Download Lecture 6

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 6, room 512 Nielsen
Chapter 4: Crystal Vibrations
Lecture in pdf format will be available at:
http://www.phys.utk.edu
1
Normal modes of two atoms per primitive basis
The harmonic potential energy is:
1
1
2
2
harm
U
= K ∑ [u1 (na ) − u2 (na ) ] + G ∑ [u2 (na ) − u1 ([n + 1]a ) ]
2 n
2 n
where u1 ( na) is the displacement around na, u2 (na ) the displacement
around na + d . If d ≤ a / 2, K ≥ G.
d 2u1 (na )
∂U harm
=−
M
= − K [u1 (na ) − u2 (na ) ] − G [u1 (na ) − u2 ([n − 1]a ) ]
2
∂u1 (na )
dt
d 2u2 (na )
∂U harm
M
=−
= − K [u2 (na ) − u1 (na ) ] − G [u2 ( na ) − u1 ([n + 1]a ) ]
2
dt
∂u2 (na )
Dai/PHYS 342/555 Spring 2006
Chapter 4-2
We seek solution
u1 (na ) = ε1 exp[i (kna − ωt )]
u2 (na ) = ε 2 exp[i (kna − ωt )]
Here ε1 and ε1 are constants that determines the relative amplitude
and phase of the vibration of the ions within each primitive cell.
[ M ω 2 − ( K + G )]ε1 + ( K + Ge − ika )ε 2 = 0
[ M ω 2 − ( K + G )]ε 2 + ( K + Ge − ika )ε1 = 0
This pair of homogeneous equations will have a solution, if the
determinant of the coefficients vanishes:
[ M ω 2 − ( K + G )]2 − | K + Ge − ika |2 = 0
K +G 1
±
K 2 + G 2 + 2 KG cos ka
Two solutions: ω 2 =
M
M
ε1
K + Ge − ika
with
=∓
| K + Ge − ika |
ε2
Dai/PHYS 342/555 Spring 2006
Chapter 4-3
For each of the N values of k there are thus two solutions,
leading to a total of 2 N normal modes. The two ω vs k curves
are two branches of the dispersion relation. Acoustic and optical
branches.
Dai/PHYS 342/555 Spring 2006
Chapter 4-4
Case 1 k
π / a Here cos ka 1 − (ka) 2 / 2, and
ω=
2( K + G )
− O(ka ) 2
M
ω=
KG
ka When k is very small, ε 2 = ∓ ε1.
2M ( K + G )
Dai/PHYS 342/555 Spring 2006
Chapter 4-5
Case 2 k = π / a
ω=
2K
, ε1 = −ε 2
M
2G
, ε1 = ε 2
M
When k = π / a, the motion in neighboring cells are 180 degrees
ω=
out of phase, and therefore the two solutions are
Dai/PHYS 342/555 Spring 2006
Chapter 4-6
Case 3 K
G To leading order in G / K we have
2K
[1 + O(G / K )], ε1 ≈ −ε 2
ω=
M
2G
1
|sin ka| [1+O(G / K ) ] , ε1 ≈ ε 2
M
2
The optical branch now has a frequency that is independent of k ,
to leading order in G / K , and equal to the vibrational frequency
ω=
of a single diatomic molecule composed of two mass M ions
connected by a spring K . The acoustic branch is just that for a
linear chain of atoms of mass 2M coupled by the weak spring G.
Dai/PHYS 342/555 Spring 2006
Chapter 4-7
An acoustic mode is one in which all ions within a primitive cell
move essentially in phase and the dynamics are dominated by the
interaction between cells; an optical mode is one in which the
ions within each primitive cell are executing molecular vibratory mode.
Case 4 K = G
In this case we are really dealing with a monatomic Bravais lattice
of lattice constant a / 2.
Dai/PHYS 342/555 Spring 2006
Chapter 4-8
Quantization of elastic waves
The energy of a lattice vibration is quantized and called a phonon. Thermal
vibrations in crystals are thermally excited phonons with energy
1
E = (n + ) ω when the mode is excited to quantum # n.
2
We can quantize the mean square phonon amplitude.
u = u0 cos kx cos ω t.
1 ∂u 2
The kinetic energy density is ρ ( ) , where ρ is the mass density. In
2 ∂t
a crystal of volume V , the average kinetic energy is:
1
1
1
1
2
2
2 2
ρV ω 〈cos kx sin ω t 〉 = ρV ω u0 = (n + ) ω ,
2
8
2
2
ω should be positive
Dai/PHYS 342/555 Spring 2006
Chapter 4-9
Phonon momentum
A phonon of wavevector k will interact with photons, neutrons &
electrons as if it has momentum k . Howver, a phonon does not
carry physical momentum.
du 1 − eikNa
Physical moment: p = M (d / dt )∑ u exp(ikna) = M
dt 1 − eika
Note that periodic condition requires eikNa = 1, this means p = 0.
At k = 0, p = MN ( du / dt ).
The crystal momentum. If the scattering of the phonon is inelastic,
with creation of a phonon of wavevector k , then
k f + k = ki + Q
Dai/PHYS 342/555 Spring 2006
Chapter 4-10
Inelastic scattering of neutrons by phonons
The kinetic energy of the incident neutron is p 2 / 2M n , where M n
is the mass of the neutron. The momentum of neutron is given by
p = k = 2π / λ. The conservation of energy requires
( ki ) 2 / 2 M n = ( k f ) 2 / 2 M n ± ω ,
where ω is the energy of the phonon created (+) or absorbed
(-) in the process.
Dai/PHYS 342/555 Spring 2006
Chapter 4-11
Dai/PHYS 342/555 Spring 2006
Chapter 4-12
Consider the normal modes of a linear chain in which the force
constants between nearest-neighbor atoms are alternately
C and 10C. Let the masses be equal, and let the nearestneighbor separation be a/2. Find ω ( K ) at K = 0 and
K = π / a. Sketch in the dispersion relation by eye.
Dai/PHYS 342/555 Spring 2006
Chapter 4-13
Chapter 5: Phonons and thermal properties
In quantum theory of specific heat, the total thermal energy of the
phonons can be written as the sum of the energies over all phonon
modes: U = ∑∑ 〈 nk , p 〉 ωk , p ,
k
p
Here 〈 nk , p 〉 denotes the thermal equilibrium occupancy of phonons
of wavevector k and polarization p. 〈 nk , p 〉 is given by the Planck
distribution function: 〈 nk , p 〉 =
1
.
exp( ω / τ ) − 1
∂U
The heat capacity is then defined as CV ≡ (
)V .
∂T
Dai/PHYS 342/555 Spring 2006
Chapter 4-14
Planck distribution function:
Consider a set of identical harmonic oscillators in thermal equilibrium.
1
The harmonic oscillator has an energy of (n + ) ω. Thus the fraction
2
of the total number of oscillators in the nth quantum state is
Nn
exp(− n ω / τ )
= ∞
∞
∑ N s ∑ exp(− s ω / τ )
s =0
s =0
The average excitation quantum # of an oscillator is:
s exp(−n ω / τ )
∑
〈 n〉 =
=
∑ exp(− s ω / τ ) exp(
1
ω /τ ) −1
Dai/PHYS 342/555 Spring 2006
Chapter 4-15
Normal mode enumeration
The energy of a collection of oscillators in thermal equilibrium
U = ∑∑
k
p
ω
k ,p
exp( ω / τ ) − 1
, If crysal had D p (ω )d ω modes of
a given plarization p between ω and ω + d ω.
U = ∑ ∫ dω D p (ω )
p
ω
, where D p (ω ) is the number
exp( ω / τ ) − 1
of modes per unit frequency, called the density of modes or
density of states.
Assume x = ω / k BT . Then ∂U / ∂T
x 2 exp( x)
.
Clat = k B ∑ ∫ d ω D p (ω )
2
[exp( x) − 1]
p
Dai/PHYS 342/555 Spring 2006
Chapter 4-16
Density of states in 1D
If each normal vibrational mode of polarization p has the
form of a standing wave, where
us = u (0) exp(−iωk , p t ) sin ska
The wavevector k is restricted by the fixed-end boundary
conditions to the values
( N − 1)π
π 2π
, ,
k= ,
L L
L
Dai/PHYS 342/555 Spring 2006
Chapter 4-17
For 1D line there is one mode for each interval Δk = π / L, so that
the number of modes per unit range of k is L / π for k ≤ π / a and
0 for k > π / a. The number of modes D (ω )dω in dω at ω
D(ω )dω =
L dk
dω , dω / dk is the group velocity.
π dω
Dai/PHYS 342/555 Spring 2006
Chapter 4-18