chapter matrix treatment of polarization

Chapter 14Matrix Treatment of Polarization

Lecture Notes for Modern Optics based on Pedrotti & Pedrotti & Pedrotti

Instructor: Nayer EradatInstructor: Nayer EradatSpring 2009

5/11/2009 1Matrix Treatment of Polarization

PolarizationPolarization of an electromagnetic wave is direction of the electric field vector E.

Mathematical presentation of polarized light: Jones VectorsMathematical presentation of polarized light: Jones Vectors

Mathematical presentation of polarizers (optical components) : Jones Matrices

Linear polarizer

Phase retarder

Rotators

5/11/2009 2Matrix Treatment of Polarization

Mathematical presentation of polarized light

Electric field of an electromagnetic wave propagating along z-direction: ( )

( )( )( )

0

0

g p p g g g

Re with and complex field components

Re

x

y

i kz txx xx

x y i kz ty yy y

E EE E eE E

E E e E E

ω φ

ω φ

− +

− +

⎧⎧ ⎫ ==⎪ ⎪ ⎪= + ⎨ ⎬ ⎨= =⎪ ⎪ ⎪⎭⎩ ⎩

E x y( )

( )0

x

y y

i kz txE e Eω φ− +

⎭⎩ ⎩

= +E x ( ) ( ) ( )

0

00 0 0Plane wave

Complex amplitude vector also contains phase

y yxi kz t i i kz t i kz ti

y x ye E e E e e eω φ φ ω ωφ− + − −⎡ ⎤= + =⎣ ⎦E

y x y E

also contains phase

State of polarization of a wave is determined by

relative amplitudes and phases of of the components of

0

00

that constitute a vector dubbed Jones vector:

xi

x xi

E eE φ

φ

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥

E

E

( ) ( )0

220 0Jones vector are normalized if 1.

yiy y

x y

E eE

E E

φ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

+ =

5/11/2009 Matrix Treatment of Polarization 3

Special cases of Jones vector

Particular forms of Jones vector:

x y 0

linearly polarized light along a line making an angle with the x axis:

φ φ

α

= =

00

0

cos cos

sin sin

x

y

ix

iy

E e AE A

AE e

φ

φ

α αα α

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

( )/ 2⎡ ⎤ ⎡ ⎤0Vertical polarization: E

( )( )

( )

cos / 2 0

1sin / 2

cos 0 1

π

π

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥

⎣ ⎦⎢ ⎥⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥

( )( )

( )

0

0

00

Horizontal polarization: 0sin 0

cos 60 11Polarized at 60 :

E

α

⎡ ⎤= =⎢ ⎥ ⎢ ⎥

⎣ ⎦⎢ ⎥⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥= = = ⎢ ⎥E

( )0

0Polarized at 60 :

2 3sin 60

: The light presented by a Jones vactor that both o

α ⎢ ⎥= = = ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦E

Conclusion1f its elements and are real (not both zero)a b


both o

( )-1

f its elements, and , are real (not both zero)

is a linearly polarized light along the angle tan /

a b

b aα =

Lissajous Figures

x yFor general case of 0 and 0 the head of the vector traces an ellipse rather than φ φ≠ ≠ E

ox oya straight line. The relative phase difference of the E and E , determines the

shape of the Lisy xφ φ φΔ = −

sajous figure and the state of polarization of the wave.


LCP and RCP0Example: consider electric field of an EM wave that has

and leads by / 2. Determine the state of polarization and x oy

x y

E E A

E E ε π

= =

=

deduce the normalized Jones vectors for this light. We write the com

y

plex amplitudes as coscos cosi t xE A tE A tE E e E A tω ωω ω− =⎧⎧ == =⎧ ⎧⎪ ⎪ ⎪ ⎪

( ) ( )

( )

0

0

2 2 2 2 2 2 2

cos cossincos cos

2

cos sin the vector traces out a circle of rad

x xx x

i ty yy yy

x y

E A tE E e E A tE A tE A t E A tE E e

E E E A t t A

ω ε

ω ωπ ωω ε ω

ω ω

− −

⎧ == =⎧ ⎧⎪ ⎪ ⎪ ⎪→ → →⎨ ⎨ ⎨ ⎨⎛ ⎞ == − = −= ⎪⎪ ⎜ ⎟ ⎩⎪ ⎪⎩⎩ ⎝ ⎠⎩

= + = + = E ius A.( )y

( )( )

0 00 / 2

* 2 2 * 2

1Finding the Jones vector: then

0, / 2

N li i 1 1 2 1 1/ 2

x

y

ix oy x

i ix y oy

E E A E e AA

iAeE e

E E A A A

φ

φ πφ φ π

⎡ ⎤= =⎧ ⎡ ⎤ ⎡ ⎤⎪ = = =⎢ ⎥⎨ ⎢ ⎥ ⎢ ⎥= = ⎢ ⎥ ⎣ ⎦⎪ ⎣ ⎦⎩ ⎣ ⎦E

( )( )2 2 * 20 0

0

Normalization: =1 1 2 1 1/ 2

11The normalized Jones vector is: 2

E E A ii A A

i

→ + = = → =

⎡ ⎤= ⎢

⎣ ⎦E We call this a left-circularly polarized ⎥

light or LCP since when we view this light head-on we see the vector tip is rotating

counterclockwise on a circle of radius 1/ 2. Figure shows the at di0

0

E

E

0

ffrent times. 11If leads by /2 the would rotate clockwiseE E π⎡ ⎤

= ⎢ ⎥E E


0If leads by /2 the would rotate clockwise. 2

We have right-circularly polarized light or RCP in this case.

y xE Ei

π = ⎢ ⎥−⎣ ⎦0E E

Elliptically polarized light0Example: consider electric field of an EM wave that has and where A and B

are positive numbers and leads/lags by / 2. Determine the state of polarizationx oy

x y

E A E B

E E ε π

= =

=

and deduce the normalized

( ) ( )0

Jones vectors for this light. coscosi t x

x xx

i

E A tE A tE E e ω ωωπ

− =⎧⎧ == ⎧⎪ ⎪ ⎪→ →⎨ ⎨ ⎨ ⎛ ⎞⎪( ) ( )0

cos cos2

cossin

i ty y yy

x

E B t E B tE E e

E A tE B t

ω ε πω ε ω

ωω

− −⎨ ⎨ ⎨ ⎛ ⎞= − = −= ⎪ ⎜ ⎟⎪ ⎪⎩⎩ ⎝ ⎠⎩=⎧⎪→ ⎨ =⎪⎩

( )( )( )* *2 2 20 0

sin

Normalization: =1 1

yE B t

E E A iB iB A B

ω=⎪⎩

→ + = + =

1 1A A⎡ ⎤ ⎡ ⎤Jones vector counte 0 / 22 2 2 2

0 / 22 2 2 2

1 1rclockwise

1 1Jones vector clockwise

i

i

iBBeA B A BA A

iBBeA B A B

π

π−

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥

+ + ⎣ ⎦⎣ ⎦⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥−+ + ⎣ ⎦⎣ ⎦

E

E2 2 2 2

: the Jones vector with elements un-equal in magnitude, one of which is pur

iBBeA B A B −+ + ⎣ ⎦⎣ ⎦

Conclusion 2e imaginary, represents an elliptically polarized light.


0 0

0 0

Figure shows the for two cases of (major axis along y) and

(major axis along x). y x

y x

E E

E E

>

<0E

Elliptically polarized light oriented at an angle relative to x‐axis

( )0Example: consider electric field of an EM wave that has and where A and B

are positive numbers and and have phase difference of 1/ 2 andx oyE A E b

E E mφ π

= =

Δ ≠ ± +( )are positive numbers and and have phase difference of 1/ 2 and

where 0, 1, 2,... Determine

x yE E m

m m

φ π

φ π

Δ ≠ ± +

Δ ≠ ± = ± ±the state of polarization and deduce the normalized Jones vectors for this light.

We can ass me and 0φ ε φ φ εΔ

00

0

We can assume and 0,

Counterclockwicos sin

x

y

x y

ix

i iy

E e A A Ab ib B iCbeE e

φ

φ ε

φ ε φ φ ε

ε ε

Δ = = =

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ +⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦

E se rotation, general case⎣ ⎦

( )( )( )( )* *2 2 2 20 0Normalization: =1 1

Jones vector of an elliptically polarized light with major axis inclined at an angle is:

E E A B iC B iC A B C

α

→ + + + = + + =

⎡ ⎤0

2 2 2

21 where tan 2AB iCA B C

α⎡ ⎤

= =⎢ ⎥++ + ⎣ ⎦E 0

2 20 0

2 2 -1

cos

and tan

x oy

x y

E EE E

CE A E B C

ε

ε

−

⎛ ⎞= = + = ⎜ ⎟0 0and , , tan

0 counteclockwiseIf 0 and

0 clockwise

x yE A E B CB

CA

C

ε= = + = ⎜ ⎟⎝ ⎠

>⎧> ⎨ <⎩


Note: polarization state represented by the Jones vector does not change if it is multiplied by a constant. So we can always make 0. A >

Usefulness and some properties of the Jones vectorsTwo properties of the polarization vector or Jones vector :

1) polarization state of a wave does not change if its Jones vector is multiplied by a constant.

It only affects the amplitude.2) polarizatio

p p p

n state of a wave does not change if its Jones vector is multiplied by a constant phase factor . ie φ) p g p y p

It promotes phase of each element by but not the phase difference .

: illustrating usefu

φ φΔ

Example 1 lness of the Jones vectors.a) Polarization state of a superposition of two waves can be found by adding the Jones vectors: 1 1 2

0i i

⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

: We can generate a linearly polarized Conclusion light with mixing equal portions of LCP and RCP light.

: superposition of horizontally and vertically linearly polarized light:0 1 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤

Example 20 1 11 0 1

: y mixing equal protions of b

⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦Conclusion

0

vertically and horizontally linearly polarized light we can get linearly polarized light at an angle 45


linearly polarized light at an angle 45 .

Summary of the polarization states and their Jones vectors


Mathematical presentation of polarizers

We can represent an optical instrument or device by its transfer matrix or abcd matrix: a bc d⎡ ⎤

= ⎢ ⎥⎣ ⎦

M

There are optical devices that affect (change) state of polarization of the light. Our goal is to rec d⎣ ⎦

presenteach of these devices with a transfer matrix such that multiplying it with the Jones vector of the original light produces the resulting light

0

light produces the resulting light.x x

y y

i ix x

i ioy y

E e E ea bc d E e E e

φ φ

φ φ

⎡ ⎤ ⎡ ⎤⎡ ⎤=⎢ ⎥ ⎢⎢ ⎥

⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣ or =⎥

⎥⎦0ME E


Linnear polarizers: it selectively removes virations in a given direction and transmits in perpendicular direction.

: sometimes the process of removing other polarizations is partial

Linear polarizer

Partial polarization and not 100% efficient: sometimes the process of removing other polarizations is partial Partial polarization

( ) ( )( ) ( )

and not 100% efficient. See the figure how the output light is polarized along the transmission axis (TA).

0 1 00 0Along the TA:

1 1 0 1 1

a ba bc d c d

⎧ + =⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎪= → ⎨⎢ ⎥ ⎢ ⎥ ⎢ ⎥ + =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎪⎩ ( ) ( )1

Perpendicular to TA: a bc d

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎪⎩

⎡ ⎤⎢ ⎥⎣ ⎦

( ) ( )( ) ( )1 0 00

0 0 1 0 0

The result for linear polarizer along the y axis (vertically) is:

a b

c d

⎧ + =⎡ ⎤ ⎡ ⎤ ⎪= → ⎨⎢ ⎥ ⎢ ⎥ + =⎣ ⎦ ⎣ ⎦ ⎪⎩p g y ( y)

0 0 1 0Linear polarizer, TA vertical . Linear polarizer, TA horizontal

0 1 0 0

:

⎡ ⎤ ⎡ ⎤→ = → =⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦M M

Exercise 0 Derive the polarization matices for the linear polarizer at 45 :

Polarization same Polarization as the polarizer's TA perpendicular to

the polarizer'

1 1 1 and

1 1 1a b a bc d c d⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

s TA

00⎡ ⎤

= ⎢ ⎥⎣ ⎦

the polarizer s TA

0 1 11Linear polarizer, TA at 45 : 1 12

The most general case of the linear polarizer:

⎡ ⎤= ⎢ ⎥

⎣ ⎦M


2

2

cos sin cosLinear polarizer, TA at :

sin cos sinθ θ θ

θθ θ θ

⎡ ⎤= ⎢ ⎥⎣ ⎦

M

Phase retarderThe pase retarder introduces a phase difference between the orthogonal polarization components. If the speed of light in each orthogonal direction is different, there would be a cumulative phase difference between the components as light emerges from the media.

(FA): the axis along which the speed of light is faster or index of refraction is lower.

(SA): the axis along which

φΔFast axis

Slow axis the speed of light is slower or index of refraction is higher.

t t i th t ill t fFi di th t i f t d( ) ( )

0 0 0 0

0

: we want a matrix that will transform

to and to y yyx xxiiii

x x y y

x

E e E e E e E e

E ea b

φ εφφ εφ ++

⎡ ⎤⎢ ⎥

Finding the matrix for retarder

( )

( )0 0

Phase retarderx xx x

ii ixE e eφ εφ ε+⎡ ⎤⎡ ⎤ ⎡ ⎤

⎢ ⎥= → =⎢ ⎥ ⎢ ⎥⎢ ⎥

Mc d⎢ ⎥⎣ ⎦

M

( )0 0

Phase retarder0

(QWP), a retarder with the net phase difference /2

SA h i l l d

y yy yi iiy y

E e eE eφ εφ ε

π

π π π

+⎢ ⎥ →⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

−

M

Quarter wave plate/ 4 0ie π⎧ ⎡ ⎤

⎪ ⎢ ⎥Mx y SA horizontal, let , and 2 4 4x yπ π πε ε ε ε− = = = − → / 4

/ 4

x y / 4

0

0 SA vertical, let , and

2 4 4 0

i

i

y x i

e

ee

π

π

π

π π πε ε ε ε

−

−

⎡ ⎤=⎪ ⎢ ⎥

⎪ ⎣ ⎦⎨

⎡ ⎤⎪ − = = − = + → = ⎢ ⎥⎪⎣ ⎦⎩

M

M

/ 4 / 41 0 1 0QWP, SA vertical, QWP, SA horizontal

0 0

(HWP): a retarde

i ie ei i

π π− ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

−

M M

Half wave plate r with the net phase difference x yε ε π− =


/ 2 / 21 0 1 0QWP, SA vertical, QWP, SA horizontal

0 1 0 1i ie eπ π− ⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦M M

Phase rotator

( )The phase rotator rotates the drection of polarization of the linearly polarized light by some angle . β

⎡ ⎤( )( )

coscossin sin

i

a bc d

θ βθθ θ β

β β

⎡ ⎤+⎡ ⎤ ⎡ ⎤= →⎢ ⎥⎢ ⎥ ⎢ ⎥ +⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦

⎡ ⎤M

cos sin Phase rotator

sin cosβ ββ β

−⎡ ⎤= ⎢ ⎥⎣ ⎦

M


Example: Production of circularly polarized light by combining a linear polarizer with a QWP

1 0 1 11 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞

0

/ 4 / 4

Linearly polarized The incident light is divided l i ll b t th l d

1 0 1 11 10 12 2

i i

QWP

e ei i

π π⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟− −⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦

0slow axis equally between the slow and light at 45horizontal fast axis and becomes

right-circularly polarized

/ 4 / 41 0 1 11 1i ie eπ π− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟⎝ ⎠

0 Linearly polarized The incident light is divided

slow axis equally between the slow light at 45vertical and fast axis a

0 12 2QWP

i i⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦

nd becomes left-circularly polarizedleft circularly polarized


Example: Left‐circularly polarized light is passing through an eighth wave plate

Eighth wave plate is a phase retarder that introduces a relative phase difference

x

x

0/ 4/ 4

of 2 /8 or /4 between the SA and FA. Assume 0

0 1 000

x

yy

i

ii

eee

εε

πε πε

π π ε

==

=

⎡ ⎤ ⎡ ⎤= ⎯⎯⎯→⎢ ⎥ ⎢ ⎥

⎣ ⎦⎢ ⎥⎣ ⎦M

Phase retarderl

00E

ee ⎣ ⎦⎢ ⎥⎣ ⎦ wave plate

/ 4 / 4 3 / 4

1 11 0 10

ighth

i i ie i ie eπ π π

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦Left-circularly wave plate Elliptically polarized light polarized light

3 / 4

0

1 1

Eighth

e i ie e

e iπ

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= − +

0

0

2 2

wherx

y

E AB iCE

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥+⎣ ⎦⎢ ⎥⎣ ⎦

2 20x 0

1 1e E 1, and , 1 and 2 2 yA B C E B C= = = − = = + =

⎣ ⎦

0 02 20 0

2 costan 2 45

0 and C>0 so they have the same sign so the elliptically polarized light has

x oy

x y

E EE E

A

εα α= → = −

−

>


0 and C>0 so they have the same sign so the elliptically polarized light has counterclockwise rotation. A >

chapter matrix treatment of polarization

Documents