Feature Selection/Extraction for Classification Problems
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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 ; OI1OI for any operation
y
OI
OI
x
OI
OI OI or from OI1OI 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 ) TdH (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 2d / .
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 )
ikr
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.
ikr
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 )
ikr
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 )
ikr
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 (2f / c)2 ( / c)2 .
If 1 with refractive index n , vg c / n, 0 / n, kc / 2f /
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 dn ( z )
1 (ik y a y )e y n ( z )ax e y
ay
( z)
dz
dn ( z )
ik y y
1
e ik yn ( z )az
ay
( z)
dz
9
y
y
y
dn ( z ) 1 ik y y
d ( z )
ik y y
1
e ik yn ( z )az
a y
e ik yn ( z )az n
ay
( z)
dz
(
z
)
dz
ax a y
az
dn ( z )
* (ik yn ( z )a x )
ay
x y
z
dz
0 0 ik yn ( z ) ,
0
ax
x
0
ay
y
dn ( z )
dz
az
2
a d n ( z )
x
z
dz2
0
1
d 2n ( z ) 1 ik y y
dn ( 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
dn ( 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 dn ( 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 dn ( z )
dz ( z ) dz
2
d 2n ( z ) d 1 dn ( z ) k y
2
1
(
z
)
(
/
c
)
n ( z )
n
( z ) dz2
dz ( z ) dz ( z )
2
1 d 1 dn ( 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 2ff 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(iz ), 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 ( xd ) (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 ky 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
ikr
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
ikr
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 uk (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 ikr
2 ik r
e uk (r ) (k ) / c e uk (r ).
(r )
1 ikr
ikr
ik r
ik r
1
e uk (r ) (e ) uk (r ) e uk (r ) e (ik ) uk (r )
(r )
(r )
ikr
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 kn also satisfies the master eq. with the same eigenvalue as H kn
i ( k R )
k k , i.e., TR (O H kn ) e
(O H kn ).
O H kn
Note) State
is the Bloch state with
Proof; We need to prove TR (O H kn ) O (T R H kn ), i.e., O TRO O OT 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( )
O1
O1T O O1
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 T1R QED. (note def . of TR : TR r r R)
y
y
i ( k 1R )
TR (O H kn ) O (T1R H kn ) O (e
H kn )
a b a b
1
1
1
k R k R k R
ei ( k R ) (O H kn ) ei (k R ) (O H kn )
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 kn 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 kn M x H kn M x / c H kn (k ) / c M x H kn . M x H kn
also satisfies the master eq. with the same eigenvalue as H kn . We thus need
to prove that TR (M x H kn ) M x (TM x1R H kn ), M x1TR M x M x1M xTM x1R TM x1R (note M x1 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 x1TR M x M x1
M x TM 1R
x
b
y
y
b
,
y
b
y
i ( k M x1R )
ik M x1R
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 kn ) 2 H kn , i.e., H kn 2 H kn
c
c
*
2
H
H kn satisfies the same eq. as kn with same n (k ).
* i ( kr )
*
i ( k r )
H kn e uk (r ) H kn e
uk (r ). H kn 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 ( kr ) i t i ( kr t)
Note) H (r , t ) H (r )e e uk (r )e uk (r )e
*
* i t i ( kr ) i t i ( kr t)
*
If we take H kn such that H (r , t ) H kn (r )e e uk (r )e uk (r )e
*
Thus taking H kn 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