Transcript Vibrational Properties of the Lattice

IV. Vibrational Properties of the Lattice
A. Heat Capacity—Einstein Model
B. The Debye Model — Introduction
C. A Continuous Elastic Solid
D. 1-D Monatomic Lattice
E. Counting Modes and Finding N()
F. The Debye Model — Calculation
G. 1-D Lattice With Diatomic Basis
H. Phonons and Conservation Laws
I.
Dispersion Relations and Brillouin Zones
J.
Anharmonic Properties of the Lattice
Having studied the structural arrangements of atoms in solids, we now turn to
properties of solids that arise from collective vibrations of the atoms about
their equilibrium positions.
A. Heat Capacity—Einstein Model (1907)
For a vibrating atom:
E1  K  U
kz
m
kx
 12 m vx2  12 m vy2  12 m vz2  12 k x x 2  12 k y y 2  12 k z z 2
ky
Classical statistical mechanics — equipartition theorem: in thermal
equilibrium each quadratic term in the E has an average energy 12 kBT , so:
E1  6(12 kBT )  3kBT
Classical Heat Capacity
For a solid composed of N such atomic oscillators:
E  NE1  3NkBT
Giving a total energy per mole of sample:
E 3Nk BT

 3N A k BT  3RT
n
n
So the heat capacity at
constant volume per mole is:
d E
   3R  25 molJ K
CV 
dT  n V
This law of Dulong and Petit (1819) is approximately obeyed by most
solids at high T ( > 300 K). But by the middle of the 19th century it was
clear that CV  0 as T  0 for solids.
So…what was happening?
Einstein Uses Planck’s Work
Planck (1900): vibrating oscillators (atoms) in a solid have quantized
energies En  n n  0, 1, 2, ...
[later QM showed En  n  12  
is actually correct]
Einstein (1907): model a solid as a collection of 3N independent 1-D
oscillators, all with constant , and use Planck’s equation for energy levels
occupation of energy level n:
(probability of oscillator
being in level n)
f ( En ) 
e  En / kT

e
classical physics
(Boltzmann factor)
 En / kT
n 0

Average total
energy of solid:

E  U  3 N  f ( En ) En  3 N
n 0
 En / kT
E
e
 n
n 0

e
n 0
 En / kT
Some Nifty Summing

Using Planck’s equation:
U  3N
 n e
n 0

e
 n / kT
Now let x 
 n / kT

kT
n 0

U  3 N
ne
n 0

e
 nx
Which can
be rewritten: U  3N
 nx
n 0
Now we can use
the infinite sum:
So we obtain:
d   nx
 e
dx n 0

e
 3N
 nx
n 0

1
x


1 x
n 0
n
for x  1 To give:
 
d  x
 e
dx n 0
n
 e 

x n
n 0
 e 

x n
n 0
d  ex 
  x 
dx  e  1  3N
3N
U  3N


e x  1 e  / kT  1
 ex 
 x 
 e 1 
1
ex

 x
x
1 e
e 1
At last…the Heat Capacity!
d U 
d  3N A 
CV 
  
  / kT 
dT  n V dT  e
1 
Using our previous definition:
Differentiating:
CV 

   3R   e
e  1
 1
 3N A e  / kT
Now it is traditional to define
an “Einstein temperature”:
So we obtain the prediction:
e
 / kT
E 
 
kT 2
 2
kT
 / kT
 / kT
2
2

k
CV (T ) 
3R
 e
E 2
e
T
E /T
 E /T

1
2
Limiting Behavior of CV(T)
High T limit:
E
T
 1
CV



1 
(T ) 
 3R
1  1
3R   e
(T ) 
 3R   e
e 
3R
E 2
E
T
T
2
E
T
T
 1
E 2
CV
These predictions are
qualitatively correct: CV  3R
for large T and CV  0 as T  0:
 E /T
T
E /T 2
E 2
T
3R
CV
Low T limit:
E
T/E
 E / T
But Let’s Take a Closer Look:
High T behavior:
Reasonable
agreement with
experiment
Low T behavior:
CV  0 too quickly
as T  0 !
B. The Debye Model (1912)
Despite its success in reproducing the approach of CV  0 as T  0, the
Einstein model is clearly deficient at very low T. What might be wrong with
the assumptions it makes?
• 3N independent oscillators, all with frequency 
• Discrete allowed energies: En  n
n  0, 1, 2, ...
Details of the Debye Model
Pieter Debye succeeded Einstein as professor of physics in Zürich, and soon
developed a more sophisticated (but still approximate) treatment of atomic
vibrations in solids.
Debye’s model of a solid:
• 3N normal modes (patterns) of oscillations
• Spectrum of frequencies from  = 0 to max
• Treat solid as continuous elastic medium (ignore details of atomic structure)
This changes the expression for CV
because each mode of oscillation
contributes a frequency-dependent
heat capacity and we now have to
integrate over all :
CV (T ) 
max
N ( ) C


E
( , T ) d
0
# of oscillators per
unit 
Einstein function
for one oscillator
C. The Continuous Elastic Solid
We can describe a propagating vibration of amplitude u along a rod of
material with Young’s modulus E and density  with the wave equation:
 2u
E  2u

2
t
 x 2
for wave propagation along the x-direction
By comparison to the general form of the 1-D wave equation:
2
 2u
2  u
v
2
t
x 2
  2f  2
we find that
v

v
E

So the wave speed is independent of
wavelength for an elastic medium!
 (k ) is called the
dispersion relation of
the solid, and here it is
linear (no dispersion!)

 kv
group velocity vg 
d
dk
k
D. 1-D Monatomic Lattice
By contrast to a continuous solid, a real solid is not uniform on an atomic
scale, and thus it will exhibit dispersion. Consider a 1-D chain of atoms:
M
a
In equilibrium:
s p
s
s 1
s 1
Longitudinal wave:
u s 1
For atom s,
Fs   c p us  p  us 
p
us
u s 1
p = atom label
p =  1 nearest neighbors
p =  2 next nearest neighbors
cp = force constant for atom p
us p
1-D Monatomic Lattice: Equation of Motion
 2u s
Fs  M 2   c p us  p  us 
t
p
Now we use Newton’s second law:
For the expected harmonic
traveling waves, we can write
Thus:
i ( kxs t )
us  ue
xs = sa = position of atom s

Mu(i)2 ei ( ksa t )   c p uei ( k ( s p) at )  uei ( ksa t )

p
Or:


 M 2ei ( ksa t )  ei ( ksa t )  c p eikpa 1
p
So:


 M 2   c p eikpa 1
Now since c-p = cp by symmetry,
p


 M 2   c p eikpa  eikpa  2   2c p cos(kpa) 1
p 0
p 0
1-D Monatomic Lattice: Solution!
2 
The result is:
2
M
 c p (1  cos(kpa)) 
p 0
4
M
2 1
c
sin
 p ( 2 kpa)
p 0
The dispersion relation of the monatomic 1-D lattice!
Often it is reasonable to make the
nearest-neighbor approximation (p = 1):
4c1
 
sin 2 ( 12 ka )
M
2

The result is periodic in k
and the only unique
solutions that are
physically meaningful
correspond to values in
the range: 


a
k
a
4c1
M
k

2
a


a
0

a
2
a
Dispersion Relations: Theory vs. Experiment
In a 3-D atomic lattice we
expect to observe 3 different
branches of the dispersion
relation, since there are two
mutually perpendicular
transverse wave patterns in
addition to the longitudinal
pattern we have considered.
Along different directions in
the reciprocal lattice the
shape of the dispersion
relation is different. But
note the resemblance to the
simple 1-D result we found.
E. Counting Modes and Finding N()

A vibrational mode is a vibration of a given wave vector k (and thus ),
E   . How many
frequency  , and energy
are found in the
 modes


interval between (, E, k ) and (  d, E  dE, k  dk ) ?
# modes

dN  N ()d  N ( E)dE  N (k )d k
3
We will first find N(k) by examining allowed values of k. Then we will be
able to calculate N() and evaluate CV in the Debye model.
First step: simplify problem by using periodic boundary conditions for the
linear chain of atoms:
We assume atoms s
and s+N have the
same displacement—
the lattice has periodic
behavior, where N is
very large.
s+N-1
L = Na
s
s+1
x = sa
x = (s+N)a
s+2
First: finding N(k)
Since atoms s and s+N have the same displacement, we can write:
us  us  N
uei ( ksa t )  uei ( k ( s  N ) at )
This sets a condition on
allowed k values:
kNa  2n 
So the separation between
allowed solutions (k values) is:
Thus, in 1-D:
k
1  eikNa
2n
Na
2
2
k 
n 
Na
Na
n  1, 2, 3, ...
independent of k, so
the density of modes
in k-space is uniform
# of m odes
1
Na
L



interval of k  space k 2 2
Next: finding N()
Now for a 3-D lattice we can apply periodic boundary
conditions to a sample of N1 x N2 x N3 atoms:
N3c
# of m odes
N a N 2 b N 3c
V
 1
 3  N (k )
volum eof k  space 2 2 2 8
Now we know from before
that we can write the
differential # of modes as:
We carry out the integration
in k-space by using a
“volume” element made up
of a constant  surface with
thickness dk:
N2b
N1 a
 V 3
dN  N ()d  N (k )d k  3 d k
8
3

d k  ( surface area ) dk 
3
 dS dk

N() at last!
Rewriting the differential
number of modes in an interval:
We get the result:
N ( ) 
dN  N ( )d 
V
dS  dk
3 
8
V
dk
V
1
dS

dS

 
8 3 
d 8 3 
k
A very similar result holds for N(E) using constant energy surfaces for the
density of electron states in a periodic lattice!
This equation gives the prescription for calculating the density of modes
N() if we know the dispersion relation (k).
We can now set up the Debye’s calculation of the heat capacity of a solid.
F. The Debye Model Calculation
We know that we need to evaluate an upper limit for the heat capacity integral:
CV (T ) 
max
N ( ) C


E
( , T ) d
0
If the dispersion relation is known, the upper limit will be the maximum  value.
But Debye made several simple assumptions, consistent with a uniform, isotropic,
elastic solid:
• 3 independent polarizations (L, T1, T2) with equal propagation speeds vg
• continuous, elastic solid:  = vgk
• max given by the value that gives the correct number of modes per polarization (N)
N() in the Debye Model
d
vg 
dk
First we can evaluate
the density of modes:
N ( ) 
Since the solid is isotropic, all
directions in k-space are the same, so
the constant  surface is a sphere of
radius k, and the integral reduces to:
Giving:
V 2
N ()  3 4k  2 3
8 vg
2 vg
V
2
k
V
1
V
dS

dS

3 
3

8
vg 8 vg
2
dS

4

k
 
for one polarization
Next we need to find the upper limit for the integral over the allowed range of
frequencies.
max in the Debye Model
Since there are N atoms in the solid, there are N unique
modes of vibration for each polarization. This requires:
max
3
Vmax
2
Giving:
 d  2 3  N
2 3 
2 vg  0
6 vg
V
 max
N ( ) d  N


0
1/ 3
 6 N 

 V 
max  vg 
2
 D
The Debye cutoff frequency
Now the pieces are in place to evaluate the heat capacity using the Debye
model! This is the subject of problem 5.2 in Myers’ book. Remember that
there are three polarizations, so you should add a factor of 3 in the expression
for CV. If you follow the instructions in the problem, you should obtain:
T 
CV (T )  9 Nk B  
 D 
3  /T
D

0
4 z
z e dz
(e z  1) 2
And you should evaluate this
expression in the limits of low T
(T << D) and high T (T >> D).
Debye Model:
Theory vs. Expt.
Better agreement
than Einstein
model at low T
Universal behavior
for all solids!
Debye temperature
is related to
“stiffness” of solid,
as expected
Debye Model at
low T: Theory vs.
Expt.
Quite impressive
agreement with
predicted CV  T3
dependence for Ar!
(noble gas solid)
(See SSS program
debye to make a
similar comparison
for Al, Cu and Pb)
G. 1-D Lattice with Diatomic Basis
Consider a linear diatomic chain of atoms (1-D model for a crystal like NaCl):
a
In equilibrium:
M1
M2
M1
M2
Applying Newton’s second law and the nearest-neighbor approximation to this
system gives a dispersion relation with two “branches”:

 M1  M 2   2  M1  M 2 
4c
2 1
  c1 
 
  c1 
sin ( 2 ka )

 M 1M 2    M 1M 2  M 1M 2
2
2
-(k)
  0 as k  0
+(k)
  max as k  0 optical modes
1/ 2
2
1
acoustic modes (M1 and M2 move in phase)
(M1 and M2 move out of phase)
1-D Lattice with Diatomic Basis: Results

optical
These two branches may
be sketched schematically
as follows:
gap in allowed frequencies
acoustic


a
0
k

a
In a real 3-D solid the dispersion relation will differ along different directions in
k-space. In general, for a p atom basis, there are 3 acoustic modes and p-1
groups of 3 optical modes, although for many propagation directions the two
transverse modes (T) are degenerate.
Diatomic Basis: Experimental Results
The optical modes generally have frequencies near  = 1013 1/s, which is in
the infrared part of the electromagnetic spectrum. Thus, when IR radiation is
incident upon such a lattice it should be strongly absorbed in this band of
frequencies.
At right is a transmission spectrum
for IR radiation incident upon a
very thin NaCl film. Note the
sharp minimum in transmission
(maximum in absorption) at a
wavelength of about 61 x 10-4 cm,
or 61 x 10-6 m. This corresponds
to a frequency  = 4.9 x 1012 1/s.
If instead we measured this spectrum for
LiCl, we would expect the peak to shift to
higher frequency (lower wavelength)
because MLi < MNa…exactly what happens!
H. Phonons and Conservation Laws
Collective motion of atoms = “vibrational mode”:
Quantum harmonic oscillator:
En  n
us ( x, t )  uei (kxs t )
n  0, 1, 2, ...
Energy content of a vibrational mode of frequency  is an integral number
of energy quanta   . We call these quanta “phonons”. While a photon is a
quantized unit of electromagnetic energy, a phonon is a quantized unit of
vibrational (elastic) energy.

Associated with each mode of frequency  is a wavevector k , which leads
to the definition of a “crystal momentum”:

k
Crystal momentum is analogous to but not equivalent to linear momentum.
No net mass transport occurs in a propagating lattice vibration, so the linear
momentum is actually zero. But phonons interacting with each other or with
electrons or photons obey a conservation law similar to the conservation of
linear momentum for interacting particles.
Phonons and Conservation Laws
Lattice vibrations (phonons) of many different
frequencies can interact in a solid. In all
interactions involving phonons, energy must be
conserved and crystal momentum must be
conserved to within a reciprocal lattice vector:

1 k1
Schematically:
1   2  3




k1  k 2  k3  G

3 k3

2 k 2
Compare this to the special case of elastic
scattering of x-rays with a crystal lattice:
   Just a special case of
k   k  G the general
Photon wave vectors
conservation law!
I. Brillouin Zones of the Reciprocal Lattice
Remember the dispersion relation of the 1-D monatomic lattice, which repeats
with period (in k-space) 2 / a :

4c1
M
k

4
a

3
a

2
a


0
a

a
2
a
3
a
4
a
1st Brillouin Zone (BZ)
2nd Brillouin Zone
3rd Brillouin Zone
Each BZ contains
identical information
about the lattice
Wigner-Seitz Cell--Construction
For any lattice of points, one way to define a unit cell is to connect each lattice
point to all its neighboring points with a line segment and then bisect each line
segment with a perpendicular plane. The region bounded by all such planes is
called the Wigner-Seitz cell and is a primitive unit cell for the lattice.
1-D lattice: Wigner-Seitz cell is the line
segment bounded by the two dashed planes
2-D lattice: Wigner-Seitz cell is the shaded
rectangle bounded by the dashed planes
1st Brillouin Zone--Definition
The Wigner-Seitz cell can be defined for any kind of lattice (direct or reciprocal
space), but the WS cell of the reciprocal lattice is also called the 1st Brillouin
Zone.
The 1st BZ is the region in reciprocal space containing
all information about

the lattice vibrations of the solid. Only
 the k values in the 1st BZ correspond
to unique vibrational modes.
Any k outside this zone is mathematically

equivalent to a value k1 inside the 1st BZ. This is expressed in terms of a
general translation vector of the reciprocal lattice:


G
4c1
M
  
k  k1  G
k

4
a

3
a

2
a


a
0

k1

a
2
a

k
3
a
4
a
1st Brillouin Zone for 3-D Lattices
For 3-D lattices, the construction of the 1st Brillouin Zone leads to a
polyhedron whose planes bisect the lines connecting a reciprocal lattice point
to its neighboring points. We will see these again!
bcc direct lattice  fcc reciprocal lattice
fcc direct lattice  bcc reciprocal lattice
I
J. Anharmonic Properties of Solids
Two important physical properties that ONLY occur because of anharmonicity
in the potential energy function:
1.
2.
Thermal expansion
Thermal resistivity (or finite thermal conductivity)
Thermal expansion
In a 1-D lattice where each atom experiences the same potential energy
function U(x), we can calculate the average displacement of an atom from its
T=0 equilibrium position:

x 
U ( x ) / kT
xe
dx



U ( x ) / kT
e
dx


I
Thermal Expansion in 1-D
Evaluating this for the harmonic potential energy function U(x) = cx2 gives:

x 
 xe


e
 cx2 / kT
 cx2 / kT
dx
dx
Now examine the numerator
carefully…what can you conclude?
x  0!
independent of T !

Thus any nonzero <x> must come from terms in U(x) that go beyond x2. For
HW you will evaluate the approximate value of <x> for the model function
U ( x)  cx2  gx3  fx4
(c, g, f  0 and gx3 , fx 4  kT )
Why this form? On the next slide you can see that this function is a reasonable
model for the kind of U(r) we have discussed for molecules and solids.
Potential Energy of Anharmonic Oscillator
(c = 1 g = c/10 f = c/100)
U = cx2 - gx3 - fx4
U = cx2
Potential Energy U (arb. units)
16
14
12
Do you know what
form to expect for <x>
based on experiment?
10
8
6
4
2
0
-5
-3
-1
1
Displacement x (arbitrary units)
3
5
Lattice Constant of Ar Crystal vs. Temperature
Above about 40 K, we see:
a(T )  a(0)  x  T
Usually we write: L  L0 1   T  T0   = thermal expansion coefficient
Thermal Resistivity
When thermal energy propagates through a solid, it is carried by lattice waves
or phonons. If the atomic potential energy function is harmonic, lattice waves
obey the superposition principle; that is, they can pass through each other
without affecting each other. In such a case, propagating lattice waves would
never decay, and thermal energy would be carried with no resistance (infinite
conductivity!). So…thermal resistance has its origins in an anharmonic
potential energy.
low T
Thermal
energy flux
(J/m2s)
high T
Classical definition of
thermal conductivity
1
3
  CV v
CV heat capacity per unit volume
dT
J  
dx
v wave velocity
 mean free path of scattering
(would be  if no anharmonicity)
Phonon Scattering
There are three basic mechanisms to consider:
1. Impurities or grain boundaries in polycrystalline sample
2. Sample boundaries (surfaces)
3. Other phonons (deviation from harmonic behavior)
deviation from
perfect crystalline
order
To understand the experimental dependence  (T ) , consider limiting values
of CV and  (since v does not vary much with T).
 T low T
3
CV
3R
high T
1

 e  / kT  1
n ph
  low T

high T
kT
Temperature-Dependence of 
The low and high T limits are summarized in this table:
CV


low T
 T3
nph  0, so
  , but then
  D (size)
 T3
high T
3R
 1/T
 1/T
How well does this match experimental results?
Experimental (T)
 T3
 T-1 ?
(not quite)
Phonon Collisions: (N and U Processes)
How exactly do phonon collisions limit the flow of heat?
2-D lattice  1st BZ in k-space:



q1  q2  q3

q1

q2
2
a
2
a

q3
No resistance to heat flow
(N process; phonon momentum conserved)
 Predominates at low T << D since 
and q will be small
Phonon Collisions: (N and U Processes)
What if the phonon wavevectors are a bit larger?
2-D lattice  1st BZ in k-space:

q3
2
a
2
a

q1

q2




q1  q2  q3  G
 
q1  q2

G
Umklapp = “flipping over” of
wavevector!
Two phonons combine to give a net phonon
with an opposite momentum! This causes
resistance to heat flow.
(U process; phonon momentum “lost” in
units of ħG.)
 More likely at high T >> D since  and
q will be larger