Chapter 3 Polarization of Light WavesLecture 1 Polarization3.1 The concept of polarizationIntroduction:1)Since F = qE, polarization determines force direction. The generation, propagation and control of lasers thus crucially depend on their state of polarization.2)The electric field E is used to define the polarization state of the light.3)In an anisotropic material the index of refraction depends on the polarization state of the light, which is used to manipulate light waves.

1

3.2 Polarization of monochromatic plane waves

For a monochromatic plane wave propagating in the z direction, the x and y components of the electric field oscillate independently.

)cos(),()cos(),(

)]](exp[)ˆˆRe[(

)]](exp[Re[),(

yyy

xxx

iy

ix

kztAtzEkztAtzE

kztieAeA

kztitzyx

ji

AE Ex

Ey

2

Ex

Ey

t

Ex

t

EEy

3

The trajectory of the end point of the electric field vector, at a fixed point as time goes, is

. ,

sincos2

) geliminatin()cos()cos(

2

22

xy

y

y

x

x

y

y

x

x

yyy

xxx

AE

AE

AE

AE

kztkztAEkztAE

Light is thus generally elliptically polarized.A complete description of the elliptical polarization needs1)Orientation angle 2)Ellipticity (shape) a/b, and3)Handedness (sense of revolution, can be combined to show the sign of ellipticity).

Ax

Ay E

y

x

x'y'

ab

They can be obtained by rotating the ellipse to its normal coordinates:

cos2

2tan

2

cos4 ,

1

22

2222222222

2'

2'

yx

yx

yxyxyx

yx

AAAA

AAAAAAba

bE

aE

4

Handedness (sense of revolution) of elliptical polarization:Looking at the approaching light, if the E vector revolves counterclockwise, the polarization is right-handed, with sin <0. If the E vector revolves clockwise, the polarization is left-handed, with sin >0. Many books use the opposite definition.

= - -3/4 -/2 -/4 0/4 /2 3/4

Linear polarization:

x

y

x

y

xy

AA

EE

,0

Circular polarization:

yx

xy

AA

2

= -/2 /2

5

3.4 Jones vector representationA plane wave can be uniquely described by a Jones vector in terms of its complex amplitudes on the x and y axes:

If we are only interested in the polarization state of the wave, we use the normalizedJones vector, with J+J=1:

y

x

iy

ix

eAeA

J

. ,angle) (auxiliary arctan ,sin

cos),( xy

x

yi A

Ae

J

Examples:1)Linearly polarized light:

2)Right- and left-handed circularly polarized light:

.10

ˆ ,01

ˆ especially ,sin

cos and

sincos

yx

. 1

21ˆ and

12

1ˆ

ii

LR

Some relations:

LRyLRx

yxLyxR

LRyx

ˆˆ2

ˆ ,ˆˆ2

1ˆ

ˆˆ2

1ˆ ,ˆˆ2

1ˆ

0ˆˆ ,0ˆˆ

i

ii

Lecture 2 Jones vector

6

Jones vectors are important when applied with Jones calculus, which enables us to track the polarization state and the intensity of a plane wave when traversing an arbitrary sequence of optical elements.

= - -3/4 -/2 -/4 0/4 /2 3/4

Examples of Jones vectors:

2

15

1)1(2

15

121

51

)1(21

51

21

51

)1(21

51

21

51

)1(21

51

21

51

iiiiii

J

Light intensity:

22** , yxiy

ixi

yi

xiy

ix AA

eAeA

eAeAIeAeA

y

xyx

y

x

JJJ

7

Solution 1: Suppose the ellipse is erect after rotating the x, y axes by an angle , that is

baccba

cbaB

cbaA

ByAx

cbayx

cbaycbax

yxyxcyxbyxacxybyax

ByAx yxyyxx

yxyyxx

2tan02cos2sin2sin

cossincossin

cossinsincos

1''

1)2cos2sin2sin(''

)cossincossin(')cossinsincos('

)cos'sin')(sin'cos'()cos'sin'()sin'cos'(

.1'' and ,cos'sin'sin'cos'

,cossin'

sincos'

22

22

22

222222

2222

22

(*Reading) What is the orientation angle of the ellipse How long are the principle axes?

? 122 cxybyax

8

2)()(

,

2tan

then,'' If

2)()(

2)()(1

)()(tan

tan1tan22tan

tan22tan

sin22sin2

1

2tan

2sin2

sin)(1cossinsincos

1

)2

or axes, principle thebe (Let

22

2222

2/12222

2

22

2

2

2

2222

2

cbabaBA

bac

ByAxcxybyax

cbabar

cbabar

ccbaba

bac

cacar

bac

cabarcba

r

r

m

mm

mm

mm

mm

mm

m

m

mmmmmmm

m

mm

2)()(

,

2tan

then,'' If

22

2222

cbabaBA

bac

ByAxcxybyax

m

9

2/1

22222222

2/1

22222222

2/1

222

22222222

2/1

2

22

2222

2/122

0'0'

22

22

2

22

2

cos4

cos4

2sin2

cos4

sin2/cos211112

)()(,

cos2

11

cos2

2tan

,sincos2 of case For the

yxyxyx

yxyxyxyx

yxyxyx

yxyxyxyx

yx

yx

yx

yxm

y

y

x

x

y

y

x

x

AAAAAA

AAAAAAAA

AAAAAA

AAAAAAcbaba

EE

AAAA

AA

AAba

c

AE

AE

AE

AE

10

Solution 2: Let us use a little linear algebra.

bacbac

cbaB

cbaA

cbabac

baccba

bcca

Rbc

caR

BA

yx

BA

yxByAx

yx

Ryx

yx

yx

bcca

yxcxybyax

2tan0cossin)()sin(cos2

cossincossin

cossinsincos

cossinsincoscossin)()sin(cos2

cossin)()sin(cos2

cossinsincos

cossinsincos

2/2/

cossinsincos

2/2/

00

1''

00

''1''

cossinsincos

''

12/

2/1

22

22

22

2222

2222

1

22

22

Solution 3: The problem is: given the condition for x and y, what is the maximum x2 + y2? This can be solved by the Lagrange multiplier method.

11

bac

cacb

cyaxcxby

xy

cxbyyyf

cyaxxxf

cxybyaxyxyxf

2tantan1tan2

tan2tan2

22tan

0)2(2

0)2(2

)1(),,(

2

2222

cossincossin

cossinsincoscossinsincos

1

.2tan02cos2sin2sin

0cossinsincos 010

.cossinsincos

11

22

22

222

222

22222

cbaB

cbaAcba

r

baccba

cbadd

rdd

ddr

cbarcxybyax

m

Solution 4: In polar coordinates, The curve is now .sin ,cos ryrx

122 cxybyax

12

indices). repeatedover (sum 1''''''

is,That

1'''

'''''''''

'''

as systems coordinate '-- and -- thein expressed is surface quadratica Surppose

333231

232221

131211

333231

232221

131211

rSrSrr TTjijijiji xSxxSx

zyx

SSSSSSSSS

zyxzyx

SSSSSSSSS

zyx

zy'x'zyx

Reading: Diagonalizing the tensor of a quadratic surfaceQuestion: What is the orientation angle of the ellipse

How long are the semiaxes?

Solution 5:Diagonalizing the tensor of a quadratic surface

? ''1 2222 ByAxcxybyax

y

x

x'y'

A-1/2

B-1/2

13

.100

,010

,001

have we,1 since Also

.' ,' ,'

'000'000'

''

, diagonizes tion tranformacoordinate othogonal theSuppose

133

123

113

132

122

112

131

121

111

1

133

123

113

31

33

123

113

132

122

112

21

32

122

112

131

121

111

11

31

121

111

3

2

1

133

132

131

123

122

121

113

112

111

133

132

131

123

122

121

113

112

111

111

RRR

RRR

RRR

RRR

SRRR

RRR

SRRR

RRR

SRRR

SS

S

RRRRRRRRR

RRRRRRRRR

RRR

RR

SSS

S

SRSRRSRSSR

a tensor. is surface quadratic thespecifies that .'

1'''')(')'()(')'()'(

then, ,' systems,

coordinate two therelates that tion tranformacoordinate othogonal theis Suppose

1

11111

1

SRSRSrSrrRSRrrRSRrrRSrRSrr

RRRrrR

TTTTTT

T

14

Therefore we conclude:If the orthogonal transformation R diagonalizes S into S',

then1) The diagonal elements of S' are the eigenvalues of S.2) The column vectors of R-1 (or the row vectors of R) are the eigenvectors of S.3) The new coordinate axes lie along the eigenvectors of S.

'000'000'

'

3

2

11

SS

SSRSR

.1 and 1

simply thenare semiaxes The .1'' in tscoefficien theare seigenvalue The 3)

. angle norientatio show the whichcurve, theof axes principle theare rseigenvecto The 2) . of seigenvalue and rseigenvecto thefind is,That . ng Diagonizi1)

:Strategy2/

2/1

2/2/

1

questionour Back to

22

22

BA

ByAx

bcca

yx

bcca

yxcxybyax

SS

S

15

bac

yyyyx

yxyx

cbabaB

rcbaba

B

cbabaA

rcbaba

A

cbabac

ycxcbaba

yx

bcca

cbababcca

m

m

2121

1121

21

11

2/122

2

22

2

2/122

1

22

1

222,1

22

22

2,1

)(222tan

2)()(1

2)()(

2)()(1

2)()(

)()( vectorsEigen

022

)()(0

2/2/

.2

)()( sEigenvalue0

2/2/

r