a uniaxial crystal with diagonal dielectric tensor
   
  only waves with D polarized along the y- or z- direction can propagate in the x-direction.  (See example.)   For these waves E is parallel to D.  Waves polarized along the y-direction propagate with v = c/n_o and waves polarized along the z-direction propagate with v = c/n_e.
  Let Δn = n_o - n_e.
  Using ω/k = c/n we have 
  E(x,t) = E₀ exp(iω(n_ox/c - t))j + iE₀ exp(iω(n_ex/c - t))k 
  or
  E(x,t) = E₀ exp(iω(n_ox/c - t))j + E₀ exp(i((π/2) - ωΔnx/c)) exp(iω(n_ox/c - t))k.
• Details of the calculation:
  (a)  When (π/2) - ωΔnx/c = ±mπ, then E is linearly polarized, the polarization direction making an angle of ±45^o with the y-axis in the yz-plane.  This happens when x = (m + ½)λ/(2Δn), m = 0, 1, 2, ... .  (Here λ is the free space wavelength.)
  (b)  The minimum thickness d of a quarter-wave plate is d = λ/(4Δn).

Problem:

Show that the quarter wave plate of the previous problem can transform linearly polarization at its "input" into circular polarization at its "output" and determine the angle of E with respect to the y-axis.

Solution:

• Reasoning:
  The thickness d of a quarter-wave plate is d = λ/(4Δn).
• Details of the calculation:
  If E = E₀j ± E₀k at x = 0, i.e. E is linearly polarized, the polarization direction making an angle of ±45^o with the y-axis in the yz-plane, then 
  E(d,t) = E₀ exp(iω(n_od/c - t))j ± E₀ exp(-i(ωΔnd/c)) exp(iω(n_od/c - t))k,  or 
  E(d,t) = E₀ exp(iω(n_od/c - t))j ± E₀ exp(-i(π/2)) exp(iω(n_od/c - t))k,  or 
  E(d,t) = E₀ exp(iω(n_od/c - t))j ± iE₀ exp(iω(n_od/c - t))k,
  i.e. E is circularly polarized at the output.