# Propagation in an Anisotropic Medium

This part continues from the Propagation in an Isotropic Medium tutorial.

In an anisotropic medium, the tensors χ and ε do not reduce to scalars. Therefore, $$\mathbf{P}\not\parallel\mathbf{E}\text{ and }\mathbf{D}\not\parallel\mathbf{E}$$. As a result, (81) [from the propagation in an isotropic medium tutorial] is not true any more, and, in general,

$\tag{104}\boldsymbol{\nabla}\cdot\mathbf{E}\ne0$

Consequently, (82) cannot be used for propagation of a monochromatic wave in an anisotropic medium. Instead, (80) has to be used.

$\tag{80}\boldsymbol{\nabla}\times\boldsymbol{\nabla}\times\mathbf{E}+\mu_0\epsilon(\omega)\cdot\frac{\partial^2\mathbf{E}}{\partial t^2}=0$

Anisotropic χ and ε

In a linear anisotropic medium, both χ and ε are second-rank tensors. They can be expressed in the following matrix forms:

$\tag{105}\pmb{\chi}=\begin{bmatrix}\chi_{11}&\chi_{12}&\chi_{13}\\\chi_{21}&\chi_{22}&\chi_{23}\\\chi_{31}&\chi_{32}&\chi_{33}\\\end{bmatrix}$

and

$\tag{106}\pmb{\epsilon}=\begin{bmatrix}\epsilon_{11}&\epsilon_{12}&\epsilon_{13}\\\epsilon_{21}&\epsilon_{22}&\epsilon_{23}\\\epsilon_{31}&\epsilon_{32}&\epsilon_{33}\\\end{bmatrix}$

The relationships $$\mathbf{P}=\epsilon_0\pmb{\chi}\cdot\mathbf{E}$$ and $$\mathbf{D}=\pmb{\epsilon}\cdot\mathbf{E}$$ are carried out as products between a tensor and a column vector.

For example,

$\tag{107}\begin{bmatrix}D_1\\D_2\\D_3\\\end{bmatrix}=\begin{bmatrix}\epsilon_{11}&\epsilon_{12}&\epsilon_{13}\\\epsilon_{21}&\epsilon_{22}&\epsilon_{23}\\\epsilon_{31}&\epsilon_{32}&\epsilon_{33}\\\end{bmatrix}\begin{bmatrix}E_1\\E_2\\E_3\end{bmatrix}$

In general, the matrix in (106) representing the tensor ε is not diagonal. It can be diagonalized by a proper choice of the coordinate system, yielding

$\tag{108}\pmb{\epsilon}=\begin{bmatrix}\epsilon_1&0&0\\0&\epsilon_2&0\\0&0&\epsilon_3\end{bmatrix}$

where εi, for i = 1, 2, 3, are the eigenvalues of ε with their corresponding eigenvectors, $$\hat{u}_\text{i}$$, being the axes of the coordinate system chosen to diagonalize ε. The characteristics of εi and $$\hat{u}_\text{i}$$ depend on the symmetry properties of ε. The two matrices representing χ and ε have the same symmetry properties because ε = ε0(1 + χ), where 1 has the form of a 3x3 identity matrix in its addition to the tensor χ. Therefore, χ and ε are diagonalized by the same set of eigenvectors that represent the axes of the chosen coordinate system.

The symmetry properties of ε, as well as those of χ, are determined by the properties of the medium.

1. Reciprocal media. Nonmagnetic materials in the absence of an external magnetic field are reciprocal media. In a reciprocal medium, the Lorentz reciprocity theorem of electromagnetics holds; consequently, the source and the detector of an optical signal can be interchanged. If such a material is not optically active, its optical properties are described by a symmetric ε tensor: εij = εji. For a symmetric tensor, the eigenvectors $$\hat{u}_\text{i}$$ are always real vectors. They can be chosen to be $$\hat{x},\hat{y},\text{ and }\hat{z}$$ of a rectangular coordinate system in real space. This is true even when ε is complex.
• (a) If a nonmagnetic medium does not have an optical loss or gain, its ε tensor is Hermitian. A symmetric Hermitian tensor is real and symmetric: εij* = εij = εji  = εji*. In this case, the eigenvalues εi have real values.
• (b) If a nonmagnetic medium has an optical loss or gain, its ε tensor is not Hermitian but is complex and symmetric: εij = εji but εij ≠ εji*. Then, the eigenvalues εi are complex.
• (c) If a nonmagnetic medium is optically active, it is still reciprocal although its ε tensor is not symmetric. In this case, the eigenvectors are complex but the eigenvalues can be real if the medium is lossless.
2. Nonreciprocal media. Magnetic materials, and nonmagnetic materials subject to an external magnetic field, are nonreciprocal media. In such a medium, no symmetry exists when the source and the detector of an optical signal are interchanged. The ε tensor describing the optical properties of such a material is not symmetric: εij ≠ εji. The eigenvectors $$\hat{u}_\text{i}$$ are complex vectors. Therefore, they are not ordinary coordinate axes in real space, as seen later in the discussion on magneto-optic devices.
• (a) For a lossless magnetic medium, ε is Hermitian: εij = εji*. In this case, the eigenvalues εi are real even though the eigenvectors are complex.
• (b) For a magnetic medium that has an optical loss or gain, ε is neither symmetric nor Hermitian. Both the eigenvectors and the eigenvalues are complex.

Most materials used for photonic devices are nonmagnetic dielectric materials that are not optically active. The properties of magnetic materials are of interest to us only in consideration of magneto-optic devices, discussed in later tutorials. Similarities and differences between magnetic and optically active materials are also briefly mentioned in later tutorials. The discussion in the rest of this part is specific to nonmagnetic dielectric materials that are not optically active.

According to the above, in a dielectric material the axes of the coordinate system in which ε is diagonal are real in space and can be labeled $$\hat{x},\hat{y},\text{ and }\hat{z}$$. Noncrystalline materials are generally isotropic, for which the choice of the orthogonal coordinate axes $$\hat{x},\hat{y},\text{ and }\hat{z}$$ is arbitrary. In contrast, many crystalline materials that are useful for photonic device applications are anisotropic. For any given anisotropic crystal, there is a unique set of coordinate axes for ε to be diagonal. These unique $$\hat{x},\hat{y},\text{ and }\hat{z}$$ coordinate axes are called the principal dielectric axes, or simply the principal axes, of the crystal. In the coordinate system defined by these principal axes, ε is diagonalized with eigenvalues εx, εy, and εz. The components of D and E along these axes have the following simple relations:

$\tag{109}D_x=\epsilon_xE_x,\qquad D_y=\epsilon_yE_y,\qquad D_z=\epsilon_zE_z,$

The values εx0, εy0, and εz0 are the eigenvalues of the dielectric constant tensorε0, and are called the principal dielectric constants. They define three principal indices of refraction:

$\tag{110}n_x=\sqrt{\frac{\epsilon_x}{\epsilon_0}},\qquad n_y=\sqrt{\frac{\epsilon_y}{\epsilon_0}},\qquad n_z=\sqrt{\frac{\epsilon_z}{\epsilon_0}}$

Note that when ε is diagonalized, χ is also diagonalized along the same principal axes with corresponding principal dielectric susceptibilities, χx, χy, and χz. The principal dielectric susceptibilities of any lossless dielectric material always have positive values; therefore, the principal dielectric constants of such a material are always larger than unity.

Because D ⊥ k due to the fact that  · D = 0, there is no D component along the direction of wave propagation. In general, D can be decomposed into two mutually orthogonal components, each of which is also orthogonal to k. In an anisotropic crystal, these two components generally have different indices of refraction, and thus different propagation constants. This phenomenon is called birefringence. Such a crystal is a birefringent crystal.

Example

At an optical wavelength of 1 µm, the permittivity tensor of the KDP crystal represented in a rectangular coordinate system defined by $$\hat{x}_1,\,\hat{x}_2,\text{ and }\hat{x}_3$$ is found to be

$\pmb{\epsilon}=\epsilon_0\begin{bmatrix}2.28&0&0\\0&2.25&-0.05196\\0&-0.05196&2.19\end{bmatrix}$

Find the principal axes and the corresponding principal indices for this crystal.

Note that ε is represented by a symmetric matrix because KDP is a nonmagnetic dielectric crystal. Diagonalization of this matrix yields the following eigenvalues and corresponding eigenvectors:

\begin{align}&\epsilon_x=2.28\epsilon_0,\hat{x}=\hat{x}_1\\&\epsilon_y=2.28\epsilon_0,\hat{y}=0.866\hat{x}_2-0.500\hat{x}_3\\&\epsilon_z=2.16\epsilon_0,\hat{z}=0.500\hat{x}_2+0.866\hat{x}_3\\\end{align}

Therefore, the principal axes of the crystal are $$\hat{x},\,\hat{y},\text{ and }\hat{z}$$, given above, and the principal indices of refraction are

\begin{align}&n_x=\sqrt{2.28}=1.51\\&n_y=\sqrt{2.28}=1.51\\&n_z=\sqrt{2.16}=1.47\end{align}

Index Ellipsoid

The inverse of the dielectric constant tensor mentioned above is the relative impermeability tensor:

$\tag{111}\pmb{\eta}=[\eta_\text{ij}]=\left(\frac{\pmb{\epsilon}}{\epsilon_0}\right)^{-1}$

where i and j are spatial coordinate indices. In a general rectangular coordinate system (x1, x2, x3), the ellipsoid defined by

$\tag{112}\sum_{\text{i, j}}x_\text{i}\eta_{\text{ij}}x_\text{j}=1$

is called the index ellipsoid or the optical indicatrix. In a nonmagnetic dielectric medium, η is a symmetric tensor, i.e., ηij = ηji, because ε is symmetric. Therefore, (112) can be written as

$\tag{113}\eta_{11}x_1^2+\eta_{22}x_2^2+\eta_{33}x_3^2+2\eta_{23}x_2x_3+2\eta_{31}x_3x_1+2\eta_{12}x_1x_2=1$

This equation is usually written as

$\tag{114}\eta_1x_1^2+\eta_2x_2^2+\eta_3x_3^2+2\eta_4x_2x_3+2\eta_5x_3x_1+2\eta_6x_1x_2=1$

using the following index contraction rule to reduce the double index ij of ηij to the single index α of ηα : (115)

The index ellipsoid equation is invariant with respect to coordinate rotation. When a coordinate system with axes aligned with the principal dielectric axes of the crystal is chosen, ε is diagonalized. Thus the tensor η is also diagonalized with the following eigenvalues:

$\tag{116}\eta_x=\frac{\epsilon_0}{\epsilon_x}=\frac{1}{n_x^2},\qquad\eta_y=\frac{\epsilon_0}{\epsilon_y}=\frac{1}{n_y^2},\qquad\eta_z=\frac{\epsilon_0}{\epsilon_z}=\frac{1}{n_z^2}$

In this coordinate system, the index ellipsoid takes the following simple form:

$\tag{117}\frac{x^2}{n_x^2}+\frac{y^2}{n_y^2}+\frac{z^2}{n_z^2}=1$

Comparing (117) with (114), we find that the terms containing cross products of different coordinates are eliminated when the coordinate system of the principal dielectric axes is used. The principal axes of the index ellipsoid now coincide with the principal dielectric axes of the crystal, and the principal indices of refraction of the crystal are given by the semiaxes of the index ellipsoid. This is illustrated in figure 8 below. Therefore, a coordinate transformation by rotation to eliminate cross-product terms in the index ellipsoid equation is equivalent to diagonalization of the ε tensor. The principal dielectric axes and their corresponding principal indices of refraction can be found through either approach. Between the two approaches, however, diagonalization of the ε tensor is better because it is more systematic and is easier to carry out. Example

Find the principal axes and their corresponding principal indices for the KDP crystal given in the previous example by using the index ellipsoid instead of diagonalizing the ε tensor as done in the previous example. Compare the two approaches.

The relative impermeability tensor in the (x1, x2, x3) coordinate system can be found by inverting the ε tensor:

$\pmb{\eta}=\left(\frac{\pmb{\epsilon}}{\epsilon_0}\right)^{-1}=\left[\begin{matrix}2.28&0&0\\0&2.25&-0.05196\\0&-0.05196&2.19 \end{matrix}\right]^{-1}\approx\begin{bmatrix}\frac{1}{2.28}&0&0\\0&\frac{1}{2.25}&0.01055\\0&0.01055&\frac{1}{2.19} \end{bmatrix}$

In the (x1, x2, x3) coordinate system, the index ellipsoid is thus described by the following equation:

$\frac{x_1^2}{2.28}+\frac{x_2^2}{2.25}+\frac{x_3^2}{2.19}+0.0211x_2x_3=1$

To find the principal axes and their principal indices of refraction, the cross-product term has to be eliminated by rotating the coordinates. From the previous example, we know that this can be done by taking

$x_1=x,\qquad x_2=0.866y+0.500z,\qquad x_3=-0.500y+0.866z$

Substitution of these relations into the above index ellipsoid equation transforms it into the following equation for the index ellipsoid in the (x, y, z) coordinate system:

$\frac{x^2}{2.28}+\frac{y^2}{2.28}+\frac{z^2}{2.16}=1$

Thus the principal indices are

\begin{align}n_x=\sqrt{2.88}=1.51\\n_y=\sqrt{2.28}=1.51\\n_z=\sqrt{2.16}=1.47\end{align}

Comparing the two approaches illustrated in this example the the previous example, it is clear that they are equivalent to one another. It is also clear that the method of diagonalizing ε described in the previous example is more systematic and straightforward than that of eliminating the cross-product terms in the equation for the index ellipsoid, particularly when there is more than one cross-product term.

Propagation along a Principal Axis

We first consider the simple case when an optical wave propagates along one of the principal axes, say $$\hat{z}$$. Then the field can be decomposed into two normal modes, each of which is polarized along one of the other two principal axes, $$\hat{x}$$ or $$\hat{y}$$. We see from (109) and (110) that each field component along a principal axis has a characteristic index of refraction ni, meaning that it has a characteristic propagation constant of ki = niω/c, which is determined by the polarization of the field but not by the direction of wave propagation. For a wave propagating along $$\hat{z}$$, the electric field can be expressed as

\tag{118}\begin{align}\mathbf{E}&=\hat{x}\mathcal{E}_xe^{\text{i}k^xz-\text{i}\omega t}+\hat{y}\mathcal{E}_ye^{\text{i}k^yz-\text{i}\omega t}\\&=\left[\hat{x}\mathcal{E}_x+\hat{y}\mathcal{E}_ye^{\text{i}(k^y-k^x)z}\right]e^{\text{i}k^xz-\text{i}\omega t}\end{align}

Because the wave propagates in the z direction, the wavevectors are $$\mathbf{k}^x=k^x\hat{z}$$ for the x-polarized field and $$\mathbf{k}^y=k^y\hat{z}$$ for the y-polarized field. Note that kx = nxω/c and k= nyω/c are the propagation constants of the x- and y-polarized fields, respectively, not to be confused with the x and y components of a wavevector k, which are normally expressed as kx and ky.

The field expressed in (108) has the following propagation characteristics.

1. If it is originally linearly polarized along one of the principal axes, it remains linearly polarized in the same direction.
2. If it is originally linearly polarized at an angle $$\theta=\tan^{-1}(\mathcal{E}_y/\mathcal{E}_x)$$ with respect to the x axis, its polarization state varies periodically along z with a period of 2π/|ky - kx|. In general, its polarization follows a sequence of variations from linear to elliptical to linear in the first half-period and then reverses the sequence back to linear in the second half-period. At the half-period position, its is linearly polarized at an angle θ on the other side of the x axis. Thus the polarization is rotated 2θ from the original direction. This is shown in figure 9(a) below. In the special case when θ = 45°, the wave is circularly polarized at the quarter-period point and is linearly polarized with its polarization rotated by 90° from the original direction at the half-period point. This is shown in figure 9(b) below. These characteristics have very useful applications. A plate of an anisotropic material that has a quarter-period thickness of

$\tag{119}l_{\lambda/4}=\frac{1}{4}\cdot\frac{2\pi}{|k^y-k^x|}=\frac{\lambda}{4|n_y-n_x|}$

is called a quarter-wave plate. It can be used to convert a linearly polarized wave to circular or elliptic polarization, and vice versa. A plate of thickness 3lλ/4 or 5lλ/4 or any odd integral multiple of lλ/4 also has the same function. In contrast, a plate of a half-period thickness of

$\tag{120}l_{\lambda/2}=\frac{\lambda}{2|n_y-n_x|}$

is called a half-wave plate. It can be used to rotate the polarization of a linearly polarized wave by any angular amount by properly choosing the angle θ between the incident polarization with respect to the principal axis $$\hat{x}$$, or $$\hat{y}$$, of the crystal. A plate of a thickness that is any odd integral multiple of lλ/2 has the same function.

Note that though the output from a quarter-wave or half-wave plate can be linearly polarized, the wave plates are not polarizers. They are based on different principles and have completely different functions.

For the quarter-wave and half-wave plates discussed here, nx ≠ ny. Between the two crystal axes $$\hat{x}$$ and $$\hat{y}$$, the one with the smaller index is called the fast axis while the other, with the larger index, is the slow axis.

Example

KDP can be used to make quarter-wave and half-wave plates. Find the thicknesses of the quarter-wave and half-wave plates made of KDP for 1µm wavelength.

From the previous example, we know that nx = ny = 1.51 and nz = 1.47 for KDP at 1µm wavelength. Because nx = ny, we cannot use nx and ny to make a wave plate that allows the beam to propagate in the z direction. Instead, the beam can propagate in any direction on the xy plane so that the difference between nz and nx = ny can be used for the function of a wave plate. Assuming that the wave propagates in the x direction, then the thickness of a quarter-wave plate for λ = 1µm is

$l_{\lambda/4}=\frac{\lambda}{4|n_y-n_z|}=\frac{1\,\mu\text{m}}{4\times|1.51-1.47|}=6.25\,\mu\text{m}$

A quarter-wave plate at 1µm wavelength can have a thickness of any odd integral multiple, such as 18.75 µm, 31.25 µm, ..., of 6.25 µm.

A half-wave plate for the 1 µm wavelength has a thickness of

$l_{\lambda/2}=\frac{\lambda}{2|n_y-n_z|}=\frac{1\,\mu\text{m}}{2\times|1.51-1.47|}=12.5\,\mu\text{m}$

A plate of a thickness that is an odd multiple, such as 37.5 µm, 62.5 µm, ..., of 12.5 µm also functions as a half-wave plate at 1 µm wavelength. For these wave plates, $$\hat{z}$$ is the fast axis and $$\hat{y}$$ is the slow axis because nz < ny.

Optical Axes

The state of polarization of an optical wave generally varies along its path of propagation through an anisotropic crystal unless it is linearly polarized in the direction of a principal axis. However, in an anisotropic crystal with nx = ny ≠ nz, a wave propagating in the z direction does not see the anisotropy of the crystal because in this situation the x and y components of the field have the same propagation constant. This wave will maintain its original polarization as it propagates through the crystal. Evidently, this is true only for propagation along the z axis in such a crystal. Such a unique axis in a crystal along which an optical wave can propagate with an index of a refraction that is independent of its polarization direction is called the optical axis of the crystal.

For an anisotropic crystal that has only one distinctive principal index among its three principal indices, these is only one optical axis, which coincides with the axis of the distinctive principal index of refraction. Such a crystal is called a uniaxial crystal.  It is customary to assign $$\hat{z}$$ to this unique principal axis. The identical principal indices of refraction are called the ordinary index, no, and the distinctive index of refraction is called the extraordinary index, ne. Thus, nx = ny = no and nz = ne. The crystal is called positive uniaxial if ne > no and is negative uniaxial if ne < no.

For a crystal that has three distinct principal indices of refraction, there are two optical axes, neither of which coincides with any one of the principal axes. Such a crystal is called a biaxial crystal because of the existence of two optical axes.

Ordinary and Extraordinary Waves

When an optical wave propagates in a direction other than that along an optical axis, the index of refraction depends on the direction of its polarization. In this situation, there exist two normal modes of linearly polarized waves, each of which sees a unique index of refraction. One of them is the polarization perpendicular to the optical axis. This normal mode is called the ordinary wave. We use $$\hat{e}_o$$ to indicate its direction of polarization. The other normal mode is clearly one that is perpendicular to $$\hat{e}_o$$ because the two normal-mode polarizations are orthogonal to each other. This normal mode is called the extraordinary wave, and its direction of polarization is indicated by $$\hat{e}_e$$.

Note that these are the directions of D rather than those of E. For the ordinary wave, $$\hat{e}_o\parallel\mathbf{D}_o\parallel\mathbf{E}_o$$. For the extraordinary wave, $$\hat{e}_e\parallel\mathbf{D}_e\nparallel\mathbf{E}_e$$ except when $$\hat{e}_e$$ is parallel to a principal axis. Both $$\hat{e}_o$$ and $$\hat{e}_e$$, being the unit vectors of Do and De, are perpendicular to the direction of wave propagation, $$\hat{k}$$. From this understanding, both $$\hat{e}_o$$ and $$\hat{e}_e$$ can be found if both $$\hat{k}$$ and the optical axis are known. For a uniaxial crystal with optical axis $$\hat{z}$$, this means that

$\tag{121}\hat{e}_o=\frac{1}{\sin\theta}\hat{k}\times\hat{z},\qquad\hat{e}_e=\hat{e}_o\times\hat{k}$

if the vector $$\hat{k}$$ is in a direction that is at an angle θ with respect to $$\hat{z}$$ and an angle φ with respect to the axis $$\hat{x}$$. Therefore, we have

$\tag{122}\hat{k}=\hat{x}\sin\theta\cos\phi+\hat{y}\sin\theta\sin\phi+\hat{z}\cos\theta$$\tag{123}\hat{e}_o=\hat{x}\sin\phi-\hat{y}\cos\phi$$\tag{124}\hat{e}_e=-\hat{x}\cos\theta\cos\phi-\hat{y}\cos\theta\sin\phi+\hat{z}\sin\theta$

The relationships among these vectors are illustrated in figure 10 below. The indices of refraction associated with the ordinary and extraordinary waves can be found by using the index ellipsoid given in (117), as is shown in figure 11 below. The intersection of the index ellipsoid and the plane normal to $$\hat{k}$$ at the origin of the ellipsoid defines an index ellipse. The principal axes of this index ellipse are in the directions of $$\hat{e}_o$$ and $$\hat{e}_e$$, and their half-lengths are the corresponding indices of refraction. For a uniaxial crystal, the index of refraction for the ordinary wave is simply no. The index of refraction for the extraordinary wave depends on the angle θ and is given by

$\tag{125}\frac{1}{n_e^2(\theta)}=\frac{\cos^2\theta}{n_o^2}+\frac{\sin^2\theta}{n_e^2}$

which can be seen from figure 11.

Because D is orthogonal to k and can be decomposed into Do and De components, we have

$\tag{126}\mathbf{D}=\hat{e}_o\mathcal{D}_oe^{\text{i}k_o\hat{k}\cdot\mathbf{r}-\text{i}\omega t}+\hat{e}_e\mathcal{D}_ee^{\text{i}k_e\hat{k}\cdot\mathbf{r}-\text{i}\omega t}$

where ko = noω/c and ke = ne(θ)ω/c.

In general, E cannot be written in the form of (126) because its longitudinal component along the wave propagation direction k does not vanish except when θ = 0° or 90°. We see that ne(0°) = no and ne(90°) = ne. The special case when the wave propagates along one of the principal axes discussed earlier belongs to one of these situations.

The normal-mode polarizations for an optical wave propagating in a biaxial crystal can be found following a similar, albeit more complicated, procedure.

Example

From the preceding three examples, we find that KDP is a uniaxial crystal with $$\hat{z}$$ being its optical axis because nx = ny ≠ nz. At 1 µm wavelength, we have no = 1.51 and ne =  1.47. KDP is negative uniaxial because no > ne. For an optical wave propagating in KDP along a direction $$\hat{k}$$ that makes an angle θ with respect to the optical axis $$\hat{z}$$, the refractive index for the extraordinary wave is a function of θ. For θ = 0°, ne(0°) = no = 1.51. For θ = 90°, ne(90°) = ne = 1.47. For 0° < θ < 90°, 1.47 < ne(θ) < 1.51. For example,

$n_e(30^{\circ})=\left(\frac{\cos^2{30^\circ}}{n_o^2}+\frac{\sin^2{30^\circ}}{n_e^2}\right)^{-1/2}=1.50$

$n_e(60^{\circ})=\left(\frac{\cos^2{60^\circ}}{n_o^2}+\frac{\sin^2{60^\circ}}{n_e^2}\right)^{-1/2}=1.48$

Spatial Beam Walk-off

Each of the normal modes has a well-defined propagation constant. Therefore, the fields of monochromatic ordinary and extraordinary waves in an anisotropic medium can be separately written in the form of (47) [as in the harmonic fields tutorial], with k = ko for ordinary wave and k = ke for the extraordinary wave.

By using (83), [from the propagation in an isotropic medium tutorial], Maxwell's equations for a normal mode, either ordinary or extraordinary, reduce to the following

$\tag{127}\mathbf{k}\times\mathbf{E}=\omega\mu_0\mathbf{H}$$\tag{128}\mathbf{k}\times\mathbf{H}=-\omega\mathbf{D}$$\tag{129}\mathbf{k}\cdot\mathbf{D}=0$$\tag{130}\mathbf{k}\cdot\mathbf{H}=0$

Note that because no ≠ ne, these relations apply to the ordinary and extraordinary normal mode separately with different values for k but not to a wave mixing the two modes.

At optical frequencies, B =  μ0H is also true in an anisotropic medium. Therefore, (127) and (130) have the same format as (86) and (89), [from the propagation in an isotropic medium tutorial], respectively.

Because (88) for a wave in an isotropic medium is now replaced by (129), we have D ⊥ k for both ordinary and extraordinary waves. For an ordinary wave, Eo ⊥ ko because Do || Eo. Therefore, the relationships shown in figure 12(a) below among the field vectors for an ordinary wave in an anisotropic medium are the same as those shown in figure 7, [from the propagation in an isotropic medium tutorial], for a wave in an isotropic medium.

However,  $$\mathbf{E}_e\not\perp\mathbf{k}_e$$ for an extraordinary wave in general, and Se is not necessarily parallel to ke because $$\mathbf{D}_e\not\perp\mathbf{E}_e$$  The only exception is when $$\hat{e}_e$$ is parallel to a principal axis. As a result, the direction of power flow, which is that of Se, is not the same as the direction of wavefront propagation, which is normal to the planes of constant phase and is that of ke. This is shown in figure 12(b) below together with the relationships among the directions of the field vectors.

Note that EeDeke, and Se lie in a plane normal to He because Be || He. Though (90) is still true according to (127), the relations between E and H in (92) are no longer valid for an extraordinary wave. If the electric field of an extraordinary wave is not parallel to a principal axis, its Poynting vector is not parallel to its propagation direction because Ee is not parallel to De. As a result, its energy flows away from the direction of its wavefront propagation. This phenomenon is known as spatial beam walk-off. If this characteristic appears in one of the two normal modes of an optical wave propagating in an anisotropic crystal, the optical wave will split into two beams of parallel wavevectors but separate, nonparallel traces of energy flow.

For simplicity, let us consider the propagation of an optical wave in a uniaxial crystal with $$\hat{k}$$, for both ordinary and extraordinary waves, at an angle θ with respect to the optical axis $$\hat{z}$$. Clearly, there is no wall-off for the ordinary wave because Eo || Do and So || $$\hat{k}$$. For the extraordinary wave, Se is not parallel to $$\hat{k}$$ but points in a direction at an angle ψe with respect to the optical axis. Figure 13(a) below shows the relationships among these vectors. The angle α between Se and $$\hat{k}$$, which is defined as α = ψe - θ, is called the walk-off angle of the extraordinary wave. Note that α is also the angle between Ee and De, as can be seen from figure 13(a).

Because neither Ee nor De is parallel to any principal axis, their relationship is found through their projections on the principal axes:

$D_z^e=n_e^2\epsilon_0E_z^e$$D_{xy}^e=n_o^2\epsilon_0E_{xy}^e$

Using these two relations and the definition of α in figure 12(b) and 13(a), it can be shown that the walk-off angle is given by

$\tag{131}\alpha=\psi_e-\theta=\tan^{-1}\left(\frac{n_o^2}{n_e^2}\tan\theta\right)-\theta$

If the crystal is positive uniaxial, α as defined in figure 13(a) is negative. This means that Se is between $$\hat{k}$$ and $$\hat{z}$$ for a positive uniaxial crystal. If the crystal is negative uniaxial, α is positive and $$\hat{k}$$ is between Se and $$\hat{z}$$. No walk-off appears if an optical wave propagates along any of the principal axes of a crystal.

A birefringent crystal can be used to construct a very simple polarizing beam splitter by taking advantage of the walk-off phenomenon. For such a purpose, a uniaxial crystal can be cut into a plate whose surfaces are at an oblique angle with respect to the optical axis, as is shown in figure 13(b). When an optical wave is normally incident upon the plate, it splits into ordinary and extraordinary waves in the crystal if its original polarization contains components of both polarizations. The extraordinary wave is then separated from the ordinary wave because of spatial walk-off, creating two orthogonally polarized beams. However, because of normal incidence, both ke and ko are parallel to the direction of $$\hat{k}$$ although they have different magnitudes. When both beams reach the other side of the plate, they are separated by a distance of d = l tan α, where l is the thickness of the plate. After leaving the plate, the two spatially separated beams propagate parallel to each other along the same direction $$\hat{k}$$ because the directions of their wavevectors have not changed, as is shown in figure 13(b).

Example

Find the spatial walk-off angle at 1 µm wavelength at a few representative propagation directions in KDP. Design a polarizing beam splitter at this wavelength using a KDP crystal.

For a KDP crystal,  no = 1.51 and ne = 1.47 at 1 µm wavelength. The spatial walk-off angle α of an extraordinary wave is a function of the angle θ between the wave propagation direction $$\hat{k}$$ and the optical axis $$\hat{z}$$ of the crystal. For example,

$\alpha=\tan^{-1}\left(\frac{1.51^2}{1.47^2}\tan30^\circ\right)-30^\circ=1.35^\circ,\qquad\text{for }\theta=30^\circ$$\alpha=\tan^{-1}\left(\frac{1.51^2}{1.47^2}\tan45^\circ\right)-45^\circ=1.54^\circ,\qquad\text{for }\theta=45^\circ$$\alpha=\tan^{-1}\left(\frac{1.51^2}{1.47^2}\tan60^\circ\right)-60^\circ=1.31^\circ,\qquad\text{for }\theta=60^\circ$

From these numerical examples, we find that the walk-off angle does not vary monotonically with θ.

A polarizing beam splitter can be made by cutting a KDP crystal at an angle, such as 45°, with respect to its optical axis for a parallel plate of thickness l. A beam at 1 µm wavelength that consists of a mix of extraordinary and ordinary polarizations is normally incident on the plate for θ = 45° and α = 1.54°. Because the ordinary wave does not have walk-off, the Poynting vectors of the extraordinary and ordinary components of the beam separate at an angle of α = 1.54°. If a minimum spatial separation of d = 100 µm between the extraordinary and ordinary components is desired on the exit surface of the KDP plate, the minimum thickness of the plate has to be l > d/tan α = 3.7 mm.

Optical Anisotropy and Crystal Symmetry

The optical anisotropy of a crystal depends on its structural symmetry. Crystals are classified into seven systems according to their symmetry. The linear optical properties of these seven systems are summarized in table 2 below. Some important remarks regarding the relation between the optical properties and the structural symmetry of a crystal are made:

1. A cubic crystal need not have an isotropic structure although its linear optical properties are isotropic. For example, most III-V semiconductors, such as GaAs, InP, InAs, AlAs, etc., are cubic crystals with isotropic linear optical properties. Nevertheless, they have well-defined crystal axes, $$\hat{a}$$, $$\hat{b}$$, and $$\hat{c}$$. They are also polar semiconductors, which have anisotropic nonlinear optical properties.
2. Although the principal axes may coincide with the crystal axes in certain crystals, they are not the same concept. The crystal axes, denoted by $$\hat{a}$$, $$\hat{b}$$, and $$\hat{c}$$ are defined by the structural symmetry of a crystal, whereas the principal axes, denoted by $$\hat{x}$$, $$\hat{y}$$, and $$\hat{z}$$, are determined by the symmetry of ε. The principal axes of a crystal are orthogonal to one another, but the crystal axes are not necessarily so. The next part continues with the Gaussian Beam tutorial