Download Lecture 8: Light propagation in anisotropic media

Lecture 8: Light propagation in
anisotropic media
Petr Kužel
• Tensors — classification of anisotropic media
• Wave equation
• Eigenmodes — polarization eigenstates
• Normal surface (surface of refractive indices)
• Indicatrix (ellipsoid of refractive indices)
Anisotropic response
Isotropic medium, linear response:
P = ε0χ E
D = ε E = ε 0 (1 + χ ) E
Anisotropic medium: the polarization direction does not coincide with
the field direction:
(
)
Py = ε 0 (χ 21E x + χ 22 E y + χ 23 E z ),
Pz = ε 0 (χ 31E x + χ 32 E y + χ 33 E z ).
Px = ε 0 χ11E x + χ12 E y + χ13 E z ,
Pi = ε 0 χ ij E j
(
)
Di = ε ij E j = ε 0 1 + χij E j
! Tensors
k⊥D 
 k is not parallel to S
S ⊥E 
Tensors
Transformation of a basis:
f j = Aij ei
Corresponding transformation of a tensor:
xk = Aki xi′
xi′ = Aik−1 xk
1st rank (vector)
tkl′ = Aik−1 A−jl1tij
tij = Aki Alj t kl′
2nd rank
pi1 ,i2!in = Ak1i1 Ak2i2 ! Aknin pk′1 ,k2!kn
nth rank
Tensors: symmetry considerations
Intrinsic symmetry: reflects the character of the physical phenomenon
represented by the tensor (usually related to the energetic considerations)
ε ik = ε∗ki
Extrinsic symmetry: reflects the symmetry of the medium
pi1 ,i2!in = Ak1i1 Ak2i2 ! Aknin pk1 ,k2!kn
ε ij = Aki Alj ε kl
Example: two-fold axis along z:
 −1 0 0


A =  0 −1 0 
 0 0 1


ε13 = A11 A33 ε13 = – ε13
⇒
ε13 = ε31 = 0
ε23 = A22 A33 ε23 = – ε23
⇒
ε23 = ε32 = 0
Tensors: symmetry considerations
Second example: higher order axis along z:
 cos α sin α 0 


A =  − sin α cos α 0 
 0
0
1 

ε11 = c 2ε11 + s 2 ε 22 − 2 scε12
ε 22 = s 2 ε11 + c 2ε 22 + 2scε12
The above system of equations leads to:
(ε11 − ε 22 )s 2 = −2scε12
(ε11 − ε 22 )sc = 2s 2ε12
Solution for a higher (than 2) order axis: sinα ≠ 0
ε11 = ε22,
ε12 = 0
(
)
ε12 = scε11 − scε 22 + c 2 − s 2 ε12
εij
Optical symmetry
Crystallographic
system
Dielectric tensor
optical system of axes
Axes systems:
ç
isotropic
cubic
e
=
ç
e
0
ç
ç
• crystallographic
• optical (dielectric)
n
0
0
2
æ
è
uniaxial
• laboratory
biaxial
0
n
0
2
0
0
n
2
ö
÷
÷
÷
÷
ø
crystallographic system
of axes
n2 0
0


2
0
ε = ε 0 0 n

2 
0
0
n


hexagonal
tetragonal
trigonal
 no2 0
0


ε = ε 0 0 no2 0 


0 ne2 
0
 n o2 0
0


ε = ε 0 0 n o2 0 


0 ne2 
0
orthorhombic
 n x2 0
0


ε = ε 0 0 n 2y 0 


0 n z2 
0
 n x2 0
0


ε = ε 0 0 n 2y 0 


0 n z2 
0
monoclinic
 n x2 0
0


2
ε = ε 0 0 n y 0 


0 n z2 
0
 ε11 ε12 0 


ε =  ε12 ε 22 0 
 0
0 ε 33

triclinic
 n x2 0
0


2
ε = ε 0 0 n y 0 

2 
0
0
n
z


 ε11 ε 12 ε 13


ε =  ε12 ε 22 ε 23
ε

 13 ε 23 ε 33
Wave equation
Maxwell equations for harmonic plane waves:
k ∧ E = ωµ0 H
k ∧ H = −ω ε ⋅ E
in the system of principal optical axes
 nx2

ε = ε0  0

0
0
n 2y
0
0

0

nz2 
Wave equation:
k ∧ (k ∧ E ) + ω2µ 0 ε ⋅ E
= k (k ⋅ E ) − k 2 E + ω2µ 0 ε ⋅ E = 0
or
ω2
s(s ⋅ E ) − E + 2 2 ε r ⋅ E = 0
k c
Wave equation: continued
We define effective refractive index:
n=
kc
ω
Wave equation in the matrix form:
(
 nx2 − n 2 s y2 + s z2


n2 sx s y

2
n
sx sz

)
n2 sx s y
(
n 2y − n 2 s x2 + s z2
n2 s y sz
)
  Ex 
 
2
  Ey  = 0
n s y sz

nz2 − n 2 s x2 + s 2y   E z 
n 2 sx sz
(
)
Solutions of this homogeneous system: det = 0.
• Effective refractive index n for a given propagation direction
• Frequency ω for a given wave vector k
• Surface of accepted wave vector moduli for a given ω (normal surface
or surface of refractive indices)
Wave equation: continued
After having developed the determinant one obtains:
2 2
2 2
2 2



 2 2
n
ny
ω
ω
 ω2 nx2



n
ω
y
2 
2
2
2
2   ω nz
2
z
 2 −k 



+
k
k
k
k
k
−
−
+
−
−
  c2
 c2
 x  c2
 c2

 c









2 2
2  ω nx
k y  2
 c
2  ω
2 2
nz
2
− k  
 c
−k
2 2
2  ω nx
 + kz  2


2

 c
 2 n 2y

2
− k 
−k = 0.
 c2



2  ω
or in terms of the effective refractive index:
(n
2
x
)(
)(
)
− n 2 n 2y − n 2 nz2 − n 2 +
[ (
)(
) (
)(
) (
)(
)]
n 2 s x2 n 2y − n 2 nz2 − n 2 + s 2y nx2 − n 2 nz2 − n 2 + s z2 nx2 − n 2 n 2y − n 2 = 0 .
This is a quadratic equation in n2, thus it provides two eigenmodes for a
given direction of propagation s.
Wave equation: continued
In the case when n ≠ ni, the wave equation can be further simplified:
s 2y
s x2
s z2
1
+
+
=
n 2 − nx2 n 2 − n 2y n 2 − nz2 n 2
Polarization of the electric field for the eigenmodes:
(
 nx2 − n 2 s y2 + s z2


n2 sx s y

2
n
sx sz

)
n2 sx s y
(
n 2y − n 2 s x2 + s z2
n2 s y sz
)
  Ex 
 
2
  Ey  = 0
n s y sz

nz2 − n 2 s x2 + s 2y   E z 
n 2 sx sz
(
)
For n ≠ ni:
 sx 
 2

2
 N − nx 
 sy 
E = 2
2 
N
n
−

y 
 sz 
 2
2 
N
n
−
z 

Isotropic case
nx = ny = nz ≡ n0
wave equation:
(n
2
0
)
2
− n2 n2 = 0
Normal surface: degenerated sphere
Eigen-polarization: arbitrary
(Isotropic materials, cubic crystals)
Uniaxial case
nx = ny ≡ no and nz ≡ ne ≠ no
wave equation:
(n
2
o
− n2
)
 1 s x2 + s 2y s 2 
 −
− z2  = 0
2
2
n
ne
no 

Normal surface: sphere + ellipsoid with one common point in the z-direction
Sections of the normal surface:
positive uniaxial crystal
sy
no
sz
no
ne
ne
sx
sx
Uniaxial case: continued
sz
no
S
s or k
E D
H
ne
sx
Uniaxial case: continued
Propagation
(z ≡ optical axis)
ordinary ray
extraordinary ray
degenerated case (equivalent to isotropic medium);
index n , E ⊥ k
o
index n ,
e
E // z
index n ,
o
E in the (xy) plane,
index n(θ),
E in the (kz) plane,
E⊥k
E⋅k≠0
k // z
k⊥z
k: angle θ with z
with
sin 2 θ cos 2 θ
1
+
=
ne2
no2
n 2 (θ )
Biaxial case
nx < ny < nz
Normal surface has a complicated form
Biaxial case: continued
sy
Sections of the normal surface by the planes xy, xz, and yz:
sx = 0:
sy = 0:
nz
nx
 1 s z2 + s y2 
 −

2
 n2
n x 

 1 s z2 s 2y 
 −
− =0
 n 2 n 2y n z2 


ny
sx
sz
ny
 1 s x2 + s z2 
 −

2
2
n
n y 

 1 s x2 s z2 
 2 − 2 − 2=0
n
nz n x 

nz
sx
sz n
y
sz = 0:
 1 s x2 + s 2y 
 −

2
2
n
nz 

 1 s x2 s 2y 
 − − =0
 n 2 n 2y nx2 


nx
nz
sy
Group velocity in anisotropic
media
Maxwell equations in the Fourier space:
k ∧ E = ω µ0 H
/⋅ H
k ∧ H = −ω ε ⋅ E
/⋅ E
One can finally obtain:
ω=
2k ⋅ (E ∧ H )
S
=k⋅
U
E⋅D+ H ⋅B
This means:
v g = ∇k ω =
S
= ve
U
Birefringence
Isotropic medium: incident plane wave, reflected plane wave
Anisotropic medium: refracted wave is decomposed into the eigenmodes
• Tangential components should be conserved on the interface:
ki sin α i = k1 sin α1 = k 2 sin α 2
• k1 and k2 are not constant but depend on the propagation direction
• Thus we get generalized Snell’s law for an ordinary and an extraordinary beam:
ni sin α i = no sin α1 = n(α 2 )sin α 2
• Analytic solution only for special cases
• Numerical solution
• Graphic solution
Birefringence: graphic solution
Birefringence: uniaxial crystals
positive (ne > no)
negative (ne < no)
O
optical axis
E
optical axis
O
E
O
optical axis
optical axis
E
E
O
optical axis
E
O
E
optical axis
O
Indicatrix: ellipsoid of indices
z
s
DII, NII
DI, NI
x2 y2 z 2
+
+
=1
nx2 n 2y nz2