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1Polarization States and Jones Vectors

We have seen in that light is an electromagnetic wave which satisfies Maxwell’s equations and the wave equation derived therefrom. Since the electric field is a vector which oscillates as function of time in a certain direction, we say that the wave has a certain polarization. In this chapter we look at the different types of polarization and how the polarization of a light beam can be manipulated.

From Maxwell’s equations and the wave equation, we know that the (real) electric field E(r,t)\mathbf{\mathcal{E}}(\mathbf{r},t) of a time-harmonic plane wave is always perpendicular to the direction of propagation, which is the direction of the wave vector k\mathbf{k} as well as the direction of the Poynting vector (the direction of the power flow). Let the wave propagate in the zz-direction:

k=(00k).\begin{align*} \mathbf{k}= \left( \begin{array}{c}0\\0\\ k \end{array}\right). \end{align*}

Then the electric field vector does not have a zz-component and hence the real electric field at zz and at time tt can be written as

E(z,t)=(Axcos(kzωt+φx)Aycos(kzωt+φy)0).\begin{align*} \mathbf{\mathcal{E}}(z,t) = \left(\begin{array}{c}{\cal A}_x \cos(kz-\omega t + \varphi_x) \\{\cal A}_y \cos(kz-\omega t + \varphi_y) \\0 \end{array}\right). \end{align*}

where Ax{\cal A}_x and Ay{\cal A}_y are positive amplitudes and φx\varphi_x, φy\varphi_y are the phases of the electric field components. While kk and ω\omega are fixed, we can vary Ax{\cal A}_x, Ay{\cal A}_y, φx\varphi_x and φy\varphi_y. This degree of freedom is why different states of polarization exist: the state of polarization is determined by the ratio of the amplitudes and by the phase difference φyφx\varphi_y-\varphi_x between the two orthogonal components of the light wave Fowles, 1989. Consider the electric field in a fixed plane z=0z=0:

(Ex(0,t)Ey(0,t))=(Axcos(ωt+φx)Aycos(ωt+φy))=Re{(AxeiφxAyeiφy)eiωt}=Re{(Ex(0)Ey(0))eiωt}\begin{align*} \left(\begin{array}{c}{\cal E}_x(0,t) \\{\cal E}_y(0,t) \end{array} \right) = \left( \begin{array}{c}{\cal A}_x \cos(-\omega t + \varphi_x) \\{\cal A}_y \cos(-\omega t + \varphi_y) \end{array}\right) =\text{Re}\left\{ \left( \begin{array}{c}{\cal A}_x e^{i\varphi_x} \\{\cal A}_y e^{i\varphi_y} \end{array}\right) e^{-i\omega t} \right\} = \text{Re}\left\{ \left(\begin{array}{c}E_x(0) \\E_y(0) \end{array}\right) e^{-i \omega t} \right\} \end{align*}

The complex vector

J=(Ex(0)Ey(0))=(AxeiφxAyeiφy),\begin{align*} {\mathbf J}= \left(\begin{array}{c}E_x(0) \\E_y(0) \end{array}\right) = \left(\begin{array}{c}{\cal A}_x e^{i\varphi_x} \\{\cal A}_y e^{i\varphi_y} \end{array}\right), \end{align*}

is called the Jones vector. It is used to characterize the polarization state. Let us see how, at a fixed position in space, the electric field vector behaves as a function of time for different choices of Ax{\cal A}_x, Ay{\cal A}_y and φyφx\varphi_y-\varphi_x.

a) Linear polarization: φyφx=0\varphi_y-\varphi_x=0 or φyφx=π\varphi_y-\varphi_x=\pi.] When φyφx=0\varphi_y-\varphi_x=0 we have

J=(AxAy)eiφx.\begin{align*} {\mathbf J}= \left(\begin{array}{c}{\cal A}_x \\{\cal A}_y \end{array}\right) e^{i \varphi_x}. \end{align*}

Equality of the phases: φy=φx\varphi_y=\varphi_x, means that the field components Ex(z,t){\cal E}_x(z,t) and Ey(z,t){\cal E}_y(z,t) are in phase: when Ex(z,t){\cal E}_x(z,t) is large, Ey(z,t){\cal E}_y(z,t) is large, and when Ex(z,t){\cal E}_x(z,t) is small, Ey(z,t){\cal E}_y(z,t) is small. We can write

(Ex(0,t)Ey(0,t))=(AxAy)cos(ωtφx),\begin{align*} \left(\begin{array}{c}{\cal E}_x(0,t) \\{\cal E}_y(0,t) \end{array}\right) = \left(\begin{array}{c}{\cal A}_x \\{\cal A}_y \end{array}\right) \cos(\omega t-\varphi_x), \end{align*}

which shows that for φyφx=0\varphi_y-\varphi_x=0 the electric field simply oscillates in one direction given by real the vector Axx^+Ayy^{\cal A}_x \hat{\mathbf{x}} + {\cal A}_y \hat{\mathbf{y}}. See Figure 1.

If φyφx=π\varphi_y-\varphi_x=\pi we have

J=(AxAy)eiφx.\begin{align*} {\mathbf J}= \left( \begin{array}{c}{\cal A}_x \\-{\cal A}_y \end{array}\right) e^{i \varphi_x}. \end{align*}

In this case Ex(z,t){\cal E}_x(z,t) and Ey(z,t){\cal E}_y(z,t) are out of phase and the electric field oscillates in the direction given by the real vector Axx^Ayy^{\cal A}_x \hat{\mathbf{x}} - {\cal A}_y \hat{\mathbf{y}}.

**b) Circular polarization: ** φyφx=±π/2\varphi_y-\varphi_x=\pm \pi/2, Ax=Ay{\cal A}_x={\cal A}_y. In this case the Jones vector is:

J=(1±i)Axeiφx.\begin{align*} {\mathbf J}= \left(\begin{array}{c}1 \\\pm i \end{array}\right) {\cal A}_x e^{i \varphi_x}. \end{align*}

The field components Ex(z,t){\cal E}_x(z,t) and Ey(z,t){\cal E}_y(z,t) are π/2\pi/2 radian (90 degrees) out of phase: when Ex(z,t){\cal E}_x(z,t) is large, Ey(z,t){\cal E}_y(z,t) is small, and when Ex(z,t){\cal E}_x(z,t) is small, Ey(z,t){\cal E}_y(z,t) is large. We can write for z=0z=0 and with φx=0\varphi_x=0:

(Ex(0,t)Ey(0,t))=(Axcos(ωt)Axcos(ωt±π/2))=Ax(cos(ωt)±sin(ωt))\begin{align*} \left( \begin{array}{c}{\cal E}_x(0,t) \\{\cal E}_y(0,t) \end{array}\right) = \left(\begin{array}{c}{\cal A}_x \cos(-\omega t)\\{\cal A}_x \cos(-\omega t \pm \pi/2) \end{array}\right) = {\cal A}_x \left(\begin{array}{c}\cos(\omega t)\\\pm \sin(\omega t) \end{array}\right) \end{align*}

At a given position, the electric field vector moves along a circle as time proceeds. When for an observer looking towards the source, the electric field is rotating anti-clockwise, the polarization is called **left-circularly polarized ** (+ sign in Eq. (9)), while if the electric vector moves clockwise, the polarization is called **right-circularly polarized ** (- sign in Eq. (9)).

c) Elliptical polarization: φyφx=±π/2\varphi_y-\varphi_x=\pm \pi/2, Ax{\cal A}_x and Ay{\cal A}_y arbitrary. The Jones vector is:

J=(Ax±iAy)eiφx.\begin{align*} {\mathbf J}= \left(\begin{array}{c}{\cal A}_x \\\pm i {\cal A}_y \end{array} \right) e^{i \varphi_x}. \end{align*}

In this case we get instead of Eq. (9) (again taking φx=0\varphi_x=0):

(Ex(0,t)Ey(0,t))=(Axcos(ωt)±Aysin(ωt)).\begin{align*} \left(\begin{array}{c}{\cal E}_x(0,t) \\{\cal E}_y(0,t) \end{array}\right) = \left( \begin{array}{c}{\cal A}_x \cos(\omega t)\\\pm {\cal A}_y\sin(\omega t) \end{array}\right). \end{align*}

which shows that the electric vector moves along an ellipse with major and minor axes parallel to the xx- and yy-axis. When the + sign applies, the field is called left-elliptically polarized, otherwise it is called right-elliptically polarized.

d) Elliptical polarization: φyφx=\varphi_y-\varphi_x= anything else, Ax{\cal A}_x and Ay{\cal A}_y arbitrary. The Jones vector is now the most general one:

J=(AxeiφxAyeiφy).\begin{align*} {\mathbf J}= \left(\begin{array}{c}{\cal A}_x e^{i \varphi_x} \\{\cal A}_y e^{i \varphi_y} \end{array}\right). \end{align*}

It can be shown that the electric field vector moves always along an ellipse. The exact shape and orientation of this ellipse varies with the difference in phase φyφx\varphi_y-\varphi_x and the ratio of the amplitude Ax,Ay{\cal A}_x,{\cal A}_y and, except when φyφx=±π/2\varphi_y-\varphi_x = \pm \pi/2, the major and minor axis of the ellipse are not parallel to the xx- and yy-axis. See Figure 3.

Remarks.

Jx2+Jy2=1.\begin{align*} |J_x|^2 + |J_y|^2 =1. \end{align*}

The normalized vector represents of course the same polarization state as the unnormalized one. In general, multiplying the Jones vector by a complex number does not change the polarization state. If we multiply for example by eiθe^{i\theta}, this has the same result as changing the instant that t=0t=0, hence it does not change the polarization state. In fact:

E(0,t)=Re[eiθJeiωt]=Re[Jeiω(tθ/ω)]\begin{align*} \mathbf{\mathcal{E}}(0,t) = \text{Re} \left[ e^{i \theta} \mathbf{J} e^{-i\omega t} \right] = \text{Re} \left[ \mathbf{J} e^{-i\omega( t- \theta/\omega)}\right] \end{align*}
Linear polarization state of electromagnetic waves. The electric field vector oscillates along a single fixed direction perpendicular to the direction of propagation, tracing out a straight line in the plane perpendicular to the wave vector.

Figure 1:Linear polarization state of electromagnetic waves. The electric field vector oscillates along a single fixed direction perpendicular to the direction of propagation, tracing out a straight line in the plane perpendicular to the wave vector.

Circular polarization

Figure 2:Circular polarization

Elliptical polarization

Figure 3:Elliptical polarization

Illustration of different types of polarization. The horizontal and vertical arrows indicate the momentary field components Ex,Ey{\cal E}_x, {\cal E}_y. The thick arrow indicates the vector E\mathbf{\mathcal{E}}. The black curve indicates the trajectory of E(t)\mathbf{\mathcal{E}}(t).

2Creating and Manipulating Polarization States

We have seen how Maxwell’s equations allow the existence of plane waves with many different states of polarization. But how can we create these states, and how do these states manifest themselves?

Natural light often does not have a definite polarization. Instead, the polarization fluctuates rapidly with time. To turn such randomly polarized light into linearly polarized light in a certain direction, we must extinguish the light polarized in the perpendicular direction.The remaining light is then linearly polarized along the desired direction. One could do this by using light reflected under the Brewster angle (which extinguishes p-polarized light), or one could let light pass through a dichroic crystal, which is a material which absorbs light polarized perpendicular to its so-called optic axis. A third method is sending the light through a wire grid polarizer, which consists of a metallic grating with sub-wavelength slits. Such a grating only transmits the electric field component that is perpendicular to the slits.

So suppose that with one of these methods we have obtained linearly polarized light. Then the question rises how the state of linear polarization can be changed into circularly or elliptically polarized light? Or how the state of linear polarization can be rotated over a certain angle? We have seen that the polarization state depends on the ratio of the amplitudes and on the phase difference φyφx\varphi_y-\varphi_x of the orthogonal components Ey{\cal E}_y and Ex{\cal E}_x of the electric field. Thus, to change linearly polarized light to some other state of polarization, a certain phase shift (say Δφx\Delta \varphi_x) must be introduced to one component (say Ex{\cal E}_x), and another phase shift Δφy\Delta \varphi_y to the orthogonal component Ey{\cal E}_y. We can achieve this with a birefringent crystal, such as calcite. What is special about such a crystal is that it has two refractive indices: light polarized in a certain direction experiences a refractive index non_o, while light polarized perpendicular to it feels another refractive index nen_e (the subscripts oo and ee stand for “ordinary” and “extraordinary”, but for our purpose we do not need to understand this terminology). The direction for which the refractive index is smallest (which can be either non_o or nen_e) is called the fast axis because its phase velocity is largest, and the other direction is the slow axis. Because there are two different refractive indices, one can see double images through a birefringent crystalHorvath et al., 2011. The difference between the two refractive indices Δn=neno\Delta n=n_e-n_o is called the birefringence.

Suppose ne>non_e>n_o and that the fast axis, which corresponds to non_o is aligned with Ex{\cal E}_x, while the slow axis (which then has refractive index nen_e) is aligned with Ey{\cal E}_y. If the wave travels a distance dd through the crystal, Ey{\cal E}_y will accumulate a phase Δφy=2πneλd\Delta \varphi_y=\frac{2\pi n_e}{\lambda}d, and Ex{\cal E}_x will accumulate a phase Δφx=2πnoλd\Delta \varphi_x=\frac{2\pi n_o}{\lambda}d. Thus, after propagation through the crystal the phase difference φyφx\varphi_y-\varphi_x has increased by

ΔφyΔφx=2πλd(neno).\begin{align*} \Delta\varphi_y-\Delta\varphi_x = \frac{2\pi}{\lambda}d(n_e-n_o). \end{align*}

2.1Jones Matrices

By letting light pass through crystals of different thicknesses dd we can create different phase differences between the orthogonal field components and in this way we can create different states of polarization. To be specific, let J\mathbf{J}, as given by Eq. (4), be the Jones vector of the plane wave before the crystal. Then we have, for the Jones vector after the passage through the crystal:

J~=MJ,\begin{align*} \mathbf{\tilde{J}}={\cal M}\mathbf{J}, \end{align*}

where

M=(e2πiλdno00e2πiλdne)=e2πiλdno(100e2πiλd(neno)).\begin{align*} {\cal M}= \left( \begin{array}{cc}e^{\frac{2\pi i}{\lambda } d n_o} & 0 \\0 & e^{\frac{2\pi i}{\lambda } d n_e} \end{array}\right) = e^{\frac{2\pi i}{\lambda } d n_o} \left( \begin{array}{cc}1 & 0 \\0 & e^{\frac{2\pi i}{\lambda } d (n_e -n_o)} \end{array}\right). \end{align*}

A matrix such as M{\cal M}, which transfers one state of polarization of a plane wave in another, is called a Jones matrix. Depending on the phase difference which a wave accumulates by traveling through the crystal, these devices are called quarter-wave plates (phase difference π/2\pi/2), half-wave plates (phase difference π\pi), or full-wave plates (phase difference 2π2\pi). The applications of these wave plates will be discussed in later sections.

Consider as example the Jones matrix which described the change of linear polarized light into circular polarization. Assume that we have diagonally (linearly) polarized light, so that

J=12(11).\begin{align*} J= \frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\1 \end{array}\right). \end{align*}

We want to change it to circularly polarized light, for which

J=12(1i),\begin{align*} J= \frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\ i \end{array}\right), \end{align*}

where one can check that indeed φyφx=π/2\varphi_y-\varphi_x=\pi/2. This can be done by passing the light through a crystal such that Ey{\cal E}_y accumulates a phase difference of π/2\pi/2 with respect to Ex{\cal E}_x. The transformation by which this is accomplished can be written as

(100i)12(11)=12(1i).\begin{align*} \left( \begin{array}{cc}1&0\\0&i \end{array}\right) \frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\1 \end{array}\right) =\frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\ i \end{array}\right). \end{align*}

The matrix on the left is the Jones matrix describing the operation of a quarter-wave plate.

Another important Jones matrix is the rotation matrix. In the preceding discussion it was assumed that the fast and slow axes were aligned with the xx- and yy-direction (i.e. they were parallel to Ex{\cal E}_x and Ey{\cal E}_y). Suppose now that the slow and fast axes of the wave plate no longer coincide with x^\widehat{\mathbf{x}} and y^\widehat{\mathbf{y}}, but rather with some other x^\widehat{\mathbf{x}}' and y^\widehat{\mathbf{y}}' as in Figure 4. In that case we apply a basis transformation: the electric field vector which is expressed in the x^\widehat{\mathbf{x}}, y^\widehat{\mathbf{y}} basis should first be expressed in the x^\widehat{\mathbf{x}}', y^\widehat{\mathbf{y}}' basis before applying the Jones matrix of the wave plate to it. After applying the Jones matrix, the electric field has to be transformed back from the x^\widehat{\mathbf{x}}', y^\widehat{\mathbf{y}}' basis to the x^ \widehat{\mathbf{x}}, y^\widehat{\mathbf{y}} basis.

Let E\mathbf{E} be given in terms of its components on the x^\hat{\mathbf{x}}, y^\hat{\mathbf{y}} basis:

E=Exx^+Eyy^.\begin{align*} \mathbf{E}=E_x \widehat{\mathbf{x}} + E_y \widehat{\mathbf{y}}. \end{align*}

To find the components ExE_{x'}, EyE_{y'} on the x^\widehat{\mathbf{x}}', y^\widehat{\mathbf{y}}' basis:

E=Exx^+Eyy^.\begin{align*} \mathbf{E}=E_{x'} \widehat{\mathbf{x}}' + E_{y'} \widehat{\mathbf{y}}'. \end{align*}

we first write the unit vectors x^\widehat{\mathbf{x}}' and y^\widehat{\mathbf{y}}' in terms of the basis x^\hat{\mathbf{x}}, y^\hat{\mathbf{y}} (see Figure 4)

x^=cosθx^+sinθy^,\begin{align*} \widehat{\mathbf{x}}' &= \cos\theta \, \widehat{\mathbf{x}} + \sin \theta \, \widehat{\mathbf{y}}, \end{align*}
y^=sinθx^+cosθy^,\begin{align*} \\ \widehat{\mathbf{y}}' &= -\sin\theta \, \widehat{\mathbf{x}} + \cos \theta \, \widehat{\mathbf{y}},\end{align*}

By substituting Eq. (23) and Eq. (24) into Eq. (22) we find

E=Exx^+Eyy^=Ex(cosθx^+sinθy^)+Ey(sinθx^+cosθy^),=(cosθExsinθEy)x^+(sinθEx+cosθEy)y^.\begin{align*} \mathbf{E} &= E_{x'} \widehat{\mathbf{x}}' + E_{y'} \widehat{\mathbf{y}}' \\ &= E_{x'} ( \cos\theta \, \widehat{\mathbf{x}} + \sin \theta \, \widehat{\mathbf{y}} )+ E_{y'} (-\sin\theta \, \widehat{\mathbf{x}} + \cos \theta \, \widehat{\mathbf{y}}), \\ &= ( \cos \theta E_{x'} - \sin\theta E_{y'} )\widehat{\mathbf{x}}+ ( \sin\theta E_x + \cos \theta E_y)\widehat{\mathbf{y}}. \end{align*}

Comparing with Eq. (21) implies

(ExEy)=(ExcosθEysinθExsinθ+Eycosθ)=Rθ(ExEy),\begin{align*} \begin{pmatrix} E_{x} \\ E_{y} \end{pmatrix}=\begin{pmatrix} E_{x'} \cos \theta - E_{y'} \sin \theta \\ E_{x'} \sin \theta + E_{y'} \cos \theta \end{pmatrix} = {\cal R}_{\theta} \begin{pmatrix} E_{x'} \\ E_{y'} \end{pmatrix}, \end{align*}

where Rθ{\cal R}_{\theta} is the rotation matrix over an angle θ\theta in the anti-clockwise direction: Hecht, 2002.

That R(θ){\cal R}(\theta) indeed is a rotation over angle θ\theta in the anti-clockwise direction is easy to see by considering what happens when Rθ{\cal R}_\theta is applied to the vector (1,0)T(1,0)^T Hecht, 2002.

This relationship expresses the components ExE_{x'}, EyE_{y'} of the Jones vector on the x^\widehat{\mathbf{x}}', y^\widehat{\mathbf{y}}' basis, which is aligned with the fast and slow axes of the crystal, in terms of the components ExE_x and EyE_y on the original basis x^\widehat{\mathbf{x}}, y^\widehat{\mathbf{y}}. If the matrix M{\cal M} describes the Jones matrix as defined in Eq. (17), then the matrix MθM_{\theta} for the same wave plate but with xx' as slow and yy' as fast axis, is, with respect to the x^\widehat{\mathbf{x}}, y^\widehat{\mathbf{y}} basis, given by:

Mθ=RθMRθ.\begin{align*} {\cal M}_{\theta}={\cal R}_{\theta}{\cal M} {\cal R}_{-\theta}. \end{align*}

This is a standard result from linear algebra involving basis transformations.

If the wave plate is rotated, the fast and slow axis no longer correspond to x and y. Instead, we have to introduce a new coordinate system (x',y').

Figure 4:If the wave plate is rotated, the fast and slow axis no longer correspond to xx and yy. Instead, we have to introduce a new coordinate system (x,yx',y').

2.2Linear polarizers

A polarizer that only transmits horizontally polarized light is described by the Jones matrix:

MLP=(1000).\begin{align*} {\cal M}_{LP}=\left( \begin{array}{cc}1&0\\0&0 \end{array}\right). \end{align*}

Clearly, horizontally polarized light is completely transmitted, while vertically polarized light is not transmitted at all. More generally, for light that is polarized at an angle α\alpha, we get

Mα=MLP(cosαsinα)=(1000)(cosαsinα)=(cosα0).\begin{align*} {\cal M}_\alpha={\cal M}_{LP}\left(\begin{array}{c}\cos\alpha\\\sin\alpha \end{array}\right)=\left(\begin{array}{cc}1&0\\0&0 \end{array}\right) \left(\begin{array}{c}\cos\alpha\\\sin\alpha \end{array}\right) = \left( \begin{array}{c}\cos\alpha\\0 \end{array}\right). \end{align*}

The amplitude of the transmitted field is reduced by the factor cosα\cos\alpha, which implies that the intensity of the transmitted light is reduced by the factor cos2α\cos^2 \alpha. This relation is known as Malus’ law.

2.3Degree of Polarization

Natural light such as sun light is unpolarized. The instantaneous polarization of unpolarized light fluctuates rapidly in a random manner. A linear polarizer produces linear polarized light from unpolarized light. It follows from Eq. (29) that the intensity transmitted by a linear polarizer when unpolarized light is incident, is the average value of cos2α\cos^2\alpha namely 12\frac{1}{2}, times the incident intensity.

Light that is a mixture of polarized and unpolarized light is called partially polarized. The degree of polarization is defined as the fraction of the total intensity that is polarized:

 degree of polarization=IpolIpol+Iunpol.\begin{align*} \text{ degree of polarization} = \frac{I_{pol}}{I_{pol} + I_{unpol}}. \end{align*}

2.4Quarter-Wave Plates

A quarter-wave plate has already been introduced above. It introduces a phase shift of π/2\pi/2, so its Jones matrix is

MQWP=(100i),\begin{align*} {\cal M}_{QWP}= \left(\begin{array}{cc}1&0\\0& i \end{array}\right), \end{align*}

because exp(iπ/2)=i\exp(i\pi/2)=i. To describe the actual transmission through the quarter-wave plate, the matrix should be multiplied by some global phase factor, but because we only care about the phase difference between the field components, this global phase factor can be omitted without problem. The quarter-wave plate is typically used to convert linearly polarized light to elliptically polarized light and vice-versaSaleh & Teich, 2007. If the incident light is linearly polarized at angle α\alpha, the state of polarization after the quarter-wave plate is

(cosαisinα)=(100i)(cosαsinα).\begin{align*} \left(\begin{array}{c}\cos\alpha\\ i\sin\alpha \end{array}\right) = \left( \begin{array}{cc}1&0\\0& i \end{array}\right) \left(\begin{array}{c}\cos\alpha\\\sin\alpha \end{array}\right). \end{align*}

In particular, if incident light is linear polarized under 45o45^o, or equivalently, if the quarter wave plate is rotated over this angle, it will transform linearly polarized light into circularly polarized light (and vice versa).

12(1i)=(100i)12(11).\begin{align*} \frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\ i \end{array}\right) = \left(\begin{array}{cc}1&0\\0& i \end{array}\right) \frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\1 \end{array}\right). \end{align*}

A demonstration is shown inBartholinus, 1669.

2.5Half-Wave Plates

A half-wave plate introduces a phase shift of π\pi, so its Jones matrix is

MHWP=(1001),\begin{align*} {\cal M}_{HWP}= \left( \begin{array}{cc}1&0\\0& -1 \end{array}\right), \end{align*}

because exp(iπ)=1\exp(i\pi)=-1. An important application of the half-wave plate is to change the orientation of linearly polarized lightFowles, 1989. After all, what this matrix does is mirroring the polarization state in the xx-axis. Thus, if we choose our mirroring axis correctly (i.e. if we choose the orientation of the wave plate correctly), we can change the direction in which the light is linearly polarized arbitrarilyGoldstein, 2011. To give an example: a wave with linear polarization parallel to the xx-direction, can be rotated over angle α\alpha by rotating the crystal such that the fast axis makes angle α/2\alpha/2 with the xx-axis. Upon propagation through the crystal, the slow axis gets an additional phase of π\pi, due to which the electric vector makes angle α\alpha with the xx-axis (see Figure 4). It is not difficult to verify that when the fast and slow axis are interchanged, the same linear state of polarization results.

Rotation of horizontally polarized light over an angle \alpha using a half-wave plate.

Figure 4:Rotation of horizontally polarized light over an angle α\alpha using a half-wave plate.

2.6Full-Wave Plates

A full-wave plate introduces a phase difference of 2π2\pi, which is the same as introducing no phase difference between the two field components. So what can possibly be an application for a full-wave plate? We recall from Eq. (15) that the phase difference is 2π2\pi only for a particular wavelength. If we send through linearly (say vertically) polarized light of other wavelengths, these will become elliptically polarized, while the light with the correct wavelength λ0\lambda_0 will stay vertically polarized. If we then let all the light pass through a horizontal polarizer, the light with wavelength λ0\lambda_0 will be completely extinguished, while the light of other wavelengths will be able to pass through at least partially. Therefore, * full-wave plates can be used to filter out specific wavelengths of light*.

3More on Jones matrices

If the direction of either the slow or fast axis is given and the ordinary and extra-ordinary refractive indices non_o and nen_e, it is easy to write down the Jones matrix of a birefringent plate of given thickness dd using the rotation matrices, see Eq. (27). Instead of using the rotation matrices, one can also write down a system of equations for the elements of the Jones matrix. Suppose that vo^=vo,x+x^+vo,yy^\hat{\mathbf{v_o}}=v_{o,x}+\hat{\mathbf{x}}+v_{o,y}\hat{\mathbf{y}} and ve^=ve,xx^+ve,yy^\hat{\mathbf{v_e}}=v_{e,x}\hat{\mathbf{x}}+ v_{e,y} \hat{\mathbf{y}}, are in the direction of the ordinary and the extra-ordinary axes, respectively. Then if the Jones matrix is

M=(abcd),{\cal M}=\left( \begin{array}{cc}a & b \\c & d \end{array}\right),

then

Mv^o=eiknodv^o,Mv^e=eiknedv^e\begin{align*} {\cal M} \hat{\mathbf{v}}_o & = e^{i k n_o d} \, \hat{\mathbf{v}}_o, \\ {\cal M} \hat{\mathbf{v}}_e & = e^{i k n_e d} \, \hat{\mathbf{v}}_e \end{align*}

which implies

avo,x+bvo,y=eiknodvo,x,cvo,x+dvo,y=eiknodvo,y,ave,x+bve,y=eiknedve,x,cve,x+dve,y=eiknedve,x.\begin{array}{cc}a v_{o,x} + b v_{o,y} & = e^{i k n_o d} v_{o,x}, \\c v_{o,x} + d v_{o,y} & = e^{i k n_o d} v_{o,y}, \\a v_{e,x} + b v_{e,y} & = e^{i k n_e d} v_{e,x}, \\c v_{e,x} + d v_{e,y} & = e^{i k n_e d} v_{e,x}. \end{array}

Similarly, for a linear polarizer it is simple to write down the Jones matrix if one knows the direction in which the polarizer absorbs or transmits all the light: use Eq. (28) in combination with the rotation matrices. Alternatively, if v^\hat{\mathbf{v}} is in the direction of the linear polarizer and w^\hat{\mathbf{w}} is perpendicular to it, we have

Mv^=v^Mw^=0,\begin{align*} {\cal M} \hat{\mathbf{v}} & = \hat{\mathbf{v}} \\ {\cal M} \hat{\mathbf{w}} & = \mathbf{0}, \end{align*}

which is a system of equation of type Eq. (37) for the elements of the Jones matrix.

Suppose now that the complex (2,2)-matrix Eq. (35) is given. How can one verify whether this matrix corresponds to a linear polarizer or to a wave plate? Note that the elements of a Jones matrix are in general complex.

1. Linear polarizer. The matrix corresponds to a linear polarizer if there is a real vector which remains invariant under M{\cal M} and all vectors orthogonal to this vector are mapped to zero. In other words, there must be an orthogonal basis of real eigenvectors and one of the eigenvalues must be 1 and the other 0. Hence, to check that a given matrix corresponds to a linear polarizer, one should verify that one eigenvalue is 1 and the other is 0 and furthermore that the eigenvectors are real orthogonal vectors. It is important to check that the eigenvectors are real because if they are not, they do not correspond to particular linear polarization directions and then the matrix does not correspond to a linear polarizer.

2. Wave plate. To show that the matrix corresponds to a wave plate, there should exist two real orthogonal eigenvectors with, in general, complex eigenvalues of modulus 1. In fact, one of the eigenvectors corresponds to the ordinary axis with refractive index non_{o}, and the other to the extra-ordinary axis with refractive index nen_e. The eigenvalues are then

eikn1d     and     eikn2d,\begin{align*} e^{i k n_1 d} \;\;\text{ and } \;\; e^{i k n_2 d}, \end{align*}

where dd is the thickness of the plate and kk is the wave number. Hence to verify that a (2,2)(2,2)-matrix corresponds to a wave plate, one has to compute the eigenvalues and check that these have modulus 1 and that the corresponding eigenvectors are real vectors and orthogonal.

3. Jones matrix for propagation through sugars In sugars, left and right circular-polarized light propagate with their own refractive index. Therefore sugars are called circular birefringent. The matrix Eq. (35) corresponds to propagation through sugar when there are two real orthogonal unit vectors v^\hat{\mathbf{v}} and w^\hat{\mathbf{w}} such that the circular polarization states

v^+iw^,      v^iw^\hat{\mathbf{v}}+ i \hat{\mathbf{w}}, \;\;\; \hat{\mathbf{v}}- i \hat{\mathbf{w}}

are eigenstates of M{\cal M} with complex eigenvalues with modulus 1.

4Decomposition of an Elliptical polarization state into sums of Linear & of Circular States

Any elliptical polarization state can be written as the sum of two perpendicular linear polarized states:

J=(AxeiφxAyeiφy)=Axeiφx(10)+Ayeiφy(01).\begin{align*} J= \left(\begin{array}{cc}{\cal A}_x e^{i \varphi_x} \\{\cal A}_y e^{i \varphi_y} \end{array}\right) = {\cal A}_x e^{i \varphi_x} \left( \begin{array}{c}1\\0 \end{array} \right) + {\cal A}_y e^{i \varphi_y} \left(\begin{array}{c}0 \\1 \end{array}\right). \end{align*}

Furthermore, any elliptical polarization state can be written as the sum of two circular polarization states, one right- and the other left-circular polarized:

J=(AxeiφxAyeiφy)=12(AxeiφxiAyeiφy)(1i)+12(Axeiφx+iAyeiφ)(1i).\begin{align*} J= \left(\begin{array}{c}{\cal A}_x e^{i \varphi_x} \\{\cal A}_y e^{i \varphi_y} \end{array}\right) =\frac{1}{2}({\cal A}_x e^{i \varphi_x} - i {\cal A}_y e^{i \varphi_y}) \left( \begin{array}{c}1\\i \end{array}\right) + \frac{1}{2} ( {\cal A}_x e^{i \varphi_x} + i {\cal A}_y e^{i \varphi}) \left(\begin{array}{c}1 \\-i \end{array}\right). \end{align*}

We conclude that to study what happens to elliptic polarization, it suffices to consider two orthogonal linear polarizations, or, if that is more convenient, left- and right-circular polarized light. In a birefringent material each of two linear polarizations, namely parallel to the o-axis and parallel to the e-axis, propagate with their own refractive index. To predict what happens to an arbitrary linear polarization state which is not aligned to either of these axes, or more generally what happens to an elliptical polarization state, we write this polarization state as a linear combination of o- and e-states, i.e. we expand the field on the o- and e-basis.

To see what happens to an arbitrary elliptical polarization state in a circular birefringent material, the incident light is best written as linear combination of left-and right-circular polarizations.

5Chapter Summary

6References

References
  1. Fowles, G. R. (1989). Introduction to Modern Optics. Dover Publications.
  2. Born, M., & Wolf, E. (1999). Principles of Optics (7th ed.). Cambridge University Press.
  3. Horvath, G., Barta, A., Pomozi, I., Suhai, B., Hegedus, R., Åkesson, S., Meyer-Rochow, B., & Wehner, R. (2011). Viking navigation using polarized skylight. Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences, 467(2127), 651–660.
  4. Hecht, E. (2002). Optics (4th ed.). Addison Wesley.
  5. Saleh, B. E. A., & Teich, M. C. (2007). Fundamentals of Photonics (2nd ed.). Wiley-Interscience.
  6. Bartholinus, E. (1669). Experimenta crystalli islandici disdiaclastici quibus mira & insolita refractio detegitur. Copenhagen: Daniel Paulli.
  7. Fowles, G. R. (1989). Introduction to Modern Optics. Dover Publications.
  8. Goldstein, D. H. (2011). Polarized Light (3rd ed.). CRC Press.