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PHONON I: The dispersion relation (by CHY)
Introduction
The static lattice model, that a solid consists of rigid, immobile periodic array of ions, can
successfully explain X-ray diffraction experiments. But, it is only an approximated picture.
There are many physical phenomena that cannot be simply explained by the static model.
For example, thermal conductivity: how does heat transmit from one end of a spoon to the
other? The other good examples include melting, inelastic scattering, resistance change as a
function of temperature. Actually, only a more accurate picture including lattice vibration
can quantitatively explain the X-ray diffraction intensity at different temperatures. So, there
is a need to better describe the lattice.
One approach is to consider that the lattice can vibrate as if the atoms are linked by springs.
We call this collective (many atoms are needed in this picture) motion phonon. The purpose of this section is to derive the dispersion relation (energy versus momentum) of this
excitation based on classical equation of motion. Many thermal properties can be derived
once the dispersion relation is known. Later improvements, including a quantum mechanical
description and an anharmonic theory of phonon decay, of this simple classical picture will
also be introduced.
REFERENCES:
1. Kittel: chapter 4 and 5
2. AM: chapter 21-26
We make the following assumptions when deriving the dispersion relation within the classical
picture:
1. Simple harmonic approximation:
The position of an atom (or, more accurately, nucleus) r(R) = R+ u(R), where R is a
Bravais lattice vector, and therefore u(R) is a measure of the deviation from the static
lattice cite. To the limit that u ≪ |R|, we assume that the force between the nearest
neighbors is linearly proportional to the spacing between them. This assumption is related
to the cohesive energy of a solid. At the bottom of the cohesive potential, the functional
shape can be well-approximated by a parabolic function, which corresponds to appling a
spring in between two atoms.
2. Adiabatic approximation:
As ion moves, electrons will respond instantaneously. Therefore, at any time, the electrons
are at their lowest states. As a result of screening, as will be discussed later in this course
when we talk about the properties owing to electrons, the electrons will only weakly modify
the interaction strength between atoms. Within this approximation, we will ignore the
complication of electron-electron and electron-ion interaction. In turn, we only focus on the
parabolic potential well (cohesive energy) in response to the lattice spacing.
3. Nearest-neighbor (N.N.) approximation:
By taking into account only the nearest neighbors, the computation can be much simplified. As we learned later from experiments, it is a pretty accurate (within a few percent)
1
approximation already. Theoretical predictions are often based on assumptions, and N.N.
approximation is a good one, based on that the theoretical results can well-explain experimental observations.
We start from Hook’s law:
Force = −C × displacement,
where C is the spring constant, and the displacement describes how much the spring is
stretched.
When there is a mass M attached to one end of the spring, the equation of motion is:
M ü = f orce.
Further, the potential energy Uharmonic is C × (displacement)2 /2.
The equation of motion can be extented to that: M ü = f orce = −dUharmonic /du.
We take a simple one-dimensional model, i.e., we only consider the case where ions are
arranged on a one-dimensional line. The movements is limited to be in this dimension only.
There is only one atom per basis, and the lattice constant is a. (â is the primitive vector.)
(So, R= nâ, where n is an integer.) At cite s, the deviation is labeled us . And, there are
two forces applied at this atom, one from the left and one from the right:
Uharmonic = C/2 × [(us − us+1)2 + (us − us−1)2 ]
= C/2 × (u2s − 2us us+1 + u2s − 2us us−1 + ...).
Keeping only the leading terms (in us ) and taking the derivative over us , we get the equation
of motion:
−dUharmonic /dus = −C(2us − us+1 − us−1) = M u¨s .
A (circular) boundary condition that can be justified is uN = u0 , where N is a large number.
This interesting boundary condition comes from the following observations/arguments:
1. The phonons will mostly travel in the bulk of the material, which means that surface (the
ends of our 1D string) is not important for now.
2. The number of atoms on the 1D string is very large, but the exact number should not be
important, for otherwise the exact nature and role of the atoms at the ends will have to be
well-defined.
3. Experimental results show that the phonon spectra are not sensitive to the surface, which
confirms that there is no need to specify the nature of the surface to great detail.
4. The phonon ‘lifetime’ is a finite number. That is to say, in real samples, phonons will
travel in a volume consisting of a finite number of atoms. The theoretical treatment therefore
should not be sensitive to the exact number of atoms, so long as the number of atoms is
large enough.
So, we find that if us ∝ ei(kna−wt) , it can satisfy the B.C. that uN = u0 . (That is, us±1 ∝
ei(k(n±1)a−wt) .)
To verify, take ei(kN a) = 1, we have kNa = 2πm, where m is an integer. That is, k =
2πm/Na. Yes, when N is a very, very large number, the separation of neighboring k ′ s
becomes very small. If we take the convention that −π/a < k < π/a, we say the k is in the
2
first Brillouin Zone (B.Z.). In fact, k can be of any value, but you can show yourself that if k
′
′
is 2π/a away from a state k in the first B.Z., then this state k of your choice will represent
a solution (deviation in motion) identical to that of the state k. The use of the first B.Z.
can therefore reduce the size of k space needed. The periodic array of atoms in real space is
the fundamental cause of such periodic behavior in k-space.
Similar argument can be expected for the electron wavefunctions in a crystal, where the
periodic potential will impose a periodic behavior in k-space. Note here that the number of
allowed (generated) states in the first B.Z. is N, since △k = 2π/Na.
Homework: Plot a phonon wave with wavelength shorter than 2a, and show that the
atomic movement can be well-represented by another wavelength much longer.
Plugging the functional form of us into the equation of motion, we have:
(i)2 M(−ω)2 ei(kna−wt) = −C(2)[1 −q
cos(ka)]ei(kna−wt) . And, this leads to:
Mω 2 = 2C[1 − cos(ka)], i.e., ω = 2C[1 − cos(ka)]/M.
A plot of ω versus k is the dispersion relation, from which the group velocity can be calculated. At the B.Z. boundary, the group velocity is zero. A state with k = π/a has a
wavelength of 2a, which means that the neighboring atoms will form a standing wave. In
comparison, for states near k = 0, the group velocity is a finite value — this is the speed of
sound in this 1D solid.
Two atoms in a basis
The previous example — 1D array of 1 atom/basis crystal — shows that we have certain
dispersion relation between the vibrational energy and the wavelength. There is only one
branch in the dispersion relation. Here, we will show a more complex situation where there
are two atoms (either similar or dissimilar) per basis.
Take two types of atoms with masses M1 and M2 , and the deviations are us and vs ,
respectively. The lattice constant is a. The equation of motion is: M1 u¨s = f orce =
−C(2us − vs − vs−1 ), and M2 v¨s = f orce = −C(2vs − us − us+1 ). Take the same assumption
on B.C., the deviations are of the form: us = uei(kna−ωt) ; vs = vei(kna−ωt) . Plugging the wave
equations into the equation of motion, we get:
(M1 ω 2 − 2C)u + C(1 + e−ika )v = 0; (M2 ω 2 − 2C)v + C(1 + eika )u = 0.
There
"
is only a non-zero solution
(of u and v) when the following determinant is zero:
#
M1 ω 2 − 2C C(1 + e−ika )
C(1 + eika ) M2 ω 2 − 2C
We have then an equation:
M1 M2 (ω 2 )2 − 2C(M1 + M2 )ω 2 + 2C 2 [1 − cos(ka)] = 0.
Homework:Derive the above relation from the determinant.
Homework: Find out limka→0 ω =?
Homework: Find out ω(k = π/a) =? and ω(k = −π/a) =?
3
At the limit ka → 0, there are two solutions: (1) optical branch, where ω 2 = 2C(1/M1 +
1/M2 ); and (2) acoustic branch, where ω 2 = Ck 2 a2 /2(M1 + M2 ).
The two values at the B.Z. boundary are: (1)
q
q
2C/Mheavier for the acoustic branch; and (2)
2C/Mlighter for the optical branch. (Mheavier is the mass of the heavier one between M1
and M2 .)
Homework: Find the phonon energies for Si, Ge, and GaAs, and Calculate the size of their
first B.Z. Plot their phonon dispersion relations.
For the same k, optical phonon energy is higher than the corresponding acoustic phonon
energy. If you write down the deviation functions (us /vs ), you can convince yourself that
the relative phase of the deviation for the N.N. atoms will be such that:
(1) for optical phonons, the deviations are out of phase; and
(2) the deviations are in phase for the acoustic phonons.
The origin of optical phonon branch is only that there are more than one atom in the basis.
If there is only one atom per basis, then the solid can only have acoustic phonon modes.
Now, we can extend the discussion to three-dimensional case, where each basis have p atoms
per basis. Each atom owns three degrees of freedom — atoms can choose to vibrate in three
independent directions. A basis will then have 3p degrees of freedom, forming 3 acoustic
phonon branches and (3p - 3) optical phonon branches. For the 3 acoustic modes, one is
longitudinal mode, and the rest two are transverse modes. Such distribution is also true
for optical modes. In longitudinal modes, the direction of the deviation is parallel to the
k vector. In transverse modes, they are perpendicular to each other. Because the spacing
between atoms is smaller for the longitudinal modes, the phonon energy will be higher.
PHONON II: Lattice contribution to specific heat
Introduction
Based on the classical picture with the help of several assumptions, we have derived the
phonon dispersion relation for three-dimensional crystals, even with a basis consisting of
multiple atoms. Although only one spring constant between neighboring atoms is used,
the system shows a spectrum of eigenenergies, each corresponding to a particular mode of
vibration. However, there is still room for improvement: this classical picture cannot explain
the temperature dependence of specific heat.
According to statistical mechanics, a solid at equilibrium will carry a thermal energy per
unit volume:
1
< Energy >=
V
R
dVdke−ǫ/kt ǫ
,
−ǫ/kt
volume k dVe
R
volume k
R
R
(1)
where ǫ is the total energy (kinetic plus potential) and the exponential term is the probability
4
that states of this energy is occupied.
In classical phonon picture using harmonic oscillators, the density of state over k-space is
constant, and as a result the specific heat due to lattice vibration is a constant of 3k per ion.
(The Boltzmann constant k = 1.38 ×10−16 erg/K.) But in fact, the specific heat goes like
AT + BT 3 . At low (below 100K or so) temperatures, the Cv begins to drop, and eventually
vanishes at absolute zero. The linear term comes from free electrons; for insulators, A would
be zero. Here we want to show that the T 3 dependence can be well-explained by the quantum
description of phonons.
We first formulate the ‘quantization’ of the wave-like vibration of atoms. In quantum language, we talk about the number of phonons at different modes, rather than its vibrational
amplitude.
Quantization of phonons
The classical description of a 1D spring-mass harmonic oscillator system is that the potential
energy V = 21 Cx2 and F orce = −Cx = mẍ, where m is mass, C is the spring force constant,
and C and m are related by the classical frequency of oscillation ω = (C/m)1/2 . The solution
is x(t) = Acos(ωt) + Bsin(ωt) = x0 cos(ωt + φ), where x0 is the amplitude and φ is the phase
of the oscillator. The energy conserved in the system is kinetic energy plus the potential
energy:
K.E.+V = 21 mẋ2 + 12 mω 2 x2 = 12 ω 2 x20 . Note here that x0 is only a number, and the oscillation
frequency is set by the spring constant and the mass. So when the kinetic energy is increased,
the amplitude will be larger.
The quantum description of a 1D simple harmonic oscillator is:
Hamiltonian H = p2 /2m + 21 mω 2 x2 ; Eigenenergy en = (n + 21 )h̄ω; Eigenstates |n > (Hermite
polynomials), n = 0, 1, 2, ....
We notice here that quantum theory tells us several new ideas:
(1) the energy is quantized;
(2) the energy spacing is constant;
(3) the lowest possible energy is not zero; and
(4) the probability of finding the particle.
The energy of a particular mode is (n + 21 )h̄ω. When the energy is increased, in quantum
language we say that the mode (ω) is excited to quantum number n, or, the mode is occupied
by n phonons. (Classically, you can imagine that the oscillatory amplitude for this mode is
increased.) This relation between (classical) amplitude and (quantum) occupation number
can be shown as follows.
We define the creation
and annihilation
operators:
q
q
annihilation: a = (mω/2h̄)x + i (1/2mωh̄)p;
q
q
creation: a∗ = (mω/2h̄)x − i (1/2mωh̄)p; where
a∗ a = n;√
a|n >= √n|n − 1 >;
a∗ |n >= n + 1|n + 1 >; and
5
x=
q
h̄/2mω(a + a∗ ).
Based on these basic properties, we find the expectation value of x2 :
< x2 >=< n|xx|n >=< n|(h̄/2mω)(a + a∗ )(a + a∗ )|n >
= h̄/2mω < n|aa + a∗ a + a∗ a∗ + aa∗ |n >
= h̄/2mω(2n + 1).
Therefore, (n + 1/2)h̄ω = mω 2 < x2 >. The classical picture is therefore related to the
corresponding quantum picture in terms of the population of phonon modes.
The value of ω for most discussions will be postive. When the temperature of a solid is
approaching melting, then in theory we invent the concept of imaginary ω.
Planck distribution
For deriving the quantum specific heat, we need to first define the Planck distribution of
phonons. Considering a large number of identical oscillators (all with the same frequency
ω) at thermal equilibrium. Based on the assumptions that we are treating indistinguishable,
identical oscillators, and that the number of available states is much larger than the number
of oscillators (the number of available phonon modes is of the order of the number of atoms,
but the number of phonons would be of the order of several thousand at most), we have the
ratio between them (higher energy state versus the lower energy state) is e−[(n+1)−n]h̄ω/kT =
e−h̄ω/kT < 1, where the number of oscillators at quantum state n + 1 is Nn+1 , and the number
of oscillators at quantum state n is Nn . (Reference: Any statistical mechanics textbook.)
The number of oscillators at the nth state
e−nh̄ω/kT
Nn
= P∞ −sh̄ω/kT .
= P∞
Total number of oscillators
s=0 Ns
s=0 e
(2)
So, the averaged occupation number would be:
P∞
1
s e−sh̄ω/kT
.
= h̄ω/kT
−sh̄ω/kT
e
−1
s=0 e
< n >= Ps=0
∞
(3)
If all oscillators were at the same energy with s = 4, then, < n >= 4. But, following the
temperature-dependent occupation probability, there is a well-defined distribution – called
Planck distribution. (This is a special case of Bose-Einstein distribution in that the total
number of particles is not conserved, but temperature-dependent. Later, when we talk about
electrons, we will talk more about it.)
At very high temperatures (h̄ω ≪ kT ), < n > is a large number:
lim
T →∞
1
eh̄ω/kT
−1
= lim
δ→0
eδ
1
1
1
kT
=
= =
.
−1
1+δ−1
δ
h̄ω
(4)
We therefore have that the population is proportional to Temperature. This result is the
classical limit.
6
At very low temperatures (h̄ω ≫ kT ), we get:
lim
1
kT ≪h̄ω eh̄ω/kT
−1
= e−h̄ω/kT ≪ 1.
(5)
Therefore, at low enough temperatures, the averaged population number (for many identical
oscillators at one frequency) is approaching zero exponentially as temperature is lowered.
This is the difference between the classical picture and the quantum picture. Since the
energy of quantum oscillators are quantized, as kT becomes less than the energy separation
of quantum states, the quantum effect will dominate.
Total energy
Given that the Planck distribution determines the average population of harmonic oscillators, then we have the equation for total (lattice) energy of a solid. We will discuss that
contribution from electrons later in the course.
U = total lattice vibrational energy per unit volume =
=
1X
1
h̄ω(k)[ns (k) + ]
V k,p
2
1 X
1 X1
1 X
h̄ω
h̄ωs
=
h̄ω
+
,
s
V k,p eh̄ω/kT − 1
V k,p 2
V k,p eh̄ω/kT − 1
(6)
(7)
where the sum is over all wavevectors k and all branches, and ω is a function of k. Here
we have separated the vibrational energy into two terms: (1) one is the zero-point energy
term, where there is no temperature dependence except that the phonon energies will indeed
change slightly due to lattice constant change as a function of temperature, (2) the other
one is the additional lattice vibration contribution at a non-zero temperature.
Note that a sum over states can be done by simply summing up one by one, or, by integration:
X
all states k
= (L/2π)
d
Z
dk =
Z
dωD(ω),
(8)
where L is the length of the sample in one dimension, d is the dimension under discussion,
and D is the density of states in one branch (polarization): how many states (in k-space)
do we have in a very small frequency window dω at frequency ω. For a typical bulk (d = 3)
sample, Ld = V . The larger the sample size is, the denser phonon states are in k-space.
Note that the sum over k-space is limited to the first Brillouin zone, following the definition
P
of reciprocal lattice. (By the way, k, allbranches =?) For each k, there are 3p normal modes
(p is the number of ions in a basis).
Therefore, the specific heat in 3D with large enough sample size can be represented as:
Cv = ∂U/∂T = ∂/∂T [
1
V
X
k,branches
h̄ω
eh̄ω/kT
−1
]=
7
X
∂
[
∂T branches
Z
dk
h̄ωs
].
3
h̄ω/kT
(2π) e
−1
(9)
High temperature limit: quantum meets classical
When kT ≫ h̄ω, Cv approaches 3Nk. The difference between the classical and quantum results is called the quantum correction. However, at yet higher temperatures, the anharmonic
component will come into the picture.
Low temperature limit: Cv has T 3 dependence
We observe that the last equation can be simplified because:
1. At very low temperatures the optical phonon modes can be ignored, since they are not
populated at all;
2. Even for acoustic phonons, only the lowest energy states are occupied, therefore the
phonon dispersion relation can be approximated to be linear in k;
3. Since only the lowest energy states with very small k’s will contribute to specific heat,
the integration can now be safely done in the whole k-space.
This is to say,
lim Cv =
h̄ωs ≫kT
∂
∂T
Z
s=LA,T A1 ,T A2
2π 2
kT 3
h̄ωs
dk
∼
k(
),
3
h̄ω
/kT
(2π) e s
−1
5
h̄vsound
(10)
where vsound is the average phase velocity of the acoustic phonon branches near k = 0. To
be more exact,
3
1/vsound
1X
=
3 s
Z
dω 1
.
4π (ω/k)3
(11)
Or, in another words,
Cv
2π 2 k 4
.
=
3
T →0 T 3
5 h̄3 vsound
(12)
lim
For materials that are insulators at low temperatures, this T 3 relation works very well.
For conductors, the contribution from free electrons dominate at low temperatures. So, in
addition to the T 3 dependence, we have another electronic term.
Debye model
In the Debye approximation, we assume that (1) there are three degenerate acoustic branches,
each with the same linear dispersion relation: ω = vsound k; (2) the number of k states allowed
is the same as the number of ions (N) in the solid.
A state in k-space will occupy a volume of (2π)3 /V , so the Debye sphere will have a radius
kDebye that makes:
4 3
πk
/[(2π)3 /V ] = N.
3 Debye
(13)
8
3
So, N/V = kDebye
/6π 2 . (kDebye is then very close to the inverse of the spacing between atoms
in the solid.)
Cv within Debye approximation would then be:
Z ΘDebye /T
9Nk
x4 ex dx
N
3
Cv =
(T /ΘDebye )
∼ 234(T /ΘDebye )3 k,
x
2
V
(e − 1)
V
0
(14)
at low temperatures. The ΘDebye is Debye temperature, a number that we chose to simplify
the integration:
kΘDebye = h̄ωDebye . For typical insulators, ΘDebye is a few hundred Kelvin.
Looking back the contribution of Debye approximation, we have the following observations:
1. If there are more than one atom in the basis, than the contribution from the optical
phonon branches are taken into account by enlarging the size of integration beyond the first
B.Z., so that the total number of wavevectors that will be summed is still N, the total
number of ions in the solid.
2. The Debye temperature is of the order of 100K to several hundred K’s. It represents a
transition between the quantum regime and the classical regime – the T 3 law works below
the Debye temperature (quantum theory becomes important).
3. The Debye temperature is usually determined by fitting the resulting equation to experimental data. Therefore, as the sample temperature increases, the Debye temperature will
decrease (in order to have a better fit). In fact, only at very low temperatures (below a few
K) would Debye approximation be physical, because then only the lowest acoustic modes
can be occupied.
Einstein model
Einstein model considers that we only have optical phonons of the same frequency ω. As
a result, each optical branch will contribute Nh̄ωEinstein /(eh̄ωEinstein /kT − 1), where N is the
number of lattice points per unit volume. In three dimensions, we have 3 times the above
expression. Taking derivative over temperature, the Cv can be found to be:
3Nk
(h̄ωEinstein /kT )2 eh̄ωEinstein /kT
.
(eh̄ωEinstein /kT )2
(15)
The Einstein picture can be specifically used to describe optical phonons. Other than that,
its contribution is to use quantum statistics to first explain the qualitative feature of the
temperature dependence of the specific heat of insulators.
Density of states
The volume in k-space that one state occupies is:
1. 1D: 2π/L, where L is the dimension of the sample;
2. 2D: (2π/L)2 ;
3. 3D: (2π/L)3 = 8π 3 /V .
9
For three dimensions, the density of states, the number of states within a narrow energy
window dE, near energy, is therefore understood to come from:
dk
)(dω), where the first term comes from the number of states that are
D(ω)dω = ( 8πV 3 )( dω
in (volume) dk, the second term comes from the volume in k-space the energy window dω
occupies, and the last term is the small energy window at ω.
Therefore,
D(ω) =
V
V 1
1
.
dω =
3
3
8π dk
8π |∇k ω| |ω
(16)
The density of state is inversely proportional to group velocity. When the dispersion relation
is flat, there is a singularity in D.
An alternative way to define the density of states is to integrate over the number of states
on a constant energy surface in k-space.
PHONON III: Anharmonic effects
Introduction
The harmonic picture cannot explain the following observations.
1. There is interaction between phonon modes.
The evidence is that there is such thing called ‘thermal equilibrium.’ So, there must be
(efficient) exchange of energy and momentum.
2. All solids have non-zero thermal expansion coefficient.
But, the expectation value of the deviation < x > within harmonic approximation is zero.
3. Thermal conductivity of a solid is finite.
If the phonon modes are perfect eigenstates of the phonon system, then the thermal conductivity would be in theory infinity: phonons are everywhere in a solid and they never
die.
4. Raman scattering and Neutron scattering line width.
There is a finite lifetime of the scattered phonon (homogeneous broadening). But, where
can a phonon turn into?
5. The specific heat does not really go to 3Nk at high temperatures.
Thermal expansion coefficient
Taking the cohesive potential of the lattice, we can define the averaged position of an atom
in a particular quantum state n as:
< x >=< n|x|n >.
If a pure harmonic oscillator picture is used, then < x > would be at the lattice cite at all
10
times. Taking higher order terms to better describe the shape of the cohesive potential, we
can write:
V (x) = cx2 + gx3 + f x4 + ..... And a new set of eigenstates can be derived, and these
eigenstates make < x >=< n|x|n > deviated from the static lattice position.
Under nonzero temperatures, there is a distribution of the phonon modes. We therefore put
the distribution function in to calculate the averaged lattice constant:
< x >=
P
n
< n|x|n > e−h̄ωn /kT
P −h̄ω /kT
n
ne
(17)
The thermal expansion coefficient can be calculated basically by taking derivative of the
above equation with respect to temperature. The leading term of < x > would be 3gkT /4c2.
Therefore, the expansion coefficient is constant at high temperature, and it will approach
zero when temperature is low, since at low temperatures only the lowest energy phonons are
occupied, and in this regime the lattice is well-approximated by a harmonic oscillator. The
dependence at low temperature is T 3 , which can be derived from thermodynamics.
Lattice thermal conductivity
A claim is often made: a perfect lattice should have perfect thermal conductivity. The
reason is that any wave packet can be represented as a linear combination of eigenstates.
A distribution of phonon excitation in real space can be represented by a distribution of
phonon modes in k-space. Since there is basically no reason for these phonons to disappear,
the thermal conductivity should be infinitely large. Well, there are good physical reasons
why a perfect insulator can never be a perfect thermal conductor. The answer is that
phonon can scatter too: they can scatter off lattice imperfections, such as vacancy, surface,
etc. But, most importantly, phonon can scatter with other phonons. (Were not for these
efficient exchange of energy and momentum between phonons, how would you define thermal
equilibrium?)
Remember that the creation and annihilation operators can be defined from x and p? That
is, x and p can be defined by creation and annihilation operators, and, so is the anharmonic
potential, e.g., x3 = xxx ∝ (a + a∗ )(a + a∗ )(a + a∗ ). From here, we learn that in terms of
actual physical processes, the x3 term will introduce a ‘three particle’ process, where one
phonon can decay into two phonons, or two phonons can merge into one phonon. (Note here
that it is not physical to have consecutive aaa or a∗ a∗ a∗ processes, since energy would not
be conserved in the system then.)
The x4 term would imply a four (one-three, three-one, two-two) phonon process. And a mere
x2 term does not introduce interaction between modes; each mode is a perfect solution.
From conservation of momentum and energy, we have:
P
P
′
h̄ωs (k)nk,s = h̄ωs (k)nk,s , for conservation of energy; and
P
P
′
knk,s = knk,s + K, for conservation of momentum,
where K is a reciprocal lattice vector.
Now, we use a simple transport picture to calculate the termal conductivity. We use the
11
Debye model (linear dispersion, ω = vsound k. Note here only acoustic phonons are in the
picture, which is in fact not too far from the fact that the optical phonons almost do not
travel at all.
The thermal conductivity coefficient κ is defined with respect to a steady state flow of heat
along a long rod with a well-defined temperature gradient dT /dx:
, where j is the energy flux: the energy (in Joule, e.g., per unit area per unit time).
j = −κ dT
dx
Take τ as the averaged scattering time, i.e., in average, each phonon will be scattered once
in time τ . At any point in the rod, we maintain thermal equilibrium. The ‘energy density’
U at a particular point in space is the equilibrium energy density at the corresponding
temperature. Then, for a point x0 in space, only phonons within the volume of radius
l = vsound τ can contribute to the net energy flow. For volumes far away, the phonons are
not capable of transmitting energy to the point x0 . The net energy flow j is therefore:
j =< vsound,xdirection U(x0 − lcosθ) >θ =
1
=
2
Z
1
−1
Z
0
π
[vsound cosθ]U(x0 − lcosθ)
µdµvsoundU(x0 − lµ),
[2πsinθ]dθ
4π
(18)
(19)
where µ = cosθ.
In the above equations, the averaging over θ is done by:
< f (θ, φ) >θ =
Rπ
0
dθ
R 2π
dφ2πsinθf (θ, φ)
R
R
=
dθ dφ2πsinθ
0
R
dθ dφsinθf (θ, φ)
4π
R
(20)
The integration over φ generates a term 2πsinθ, and the integration over angles at the
R
R
R
denominator, dθ dφ becomes dθ2πsinθ = 4π.
Taking the leading linear term only, we get:
∂U 1
j = −vsound l
∂x 2
∂U ∂T
1
(− )
µ2 dµ = vsound l
3
∂T
∂x
−1
Z
1
(21)
By the definition of thermal conductivity, we therefore arrive at:
2
κ = 31 vsound lCv = 31 Cv vsound
τ.
High temperature limit: κ goes like 1/T x
Cv has no dependence on temperature. The number of phonons follows the Planck distribution, so n(k) = 1/[eh̄ω/kT − 1] ∼ kT /h̄ω. When the number of phonons increase, there
are more phonons to scatter with, so that scattering rate 1/τ should go at least like T. The
experimental value is that κ ∝ 1/T x , where x is in between 1 and 2, consistent with theory.
12
Low temperature limit: umklapp and normal processes
We first note that: 1. only scattering between the phonons near the zone center is possible,
2. the population of phonon goes like n ∼ e−h̄ω/kT .
At low temperature, under equilibrium,
P
kn(k) = 0.
When a solid transmits heat, it is NOT in thermal equilibrium anywhere, therefore, around
P
a local point,
kn(k) would not be zero. If the conservation of momentum is strictly
conserved, this net momentum is not going to change, and therefore there is perfect thermal
conductivity.
However, there is this umklapp process (as opposed to normal processes), where the if the
scattered phonon momentum is beyond the first B.Z., then since all the physical properties
this state represent are the same as a translated (by K) state in the first B.Z., the effective
momentum can be changed. (k1 + k2 = k3 + K, where K is a lattice vector in the reciprocal
lattice.) As a result, the net (crystal) momentum will NOT be conserved.
There is always some small (no matter how small it is) probability that the phonon scattering can have this umklapp process. The phonon(s) that can have umklapp process must
then have a wavevector close to the kDebye and we can use Debye temperature to show its
population:
n(kD ) ∼ e−ΘDebye /T .
So, at low temperature limit, the number of phonons capable of introducing the umklapp
process reduces, so we get:
1/τ ∼ e−ΘDebye/T .
The κ is therefore proportional to:
T 3 × eΘDebye /T , in the way that when the exponential term is so large, the T 3 law will
dominate.
Note the features of κ versus temperature (page 505 of AM). As the size of the sample is
smaller, the κ will also be smaller, due to boundary scattering.
Second sound
For sound to transmit in gas, we observe that the frequency of sound should be much less
than the ‘scattering rate’ of inter-molecular scattering.
To allow sound to transmit in a solid, we have similar constraint: 1/τumklapp < ωsound <
1/τnormal . That is, the normal process must be fast enough so that phonons can maintain
local equilibrium fast enough to show the (1) compressibility and (2) elasticity. However,
on the other hand, the umklapp process can ‘damp’ the transmit of sound. Therefore,
1/τumklapp < ωsound. The frequency less than 1/τumklapp is the SECOND SOUND frequency,
at which only a temperature wave can transmit, or, a wave of local energy density. The
physics is different from simply a temperature gradient in a solid.
Measurement of phonon dispersion relations
13
1. Neutron scattering: mass = 1838.65mo , K.E. = p2 /2m. Single-phonon scattering generate
peaks, while multiple-phonon scattering will generate a broad background. The linewidth
implies a finite lifetime: the phonon probed will eventually decay into other harmonic or
anharmonic states.
2. Inelastic light scattering: Stokes (emission) and anti-Stokes (absorption) processes for
optical phonon scattering (Raman); or acoustic phonon scattering (Brillouin scattering).
14