Nano Photonics Materials Lecture Notes
Nano Photonics Materials Lecture Notes
Nano Photonics Materials Lecture Notes
Materials Chemistry III
(Nano‐Photonics Materials)
3 4
Contents Magnetic moment of one atom
1. Materials and Devices for Optical Communication B m
B( z ) Let’s calculate the magnitude of
2z 3 magnetic moment m of one atom!!
2. Nature of Optical Wave
N S v
3. Diffraction and Optical Thin Film I e S a 2
z 2a
4. Optical Intensity and Surface Plasmon v
-e m ( e )(a 2 )
2a
5. Fundamentals of Magnetism m
Today a e e
( )aM e v ( )L
6. Electron Spin and Paramagnetism Fe 2M e 2M e
I
Spin Angular momentum
7. Exchange Interaction and Ferromagnetism
According to quantum theory,
8. Spintronics and Its Application
Spin “up”
m IS an angular momentum L is an
(E-B) ( h / 2 )
integer multiple of .
L
Then we put ,
Spin “down ”
Quantum well
5 6
Angular momentum Origin of magnetism
Conservation of angular momentum
Lrp in central force field Revolution Rotation
p( M e v ) dL dr dp
pr B B
-e dt dt dt B B
r
v M ev r F -e
+e F
0 ( r F ) N N
+e = m -e = m
Angular momentum is conserved!! S S
F : Central force Spin
Quantum number!!
cf. | L | l (l 1) Lz ml
l : Azimuthal quantum number ml : Magnetic quantum number
7 8
Angular momentum Spin
Schrödinger equation Schrödinger equation
2 2 2 2
V (r ) (r ) (r ) V (r ) (r ) (r )
2M e 2M e
n ,l ,m (r , , ) Rn ,l (r )Yl m ( , ) n ,l ,m (r , , ) Rn ,l (r )Yl m ( , )
Spherical function Yl m Lz i LzYl m ml Yl m Spin function SmS
S z SmS mS SmS ①
s l 0 L2Yl m 2l (l 1)Yl m 1 S 2 SmS 2 S ( S 1) SmS ②
p l 1 S
[ Lx , Ly ] iLz 2 [ S x , S y ] i S z
d l2
[ S y , S z ] i S x ③
f l 3 [ Ly , Lz ] iLx 1 1
mS mS [ S z , S x ] iS y
m 3, 2, 1, 0, 1, 2, 3 [ Lz , Lx ] iLy 2 2
m l , l 1, , 0, , l 1, l 2l 1 mS S , S 1, , 0, , S 1, S 2S 1
9 10
Spin Commutation relation
Notice !!
① S z mS
mS mS
Sz , [ px , x] px x xpx i
S S
2 2 [ A, B] AB BA
3 px x / 2
② S 2 SmS 2 S ( S 1) SmS | S | S ( S 1)
2 A: To measure the height AB
[ px , x] px x xpx
We cannot determine Sx and Sy
③ [ S x , S y ] i S z B: To measure the weight
at the same time.
(i ) x x(i )
z We can only determine one x x
3 Spin precession
parameter, for example Sz.
2 [ px , x] f ( px x xpx ) f
173 cm
2 Notice !!
(i ) xf x(i ) f 63 kg
y [ px , x] px x xpx i x x
px x / 2 f f Commutable
2 i( f x ) ix if [ A, B] 0
x x x
2
11 12
Commutation relation Quiz 1
Notice !!
Prove the following commutation relation.
[ px , x] px x xpx i
[ A, B] AB BA
px x / 2 [ S x , S y ] iS z
A: To study for exam.
Here Sx, Sy, and Sz are given as follows:
[ px , x] px x xpx B: To take the exam.
1 i 1
(i ) x x(i ) 0 0 2 0
x x 2
Sx Sy Sz 2
[ px , x] f ( px x xpx ) f 1 0 i 0 0 1
BA ( A B) High score 2 2 2
(i ) xf x(i ) f AB ( B A) Low score
x x
f f Not Commutable
i( f x ) ix if [ A, B] 0
x x
13 14
Magnetic ordering Magnetic ordering
N
N ① Paramagnetism ② Ferromagnetism
m
Electron -e S
Spin Magnetic moment Simplified
S version
H
Random Magnetic ordering
M = 0 M≠0 M < 0 C, Cu, Pb M = 0 Cr, CoO, Fe2O3
N
( 0)
Magnetization (Wb/m2) M m
i
i
M H : Susceptibility
15 16
Magnetic ordering Diamagnetism
Compound,
oxide materials Faraday's law
H
d d ( BS )
④ ② V
AFM FM I dt dt
① ③ r
DM
S BS
PM
-e
dH B
2 rEs r 0 2
dt I
http://www.jst.go.jp/sicp/ 0 dH + -
ws2011_eu/presentation/p Es r
resentation_02.pdf H2O, superconductor
2 dt V
17 18
Diamagnetism Diamagnetism
v
H Ms
t
eEs H m
02 e 2
4m
r2H
4m
x
02 e 2 2
y2 H
e 02 e 2 2 N 02 e 2 Z 2
I v Es t I a H a
6m 6m
r Ms r M ZNm
-e -e M H
e 0
r H z
2M s H
dH x2 y 2 z 2 a2
2 rEs r 2 0
dt ev r
m 0 ( r )
2 x2 y 2 z 2
0 dH 2 r x
y
y
Es r 2 2
2 dt 02 e 2 2 x2 y 2 a
r H x 3
4m a: Radius
19 20
Quiz 2 Diamagnetism
Calculate the relative diamagnetic susceptibility for copper,
in which atomic number is 29, atomic mass is 63.54, density
is 8.94 g/cc, and average orbital radius is 0.5A. Here,
Avogadro’s number NA is 6.02 x 1023 mol‐1, electric charge of
electron e is 1.6 x 10‐19 C, and electron mass is 9.11 x 10‐31 kg.
A. = 3.6 x 10‐5
N 02 e 2
Z a2
6m
0 4 107 H / m Naphthalene Graphite
21 22
Susceptibility of Fe-based superconductors Magnetic ordering
Meissner effect
Pauli paramagnetism ① Paramagnetism ② Ferromagnetism
0 = 4 x 10‐7 H/m Si S j
/ 0 = 3.8 x 10‐9 H/m
Tc
0.0040
M = 0 Na, Pd, Al
0.0
Jij > 0
SmFeAsO1‐xFx
0.0035 i) Pauli exclusion principle M ≠ 0 Fe, Co, Ni
-0.5 0.00
-0.05
-0.10
57.4 K ii) Many‐body effect
4
4
-1.0
4
-0.15
0.0030
-0.20
-0.25 ④ Antiferromagnetism
-1.5 -0.30
48 50 52 54 56 58
0.0025
Temperature (K)
Exchange interaction
-2.0
0 10 20 30
Temperature (K)
40 50 60 70
0.0020
0 10 20 30 40 50 60 70 wij 2 J ij Si S j
Temperature (K) Jij < 0
Perfect diamagnetism
Jij: Exchange integral M = 0 Cr, CoO, Fe2O3
M H : Susceptibility Courtesy of Prof. M. Fujioka
23 24
Paramagnetism Torque and potential energy
Langevin theory For a spin which makes an Torque T r F qm
angle with H, the potential l qm H
energy is given by | T || r || F | sin
d qm H
U mH cos ,
H
so that the probability for a qm H
m
spin to take this direction is
| T | lqm H sin
U mH sin
P exp( )
Angular distribution of k BT
paramagnetic spins in a mH cos U mH sin d
magnetic field H P A exp( )
k BT mH cos
25 26
Paramagnetism Paramagnetism
Langevin theory For a spin which makes an Langevin theory The probability for a spin to
angle with H, the potential make an angle between
energy is given by and d is given by
d d
mH cos
U mH cos , p ( )d A exp( ) 2 sin d
k BT
H H d
m so that the probability for a m sin
spin to take this direction is
U
P exp( )
Angular distribution of k BT Angular distribution of mH cos
A exp( ) 2 sin d 1
mH cos k BT
0
paramagnetic spins in a paramagnetic spins in a
magnetic field H P A exp( ) magnetic field H A
1
k BT mH cos
0
A exp(
k BT
) 2 sin d
27 28
Paramagnetism Paramagnetism
mH cos 1
) 2 sin d
e xdx
exp( x
Langevin theory k BT Langevin theory M Nm 1
p ( )d
mH cos 1
e dx
x
0 exp( k BT ) 2 sin d 1
d d 1 1
e
x
M Nmcos dx e )
(eQ3(2)
1
H Nm cos p( )d H
m 0
m
1
e
x
xdx
1
(e e Q3(3)
1
) 2 (e e )
mH cos
1
0
exp(
k BT
) cos sin d
Nm e e 1
mH cos M Nm
0 exp( k BT ) sin d e e
Angular distribution of Angular distribution of
1 mH 1
paramagnetic spins in a
Nm 11
e x xdx
k BT
paramagnetic spins in a Nm coth
Q3(4) NmL( )
magnetic field H Q3(1) magnetic field H
1e dx
x
cos x Langevin function
29 30
Paramagnetism Paramagnetism
mH
Langevin theory M NmL( ) Langevin theory Nm 2
M: Large k BT
1 Curie law 3k BT
0.8
d d
0.6 CuSO4K2SO46H2O Paramagnetic salts
L()
e e 1
H 0.4
L( )
m 0.2 e e m Nm 2
3k B
0
0 2 4 6 8 10
1 H: Large 1
Angular distribution of Angular distribution of
L( ) ~ T: Low T
paramagnetic spins in a 3 paramagnetic spins in a
magnetic field H Nm Nm 2 magnetic field H Henry, PR 88, 559 (1952)
M H H
3 3k BT Susceptibility Magnetic moment
31 32
Paramagnetism Paramagnetism
In the previous calculation, we assumed that the spin can Curie law
take all possible orientation. In reality, a spin can have only Nm 2
discrete orientations because of spatial quantization. Classical theory
3k BT
M NmL( ) (Classical picture)
Paramagnetic salts Ng 2 J ( J 1) B2 Paramagnetic salts
Quantum theory
3k BT
M NgJ B BJ ( ) (Quantum theory)
Brillouin function
2J 1 2J 1 1 m g J ( J 1) B2 Nm 2
BJ ( ) coth coth 3k B
2J 2J 2J 2J
gJ B H 1 J: Total angular momentum
J 1 L: Orbital angular momentum
1
k BT BJ ( ) ~ Ng 2 J ( J 1) B2 T
3J Henry, PR 88, 559 (1952) S: Spin angular momentum Henry, PR 88, 559 (1952)
3k BT
Russell-Saunders interaction: J=L+S
33
Quiz 4
Calculate the paramagnetic susceptibility for an ideal gas,
in which each molecule has a magnetic moment with J = 1, g
= 2, at 1 atmosphere pressure and 0oC. The 1 mol ideal gas
takes a volume of 22.4 l at the above‐mentioned
temperature and pressure. Avogadro’s number NA is 6.02 x
1023 mol‐1. Boltzmann constant kB is 1.38 x 10‐23 J/K. Bohr
magneton B is 1.17x 10‐29 Wb・m.
A. = 2.6 x 10‐12 H/m