Transcript Chapter 4: Electromagnetic Propagation in Anisotropic Media

Chapter 4 Electromagnetic Propagation in
Anisotropic Media
Lecture 1 Light propagation in anisotropic media
Introduction:
1) The optical properties (e.g., refractive indices and absorption) of an anisotropic
medium remarkably depend on the propagation direction and the polarization state of
the incident light, as well as on the external forces (electric, acoustic and mechanic)
exerted on the materials.
2) Anisotropic media exhibit many interesting and important phenomena, including
birefringence, double refraction, conic refraction, optical rotation, Faraday effect, and
electro-optical phenomena. These effects are employed to design and fabricate
various optical elements, including polarizers, filters, beam splitters, rotators, and
many electro-optical devices. A significant part of laser physics and nonlinear optics
deals with the generation and control of light using the optical properties of
anisotropic crystals.
3) We therefore conclude that studying the propagation of light wave in anisotropic
media is essential in understanding the physics of light and its application in the lab.
1
4.1 The dielectric tensor of an anisotropic medium
Crystals are made up of regular periodical arrays of molecules. In an anisotropic crystal,
the polarization induced by an electric field is in general not in the electric field direction.
This can be seen if we consider that the electrons are anisotropically bonded in the crystal.
Pi   0  ij E j (sum assumed,  ij is the susceptibi lity tenso r.)
Di   0 Ei  Pi   0 ( ij   ij ) E j   ij E j .
 ij is called the dielectric tensor .
We currently make the following assumptions on the media:
1) Homogeneous. The medium is identical if translated internally.
2) Nonabsorption.  is real. Complex permittivity tensors will be considered later when
we study the optical activity of crystals.
3) Linear.  does not depend on the strength of the external electric field. Nonlinear
polarization is the basis of nonlinear optics, which we will learn a little in this course.
4) Nonmagnetic. m  m0. Our results can be easily extended to magnetically isotropic
media.
2
The dielectric tensor ij is symmetric in a nonabsorptive linear medium. Proof:
B
D

E  
,H 

  HB

t
t
    ( E  H)  E  D
  (E  H)  H  (  E)  E  (  H) 
  HB
  U  U  U  E  D
  E  E

 S  ED
e
m
e
i ij j

   ij   ji
1
1
1




U e (linear medium )  E  D  U e   ij ( Ei E j  Ei E j )  ( ji   ij ) Ei E j 

2
2
2
This is an intrinsic symmetry. It comes from the nature of the thermodynamic
requirement that Ue is a state function of E, which takes all Ei as independent variables.
It does not require the symmetry of the crystal or the linearity of the medium. This can be
generalized as follows.
Yi  Yi ( X 1 , X 2 ,)


U  U ( X 1 , X 2 ,)
 2
 2
Yi Y j




  d   Yi dX i ,
dU  X i dYi e.g., F  dr, E  dD 
X i X j X j X i
X j X i
Define  ( X 1 , X 2 ,)  U  X iYi 
 Dx    x
The real and symmetric dielectric
  
tensor ij can then be diagonalized in  D y    0
the principal dielectric axis system:  D   0
 z 
0
y
0
 nx2
0  E x 

 
0  E y    0  0
0
 z  E z 

0
n y2
0
0  E x 
 
0  E y 
nz2  E z 
Relative permittivity
3
4.2 Plane wave propagation in anisotropic media
In an anisotropic medium, the phase velocity of light depends on its polarization state and
its propagation direction. For a given propagation direction, there exist in general two
eigenwaves, each has its own eigen refractive index (or equivalently eigen phase velocity )
and eigen polarization. All light traveling in that direction can be decomposed into the two
eigenwaves.
Question: For a given wave normal direction s in the crystal, what are the eigen refractive
indices and eigen polarizations?
This is answered by solving the Maxwell’s equations with an anisotropic dielectric tensor.
Suppose the phase of all the electromagnetic fields (E, H, D, and B) varies in the form:
  n
k
in 

exp i t  k  r   exp i  t  s  r , where s  . Then   
s,  i 
c
k
c
t

 
B
n
E  
H
sE
E
t
mc
D
a
D
n
H 
 D   sH
t
c
k (s)
a
H (B)
Relations between the directions of the vectors:
S=E×H (t)
1) D, H, and s are mutually perpendicular.
2) D, E, s, and E×H (energy flow) lie in the same plane.
3) The Poynting vector S=E×H is generally not along s.
4
Wave normal direction s and energy transfer direction t
Wave normal direction s 
k
is the direction cosine of the wave vector.
k
n
s is perpendicu lar to the wavefront s.
c
S EH
S
Energy tra nsfer direction is t  
, with the velocity .
S EH
U
Wave vector k 
c
s is the velocity for phase transfer.
n
kc k
Refractive index n 

, where  is the wavelengt h in vacuum.
 2
Phase velocity v p 
D
H (B)
E
a
a
k (s)
S=E×H (t)
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Lecture 2 Eigenwave equation
We continue to search for the eigenwaves propagating in a given direction s in a crystal.
B
n

H
s  E
t
mc

22
2
  D  n  0s  (s  E)  E    nn 00s ssEE00
D
n
H 
 D   sH 

t
c
This is the eigenwave equation for determining the eigen refractive indices (eigen values)
and polarization (eigen states) of a plane wave propagating in a prescribed direction s. We
now realize it in the matrix form.
E  
  n  s  s  E  0

2
2
2
  E  n  0 E  s(s  E)  0  (ni / n  1) Ei   si s j E j  0
s  (s  E)  s(s  E)  E
j
  si s j  (1  ni2 / n 2 ) ij E j  0 
2
0


j
 s x2  (1  nx2 / n 2 )
 E x 
sx s y
sx sz

 
2
2
2
s y sx
s y  (1  n y / n )
s y sz

 E y   0

2
2
2 
s
s
s
s
s

(
1

n
/
n
)  E z 
z
x
z
y
z
z

Fun math tricks:
 0  sz

s   sz
0
 s
 y sx
s  s  ss  1
sy 

 sx 
0 
6
Solving the eigenwave equation (linear algebra method):
For a nontrivial solution of E, the determinant of the matrix must be 0.
 sx2  (1  n x2 / n 2 )

sx s y
s x sz


2
2
2
det 
s y sx
s y  (1  n y / n )
s y sz
0
2
2
2 

s
s
s
s
s

(
1

n
/
n
)
z
x
z
y
z
z

s 2y
s x2
sz2
1
This can be simplified into 2



.
2
2
2
2
2
2
n  n x n  n y n  nz n
Fresnel’ ss equation
equation of
of wave
wave normals
normals . It is a quadratic equation of n 2 ,
This is called the Fresnel’
which in general supports two refractive indices for a given direction ( sx , s y , sz ).
Another form of the same equation is
sx2
1
1

n 2 n x2

s 2y
1
1

n 2 n 2y

sz2
1
1

n 2 nz2
 0.
The correspond ing electric field E of each eigenstate is then found to be given by
sy
sx
sz
Ex : E y : Ez  2
:
:
n  n x2 n 2  n 2y n 2  nz2
7
Question: For a given wave normal direction s in the crystal, what are the eigen refractive
indices and eigen polarizations?
Answer:
1)In general two refractive indices, n1 and n2, are given by solving the Fresnel’s equation
of wave normals.
s y2
sx2
sz2
1



n 2  nx2 n 2  n y2 n 2  nz2 n 2
2)The polarization direction of E of each eigen state is given by
sy
sx
sz
Ex : E y : Ez  2
:
:
n  nx2 n 2  n y2 n 2  nz2
One other and easier way to solve the eigenwave equation is given below:
  n  s  s  E  0

2
2
  E  n  0 E  s(s  E)  0   i Ei  n  0 Ei  si (s  E)
s  (s  E)  s(s  E)  E
s y2
n 2 si (s  E)
n 2 si2
s x2
s z2
1


 Ei  2
eq.
1

s

E

(
s

E
)




 2 2
n  ni2
n 2  nx2 n 2  n y2 n 2  nz2 n 2
i  x , y , z n  ni
2
0
eq. 1  E x : E y : E z 
sy
sx
sz
:
:
.
2
2
2
2
2
2
n  nx n  n y n  nz
8
4.4 Phase velocity, group velocity, and energy velocity
n( )
.
k
c
Therefore in general each of the two eigenwaves in a crystal has its own phase velocity.
Group velocity is the velocity of energy flow of a laser pulse in a dispersive medium.
We now generalize the concept of group velocity, where the light pulse moves in a bundle
of directions centered at k0. We decompose the light in the k-space (momentum space).
The phase velocity of a plane wave is v p 

s, with k 
A(k ) exp i (k )t  k  r d k 
 k

 ( k )   ( k 0 )   k ( k ) k  ( k  k 0 )   
0

E(r, t )  
E(r, t )  exp i  (k 0 )t  k 0  r 


 k



A(k ) exp i k (k ) k t  r  (k  k 0 ) d k
0
 f k (k ) k t  r exp i  (k 0 )t  k 0  r 
0
    
.
 Group velocity v g  k (k )  
,
,
 k k k 
y
z 
 x
In the k-space, (k) = constant is called a wave normal surface. Group velocity is the
gradient of angular frequency in the k-space, and is therefore always perpendicular to the
wave normal surface.
9
The energy velocity is defined as v e 
S
E H

.
U 1 E  D  H  B 
2
For a non-absorptive medium, vg= ve. A neat proof is given in our textbook.
group velocity
kz
(k) = const
wave normal surface
k-space
ky
kx
10
Lecture 3 Wave normal surface in the k-space
We now express the Fresnel’s equation of wave normals in terms of k=(kx, ky, kz), that is,
in the k-space. This is not mathematically new compared to what we have done by using
n and s to express k. However, the k-space gives us more convenience.

  2

 2 2 2  
  n  0s  s  E  0   2   0k  k  E  0   ki k j   k  2 ni  ij  E j  0 
c
j 
 c


 

2

 2 2
 2 nx  k y2  k z2
c

k ykx



kzkx


kxk y
2
c2
 2 2
 2 nx  k y2  k z2
c

det 
k ykx


kzkx


n y2  k x2  k z2
kzk y


 E x 
 
k ykz
 E y   0 
 E z 
2
 2 2 2
nz  k x  k y 
c2

kxk y
2
c
2
n y2  k x2  k z2
kzk y
kxkz




k ykz
  0.

2 2 2 2 
nz  k x  k y 
2
c

kxkz
11
This determinate can be simplified into
2 
 2 1
 2 4
 k x2



k y2


k
1
1
1
1
1
2
2
2
2
2

 2 2  2 z 2 k x  k y  k z   k x  2  2   k y  2  2   k z  2  2  2  4  0.
2
2
n n

n

c
  n y nz 
 nx nz 
 y z nx nz nx n y 
 x n y  c
This is an (k) = constant wave normal surface. The surface is composed by all the wave
normals k that have the same frequency . Note that the relative permittivities also depends
on frequency. For future use, we simplify the equation for some special cases.
1) If nx  n y  nz , then
k x2
k2 
2
2
k y2

nx2
k2 
2
2

n y2
k z2
k2 
2
2
 1.
nz2
c
c
c
This is the general form of Fresnel’ s equation of wave normals.
2) If k y  0, then
 k x2  k z2  2  k x2 k z2  2 

 2  2  2  2   0
 n2
c  nz nx c 
y

3) If nx  n y  no , nz  ne , then
 k x2  k y2 k z2  2  k 2  2 

 2  2  2  2   0.
 n2
no c  no c 
e

12
Wave normal surface in the k-space:
The wave normal surface (k) = constant contains the end points of all the k vectors for a
given frequency . Here are the general characteristics of the wave normal surface.
1)The wave normal surface consists of two shells, with only four points in common.
2)The two lines that go through the origin and the four common points are called the
optic axes. When light is propagating in the direction of one of the two optic axes, there
is only one possible k value, and thus only one refractive index.
3)Any other light propagating direction intersects the two shells of the wave normal
surface at two different points, giving two possible k values. This results in two refractive
indices for the two eigenwaves, proportional to the k values.
4)For each eigenwave, the energy flow (group velocity or energy velocity) of the light is
perpendicular to the wave normal surface at its k point.
Once the wave vector k is known, the polarization of the corresponding E field of the
eigenwave is given by
ky
kx
kz
Ex : E y : Ez  2
:
:
k  nx2 2 / c 2 k 2  n y2 2 / c 2 k 2  nz2 2 / c 2
13
14
15
16
Spatial relations between the fields of the two eigenwaves:
s  ( s x , s y , sz )

 s
s
s
e 2 x 2, 2 y 2, 2 z 2
n n n n n n 
z 
y
x

2
 nx2 sx
ny sy
sz 
n

z

 n 2e  s
, 2
, 2
d e 2
2
2
2
n n n n n n 
0
z 
y
x

D1
E1
a
2
d1  d 2  n12e1  s  d 2  n12e1  d 2  n22e 2  d1 

e1  d 2  e 2  d1

 d1  d 2  e1  d 2  e 2  d1  0, if n1  n2 .
e1  e 2 
H1 (B1)
D2
b
a
S2(t2)
b
k (s)
S1(t1)
E2
H2 (B2)
1
d1  s d 2  s
0


n12n22
n22
n12
Orthogonality of the two eigenwaves:
1
E1  D2  E2  D1  0  U e  (E1  E2 )  (D1  D2 )  U1  U 2
2
s  (E1  H2 )  s  (E2  H1 )  0  s  S  s  (E1  E2 )  (H1  H2 )  s  (S1  S2 )
17
Lecture 4 The index ellipsoid
4.3 The index ellipsoid
When light propagates in a crystal, the D vectors of the two eigenwaves and the wave
normal direction s form a mutually perpendicular triad. It is therefore more convenient to
present the field vectors in the D space.
The electric energy density Ue is given by
2
2
1
1  Dx2 Dy Dz2 
1  Dx2 D y Dz2 
Ue  E  D 





.
2
2   x  y  z  2 0  nx2 n y2 nz2 
2
Dx2 Dy Dz2
The constant energy density surface in D space is then 2  2  2  2 0U e .
nx n y nz
If we denote D / 2 0U e  r  ( x, y, z ) then we have
x2 y2 z 2
 2  2  1.
2
nx n y nz
This surface is called the index ellipsoid (or the optical indicatrix).
18
The role of the index ellipsoid:
For a given arbitrary wave normal direction s, the index
ellipsoid can be used to
1)Find the indices of refraction of the two eigenwaves.
2)Find the corresponding directions of the D vectors of
the two eigen waves.
s
D2
n2
D1
n1
The prescription is as follows:
1)Draw a plane that is through the origin and is
perpendicular to s. This plane intersects the index
ellipsoid surface with a particular intersection ellipse.
2)The lengths of the two semiaxes of the intersection
ellipse, n1 and n2, are the two indices of refraction of
the eigenwaves.
3)The D vectors of the eigenwaves are each parallel to
the two axes of the intersection ellipse.
19
Proof of the prescription of using the index ellipsoid:
 x2 y2 z 2
 2  2  2 1
The intersection ellipse is given by  nx n y nz
 xs  ys  zs  0
y
z
 x
The lengths (squared) of the semiaxes of the ellipse is given by the extrema of
r 2  x 2  y 2  z 2 subject to the above two auxiliary conditions. This can be solved by
the Lagrange undetermined multipliers method. Construct
 x2 y2 z 2 
2
2
2
F ( x, y, z, 1 , 2 )  x  y  z  1 xsx  ys y  zs z   2  2  2  2  1
n

 x n y nz

F F F F F
The extrema of r 2  x 2  y 2  z 2 is then given by solving




 0.
x y z 1 2
  2 1 xi si 2 xi2 

2


x



0

r


  i

2
2 
2
n
i
F F F
s  x
i
 




 0  xi  1 i  2 2 i  0  

2
x y z
2
ni
si xi 
  s x  1si  2 si xi   0  1  
2
 i i

2
ni2 
2
ni2 
i
 i 

2   r 2


 r 2  2  sx x s y y sz z 


s y y sz z    xi 1  2   r si  2  2  2   0
2 sx x
ny
nz 
 ni 
1  2r  n 2  n 2  n 2  
 nx
y
z 
 x


20

 rm2  2  sx xm s y ym sz zm 


xmi 1  2   rm si
 2  2  0 (Here m  1, 2 meams extrema.) 
 n2
n
ny
nz 
i 


 x

If we according to the prescripti on assume the extrema rm2  n 2 ,


D
Di
 i Ei

and assume rm // D, that is rm  n
, then xmi  n
n
|D|
|D|
|D|


s  E
s E
s E 
n 2 si  x x2 x  y y2 y  z z2 z 
nx
ny
nz  n 2 si 0ni2 (s  E)

 i Ei 


2
2
2
2
n / ni  1
n  ni
sy

sx
sz
E
:
E
:
E

:
:
x
y
z

n 2  n x2 n 2  n 2y n 2  nz2
n 2 si (s  E)

Ei  2

2
2
2
n  ni2
 s x  s y  sz  1
 n 2  n x2 n 2  n 2y n 2  nz2 n 2

That is, if we assume the extrema of the intersection ellipse be rm  n and rm // D , then the
resultant E fields will have the correct polarizations, and the resultant n1 and n2 will satisfy
the Fresnel’s equation of wave normals.
21
Comparison of the wave normal surface method and the index ellipsoid method
We have studied the wave normal surface method and the index ellipsoid method on
light propagating in anisotropic crystals. Let us compare them here.
1) The wave normal surface shows us the allowed two wave vectors, and thus the
refractive indices, at a given light propagation direction. The index ellipsoid does
similar things, plus it also shows us the allowed direction of the D vectors.
2) The index ellipsoid displays the refractive indices and the directions of the D
vectors in a convenient visual way. It also involves easier mathematics. Because of
this simplicity, it is often first considered in solving problems.
3) However, the knowledge in the wave normal surface is much more profound. For
example, it displays the optic axes. It shows the wave vector variation in the kspace. It shows the energy transfer direction. It has more mathematical base, and
therefore allows for deeper theoretical derivations.
4) We tentatively summarize that for solving problems of light propagating in
anisotropic crystals, we may first try the index ellipsoid method. However, we need
to keep in mind that we have a more powerful wave normal surface method. This is
particularly true when we are directly dealing with the wave vectors rather than
merely the refractive indices in the problems.
22
Lecture 5 Light propagation in uniaxial crystals
4.5 Classification of anisotropic media (crystals)
Crystals are optically classified into 3 groups, namely the isotropic (or cubic) , the
uniaxial, and the biaxial crystals, according to the number of independent elements of
their dielectric tensors in the principle axis systems.
Our text shows many examples of isotropic and anisotropic crystals commonly used in
making optical devices.
Dielectric tensor
Examples
Isotropic
Uniaxial
 0 0 


0  0
0 0  


 x 0 0 


0

0


x
0 0  
z

NaCl, diamond
Biaxial
 x

0
0

0
y
0
0

0
 z 
Positive crystal
when z>x,
negative crystal
when z<x.
Usually choose
x<y<z
Quartz (positive)
Calcite (negative)
BBO (Beta-Barium
Borate, negative)
Mica
Topaz
23
4.6 Light propagation in uniaxial crystals
z
The index ellipsoid:
The equation of the index ellipsoid of a uniaxial crystal is
x2  y2 z 2
 2 1
no2
ne
s
De n q)
q
e
no   x /  0   y /  0 , ordinary refractive index
ne   z /  0 , extraordin ary refractive index
Do
y
no
ne  no : positive uniaxial crystal  prolate spheroid
ne  no : negative uniaxial crystal  oblate spheroid
The index ellipsoid is rotationally symmetric around the z-axis. Let s be in the y-z plane
with a polar angle q. The two polarization directions of the D vectors are: Do is parallel
to the x-axis, De is in the y-z plane and is perpendicular to s.
The corresponding refractive indices are:
no  no ,

1/ 2
2
2
 cos 2 q sin 2 q 
 ne (q ) cos q
ne (q ) sin q

 1  ne (q )   2  2  .

2
2
no
ne
ne 
 no


 

When s is on the z direction, ne(0°) = no. Therefore the z-axis is the optic axis.
24
The wave normal surface:
If nx  n y  no , nz  ne , then the wave normal
k z  / c 
De(q)
s
surface is reduced to
q
 k x2  k y2 k z2  2  k 2  2 

 2  2  2  2   0.
 n2
no c  no c 
e

The refractive indices are given by solving the
two factors:
no  no ,

1 / 2
 cos 2 q sin 2 q 

 .

ne (q )  
2
2
ne 
 no

Notes to uniaxial crystals:
1)At a given propagation direction, there are in general
two eigen refractive indices, each has its own eigen
polarization direction. This is called birefringence.
2)The E field of the o-wave is always polarized
perpendicular to the plane that contains s and the optic
axis (principal plane), while the E field of the e-wave is
polarized parallel to that plane.
ne q)
no
k y  / c 
no n e
Do
Positive uniaxial crystal
k z  / c 
De(q)
s
n
q ne oq)
Do
no
k y  / c 
ne
Negative uniaxial crystal
25
Lecture 6 Double refraction
4.7 Double refraction at a boundary
Up to now we discussed the light propagation in an anisotropic crystal when it is already
inside the crystal, but how is the light refracted into the crystal?
We recall that the boundary condition (B2n  B1n , D2 n  D1n , E2t  E1t , H2t  H1t ) requires
1) The wave vectors of the incident, reflected, and refracted light ( k 0 , k r , and k t ) lie in
the plane of incidence.
2) The tangential components of all three wave vectors on the boundary interface is the
same: k0 sin q0  kr sin q r  kt sin qt  const.
We know that in the crystal the length of kt
depends on the angle of refraction qt , as well as on
the orientation of the optic axes. Each qt supports
two kt because of the double shell structure of the
wave normal surface.
Also because of the double shell structure of the
wave normal surface, in general there are two
refraction angles, q1 and q2, both satisfy
k1 sin q1  k 2 sin q 2  k0 sin q 0  const , or
n1 sin q1  n2 sin q 2  n0 sin q 0
k0 sin q 0
k0
 k1 sin q1
q0
 k 2 sin q 2
boundary
q2
q1
wave normal
surfaces
k1
k2
26
The refraction at a boundary can be explained on the intersection between the wave
normal surface and the plane of incidence.
At the boundary, the k0 beam is uniquely decomposed into the reflected beam and the two
eigenstates of the refracted k1 and k2 beams, according to the boundary conditions:
B2n  B1n , D2 n  D1n , E2t  E1t , H2t  H1t
Each refracted beam then propagates separately. This is called double refraction.
Please note that in general the polarization of the k1 and k2 beams are not orthogonal since
they are in different directions. The incident light still can be uniquely decomposed into the
reflected beam and the two refracted beams according to the boundary conditions.
Internal double reflection:
When light is internally reflected from the surface
of an anisotropic material, double reflection may
occur. It can be discussed similarly on the
intersection between the wave normal surface and
the plane of incidence.
k0 sin q 0
k0
 k1 sin q1
q0
 k 2 sin q 2
boundary
q2
q1
wave normal
surfaces
k1
k2
27
Double refraction at the boundary of a uniaxial crystal: Examples
The directions of the D vectors are shown. There exist an ordinary wave with the
refractive index no, and an extraordinary wave whose refractive index ne depends on its
direction of propagation.
Positive
uniaxial
crystal
optic
axis
optic
axis
ki
ki
Negative
uniaxial
crystal
ke
optic
axis
ki
ko
ke
ko
ke
optic
axis
ki
ki
ko
ko
ke
optic
axis
optic
axis
ke
ko
ki
ke
ko
28
o e
29
Deviation angle between the energy flow of the o-ray and e-rays
Suppose the light is incident normally on the surface of a uniaxial crystal. Then the wave
vectors koand ke are in the same direction. Let us see how much the energy flow of the eray is deviated away from that of the o-ray.
z
te
Do  ( Dox ,0,0)  E o  ( Dox /  x ,0,0)
s (to)
t o is in the E o - s plane and is perpendicular to E o  t o  s.
a
De  (0, De cos q , De sin q )  E e  (0, De cos q /  x , De sin q /  z )
De n q) b
e
q
Ee
y
t e is in the E e - s plane and is perpendicular to E e .
no
Suppose t e has a polar angle b .
D (E )
2
tanb 
o
o
De sin q /  z  no 
   tan q
De cos q /  x  ne 
 n  2

o
The angle between t e and t o is then a  q  b  q  arctan   tan q 
 ne 


2 tan q  ne

, and the maximum deviation
If ne is very close to no , we have a 

1
1  tan 2 q  no

occurs at q  45 with a  (ne  no ) / no .
Example:
For KDP (KH2PO4) crystal, no=1.50737, ne=1.46685. At q =45°, the angle between the eray and the o-ray is a =1.56°.
30
Lecture 7 Light propagation in biaxial crystals
z
4.8 Light propagation in biaxial crystals
The index ellipsoid:
The equation of the index ellipsoid of a biaxial crystal is
x2 y2 z 2


 1 (suppose nx  n y  nz )
nx2 n y2 nz2
s
n(b)=ny b
D2
D1
ny
x
Generally for a wave normal direction s (q, j) there exist two
allowed vectors D1 and D2 with their specific refractive indices
n1 (q, j) and n2 (q, j).
If s is in the x-z plane, we have n1(q)= ny always, and n x<n2 (q)< nz. Since n x<ny< nz,
there exists a special angle b which makes n2 (b)= n1 (b)= ny. That is, the intersection
ellipse is a circle and there is no difference between the two refractive indices. This
special direction b is called the optic axis of the crystal.
For a biaxial crystal, there are two such optic axes, both are on the x-z plane. They are
located symmetrically on each side of the z-axis. Thus comes the name biaxial crystal.
It is not difficult to find that the polar angle of the two optic axes is given by
 1
1 
tan b    2  2 
n

 x ny 
 1

 2  12    nz
n

nx
 y nz 
n y2  nx2
n n
2
z
2
y
.
31
When an unpolarized light is propagating along the optic axis of a biaxial crystal, the
allowed D vector is in any direction perpendicular to the s vector. This results in the
allowed E vectors locating in one plane. The ray direction t is then found to form a cone
on one side of the light propagation direction s. Any plane perpendicular to s intersects
the cone with a circle. This phenomenon is called conical refraction.
Conical refraction is predicted by William Hamilton in 1832, and was confirmed
experimentally by Humphrey Lloyd the next year. Both were Irish scientists.
s  ( s x ,0, s z )
D  s  Dx s x  Dz s z  0  E x x s x  Ez z s z  0  E  ( x s x ,0,  z s z )
t is in the plane of E, s, D, and t  E.
We discuss the details of conical refraction later using the wave normal surfaces.
t1
s
t2
D1
t3
E1
E2
D2
y
E3(D3)
32
Wave normal surfaces:
We already know that the wave normal surface of a biaxial crystal consists of two
complex shells, with only four points in common. Let us consider the intersections
between the wave normal surface and the three coordinate planes. If we set ky=0, then
 k x2  k z2  2  k x2 k z2  2 

 2  2  2  2   0
 n2
c  nz nx c 
y

This consists of a circle with the radius ny/c, and an ellipse with semiaxes nz/c and
nx/c. The intersections with the other two coordinates planes are similar, each has a circle
and an ellipse. The optic axes lie in the x-z plane.
k y  / c 
k z  / c 
k z  / c 
nz
ny
ny
nx
nx
nx
ny
k x  / c 
nz
ny
k x  / c 
nz
k y  / c 
nx
nz
33
Lecture 8 Conical refraction
Conical refraction:
The group velocity of light is the gradient of the wave normal surface in the k-space, and
is thus perpendicular to the wave normal surface. However, on the optic axis, the two
shells of the wave normal surface degenerate into a point, which is a singular point where
the gradient is not well defined. This is like the stem scar of a water melon. The energy
flow of light at this point is governed by the nature of the singularity. We then need to
examine the shape of the wave normal surface at that point. The wave vector at the
singular point is given by
k 0  ( k0 x , k0 y , k0 z ), with
k0 x  n y

c
sin b 
nz
n 2y  n x2
c
n n
2
z
2
x
k0 y  0
k0 z  n y

c
n
tan b  z
nx
cos b 
nx
n n
c
n n
2
z
2
z
2
y
2
x
k z  / c 
ny
nx
b
k0
ny
k x  / c 
nz
n 2y  nx2
nz2  n 2y
34
We are interested in the neighborhood of this singular point. We therefore do a Taylor
expansion of the wave normal surface at this point.
k x  k 0 x  x

Let k y  k0 y  y ,
k  k  z
0z
 z
substitute it into the equation of the wave normal surface
2 
 2 1
 k x2
 2 1
  2  4
k y2
 2 1
k
1
1
1
2
2
2
z

k  k y  k z   k x     k y     k z   


 4  0,
2
2 
2
2 
2
 n2 n2 
 n2n2 n2n2 n2n2  x


n
n
n
n
c
c
  y
x z
x y 
z 
z 
y 
 x
 y z
 x

and keep up to the second power of x, y, and z, we have
(I confirmed it by Mathematica.)
4(k0 x x  k0 z z)( nx2k0 x x  nz2k0 z z)  y 2 (n2y  nx2 )( n2y  nz2 )  0.
This is a cone with its vertex at k0. The cone is
symmetric about the y =0 plane.
k z  / c 
ny
nx
2
b
q2
q1
ny
k x  / c 
nz
35
The cone intersects the y=0 plane by two lines, which make an angle 2 given by
2   q 2  q1


k0 x

tan q1  
  tan b
k0 z


(n 2y  nx2 )( nz2  n 2y )
2
2
nk
n
2 2
tan q 2   x2 0 x   x2 tan b   tan 2  
n
nz
x

n z k0 z
nz

2
2

n n y  nx
tan b  z

nx nz2  n 2y

We then rotate the x-z axes by an angle of /2−b + to the x'-z' axes. The cone will be
erect in the new coordinates ( I confirmed it by Mathematica):
1
x'2
2
z' 
y' 
2
(1  tan  )
tan 2 
2
The energy flow is everywhere perpendicular to
this cone surface, which form a light cone of
k z  / c 
ny z'
2
2
nx
b
k0
ny
x'
k x  / c 
nz
z '2 (1  tan 2  ) y'2  x'2 tan 2 
for conical refraction.
36
The light cone due to conical refraction z '2 (1  tan 2  ) y'2  x'2 tan 2 
has the following characters:
1)The cone contains the optic axis as z '   x' tan  .
2)All other lights on the cone are above the optic axis
(if the z axis is upward).
3)The cone has an apex angle of 2 in the y'=0 plane.
4)The cone intersects any plane that is perpendicular
to the optic axis with a circle.
k z  / c 
ny z'
2
2
nx
b
k0
ny
x'
k x  / c 
nz
Proof of 4): We rotate x'-z' axes by an angle of – to
the x"-z" axes. The cone will be
z"2  x" z" tan 2  y"2  0
It is a circle at any x"=const.
37
Lecture 9 Phase-matching in second harmonic generation
Second harmonic generation
When an intense laser beam passes through a crystal, the molecules of the crystal can be
nonlinearly polarized, which may cause the crystal radiate at the doubled frequency of the
incident light. This is called second harmonic generation.
Suppose the incident light has a frequency of . It passes through a crystal with thickness
L in the x direction. Suppose the E-field of the incident light (which is called the
fundamental beam) is E (, x, t )  E0 ( ) expik ( ) x  t .
We suppose the conversion to second harmonic is small so that the amplitude of the
fundamental beam is almost constant. We take care of the polarization issue later.
The second order nonlinear polarization of the material is
P ( 2) (2, x, t )   ( 2) (2 ; ,  ) E (, x, t ) E (, x, t ).
Here (2) is the second order nonlinear optical susceptibility .
According to the theory of radiation, the complex field for
the second harmonic that is generated by the crystal inside dx
but observed at L is
e(2 , x, L, t )dx  P ( 2) (2 , x, t )dx  exp ik (2 )( L  x)
e(2 , x, L, t )dx  aE ( , x, t ) E ( , x, t ) exp ik (2 )( L  x)dx
dx
x
L
Here a is just a constant.
38
The total field of the second harmonic at the output surface of the crystal is then
E (2 , L, t )  a  E ( , x, t ) E ( , x, t ) exp ik (2 )( L  x)dx
L
0
 aE ( ) exp  i 2t  exp i 2k ( ) x  ik (2 )( L  x)dx
L
2
0
0
sin kL / 2
L

exp i2k ( )  k (2 )  exp  i 2t 
kL / 2
2

4
n(2 )  n( ) is the phase mismatch.
Here k  k (2 )  2k ( ) 
 aLE02 ( )

(I believe it should be called the wave-vector mismatch, and kL should be called the
phase mismatch. This mistake was made by somebody many years ago.)
The intensity of the produced second harmonic is
sin 2 kL / 2
I (2 )  E (2 , L, t )  a L I ( )
.
2
kL / 2
2
2
2 2
For efficient second harmonic generation:
I (2 )
 I ( ).
I ( )
2)Phase-matching condition k  0, that is n(2 )  n( ).
This requirement is critical due to the narrow shape of the sinc function.
1)Conversion efficiency
39
Phase-matching in BBO crystal
BBO (Beta-Barium Borate, β-BaB2O4) is a negative
uniaxial optical crystal, i.e., ne< no.
x2  y2
z2
The index ellipsoid is 2

1
no ( ) ne2 ( )
with the z-axis as the optic axis.
Let q be the direction of the light wave, the refractive
index of the e light, ne(, q), is then
 cos q sin q 
cos q sin q
1



n
(

,
q
)

 2
e
 2

no2 ( ) ne2 ( ) ne2 ( ,q )
n
(

)
ne ( ) 
 o
2
2
2
2

1
2
Phase-matching condition requires that n(2 )  n( ) . This is impossible if both the
fundamental and the second harmonic are o-rays, or both are e-rays, because of the
dispersion of the material. However, since ne< no, and normally no ,e (2 )  no,e ( ) , we
can let  be o-ray and 2 be e-ray, and hopefully to accomplish the phase-matching
condition. That is o + o  e, which is called type-I phase-matching. For contrast, typeII phase matching refers to o + e  o or e.
40
To achieve type-I phase-matching in BBO, the phase-matching angle qm, where k=0,
is given by
ne (2 , q m )  no ( )

1
 
2
2
 cos q m sin q m  2   q m  arcsin
ne (2 , q m )   2
 2
 
n
(
2

)
n
(
2

)
e
 o
 
no 2 ( )  no 2 (2 )
.
ne 2 (2 )  no 2 (2 )
This is not hard to calculate because we have the analytic Sellmeier equations for the
refractive indices as a function of the wavelength or frequency. The Sellmeier
equations of BBO are ( in mm):
no2() = 2.7359+0.01878/( 2-0.01822)-0.01354  2
ne
2()
= 2.3753+0.01224/(
2-0.01667)-0.01516
2
m
degree
Phase-matching angle
in BBO crystal
80
60
We then find that the BBO crystal can achieve
the phase-matching condition for input
wavelengths down to a little more than 400 nm.
40
20
0.4
0.6
0.8
m
1
41
Angular sensitivity of phase-matching:
k ( , q ) 
4

ne (2 , q )  no ( )
k ( , q )  k ( , q m ) 
I (2)
k
k
q  ... 
q .
q q m
q q m
1.0
0.5
 1
k
2 3
1 

no ( )  2
 2
 sin 2q m
q q m

n
(
2

)
n
(
2

)
e
 o

sin kL / 2 1
k



kL

2
.
783

q  L  2.783 
q q m
kL / 22 2
2.783
kL
2

 1
2.783  3
1 
2q 
  no ( )  2
 2
sin
2
q
m

L

n
(
2

)
n
(
2

)
o
e




1
This is only 0.11°for a BBO crystal of 1-mm long with an input beam at 600 nm.
Therefore the phase-matching condition is extremely angular-sensitive.
In practice the experimental angular width of phase matching can be larger, because 1)
The angle is measured outside the crystal. 2) The laser has a bandwidth. 3) Intensity
saturation. 4) A Gaussian beam has a divergence in its propagation directions.
Similarly we can also calculate the wavelength sensitivity for phase-matching. It is
found to be about 1 nm for 1-mm thick BBO crystal, which is again quite sensitive.
42
Angular and wavelength sensitivity
of phase-matching in BBO crystal
m,
,
40.7
degree
kL=0
40.6
2q
40.5
2
kL=±2.783
40.4
40.3
0.599
0.6
0.601
0.602
m
43
Lecture 10 Optical activity
4.9 Optical activity
Optical activity refers to the phenomena that when a linearly polarized light is passing
through a medium, the polarization plane is rotated. It is thus also called optical rotation.
When this happens, the medium is said to be optically active. The rotation of the
polarization plane is proportional to the path length of the light, and can be measured in
degree/centimeter. Common optically active media include quartz, sugar and syrup. It can
be used to measure blood sugar concentration in diabetic people.
Optical activity occurs in a chiral molecule, where the molecule does not overlap with its
mirror image, and the electron cloud takes some kind of helical shape. Left- and rightcircularly polarized light is different in polarizing a chiral molecule, which causes different
refractive indices. This is called circular birefringence (or circular double refraction).
Most amino acids (the building blocks of life) are left-handed, which makes optical activity
prominent in nature.
If when observed facing the light the
rotation of the polarization plane is
counterclockwise, the substance is
dextrorotatory (right-handed). When
the rotation is clockwise, it is called
levorotatory (left-handed).
44
Explaining optical activity using circular birefringence:
Suppose the incident light is linearly polarized in the x-axis, and is propagating along
the z-axis. Suppose the refractive indices for circularly polarized light is nr and nl.
At z  0 : E(0)  E0 xˆ exp it  
At distance z : E( z ) 
E0
2


E0 ˆ ˆ
R  L exp it 
2
ˆ
 nr
R
exp
i

c


 n
 ˆ
z  L
exp   i l
c



z  exp it 

E0 
 nl 
 nr 
ˆ
ˆ
ˆ
ˆ




x

i
y
exp

i
z

x

i
y
exp
z  exp it 
i


2 
c
c




  n 
E    n 
 n  
 n   
 0 xˆ exp   i r z   exp   i l z   iyˆ exp   i l z   exp   i r z   exp it 
2   
c 
c 
c 
c  


 

E  
 (nl  nr ) z
 (nl  nr ) z
 (n  nr )z 
 (n  nr )z  
 0 xˆ 2 cos
exp   i l
i exp   i l
  iyˆ  2 sin
  exp it 
2  
2c
2c
2
c
2
c



 



 (nl  nr ) z
 (nl  nr ) z 
(nl  nr )z  

 E0 xˆ cos
 yˆ sin
exp
i

t




2c
2c
2c




The specific rotary power is

 (nl  nr )
2c


(nl  nr ),

which is very sensitive in measuring circular birefringence.
45
Theory on optical activity:
A helical molecule can be polarized along its axis by a circulating current driven by an

circular electric field produced by a time-varying magnetic field: p  aE  b H
The constitutive equation of a medium driven by a plane wave is then revised to
    E, and   is
D  E  i 0Gs  E, because H
Gs=G is called the gyration vector. The length G of the gyration vector describes the
rotation power of the medium. It varies with the direction of the wave, and can be
expressed by a gyration tensor g of the medium as G  s  gs  g ij si s j .
Now our original eigenwave equation   n 2 0s  s  E  0 is revised to
  i Gs  n  s  s  E  0
2
0
0
After realizing it into the matrix form, we have
G
G
 2

2
2
sx s y  i 2 sz
sx sz  i 2 s y 
 s x  (1  nx / n )
n
n

 E x 
 
G
2
2
2
 s s i G s

s y  (1  n y / n )
s y sz  i 2 sx  E y   0
 y x n2 z

n

 E z 
G
G
2
2
2
sz s y  i 2 sx
s z  (1  nz / n ) 
 sz sx  i 2 s y
n
n


 0  sz

s   sz
0
 s
 y sx
s  s  ss  1
sy 

 sx 
0 
46
For a nontrivial solution of E, the determinant of the matrix must be 0, which is
simplified to


s y2
G 2 nx2 sx2  n y2 s y2  nz2 sz2
sx2
sz2
1


 
n 2  nx2 n 2  n y2 n 2  nz2 n 2 n 2 n 2  nx2 n 2  n y2 n 2  nz2




(*1)
The corresponding wave normal surface is
2  
G
nx2 k x2  n y2 k y2  nz2 k z2


2
2
2
ky
kx
kz
c



1

2 2
2 2
2 2
 2 2  2  2 2  2  2 2 
2
2
2
2 2
k  2 nx k  2 n y k  2 nz
k  k  2 nx  k  2 n y  k  2 nz 
c
c
c
c
c
c




6


I believe this surface is formidable for anyone attempting to draw using Mathematica.



Fortunately equation 1 can be simplified to n 2  n12 n 2  n22  G 2 , where n1 and n2
are the solutions of the equation with G=0, i.e., when there is no optical activity.
On the optic axis, we have n1  n2  n , the eigen refractive indices are then
G
n 2  n 2  G, and n  n 
when G is small.
2n
G


.
These are two circularly polarize waves, with a rotary power of
n
47
The eigen polarization state in an optically active medium:
If we solve equation 1 for the two general eigen refractive indices, and then substitute into
the eigenwave equation, we may have the eigen polarization states for light in an optically
active medium. Surely it is too complicated. Our textbook did it in a simple way. We can
rewrite the eigenwave equation into that for the D vector, and express it in the (D1, D2, s)
coordinate system, thus reducing one dimension. The corresponding eigen polarization,
when G is small, and in terms of D, is
2
2 
 





1
1
1
1
1
1
G
 










2 
  2
4  n12 n22   n12 n22  
J    2  n1 n2 

iG



2 2


n1 n2


D2
D1
We conclude that
1)The eigen polarization states are generally two orthogonal elliptically polarized
waves, oriented along the D1 and D2 axes.
2)In an isotropic medium, or when the light is propagating in the optic axis of an
anisotropic medium, the eigen polarizations are left- and right- circularly polarized light.
3)In an anisotropic medium, and when the light is not close to the optic axis, because G
is usually much smaller than
, the eigen polarizations are almost linear.
n12  n22
48
4.10 Faraday rotation
Faraday rotation refers to the phenomena that when a linearly polarized light is passing
through a medium placed in an external magnetic field, which is along the light
propagation direction, the polarization plane is rotated.
The origin of the effect is as follows. The electrons are moving in a molecule, driven by the
electric field of the light. The external magnetic field will displace the electrons laterally
due to the Lorentz force qv×B. The induced dipole momentum then includes a term
proportional to E×B. The constitutive equation is then
D  E  i 0gB  E
Here g is the megnetogyration coefficient, gB is the gyration vector. The discusses
followed should be similar to optical activity, with a specific rotation given by   VB ,
where V is called the Verdet constant, can be measured by deg/Gauss·mm.
One distinct difference between optical activity and Faraday effect is the reversibility of
the effect when the light is going back. Optical activity is reversible, while in Faraday
effect when the light is reversed the rotation is doubled. This is ready to explain by the
two constitutive equations:
b
k
D  E  i 0 Gs  E, optical activity
B
k
D  E  i 0gB  E, Faraday effect
B
k
B
d
49
Optical isolator:
An optical isolator, or optical diode, is an optical device that allows the transmission of
light in only one direction. It is used to prevent unwanted light feedback into a laser
cavity. The operation of the device depends on the non-reversibility of Faraday effect.
The optical isolator in the figure consists of three
parts. An input polarizer allows only vertically
polarized light to pass. A Faraday rotator rotates
the polarization by 45°. An output polarizer
(analyzer) just let the 45° polarized light to pass
by. If light is fed back from somewhere later in
the beam path, the Faraday rotator will turn it
into horizontally polarized light, which is then
absorbed (or deflected) by the first polarizer.
50