Transcript Feature Selection/Extraction for Classification Problems

Lecture 3. Symmetries & Solid State Electromagnetism
2009. 04.
Hanjo Lim
School of Electrical & Computer Engineering
[email protected]
1


Photonic crystals ; the structures having symmetric natures
important symm.; translat., rotat., mirror, inversion, time-reversal
symmetry of a system => general statements on system’s behavior,
classify normal modes.
Example : a two-dimensional metal cavity with an inversion symm.
 
 


If H (r ) is found, H (r ) is an eigenmode of freq. 1, i.e., H (r)  H (r, 1 )
 
 
 
 H (r ,  );even mode
 Then H (r ) m ay be H (r , 1 )     1
 H (r , 1 ); odd mode


m(r )  m(r )
proof) Inversion symm.
means
that
metal
pattern;









 if m(r )  H (r , ), H (r , )  m(r )  m(r )  H (r , )
 
 
 
If H (r ) is nondegenerate, H (r , ) must be the same mode with H (r ).
 
 
 H (r )  H (r ) with the eigenvalue  of the inversion operation.
 
 


2
2
Likewise H (r )  H (r ).   H (r)  H (r)   1

Formal(mathematical) treatment of inversion symmetry


operator
that
inverts
vector,
I  
 I;

Let 
OI ; operator that inverts vector field f and its argument

r
 




 def ) OI f (r )  If (I r)   f (r) for a system f (r)   f (r)



property of inversion symmetry ;   OI1OI for any operation 
y
OI

OI


x



OI
 
 OI   OI or from OI1OI  1,


1
  OI  OI
OI
def) commutator operator A, B  AB  BA of two operators A and B


Symm. system under inversion means OI ,   0 with OI and any  .
3





 
 

1
Let      , then OI , H  OI (H )  (OI H )  0 means that
 (r )
2 
2
 








 (OI H )  OI ( H )  OI  2 H   2 (OI H ), ie, OI H is an eigenfunction  operator.
c  c
 

∴ If H (r , 1 )is a harmonic mode, OI H is also a mode with freq 1.
If there is no degeneracy, there can only be one mode per frequency


 OI H  H , and   1
General aspect ; With two commuting operators, simultaneous
eigenfunctions of both operators can be constructed.



Ex) eigenvalue of OI ;1,  (OI H )  H , classify eigenmode parity
 
 
What if there is degeneracy ? If H 2 (r , 1 )  H1 (H1, 1 ), then?
4
Continuous Translational Symmetry
def) A system with transl. symm.;
unchanged by a translation operation


through a displacement d . Let translational operator is Td for each d ,

 

If a system is invariant under Td operation, Td f (r )  f (r  d )  f (r )

 

Let our system is translationally invariant, or Td  (r )   (r  d )   (r ).

 


Then, Td ,   Td   Td  0  Td transforms to itself , i.e., Td f (r )  f (r ).
  
    
proof: Td , H (r )  TdH (r ) Td H (r )

2
  
   

 



1
 Td  2  H (r )         H (r  d )   (r )   (r  d )
 (r )
c 



2


   





1
  2  H (r  d )          H (r  d )
c 
 (r  d ) 

2
    2    



  2  H (r  d )   2  H (r  d )  0
c 
c 

∴ The modes of  can be classified by how they behave under Td . 5
def) a system with continuous translat. symm. ; invariant under all Td ' s
ex) If  (z ) is invariant under all the Td ' s in the z-direction,
H ( z)  H ( z  d ) for any d .  uniform plane wave propagation in z-direction
  ikz
 H  H o e is the eigenfunction. Then Td eikz  eik ( z d )  (eikd )eikz.
∴
Eigenvalue of operator Td ; eikd .∴ We can classify the waves by
Note eikz  eik ( z  d ) .
 Td
k  2 / 
changes the phase only by kd  2d / .
ex) If a system is invariant in all three directions (ex: free space),
 
 


eigenmodes; H k (r )  H o exp(ik  r ) with any constant vector H o .
 
Note) The eigenfunctions exp(ik  r ) can be classified by their particular
values for wave vector k ( direction,  ,  )
 
Implication of   H  0 on the plane wave : transversality condition 6


 

   

   
  H    Ho exp(ik  r )  0    ( fG)  f  G  G  f



ik y
   
   (ax   a y   az  )(eikx x  e y  eikz z )
 
  ikr

x
y
z 
  H  exp(ik  r )  H 0  H 0  (e )   exp(ik  r )
 


 ik r 
 

 i(k x ax  k y a y  k z az )e  ik exp(ik  r )
 

 Ho  ik exp(ik  r )  0
 Ho  k
 
 

In conclusion,
the
plane
waves
H k (r )  H o  exp(ik  r ) are the solutions of

 
   
2
master eq.   1/  (r )  H (r )   / c H (r ) with 3-dimensional continuous


translational symmetry.

 
   
 
  ikr 
ik r
Proof;   H 0e(ik  r )  e   H 0  (e )  H 0  exp(ik  r )(ik )  H 0
 



   
ik r
   1 /  (r )  H (r )  (1 /  )  (e ik  H 0 )
 

 
 


 ikr
ik r
ik r
  (e ik  H 0 )  e   (ik  H 0 )  (e )  (ik  H 0 )
 
 
        
ik r
 e (ik  ik  H 0 )  a  b  c  b (a  c )  c (a  b )
   
  
 ikr
ik r
2
 e k (k  H 0 )  H 0 (k  k )  k H 0e
 
 
   k2 

2
H exp(ik  r )  ( / c) H 0 exp(ik  r )
∴ Master eq. becomes  H (r ) 
 0




7
If   1 ( free space), c  f  k 2  (2 / )2  (2f / c)2  ( / c)2 .
If   1 with refractive index n   , vg  c / n,   0 / n, kc /   2f /   
 
 


2
Holds.  k /    / c and H0 exp(ik  r )  ( / c) H0 exp(ik  r ) .

∴ Spectifying k; propagation direction &  => how mode behaves

z
ex) An infinite plane of glass with  (r )   ( z);
Invariant under the continuous translation
y
x
operation in the x- or y-direction.
 
 
ik  
The eigenmodes should have the form H k (r )  e h ( z ) with the in-plane




wave vector k  k x ax  k y a y and h (z) that would not be determined by
the symmetry consideration only.

 
But a constraint on h (z) exists from the transversality condition   H  0.
8




 
 
 
 
ik 
  h ( z)e  0    h ( z)  ik  h ( z)  0

∴ We can classify the modes by their values of k and the band number


n, i.e, by (k , n) if there are many modes for a given k .

Assume a glass plane of width a and a mode with the x-polarized H field



H k ,n (r )  exp( ik y y )n ( z )a x , and  (r )   ( z )  1 for 0  z  a.

 ik y


2 ik y
1




e

(
z
)
a

(

/
c
)
e

(
z
)
a
Then master eq. becomes  ( z )
n
x
n
x

 
ax a y az

 ik y y





ik y y
1
 

 ( e )  n ( z ) a x  e   n ( z ) a x    n ( z ) a x  
 ( z)
x y z
n ( z ) 0 0

 ik y

ik y dn ( z )  

   1 (ik y a y )e y  n ( z )ax  e y
ay 
 ( z) 
dz


 dn ( z )  
ik y y 
1
 
e  ik yn ( z )az 
ay 
 ( z)
dz


9

y


y

y
 
 dn ( z )     1 ik y y  
 d ( z )  
ik y y
1

e    ik yn ( z )az 
a y   
e    ik yn ( z )az  n
ay 
 ( z)
dz

(
z
)
dz

 



 

ax a y
az

  dn ( z )  




*   (ik yn ( z )a x ) 

ay  
x y
z
dz


0 0  ik yn ( z ) ,
0

ax

x
0

ay

y
dn ( z )
dz

az
2
  a d n ( z )
x
z
dz2
0
 1  
 d 2n ( z )   1  ik y y

dn ( z )  
ik y y 
ik y y
1


e   ax


e

e



ik

(
z
)
a

ay 
 
y n
z
2
   ( z )
 ( z)

(
z
)
dz
dz
 



2
 1  


dn ( z )  
ik y y d n ( z ) 
ik y y 
1
1

e
a

e
(
ik
)
a




ik

(
z
)
a

ay 
  ( z) y y
 
x
y n
z
 ( z)

(
z
)
dz
dz2

 






* 1/  ( z )    a x   a y   a z   1  a z d 1
y
z   ( z )
dz  ( z )
 x
2
2

 d  1  dn ( z )  

ik y y d n ( z ) 
ik y y  k y
2 ik y y
1



e
a

e

(
z
)
a

a

(

/
c
)
e

(
z
)
a
  ( z )  dz  
x
x
x
n
x
  ( z) n
 ( z)
dz
dz2






10
d  1 dn ( z ) 


dz   ( z ) dz 
2
d 2n ( z ) d  1  dn ( z )  k y
2
1




(
z
)

(

/
c
)
n ( z )




n
 ( z ) dz2
dz   ( z )  dz   ( z )
2 
  1   d  1 dn ( z )   k y2


∴ Master eq. becomes        
 2 n ( z )


(
r
)
dz

(
z
)
dz

(
z
)
c 



 
2
2
k
Let  (yz )  c 2   2 . If  2  0, i.e., 2  k y    / c  2ff   2  2 / glass , that is,
0 / 
y
0
d 
 y  glass , d  1
  2  0.   ( z ); exp(z), evanescent wave to the air, i.e.,

dz   ( z ) dz 
confined wave in the glass => discrete modes(bands)
d  1 d    2   0.   ( z ); exp(iz ), traveling wave
If  2  0, i.e.,  y  glass , dz
  ( z ) dz 


extending both in the glass and the air region.
The separation of continuous states and
discrete bands at   ck y /  : light line.
11
If k y is large, i.e, the wave of short 0 or propagation more in z-direction,
d  1 d    2 , and the wave is well confined in the glass (0  z  a).
dz   ( z ) dz 
d 2
2
Let  ( z )   ( glass), then dz2     0, with  (0)   (a)  0
   Asin  z  B cos  z    Asin  z.
2
2 2
2
2 2
k
y
2
n


n
 (a)  0     n / a.    2 , 
 2  2 .
 c
a
a
k y2
2

 .
These modes decay ever more rapidly as k y increase, since   
c2
 Discrete translational symmetry
Photonic crystals actually have discrete translational symmetry(DTS).
ex) 1D PhC : DTS for 1D and CTS for 2D, 2D PhC : DTS for 2D and
CTS for 1D, 3D PhC : DTS for 3D.
 

 

ex) Fig. 4: 2D PhC with primitive lattice vector a  ayˆ ,  (r  R)   (r ) for R  la
with integer l , and unit cell: xz slab with the width a in the y-direction
12
2


z
y
x
That is, photonic crystals are composed of repetition of unit cells.



Tanslational symmetry means that  ,Tdxˆ  0 and  ,TR  0 with R  layˆ

Eigenmodes of  ; simultaneous eigenfunctions of Tdxˆ and TR
Tdxˆ eikx x  eikx ( xd )  (eikx d )eikx x , TR e
ik y y
e
ik y ( y la)
 (e
ik y la
ik y y
)e
, (
); eigenvalue.
∴
Modes can be classified by specifying kx and k y values.
But not all the values of k y yield different eigenvalues. With k y  k y  2 ,
a
2 la
i

ik  y
ik  la
ik  y
ik
la
ik
la
i
k
y


TR e  (e )e   e  e a eik y  (e )e .  exp(ik y y) and exp(ik y ) have
y
y
y
y

y
y

exp(ik yla)
the same eigenvalue
for TR if ky  k y  2 / a.
ik la
∴ Any k y  k y  m  2 with integer mgives identical TR eigenvqlue of (e ),
a
i.e., k y and k y is a degenerate set. That is, the addition of an integral
multiple of b  2 on k y leaves the state unchanged.
a

 b  b yˆ : called as “primitive reciprocal lattice vector”.
13
y
 Bloch theorem, Bloch ftunction and Brillouin zone
Any linear combination of degenerate eigenmodes for k y and k y  k y  mb
is an eigenfunction. Therefore, the general solution of a system having a
DTS(CTS)
 in the y-direction (x-direction) is
 
H k x ,k y (r )  aeikx x


i ( k  mb ) y
ik y
ck y , m ( z ) e y
 eikx x  e y
m
m
polarization direction

u ( y  la, z ) 


ck y ,m ( z )eimb ( y la) 


ck y ,m ( z )eimby

im 2 la
a

Let uk y ( y, z); periodic ft. in y


ck y ,m ( z )eimby  u ( y, z ); periodic ft.
in y-dir.
m
m
Bloch theorem ; a wave propagating through the periodic material in the

ik y 
H ( x, y, z )  e uk ( y, z ,  ), or more
y-direction can always
be
expressed
as
    ikr
generally  (r )  uk (r )e if dielectrics periodic in 3D. Bloch ft.

Note) Discrete periodicity in the y-direction gives H ( x, y, z) that is simply
the product of a plane wave in the y-direction and a y-periodic ft.


and thus, H k  mb  H k . Thus mode frequencies must also be periodic in14
e
y
y
y
y
that (k y  mb)  (k y ). ∴ Knowl. about    k y  (1st BZ) is sufficient.
a
a
When the dielectric is periodic in 3D, the eigenmodes have the form of


 
 
ik  r  
H k (r )  e uk (r ) with the k inside the first BZ and a periodic ft. uk (r )

    
satisfying uk (r  R)  uk (r ) for all lattice vector R.
 Photonic band structures
    ikr
EM modes of a photonic crystal should have a Bloch form H k (r )  uk (r )e
and all the informations about such a mode is given by k y and uk (r).


2 
 

To solve for uk (r ), let’s start from  H k   (k ) / c  H k , i.e., from the master eq.
k y , so



  1  ikr   
2 ik r  
      e uk (r )   (k ) / c e uk (r ).
  (r )


 
  1  ikr    
 ikr  
  


ik r
ik r
1



      e uk (r )     (e )  uk (r )  e   uk (r )     e (ik  )  uk (r )
 (r )
  (r )


 ikr
   
   
ik  r
1
1

 ik e   (ik  )uk (r )  e    (ik  )u k (r )
 (r )
 (r )
15




   1    

2 

∴ Master eq. becomes (ik  )    (ik  )  uk (r )   (k ) / c uk (r ),or
  (r )

   1   
 

 
2 
 k uk (r )   (k ) / c uk (r ) with  k  (ik  )    (r) (ik  ) 

eigenfunction u  (r) satisfying u  (r  R)  u  (r)
k
k
k



∴
Solving this eigenvalue problem for the unit cell & for each value of

=> photonic band structure n (k )
Restricting an eigenvalue problem to a finite volume leads to a discrete
spectrum of eigenvlues (ex: nearly free electronics in the 1st BZ).

∴ For each value of k , an infinite set of modes =>band index n.




 k has the k only as a parameter in it. Thus n (k ) is a continuously

varying ft. with k for a given n.
16

k
 Rotational Symmetry and Irreducible BZ.
- Phonic crystal : usually have rotational, reflection, inversion symmetry.
 
f (r )
ex) Assume a PhC with a 6-fold rotational symmetry.



r
r
 f
 
f

Let the operator (nˆ, ) rotates vectors f
f
(r )
r


r
r 


r
by an angle  about the nˆ  axis .
r

2
2
3
1
1
1
4
5
6

r6
To rotate a vector field

r3
 
f (r ),we
 
f 2 (r2 )

r2
need to transform so that
 
f1 (r1 )

r1
 
f 2 (r2 )


f   f

f (r6 )


and r   1r .
 
f1 (r1 )

r1

r2

r6

r6
before field rotation
after field rotation
17
def) vector field rotational operator O



 
 1 
: O f (r )  f ( r )

  , O  0 if the system is variant to the rotation
then  




2

 (O H kn )  O ( H kn )  n (k ) / c (O H kn )


 O H kn also satisfies the master eq. with the same eigenvalue as H kn


 




i ( k  R )


k   k , i.e., TR (O H kn )  e
(O H kn ).

O H kn
Note) State
is the Bloch state with


Proof; We need to prove TR (O H kn )  O (T R H kn ), i.e., O TRO  O OT R ( T R ).
(sub proof) Without loss of generality (WLOG), let   (0, ), rotation
about the origin through the angle  in the xy-plane.
1
Let
1
1
1
1
 x 
 x   cos sin   x   x cos  y sin  
   (0, )   
   


y 
 y    sin  cos  y    x sin   y cos 
 a
displacement vector is R   , then with the translation operator TR
b
18
 x
 x cos  y sin   a 
 cos( ) sin( ) 
  O1  

O1T O    O1 
y

x
sin


y
cos


b

sin(


)
cos(


)
 





R
 cos  sin    x cos  y sin   a 



sin

cos


x
sin


y
cos


b



Let M
1
 x a
M  
 y b

 x
a
1  a 
1
1
    M    M corresponds to operator  and   to vector R.
 y
b
b




 x
 x
1
     R  T1R   QED. (note def . of TR : TR r  r  R)
 y
 y

 


i ( k  1R )



 
 
 TR (O H kn )  O (T1R H kn )  O (e
H kn )
a  b  a  b


 



 1 
 
1
1


 k   R  k   R  k  R
 ei ( k  R ) (O H kn )  ei (k R ) (O H kn )
Since O H  is the Bloch state with k and same eigenvalue as H  ,
 kn
kn


it follows that n (k )  n (k ).
19
In general, whenever a photonic crystal has a rotation, mirro-reflection,

or inversion symmetry (point group) n (k ) have that symmetry as well.
Full symmetry of the point group => some regions of BZ have repeated
pattern => irreducible BZ( the smallest region not related by symmetry).
ex)
-Real lattice has 4-fold symmetry and
reflection symmetries


-Field patterns H kn in real space or n (k )
in the rest of the BZ is just the copies
of the irreducible BZ.
 Mirror symmetry and Seperation of Modes
Mirror reflection symmetry => Separation of the eigenvalue equation for

 
 k into two separate equations ( Ek , H k , , // to mirror plane) => Provides
immediate information about the mode symmetries (ex: Fig. 4).
20
Mirror reflection M x in the yz  plane changes xˆ to - xˆ and leaves yˆ and zˆ.
Mirror reflection M y in the xz  plane changes yˆ to - yˆ and leaves xˆ and zˆ.
For a system to have mirror symmetry,
it should be invariant under the


simultaneous reflection of f and r .
 


O
;
O
f
(
r
)

M
f
(
M
r
Def) mirror reflection operator M
M
x
x )
   
 
 
ex)
Note
1)
OM OM f (r )  f (r ).  OM f (r )  f (r )

x
M xr
x
x
x
x
with   1, and thus   1

f (r ) Note 2) , O   0  O H   ei H  with the reflected
M
M
k
M k
2

M x f (M x r )

r
Mx
x

wave vector M x k and an arbitrary phase 
x
x

 2

 



2
Proof : Since  , OM x  0,  M x H kn  M x H kn  M x  / c  H kn   (k ) / c M x H kn .  M x H kn

also satisfies the master eq. with the same eigenvalue as H kn . We thus need





to prove that TR (M x H kn )  M x (TM x1R H kn ), M x1TR M x  M x1M xTM x1R  TM x1R (note M x1  M x ).




21
 1 0 
Since
transforms
to
we may take M x  

0
1


 a
x
 x  a  x
 x
1  a 

in 2D space. Let R    then M x1TR M x    M x1
     M x    TM 1R  
x
b
y
y

b
 ,
 

  y
b
 y


 
 





i ( k M x1R )
ik M x1R
i ( M x k R )
 TR (M x H k )  M xTM 1R H k  M x e
H k  e
(M x H k )  e
(M x H k ).
x


Thus OM x H k is the Bloch state with the reflected wave vector M x k .
 
Note that we can always take a mirror plane so that M x r  r , since our
dielectric has CTS in x -direction. But M y r  r only for a certain r and M y .

 
 
i ( k r )  
If M x k  k , i.e., the Bloch wave H k (r )  e uk (r ) propagates in the yz  plane,
 

   

 
from M x r  r , OM x H k (r )  M x H k (M x r )   H k (r ). Ek (r ) : obeys similar eq.


∴ Both Ek and H k must be either even or odd under the OM x operation.


But, E is a vector, H is pseudovector. Thus OM x -even mode must be H x , Ey
( x, y, z)  ( x, y, z ),
( x, y, z)
Mx


and E z , while the OM –odd modes must have the components Ex ,
x
H y and H z .
22
Difference of behaviors between vector and pseudovector under inversion
operation (coordinate transform) and mirror operation (world transform).
In general, for a given mirror operator M such that  , OM  0, mode
 
 
separation is possible at the position where Mr  r for Mk  k , according to



the polarization depending on whether Ek or H k is parallel to mirror M
(ex: TE and TM modes in 2D PhCs).
But this mode separation concept is not so useful for 3D PhCs.
 Time-Reversal Invariance


2
Since  k is Hermitian2 and n (k ) is real, 2complex
conjugate of master eq.

  * n ( k )  *
  * n ( k )  *
is given by ( H kn )  2 H kn , i.e.,  H kn  2 H kn
c
c
 *


2


H
 H kn satisfies the same eq. as kn with same n (k ).



 * i (  kr )  
 *
i ( k r )  
H kn  e uk (r )  H kn  e
uk (r ).  H kn is just the Bloch state at (k , n), and thus,
23



n (k )  n (k ) holds independent of the photonic crystal structure.
 
  i t i ( kr )   i t   i ( kr  t)
Note) H (r , t )  H (r )e  e uk (r )e  uk (r )e
* 
 *  i t i ( kr )   i t   i ( kr  t)
 *
If we take H kn such that H (r , t )  H kn (r )e  e uk (r )e  uk (r )e


*
Thus taking H kn is equivalent to taking time as (t ).  n (k )  n (k ) is
a
consequence of the time-reversal symmetry of the Maxwell eqs.
 Electrodynamics in PhCs and electrons in crystals
Formation of energy bands and energy gap Eg in semiconductors: related
to the periodicity of crystals.



 

2
2



Schroedinger eq. is ( / 2m)  V (r )  (r )  E (r ) with V (r  R)  V (r ) for any



translational vector R. Hamiltonian H (r )  (2 / 2m)2  V (r ) has the
 

translational property such that H (r  R)  H (r ), since the kinetic term is
 

 


i ( R)
ik R


(
r

R
)

e

(
r
)

e

(
r
)  Table 1
invariant under any translation.
24