In a material with polarization P and magnetization M Maxwell’s equations in macroscopic form are

,

.

Let us assume that the material is nonmagnetic and contains no free charges.

M_f = 0,   r_f = 0,   j_f = 0.

Then B = m₀H, and

If we are looking for plane wave solutions that satisfy these equations,

E(r,t) = E₀exp(i(k×r-wt)),   D(r,t) = D₀exp(i(k×r-wt)),   H(r,t) = H₀exp(i(k×r-wt)),

then we need

What is the relationship between D and E?

In a lossless macroscopic medium Maxwell's equations require that for any direction of propagation of a sinusoidal plane electromagnetic wave, D is perpendicular to . 

k×D = 0.

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.  So 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 a 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.

If we choose the coordinate system s shown in the figure below, and apply a force F = F_h + F_v in the x-direction,  then this force produces a displacement Dx = 3d/Ö2 in the x-direction and a displacement Dy = d/Ö2 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 en 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,

.

If we use the Einstein summation convention that, unless explicitly stated, repeated indices are automatically summed over, then

P_i = e₀ c_ij E_j,    i,j = 1, 2, 3, = x, y, z.

In vector form we write P = c × E, where c denotes the susceptibility tensor.

We have

D_i = e₀E_i + P_i = e₀ (d_ij + c_ij )E_j = e_ij E_j,   or   D = e × E,

where e denotes the dielectric tensor.

.

D and E are not necessarily parallel vectors.

For a lossless (non-absorbing) medium e_ij = e_ji^*, e is a Hermitian tensor.  If the medium has no optical activity, then the elements of e are real and e is a real, symmetric tensor.

For Hermitian tensor we can always find a Cartesian coordinate system in terms of which its matrix is diagonal.

(e × E = lE.  Any Hermitian matrix has real eigenvalues and its eigenvectors can form an orthonormal basis for the vector space.)

In that coordinate system e has real elements and is of the form

.

The quantities e_x, e_y, and e_z are called the principal values of e.  They are all non-negative and in general are functions of the frequency w.  The directions of the eigenvectors of the matrix e are the principal axes.

Now consider a plane wave with wave vector k propagating through this medium.  In a coordinate system with its axes along the principal axes, the equation D = e₀n²E_^ for a component of D along one of the principal axes becomes

e_iE_i = e₀n²(E_i – (1/k²)k_i(k×E)),

since E_^ = E - (×E) = E – (1/k²)k(k×E)).

This can be rewritten as e_iS_j(d_ijE_j)  = e₀n²(S_j(d_ijE_j) – (1/k²)k_iS_j(k_jE_j)), or

.

In matrix form this may be written as ME = 0, with

.

Here we have used

.

If the e_i are known, and we choose a particular direction of propagation for the wave front, then we can solve for E.

ME = 0 only if the characteristic equation is satisfied and det(M) = 0.  The characteristic equation appears cubic in n², but when written in terms of a polynomial in n² the coefficients of the n⁶ terms vanish, leaving a quadratic equation in n².  This equation has two positive roots, n₁² and n₂².

For a given direction of propagation , there are in general two values for the refractive index, n₁ and n₂.  Substituting each of these values back into the equation ME = 0, we can solve for corresponding components of E up to a multiplicative constant.

Let the solutions E₁ and E₂ correspond to n₁ and n₂, respectively.

We may resolve E into components parallel and perpendicular to .

If n₁² is not equal to n₂², then D₁ and D₂ are perpendicular to each other.

To show this, consider

.

Therefore

.

But parallel and perpendicular all refer to .

Therefore

If n₁² is equal to n₂², then the directions of D₁ and D₂ are arbitrary as long as they are perpendicular to 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|| ¹ 0.  The parallel components of E₁ and E₂ are only zero if the wave propagates along one of the principal axes of the crystal and the matrix M is diagonal.

Example:

Let be the z-direction, i.e. the wave front propagates along the z-direction. Then k = k_z and

The characteristic equation det(M) = 0, or (n²-n_x²) (n²-n_y²) n_z² = 0 implies that n² = n_x², or n² = n_y², or n_z² = 0.  But n_z² = 0 is not a physical solution, since e_z > 0.  We therefore have two solutions, n₁² = n_x² and n₂² = n_y².

The direction of E₁ = D₁/( e₀n₁²) corresponding to n₁ is the x-direction and the direction of E₂ = D₂/( e₀n₂²) corresponding to n₂ is the y-direction.  E is parallel to D.

When D is not polarized along the x- or y-direction, then E is not parallel to D as required.  Such a wave does not propagate through the material.  A wave propagating along a principal axis must be linearly polarized along one of the two remaining principal axes of the dielectric tensor.  An arbitrarily polarized wave must be treated a linear superposition of two waves each linearly polarized along one of the two remaining principal axes, and these two waves propagate independently with different phase velocities.

The phase velocity of the wave polarized along the x-direction is c/n₁, and the phase velocity of the wave polarized along the x-direction is c/n₂.

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:

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_p = v_g = c/n_o and waves polarized along the z-direction propagate with v_p = v_g = c/n_e.  Let Dn = n_o - n_e.

We have 

E(x,t) = E₀ exp(iw(n_ox/c - t))j + iE₀ exp(iw(n_ex/c - t))k 

or

E(x,t) = E₀ exp(iw(n_ox/c - t))j + E₀ exp(i((p/2) - wDnx/c)) exp(iw(n_ox/c - t))k.

(a)  When (p/2) - wDnx/c = ±mp, 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+1/2)l/(2Dn), m = 0, 1, 2, ... .  (Here l is the free space wavelength, k = 2p/l, and w/k = c.)
(b)  The thickness d of a quarter-wave plate is d = l/(4Dn).

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:

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(iw(n_od/c - t))j ± E₀ exp(-i(wDnd/c)) exp(iw(n_od/c - t))k,  or 

E(d,t) = E₀ exp(iw(n_od/c - t))j ± E₀ exp(-i(p/2)) exp(iw(n_od/c - t))k,  or 

E(d,t) = E₀ exp(iw(n_od/c - t))j iE₀ exp(iw(n_od/c - t))k,

i.e. E is circularly polarized at the output.