Download Lecture 18:

PHYS 342/555
Introduction to solid state physics
Instructor: Dr. Pengcheng Dai
Associate Professor of Physics
The University of Tennessee
(Room 407A, Nielsen, 974-1509)
(Office hours: TR 1:10AM-2:00 PM)
Lecture 18, room 512 Nielsen
Chapter 10: Plasmons, Polaritons, and polarons
Lecture in pdf format will be available at:
http://www.phys.utk.edu
1
Measuring the Fermi surface
Fermi surface measurements require pure single crystal at low
temperatures and is frequently perform in very strong magnetic
fields. The shape of the Fermi surface is intimately involved
in the transport coefficients of a metal as well as in the equilibrium
and optical properties. The most powerful method to deduce
the Fermi surface geometry is the de Haas-van Alphen effect.
χ = dM / dH ,
Landau accounted the
oscillations in free
electron theory.
Dai/PHYS 342/555 Spring 2006
Chapter 10-2
The change in 1/ H through a single
period of oscillation, Δ (1/ H ) , was
determined by:
⎛ 1 ⎞ 2π e 1
Δ⎜ ⎟ =
c Ae
⎝H⎠
where Ae is any extremal cross-section
area of the Fermi surface in a plane
normal to the magnetic field.
Dai/PHYS 342/555 Spring 2006
Chapter 10-3
The de Haas-van Alphen effect is an indication
of the failure of the semiclassical model. The
failure arises whenever the semiclassical theory
predicts closed orbits for the electron motion
projected on a plane perpendicular to the field.
When this happens, the energies of motion
perpendicular to H are quantized.
Dai/PHYS 342/555 Spring 2006
Chapter 10-4
Free electrons in a uniform magnetic field
The orbital energy levels of an electron in a cubic box with sides
of length L parallel to the x-, y -, and z -axes are determined in
the presence of a uniform magnetic field H along the z -direction
by two quantum numbers, ν and k z :
2
1
eH
εν ( k z ) =
k + (ν + ) ωc , ωc =
.
2m
2
mc
ν runs through all nonmagnetic integers, and k z takes on the
2
z
same values as in the absence of a magnetic field:
k z = 2π nz / L for any integral nz .
The energy of motion perpendicular to the field, which would
be
2
(k x2 + k y2 ) / 2m if no field were present, is quantized in
steps of ωc (ωc = eH / mc). This is orbit quantization.
Dai/PHYS 342/555 Spring 2006
Chapter 10-5
Dai/PHYS 342/555 Spring 2006
Chapter 10-6
Semi-classical motion in a uniform magnetic field
In the semiclassical equation of motion:
1 ∂ε (k ) d ( k )
1
v (k ) =
,
= ( − e) v ( k ) × H .
dt
c
∂k
Electrons move along curves given by the intersection of
surfaces of constant energy with planes perpendicular
to the magnetic field.
Dai/PHYS 342/555 Spring 2006
Chapter 10-7
The projection of the real space orbit in a plane perpendicular to the
field, r = r − Hˆ ( Hˆ ⋅ r ), can be found by taking the vector product
⊥
of both sides with a unit vector parallel to the field.
eH
eH
d( k )
ˆ
ˆ
ˆ
=−
(r − H ( H ⋅ r ) = −
r⊥
H×
dt
c
c
c ˆ
r⊥ (t ) − r⊥ (0) = −
H × (k (t ) − k (0)).
eH
Dai/PHYS 342/555 Spring 2006
Chapter 10-8
The time taken to traverse the orbit between k1 and k2 is:
t2
k2
t1
k1
t2 − t1 = ∫ dt = ∫
dk
i
|k |
2
2
c k2
dk
c 1
=
=
eH ∫k1 (∂ε / ∂k ) ⊥ eH Δε
∫
k2
k1
( )
Δ k dk
2
c ⎛ ∂A1,2 ⎞
c ⎛ ∂A(ε , k z ) ⎞
=
⎜
⎟=
⎜
⎟
eH ⎝ ∂ε ⎠ eH ⎝ ∂ε
⎠
2
Δε =
∂ε
⎛ ∂ε ⎞
⎛ ∂ε ⎞
⋅Δ k = ⎜ ⎟ ⋅Δ k = ⎜ ⎟ Δ k
∂k
⎝ ∂k ⎠ ⊥
⎝ ∂k ⎠ ⊥
( )
( )
Dai/PHYS 342/555 Spring 2006
( )
Chapter 10-9
Levels of Bloch electrons in a uniform magnetic field
In free electron theory, a level of ε F must have a quantum number ν
whose order of magnitude is ε F / ωc = ε F /[(e / mc) H ].
e / mc = 1.16 ×10−8 eV/G. Typically, ε F is several eV, so quantum
number ν will be of order 104 .
Energies at two adjacent levels is determined by hν . Let εν (k z ) be
the energy of the ν th allowed level at a given k z .
( εν +1 (k z ) − εν (k z ) ) = hν = h / T (εν (k z ), k z ),
∂A ( εν ) 2π eH
−
=
(
)
(
)
.
ε
k
ε
k
( ν +1 z ν z )
∂ε
c
Because we are interested in εν of order ε F , εν +1 ( k z ) − εν ( k z )
A ( εν +1 ) − A ( εν ) = 2π eH / c = ΔA
εF.
A(εν (k z ), k z ) = (ν + λ )ΔA, where λ is independent of ν .
Dai/PHYS 342/555 Spring 2006
Chapter 10-10
Origin of the oscillatory phenomena
Most electronic properties of metals depend on the density of levels
at the Fermi energy, g ( ε F ) . It follows that g ( ε F ) will be singular
whenever the value of H causes an extramal orbit on the Fermi surface
to satisfy the quantization condition (ν +λ )ΔA = Ae (ε F ).
⎛1
Δ⎜
⎝H
e
⎞ 2π e 1
−4
=
.
=
1.34
×
10
K / G.
⎟
c Ae ( ε F ) mck B
⎠
Dai/PHYS 342/555 Spring 2006
Chapter 10-11
Alternative interpretation
The area between successive orbits is
ΔS = S n − S n −1 = 2π eB / c.
The area in k-space occupied by a single orbital is (2π / L) 2 .
The number of free electron orbitals that coalesce in a single
magnetic level is
D = (2π eB / c)(2π / L) 2 .Dai/PHYS 342/555 Spring 2006
Chapter 10-12
For a system of N electrons at absolute zero the Landau levels
are filled to s. Orbitals at the next higher level s + 1 will be partially
filled. The Fermi level will lie between s and s + 1. As the magnetic
field is increased the electrons move to lower levels because the area
between successive circles are increased
πΔ(k 2 ) = (2π k )(Δk ) = 2π eB / c.
Dai/PHYS 342/555 Spring 2006
Chapter 10-13
Fermi surface of copper
The electron concentration in a monovalent metal with an fcc
structure is n = 4 / a 3 . The radius of a free electron Fermi sphere
is k F = (3π 2 n)1/ 3
(4.90 / a ).
Dai/PHYS 342/555 Spring 2006
Chapter 10-14
Chapter 10 Plasmons, Polaritons, and Polarons
The dielectric constant ε of electrostatics is defined in terms of the
electric field E and the polarization P, the dipole moment density:
D = E + 4π P = ε E; ε is the relative permittivity.
Suppose a positively charged particle is placed at a fixed position
inside the electron gas. It will then attract electrons and create a
surplus negative charge around it, thus reducing its field.
If ρ ext is the particle's charge density without screening, the full
charge density should be ρ = ρ ext + ρind .
D is related to the external applied charge density ρ ext and E is
related to the total charge density ρ = ρ ext + ρind .
divD = divε E = 4πρ ext
divE = 4π ( ρ ext + ρind ).
Dai/PHYS 342/555 Spring 2006
Chapter 10-15
Define ε ( K ) such that:
D( K ) = ε ( K ) E ( K ); ε ( K ) is the relative permittivity.
Then we have
divE = div∑ E ( K )eiK ⋅r = 4π ∑ ρ ( K )eiK ⋅r .
divD = div∑ ε ( K ) E ( K )eiK ⋅r = 4π ∑ ρext ( K )eiK ⋅r
ε ( K ) = ρext ( K ) / ρ ( K ) = 1 − ρind ( K ) / ρ ( K )
The electrostatice potentials satisfy the Poisson equation
∇ 2ϕext = −4πρext ; and ∇ 2ϕ = −4πρ .
ϕext ( K ) ρext ( K )
=
= ε ( K ).
ϕ (K )
ρ (K )
Dai/PHYS 342/555 Spring 2006
Chapter 10-16
Plasma optics
The equation of motion of a free electron in an electric field:
d 2x
m 2 = −eE. if x and E have the time dependence e − iωt , then
dt
−ω 2 mx = −eE ; x = eE / mω 2 .
The dipole moment of one electron is − ex = −e 2 E / mω 2 , and
the polarization defined as the dipole moment per unit volume is
P = − nex = − ne 2 E / mω 2 , the dielectric function is then
ωp
4π ne 2
D(ω )
P(ω )
2
1
,
4
/ m.
≡ 1 + 4π
= 1−
≡
−
=
ε (ω ) ≡
ω
π
ne
p
2
2
E (ω )
E (ω )
mω
ω
If positive ion core background has a dielectric constant ε (∞),
2
ε (ω ) = ε (∞) − 4π ne 2 / mω 2 = ε (∞)[1 − ω p2 / ω 2 ].
Dai/PHYS 342/555 Spring 2006
Chapter 10-17
Dispersion relation for electromagnetic waves
Maxwell's equation for a nonmagnetic isotropic medium,
∂2D
2 2
=
c
∇ E.
2
∂t
If E ∝ eiK ⋅r e − iωt and D = ε (ω , K ) E; then
ε (ω , K )ω 2 = c 2 K 2 ;
Dai/PHYS 342/555 Spring 2006
Chapter 10-18
If ε is real and >0: K is real and a transverse electromagnetic wave
propagates with the phase velocity c / ε 1/ 2 .
If ε is real and <0: K is imaginary and the wave is damped with a
characteristic length 1/|K | .
If ε is complex. For ω real, K is complex and the waves are damped
in space.
ε = ∞. This means the system has a finite response in the absence of
an applied force; thus the poles of ε (ω , K ) define the frequencies of
the free oscillations of the medium.
ε = 0. Longitudinally polarized waves are possible only at the
zero of ε .
Dai/PHYS 342/555 Spring 2006
Chapter 10-19
Dispersion relation for electromagnetic waves
The dispersion relationship
ε (ω )ω 2 = ε (∞)[ω 2 − ω p2 ] = c 2 K 2 .
For ω < ω p we have K 2 < 0, so that K is imaginary. Waves incident
on the medium in the frequency range 0 < ω < ω p do not propagate,
but will be totally reflected.
An electron gas is transparent when ω > ω p , the dispersion relation
is then
ω 2 = ω p2 + c 2 K 2 / ε (∞); this describes transverse electromagnetic
waves in a plasma.
Dai/PHYS 342/555 Spring 2006
Chapter 10-20
Dai/PHYS 342/555 Spring 2006
Chapter 10-21
Longitudinal plasma oscillations
The zero of the dielectric function determine the frequencies of the
longitudinal modes of oscillation. ε (ωL )=0
ε (ωL ) = 1 − ω p2 / ωL2 = 0. When ωL = ω p , there is a free longitudinal
oscillation mode of an electron gas at the plasma frequency as the
low-frequency cutoff of transverse electron magnetic waves.
The equation of motion is
d 2u
nm 2 = − neE = −4π n 2 e 2u,
dt
1/ 2
2
⎛
⎞
4
π
d u
ne
2
+ ω p u = 0; ω p = ⎜
⎟ .
2
dt
⎝ m ⎠
2
Dai/PHYS 342/555 Spring 2006
Chapter 10-22
The displacement establishes a surface charge density − neu on
the upper surface of the slab and +neu on the lower surface. An
electric field E = 4π neu is produced inside the slab.
Dai/PHYS 342/555 Spring 2006
Chapter 10-23
Dai/PHYS 342/555 Spring 2006
Chapter 10-24
We consider the plane interface z = 0 between metal 1 at Z > 0 and
metal 2 at Z < 0. Metal 1 has bulk plasmon frequency ω p1 ; metal
2 has ω p 2 . The dielectric constants in both metals are those of free
electron gases. Show that surface plasmons associated with the
interface have the frequency
1/ 2
⎡1 2
⎤
ω = ⎢ (ω p1 + ω p2 2 ) ⎥
⎣2
⎦
.
Dai/PHYS 342/555 Spring 2006
Chapter 10-25