17. Jones Matrices & Mueller Matrices Jones Matrices

Rotation of coordinates - the rotation matrix

Stokes Parameters and unpolarized light

Mueller Matrices R. Clark Jones (1916 - 2004)

Sir George G. Stokes (1819 - 1903) Hans Mueller (1900 - 1965) Jones vectors describe the polarization state of a wave

Define the polarization state of a field as a 2D vector— “Jones vector” —containing the two complex amplitudes:

EExx 1 1  EE  x  22  EEEyyx E y   EExy  (normalized to length of unity)

1 A few examples: 0  1      0° linear (x) polarization: Ey /Ex = 0 tan  linear (arbitrary angle) polarization: Ey /Ex = tan 

right or left circular polarization: Ey /Ex = ±j  1     j To model the effect of a medium on light's polarization state, we use Jones matrices.

Since we can write a polarization state as a (Jones) vector, we use

matrices, A, to transform them from the input polarization, E0, to the output polarization, E . 1 aa EE A  11 12  10 A    aa21 22  This should be This yields: EaEaE1110120x  xy thought of as a transfer function. EaEaE1210220yxy

10 For example, an x-polarizer can be written: Ax    00 10E E So:  0x 0x EE10Ax    00E0 y 0 Other Jones matrices

00 A y-polarizer:   A y    01

R. Clark Jones A half-wave plate: 10 (1916 - 2004) AHWP    01  101   1       011    1 

A half-wave plate rotates 45-degree- 10 1   1       polarization to -45-degree, and vice versa. 011   1 

A quarter-wave plate: 10 101   1  AQWP         0  j 01 jj    The orientation of a wave plate matters. 0° or 90° Polarizer Remember that a quarter-wave plate only converts linear to circular if the input polarization is ±45°. Wave plate w/ axes at If it sees, say, x polarization, 0° or 90° the input is unchanged.

101   1       000 j   

AQWP

Jones matrices are an extremely useful way to keep track of all this. A wave plate example

What does a quarter-wave plate do if the input polarization is linear but at an arbitrary angle? 10 11     0 j tan j tan   

AQWP Ein Eout

For arbitrary , this is an elliptical polarization.

 = 30°  = 45°  = 60° Jones Matrices for standard components Rotated Jones matrices

What about when the polarizer or wave plate responsible for the transfer function A is rotated by some angle, ? Rotation of a vector by an angle  means multiply by the rotation matrix: rotated Jones vector of the input ERE0011' and ERE ' rotated Jones vector where: of the output   cos( ) sin( ) R   sin( ) cos( )  -1 Rotating E1 by  and inserting the identity matrix R() R(), we have:

ERER' AA ER   R1 R E 11  0   0  R AAARRERREE11   ' ' '  000   

1 Thus: AA'  RR   Rotated Jones matrix for a polarizer

1 Example: apply this to an x polarizer. AA' RR   

cos( ) sin( ) 1 0 cos( ) sin( ) Ax   sin( ) cos( ) 0 0 sin( ) cos( )

cos( ) sin( ) cos( ) sin( )   sin( ) cos( ) 0 0

 cos2 ( ) cos(  )sin( )   2  cos( )sin( ) sin ( )  So, for example:

 1/2 1/2  1   for a small Ax 45   Ax     1/2 1/2  0  angle   To model the effect of many media on light's polarization state, we use many Jones matrices.

The aggregate effect of multiple components or objects can be described by the product of the Jones matrix for each one.

inputtransfer function output   

E0 E1 A1 A2 A3

EE13210 AAA

The order may look counter-intuitive, but order matters! Multiplying Jones Matrices x y z Crossed polarizers: x-pol E0 E1 E10 AAyxE y-pol 00 10 00 AAyx so no light leaks through. 01 00 00

rotated Uncrossed polarizers x-pol

(by a slight angle ): E0 E1  00 1 00 y-pol AAyx 01 0 0

EExx00  0  AA  So I ≈ 2 I yx EE   out in,x yy0  Ex  

 Multiplying Jones Matrices x y z x-pol Now, it is easy to compute how E inserting a third polarizer 0 E between two crossed polarizers 1 leads to larger transmission. 45º-pol y-pol

E1450 AAyx AE

11 00 0022  10   AAA   yx45  1 01 1 1  00  0 22  2 

0 00E  xin,  Thus: E1  1 1 E  0 yin, Ex,in 2 2 The third polarizer, between the other two, makes the transmitted wave non-zero. Natural light (e.g., sunlight, light bulbs, etc.) is unpolarized

The direction of the E vector is randomly changing. But, it is always perpendicular to the propagation direction.

polarized light natural light Light with very complex polarization vs. position is "unpolarized."

Light that has scattered multiple times, or that has scattered randomly, often becomes unpolarized as a result.

Here, light from the blue sky is polarized, so when viewed through a polarizer it looks much darker. Light from clouds is unpolarized, so its intensity is reduced by only 50%.

If the polarization vs. position is unresolvable, we call this “unpolarized.” Otherwise, we refer to this light as “locally polarized” or “partially polarized.” When the phases of the x- and y-polarizations fluctuate, we say the light is "unpolarized."

Ezt(,) Re E exp jkzt t xx 0    x Ezt(,) Re E exp jkzt t yy0   y

where x(t) and y(t) are functions that vary on a time scale slower than the period of the wave, but faster than you can measure.

The polarization state (Jones vector) is:

 1  In practice, the   E0 y amplitudes are also  exp  jtjt   yx functions of time!  E0 x 

As long as the time-varying relative phase, x(t)–y(t), fluctuates, the light will not remain in a single polarization state and hence is unpolarized. Stokes Parameters We cannot use Jones vectors to describe something that is rapidly fluctuating like this. So, to treat fully, partially, or unpolarized light, we use a different scheme. We define "Stokes parameters."

Suppose we have four detectors, three with polarizers in front of them:

#0 detects total irradiance...... I Note that these 0 quantities are time- #1 detects horizontally polarized irradiance...... …...I1 averaged, so even randomly polarized #2 detects +45° polarized irradiance...... I2 light will give a well- #3 detects right circularly polarized irradiance.....…….I3 defined answer.

The Stokes parameters:

S0  I0 S1  2I1 – I0 S2  2I2 – I0 S3  2I3 – I0 Interpretation of the Stokes Parameters

The Stokes parameters:

S0  I0 S1  2I1 – I0 S2  2I2 – I0 S3  2I3 – I0

S0 = the total irradiance

S1 = the excess in intensity of light transmitted by a horizontal polarizer over light transmitted by a vertical polarizer

S2 = the excess in intensity of light transmitted by a 45° polarizer over light transmitted by a 135° polarizer

S3 = the excess in intensity of light transmitted by a RCP filter over light transmitted by a LCP filter

What we mean when we say ‘unpolarized light’: All three of these excess quantities are zero Degree of polarization

If any of the excess quantities (S1, S2, or S3) are non- zero, then the wave has some degree of polarization. We can quantify this by defining the “degree of polarization”:

2221/2 = 1 for polarized light Degree of polarization = S123 + S + S / S 0= 0 for unpolarized light

Note that this quantity can never be greater than unity,

since S0 is the total intensity. This is not the same as the ‘degree of polarization’ defined in the homework problem, which was only defined for fully polarized light. The Stokes vector

S0  We can write the four Stokes parameters in vector form: S  S   1  S2    S3 

The Stokes vector S contain information about both the polarized part and the unpolarized part of the wave.

S = S(1) + S(2) unpolarized part: polarized part:

222 SSSS222  SSS 0123  123 1 0 2  S  S   S  1   S  0  2    0  S3  Stokes vectors (and Jones vectors for comparison)

Sir George G. Stokes (1819 - 1903) Mueller Matrices multiply Stokes vectors

We can define matrices that multiply Stokes vectors, just as Jones matrices multiply Jones vectors. These are called Mueller matrices.

Sin Sout M1 M2 M3

To model the effects of more than one medium on the polarization state, just multiply the input polarization Stokes vector by all of the Mueller matrices:

Sout = M3 M2 M1 Sin

(just like Jones matrices multiplying Jones vectors, except that the vectors have four elements instead of two) Mueller Matrices (and Jones Matrices for comparison) With Stokes vectors and Mueller matrices, we can describe light with arbitrarily complicated combination of polarized and unpolarized light.

Hans Mueller (1900 - 1965)