2Polarized Light

2.1The Description of Polarized Light

This chapter starts with the description of monochromatic plane waves propagating in arbitrary directions and the description of their electric and magnetic fields. Table 2.1 lists several common and useful methods for the description of polarized light.1–3 In this chapter, first the polarization vector is used to describe the polarization state of plane waves. Then, Jones vectors, which describe monochromatic plane waves propagating along the z-axis, are treated in detail, analyzing the polarization state and ellipse, phase, polarized flux components, and orthogonality of plane waves. Such a monochromatic wave can be generated with a laser.1 Finally, considering the polarization of sources, two models for the polarization of light beams entering optical systems are defined, the dipole spherical wave and the double pole spherical wave. Chapter 3 continues with the description of incoherent light and polychromatic light, light that is not monochromatic, using the Stokes parameters and the Poincaré sphere, a graphical representation of the polarization state.

Table 2.1Common Methods for Polarized Light Calculations

Light Representation	Properties	Polarization Element Representation
Jones vector	Monochromatic plane wave along z-axis Two complex elements	Jones matrix
Polarization vector	Monochromatic plane wave in arbitrary direction Three complex elements	Polarization ray tracing matrix
Stokes parameters	Incoherent light along z-axis Four real elements	Mueller matrix

This book principally uses Jones vectors for plane waves and polarization vectors (three-dimensional electric field vectors) on spherical (or nearly spherical) wavefronts for describing polarized light and uses Stokes parameters where they have advantages.

2.2The Polarization Vector

Light is a transverse electromagnetic wave, an oscillating electric and magnetic field propagating in vacuum at the speed of light.4–6 The simplest light wave is the monochromatic plane wave with a flat wavefront, representing a collimated beam of light, as shown in Figure 2.1 for a linearly polarized electromagnetic transverse wave. Consider a monochromatic plane wave with wavelength λ, propagating in the direction of unit propagation vector k, with an angular frequency ω in radians per second.2 The electric field E (r, t) of the monochromatic plane wave in space, r, and time, t, is

Figure 2.1The electric field vectors (red) and magnetic field vectors (blue) for a linearly polarized electromagnetic wave. Three wavefronts, surfaces of constant phase are shown, separated by one wavelength.

$$\mathbf{E}(\mathbf{r},t)=ℛe\left[\mathbf{E}{e}^{i\left(\frac{2\pi }{\lambda }k\cdot r-\omega t-{\varphi }_o\right)}\right]=ℛ\mathit{e}\left[\left(\begin{array}{l}{E}_x\\ E_y\\ E_z\end{array}\right)e^{i\left(\frac{2\pi }{\lambda }k\cdot r-\omega t-{\varphi }_o\right)}\right]=ℛ\mathit{e}\left[e^{i\left(\frac{2\pi }{\lambda }k\cdot r-\omega t-{\varphi }_o\right)}\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\\ A_z{e}^{-i{\varphi }_x}\end{array}\right)\right].$$2.1

2.1

The polarization vector E describes the polarization state. E is the complex amplitude with units of volts per meter. The three field components of E are shown in three different forms, from left to right in Equation 2.1, where $(\begin{array}{ccc}{E}_x& E_y& E_z\end{array})$ are the complex amplitude components along the three coordinate axes x, y, and z. Ax, Ay, and Az are the magnitudes of the complex amplitudes and ϕx, ϕy, and ϕz are their phases. The surfaces of constant phase fronts are shown as blue planes separated by λ in Figure 2.1.

E is an electric field vector, but since it fully characterizes the polarization ellipse in three dimensions, E is also called the polarization vector. The propagation vector7k

$$\mathbf{k}=(k_x,k_y,k_z),\left|\mathbf{k}\right|=1,$$2.2

2.2

is normalized and describes the direction of propagation.3k points along the light rays associated with the wavefront. The three components (kx, ky, kz) are the direction cosines, the cosines of the angles between k and the x-, y-, and z-axes. The vector r identifies position in a global coordinate system,

$$\mathbf{r}=(r_x,r_y,r_z).$$2.3

2.3

At the origin of the chosen coordinate system, r = (0, 0, 0) and t = 0, the E-field is

$$\mathbf{E}=\left(\begin{array}{l}{E}_x\\ E_y\\ E_z\end{array}\right).$$2.4

2.4

For propagation in isotropic media, such as vacuum, air, water, and glass, the polarization vector is orthogonal to the propagation vector, so their dot product is zero,

$$\mathbf{E}\cdot \mathbf{k}=\left(\begin{array}{l}{E}_x\\ E_y\\ E_z\end{array}\right)\cdot \left(\begin{array}{l}{k}_x\\ k_y\\ k_z\end{array}\right)=0.$$2.5

2.5

The term $\frac{2\pi }{\lambda }\mathbf{k}\cdot \mathbf{r}-\omega t-{\varphi }_o$ is the phase of the plane wave, which is specified for all space and time, and is usually expressed in radians, sometimes in degrees. Phase specifies a location within a periodic phenomenon. ϕo is the absolute phase of the light, which is the phase of the light at the origin (r = (0, 0, 0) and t = 0). Phase can also be expressed in waves, Φ,

$$\Phi =\frac{\mathbf{k}\cdot \mathbf{r}}{\lambda }-\frac{\omega t+{\varphi }_o}{2\pi }.$$2.6

2.6

Φ is constant on a surface of constant phase, a plane wavefront in this case. The phase can be separated into an integer part and a fractional part. The integer part of Φ provides a method to number wavefronts, such as wavefront −2, −1, 0, 1, 2, …. The fractional part of Φ identifies where within a cosinusoidal wave a point in r and t is located, or for polarized light, where on the polarization ellipse the electric field is located. The phases of the individual x, y, and z components of the field in Equation 2.1 are ϕx + ϕo, ϕy + ϕo, and ϕz + ϕo.

2.3Properties of the Polarization Vector

E is a complex vector, so each complex number can be expressed in polar coordinate form,

$$\mathbf{E}=\left(\begin{array}{l}{E}_x\\ E_y\\ E_z\end{array}\right)=\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\\ A_z{e}^{-i{\varphi }_z}\end{array}\right).$$2.7

2.7

Ax, Ay, and Az are the real amplitudes, the maximum value each component reaches during a period. The phases ϕx, ϕy, and ϕz are provided with minus signs due to the decreasing phase sign convention, $\left(\frac{2\pi }{\lambda }\mathbf{k}\cdot \mathbf{r}-\omega t\right)$, see Section 2.17. Multiplying E by the temporal phase e−iωt and taking the real part at r = (0, 0, 0) yields E(t), the polarization ellipse, oriented in three dimensions,

$$\mathbf{E}(t)=\mathit{ℛ}\mathit{e}\left[e^{-i\omega t}\left(\begin{array}{l}{E}_x\\ E_y\\ E_z\end{array}\right)\right]=\mathit{ℛ}\mathit{e}\left[e^{-i\omega t}\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\\ A_z{e}^{-i{\varphi }_z}\end{array}\right)\right].$$2.8

2.8

Figure 2.2 shows the three-dimensional polarization ellipse for left circularly polarized light.1 The symbol E will be used for both Jones vectors and polarization vectors, since both are electric field amplitudes. The electric and magnetic vectors must oscillate only in the plane perpendicular to the propagation vector k, the transverse plane, shown in Figure 2.3.

Figure 2.2(Left) A left circularly polarized polarization ellipse (red) shown in 3D with propagation vector k = (6, 2, 9)/11 (black). (Right) Ellipse drawn on the unit sphere with k emerging normal to the sphere. Two basis vectors defining the transverse plane are shown in blue.

Figure 2.3With respect to a propagation vector k, the transverse plane contains the two orthogonal directions perpendicular to k with basis vectors b1 and b2, which can be chosen in arbitrary directions.

For light to be linearly polarized, all three E-field components should pass through zero at the same time, twice per period. This requires that the three phases in Equation 2.8 differ from each other by either 0 or π,

$${\varphi }_x-{\varphi }_y=0\text{or}\pi ,{\varphi }_x-{\varphi }_z=0\text{or}\pm \pi .$$2.9

2.9

For circularly polarized light, the magnitude of the electric field vector is constant. This can be tested by evaluating the magnitude of the real part of E at several times, for example, at t = 0 and a quarter period later (t = π/(2ω)) when the imaginary part of E yields a real E-field. Hence, E describes circularly polarized light when these two vectors E(0) and E(π/2) are orthogonal and equal in magnitude,

$$\mathit{ℛ}\mathit{e}(E)\cdot \mathcal{J}\mathit{m}(E)=0\text{and}|\mathit{ℛ}\mathit{e}(E)|=|\mathcal{J}\mathit{m}(E)|.$$2.10

2.10

If the light is not linearly or circularly polarized, it is elliptically polarized light. To determine the helicity of the polarization state, right or left, one can evaluate the cross product of the electric field vector at time t = 0 and t = π/(2ω) = T/4. If the cross product is antiparallel to the propagation vector, the light’s helicity is right handed. If the cross product is parallel to the propagation vector, the light’s helicity is left handed. Thus, the helicity (handedness) is determined by the sign of

$$[\mathbf{E}(0)\times \mathbf{E}(\pi \text{/}(2\omega ))]\cdot \mathbf{k}\text{.}$$2.11

2.11

If Equation 2.11 is positive, the electric field is rotating counterclockwise (i.e., left circular), and if it is negative, the electric field is rotating clockwise (i.e., right circular). To determine the ellipticity and orientation of the major axis, consult Section 2.16.

Math Tip 2.1Adjoint of a Vector

The dagger superscript† indicates the adjoint of a vector, the complex conjugate of the transpose. For example, the adjoint of $\mathbf{E}=\left(\begin{array}{c}{A}_x{e}^{-i{\varphi }_x}{A}_y{e}^{-i{\varphi }_y}{A}_z{e}^{-i{\varphi }_z}\end{array}\right)$ is

$${\mathbf{E}}^{†}=\left(\begin{array}{l}{A}_x{e}^{i{\varphi }_x}{A}_y{e}^{i{\varphi }_y}{A}_z{e}^{i{\varphi }_z}\end{array}\right).$$2.12

2.12

The adjoint provides a shortcut to get the magnitude squared of a complex vector ${\left|\mathbf{E}\right|}^2$,

$${\left|\mathbf{E}\right|}^2={\mathbf{E}}^{†}\cdot \mathbf{E}=\left(\begin{array}{ll}{A}_x{e}^{i{\varphi }_x}& A_y{e}^{i{\varphi }_y}{A}_z{e}^{i{\varphi }_z}\end{array}\right)\cdot \left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\\ A_z{e}^{-i{\varphi }_z}\end{array}\right)=A_x^2+A_y^2+A_z^2.$$2.13

2.13

2.4Propagation in Isotropic Media

The description of E above assumed propagation in vacuum. In glass, water, and other media, the velocity of light, V, is reduced and the medium is characterized by its refractive index n,

$$n=\frac{c}{V}.$$2.14

2.14

The light is slowed down because as its E- and H-fields propagate through transparent materials, the fields induce motion in the atom’s charges at ω, and this charge motion gives rise to light in the same direction at the same frequency but slightly delayed. Thus, the refractive index characterizes the strength of the interaction of light of a particular frequency with a material. A material is isotropic if the strength of this interaction is independent of the direction of E and H; that is, n is the same for all propagation directions and polarization states. Anisotropic materials such as calcite and quartz are discussed in Chapter 19 (see discussion on anisotropic materials).

2.5Magnetic Field, Flux, and Polarized Flux

Light is a transverse electromagnetic wave. The light’s magnetic fieldH (r, t) oscillates perpendicular to and in phase with the electric field. For light propagating in vacuum, the associated magnetic field is

$$\mathbf{H}(\mathbf{r},t)=\mathit{ℛ}\mathit{e}\left[\mathbf{H}{e}^{i\left(\frac{2\pi }{\lambda }\mathbf{k}\cdot \mathbf{r}-\omega t-{\varphi }_o\right)}\right]=\mathit{ℛ}\mathit{e}\left[\left(\begin{array}{l}{H}_x\\ H_y\\ H_z\end{array}\right)e^{i\left(\frac{2\pi }{\lambda }\mathbf{k}\cdot \mathbf{r}-\omega t-{\varphi }_o\right)}\right],$$2.15

2.15

which is specified in units of amperes per meter. Figure 2.4 shows the electric and magnetic fields as a function of time for several polarization states.

Figure 2.4Electric (red) and magnetic (blue) fields rendered in three dimensions for three polarization states. (Left) Vertical linearly polarized light. (Middle) 45° linearly polarized light. (Right) Right circularly polarized light.

The Poynting vectorS describes the instantaneous flow of energy of an electromagnetic wave in watts per meter squared,2

$$\mathbf{S}(\mathbf{r},t)=\mathbf{E}(\mathbf{r},t)\times \mathbf{H}(\mathbf{r},t).$$2.16

2.16

The Poynting vector has the same units as irradiance, energy per area per second, or kilogram per second.3,8 For linearly polarized light, the Poynting vector oscillates twice per period; for circularly polarized light, the Poynting vector is constant. Optical detectors make measurements that average over many periods of the light, because detectors cannot respond at optical frequencies of hundreds of terahertz. Thus, optical detectors measure a time-averaged flow of energy, called the flux or irradiance P. The flux of a beam of light is the time average of the energy crossing a unit area perpendicular to the direction of energy flow per unit time. The flux is often referred to as the intensity, but in radiometry, intensity specifically refers to the watts per steradian from a point source. Nonetheless, this use of the term intensity is widespread and well understood, even if it doesn’t follow official definitions.9,10 The irradiance of a light beam, P, is the power per unit area transported by the light’s electromagnetic fields and is measured in watts per meter squared. The flux P of our monochromatic plane wave Equation 2.1 is

$$P=\frac{{\epsilon }_{0}c}{2}{\mathbf{E}}^{†}\cdot \mathbf{E}=\frac{{\epsilon }_{0}c}{2}\left(\begin{array}{lll}{A}_x{e}^{i{\varphi }_x}& A_y{e}^{i{\varphi }_y}& A_z{e}^{i{\varphi }_z}\end{array}\right)\cdot \left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\\ A_z{e}^{-i{\varphi }_z}\end{array}\right)=\frac{{\epsilon }_{0}c}{2}\left(A_x^2+A_y^2+A_z^2\right).$$2.17

2.17

Frequently, the constant ε0c/2 is dropped and calculations such as Jones vector problem sets are worked in “normalized flux” E† · E. The ε0c/2 is only necessary to work in MKS units and produces answers in radiometric units.

2.6Jones Vectors

In many polarization problems, light propagation is often restricted to a single direction. When light propagates through a sequence of sheet polarizers and sheet retarders, the light propagates in a single direction. Our convention is to propagate light along the z-axis, from negative toward positive. Such light is frequently modeled as a plane wave with a propagation vector

$$\mathbf{k}=(0,0,1).$$2.18

2.18

Such light has no E-field component along the direction of propagation, since E always oscillates in the transverse plane of k for all isotropic media. In this configuration, the light’s z electric field component Ez is zero, so it can be dropped and the monochromatic plane wave’s description is thus simplified to only the x and y electric field components. The two-element vector is the Jones vector.11,13 The Jones vector E has two complex elements and represents the polarization state of this z-propagating monochromatic light field as

$$\mathbf{E}=\left(\begin{array}{l}{E}_x\\ E_y\end{array}\right)=\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\end{array}\right).$$2.19

2.19

The Jones vector has four degrees of freedom that define the polarization state of the wave, two arbitrary electric field amplitudes, Ax and Ay, and two arbitrary phases, ϕx and ϕy. The units of the Jones vector are the same as the units of the electric field, volts per meter. A normalized Jones vector$\widehat{\mathbf{E}}$, indicated with a caret above, has been scaled to a magnitude of one,

$$\left|\widehat{\mathbf{E}}\right|={\mathbf{E}}^{†}\cdot \mathbf{E}=A_x^2+A_y^2=1.$$2.20

2.20

Here, “normalized” polarized fluxes mean the flux is one. Table 2.2 lists the normalized Jones vectors for six polarization states. These six states are considered as our basis polarization states, those that most frequently occur in defining polarization element properties.

Table 2.2Normalized Jones Vectors for the Six Basis Polarization States

Polarization State	Jones Vector	Polarization Ellipse
Horizontal, 0°	$\mathbf{H}=\left(\begin{array}{c}1\\ 0\end{array}\right)$
Vertical, 90°	$\mathbf{V}=\left(\begin{array}{c}0\\ 1\end{array}\right)$
45°	$\mathbf{45}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$
135°	$\mathbf{135}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -1\end{array}\right)$
Right circular	$\mathbf{R}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -i\end{array}\right)$
Left circular	$\mathbf{L}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ i\end{array}\right)$

The orientation is measured from horizontal x-axis and increasing counterclockwise.

By definition, monochromatic light is periodic with a single frequency. Because of this periodicity, its electric field vector traces a simple figure, the polarization ellipse, the most iconic representation of polarized light. Looking into the beam, the polarization ellipse is defined by the tip of the electric field vector, which traces an ellipse as a function of time. To generate the polarization ellipse, the Jones vector is multiplied by the temporal phase e−iωt and the time is varied over one period, yielding

$$\mathbf{E}(t)=\mathit{ℛ}\mathit{e}\left[e^{-i\omega t}\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\end{array}\right)\right],$$2.21

2.21

as shown in Figure 2.5. Overall, the electric field exerts a much larger force than the magnetic field in most light–matter interactions; thus, the electric field, not the magnetic field, is used to define the direction of polarization and the polarization ellipse.1 This is the essence of polarization, the transverse nature of E. One important task of the polarization calculi, the Jones calculus, the Mueller calculus, and the polarization ray tracing calculus, is to describe these transverse properties and the associated polarization state transformations in many different types of optical systems.

Figure 2.5(a) The electric field vectors for an elliptical polarization state over one period. (b) The electric field vector as a function of time during one period. (c) The electric field vectors are shown with their bounding ellipse as a function of time. (d) Usually, just the ellipse is drawn to indicate the polarization state. This is the polarization ellipse. The location of the arrow indicates the absolute phase of the wave. This polarization ellipse has a left circularly polarized component because it circulates counterclockwise when looking into the beam.

The two complex Jones vector components, Ex and Ey, can be expressed in polar coordinate form with an amplitude A and phase ϕ, or (on the right) with the x-phase ϕx factored out,

$$\mathbf{E}=\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\end{array}\right)=e^{-i{\varphi }_x}\left(\begin{array}{l}{A}_x\\ A_y{e}^{-i\varphi }\end{array}\right).$$2.22

2.22

In this form, the relative phase ϕ, the phase difference between the x and y field components, is evident,

$$\varphi ={\varphi }_y-{\varphi }_x.$$2.23

2.23

Example 2.1A Family of Polarization Ellipses

Figure 2.6 shows the polarization ellipses for a family of Jones vectors

$$\mathbf{E}=\left(\begin{array}{l}1\\ 0.5e^{-i\varphi }\end{array}\right).$$2.24

2.24

Figure 2.6Polarization ellipses for Ex = 1 and Ey = ½ for several relative phases.

The ellipses vary as a function of relative phase ϕ. Here the y-amplitude is set to one-half the x-amplitude.

The polarization ellipse fits inside a rectangular box twice the size of its amplitudes Ax and Ay. The relative phase determines how the ellipse fits in the box:

1. When ϕ = 0° or 180°, the state is linear along one of the diagonals.

2. When ϕ = ±90°, the ellipse touches the middle of all four sides and has the maximum ellipticity within this set.

3. For 0° < ϕ < 180°, the ellipse is right elliptically polarized.

4. For −180° < ϕ < 0° or 180° < ϕ < 360°, the ellipse is left elliptically polarized.

The location of the ellipse arrow indicates the absolute phase of the polarization state. In this set, Ex was chosen to equal to one at t = 0; hence, the arrow is at the +x side. Multiplying the Jones vector by a complex phase e−iϕ advances the arrow to another point on the ellipse.

Normalizing a Jones vector is the operation of adjusting the amplitude; thus, the normalized flux P is one. For the Jones vector of Equation 2.22, the normalized Jones vector $\widehat{\mathbf{E}}$ is

$$\widehat{\mathbf{E}}=\frac{1}{\sqrt{A_x^2+A_y^2}}\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\end{array}\right),$$2.25

2.25

so that $\left|\widehat{\mathbf{E}}\right|=1$.

Math Tip 2.2Matrix Vector Multiplication

Matrix multiplication performs a linear transformation on a vector producing a new vector from a linear combination of the original vector’s elements. Given an N-element vector A and an M × N-element matrix C (M rows and N columns), the elements of the resulting matrix B are

$$b_j={\displaystyle \sum _{n=1}^N{c}_{j,n}{a}_n},$$2.26

2.26

where a, b, and c are components of vector A, matrix B, and vector C. For example, the general equation for 3 × 3 matrix vector multiplication is

$$\mathbf{C}\cdot \mathbf{A}=\mathbf{B}=\left(\begin{array}{l}{b}_1\\ b_2\\ b_3\end{array}\right)=\left(\begin{array}{lll}{c}_{11}& c_{12}& c_{13}\\ c_{21}& c_{22}& c_{23}\\ c_{31}& c_{32}& c_{33}\end{array}\right)\left(\begin{array}{l}{a}_1\\ a_2\\ a_3\end{array}\right)=\left(\begin{array}{l}{c}_{11}{a}_1+c_{12}{a}_2+c_{13}{a}_3\\ c_{21}{a}_1+c_{22}{a}_2+c_{23}{a}_3\\ c_{31}{a}_1+c_{32}{a}_2+c_{33}{a}_3\end{array}\right).$$2.27

2.27

2.7Evolution of Overall Phase

The phase of a polarization state is changed by multiplying a Jones vector by $e^{-i{\varphi }_o}$,

$$e^{-i{\varphi }_o}\mathbf{E}=e^{-i{\varphi }_o}\left(\begin{array}{l}{E}_x\\ E_y\end{array}\right).$$2.28

2.28

This operation advances the electric field and moves the polarization ellipse’s arrow but the shape of the polarization ellipse is unchanged by the change of overall phase, as shown in Figure 2.7.

Figure 2.7Polarization ellipses for Ex = 1 and Ey = e−i π/4/2 for several phases ϕo.

2.8Rotation of Jones Vectors

One application of matrix multiplication is performing rotations on polarization states as shown in Figure 2.8. The orientation of a Jones vector is rotated by multiplying the vector by the Cartesian rotation matrixR(α),

Figure 2.8An elliptically polarized state shown on the left is rotated counterclockwise by an angle α. The state after rotation is shown on the right.

$$\mathbf{R}(\alpha )=\left(\begin{array}{ll}\text{cos}\alpha & -\text{sin}\alpha \\ \text{sin}\alpha & \text{cos}\alpha \end{array}\right).$$2.29

2.29

This rotation operation rotates the major axis of the ellipse, leaving the ellipticity the same. The phase remains the same with respect to the major axis. For the example of Figure 2.8, the rotation equation is

$$\left(\begin{array}{ll}\text{cos}\alpha & -\text{sin}\alpha \\ \text{sin}\alpha & \text{cos}\alpha \end{array}\right)\left(\begin{array}{l}1\\ i/2\end{array}\right)=\left(\begin{array}{l}\text{cos}\alpha -i\text{sin}\alpha /2\\ \text{sin}\alpha +i\text{cos}\alpha /2\end{array}\right).$$2.30

2.30

2.9Linearly Polarized Light

The electric field vector associated with linearly polarized light oscillates in a single direction between the positive and negative direction. The magnitude of the magnetic field oscillates along the orthogonal direction and in phase with the electric field as shown in Figure 2.9. The magnitude of the electric field goes to zero twice per period. With elliptically and circularly polarized light, the magnitude never goes to zero as is seen in Figure 2.5. These zero magnitudes require the relative phase between the x- and y-components ϕx − ϕy be either 0° or 180°. The Jones vector for horizontal linearly polarized light (0°) is

Figure 2.9The electric field (horizontal) and magnetic field (vertical) for a linearly polarized state.

$$\mathbf{E}=A{e}^{-i{\varphi }_o}\left(\begin{array}{l}1\\ 0\end{array}\right),$$2.31

2.31

where A is the amplitude and ϕo is the absolute phase. ϕo is associated with a negative sign due to the decreasing phase sign convention. The Jones vector for normalized horizontal linearly polarized light (with unit amplitude and zero absolute phase) has the symbol H,

$$\mathbf{H}=\left(\begin{array}{l}1\\ 0\end{array}\right),$$2.32

2.32

where H is not to be confused with the magnetic field vector. The Jones vector for normalized linearly polarized lightLP(α) polarized at an angle α, measured counterclockwise from the x-axis, is obtained via the Jones vector rotation operationR (α) (Equation 2.29), where

$$\mathbf{LP}(\alpha )=\mathbf{R}(\alpha )\cdot \mathbf{H}=\left(\begin{array}{ll}\text{cos}\alpha & -\text{sin}\alpha \\ \text{sin}\alpha & \text{cos}\alpha \end{array}\right)\cdot \left(\begin{array}{l}1\hfill \\ 0\hfill \end{array}\right)=\left(\begin{array}{l}\text{cos}\alpha \\ \text{sin}\alpha \end{array}\right).$$2.33

2.33

The Jones vector for an arbitrary beam of linearly polarized light is

$$\mathbf{E}=A{e}^{-i{\varphi }_o}\left(\begin{array}{l}\text{cos}\alpha \\ \text{sin}\alpha \end{array}\right),$$2.34

2.34

where A is the amplitude and ϕo is the absolute phase at t = 0.

2.10Circularly Polarized Light

Monochromatic circularly polarized light has a constant electric field amplitude whose orientation uniformly rotates in the transverse plane. Circularly polarized light occurs in two forms, right circularly polarized and left circularly polarized, depending on the direction of rotation or helicity of the electric and magnetic field vectors as shown in Figure 2.6. By convention, as shown in Figure 2.10, when looking into the beam toward the negative z-direction through time, the electric field vector for right circularly polarized light rotates clockwise.13 If you align the thumb of your left hand along the direction of propagation, out of the page, then the fingers of your left hand point the direction of motion of the electric field. Thus, right circularly polarized light obeys the left hand rule. Similarly, the electric field vector for left circularly polarized light rotates counterclockwise when looking into the beam and obeys the right hand rule.1

Figure 2.10(Left) Looking into a left circularly polarized beam, the electric and magnetic fields circulate counterclockwise in time. (Right) Looking into a right circularly polarized beam, the electric and magnetic fields circulate clockwise in time.

Looking into a right circularly polarized beam, the electric and magnetic fields rotate clockwise in time. The right circularly polarized Jones vector R for a normalized beam is

$$\mathbf{R}=\frac{1}{\sqrt{2}}\left(\begin{array}{r}\hfill 1\\ \hfill -i\end{array}\right).$$2.35

2.35

For a left circularly polarized beam, the electric and magnetic fields circulate counterclockwise and the normalized Jones vector L is

$$\mathbf{L}=\frac{1}{\sqrt{2}}\left(\begin{array}{l}1\\ i\end{array}\right).$$2.36

2.36

For each state, a time helixE(x,y,t) and a space helixE(x,y,z) can be drawn as a three-dimensional space curve. The space helix for right circularly polarized light is found by setting t = 0, yielding

$$ℛe[\mathbf{E}(z,0)]=A\left[\text{cos}\left(\frac{2\pi }{\lambda }z\right)\widehat{x}+\text{cos}\left(\frac{2\pi }{\lambda }z-\frac{\pi }{2}\right)\widehat{y}\right]=A\left[\text{cos}\left(\frac{2\pi }{\lambda }z\right)\widehat{x}+\sin \left(\frac{2\pi }{\lambda }z\right)\widehat{y}\right].$$2.37

2.37

This helix is right handed; when the fingers of the right hand curl in the direction the vector is advancing, the thumb points toward increasing z. The time helix for right circularly polarized light is found by setting z = 0, yielding

$$ℛe[\mathbf{E}(0,t)]=A\left[\text{cos}\left(\frac{2\pi }{\lambda }z\right)\widehat{\mathbf{x}}+\cos \left(\frac{2\pi }{\lambda }z+\frac{\pi }{2}\right)\widehat{\mathbf{y}}\right]=A\left[\text{cos}\left(\frac{2\pi }{\lambda }z\right)\widehat{\mathbf{x}}-\text{sin}\left(\frac{2\pi }{\lambda }z\right)\widehat{\mathbf{y}}\right].$$2.38

2.38

This right circularly polarized time helix is left handed. Note that the space helix and the time helix have opposite helicity, due to the minus sign in $\frac{2\pi }{\lambda }z-\omega t$. The left side of Figure 2.11 shows the time helix for right circularly polarized light and the right side shows the space helix. Note the opposite helicity. Thus, the terms left and right circular polarization are named after the corresponding space helices. For left circularly polarized light, the helicity is reversed in the two figures.

Figure 2.11The time helix (left) and space helix (right) for right circularly polarized light rotate in opposite directions. The time helix is the space curve drawn through the ends of the electric field vectors in x, y, and t, showing that the time helix is left handed. The space helix in x, y, and z, for right circularly polarized light, is right handed.

2.11Elliptically Polarized Light

Figure 2.12 graphs the key geometrical features of a polarization ellipse. The ellipticity ε of the ellipse is defined as the length of the minor axis b divided by the length of the major axis a,

Figure 2.12a is the length of the major axis of the polarization ellipse; b is the length of the minor axis. ψ is the orientation of the major axis.

$$\epsilon =\frac{b}{a}.$$2.39

2.39

The orientation of the major axis ψ is measured counterclockwise from the x-axis. Figure 2.13 shows a family of ellipses of increasing ellipticity.

Figure 2.13Ellipses with ellipticity of ε = 0, 0.2, 0.4, 0.6, 0.8, 1.

With polarized light, the ellipses are associated with the oscillation of the electric and magnetic field vectors; both clockwise and counterclockwise ellipses occur. Therefore, the ellipticity of the polarization ellipse is generalized to positive and negative values and varies from −1 for right circularly polarized light to 1 for left circularly polarized light. Ellipticity of 0 describes linearly polarized light. Similarly, when ε < 0, the light has a right helicity; when ε > 0, it has a left helicity. For a horizontal major axis, ψ = 0, with major axis amplitude a along x, minor axis b along y, the normalized elliptically polarized Jones vector is12

$$\mathbf{E}(\epsilon ,0)=\frac{1}{\sqrt{a^2+b^2}}\left(\begin{array}{l}a\\ ib\end{array}\right)=\frac{1}{\sqrt{1+{\epsilon }^2}}\left(\begin{array}{l}1\\ -i\epsilon \end{array}\right).$$2.40

2.40

The imaginary i occurs on the y-component because the field components along the major and minor axes are always 90° out of phase for elliptically polarized light. The normalized Jones vector for an arbitrary major axis orientation is obtained by the rotation operation Equation 2.29 through an angle ψ,

$$\mathbf{E}(\epsilon ,\psi )=\frac{1}{\sqrt{1+{\epsilon }^2}}\left(\begin{array}{l}\text{cos}\psi +i\epsilon \text{sin}\psi \\ -i\epsilon \text{cos}\psi +\text{sin}\psi \end{array}\right).$$2.41

2.41

The inverse problem determines the polarization ellipse parameters for an arbitrary Jones vector E. With the components of E specified in polar coordinate form

$$\mathbf{E}=\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\end{array}\right),$$2.42

2.42

the polarization ellipse must fit in a rectangle 2Ax × 2Ay and be tangent to the four sides as shown in Figure 2.15. The major axis must lie along one of the two diagonals as shown in Figure 2.14. The axis lies in the first and third quadrants if the relative phase in Equation 2.23 meets the following conditions:

$$\begin{array}{rrr}\hfill -\pi /4<\varphi <\pi /4& \hfill \text{or}& \hfill 3\pi /4<\varphi <5\pi /4.\end{array}$$2.43

2.43

Figure 2.14The bounding box for a polarization ellipse is a rectangle with sides equal to twice the x- and y-amplitudes.

Otherwise, the major axis lies in the second and fourth quadrants.

The major axis orientation, ψ, is related to the Jones vector as

$$\begin{array}{rrr}\hfill \text{tan}(2\psi )=\frac{2A_x{A}_y\text{cos}\varphi }{A_x^2-A_y^2}& \hfill \text{or}& \hfill \psi =\frac{1}{2}\text{arctan}\left(\frac{2A_x{A}_y\text{cos}\varphi }{A_x^2-A_y^2}\right).\end{array}$$2.44

2.44

The semi-major axis with a length of one-half the major axis, a, from the origin to the furthest distance, has a complicated expression,

$$a=\sqrt{A_x^2{\text{cos}}^2\psi +A_y^2{\text{sin}}^2\psi +2A_x{A}_y\text{cos}\psi \text{sin}\psi \text{cos}\varphi }.$$2.45

2.45

Similarly, the semi-minor axis is

$$b=\sqrt{A_x^2{\text{sin}}^2\psi +A_y^2{\text{cos}}^2\psi -2A_x{A}_y\text{cos}\psi \text{sin}\psi \text{cos}\varphi }.$$2.46

2.46

Hence, the equation for the ellipticity is quite involved,

$$\epsilon =\frac{b}{a}=\sqrt{\frac{A_x^2{\text{sin}}^2\psi +A_y^2{\text{cos}}^2\psi -2A_x{A}_y\text{cos}\psi \text{sin}\psi \text{cos}\varphi }{A_x^2{\text{cos}}^2\psi +A_y^2{\text{sin}}^2\psi +2A_x{A}_y\text{cos}\psi \text{sin}\psi \text{cos}\varphi }}.$$2.47

2.47

Note that a and b obey the relation

$$a^2+b^2=A_x^2+A_y^2.$$2.48

2.48

Example 2.2Elliptically Polarized Jones Vector

The Jones vector for a normalized state with major axis orientation ψ = π/4 and ellipticity ε = ½ shown in Figure 2.15 is

$$\mathbf{E}=\frac{1}{\sqrt{10}}\left(\begin{array}{l}2+i\\ 2-i\end{array}\right).$$2.49

2.49

Figure 2.15The polarization ellipse of Equation 2.49 looking into the beam (left) and a 3D view of the time helix (right).

E can be multiplied by any arbitrary phase e−iϕ.

2.12Orthogonal Jones Vectors

Polarization states are orthogonal when their major axes are 90° apart, their helicities are opposite, and the ellipticities have equal magnitude. Three pairs of orthogonal polarization states are shown in Figure 2.16. Horizontal (0°) and vertical (90°) linearly polarized light are orthogonal to each other, as are right and left circularly polarized light. The phases of orthogonal polarization states are not specified and can assume arbitrary values. For two Jones vectors E1 and E2 to represent orthogonal polarizations, the dot product of the adjoint of E1 with E2 is zero,

Figure 2.16Orthogonal polarization states have major axes 90° apart, opposite helicities, and equal ellipticity magnitudes.

$${\mathbf{E}}_1^{†}\cdot {\mathbf{E}}_2=\left(\begin{array}{ll}{E}_{1x}^{*}& E_{1y}^{*}\end{array}\right)\cdot \left(\begin{array}{l}{E}_{2x}\\ E_{2y}\end{array}\right)=0.$$2.50

2.50

Example 2.3Orthogonality of Two Jones Vectors

The Jones vector F orthogonal to

$$\widehat{\mathbf{E}}=\frac{1}{\sqrt{10}}\left(\begin{array}{l}2+i\\ 2-i\end{array}\right)$$2.51

2.51

can be found easily in non-normalized form by setting either element of F to any numerical value and solving for the other element. Here, Fy has been set to one, and the equation solved for Fx,

$${\widehat{\mathbf{E}}}^{†}\cdot \mathbf{F}=\frac{\left(\begin{array}{ll}2-i& 2+i\end{array}\right)\text{*}}{\sqrt{10}}\left(\begin{array}{l}{F}_x\\ 1\end{array}\right)=0,F_x=\frac{-3-4i}{5}.$$2.52

2.52

Then, F can be normalized if desired and an arbitrary phase e−iϕ applied,

$$\widehat{\mathbf{F}}=\frac{e^{-i\varphi }}{5\sqrt{2}}\left(\begin{array}{l}-3-4i\\ 5\end{array}\right).$$2.53

2.53

2.13Change of Basis

The Jones vector is most commonly expressed in terms of its x- and y-components, but other orthogonal polarization bases can be used. The following matrix transforms a Jones vector from the xy-basis into a basis with the normalized and orthogonal Jones vectors A and B,

$$\left(\begin{array}{l}{E}_A\\ E_B\end{array}\right)=\left(\begin{array}{ll}{A}_x^{*}& A_y^{*}\\ B_x^{*}& B_y^{*}\end{array}\right)\cdot \left(\begin{array}{l}{E}_x\\ E_y\end{array}\right).$$2.54

2.54

This is an example of a unitary change of basis, a generalized rotation (the rotation matrix may have real or complex values). An “ordinary rotation” should have a real-valued rotation matrix. If the unitary change of basis is applied to two vectors, then the inner product between the two, the dot product, is preserved; neither the angle between the vectors nor their lengths have changed.

Example 2.4Circularly Polarized Basis

One useful basis for Jones vectors uses left and right circularly polarized basis states instead of x- and y-basis states, and is obtained from the xy-basis as14

$$\left(\begin{array}{l}{E}_{\text{L}}\\ E_{\text{R}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{ll}1& -i\\ 1& i\end{array}\right)\cdot \left(\begin{array}{l}{E}_x\\ E_y\end{array}\right).$$2.55

2.55

2.14Addition of Jones Vectors

The combination of two plane waves of the same frequency traveling in the same direction is simulated by the addition of their Jones vectors. Two such monochromatic beams can be input into two different faces of a beam splitter and adjusted so that, after exiting, they propagate in the same direction. If the two Jones vectors following the beam splitters individually have Jones matrices EA and EB, then the combined beam has Jones vector E,

$${\mathbf{E}}_A+{\mathbf{E}}_B=\left(\begin{array}{l}{E}_{x,A}\\ E_{y,A}\end{array}\right)+\left(\begin{array}{l}{E}_{x,B}\\ E_{y,B}\end{array}\right)=\mathbf{E}.$$2.56

2.56

Such beam combination occurs at the output of interferometers. One must be careful when adding Jones vectors to ensure that the phases are specified properly, since a family of polarization ellipses results when either phase is varied.

Example 2.5Addition of Circularly Polarized Beams

The combination of the monochromatic right and left circularly polarized light with equal amplitudes (here set to 1/2) and an adjustable phase ϕ on the right circularly polarized component is

$$\mathbf{E}=\frac{e^{i\varphi }}{2}\left(\begin{array}{l}1\\ -i\end{array}\right)+\frac{1}{2}\left(\begin{array}{l}1\\ i\end{array}\right)=\frac{e^{i\varphi /2}}{2}\left(\begin{array}{l}{e}^{i\varphi /2}+e^{-i\varphi /2}\\ -i\left(e^{i\varphi /2}-e^{-i\varphi /2}\right)\end{array}\right)=e^{i\varphi /2}\left(\begin{array}{l}\text{cos}\left(\frac{\varphi }{2}\right)\\ \text{sin}\left(\frac{\varphi }{2}\right)\end{array}\right),$$2.57

2.57

yielding linearly polarized light. The orientation of the linearly polarized light ϕ/2 depends on the relative phase between the circularly polarized beams. Thus, the interference of left and right circularly polarized beams of equal amplitude and a variable phase provides one method to generate an adjustable linear polarization orientation.

2.15Polarized Flux Components

The flux P of a Jones vector E is calculated using Equation 2.17. The polarized flux IA of E is the part of the flux in a normalized state ${\widehat{\mathbf{E}}}_A$, the flux transmitted by an ideal polarizer transmitting ${\widehat{\mathbf{E}}}_A$,

$$I_A={\left|{\widehat{\mathbf{E}}}_A^{}\cdot \mathbf{E}\right|}^2={\left|\left(\begin{array}{ll}{E}_{A,x}^{*}& E_{A,y}^{*}\end{array}\right)\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\end{array}\right)\right|}^2.$$2.58

2.58

The fraction of E’s flux in state ${\widehat{\mathbf{E}}}_A$ is IA/P. The total flux in state E is the sum of the polarized flux in any two orthogonal polarization states. For example, the sum of the polarized flux in x and the polarized flux in y is

$${\mathbf{E}}^{†}\mathbf{E}=\left(\begin{array}{ll}{E}_x^{*}& E_y^{*}\end{array}\right)\left(\begin{array}{l}{E}_x\\ E_y\end{array}\right)=\left(\begin{array}{ll}{A}_x{e}^{i{\varphi }_x}& A_y{e}^{i{\varphi }_y}\end{array}\right)\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\end{array}\right)=A_x^2+A_y^2.$$2.59

2.59

A monochromatic beam in a polarization state E can be divided by an ideal polarizing beam splitter into two orthonormal (orthogonal and normalized) polarization states ${\widehat{\mathbf{E}}}_A$ and ${\widehat{\mathbf{E}}}_B$ with complex amplitudes α and β given by

$$\mathbf{E}=\left({\widehat{\mathbf{E}}}_A^{†}\cdot \mathbf{E}\right){\mathbf{E}}_A+\left({\widehat{\mathbf{E}}}_B^{†}\cdot \mathbf{E}\right){\mathbf{E}}_B=\alpha {\mathbf{E}}_A+\beta {\mathbf{E}}_B.$$2.60

2.60

For an arbitrary Jones vector E,

$$\mathbf{E}=e^{-i{\varphi }_x}\left(\begin{array}{l}{A}_x\\ A_y{e}^{-i\varphi }\end{array}\right),$$2.61

2.61

the component of flux into each of the basis polarization states, IH, IV, I45, I135, IR, and IL, are listed in Table 2.3. These equations are used in Chapter 3 for conversion between Jones vectors and Stokes parameters.

Table 2.3The Flux of E in the Basis Polarization States

Polarization State	Basis Jones Vector	Polarized Flux I
Horizontal, 0°	$\mathbf{H}=\left(\begin{array}{c}1\\ 0\end{array}\right)$	$I_{\text{H}}={\left|E_x\right|}^2=A_x^2$
Vertical, 90°	$\mathbf{V}=\left(\begin{array}{c}0\\ 1\end{array}\right)$	$I_{\text{V}}={\left|E_y\right|}^2=A_y^2$
45°	$\mathbf{45}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$	$I_{45}=\frac{1}{2}{\left|E_x+E_y\right|}^2=\frac{A_x^2+A_y^2}{2}+A_x{A}_y\text{cos}\varphi$
135°	$\mathbf{135}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -1\end{array}\right)$	$I_{135}=\frac{1}{2}{\left|E_x-E_y\right|}^2=\frac{A_x^2+A_y^2}{2}-A_x{A}_y\text{cos}\varphi$
Right circular	$\mathbf{R}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -i\end{array}\right)$	$I_{\text{R}}=\frac{1}{2}{\left|E_x+i{E}_y\right|}^2=\frac{A_x^2+A_y^2}{2}+A_x{A}_y\text{sin}\varphi$
Left circular	$\mathbf{L}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ i\end{array}\right)$	$I_{\text{L}}=\frac{1}{2}{\left|E_x-i{E}_y\right|}^2=\frac{A_x^2+A_y^2}{2}-A_x{A}_y\text{sin}\varphi$

2.16Converting Polarization Vectors into Jones Vectors

The polarization ellipse for polarization states defined in three dimensions can be converted into (two-dimensional) Jones vectors by choosing a basis in the transverse plane and calculating the amplitudes along the new basis vectors. Then, the state can be analyzed for ellipticity, flux components, and other metrics by applying the Jones matrix analysis methods of the last few sections. Their selection of two basis states in the transverse plane is arbitrary. For a given k, two normalized real-valued vectors $({\widehat{\mathbf{v}}}_1\text{and}{\widehat{\mathbf{v}}}_2)$ are selected, which are orthogonal to each other and perpendicular to k,

$${\widehat{\mathbf{v}}}_1\cdot {\widehat{\mathbf{v}}}_2=0,\mathbf{k}\cdot {\widehat{\mathbf{v}}}_1=0,\text{and}\mathbf{k}\cdot {\widehat{\mathbf{v}}}_2=0.$$2.62

2.62

Then, any polarization vector E transverse to k can be written as a superposition of ${\widehat{\mathbf{v}}}_1\text{ and }{\widehat{\mathbf{v}}}_2$,

$$\begin{array}{l}\mathbf{E}(\mathbf{r},t)=\mathit{ℛ}\mathit{e}\left[\mathbf{E}{e}^{i\left(\frac{2\pi }{\lambda }k\cdot r-\omega t-\varphi \right)}\right]=ℛe\left\{\left[(E\cdot {\widehat{v}}_1){\widehat{v}}_1+(E\cdot {\widehat{v}}_2){\widehat{v}}_2\right]e^{i\left(\frac{2\pi }{\lambda }k\cdot r-\omega t-\varphi \right)}\right\}\\ \phantom{E(r,t)}=\mathit{ℛ}\mathit{e}\left[\left(A_{v1}{e}^{-i{\varphi }_{v1}}{\widehat{v}}_1+A_{v2}{e}^{-i{\varphi }_{v2}}{\widehat{v}}_2\right)e^{i\left(\frac{2\pi }{\lambda }k\cdot r-\omega t-\varphi \right)}\right],\end{array}$$2.63

2.63

where Avi are real amplitudes along ${\widehat{\mathbf{v}}}_i$. This defines a Jones vector

$$\mathbf{E}=\left(\begin{array}{l}{A}_{v1}{e}^{-i{\varphi }_{v1}}\\ A_{v2}{e}^{-i{\varphi }_{v2}}\end{array}\right),$$2.64

2.64

in a $\left({\widehat{\mathbf{v}}}_1,{\widehat{\mathbf{v}}}_2\right)$ basis. The major axis orientation ψ in local coordinates is calculated using Equation 2.44, where ψ is measured from ${\widehat{\mathbf{v}}}_1$ to ${\widehat{\mathbf{v}}}_2$ as shown in Figure 2.17.

Figure 2.17The major axis orientation ψ measured from ${\widehat{\mathbf{v}}}_1$to$\overrightarrow{\mathbf{a}}$. The axis vector is shown in red arrow.

The selection of local coordinates is discussed further in Chapter 11 (see discussion on the Jones pupil and local coordinates systems).

Example 2.6Conversion into Jones Vectors

As an example, a left circularly polarized beam propagating along k = (6, 6, 7)/11 will be expressed in three different Jones vector basis sets to appreciate how arbitrary basis sets are and show how they interrelate. The Jones vector basis sets use three different dipole (latitude and longitude) systems, seen in Figure 2.18, to generate the ${\widehat{\mathbf{v}}}_1\text{ and }{\widehat{\mathbf{v}}}_2$ basis vectors. The polarization vector for this left circularly polarized beam is

$$\mathbf{E}=\frac{1}{22}\left(\begin{array}{l}11+7i\\ -11+7i\\ -12\end{array}\right).$$2.65

2.65

Figure 2.18A left circularly polarized state (red circle with arrow indicating phase) with propagation vector (black arrow) is shown with basis vectors in three possible local coordinate systems for Ex and Ey. On the left, the first basis vector (green) is along a latitude line and the second basis vector (blue) along a longitude line with the axis along x. In the middle, the same polarization state is shown but with basis vectors chosen along latitude and longitude with the axis along y, and in the right, the axis is along z.

The basis vector along latitude is generated by taking the normalized cross product between k and the dipole axis d as

$${\widehat{\mathbf{v}}}_1=\frac{\mathbf{d}\times \mathbf{k}}{\left|\mathbf{d}\times \mathbf{k}\right|},{\widehat{\mathbf{v}}}_2=\mathbf{k}\times {\widehat{\mathbf{v}}}_1.$$2.66

2.66

For the first example, a Jones vector basis uses latitude and longitude with the pole along x; the basis vectors are

$${\widehat{\mathbf{v}}}_1=\frac{1}{\sqrt{85}}\left(\begin{array}{l}0\\ -7\\ 6\end{array}\right),{\widehat{\mathbf{v}}}_2=\frac{1}{11\sqrt{85}}\left(\begin{array}{l}85\\ -36\\ -42\end{array}\right)$$2.67

2.67

and the Jones vector Ejx becomes

$${\mathbf{E}}_{jx}=\left(\begin{array}{l}\mathbf{E}\cdot {\widehat{\mathbf{v}}}_1\\ \mathbf{E}\cdot {\widehat{\mathbf{v}}}_2\end{array}\right)=\frac{1}{2\sqrt{85}}\left(\begin{array}{l}-7+11i\\ -11-7i\end{array}\right)=\frac{-7+11i}{2\sqrt{85}}\left(\begin{array}{l}1\\ i\end{array}\right)\approx \frac{e^{2.138i}}{\sqrt{2}}\left(\begin{array}{l}1\\ i\end{array}\right).$$2.68

2.68

Choosing a latitude and longitude basis with the pole along y yields basis vectors

$${\widehat{\mathbf{v}}}_1=\frac{1}{\sqrt{85}}\left(\begin{array}{l}7\\ 0\\ -6\end{array}\right),{\widehat{\mathbf{v}}}_2=\frac{1}{11\sqrt{85}}\left(\begin{array}{l}-36\\ 85\\ -42\end{array}\right)$$2.69

2.69

and the Jones vector Ejy becomes

$${\mathbf{E}}_{jy}=\left(\begin{array}{l}\mathbf{E}\cdot {\widehat{\mathbf{v}}}_1\\ \mathbf{E}\cdot {\widehat{\mathbf{v}}}_2\end{array}\right)=\frac{1}{2\sqrt{85}}\left(\begin{array}{l}-7-11i\\ 11-7i\end{array}\right)=\frac{-7-11i}{2\sqrt{85}}\left(\begin{array}{l}1\\ i\end{array}\right)\approx \frac{e^{-2.318i}}{\sqrt{2}}\left(\begin{array}{l}1\\ i\end{array}\right).$$2.70

2.70

Finally, using a latitude and longitude basis with the pole along z yields basis vectors

$${\widehat{\mathbf{v}}}_1=\frac{1}{\sqrt{2}}\left(\begin{array}{l}-1\\ 1\\ 0\end{array}\right),{\widehat{\mathbf{v}}}_2=\frac{1}{22}\left(\begin{array}{l}-11-7i\\ 11-7i\\ 6i\end{array}\right)$$2.71

2.71

and the Jones vector Ejz becomes

$${\mathbf{E}}_{jz}=\left(\begin{array}{l}\mathbf{E}\cdot {\widehat{\mathbf{v}}}_1\\ \mathbf{E}\cdot {\widehat{\mathbf{v}}}_2\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{l}1\\ i\end{array}\right),$$2.72

2.72

which is the standard Jones vector for left circularly polarized light (Equation 2.36). In the case of circularly polarized light, the only difference between the three Jones vectors Ejx, Ejy, and Ejz is their absolute phase,

$${\mathbf{E}}_{jz}=e^{-2.138i}{\mathbf{E}}_{jx}=e^{2.138i}{\mathbf{E}}_{jy}.$$2.73

2.73

2.17Decreasing Phase Sign Convention

Electromagnetic waves are commonly written with one of two different sign conventions. Either the phase decreases with time and increases in space, $\frac{2\pi }{\lambda }\mathbf{k}\cdot \mathbf{r}-\omega t-{\varphi }_o$, the convention adopted here, or the phase increases with time and decreases in space, $\omega t-\frac{2\pi }{\lambda }\mathbf{k}\cdot \mathbf{r}+{\varphi }_o$. Depending on the choice, various plus and minus signs must be adjusted in the mathematical descriptions for circularly and elliptically polarized light. Both conventions, decreasing phase and increasing phase, are in widespread use; thus, care is necessary when using Jones calculus equations from different sources. The student should learn to identify the sign convention in manuscripts from the context, since the sign convention is often not specified.

The two conventions for phase occur because cosine is an even function,

$$\text{cos}(\theta )=\text{cos}(-\theta ).$$2.74

2.74

Therefore, the choice of phase convention would not appear to matter since

$$\text{cos}\left(\frac{2\pi }{\lambda }z-\omega t-{\varphi }_o\right)=\text{cos}\left(\omega t-\frac{2\pi }{\lambda }z+{\varphi }_o\right),$$2.75

2.75

and

$$\mathit{ℛ}\mathit{e}\left[e^{i\left(\frac{2\pi }{\lambda }z-\omega t-{\varphi }_o\right)}\right]=\mathit{ℛ}\mathit{e}\left[e^{i\left(\omega t-\frac{2\pi }{\lambda }z+{\varphi }_o\right)}\right]$$2.76

2.76

for all z and t. In our decreasing phase convention, a monochromatic plane wave propagating along z has the form

$$\mathbf{E}(z,t)=\mathit{ℛ}\mathit{e}\left[\left(\begin{array}{l}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\\ 0\end{array}\right)e^{i\left(\frac{2\pi }{\lambda }z-\omega t-{\varphi }_o\right)}\right]=\left(\begin{array}{l}{A}_x\text{cos}\left(\frac{2\pi }{\lambda }z-\omega t-{\varphi }_o-{\varphi }_x\right)\\ A_y\text{cos}\left(\frac{2\pi }{\lambda }z-\omega t-{\varphi }_o-{\varphi }_y\right)\\ 0\end{array}\right).$$2.77

2.77

Here, a wave is advanced through time by subtracting from the phase. A wave is delayed or retarded by adding to the phase. In our decreasing phase convention, the equation for a right circularly polarized plane wave of amplitude A and absolute phase ϕo is

$$\mathbf{E}(z,t)=\mathit{ℛ}\mathit{e}\left[e^{i\left(\frac{2\pi }{\lambda }z-\omega t-{\varphi }_o\right)}\frac{A}{\sqrt{2}}\left(\begin{array}{l}1\\ -i\\ 0\end{array}\right)\right].$$2.78

2.78

Thus, the Jones vector for right circularly polarized light in this book is

$$\mathbf{R}=\frac{A{e}^{-i{\varphi }_o}}{\sqrt{2}}\left(\begin{array}{l}1\\ -i\end{array}\right).$$2.79

2.79

Similarly, in our decreasing phase convention, left circularly polarized light has the Jones vector

$$L=\frac{A{e}^{-i{\varphi }_o}}{\sqrt{2}}\left(\begin{array}{l}1\\ i\end{array}\right).$$2.80

2.80

2.18Increasing Phase Sign Convention

The other sign convention, the increasing phase sign convention, is not used in this work except in this section. In the increasing phase sign convention, the phase increases with time so the sign of ϕ changes. Subscript M is used in this section to indicate that quantities expressed are using the increasing phase sign convention. The monochromatic plane wave in the increasing phase convention takes the form

$${\mathbf{E}}_M(z,t)=\mathit{ℛ}\mathit{e}\left[\left(\begin{array}{l}{A}_x{e}^{i{\varphi }_x}\\ A_y{e}^{i{\varphi }_y}\\ 0\end{array}\right)e^{i\left(\omega t-\frac{2\pi }{\lambda }z+{\varphi }_o\right)}\right]=\left(\begin{array}{l}{A}_x\text{cos}\left(\omega t-\frac{2\pi }{\lambda }z+{\varphi }_o+{\varphi }_x\right)\\ A_y\text{cos}\left(\omega t-\frac{2\pi }{\lambda }z+{\varphi }_o+{\varphi }_y\right)\\ 0\end{array}\right).$$2.81

2.81

The only change in the plane wave equation is the sign of the exponent. Note that since cosine is an even function, the sign of the argument actually does not matter. Hence, in the increasing phase sign convention, the left and right circularly polarized light Jones vectors are

$${\mathbf{L}}_M=\frac{A{e}^{i{\varphi }_o}}{\sqrt{2}}\left(\begin{array}{l}1\\ -i\end{array}\right),{\mathbf{R}}_M=\frac{A{e}^{i{\varphi }_o}}{\sqrt{2}}\left(\begin{array}{l}1\\ i\end{array}\right).$$2.82

2.82

The Jones vector for left circularly polarized light in the increasing phase convention is the Jones vector for right circularly polarized light in the decreasing phase convention and vice versa. This could lead to confusion! The choice of sign convention also affects the signs of Jones matrix elements as is documented in Chapter 5.

2.19Polarization State of Sources

The polarization state exiting a source or entering an optical system is determined by many factors: whether the light is emitted, scattered, or reflected; the temperature and roughness of the source; the polarization and direction of the illuminating light; and many other factors. The polarization can vary in simple or complex ways; for example, Figure 2.19 contains an example of a complicated polarization variation across a spherical wavefront. Very often, the polarization of the incident light is not well understood, or may vary from measurement to measurement. Two basic models for source polarization will be briefly reviewed to provide an introduction: the dipole model and the double pole model.

Figure 2.19An arbitrarily polarized wavefront with k vectors (black) and polarization ellipses (red) representing a grid of wavefront patches.

The dipole electromagnetic wave is the most commonly encountered wave after the plane wave because of its simple form. Dipole radiation is generated by charges oscillating along a straight line in simple harmonic oscillation. Dipoles are important in modeling many sources of electromagnetic waves such as radio antennae and the emission from excited atoms. The dipole wave is a spherical wave that is linearly polarized along lines of longitude.2,15,V16 The maximum amplitude is radiated in the plane perpendicular to the axis of charge oscillation. The dipole source does not radiate along the axis of charge oscillation. Dipole waves are radiated by charges undergoing simple harmonic oscillation along a line, chosen here in Figure 2.20 as the y-axis. Let θ be the longitude, measured from the z-axis, coming out of the page, and ϕ be the latitude. In this polar coordinate system, the normalized propagation vector k and the dipole wave’s polarization vector E are

Figure 2.20The polarization vector for a dipole wavefront looking into the z-axis. The axis of charge oscillation is the vertical y-axis.

$$\mathbf{k}=\left(\begin{array}{l}\text{cos}\varphi \text{sin}\theta \\ \text{sin}\varphi \\ \text{cos}\varphi \text{cos}\theta \end{array}\right),\mathbf{E}=\left(\begin{array}{l}-\text{sin}\varphi \text{sin}\theta \\ \text{cos}\varphi \\ -\text{sin}\varphi \text{cos}\theta \end{array}\right).$$2.83

2.83

The dipole E-field is shown in Figure 2.20 on a 4π steradian spherical wavefront. A dipole field can be expected for the scattered light from an isolated atom illuminated with linearly polarized light, and many other sources.

The double pole spherical wave in Figure 2.21 is another very common linearly polarized spherical wave in optics. The double pole spherical wave is generated when a lens is illuminated by collimated linearly polarized light. The incident polarization state is rotated at each lens interface as the light refracts and changes direction. When the light exits the lens, the polarization ellipse has rotated about an axis perpendicular to the incident and exiting rays; the rotation axis is their cross products, shown in Figure 2.22, where vertical linearly polarized light has its propagation direction changed by a lens, represented as the vertical line. Thus, the polarized wavefront exiting the lens is different from the dipole spherical wave. When this rotation is performed for every exiting ray in a spherical wavefront, the double pole polarization pattern results as shown in Figure 2.21. This form of polarized wavefront occurs for both positive and negative lenses. Great circles have been drawn out from the z-axis. Along each of these great circles, the polarization state forms a constant angle with the great circle; this is the result of the rotation operation shown in Figure 2.22. The double pole wave occurs for the ideal non-polarizing lenses. Deviations from this pattern due to the s- and p-Fresnel reflection and transmission coefficients occur, but typically the exiting wavefronts are in nearly the double pole form. Thus, the double pole polarized wavefront is a useful and common source model. This polarization pattern is also used as a basis for describing spherical waves and flattening them onto computer screens and pages. This topic is treated in depth in Chapter 11 (see discussion on the Jones pupil and local coordinates systems).

Figure 2.21A spherical wavefront linearly polarized in the double pole polarization pattern is the basic polarized wavefront that approximates light exiting optical lenses, both positive and negative lenses (top and bottom). The polarization forms a constant angle with great circles through the axis.

Figure 2.22Two ideal non-polarizing lenses, (top) positive lens and (bottom) negative lens, rotate the plane of polarization of the incident light (left, red lines) when the light changes direction refracting through and out of the lens.

2.20Problem Sets

1. Estimate the Jones vectors for the following polarization ellipses. Find the phases to place the arrow correctly at t = 0.

   

2. Which of the following Jones vectors are (a) linearly polarized, (b) circularly polarized, 90° or π/2 out of phase with equal amplitudes, (c) elliptically polarized, with arbitrary phase relationship?

   a.	(2, 2)	b.	(i/2, 1)	c.	(i, −i)	d.	(1, −4)	e.	(2 + 2i, −2 + 2i)
   f.	(2 + 2i, −3 + 2i)	g.	(0, 1 + i)	h.	(3, −6i)	i.	(2 + 3i, −3 + 2i)	j.	(2, −i)

3. Consider the plane wave $\mathbf{E}(\mathbf{r},t)=\mathit{ℛ}\mathit{e}\left[e^{-i\left(\frac{2\pi }{\lambda }k\cdot r-\omega t\right)}\left(\begin{array}{c}{a}_x+i{b}_x\\ a_y+i{b}_y\\ a_z+i{b}_z\end{array}\right)\right]$.

   Find the electric field and the Poynting vector at the following times and locations.

   1. t = 0, r = (0, 0, 0)

   2. t = 0, r =λ2k/(4π)

   3. t = π, r = (0, 0, 0)

   4. t = 4π/ω, r =8λ2k/π

4. 1. Find the general equation for the normalized Jones vector orthogonal to $\left(\begin{array}{l}{E}_x\\ E_y\end{array}\right)=\left(\begin{array}{c}{A}_x{e}^{-i{\varphi }_x}\\ A_y{e}^{-i{\varphi }_y}\end{array}\right)$.

   2. Verify the equation with right circularly polarized light.

   3. Why is the phase of the orthogonal Jones vector a free parameter that can be chosen arbitrarily?

   4. Given a propagation direction k = (kx, ky, kz) and a polarization vector $\mathbf{F}=\left(A_x{e}^{-i{\varphi }_x},A_y{e}^{-i{\varphi }_y},A_z{e}^{-i{\varphi }_z}\right)$, find the polarization vector orthogonal to F, normalized or non-normalized.

5. Rotate the following Jones vectors 45° counterclockwise (from +x toward +y).

   a. v1 = (1, i)

   b. v2 = (3, 3)

   c. v3 = (0, −2)

   d. v4 = (1 + i, 1 − i)

6. 1. What is the normalized flux of the Jones vector E3 = (w + ix, y + iz)?

   2. What is the flux of E3 in W/m2 (watts per meter squared)?

   3. What are the units of the Jones vector elements w + ix and y + iz?

7. 1. Find the matrix for a change of basis from the left and right circularly polarized basis states to the xy-basis.

   2. Convert the Jones vector

      $$\left(\begin{array}{l}{E}_{\text{L}}\\ E_{\text{R}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{l}{\text{e}}^{i\eta }\\ {\text{e}}^{-i\eta }\end{array}\right)$$

      

      from the circular basis into the xy-basis and identify the type of polarization state.

8. Convert the six basis polarization states (Table 2.2) from Jones vectors in the ordinary linear xy-basis into the LR circular basis (Equation 2.55).

9. Given a Jones vector E = (1, 1) in the circular Jones vector basis, perform a change of basis to the xy-basis.

10. What are the two directions that light propagating with polarization vector (4i, 6, 4i) might be propagating? This can be determined by taking the cross product between the E-field at two different times using the real field representation, not the exponential form.

11. Given the polarization vector E = (6i, 10, −8i):

    1. Show that E is circularly polarized.

    2. Find the axis of light propagation. The direction along the axis is undetermined.

    3. Which direction would the light be propagating to be left circularly polarized?

12. The equation for a circularly polarized monochromatic plane wave electric field traces a helix for a given point in space or time. Graph this helix for right circularly polarized light with amplitude 1, $\mathbf{E}(z,t)=\mathit{ℛ}\mathit{e}\left[e^{i\left(\frac{2\pi }{\lambda }kz-\omega t-\varphi \right)}\frac{A}{\sqrt{2}}\left(\begin{array}{c}1\\ -i\\ 0\end{array}\right)\right]$.

    1. Make plots of the helix in space at t = 0, (Ex(z, 0), Ey(z, 0), z), looking toward –Ez, front view in three-dimensional perspective plot.

    2. Is the space helix a right-handed or left-handed helix?

    3. Next, make the same set of plots of the helix at a fixed point as a function of time, x = y = z = 0, (Ex(0, t), Ey(0, t), t).

    4. Is the time helix a right-handed or left-handed helix?

13. For each of the following Jones vectors

    $$\begin{array}{l}{\mathbf{E}}_1=\left(\begin{array}{l}1\\ 1-i\end{array}\right),{\mathbf{E}}_2=\left(\begin{array}{l}-i\\ -i\end{array}\right),{\mathbf{E}}_3=\left(\begin{array}{l}-i\\ 1\end{array}\right),{\mathbf{E}}_4=\left(\begin{array}{l}5/4\\ 3i/4\end{array}\right),{\mathbf{E}}_5=\frac{1}{2}\left(\begin{array}{l}i\\ 1+i\sqrt{2}\end{array}\right),\\ \text{and}{\mathbf{E}}_6=\left(\begin{array}{l}-i\\ -\pi /3\end{array}\right):\end{array}$$

    

    1. Plot the polarization ellipse and indicate the direction the electric field is rotating.

    2. Calculate the phase difference between the x- and y-components, δ(ϕ) = ϕx − ϕy, orientation of the major ellipse, ψ, and the normalized flux, P. For circularly polarized light, the orientation may be undefined.

    3. Calculate the degree of circular polarization (DoCP) defined as |PL − PR|/(PL + PR).

14. 1. Compute the normal vector ŵ to complete the right-handed orthonormal basis set $\left(\widehat{\mathbf{u}},\widehat{\mathbf{v}},\widehat{\mathbf{w}}\right)$, where $\widehat{\mathbf{u}}=\frac{1}{\sqrt{3}}\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),\widehat{\mathbf{v}}=\frac{1}{\sqrt{6}}\left(\begin{array}{c}1\\ 1\\ -2\end{array}\right).$

    2. Write the expression for a rotating unit vector ŝ(t) perpendicular to û, rotating clockwise about the û-axis when looking into û, at an angular velocity of ω rad/s, such that ŝ(0) = $\widehat{\mathbf{v}}$. This is an expression for the electric field at a point associated with right circularly polarized light propagating in the û direction.

    3. Calculate the polarization vector E for this wave.

References

1W. A. Shurcliff, Polarized Light. Production and Use, Cambridge, MA: Harvard University Press (1966).

2D. Goldstein, Polarized Light, Revised and Expanded, Vol. 83, Boca Raton, FL: CRC Press, (2011).

3C. Brosseau, Fundamentals of Polarized Light: A Statistical Optics Approach, New York: Wiley, (1998).

4M. Born and E. Wolf, Principles of Optics, 6th edition, Cambridge, UK: Cambridge University Press, (1980).

5E. Hecht, Optics, Reading, MA: Addison-Wesley, 1987.

6R. M. A. Azzam and N. M. Bashara, Ellipsometry and Polarized Light, 2nd edition, Amsterdam: Elsevier (1987).

7G. Yun, K. Crabtree, and R. Chipman, Three-dimensional polarization ray-tracing calculus I: Definition and diattenuation, Appl. Opt. 50 (2011): 2855–2865.

8M. Born and E. Wolf, Principles of Optics: Electromagnetic Theory of Propagation, Interference and Diffraction of Light, CUP Archive (1999).

9E. L. Dereniak and G. D. Boreman, Infrared Detectors and Systems, New York: Wiley (1996).

10G. J. Zissis and W. L. Wolfe, The Infrared Handbook, Ann Arbor, MI: Infrared Information and Analysis Center (1978).

11R. C. Jones, A new calculus for the treatment of optical systems: I. Description and discussion of the calculus, J. Opt. Soc. Am. 31 (1941): 488–493.

12R. M. A. Azzam and N. M. Bashara, Ellipsometry and Polarization Light, New York: North-Holland (1977), p. 97.

13W. Swindell. Handedness of polarization after metallic reflection of linearly polarized light, JOSA 61.2 (1971): 212–215.

14J. P. McGuire Jr. and R. A. Chipman, Diffraction image formation in optical systems with polarization aberrations. I: Formulation and example, JOSA A 7.9 (1990): 1614–1626.

15J. D. Jackson, Classical Electromagnetism, New York: Wiley (1975).

16A. Shadowitz, The Electromagnetic Field, New York: McGraw-Hill (1975).

1 Laser light is almost monochromatic but always has a small wavelength spread.

2 ω is 2π times the frequency in hertz.

3 In most references, the magnitude of the wavevector k is 2π/λ. In this book, k is defined as the normalized wavevector; thus, the wavevector is 2π/λ k and |k| = 1.