17 phonons and conduction electrons in solids (hiroshi...
TRANSCRIPT
1
17 Phonons and conduction electrons in solids (Hiroshi Matsuoka)
In this chapter, we will discuss a minimal microscopic model for phonons in a solid and a minimal microscopic model for conduction electrons in a simple metal. In both of these models, the constituent quantum particles, phonons or electrons, are regarded as non-interacting identical particles. We will apply the grand canonical ensemble approach developed in Ch.16 to these models to calculate their state variables, especially their molar heat capacities at constant volume. More specifically, the models we will discuss below are:
1. The ideal “phonon” (boson) gas model for lattice waves in a solid
Our main goal in studying this model is to calculate its molar heat capacity at constant volume:
cv
=
12! 4
5"R T
#$
% &
'
( )
3
T < < #( )
3"R T > > #( )
*
+ ,
- ,
where R is the universal gas constant, ! is the number of atoms assigned to each crystalline lattice point (e.g., for NaCl, ! = 2 ), and ! is the characteristic temperature for the phonons or the Debye temperature that separates the low-temperature regime from the high-temperature regime for the phonons. As c
v in the ideal phonon gas model is a function of temperature T and
the molar volume v of the solid, the Debye temperature ! must be a function of the molar volume. We will also find that ! depends on the speed of sound w in the solid so that ! = ! v,w( ) . 2. The free “electron” (fermion) gas model for conduction electrons in a simple metal
Our main goal in studying this model is to calculate its molar heat capacity at constant volume:
cv
=
!2
2R
T
TF
"
# $ $
%
& ' ' T < <T
F( )
3 2( )R T >> TF( )
(
) *
+ *
where T
F is the characteristic temperature for the conduction electrons or the Fermi temperature
that separates the low-temperature regime from the high-temperature regime for the electrons. As c
v in the free electron gas model is a function of temperature T and the molar volume v of the
2
metal the Fermi temperature TF
must be a function of the molar volume. We will also find that TF
depends on the mass of electron m so that TF= T
Fv,m( ) .
The molar heat capacities for phonons and electrons at low temperatures
Note that the heat capacities for the both models approach zero as the temperature is decreased toward the absolute zero temperature, which is a consequence of the third law of thermodynamics as discussed in Sec.8.3.2. Note also that at low temperatures the heat capacities for the both models are expressed in universal scaling forms: for the phonons, we find
cv
!= Rgph
T
"#
$ %
&
' ( ,
where gph is the universal scaling function for the phonons,
gph x( ) =12!
4
5x3 ,
while for the electrons, we find
cv = RgeT
TF
!
"
# #
$
%
& & ,
where ge is the universal scaling function for the electrons,
gph x( ) =!2
2x .
At low temperatures, these heat capacities thus depend on the temperature quite differently, mainly because the phonons are bosons whereas the electrons are fermions. To show this difference between the phonons and the electrons is our main goal in this chapter.
The molar heat capacities for phonons and electrons at high temperatures
At high temperatures, both of the heat capacities approach different constant values, 3R = 6 2( )R for the phonons and 3 2( )R for the electrons. According to the theorem of equipartition of energy, “3” in 3 2( )R for the electrons corresponds to “three” terms in the kinetic energy of each electron,
3
1
2m! v 2
=1
2mvx
2
+1
2mvy
2
+1
2mvz
2 .
Note that at high temperatures, electrons behave like atoms in the monatomic ideal gas model.
At T >> ! , each atom in a solid behaves like a harmonic oscillator so that “6” in 3R = 6 2( )R is a sum of “3” and “3”, which correspond to “three” terms in the kinetic energy of each atom and three terms in its potential energy,
1
2m!
2 ! r 2
=1
2m!
2
x2
+1
2m!
2
y2
+1
2m!
2
z2 ,
where
! r is the displacement vector for the atom.
17.1 One-particle states of quantum particles The momentum vector and the wave vector of a quantum particle
In this section, we will show that each one-particle state of non-interacting bosons or fermions can be labeled by a wave vector
!
k and that the energy eigenvalue ! of the one-particle state is determined by the wave vector:
! = !
!
k ( ) .
For non-interacting quantum particles, whether bosons or fermions, the de Broglie
wavelength ! of each particle is determined by its momentum vector
! p through
! "h! p =
h
p,
where h is the Planck constant. We can then define the corresponding wave vector
!
k by having
!
k pointed in the direction of
! p and its magnitude directly related to ! by
k =
2!
".
The wave vector
!
k is then related to its momentum vector
! p by
! p = "
! k .
4
since
p =h
!= h
k
2"= !k ,
where
! = h 2!( ) .
The wave vector eigenvalues for one-particle states with the periodic boundary condition
As discussed above, the wave vector !
k of one-particle state is related to its momentum vector
! p by
! p = "
! k . In quantum mechanics, the classical momentum vector
! p is replaced by an
operator,
! p !
"
i
! " ,
so that the classical wave vector must be also replaced by the following operator:
! k =
! p
"!1
i
! " .
We can then find the eigenvalues of this wave vector operator for a particle in a cubic box of volume V = L
3 by solving the following eigenvalue equation:
1
i
!
! "! r ( ) =!
k "! r ( )
together with a boundary condition for the wave function
!! r ( ) . As long as the boundary
condition does not affect the wave function deep inside the cubic box, we can choose almost any condition. In condensed matter physics, the most widely used boundary condition for phonons and electrons is the periodic boundary conditions, where
!! r ( ) . satisfies
! L, y,z( ) = ! 0, y, z( )
! x,L, z( ) = ! x,0, z( )
! x,y,L( ) = ! x, y, L( )
"
# $ $
% $ $
In other words, the wave function takes the same values on each pair of the opposing faces of the cube. This boundary condition minimizes the effect of the boundary or the surface of the cube on the wave function since it is equivalent to divide the entire three-dimensional space into an
5
array of cubes each with volume V = L3 and to assume that the wave function behaves in the
same way in every cube. The eigenstates of the wave vector are then “plane waves,”
! !
n
! r ( ) = exp i
!
k ! n "! r ( ) V ,
and the corresponding eigenvalues are
!
k ! n =2!
Lnx,ny ,nz( ) =
2!
L
! n ,
where each eigenstate and eigenvalue are labeled by three integers
! n or three quantum numbers
whose range includes both positive and negative integers as well as zero.
! n = nx,ny, nz( ) . (n
!= 0, ±1, ± 2, ± 3,.. .)
Note that
!
k ! n = 0 is also an eigenvalue of the wave vector and the corresponding eigenstate is
constant throughout the volume V: ! !
n = 0,0,0( )
! r ( ) =1 V .
In the !
k -space, all the possible wave vectors
!
k ! n form a discrete lattice, where each lattice
point is located at a corner of a “small” cube of volume
!k( )3
=2"L
#
$ %
&
' (
3
=2"( )3
V.
The intrinsic quantum number: polarization index and spin
Besides the three quantum numbers ! n , a quantum particle may have an intrinsic quantum
number such as the polarization index for phonons, ! = 1, 2, and 3 ,
which labels “two” transverse polarization directions and ”one” longitudinal polarization direction for lattice waves, and the spin quantum number for electrons, ! = ±1 / 2 .
6
The energy eigenvalue for a one-particle state Generally, the energy eigenvalue
! !
n of a one-particle state for non-interacting quantum
particles depends on the wave vector
!
k ! n :
! !
n = !
!
k ! n ( ) .
1. For quantum particles such as phonons that we observe as waves on the macroscopic level,
the energy eigenvalue ! !
n is given by
! !
n = "" !
n ,
where
! !
n is the angular frequency of the wave and is related to
!
k ! n by what is called the
“dispersion relation”:
! !
n = !!
n
!
k ! n ( ) .
Note that as
k!
n approaches zero at the long wave length limit, the angular frequency
! !
n and
the energy eigenvalue ! !
n also goes to zero.
2. For quantum particles such as electrons that we observe as particles on the macroscopic
level, the energy eigenvalue ! !
n is given by
! ! n =
! p ! n
2
2m="2k! n
2
2m,
where m is the mass of each particle.
For both phonons in solids and electrons in simple metals, the lowest energy eigenvalue for their one-particle states is set at zero. In other words, we set their internal energy scales so that their internal energies take the value of zero when a single phonon or electron is found in the lowest energy eigenstate.
7
17.2 The ideal phonon gas model In this section, we discuss the ideal phonon gas model, where phonons are assumed to be
non-interacting quantum particles.
The number of phonons depends on T and V: the chemical potential of a phonon gas is zero The chemical potential of a phonon gas is identically zero. Physically, it is because phonons
are continuously created and destroyed in a solid so that the number Nph of phonons is not an
independent variable and is rather a function of T and V of the solid. The Helmholtz free energy of the phonon gas is then a function of T and V: F = F T ,V( ) . On the other hand, using the fundamental equation of thermodynamics, dU = TdS ! PdV + µdNph , we find
dF = dU ! d TS( ) = TdS ! PdV + µdNph( ) ! TdS + SdT( )
= !SdT ! PdV + µdNph
so that
µ =!F!Nph
"
#
$ $
%
&
' ' T ,V
= 0 .
The internal energy and the number of phonons depend on T and V
As the Helmholtz free energy depends only on T and V, so do all other state variables of the phonon gas. Being extensive, the internal energy and the number of phonons should be proportional to the volume:
U = V ˜ u T( ) and Np h = V ˜ n p h T( ) ,
where ˜ u and ˜ n p h are the internal energy and the number of phonons per unit volume and they
depend only on the temperature. As the volume V is related to the mole number n of the solid by V = nv , where v is the molar volume of the solid, the molar internal energy u of the solid due to the phonons is given by u = v ˜ u T( ) so that the molar heat capacity at constant volume of the solid is then given by
8
cv
=!u
!T
"
# $
%
& '
v
= vd˜ u
dT
"
# $
%
& ' .
The internal energy and the number of phonons in the ideal phonon gas model
Because µ = 0 , the thermodynamic potential of a phonon gas is identical to its Helmholtz free energy:
! =U " TS " µN =U " TS = F .
This implies that the grand canonical ensemble theory of the ideal phonon gas is the same as its canonical ensemble theory. In fact, the Gibbs sum of the ideal phonon gas is nothing but its partition function:
! p h T ,V( ) =" ! # n ,$ ! # n { }% exp &'E "! n '{ }( )[ ] = Z p h T, V( ) =
"! n
%! n
()
*
+ +
,
-
.
. 3exp &' / ! n '"! n '! n '
%)
* + +
,
- . .
= 3 exp &'/ ! n "! n ( )" ! n
%0
1
2 2
3
4
5 5 !
n
( = 3 e&' / ! n ( )
" ! n
" ! n = 0
6
%0
1
2 2
3
4
5 5 !
n
( =3
1 & e&'/ ! n !
n
(
where the factor of 3 is the number of polarizations for each
! n . We can then derive the
Helmholtz free energy of the ideal phonon gas as
F T,V( ) = !kBT ln Zp h = 3kBT ln 1! e! " ! n / k BT( )[ ]
! n
# .
We can find the internal energy either by setting the chemical potential to zero in the
expression derived in Ch.16:
U = 3! !
n
e" ! !
n #µ( )
#1! n
$ = 3! !
n
e"! !
n # 1! n
$
or by using the Helmholtz free energy just derived:
U = !T2 ""T
F
T
#
$ %
&
' (
) * +
, - .
V
= 3/!
n
e0 / !
n !1! n
1 .
Comparing this expression for the internal energy with
9
U = 3 !! n '" !
n '! n
# = 3 ! ! n '"!
n '! n
# ,
we find
!! n '
=1
e"# !
n $1
so that we can find the average number of the phonons by
Np h = 3 !! n '! n
" = 31
e#$ ! n %1!
n
" .
The energy eigenvalue for the phonon one-particles state
For phonons in a solid, at long wavelengths ! !
n >> a or small wavenumbers
k!
n << 2! a ,
where a is the distance between a pair of nearest-neighbor atoms inside the solid, the dispersion relation is well approximated by
! !
n " wk !
n (for
k!
n << 2! a ),
where w is the speed of sound, so that
! !
n " "wk !
n . (for
k!
n << 2! a )
Derivation
! ! n = 2"f ! n # 2"w
$ ! n
= wk ! n ,
where
f ! n !
! n " w ,
which is valid only at long wavelengths.
As the wavenumber
k!
n increases away from the long wavelength limit, the dispersion
relation starts to deviate from the linear relation ! !
n " wk !
n obtained above. As we will see, to
find the molar heat capacity at both low and high temperatures, we do not need to know exactly how the dispersion relation behaves outside the long wavelength limit.
10
The first Brillouin zone: the allowed region for the phonon wave vectors The allowed region for the wave vectors for the phonons in the
!
k -space is actually bounded since the wavelength
! !
n of a phonon cannot be less than the lattice constant a or the distance
between a pair of nearest-neighbor atoms:
! !
n " a .,
which implies
k! n =2!
" ! n
#2!
a.
This bounded region for
k!
n is called “the first Brillouin zone,” whose shape in the
!
k -space depends on the crystalline structure of the atoms in the real space. The maximum phonon energy and the characteristic temperature for phonons
As the phonon energy ! !
n is a smooth function of the wave vector
k!
n , there must be a
maximum energy !max
for the phonon one-particle states. The maximum phonon energy thus sets the phonon energy scale and we can roughly estimate its order-of-magnitude as follows. The maximum wavenumber k
max in the first Brillouin zone must be on the order of 1 a and the
maximum phonon energy !max
must be on the order of !wk
max so that
!max
~ !wkmax
~!w
a~ O
10"34
J # s( ) 103 m s( )
10"10
m( )
$
%
& &
'
(
) )
~O 10"21
J( )
~O10
"2 1 J
10"19
J eV
*
+ , ,
-
. / / ~ O 10
"2 eV( )
where a ~ O 10
!10
m( ) , and w ~ O 103 m s( ) .
A characteristic temperature Tph for the phonons, which separates their low-temperature
regime from their high-temperature regime, can be then defined by
Tph !"
max
kB~O
10#21
J
10#23
J K
$
% & &
'
( ) ) ~ O 100 K( ) .
In the Debye approximation of the ideal phonon gas model, we will define a slightly different characteristic temperature for the phonons called the Debye temperature ! by
11
! "!w
kB
6#2Na
V
$
% &
'
( )
1/3
,
where the volume per atom V N
a is on the order of a3 so that
! ~!w
kB
Na
V
"
# $
%
& '
1/ 3
~!w
kBa~!wk
max
kB~ Tph .
The total number of the phonon one-particle states in the first Brillouin zone
At high temperatures, Na atoms in the solid must be vibrating fast enough that the atoms
behave like Na three-dimensional harmonic oscillators, which must be described by 3N
a
quantum numbers. This implies that the total number of the allowed phonon one-particle states must be also 3N
a so that
3Na= 3 1
! k !
n !BZ
" ,
where the summation is over the wave vectors in the first Brillouin zone. This equation becomes useful below when we discuss the high-temperature behavior of the internal energy of the phonon gas. The internal energy and the average number of phonons
As the energy eigenvalue ! !
n of the phonon one-particle state depends on its wave vector
!
k ! n :
! !
n = !
!
k ! n ( ) ,
we can express the internal energy and the average number of the phonons as
U = 3!! k ! n ( )
e" !! k ! n ( )
#1! k ! n $BZ
% = 3 !!
k ! n ( ) fPlanck
!!
k ! n ( ),T( )! k ! n $BZ
%
and
Np h = 31
e! "! k ! n ( )
#1! k ! n $BZ
% = 3 fPlanck
"!
k ! n ( ),T( )! k ! n $BZ
% ,
where the summation is over the wave vectors in the first Brillouin zone and we have defined the Planck distribution function or the Bose-Einstein distribution function with µ = 0 by
12
fPlanck
!,T( ) "1
e#!$1
.
The Planck distribution function
The Planck distribution function is shown below as a function of !" .
For !" <<1 or ! << k
BT , we can approximate e!" in the Planck distribution function by 1 + !"
or e!" #1 + !" so that f
Planck!
1
1 +"#( ) $1=1
"#=kBT
# ! << k
BT( ) ,
whereas for !" >>1 or ! >> k
BT , we find e!" >> 1 so that
f
Planck! e
"#$ ! >> kBT( ) .
Note that for each phonon one-particle state whose energy ! is much lower than k
BT , its
average occupation number or the value of the Planck distribution function becomes large as fPlanck
!,T( ) " kBT ! >> 1 , but the contribution to the internal energy by all these phonons in this one-particle state altogether amounts to just k
BT since !f
Planck! ,T( ) " kBT , which is independent
of the one-particle state energy ! .
13
The “universal” internal energy of the phonon gas at high temperatures and the Dulong-Petit law
When the temperature T of the phonon gas is much higher than !max
kB
so that kBT >> !
max
or !"max
<< 1, all the phonon one-particle states inside the first Brillouin zone have their energies much lower than k
BT so that as we found in the previous sub-section, we can approximate the
Planck distribution function for each phonon one-particle state with energy ! by fPlanck
!,T( ) " kBT ! and consequently every one-particle state contribute an energy of kBT to
the internal energy of the phonon:
U = 3 !!
k ! n ( ) fPlanck
!!
k ! n ( ),T( )! k ! n "BZ
# $ 3kBT 1! k ! n "BZ
# = 3NakBT = 3n%RT
where n is the mole number of the solid and Na = !nNAvogadro,, where ! is the number of atoms
assigned to each crystalline lattice point (e.g., for NaCl, ! = 2 ). We have also used
3Na= 3 1
! k !
n !BZ
" ,
The molar internal energy at high temperatures then becomes u =
U
na
! 3"RT
The molar heat capacity at constant volume at high temperatures is then given by
cv
=!u!T
"
# $
%
& '
v
( 3)R ,
which is the Dulong-Petit law we have discussed in Sec.7.4.2.
Note that this result does not depend on the specific detail of the dispersion relation, which varies from one solid to another, and the Dulong-Petit law is therefore a universal property of the ideal phonon gas. The two main ingredients for this universal property are the behavior of the Planck distribution function at phonon energies much lower than k
BT and the presence of the
cutoff energy !max
. At high temperatures, where kBT >> !
max, the Planck distribution function
for all the phonon one-particle states in the first Brillouin zone is approximately k
BT !
!
k ! n ( ) >> 1
so that all the phonon one-particle states below the cutoff energy are excited and contribute to the internal energy. Moreover, all the phonons in a phonon one-particle state with energy ! together
14
contribute !f Planck !( ) " kBT of energy to the internal energy. As this energy due to the phonons in each one-particle state is independent of the phonon energy ! , the internal energy is simply the product of the total number of the phonon one-particle states, 3N
a, and k
BT .
The number of phonons at high temperatures
We can also calculate the number of phonons by
Np h = 3 fPlanck
!!
k ! n ( ),T( )! k ! n "BZ
# $ 3kBT1
!!
k ! n ( )! k ! n "BZ
# % T ,
which increases linearly with the temperature but is not universal because the sum of
1 !
!
k ! n ( )
does depend on the detail of the dispersion relation. Approximating the summation over phonon one-particle states by an integral
To calculate the internal energy and the average number of the phonons at low temperatures, it is convenient to approximate the summation over the wave vector
!
k ! n in the equations for U
and Nph by an integral. For two neighboring wave vectors that satisfy
k! ! n " k!
n =2#
L,
we find
!" k! # n ( ) $ !" k! n ( ) ~ !"w k! # n
$ k! n
=2%L
"w
kBT
&
' ( (
)
* + + ~ O
10$3 4
J ,s( ) 103 m s( )
1 m( ) 10$23
J K( )T
-
.
/ /
0
1
2 2
~ O10
$8 K
T
&
' (
)
* +
where we have used the expression for the phonon energy for the long wavelengths. For temperatures ranging from 1 mK to 1000 K, we therefore find
!" k! # n ( ) $ !" k!
n ( ) < < 1. We can then approximate the summation over the wave vector
!
k ! n in the above equations for U and Nph
by an integral as follows.
3! k !
n !BZ
" =3
#k( )3 #k( )
3
! k !
n !BZ
" =3V
2$( )3d
!
k ! k !BZ
% ,
where we have used
15
!k( )3
=2"L
#
$ %
&
' (
3
=2"( )3
V,
so that
U =3V
2!( )3d!
k "! k ( )
e#"! k ( ) $1!
k %BZ& =
3V
2!( )3d!
k "!
k ( ) fPlanck
"!
k ( ),T( )! k %BZ&
and
Np h =3V
2!( )3d!
k 1
e"#! k ( ) $1!
k %BZ& =
3V
2!( )3d!
k fPlanck
#!
k ( ),T( )! k %BZ& .
The “universal” temperature dependence of the internal energy of the phonon gas at low temperatures and the T 3 law for the molar heat capacity at constant volume
When the temperature T of the phonon gas is much lower than !max
kB
so that kBT << !
max
or !"max
>> 1, for phonon energy ! that satisfies that satisfy ! >> kBT or !" >>1 , the
exponential factor e!" becomes so large that the Planck distribution function becomes negligible. Physically, only the one-particle states with energy ! " !wk less than k
BT contribute to the
internal energy. Roughly, the number of these one-particle states is
N ph k !kBT
!w
"
# $
%
& ' = 3( )
4(3
kBT
!w
"
# $
%
& '
3
)k( )3 =
V
2( 2
kBT
!w
"
# $
%
& '
3
,
and, on the average, each of these one-particle states contribute energy on the order of k
BT to the
internal energy so that
U ~O N phkBT( ) = O1
2! 2 !w( )3 V kBT( )
4"
# $ $
%
& ' ' ( T
4 .
More precisely, in the integral for U, we can integrate over the entire k-space instead of the
inside the first Brillouin zone:
U =3V
2!( )3 d
!
k
"! k ( )
e#"! k ( ) $1!
k %BZ
& '3V
2!( )3 d
!
k
"! k ( )
e#"! k ( ) $ 1Entire
! k $space
& ,
without affecting the value of the integral. We can also use the long-wavelength expression for the phonon energy to obtain
16
U !"2
10 !w( )3 V k
BT( )
4
so that the molar internal energy at low temperatures becomes
u =U
n!
"2
10 !w( )3 v kBT( )
4 ,
where v is the molar volume of the solid.
Derivation
U ! 3V
2"( )3 d
! k
#! k ( )
e$#! k ( ) %1Entire
! k % space
& ! 3V
2"( )3 4"k 2dk
"wk
e$"wk %1
0
'
&
=3V
2" 2"w( )
3
1
$ 4
(
) * *
+
, - - d $"wk( )
$"wk( )3
e$"wk %1
0
'
& =3V
2" 2"w( )
3 kBT( )4
dqq3
eq %1
0
'
&
=3V
2" 2"w( )
3 kBT( )4 " 4
15
(
) *
+
, - =
" 2
10 "w( )3 V kBT( )
4
where
dqq3
eq !1
0
"
# =$ 4
15.
The universal molar heat capacity at constant volume at low temperatures
The molar heat capacity at constant volume at low temperatures is then given by
cv
=!u!T
"
# $
%
& '
v
(2) 2
5kBvkBT
!w
"
# $
%
& '
3
* T3.
Note that the only quantities in this equation that are specific to a particular solid are the molar volume v and the speed of sound w. We also know that the molar heat capacity should be expressed in units of the universal gas constant R so that we should expect c
v to have the
following scaling form:
17
cv = RgT
Tph v,w( )
!
"
# #
$
%
& & ,
where Tph is the characteristic temperature for the phonons defined earlier by
Tph !"max
kB~!w
kB
Na
V
#
$ %
&
' (
1/ 3
and g is a scaling function that assigns a dimensionless number for a dimensionless number. Substituting
v =V
n=V!NAvogadro
n!NAvogadro
= !NAvogadro
V
Na
,
where Na = !nNAvogadro, into the above expression for c
v, we indeed find c
v in the scaling form:
cv !2" 2
5kBv
kBT
!w
#
$ %
&
' (
3
=2" 2
5) NAvogadrokB( )
V
Na
kBT
!w
#
$ %
&
' (
3
=2"2
5)R
T
!w kB( ) Na V( )1/ 3
*
+
, ,
-
.
/ /
3
,
which implies that c
v! is a universal function of T Tph .
cv
!" R
T
Tph
#
$
% %
&
'
( (
3
.
In the next sub-section, we will define a characteristic temperature for the phonons called the
Debye temperature by
! "!w
kB6#2
Na
V
$
% &
'
( )
1/3
~!w
kB
Na
V
$
% &
'
( )
1/ 3
~ Tph
and express c
v! in terms of T ! as
cv
!"12# 4
5RT
$%
& '
(
) *
3
18
The number of phonons and the internal energy at low temperatures The number of phonons can be also found to be
Np h =3V
2!( )3 d! k
1
e"#! k ( ) $1!
k ! n %BZ
& ' 3V
2!( )3 d! k
1
e"#! k ( ) $1Entire
! k $ space
& ' 3V
2!( )3 4!k 2dk1
e""wk $10
(
&
=3V
2! 2"w( )
3
1
" 3
)
* + +
,
- . . d ""wk( )
""wk( )2
e" "wk $1
0
(
& =3V
2! 2"w( )
3 kBT( )3
dqq2
eq $1
0
(
&
=3V
2! 2"w( )3 kBT( )
3
2/ 3( ){ } =3 1.202( )
! 2"w( )3 V kBT( )
3
=3.606
!2"w( )3 V kBT( )
3
so that
U !" 2
10 !w( )3 V kBT( )
4
=" 4
30# 3( )NphkBT =
" 4
36.06NphkBT ,
which implies that the internal energy of the phonons is roughly the number of phonons times the “typical” value of phonon energy, k
BT .
The Debye approximation
In the Debye model for the ideal phonon gas, we extend the long-wavelength expression for the phonon energy to the maximum cutoff energy !
max, which we now call the Debye energy,
!D" !
max, so that for ! " !
D= !
max,
! " !wk . The first Brillouin zone is then replaced by a sphere of radius k
D defined by
kD!"D
!w
so that
3Na
= 3 1! k !
n !BZ" =
3V
2#( )3d
!
k ! k $k
D
% =3V
2#( )34#3
kD
3&
' (
)
* + =
V
2# 2k
D
3 ,
from which we find
kD
= 6! 2 Na
V
"
# $
%
& '
1/ 3
19
and
!D
= !wkD
= !w 6" 2Na
V
#
$ %
&
' (
1/ 3
.
We then define the Debye temperature by
! "#DkB
=!w
kB6$ 2 Na
V
%
& '
(
) *
1/ 3
=!w
kB6$ 2
+NAvogadro
v
%
& '
(
) *
1/ 3
,
where we have used Na = n!NAvogadro. Note that ! depends on the molar volume v of the solid
and the speed of sound w . To calculate the internal energy in terms of ! , it will be convenient to use the following
expression for the volume in terms of ! :
V = 6! 2Na
!w
kB"
#
$
% %
&
'
( (
3
so that
3V
2!2!w( )
3 =9N
a
kB"( )
3 .
The internal energy of the phonon gas in the Debye model is then given by
U ! 3V
2"( )3
d! k
#! k ( )
e$#! k ( ) %1!
k & k D
' =3V
2"( )3
4"k 2dk"wk
e$"wk % 1
0
k D
' =3V
2" 2"w( )
3
1
$ 4
(
) * *
+
, - - dq
q3
eq %1
0
. T
'
=9Na
kB.( )3
kBT( )4
dqq
3
eq %10
. T
' = 9n/R. T
.(
) *
+
, -
4
dqq
3
eq %10
. T
'
so that
u =U
n= 9!R"
T
"#
$ %
&
' (
4
dqq3
eq )1
0
" T
*
and
cv =!u!T
"
# $
%
& '
v
= 9(R 4T
)"
# $
%
& '
3
dqq3
eq *10
) /T
+,
-
.
. *
)T
"
# $
%
& '
1
e) /T *1
/
0 1 .
which implies that c
v is a universal function of T / ! .
20
The number of the phonons is also given by
Np h !3V
2"( )3d! k
1
e#$! k ( ) %1
! k & k D
' =3V
2"( )34"k 2dk
1
e#"wk % 10
k D
' =3V
2" 2"w( )3
1
# 3
(
) * *
+
, - - dq
q2
eq %10
. T
'
=9Na
kB.( )3
kBT( )3
dqq2
eq %10
. T
' = 9Na
T
.(
) *
+
, -
4
dqq2
eq % 10
. T
'
(a) At low T T << !( ) : by replacing the integration interval 0, ! T[ ] by 0,![ ] in the integral for
U and Nph , we find
u =U
n! 9"R#
T
#$
% &
'
( )
4
dqq3
eq * 1
0
+
, =3- 4
5"R#
T
#$
% &
'
( )
4
cv
=!u!T
"
# $
%
& '
v
(12) 4
5*R
T
+"
# $
%
& '
3
Nph ! 9Na
T
"#
$ %
&
' (
3
dqq2
eq )1
0
*
+ =18, 3( )Na
T
"#
$ %
&
' (
3
U !3" 4
5NakBT
T
#$
% &
'
( )
3
=" 4
30* 3( )NphkBT
(b) At high T T >> !( ) : by using eq !1 + q in the integral for U and Nph , we find
u =U
n! 9"R#
T
#$
% &
'
( )
4
dqq3
1+ q( ) *10
# T
+ = 9"R#T
#$
% &
'
( )
41
3
#T
$
% &
'
( )
3
= 3"RT
cv
=!u!T
"
# $
%
& '
v
( 3)R (the Dulong-Petit law)
Nph ! 9Na
T
"#
$ %
&
' (
3
dqq2
1+ q( ) )10
" T
* = 9Na
T
"#
$ %
&
' (
31
2
"T
#
$ %
&
' (
2
=9
2Na
T
"#
$ %
&
' (
U ! 3NakBT =
2
3NphkB"