Polarization
Jones vector & matrices
Phys 375
1
Matrix treatment of polarization

Consider a light ray with an instantaneous Evector as shown

E z, t   iˆEx z, t   ˆjE y z, t 
y
Ey
x
E x  Eoxe
i  kz t  x 
Ex
E y  Eoy e

i kz t  y

2
Matrix treatment of polarization

Combining the components

i kz t  y 
i  kz t  x 
ˆ
ˆ
E  i Eox e
 jEoy e

i y
i x
ˆ
ˆ
E  i Eox e  jEoy e e i  kz t 
 ~ i  kz t 
E  Eo e



The terms in brackets represents the complex
amplitude of the plane wave
3
Jones Vectors

The state of polarization of light is determined by

the relative amplitudes (Eox, Eoy) and,
 the relative phases ( = y - x )
of these components

The complex amplitude is written as a twoelement matrix, the Jones vector
~
i x



Eox 
Eox e 
Eox
~
i x 
Eo   ~   
e 
i 
i y 
 Eoy   Eoy e 
 Eoy e 
4
Jones vector: Horizontally polarized light


The electric field oscillations
are only along the x-axis
The Jones vector is then
written,
~

Eox   Eox e i x   A
1
~
Eo   ~   
     A 
 Eoy   0   0 
0 
where we have set the phase
x = 0, for convenience
The arrows indicate
the sense of movement
as the beam
approaches you
y

E
x
The normalized form
is
1 
0 
 
5
Jones vector: Vertically polarized light


The electric field
oscillations are only along
the y-axis
The Jones vector is then
written,
~

  0  0
E
0 
~
ox
Eo   ~   
i y      A  
 Eoy   Eoy e   A
1

Where we have set the
phase y = 0, for
convenience
y

E
x
The normalized form
is
0 
1 
 
6
Jones vector: Linearly polarized light at
an arbitrary angle


If the phases are such that  = m for
m = 0, 1, 2, 3, …
Then we must have,
Ex
m Eox
  1
Ey
Eoy
y

x
and the Jones vector is simply a line
inclined at an angle  = tan-1(Eoy/Eox)
since we can write
~


E
~
m cos  
ox
Eo   ~   A 1 

sin

E
 oy 


7
Circular polarization
y



Suppose Eox = Eoy = A
and Ex leads Ey by
90o=/2
At the instant Ex
reaches its maximum
displacement (+A), Ey
is zero
A fourth of a period
later, Ex is zero and
Ey=+A
x
t=0, Ey = 0, Ex = +A
t=T/8, Ey = +Asin 45o, Ex = Acos45o
t=T/4, Ey = +A, Ex = 0
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Circular polarization

The Jones vector for this case – where Ex leads Ey is
i
~  Eoxe x   A 
Eo  
i  
i y   
2
 Eoy e   Ae 
1
A 
i 

The normalized form is,

This vector represents circularly polarized light, where
E rotates counterclockwise, viewed head-on
This mode is called left-circularly polarized light
What is the corresponding vector for right-circularly
polarized light?


1
2
1
i 
 
Replace /2 with -/2 to get
1
2
1
 i 
 
9
Elliptically polarized light


If Eox  Eoy , e.g. if Eox=A and Eoy = B
The Jones vector can be written
 A
iB 
 
counterclockwise
 A 
  iB 


clockwise
Here A>B
10
Jones vector and polarization

In general, the Jones vector for the arbitrary
case is an ellipse ( m; (m+1/2))
y
Eoy
 Eox  
A

~
Eo  

i 
 Bcos   i sin  
E
e
oy


 
b
a
tan 2 
2 Eox Eoy cos 
E ox2  E oy2

x
Eox
11
Optical elements: Linear polarizer

Selectively removes all or most of the Evibrations except in a given direction
TA
y
x
Linear polarizer
12
Jones matrix for a linear polarizer
Consider a linear polarizer with transmission axis along the
vertical (y). Let a 2X2 matrix represent the polarizer
operating on vertically polarized light.
The transmitted light must also be vertically polarized. Thus,
 a b   0  0 
 c d  1  1

   
Operating on horizontally polarized light,
a b  1 0
 c d   0   0 

   
Thus,
0 0
M 

0
1


Linear polarizer with TA
vertical.
13
Jones matrix for a linear polarizer

For a linear polarizer with a transmission
axis at 
 cos 
M 
sin  cos 
2
sin  cos 

2
sin  
14
Optical elements: Phase retarder


Introduces a phase difference (Δ) between
orthogonal components
The fast axis(FA) and slow axis (SA) are shown
FA
y
x
SA
Retardation plate
15
Jones matrix of a phase retarder

We wish to find a matrix which will transform the
elements as follows: Eoxe i x int o Eoxei  x  x 
i y
i  y  y 
Eoy e
int o Eoy e

It is easy to show by inspection that,
ei x
M 
 0

0 
i y 
e 
Here x and y represent the advance in phase of the
components
16
Jones matrix of a Quarter Wave Plate




Consider a quarter wave plate for which |Δ| =
/2
For y - x = /2 (Slow axis vertical)
Let x = -/4 and y = /4
The matrix representing a Quarter wave plate,
with its slow axis vertical is,
e i 4
M 
 0
i 1
0 
4

e


0
i
4
e 

0
i 
17
Jones matrices: Half-wave Plate

For |Δ| = 
 e  i 2
M 
 0
e i 2
M 
 0
 i 1
0 
2

e


0
i
e 2 

i 1
0 
2

e


0
i
2
e


0
 1
0
 1
HWP, SA vertical
HWP, SA horizontal
18
Optical elements:
Quarter/Half wave plate

When the net phase difference
Δ = /2 : Quarter-wave plate
Δ =  : Half-wave plate

/2
19
Optical elements: Rotator

Rotates the direction of linearly polarized
light by a particular angle 
y

x
SA
Rotator
20
Jones matrix for a rotator


An E-vector oscillating linearly at  is rotated by
an angle 
Thus, the light must be converted to one that
oscillates linearly at ( +  )
a
c

b  cos  cos   

d   sin    sin    
cos 
 One then finds M  
 sin 
 sin  
cos  
21