Download Chapter 10 (Kittell): Lecture Notes

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Phys. 555/342:Kittell Ch 10
Plasmons, Polaritons, Polarons, and Excitons
Chapter 10 in Kittell: Be carefull he uses CGS
To map onto SI replace 4π with 1/ε0
We will discuss ε(ω,K)--energy and wavelength dependent
ε(ω,0)---describes collective excitation: Plasmons, etc.
ε(ω,K)---describes electrostatic screening, e-e, e-P, eimpurity
D-displacement
D = E + 4"P = #E E-field
P-polarization
!
" • D = " • #E = 4$% ext
" • E = 4$% = 4$( % ext + % ind )
Phys. 555/342:Kittell Ch 10
Dielectric Function for Electron Gas
Plasma Optics
"(#,k) $ "(#,o) % "(# )
& $'
with
d2x
m 2 = "eE
dt
!
2
"# mx = "eE
!
x " e i#t
E " e i#t
eE
x= 2
" m
! "ex = "e 2 E /# 2 m
2
ne
P =!"nex = " 2 E
# m
Dipole moment
!
!
Phys. 555/342:Kittell Ch 10
Dielectric Function for Electron Gas
Plasma Optics
ne 2
P = "nex = " 2 E
# m
D(# )
P(# )
"(# ) =
$ 1+ 4%
E(# )
E(# )
4$ne 2
"(# ) = 1+
m# 2
!
Plasma Frequency
2
4#ne
" 2p =
m
# 2p
"(# ) = 1+ 2
#
!
ω=ωp when ε=0
Phys. 555/342:Kittell Ch 10
Dielectric Function for Electron Gas
Plasma Optics
But the positive background also has a dielectric constant
labeled "(#)
2*
'
˜
#
4&ne
p
"(# ) = "($) %
=
"
($)
1+
)
2
2,
m#
( # +
2
!
2
4#ne
"˜ =
m$(%)
2
p
!
Again---
!
"˜ p = "
#=0
Phys. 555/342:Kittell Ch 10
Dielectric Function for Electron Gas
Dispersion
EM wave equation
" 2D
2 2
=
c
#E
2
"t
and " "D = #($,K)E
But " "E # e
2
2 2
"(#,K)# = c K
i(K •r"$t )
!
• ε real and >0: For ω real, K is real and a transverse EM wave propagates with
phase velocity c/ε1/2
• ε real and <0: For ω real, K is imaginary and the wave is damped with a
characteristic length 1/|K|
• ε is complex: For ω real, K is complex and the waves are damped in space.
• ε is infinity: This means the system has a finite response in the absence of an
applied force; thus the poles of ε(ω,K) define the frequesncies of the free
oscillations of the medium.
• ε=0: We shall see that longitudinally polarized waves are possible only at the
zeros.
!
!
Phys. 555/342:Kittell Ch 10
Transverse Optical Modes in a Plasma
2
2
"(#,K)# = c K
For " "# < #˜ p
For " "# > #˜ p
!
2
2
2
2
2 2
˜
"(# )# = "($)(# % # p ) = c K
K2 <0 so K is imaginary
K2 >0 so K is real
E "e
#|K |x
" 2 = "˜ 2p + c 2K 2 /#($)
!
This describes a transverse
electromagnetic wave in a
plasma.
!
Phys. 555/342:Kittell Ch 10
Plasmons
Phys. 555/342:Kittell Ch 10
Transparency of Alkalki Metals in the Ultraviolet
Metal should reflect below the plasma frequency
Metals should transmit above the plasma frequency
The Reflectance of Indium Antimonide with
n=4x1018 cm-1
Phys. 555/342:Kittell Ch 10
Longitudinal Plasma Oscillations
Zeros of dielectric function determine the frequency of
the longitudinal modes of oscillation.
"(# L ) = 0
2
#p
"(# L ) = 1$ 2 = 0
#
!
There is a free longitudinal oscillation
mode (fig) of an electron gas at the
plasma frequency. This is the low
frequency cut off of the transfer mode
!
Phys. 555/342:Kittell Ch 10
Plasma Oscillations
A plasm oscillation is a collective longitudinal excitation of the
conduction electrons. A plasmon is a quantum of plasm
oscillations.
Phys. 555/342:Kittell Ch 10
Plasmons
Phys. 555/342:Kittell Ch 10
Kittell Problem 10.1
Phys. 555/342:Kittell Ch 10
Solution 10.1
∂ϕ
= kA sin kx e kz , and at the boundary this is equal to Exi. The normal
∂x
component of D at the boundary, but outside the medium, is ( )kA cos kx, where for a
plasma ( ) = 1 – p2/ 2. The boundary condition is –kA cos kx = ( )kA cos kx, or
( ) = –1, or p2 = 2 2. This frequency ω = ωp 2 is that of a surface plasmon.
1. E x0 = −
Phys. 555/342:Kittell Ch 10
Kittell: Problem 10,2
Phys. 555/342:Kittell Ch 10
Kittell, solution 10.2
2. A solution below the interface is of the form ϕ(−) = A cos kx e kz , and above the
interface ϕ(+ ) = A cos kx e − kz , just as for Prob. (1). The condition that the normal
component of D be continuous across the interface reduces to 1( ) = – 2( ), or
1−
ω2p1
ω2
= −1 +
ω2p2
ω2
, so that ω2 =
1 2
(ωp1 + ω2p2 ) .
2
Phys. 555/342:Kittell Ch 10
Kittell, Problem 10.5
Phys. 555/342:Kittell Ch 10
Kittell, solution 10.5
5. md 2r/dt 2 = ! m"2r = !eE = 4#eP/3 = !4#ne 2r/3 . Thus "2$ = 4#ne 2 3m.
Phys. 555/342:Kittell Ch 10