chapter matrix treatment of polarization
TRANSCRIPT
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
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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⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
+ =
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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
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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.
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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
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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.
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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= = + = ⎜ ⎟⎝ ⎠
>⎧> ⎨ <⎩
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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
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linearly polarized light at an angle 45 .
Summary of the polarization states and their Jones vectors
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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
5/11/2009 Matrix Treatment of Polarization 11
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
5/11/2009 Matrix Treatment of Polarization 12
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− =
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/ 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
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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
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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= → = −
−
>
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0 and C>0 so they have the same sign so the elliptically polarized light has counterclockwise rotation. A >
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