# pol-xtal-EO-lab12 ```Jones Vectors:
for describing coherent monochromatic fields
Consider an elliptically polarized wave
Right handed ⇒ clockwise looking towards source.
(seems backwards: opposite from RF convention)
Ex(z, t) = 12 E0xei(ωt−kz+δx) + cc
Ey (z, t) = 21 E0y ei(ωt−kz+δy ) + cc
Horizontal V ertical
↔
l
1
0
0
1
y
χ
Ex = a1 cos ωt
Ey = a2 cos(ωt + δ)
δ phase difference
Ψ
b
2a2
α
a
2a1
a major axis
2
b minor axis
2
a +b =
tan α =
a2
a1
a21
Intensity
+ a22
−45
RHC LHC
ց
տ
1
1
1
1
1
1
√
√
√
2 −1
2 −i
2 i
Ex
† ~
∗
∗
2
2
~
I = E · E = [Ex Ey ] ·
= E0x
+ E0y
≡1
Ey
diagonal in x, y frame
45
ր
ւ
1
1
√
2 1
Left Circular Polarization CCW when viewed from +z ⇒ LHC
eiωt − e−iωt E0 iωt
Ex = −E0 sin ωt = −E0
=
ie − ie−iωt
E0
1 iωt
1 −iωt
2i
2
e
−
e
=
i
−i
i
2
eiωt + e−iωt E0 iωt
Ey = −E0 cos ωt = −E0
=
e + e−iωt
2
2
x
(-a1,a2cosδ)
tan 2ψ = tan 2α cos δ
sin 2χ = sin 2α sin δ
b
tan χ = ±
a
χ Ellipticity Angle
ψ Orientation Angle
q
iδx
a1
~ = E0xeiδ = E 2 + E 2 eiδx
E
0x
0y
a2eiδ
E0y e y
1
3
Polarization Vector Evolution with
Propagation : Linear Polarization
Jones Vectors
1
cos θ − sin θ 1
cos θ
θ rotated linear polarization R(−θ)
=
=
0
sin θ cos θ 0
sin θ
Right Hand Circular CW
E0eiδ0
1
√1
π
2 −i
E0eiδ0+ 2
Left Hand Circular CCW
E0eiδ0
1
√1
π
2 i
E0eiδ0− 2
0
1
Orthogonality
0
~ †B
~ =A
~∗ · B
~ = 0 ⇒ orthogonal (not A
~ · B)
~
=0
A
[1 0]∗
1
1
1
[1 i]∗
=1−1=0
not
[1 i]
=1+1=2
−i
−i
Superimposing
1
E0x E0x=E0y √1 1
1
0
⇒
E0x
+ E0y
=q
+ 45◦ =↑ + →=ր
2 1
E
0
1
2
2
0y
E0x + E0y
2 1
1
1
√1
√1
√
+
=
+ =l
2 i
2 −i
2 0
1
0
4
√ 1
12+.62
√1
2
1
.6
1
1
5
Polarization Vector Evolution with
Propagation : Circular and Elliptical
Polarizations
√1
2
1
i
√1
2
1
−i
√ 1
12+.52
1
.5i
Polarization Vector Evolution with
Propagation : Elliptical Polarizations
√ 1
12+.52
1
−.5i
6
√1
2
√1
2
1
ei15
1
ei45
√ 1
12+.12 1
.1ei15
√1
2
Polarization Vector Evolution with
Propagation : Elliptical Polarizations
1
e−i45
7
Jones Vector of Elliptical Polarization
1
.5ei45
√ 1
12+.52 1
.5e−i45
√ 1
12+.52
1
.5ei15
a 2 + b2 = 1
sin χ
tan χ = cos
χ =
a
b
cos χ
i sin χ
rotated by ψ into x, y system
~Jx′,y′ =
x’
y
ψ
x
cos χ
~Jx,y = R(−ψ) cos χ = cos ψ − sin ψ
i sin χ
sin ψ cos ψ i sin χ
cos α iφ0
cos ψ cos χ − i sin ψ sin χ
e
=
=
sin ψ cos χ + i cos ψ sin χ
sin αeiδ
√ 1
.12+12
.1
ei45
y’
y’
√ 1
12+.52
x’
Unit Amplitude elliptical light
Major and minor axis along x′ and y ′ with lengths a and b
For a beam with nonunity amplitude and phase
(note additional phase) φ0 = − tan−1(tan ψ tan χ)
cos α iφ0
~J = Aeiφ
e
sin αeiδ
8
9
Jones Matrix - Polarization Transformation:
complex valued 2 × 2 matrix
Ex′
jxx jxy
Ex
=
Ey′
jyx jyy
Ey
Polarizer typically px ≈ 1, py ≈ 0
′ Ex
px 0
1 0
Ex
H
=
MP =
Ey′
0 py
0 0
Ey
Waveplates: Fresnel Rhomb or Crystal Optics
y
z
M VP
=
0 0
0 1
fast axis
+45
ns +nf 2π
2
λL
10
po
x
12
λ/4
x
cos θ − sin θ
1 0
cos θ sin θ 0
cos θ +i sin θ
sin θ
=
=
y
sin θ cos θ
0 −i
− sin θ cos θ 1
sin θ −i cos θ
cos θ
sin θ cos θ + i sin θ cos θ
(1 + i) 12 sin 2θ
=
=
sin2 θ − i cos2 θ
1 − (1 + i) cos2 θ
λ
i 0
1 0
M
=
=i
0 −i
0 −1
2
pe
Rotated Waveplate Jones Matrix
ik L
iΓ/2
Ex′
Ex
e x
.
.
Ex
iφ0 e
=e
=
Ey′
Ey
. eiky L
Ey
. e−iΓ/2
Quarter Wave Plate
λ π
1 0
,
= eiπ/4
M
0 −i
4 2
λ
−iπ/4 1 0
,0 = e
M
0 i
4
x
Eigenmodes p̂o, p̂e
input polarization p̂
~
~
Eout = eiko·x̂Lp̂o(p̂o · p̂) + eike·x̂Lp̂e(p̂e · p̂) E0 = p̂o(p̂o · p̂)ei∆kL + p̂e(p̂e · p̂) E0eikeL
h
i
ke +ko
= ei∆kL/2p̂o(p̂o · p̂) + e−i∆kL/2p̂e(p̂e · p̂) E0ei 2 L

π 0th order half wave plate



2π
(2m + 1)π mth order half wave plate
∆kL = (ko − ke)L = (no − ne)L =
π

λ
2 0th order quarter wave plate


1
(m ± 4 )2π mth order quarter wave plate
Γ = (ns − nf ) 2π
λL
Half Wave Plate
po
2π n
λ o
p comes out advanced by 90
Left Circularly polarized
Retarder Jones Matrix
φ0 =
pe
LHC
Intensity transmission through rotated polarizer Malus law
1
cos2 θ
M P (θ)
=
0
sin θ cos θ
2
cos θ
I = (cos2 θ sin θ cos θ)∗ ·
= cos4 θ + sin2 θ cos2 θ = cos2 θ
sin θ cos θ
y
x
cos θ − sin θ
px 0
cos θ sin θ
R(θ)
=
M P (θ) = R(−θ)M H
P
sin θ cos θ
0 py
− sin θ cos θ
2
1
1
1 −1
cos θ sin θ cos θ
+45
−45
1
1
MP = 2
=
MP = 2
1 1
−1 1
sin θ cos θ sin2 θ
z
2π n
λ e
λ/2
x
cos θ − sin θ
1 0
cos θ sin θ 0
cos θ
sin θ
sin θ
=
=
y
sin θ cos θ
0 −1
− sin θ cos θ 1
sin θ − cos θ
cos θ
2 sin θ cos θ
sin 2θ
=
=
sin2 θ − cos2 θ
− cos 2θ
fast axis vertical
fast axis horizontal
13
Operator
cos θ − sin θ
sin θ cos θ
1 0
0 −1
2
cos θ sin θ
cos θ−sin2 θ 2 sin θ cos θ
cos 2θ sin 2θ
=
=
− sin θ cos θ
2 sin θ cos θ sin2 θ−cos2 θ
sin 2θ −cos 2θ
14
Rotated Quarter Waveplate
Two wave plates to produce arbitrary SOP
y
1 0
cos2 β − i sin2 β (1 + i) sin β cos β 45◦ 1 1 − i 1 + i
R(−β)
R(β)=
= 2
0 −i
(1 + i) sin β cos β sin2 β − i cos2 β
1+i 1−i
Arbitrary phase delay Γ at angle θ
iΓ/2
cos θ − sin θ eiΓ/2
.
cos θ sin θ
cos θ − sin θ
e
cos θ eiΓ/2 sin θ
=
sin θ cos θ
sin θ cos θ −e−iΓ/2 sin θ e−iΓ/2 cos θ
. e−iΓ/2 − sin θ cos θ
#
" iΓ/2 2
e
cos θ + e−iΓ/2 sin2 θ (eiΓ/2 − e−iΓ/2) sin θ cos θ
= (eiΓ/2 − e−iΓ/2) sin θ cos θ eiΓ/2 sin2 θ + e−iΓ/2 cos2 θ
{z
} | {z }
|
2i sin Γ/2
sin 2θ/2
iΓ/2
−iΓ/2
eiΓ/2 +e−iΓ/2
−iΓ/2 1
iΓ/2 1
+ cos 2θ e −e
e 2 (1 + cos 2θ) + e
2 (1 − cos 2θ) =
2
2
= cos Γ2 + i sin Γ2 cos 2θ
cos Γ2 + i sin Γ2 cos 2θ
i sin Γ/2 sin 2θ
=
i sin Γ/2 sin 2θ
cos Γ2 − i sin Γ2 cos 2θ
15
S
y
Fγ
unless
Jf = J3 J2 J1
Jb =
nonrecip
J3F =
JfF
x
Reflection off mirror preserves sense of rotation
Retracing circuit
note: This is different than Jones original convention in which
he used an additional notation for direction of propagation
b
ψ
a
α
x
χ
Need to invert this to determine γ and β in order to produce a desired polarization
expressed in terms of either δ and α or using χ and ψ or using a, b and α or a1, a2, δ
16
Orthogonally polarized components (narrowband case)
Ex(t) = Eox(t) cos ωt
Ey (t) = Eoy (t) cos(ωt + δ(t))
Time average polarization ellipse [δ(t) ≈ δ]
< Ex2(t) > < Ey2(t) > 2 < Ex(t)Ey (t) >
cos δ = sin2 δ
+
−
2
2
Eox
Eoy
EoxEoy
RT
Where the time averages are < Ei(t)Ej (t) >= limT →∞ T1 0 Ei(t)Ej (t)dt
2
2
< Ex2(t) >= 21 Eox
< Ey2(t) >= 21 Eoy
< Ex(t)Ey (t) >= 12 EoxEoy cos δ substi2
2
tute in, multiplt by 4EoxEoy , complete square get eqn for ellipse
2·4 2 2
2
2
2
2
4
4
4
4
1
4Eox
Eoy
Eoy
+ 21 4Eox
−
E E cos2 δ + Eox
+ Eoy
− Eox
− Eoy
2
2 ox oy
2
2 2
2
2 2
= (Eox
+ Eoy
) − (Eox
− Eoy
) − (EoxEoy cos δ)2 = (EoxEoy sin δ)2
Stokes parameters
2
2
2
2
S0 = Eox
+ Eoy
S1 = Eox
− Eoy
J = J1F J2F J3F M m J3 J2 J1
β
Stokes Parameters
On reflection, sense of rotation is preserved, which flips the handedness of helix
To maintain right handed
coordinate system, one of the axis must change sign
1 0
y
Jones Matrix of Mirror
Mm =
0 −1
x
Matrix transformation from a forward to a backwards matrix
z
jxx −jyx
jxx jxy F
⇒
−jxy jyy
jyx jyy
y
J2F
y
x
cos θ iφ0
cos2 β − i sin2 β (1 + i) sin β cos β cos 2γ
sin 2γ 0
jx
=
e
=
jy
sin θeiδ
(1 + i) sin β cos β sin2 β − i cos2 β
sin 2γ − cos 2γ 1
2
2
cos β − i sin β (1 + i) sin β cos β
sin 2γ
=
(1 + i) sin β cos β sin2 β − i cos2 β
− cos 2γ
1+i
2
2
sin 2γ(cos β − i sin β) − cos 2γ 2 sin 2β
=
sin 2β − cos 2γ(sin2 β − i cos2 β)
sin 2γ 1+i
2
Reflection and Reversal
J1F
x
S
F
N.C. Pistori, Simplified approach to the Jones calculus in retracing optical circuits, Appl. Opt. vol.34(34), p. 7870, Dec 1 1995.
non-reciprocal elements too, unless mixed with optical activity. Train of components too (unless it contains non- z
reciprocal elements). When in doubt
x
y
S2 = EoxEoy cos δ
S3 = EoxEoy sin δ
S02 = S12 + S22 + S32
17
18
Spherical Coordinates and the Stokes
parameters
Stokes Vectors
Hor
 
1
1

I0 
0
0
RHC
CW
 
1
0

I0 
0
1
Vert
+45


1
−1

I0 
0
0
 
1
0

I0 
1
0
LHC


1
0

I0 
0
−1


1
cos 2α

I0 
 sin 2α 
0
tan 2ψ = tan 2α cos δ

1
0

I0 
−1
0
S02 = S12 + S22 + s23


1
 cos 2α 

I0 
sin 2α cos δ 
sin 2α sin δ
S1 = a21 − a22
19
R
S3
S2
θ
2ψ=φ
2χ
S1
S3 = 2a1a2 sin δ = S0 sin 2χ
Both State of Polarization (SOP)
χ, ψ
L
and Degree of Polarization (DOP)
2χ
S1
20
Partially Polarized Light
= S0 cos 2χ cos 2ψ
S2 = 2a1a2 cos δ = S0 cos 2χ sin 2ψ
2ψ=φ
S1 = r cos 2χ cos 2ψ
S1 = r cos 2χ sin 2ψ
S1 = r sin 2χ
Stokes parameters
S0 = a21 + a22 = a2 + b2
q
= S12 + S22 + S32 ∝ Intensity
θ
L
tan α = xy = aa21
δ = phase shift
R
S3
S2
x = r sin θ cos φ
y = r sin θ sin φ
x = r cos θ
For the Stokes parameter coordinates
θ = 90◦ − 2χ
φ = 2ψ
α
sin 2χ = sin 2α sin δ



2
2
+ Eoy
Eox
1
2
2 
 Eox
cos 2χ cos 2ψ 
− Eoy



S=
EoxEoy cos δ  = S0  cos 2χ sin 2ψ 
sin 2χ
EoxEoy sin δ

α
CW

-45
21
S02 ≥ S12 + S22 + S32
= polarized
> partially polarized
Degree of Polarization (DOP)
p
S12 + S22 + S32
Ipol
=
0≤P ≤1
P =
Itot
S0
Stokes parameters add for two completely independent beams S = S (1) + S (2)
P
Parially polarized light =
polarized +unpolarized
 ′
 
 
S0
S0
S0
S1
0
S1





S =   = (1 − P )   + P  
S2
0
S2 
S3
0
S3
p
or decompose into two opposite polarizations (any basis) P S0 = S12 + S22 + S32
 




S0
P S0
P S0
S1 1 + P  S1  1 − P −S1
 =
+



S 2 
2P  S2 
2P −S2
S3
S3
−S3
22
Paritally Polarized Light :
Time average on the Poincare sphere
S3
S3
b
S
a
S
Paritally Polarized Light :
rapidly spinning around LHC with constant χ
S2
S2
S
S1
S1
End on view of time averaged Electric field polarization ellipses
Consider a wave periodically jumping between two polarization states Sa and Sb
Duty factor d leads to Measured Stokes Parameters S = dSa + (1 − d)Sa
Random jittering around cap on P.S. leads to a Stokes vector internal to sphere with
When uniformly distributed across surface of P.S.: Unpolarized. S12 + S22 + S32 = 0
23
Paritally Polarized Light :
randomly perturbed away from LHC
24
Depolarization: How can unpolarized light
Example depolarizers:
• Polarization randomly
jitters away from nominal value
• In this case remains LH
.4, .8
End on view of time average state of polarization
• Time averaged polarization ellipse
• Stochastic character
described by statistical
averaging definition of
Stokes parameters
25
• Viewgraph screen, Integrating sphere
• Milk, Opal, Wax paper
• Multimode optical fiber
• Pseudo-depolarizers: Babinet compensator, Lyot depolarizer (2 thick crystals axis
at 45◦), Cornu depolarizer (optically active crystal wedge biprism)
Spatially coherent Monochromatic laser input:
• Make polarization randomly vary in space across the beam:
No longer a plane wave, eg reduces spatially coherent
• Make polarization randomly vary in time:
–No longer Monochromatic, eg reduces temporal coherence
Spatially incoherent Polychromatic input beam:
• Make polarization randomly vary in space across the beam
• Make polarization randomly vary in time faster than measurement time
• Make polarization randomly vary in wavelength
26
Alternative use of the Poincare sphere for
purely polarized light (non-standard)
Measuring the Stokes Parameters
Polarizer projects entire sphere on radius line. Provides simple geometry of Malus law
on P.S.
t = cos θ
T = |t|2 = cos2 θ
4 filters
←→
ւ
ր
Degree of Polarization
I0
I1
I2
I3
ND .3
Hor pol
+45◦ pol
RHC pol
R
Isophot
S0 = 2I0
S1 = 2I1 − 2I0 = Ix − Iy
S2 = 2I2 − 2I0 = I+45◦ − I−45◦
S3 = 2I3 − 2I0 = IRHC − ILHC
S2
y
Polarizer
Block axis
p
S12 + S22 + S32
V =
S0
V = 1 for completely polarized light
Top view of Poincare sphere
S3
2χ
2θ
x<0
x>0
a 2θ2ψ
y
S1
2ψ
Polarizer
Transmission
axis
1
0%
L
1
25%
50%
75%
100%
y 1 + x 1 + cos 2θ cos2 θ + sin2 θ + (cos2 θ − sin2 θ)
=
=
=
= cos2 θ
2
2
2
2
27
Mueller matrices: Transformation of Stokes
vectors
28
Mueller matrix of a polarizer
Diattenuator with large absorption for x pol, small absorption for y

M0,0
M1,0

M2,0
M3,0
M0,1
M1,1
M2,1
M3,1
M0,2
M1,2
M2,2
M3,2

Ex′ = pxEx = e−αxLEx
M0,3
M1,3

M2,3
M3,3
Ey′ = py Ey = e−αy LEy
S0′
S1′
S2′
S3′
4 × 4 matrix of real numbers describing transformation of Stokes parameters
• Transmittance, 1 DOF
• Diattenuation, 3 DOF
• Retardance, 3 DOF (phase delay, transverse orientation)
• Depolarization, 9 DOF
∗
Hor
pol
Allows description of the transformation of both SOP and DOP
Describe depolarizing elements on polarized and partially polarized light
∗
= Ex′ Ex′ + Ey′ Ey′ = p2xEx2 + p2y Ey2 = 21 [(p2x + p2y )(Ex2 + Ey2) + (p2x − p2y )(Ex2 − Ey2)]
∗
∗
= Ex′ Ex′ − Ey′ Ey′ = p2xEx2 − p2y Ey2 = 12 [(p2x + p2y )(Ex2 + Ey2) + (p2x + p2y )(Ex2 − Ey2)]
′ ′∗
′ ′∗
= ExEy + Ey Ex = pxpy (ExEy ∗ + Ey Ex∗)
∗
∗
= i(Ex′ Ey′ − Ey′ Ex′ ) = pxpy (i(ExEy ∗ − Ey Ex∗))
 ′  2
 
px + p2y p2x − p2y 0
0
S0
S0
S1′  p2x − p2y p2x + p2y 0
 
0 
 ′ = 
 S 1 
S2  0
0
2pxpy 0  S2
′
S3
S3
0
0
0 2pxpy
p2x
2
29
1
1
0
0
1
1
0
0
Rotated
α
pol
Vert
pol
0
0
0
0
0
0
0
0
p2y
2
1
−1
0
0
−1
1
0
0
0
0
0
0
0
0
0
0
P (θ) =
1
0
0
0 cos 2θ − sin 2θ
0 sin 2θ cos 2θ
0 0
0
0
0
0
1
1
1
0
0
1
1
0
0
0
0
0
0
0
0
0
0
1
0
0 0
0 cos 2θ sin 2θ 0
0 − sin 2θ cos 2θ 0
0
0
0 1
30
Rotated Polarizer Transmission
Rotated
Polarizer
θ
x-pol
input
 ′
S0
S1′ 
 ′ =
S 2 
S3′

1
cos 2θ
sin 2θ
2

cos
2θ
cos
2θ
cos
2θ sin 2θ
1
2  sin 2θ sin 2θ cos 2θ
sin2 2θ
0
0
0
Mueller matrices for a retarder
In waveplate eigenaxis coordinates
Vertical
Polarizer
Ex′
Ey′
 
0 1
 
0
 1 =
0 0
0 0


1 + cos 2θ
2


1  cos 2θ + cos 2θ 
2 sin 2θ + sin 2θ cos 2θ 
0
31
Rotated Half waveplate pseudo-rotator
and Rotated quarter waveplate
φ = π(180◦) half wave plate case, reduces to
 ′ 
S0
1
0
0
S1′  0 cos 4θ sin 4θ
 ′ = 
S2 0 sin 4θ − cos 4θ
S3′
0
0
0
= e−iφ/2
Half Wave plate φ = π

1 0 0
0 1 0

0 0 −1
0 0 0
1
0
0
0
0 cos 2θ − sin 2θ 0
0 sin 2θ cos 2θ 0
0 0
0
1
1 0
λ/2

:
0
0

0
−1
0
0
0 1 0
0
0 0 cos φ − sin φ
0 0 sin φ cos φ
1
0
0 0
0 cos 2θ sin 2θ 0
0 − sin 2θ cos 2θ 0
0
0
0 1
32
S3
S out
S in
δ
S
δ =180 deg.
S
Clockwise rotation of waveplate causes polarization ellipse of emerging light to rotate
counter clockwise
S2
2θ
Mechanical rotation of θ leads to rotation of emerging ellipse by −2θ
 
0 0
0
S0
 
1 0
0 
 S 1 
0 cos φ − sin φ S2
0 sin φ cos φ
S3
S3
reminiscent of the Mueller matrix of a rotator, but has negative signs on diagonal which
reverses ellipticity and orientattion of emerging beam.
Mueller matrix
Propagation through a birefringent waveplate
gives rotation on the Poincare Sphere
 
0
S0
S 1 
0
 
0  S 2 
−1 S3
φ = π(90◦) quarter wave plate case, reduces to
 ′ 
 
S0
1
0
0
0
S0
S1′  0 cos2 2θ
 S1
sin
2θ
cos
2θ
sin
2θ
 ′ = 
 
S2 0 sin 2θ cos 2θ
sin2 2θ
− cos 2θ S2
S3′
0 − sin 2θ
cos 2θ
0
S3
λ/4
Corresponding
 ′ 
S0
1
S1′  0
 ′ = 
S2 0
S3′
0
Rotated Waveplate M = M R(−2θ)M φM R(2θ) =


1
0
0
0

0
cos2 2θ + cos φ sin2 2θ
sin 4θ sin2 φ2
sin φ sin 2θ 

=
φ

0
sin 4θ sin2 2
sin2 2θ + cos φ cos2 2θ − sin φ cos 2θ 
0
− sin φ sin 2θ
sin φ cos 2θ
cos φ
1 − cos2 2θ sin2 2θ
=
4
4
=e
Quarter Wave plate φ = π/2 :


1 0 0 0
0 1 0 0 


0 0 0 −1
0 0 1 0
2θ
= cos2 θ
Malus’ law only requires calculation of top row I ′ = 1+cos
2
 ′′



 

S0
1 − cos2 2θ
1 −1 0 0
1 + cos 2θ
S1′′ 1 −1 1 0 0 1  cos 2θ + cos2 2θ  1 

 ′′ = 

 


S2  2  0 0 0 0 2 sin 2θ + sin 2θ cos 2θ = 4 

S3′′
0 0 0 0
0
I ′′ =
+iφ/2
2θ
F
F
S in
S out
S1
33
S1
34
Poincare sphere visualization of operations of a
half waveplate
R
Poincare sphere visualization of operations of a
quarter waveplate
R
S3
R
S3
R
S3
S2
S3
S2
S2
S1
Fast
Slow
S1
S1
Sl
ow
S1
Sl
Slow
ow
Fast
Fa
st
Fa
st
S2
L
L
◦
Fast Axis Horizontal, Slow Axis Vertical
Fast Axis +45 , Slow Axis -45◦
for Poincareé sphere with RHC on N pole:
CCW rotation about fast axis – right hand rule: thumb towards fast
L
L
Fast Axis Horizontal, Slow Axis Vertical
35
Poincare sphere visualization of propagation
through a quarter waveplate and rotation of
the waveplate with various initial polarizations
◦
Fast Axis +45 , Slow Axis -45◦
36
Poincare sphere visualization of operations of
an optically active rotator
Rotating Linear
Polarization
R
S3 Slow
S2
S1
 ′ 
 
S0
1
0
0 0
S0
S1′  0 cos 2θ sin 2θ 0 S1
 ′ = 
 
S2 0 − sin 2θ cos 2θ 0 S2
S3′
0
0
0 1
S3
Angle of rotation
ψ′ = ψ − θ
Fast
L
37
38
Optoisolator
Reflection and Reversal
On reflection, sense of rotation is preserved, which flips the handedness of helix
To maintain right handed coordinate system, one of the axis must change sign
y
1 0
x
Jones Matrix of Mirror
JM =
0 −1
PBS
S3
slow axis y
fast
axis
z
Relation to Mueller
~ †σj V~
Sj = V
†
~ †JM †σi JM V~ = Mij Sj
Si′ = JM V~ σi JM V~ = V
Mueller matrix for a mirror following improved
tion

1 0 0
0 1 0
†
M
1

Mij = 2 T r{JM σj JM σi } = 
0 0 −1
0 0 0
Pistori, AO 34(34) p 7870 Dec 1 1995
y
45 deg.
z
sign conven-
• p-polarized light transmitted through PBS prism becomes LHC at 45◦ waveplate
0
0

0
−1
• any specular reflection off reflective surface switches to RHC
• Propagating back through -45◦ waveplate (in flipped backwards coordinate system)
• Right hand rotation around fast axis on P.S. transforms to vertical polarization
which reflects off PBS, giving optical isolation
41
β
b
F
ψ
a
eiφ0
α
tan χ =
Top view of Poincare sphere
S3
γ=
b
a
λ/2
slow
axis
2ψ
π
β = ψ+
2
L
2χ
S1
λ/4
slow
axis
ψ−χ 2ψ−2χ
ψ−χ
λ/2
fast or slow
axis
2ψ
2χ
λ/4
slow for R
fast for L
γ=
42
cos α
cos θ − sin θ eiΓ/2
.
cos θ sin θ 1
=
sin θ cos θ
sin αeiδ
. e−iΓ/2 − sin θ cos θ 0
# " φ
φ
cos φ2 + i sin φ2 cos 2θ
ei 2 cos2 θ + e−i 2 sin2 θ
=
=
i sin φ2 · sin 2θ
i sin φ2 · sin 2θ
need to solve for orientation θ and phase delay φ as a function of α, δ
Take ratio of x and y components
ψ−χ
2
β=ψ
S2
Variable Waveplate to produce arbitrary SOP
χ
λ/4 angle β
R
S1
reflective surface
Reflection off mirror preserves sense of rotation
S
λ/2 angle γ
S in
quarter−wave plate
Fγ
F
( S’) S 2
x
Two wave plates to produce arbitrary SOP
S
x
z

S out
S
( F’)
L
R
β−χ
2
χ = β − 2γ
ψ=β
43
φ
φ
cot φ2
Ex cos α
−iδ cos 2 + i sin 2 cos 2θ
=
=cot
αe
=
=cot
2θ
−
i
Ey sin αeiδ
sin 2θ
i sin φ2 · sin 2θ
h
1
1 tan 2θ = x
cot 2θ = x
x
1
φ
√ 1
=
sin
2θ
=
cot 2
h
1+x2
cot 2θ = cot α cos δ
= cot α sin δ
1
= √1+cot
sin 2θ
2 2θ
φ
cot α sin δ
cot α sin δ
cot = cot α sin δ sin 2θ = √
=√
2
1 + cot2 2θ
1 + cot2 α cos2 δ
cos α
cos α sin δ
φ
sin δ
= 1 p 2 sin α
= p
= cot
2
2
2
2
2
sin α + cos α(1 − sin δ)
1 − cos α sin δ
sin
α
Seperate into real and imaginary e−iδ = cos δ − i sin δ
2θ
44
Rotatable Variable Waveplate
retardation
φ
θ
Specify polarization in terms of orientation=ψ and ellipticity= tan χ =
Where relation with Jones field ratio angle α and phase shift δ
b
ψ
a
b
a
ψ = 12 tan−1(tan 2α cos δ)
α
χ = 12 sin−1(sin 2α sin δ)
χ
Solving for δ and α tan 2χ
δ = tan−1
= 12 tan−1(S3, S2)
sin 2ψ
α = 21 cos−1(cos 2χ cos 2ψ) = 12 cos−1(S1)
Variable
Waveplate
S1 = cos 2χ cos 2ψ
S2 = cos 2χ sin 2ψ
S3 = sin 2χ
Now plug in to eqn for orientation and ellipticity
1 −1
−1 tan 2χ
−1
−1
1
1
cot 2 cos (cos 2χ cos 2ψ) cos tan
θ = 2 cot (cot α cos δ) = 2 cot
sin 2ψ

q
1
21
−1
−1
2
cos α sin δ
 1 − cos 2 cos (S1) sin 2 tan (S3, S2) 
p
=2 tan−1 

cos 12 cos−1(S1) sin 21 tan−1(S3, S2)
1 − cos2 α sin2 δ
!
φ=2 cot−1
45
Variable wave plate to produce arbitrary SOP
using Poincare sphere geometry
Variable wave plate to produce arbitrary SOP
using Poincare sphere geometry
Poincare Sphere
Poincare Sphere
S3
z=sin2χ
Top View
S2
Poincare Sphere
1
p
2χ
s2χ
x=co q 2ψ
S2
Top view
S1
rotation angle
p
γ
S1
z
βy
z=sin2χ
x
1
γ
βy
x
η
q
p
p
η
p
p
γ
q
p
rotation around axis
p
Since 2ψ = β + η ′ use law of cosines
p
(p + q) = cos2 2χ + 12 − 2 · cos 2χ cos 2ψ
2
x +1 −2·x·1·cos 2ψ
tan γ = zq so γ = tan−1 zq
−1
δ = γ + 90 = 90 + tan
sin η ′ = 1p ⇒ η′=sin−1 p
axis of
rotation
Given ellipticity tan χ can find
2 tan χ
tan 2χ =
1 − tan2 χ
right triangle gives x = cos 2χ, z = sin 2χ
Rotation circle radius p so z 2 + q 2 = p2
p2 − q 2 = (p + q)(p − q) = z 2 = sin2 2χ
2χ
= √ 2 sin
Combine to get p − q = (p+q)(p−q)
(p+q)
2 −2·x·1·cos 2ψ
x
+1
√
2 2χ
1 √ 2 2
p=21 x2+12−2·x·1·cos 2ψ + √ 2 sin
x +1 −2·x·1·cos 2ψ − √
q
=
2
2
q
γ
axis of
rotation
rotation angle
p
z
η’
η’
p
p
q
p
2χ
s2χ
x=co q 2ψ
Top view
46
47
1
2
√
cos2 2χ+12−2·cos 2χ cos 2ψ
sin 2χ
x2 +12−2·x·1·cos 2ψ
−√
sin 2χ
cos2 2χ+12 −2·cos 2χ cos 2ψ
√
θ = 12 η ′ = 12 sin−1 21 cos2 2χ+12−2 cos 2χ cos 2ψ + √
sin2 2χ
sin 2χ
cos2 2χ+12 −2 cos 2χ cos 2ψ
48
Fresnel Reflection Coefficients: Glass to Air
n1 > n2: Critical Angle
2πn1
λ
2πn2
λ
Producing a known helicity with a Fresnel
Rhomb
n1 TIR region
n2
n2
θc
n1
θc
fast axis
fast axis
LHC
RHC
+45
-45
p comes out advanced by 90
Left Circularly polarized
Critical angle θc = sin−1 nn21
F
S
like a waveplate with
fast propagation along p
slow along s
Phase in TIR region
p
sin2 θi − sin2 θc
φp
=
tan
2
cos θi
p
2
sin θi − sin2 θc
φs
tan =
2
cos θi sin2 θc
R
Right handed rotation around fast axis
(when R is on N pole)
takes +45 linear (S2=1) to LHC (S3=-1)
fast V
L
49
Polarimetry
3 families of techniques
• PSG - Polarization State generator
2. Rotating or variable wave plates or pol.
Fourier analysis or Parameter fit
• Transmittance, 1 DOF
3. Adjustable components searching for a null
Extensions of basic polarimeters
• PA - Polarization Anlayzer
Retarder rotated in beam
Two polarizer
poisitions give
zero intensity
No intensity
variation
• 16 DOF in Mueller matrix
No polarizer
poisitions give
zero intensity
Intensity
variation
Two polarizer
poisitions give
zero intensity
• Diattenuation, 3 DOF
repeat
polarizer rotation
preceeded with λ/4 with axis
parallel to polarizer axis with
maximum intensity
No polarizer
poisitions give
zero intensity
• Retardance, 3 DOF
Two polarizer
poisitions give
zero intensity
• Depolarization, 9 DOF
No polarizer
poisitions give
zero intensity
Some example approaches
• spectroplarimetry
• Rotating polarizer after sample
• Full Mueller matrix ellipsometry
• Pol, rot ret, sample, 5:1 rot ret, pol
Intensity
variation
Polarizer rotated in beam
Measuring the Mueller matrix
1. Bank of fixed filters
eg x-pol,45◦ pol, circ pol, ND .5
Sequence of out of plane tilted detectors
• imaging polarimeter
50
Decision scheme for detecting SOP and DOP
No intensity
variation
• Polarimetry vs time
H
• Rotating sample beteen polarizers
51
Unpolarized
Circularly
Polarized
Partial
Circularly
Polarized
Linearly
Polarized
Elliptically
Polarized
Maximum intensity
obtained at same
setting of polarizer
as without retarder
Maximum intensity
obtained at different
setting of polarizer
as without retarder
Partially
Linearly
Polarized
Partially
Elliptically
Polarized
52
Rotating Waveplate polarimeter
using Fourier analysis
Rotating waveplate and/or polarizer
Unknown
Polarization
λ/4
S

Polarizer
Fβ
α
Detector
γ=α−β
 
I
 
 
 =
 

1

1 1
2 .

.
1
1
.
.
.
.
.
.


1
.
1
.
.
.


.  . cos 2γ sin 2γ .  .


.  . − sin 2γ cos 2γ .  .
.
.
.
1
.
.

 
S0
.
.
.
1
.
.
.

 
1
.
.
 . cos 2β sin 2β .  S1 

 
. cos ∆ − sin ∆  . − sin 2β cos 2β .  S2 
S3
. sin ∆ cos ∆
.
.
.
1
I = 12 {S0 + [S1 cos 2β + S2 sin 2β] cos 2(α − β) + [(S2 cos 2β − S1 sin 2β) cos ∆ + S3 sin ∆] sin 2(α − β)}
vary
vary
vary
vary
vary
α, ∆ = 0
α, while β and ∆ held constant
β, while α, ∆ held constant
α + β, while ∆ and (α − β) const
∆, while α, β constant
rotate polarizer , no waveplate S1, S2
waveplate + spinning polarizer S1, S2, S3
rotating waveplate
S1 , S2 , S3
S1 , S2 , S3
S1 , S2 , S3
1 1 .

1 1 1 .
MH Mλ/4(θ = ωt) = 2
. . .
. . .

1 cos2 2ωt sin 2ωt cos 2ωt

1 cos2 2ωt sin 2ωt cos 2ωt
= 21 
0
0
0
0
0
0
Multiplying by an unknown Stokes vector
I(t) = 12 (S0 + S1 cos2 2ωt + S2 sin 2ωt cos 2ωt + S3 sin 2ωt) = 21 (A + B sin 2ωt + C cos 4ωt + D sin 4ωt)
A = S0 + S1/2
A=
53
C = S1/2
D = S2/2
2
T
Z
S0 = A − C
T
I(t)dt
0
B=
2
T
Z
S1 = 2C
S2 = 2D and S3 = B
T
I(t) sin 2ωtdt
C=
0
2
T
Z
T
I(t) cos 4ωtdt
D=
0
2
T
Z
T
I(t) sin 4ωtdt
0
54
Jones vs Mueller vs Poincare
Variable waveplate followed by rotatable polarizer. Start with polarization α − δ



 


1
1
1 . .
.
1
 cos 2α 

 . 1 .
  cos 2α 
cos 2α
.


 


S ′ = I0
 − sin 2α  = I0 sin 2α cos(φ + δ)  =  . . cos φ − sin φ   sin 2α cos δ 
0
sin 2α sin(φ + δ)
. . sin φ cos φ
sin 2α sin δ
S3′
Transform to linear polarization by varying φ to make
= 0. Gives φ0 + δ = π
Pass through rotatable linear polarizer to find orientation. Could use rotated polarizer
Mueller matrix but final rotation wont change intensity (S0) so dont bother
 


 ′ 
I
1 1 . .
1
0
.
.
S0
′
 





1
1
.
.
0
cos
2θ
sin
2θ
.

=R (−θ)Px R (θ) = 1 R (−θ)
 S1′ I0
R (−θ)
2
 
 . . 0 0  . − sin 2θ cos 2θ 0  S2 
. . 0 0
.
.
0 1 S3′
I0
[1 + cos 2θ cos 2α + sin 2θ sin 2α cos(φ + δ)]
2
I0
I(θ, π − δ) = [1 + cos 2(θ + α)]
2
null when
φ0 + δ = π
θ0 + α = π/2
I(θ, φ) =
B = S3
Inverting gives
Polarimetry: Null Intensity method


.
1
0
0
0

.
cos2 2ωt
sin 2ωt cos 2ωt sin 2ωt 
 0



.
0 sin 2ωt cos 2ωt
sin2 2ωt
− cos 2ωt 
.
0
− sin 2ωt
cos 2ωt
0

sin 2ωt
Unknown
λ/4
Polarizer
Polarization
θ=ωt
S F
sin 2ωt 
Detector

0 
0
• Jones 2 × 2 complex matrix formalism is simplest
– Best for quick simple forward calculations
– Complicated to invert complex expressions (separating real and imaginry parts)
– Required in interferometric applications since it represents field
– Elegant for expressing Doppler shifting of rotating components
– Can only be used for pure state of polarization
• Mueller calculus use 4 × 4 real matrices operating on Stokes vectors
– Seems intimidating, but often only need 1st row (intensity)
– Simpler to invert real expressions to find set angles for polarization components
– Has not only SOP but DOP built in. Depolarization through temporal or spectral averaging
• Poincare sphere is geometrical representation of Stokes vectors
– Retardation corresponds to rotating about Fast-Slow axis of retarder
– Rapidly understand and explain polarization transformations intuituively
– Geometric guide to perform polarization calculation and verify answers
– When representing partial polarization, the inside of the PS represents DOP
– Nonstandard interpretation uses inside of PS to represent intensity (center I = 0)
∗ Geometric calculation of intensity using ruler across PS along axis of analyzer
∗ Only useful for DOP=1 (same as Jones) but provides useful geometric visualization
55
60
Anisotropic Propagation:
Monochromatic Plane Waves
Crystal Optics
~
~ = ǫ0χE
P
~ = ǫE
~ = ǫ0εr E
~
D
is the relative dielectric permitivity and χ is the susceptibility tensor
εrij = 1 + χij
Lossless conservation of energy implies Hermitian tensors
ǫij =
Time reversal symmetry
1 ~
(E
2
~ =
· D)
Similarly magnetic energy density
1~
E
2
∇× ⇒ −i~k ×
Maxwell’s Eqns:
2
61
ω2 2 ω2
n − 2 εr ê = 0
c2
c
n2mi(m̂ · ê) − (n2 − εrii)ei = 0
=⇒
mi =direction cosines
n2m̂(m̂ · ê) − (n2 − εr )ê = 0
=⇒
ei =
Multiply by mi and sum over i
X
X n2(m̂ · ê)m2
i
miei = m̂ · ê =
2 − εr
n
ii
i
i
y
nontrivial soln det(Γ ) = det |kikj − δij (k 2 + k02εij )| = 0
Eigen eqn leads to 2 solutions for each direction of propagation k̂ = m̂
n2(m̂ · ê)mi
n2 − εrii
divide by n2(m̂ · ê) to arrive at index equation
~k1(m̂), ~k2(m̂) distinct phase velocities ~ki = ni(m̂) ω m̂
c
ê1(m̂), ê2(m̂) eigenpolarizations
m̂ = (sin θ cos φ, sin θ sin φ, cos θ)
Component notation in pricnciple dielectric coordinates where ε is diagonal
 
ω2
ε − ky2 − kz2
ky kx
kz kx
ex
c2 11
 

ω2
2
2
ey = 0
Γ ê = 
ε
−
k
−
k
k
k
kx ky

22
z
y
2
x
z
c
ω2
2
2
ez
kx kz
ky kz
2 ε33 − k − k
62
ω2
E0 ~k(~k · ê) − |~k|2ê + 2 εr ê = 0
c
ω2 2
n m̂(m̂ · ê) −
c2

x
2
~k(~k · E)
~ − |~k|2E
~ · + ω εr E
~ =0
c2
~k = ω n(m̂)m̂
c
z
c
=⇒
Index Eqn
In principle coordinate system ǫ is diagonal. Can write a component or matrix eqn
ω2
E0 ~k(~k · ê) − |~k|2ê + 2 εr ê = 0
c

  

  2

ex
kx kx ky kx kz kx
k
2 ε11
ω

 ey  = 0
 kx ky ky ky kz ky  −  k 2  +
ε
22
c2
ε33
ez
kx kz ky kz kz kz
k2
2
2
2
2
2
~
k = |k| = k + k + k
y
~k × E
~ = ωµ0H
~
~k × H
~ = −ωǫ0εr E
~
=⇒
2
~k(~k · E)
~ −E
~ · (~k · ~k) + ω εr E
~ =0
c2
⇒ iω
~k × ~k × E
~ + ω εr E
~ =0
c2
~×B
~ ×C
~ = B(
~ A
~ · C)
~ − C(
~ A
~ · B)
~
Vector identity A
~
Determinental Matrix eqn for ~k, n, E
x
∂
∂t
Take curl of Faraday’s law by ~k×
~k × ~k × E
~ = −ω 2µ0ǫ0εr E
~
~ = ωµ0~k × H
~ = ωµ0(−ωǫ0εr E)
X
~ = ǫ0
εr Ei Ej
· ǫE
2 i,j ij
~ 2
~ · H)
~ = 1 µH
~ ·H
~ = µ0 |H|
Wm = 21 (B
2
2
Poyntings vector for power flow
~S = E
~ ×H
~
~ r, t) = ĥH0ei(ωt−~k·~r)
H(~
~ = −iω B
~ = −iωµ0H
~
−i~k × E
~ = iω D
~ = iωǫ0εr E
~
−i~k × H
ǫij (−~k) = ǫji(~k)
Electric Energy per unit volume
We =
ǫ∗ji
~k = ω n(k̂)k̂
c
~ r, t) = êE0ei(ωt−~k·~r)
E(~
X m2
m2y
1
m2x
m2
i
=
=
+
+ 2 z 2
r
2
2
2
2
2
2
n
n − εii n − nx n − ny n − nz
i
63
64
Index Eqn
Example 1: Uniaxial
(n2−n2x)(n2−n2y )(n2−n2z ) = n2(n2−n2y )(n2−n2z )m2x +n2(n2−n2x)(n2−n2z )m2y +n2(n2−n2x)(n2−n2y )m2z
Dyadic representation of uniaxial dielectric tensor is
ǫ = ǫo(I − ~v~v ) + ǫe~v~v
6th power terms cancel leading to a quadratic eqn in n2 for each direction m̂ = k̂
Biaxial case
m2x
n2 − εrx
+
m2y
n2 − εry
+
m2z
n2 − εrz
Uniaxial case εx = εy
kx2 + ky2 kz2 ω 2
+ r − 2
εrz
εx c
no(θ, φ) = no
ne(θ, φ) = r
!
=
1
n2
2
k2 ω
−
εx c 2
Polarization eignevectors
 mx 
n2(m̂ ·
=0
1
s^ e
n2−εr
 my x 
ê)  n2−εry 
mz
n2−εrz
θ
ke
ko
In the y-z plane ordinary is x-pol.
extraordinary pol is


0
1
 n2e cos θ 
p
2
4
ne cos θ + n4o sin2 θ −n2 sin θ
o
cos2 θ + sin2 θ
n2o
n2e
65
66
Energy flow is perpendicular to ~k-surface
Example 2: Biaxial
Consider energy flow due to superposition of 2 waves
~k2 − ~k1 = δk
~ tangent to momentum surface
~k1 × E
~k2 × E
~ 1 = ωµH
~1
~ 2 = ωµH
~2
~k1 × H
~k2 × H
~ 1 = −ωǫ E
~ 2 = −ωǫ E
~1
~2
Momentum
Surface
k2
δk
k1
S
δk
δω
~ × (E
~ 1) = (ω + δω)µ(H
~
~ 1 + δE
~ 1 + δH)
(~k1 + δk)
~
~ × (H
~ 1) = (ω + δω)ǫ (E
~ 1 + δE)
~ 1 + δH
(~k1 + δk)
~ 1∗· and conjugate of second by E
~ 1·
Expand then multiply first by H
~
~ ~
~ ~
~ ~
~
~
~ ∗ · ~k1×
~
~
E
H
H1 + ωµδH + δωµH1 + δωµδH
1 + δk×E1 + k1×δE + δk×δE = ωµ
Dyadic representation of biaxial dielectric tensor is
ǫ = ǫa~a~a + ǫb~b~b + ǫc~c~c
~a, ~b, ~c are the principal dielectric axis (not xtal unit cell)
and can rotate with λ, T
1
∗
~
~
~
~ H
~ + δk×
~ δH
~ = −ωǫ
~
~1 + δk×
~ 1 + ~k1×δH
~ 1· ~k1×
H
E
E1 − ωǫ δE − δωǫ E1 r − δωǫ δH
Neglecting second order terms and canceling wave eqn
~ · (E
~ ×H
~
~ ×H
~ ∗) + ~k · (δE
~ ∗) = δω H
~ ∗ · µH
~ + ωH
~ ∗ · µδH
δk
~ · (H
~ ∗ × E)
~ ∗
~ ∗ × E)
~ + ~k · (δH
~ = −δω E
~ · ǫE
~ ∗ − ωE
~ · ǫ δE
δk
where we used A · (B × C) = B · (C × A) = C · (A × B) = −B · (A × C), Now
subtracting
67
77
Energy flow is perpendicular to ~k-surface
Uniaxial crystal polarizations
~ · (E
~ ×H
~ ∗) − δω(E
~ · ǫE
~∗ + H
~ ∗ · µH)
~
2δk
∗
∗
~ ∗
~ · (~k × E)
~ + δE
~ · (~k × H
~ + ωH
~ · µδH
~ ∗) + ω E
~ · ǫ δE
= −δH
~ ∗ · (ωµH
~ · (ωǫ E
~ − ~k × E)
~ + δE
~ ∗ + ~k × H
~ ∗) = 0
= δH
Ordinary wave always orthogonal to ẑ (uniaxis) as well as k̂
p̂o · ẑ = 0
Extraordinary wave ⊥ to p̂o but not to ẑ or k̂, but in momentum tangent plane
=0 from Maxwell’s eqn. Now because ǫ an µ are symmetric
~ = δH
~ ∗ · µH
~ ∗ = δE
~ · ǫE
~ ∗ · µδH
~
~ · ǫ δE
~∗
ωH
E
~ ·S
~ · (E
~ = δk
~ ×H
~ ∗) = δω 1 (E
~∗ + H
~ ∗ · µH)
~ = δωU
~ · ǫE
δk
2
~ is the Poynting vector and U is the EM energy density EM energy
Where S
density
~
~ · ~ve = δω
~ ·S
~ · S = δk
~ = δωU
δk
=⇒ δk
U
U
h
i ~
J/s/cm2
cm
Since the velocity of energy flow is ~ve = US
=
s
J/cm3
~
~
Definition of group velocity vg = ∇~ (ω(k)) δω = δk · ∇~ (ω(~k))
no kz
ky
ky
ne
kx
S
Poynting
vector
ne
kx
no
78
o
oe
81
nz
1/nz
S
D2
1/nx
E1 1/ny
Fresnel’s
Ellipsoid
εxx=1
Velocity
Ovaloid
s^ e
Normal
Velocity
Surface
ke
ko
Pedal
Envelope of
wave planes
Dual
Ray Velocity
Dual
Surface
(Huygen’s Principal)
Index
Ovaloid
Index
Surface
Reciprocal
Ray Velocity
(Snell’s Law)
x 2π
unit sphere
Ray
surface
Index
Ellipsoid
ηxx=1
Single
Sheeted
ǫ = ǫo(I − ~v~v ) + ǫe~v~v
nx
ny D1
Double
Sheeted
Dyadic representation of uniaxial dielectric tensor is
λ
Momentum FT
Surface
normal
surface
(Phase Matching)
k
svr
S
p
S
k
kvp
e
o e
|ǫeˆe |
E2
kx
no
o e
Looking down on top of ẑ axis shows a degeneracy of the polarizations
m̂ = (sin θ cos φ, sin θ sin φ, cos θ)
eˆo = φ̂ = (− sin φ, cos φ, 0)
dˆo = |ǫǫeeˆˆoo| = φ̂
q
2
2
θ
/ cosǫ2 θ + sinǫ2 θ
eˆe = cos θǫxcos φ , cos θǫxsin φ , − sin
ǫz
x
z
dˆe = ǫeˆe = θ̂ = (cos θ cos φ, cos θ sin φ, − sin θ)
Example 1: Uniaxial
θ
p̂e · ŝ = 0
e o
When δω = 0 Ponting vector is normal to tangent of momentum surface.
~ · S~ = 0
δk
U
~ is orthogonal to S,
~ E
~ is always in plane of tangent of momentum surface.
Since E
p̂e · p̂o = 0
kz
k
k
p̂o · k̂ = 0
82
Real
Space
FT
Momentum Surface is scaled by optical
frequency and dispersion
Anisotropic Snell’s law refraction
1/ne
kz
ne
no
Uniaxial
1/no
c
αt
Slowness Surface
(Index surface)
λt
αt
αt
αi
Wave Quickness
αi
neω/c
noω/c
kT
λT
αr
λi
Isotropic
a) Real space
b) k space
Momentun Surface
86
87
Experimental Birefringent Crystal
interferometers: Wedge and conoscope
illuminated by white light
Tilted uniaxial crystal interface
k space
real space
kz
s1
k1
s1
s2
k2
k1
k2
k3
s2
s3
k3
k4
s3
s4
k4
kT
s4
a)
b)
88
91
Conoscopic patterns in Uniaxial Crystals
kz
no kz
ky
e o
ky
kx
White light Conoscopic patterns
ne
kx
no
ne
kx
no
e
o
o e
o e
o
e
x-polarizer
• Focusing beam probes angular variation of crystal optics
is
Uniaxial Crystal
c-
ax
• Near z axis propagation (or ⊥ z)
• Separation of sphere and ellipsoid
• no. Fringes ∝ L, ∆n, N A
y-polarizer
• Polarization interferometer requires
output polarizer
• Input polarization decomposed into
eigenmodes
Focused
light
• eigenmodes depend on azimuth angle
• eigenmodes recombine at output pol
92
Modern Application: Concentric Cross-sight
and Ringsight for skydiving cinematography
93
Conoscopy
no
ne
1
1 · sin θ = no sin θo = ne(θe) sin θe
kT
θ
θo θe
k0=ω/c
ke
eπ
(d(θo)no(θo) − d(θe)ne(θo) + text(θ))
λ
= (~ko − ~ke) · n̂L ≈ bθ 2
∆φ =
ko
Cross pattern modulated by Fresnel zone plate fringes
vertical input in terms of local eigenmodes
o e
o e
e o
~ in = E0 (v̂ · ê)ê + (v̂ · ô)ô)
E
• Brent Finley,
e
• Schumacher,
o
o
e
• Larsen and Brusgaard,
96
phase delay ê by ∆φ, project onto horizontal output
1 . ~
i∆φ
EHout = ĥ ·
+ (ĥ · ô)(v̂ · ô)
i∆φ Ein = Ein (ĥ · ê)(v̂ · ê)e
. e
97
Conoscopy
o e
e o
e
Cross pattern
o e
o
ô = (− cos θ, sin θ, 0)
ô · ê = 0
oe
Using vertical input polarization and horizontal output polarization
~ out · ĥ = [ŷ · ê(θ, φ)][ê(θ, φ) · x̂]eiδ(θ,φ) + [ŷ · ô(θ, φ)][ô(θ, φ) · x̂]
E′ = E
ê = (β
sin φ, β cos φ, α)
θ
θ
cos θ
sin φ, cos
cos φ, − sin
n2o
n2o
n2e
q
=
2
cos2
+ sin
n2
n2
ô · k̂ = 0
ô · ẑ = 0
o
= (β sin φ)(β cos φ)eiδ − cos φ sin φ
≈ sin φ cos φ(eiδ − 1)
sin 2φ iδ/2 eiδ/2 − e−iδ/2
e
2
=
2
2
δ
= sin 2φ sin eiδ/2
2
e
Project input polarization into eigenmodes of propagation which vary with propagation
angle θ, φ
~ in = [ŷ · ê(θ, φ)]ê(θ, φ) + [ŷ · ô(θ, φ)]ô(θ, φ)
E
Extraordinary component is phase delayed by δ
~ out = [ŷ · ê(θ, φ)]ê(θ, φ)eiδ(θ,φ) + [ŷ · ô(θ, φ)]ô(θ, φ)
E
I ′ = |E ′|2 = sin2 2φ sin2
δ
2
Output passed through horizontal polarizer
98
Conoscopy
sin θ = no sin θo = ne(θe ) sin θe = ne sin φ
no
ne
1
k0=ω/c
ke
ko
kT
θo θe
θo
θe
L
do de
de
t
do
x
xo xe
t
∆φ =
=
Two equivalent models for phase delay
do = cosLθo de = cosLθe xo = L tan θo xe = L tan θe t = x sin θ = L(tan θe − tan θo) sin θ
2π
no
ne(θe)
2π
[t · 1 + nodo − ne(θe)de] = L (tan θe − tan θo) sin θ +
+
∆φ =
λ
λ
cos θo cos θe
n
sin
θ
n
(θ
)
sin
θe
o
o
e
e
z}|{
z}|{
1
2π
1
L
=
(no − sin θo sin θ −
(ne(θe) − sin θe sin θ
λ cos θo
cos θe
no
2π
ne(θe) 2
2π
=
L
cos2 θo −
cos θe = L [no cos θo − ne(θe) cos θe]
λ
cos θo
cos θe
λ
= (~ko − ~ke) · n̂L = L(~ko · n̂ − ~ke · n̂)
2πno/λ
where fictitious angle is in this geometry only
t
1 · sin θ = no sin θo = ne(θe) sin θe
99
Solving for the angles
θ
kez
koz
100
=
=
no cos φ = ne(θe ) cos θe
sin θ
−1 sin θ
φ = sin
θo = sin−1
ne
ne
2πL
2πL
[no cos θo − ne(θe) cos θe ] =
[no cos θo − no cos φ]
λ
λ
sin θ
sin θ
2πL
no cos sin−1
− cos sin−1
λ
ne
ne
2πL
θ2
θ2
2πL no 1
1
no 1 − 2 − 1 − 2
=
− 2 θ2
2
λ
2no
2ne
λ 2 ne no
2πL (n2o − n2e ) 2 2πL (no − ne )(no + ne) 2
θ =
θ = bθ2
λ 2non2e
λ
2non2e
2πne/λ
kx
θ
φ θe
kz
Now for Calcite no = 1.658, ne = 1.486, at λ = .589µm with L = 1cm
2πL (no − ne)(no + ne) 2
2
2
= 2.4θdeg
λ
2non2e
q
Dark rings where
∆φ
2π
D
=
= m ⇒ θdeg
2πm
2.4
= 0, 1.62, 2.28, 2.8, 3.23, 3.62, 3.96
q
B
= 1.14, 1.98, 2.55, 3.02, 3.43, 3.79, 4.12
= 2π(m+.5)
bright rings where ∆φ = (2m + 1)π ⇒ θdeg
2.4
D
LiNbO3 at 632.8nm no = 2.288, ne = 2.2 with L = 1.2cm b = .635 θdeg
= 0, 3.14, 4.45, 5.45, 6.3, 7.0
101
Beam Splitting Prism
Wollaston
Glan Prisms
k-space description
β
α
Filled with
optical cement n c
optic axis
n e=1.486
n o=1.658
n c=1.54
nc
optic axis
no
Wollaston Prism
ne
Ordinar ray TIR
in this region
n air
optic axis
ne
n air
no
air
• Calcite Polarizing prism
– Glan-Thomson: cemented
– Glan-Taylor: air-spaced
– Glan laser: Brewster
• Can be made from Calcite
parallel piped
plane of
interface
nc
no
ne
n air
β
α
Limiting
Field Angles
• Cut along diagonal
Two prisms glues together with perpendicular optical axis orientation
OA perpendicular to direction of propagation in both halves
• Uses a lot of calcite
What would happen if you go through the prism from the side?
• Different forms (L/A ratio)
102
Beam Splitting Prism
Rochon
104
Compensators – Variable Retarders
Babinet
OA
OA1
OA2
2nd half
α
d2
d
α
OA
w
1st half
• Rays along OA collinear in first half and half equal phase velocities
– they only separate due to birefringence in second half
– first half could be replaced by glass with ng = no
d1 = (x − w)
d
d1
x
d
L
d2 = d − x ·
d
L
d
d1 − d2 = (2x − 2) − d
L
2π
(no − ne)d1
λ
2π
∆2 = (ne − no)d2
λ
2π
2π
d
∆ = ∆1 + ∆2 = (no − ne)(d1 − d2) = (no − ne) (2x − w) − d
λ
λ
L
Can we make it uniform across the device aperture (no x dependence)?
∆1 =
Find angle of beam splitting: sin θ ′ = ne sin θ
sin S sin(π − S − θ) sin(S + θ) sin S cos θ + sin θ cos S)
=
=
=
ne
no
no
no


s
n2o
sin 2Sno
no
2

± tan S 1 − 2 + 1 ⇒ θ ≈ tan S
− 1 = 1.17◦ forS = 10◦
sin θ =
2
ne
ne
ne
movable by
micrometer
105
106
Compensators – Variable Retarders
Soleil-Babinet
Electro-optic effect
Pockels originally defined EO effect in terms of impermeability pertubation rather than
dielectric pertubation. Linear (Pockels) and quadratic (Kerr) effects
movable by
micrometer
w
OA
D2
d2
d1
D0
OA
~ (0) + sE
~ (0)E
~ (0) = η 0 + ∆η
η = η 0 + rE
D1
OA
In terms of subscript notation
d0
~ (0)) − ηij (0) = ∆ 1
= ∆ηij = rijk Ek + sijkl Ek El
ηij (E
ε ij
definitions of these
coefficients
∂η rijk = ∂Eij k E=0 ∂ηij sijkl = 12 ∂E ∂E
indices k and l can be permuted
k
l
x
D2
D1
d2 = D2 − (x + w)
L
L
2π
2π
2π
∆1 = (ne − no)d1
∆2 = (ne − no)d2
∆0 = (no − ne)d0
λ
λ λ
2π
D2
D1
∆ = ∆1 + ∆2 + ∆3 = (ne − no) D2 − (x + w)
+x
− D0
λ
L
L
d 0 = D0
d1 = x
2π
D
D1 = D2 = D0 ⇒= (ne − no)w
λ
L
Uniform retardance across field (no wedge)
rotatable to arbitrary angle
Poincare Sphere
S3
E=0
since ǫ, η, χ are symmetric (really Hermitian) i, j can be permuted
⇒ rijk = rjik
⇒ sijkl = sjikl = sijlk = sjikl
Top View
S2
S2
S1
S1
Permutation symmetry of i, j reduces rijk (27 components) to 18 components independent values in 6 × 3 (18 coefficients)
sijkl (81 coefficients) ⇒ SIJ (6 × 6 = 36)
107
Linear Electro Optic Effect
reduced subscript
I = 1 , 2 , 3 , 4 , 5 , 6
ij = 11 , 22 , 33 , 23 , 13 , 12
EO modulators using KDP 4̄2m













11

12 ← 13 

ց
↑ 
22
23 

ց ↑ 
33
3x3x3 tensor becomes 6x3 matrix
tensor multiplication becomes matrix product and unfolding




∆η1
.
.
.
∆η2
 .


 
.
. 



 E1
∆η1 ∆η6 ∆η5
∆η3
 .

.
.    unfold 


E2 ⇒ ∆η6 ∆η2 ∆η4
∆ηI = 
∆η4 = rIk Ek =  .
.
. 



 E3
∆η5 ∆η4 ∆η3
∆η5
 .
.
. 
∆η6
.
.
.
C2
z
mxy
3 C2 about x, y, z
1 C4 about z consequence of C2’s and m
y C2
C2 x
mxy
Apply symmetry restrictions to zero out most of EO tensor. EO tensor for KDP


. r63Ez r41Ey
~ = r63Ez
. r41Ex
⇒
rIk Ek =
∆η = rE
r41Ey r41Ex
.
perturbed index ellipsoid for general case of arbitrary oriented applied DC field
However, reduced subscript forms no longer obey tensor transformation laws (but can
use equivalent matrix transformations)
108
109
.
.
.
r41
.
.


. E
x
.
.
 Ey 
.
r41 .
. r63
Ez
.
.
.
.
x2 y 2 z 2
+
+
+ +2r41Exyz + +2r41Ey xz + +2r63Ez xy = 1
n2o n2o n2e
112
Longitudinal EOM, z-cut KDP plate,
thickness d
KDP modulator: DC Field applied along z
rotate by 45◦ in x, y
x2 + y 2 z 2
+ 2 + 2r63Ez xy = 1
n2o
ne
V

 ′ 
cos 45 sin 45 0
x
y ′  = − sin 45 cos 45 0
0
0 1
z′
1
1
z2
′2
′2
+
r
E
−
r
E
=1
x
+
y
+
63
z
63
z
n2o
n2o
n2e
Two new principal indices
1
1
1
1
= 2 + r63Ez
= 2 − r63Ez
′2
′2
nx
no
ny
no
−2
⇒
dn = − 12 n3d n12
Now use relationship d(ndn ) = −2n−3
Birefringence seen by a wave propagating in z
V
∆n = n′y − n′x = n3o r63Ez = n3o r63
d
phase retardation between x and y
2π
2π
Γ = ∆nd = n3o r63V
λ
λ
λ
2n3o r6 3
y’
y
x’
x’
RHC
≈ 3kV
RHC
Poincare
Sphere
y’
x
y
x’
113
Amplitude Modulation
V=V0 sinωt
V
y’
x
n′x = no − 12 n3o r63Ez
n′y = no + 12 n3o r63Ez
nz = ne
x’
r63 = 26.8 pm/V in KDP
unperturbed uniaxial crystal is now weakly biaxial
π out of
phase
y’
Halfwave when Γ = π
⇒ Vπ =
z
π/2 out of
phase
114
Jones Calculus of an Amplitude Modulator
Transmittance
retarder
at 45
V=V0 sinωt
V
T
Fast
Ii
freq doubled
Io
Iout
2 Γ
2 π Vo sin ωt
= sin
= sin
Iin
2
2 Vπ
Add a bias voltage or a quarter waveplate to shift to linear modulation regime
T
Fast
Slow
Ii
y’
Polarizer
y x’ x
z
Io
V
λ/4 Analyzer
Γ=π/2
π π Vo
Iout
+
== sin2
sin ωt = 12 (1 + sin(Γm sin ωt)) ≈ 12 (1 + Γm sin ωt)
Iin
4 2 Vπ
Io
V
y x’ x
y’
z
0 0
Polarizer
Analyzer
λ/4
Variable retarder
Γ=π/2
0 1
out z }| {
in −1 1 0
−1 eiΓ/2
−1 eiδ/2
−1 1 0
Ex
Ex
0
0
R
R
R
R
R
R
R
R
=
90 45
90 0 0
Eyout
0 e−iΓ/2 | 45{z 45} 0 e−iδ/2 45 0 0 0 0 Eyin
0
=
−i sin Γ+δ
2
Analyzer
V=V0 sinωt
V
Slow
V
y x’ x
y’
z
Polarizer
Ii
Γm ≪ 1
115
As is clear by inspection and the Poincare sphere anyway
Γ+δ
|Eyout|2 = sin2
2
δ = 90 Quarter wave plate to bias at linear point
π Γ(t)
Io
π π V0
=sin2 +
sin ωt =21 (1+sin(Γm sin ωt))≈12 (1+Γm sin ωt)
=sin2 +
Ii
4
2
4 2 Vπ
Γm ≪ 1
116
Transverse EO
Phase Modulator
y’
l
z’
Fourier Component
Amplitude
y’ input
polarization
x’
-4
z
0
d
-3
-2
-1
0
1
2
3
4
~ field with modulated principal axis
Align incident E
no polarization rotation. Dont need analyzer.
For KDP longitudinal
πn3o r63Vm
λ
Ey′ (l, t) = Aei[ω0t−δ sin ωt] + cc = A
e
X
 
∆η1
∆η2 

 
∆η3 

 
∆η4 = 

 
∆η5 
∆η6
.
.
.
.
.
.
.
.
.
.
.
.
r13
r23
r33
r43
r53
r63

 
 0

 0 

 E3

n′x = no − 12 n3o r63Ez
n′y = no + 12 n3o r63Ez
nz = ne
117
Thermal compensation of an EOM
+45 input
polarization
2π
π
V
∆nl = − n3o r63 l
λ
λ
d
so really long and skinny, eg integrated optics.
Phase retardance of light polarized along x′ (Ez = Vd ) Γx =
Jn(δ)ei[ω0+nωm]t
n
Bessel Function Identity
∆ηI = rIk Ek

eg KDP oriented 45 degrees rotated around z X = x′, Y = y ′, Z = z ′
∆η6 = r63Ez
As before, but in this ny′ is irrelevant since this is prop direction
For a sinusoidally modulated applied field
Vm(t) Vm
=
sin ωt
Ez (t) =
l
l
Modulation depth δ =
y
Electric field normal to direction of
light propagation
~
∆η = rE
n3
−i 2λπ l 2o r63 Ez
−i 2π
λ no l
2π
x’
y’
V
l
Ey′ (l) = Ey′ (0)e−i λ ny′ l = Ey′ (0)e
x
−45 polarizer
y
l
d
118
Liquid Crystal Phase Modulator
KDP:
o
ne = 1.46239 + dn
dT ∆T
e
no = 1.49190 + dn
∆T
dT
x
z
no − ne = δn + dδn
dT ∆T
lab frame
X
Z
xtal 1
Z
xtal 2
Y
Y
dno
dT
dne
dT
≈ −3.7 × 10−5/K
≈ −1.9 × 10−5/K
e
EO phase modulation
= no +
−
= ne + dn
∆T
dT
i
h
′
1 1
1 −1
ei2πnx L/λ
Single device, Jones gives intensity modulation 21 −1
′ L/λ
1
i2πn
2 
z
e
i2
i2 h 2π no+ne π
h
2π ′
π
π
′
1 i 2π
2 L ei λ ∆nL−e−i λ ∆nL
λ nx L−ei λ nz L ) = 1 ei λ
(e
= sin2 δn− 21 n3o r63Ez + dδn
2
2
dT ∆T L
λ
X
dno
∆T
dT
n′x
1
2
h
−1
1 3
nr E
2 o 63 z
′
ei2πnx L/λ
n′z
ih
ei2πne L/λ
i
1 1
2 
1
For compensated device
′′
−1 1
ei2πnx L/λ
ei2πne L/λ
i
h
2
dno dne
2π
π 3
π 3
π
= sin2 n3o r63Ez L
I Edz == 21 ei λ (no+ne+ dT + dT )L ei λ no r63Ez L−e−i λ no r63Ez L
λ
• Liquid crystals require an AC bias to first induce a dipole moment, that can then
be torqued to orient with an applied field
– Varies birefringence with amplitude of switched AC field
– Somewhat slow to respond and relax back (1-10ms typical)
• Many different surface alignments, orientational forces, and crystal phases
– parallel or perpendicular, ± dielectric anisotropy, disks and rods, chiral
• Phase modulators use parallel alignment, then rotate to orthoganol with field
– Starts out as a waveplate, reduces birefringence as molecules go perpendicular
119
• Some residual birefringence remains, can be removed with fixed compensator.