Quinn J.J., Yi K.-S. Solid State Physics: Principles and Modern Applications
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Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
68 2 Lattice Vibrations
Here, g and f are positive constants. The fu
4
term simply accounts for the fact 668
that the harmonic approximation rises too quickly for large u.Thegu
3
term 669
accounts for the asymmetry in the potential for u greater than or less than 670
zero. When u is negative, −gu
3
is positive making the short range repulsion 671
larger; when u is positive, −gu
3
is negative softening the interatomic repulsion 672
for large R. 673
Now let us evaluate the expectation value of u at a temperature k
B
T = β
−1
. 674
u =
"
∞
−∞
duue
−βV
"
∞
−∞
du e
−βV
. (2.146)
But, V = V
0
+ cu
2
− gu
3
− fu
4
,andwecanexpande
β
(
gu
3
+fu
4
)
, for small
675
values of u,toobtain 676
e
−βV
=e
−β
(
V
0
+cu
2
)
1+βgu
3
+ βfu
4
. (2.147)
The integrals in the numerator and denominator of (2.146) can be evaluated.
677
Because of the factor e
−βcu
2
, we do not have to worry about the behavior of 678
the integrand for very large values of |u| so there is little error in taking the 679
limit as u = ±∞. We can easily see that 680
∞
−∞
du e
−βV
=e
−βV
0
∞
−∞
du e
−βcu
2
1+βgu
3
+ βfu
4
. (2.148)
The βgu
3
term vanishes because it is an odd function of u;theβfu
4
gives a 681
small correction to the first term so it can be neglected. This results in 682
∞
−∞
du e
−βV
e
−βV
0
π
βc
1/2
. (2.149)
In writing down (2.149) we have made use of the result
"
∞
−∞
dz e
−z
2
=
√
π. 683
The integral in the numerator of (2.146) becomes 684
∞
−∞
duue
−βV
=e
−βV
0
∞
−∞
duue
−βcu
2
1+βgu
3
+ βfu
4
. (2.150)
Only the βgu
3
term in the square bracket contributes to the integral. The 685
result is 686
∞
−∞
duue
−βV
e
−βV
0
3
√
π
4
βg (βc)
−5/2
. (2.151)
In obtaining (2.151) we have made use of the result
"
∞
−∞
dzz
4
e
−z
4
=
3
√
π
4
. 687
Substituting it back in (2.146) gives 688
u =
1
β
3g
4c
2
=
3g
4c
2
k
B
T. (2.152)
689
The displacement from equilibrium is positive and increases with temperature. 690
This suggests why a crystal expands with increasing temperature. 691
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
2.8 Anharmonic Effects 69
2.8 Anharmonic Effects 692
To get some idea about how one would go about treating anharmonic effect, 693
let us go back to the simple one-dimensional model and include terms that 694
we have neglected (up to this time) in the expansion of the potential energy. 695
We can write H = H
Harmonic
+ H
,whereH
is given by 696
H
=
1
3!
lmn
d
lmn
u
l
u
m
u
n
+
1
4!
lmnp
f
lmnp
u
l
u
m
u
n
u
p
+ ··· . (2.153)
As a first approximation, let us keep only the cubic anharmonic term and make
697
use of 698
u
m
=
k
¯h
2MNω
k
1/2
a
k
+ a
†
−k
e
ikma
. (2.154)
Substituting (2.154) in (2.153) gives
699
H
3
=
1
3!
lmn
d
lmn
kk
k
¯h
2MN
3/2
(ω
k
ω
k
ω
k
)
−1/2
(2.155)
×
a
k
+ a
†
−k
a
k
+ a
†
−k
a
k
+ a
†
−k
e
ikna
e
ik
ma
e
ik
la
.
As before, d
lmn
does not depend on l, m, n individually, but on their relative 700
positions. We can, therefore, write d
lmn
= d(n − m, n − l). Now introduce 701
g = n − m and j = n − l and sum over all values of g, j,andn instead of l, 702
m,andn. This gives for the cubic anharmonic correction to the Hamiltonian 703
H
3
=
1
3!
ngj
d(g,j)
kk
k
¯h
2MN
3/2
(ω
k
ω
k
ω
k
)
−1/2
(2.156)
×
a
k
+ a
†
−k
a
k
+ a
†
−k
a
k
+ a
†
−k
e
ikna
e
ik
(n−g)a
e
ik
(n−j)a
.
The only factor depending on n is e
i(k+k
+k
)na
,and 704
n
e
i(k+k
+k
)na
= Nδ(k + k
+ k
,K) . (2.157)
Here, K is a reciprocal lattice vector; the value of K is uniquely determined
705
since k, k
, k
must all lie within the first Brillouin zone. Eliminate k
remem- 706
bering that if −(k + k
) lies outside the first Brillouin zone, one must add a 707
reciprocal lattice vector K to k
to satisfy (2.157). With this H
3
becomes 708
H
3
= N
kk
1
3!
gj
d(g,j)e
−ik
ga
e
|rmi(k+k
)ja
¯h
2MN
3/2
(2.158)
×(ω
k
ω
k
ω
k+k
)
−1/2
a
k
+ a
†
−k
a
k
+ a
†
−k
a
−(k+k
)
+ a
†
k+k
.
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
70 2 Lattice Vibrations
'
– (
)
¢
¢
+
¢
'
+
'
-
'
'
Fig. 2.17. Scattering of phonons: (a) annihilation of three phonons, (b) annihilation
of two phonons and creation of a third phonon, (c) annihilation of one phonon and
creation of two phonons, (d) creation of three phonons
Now define 709
G(k, k
)=
1
3!
gj
d(g,j)e
ikja
e
ik
(j−g)a
¯h
3
2
3
M
3
Nω
k
ω
k
ω
k+k
1/2
. (2.159)
Then, H
3
is simply 710
H
3
=
kk
G(k, k
)
a
k
+ a
†
−k
a
k
+ a
†
−k
a
−(k+k
)
+ a
†
k+k
. (2.160)
711
Feynman Diagrams 712
In keeping track of the results obtained by applying H
to a state of the har- 713
monic crystal, it is useful to use Feynman diagrams. A wavy line will represent 714
a phonon propagating to the right (time increases to the right). The interac- 715
tion (i.e., the result of applying H
3
) is represented by a point into (or out of) 716
which three wavy lines run. There are four fundamentally different kinds of 717
diagrams (see Fig. 2.17): 718
1. a
k
a
k
a
−(k+k
)
annihilates three phonons (Fig. 2.17a). 719
2. a
k
a
k
a
†
k+k
annihilates two phonons and creates a third phonon (Fig. 2.17b). 720
3. a
k
a
†
−k
a
†
k+k
annihilates a phonon but creates two phonons (Fig. 2.17c). 721
4. a
†
−k
a
†
−k
a
†
k+k
creates three phonons (Fig. 2.17d). 722
Due to the existence of anharmonic terms (cubic, quartic, etc. in the dis- 723
placements from equilibrium) the simple harmonic oscillators which describe 724
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
2.9 Thermal Conductivity of an Insulator 71
the normal modes in the harmonic approximation are coupled. This anhar- 725
monicity leads to a number of interesting results (e.g., thermal expansion, 726
phonon–phonon scattering, phonon lifetime, etc.) We will not have space to 727
take up these effects in this book. However, one should be aware that the har- 728
monic approximation is an approximation. It ignores all the interesting effects 729
resulting from anharmonicity. 730
2.9 Thermal Conductivity of an Insulator 731
When one part of a crystal is heated, a temperature gradient is set up. In 732
the presence of the temperature gradient heat will flow from the hotter to the 733
cooler region. The ratio of this heat current density to the magnitude of the 734
temperature gradient is called the thermal conductivity κ
T
. 735
In an insulating crystal (i.e., one whose electrical conductivity is very small 736
at low temperatures as a result of the absence of nearly free electrons) the 737
heat is transported by phonons. Let us define u(x) as the internal energy per 738
unit volume in a small region about the position x in the crystal. We assume 739
that u(x) depends on position because there is a temperature gradient
∂T
∂x
in 740
the x-direction. Because the temperature T depends on x, the local thermal 741
equilibrium phonon density ¯n
qλ
=
e
¯hω
qλ
/Θ
− 1
−1
will also depend on x. 742
This takes a little explanation. In our discussion of phonons up until now, a 743
phonon of wave vector k was not localized anywhere in the crystal. In fact, all 744
of the atoms in the crystal vibrated with an amplitude u
k
and different phases 745
e
ikna−iω
k
t
. In light of this, a phonon is everywhere in the crystal, and it seems 746
difficult to think about difference in phonon density at different positions. In 747
order to do so, we must construct wave packets with a spread in k values, Δk, 748
chosen such that (Δk)
−1
is much larger than the atomic spacing but much 749
smaller than the distance Δx over which the temperature changes appreciably. 750
Then, by a phonon of wavenumber k we will mean a wavepacket centered at 751
wavenumber k. The wavepacket can then be localized to a region Δx of the 752
order (Δk)
−1
. If the temperature at position x is different from that at some 753
other position, the phonon will transport energy from the warmer to the cooler 754
region. The thermal current density at position x can be written 755
j
T
(x)=
dΩ
4π
s cos θu(x − l cos θ). (2.161)
In this equation u(x) is the internal energy per unit volume at position x,
756
s is the sound velocity, l is the phonon mean free path (l = sτ,whereτ is 757
the average time between phonon collisions), and θ is the angle between the 758
direction of propagation of the phonon and the direction of the temperature 759
gradient (see Fig. 2.18). A phonon reaching position x at angle θ (as shown 760
in Fig. 2.18 had its last collision, on the average, at x
= x −l cos θ.Butthe 761
phonons carry internal energy characteristic of the position where they had 762
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
72 2 Lattice Vibrations
θ
θ
phonon propagating
x
x
in angle
Fig. 2.18. Phonon propagation in the presence of a temperature gradient in the
x-direction
their last collision, so such phonons carry internal energy u(x−l cos θ). We can 763
expand u(x − l cos θ)asu(x) −
∂u
∂x
l cos θ, and integrate over dΩ = 2π sin θdθ. 764
This gives the result 765
j
T
(x)=−
1
3
sl
∂u
∂x
. (2.162)
Of course the internal energy depends on x because of the temperature gra-
766
dient, so we can write
∂u
∂x
=
∂u
∂T
∂T
∂x
. The result for the thermal conductivity 767
κ
T
= −j
T
∂T
∂x
−1
is 768
κ
T
=
1
3
s
2
τC
v
. (2.163)
769
In (2.163), we have set l = sτ and
∂u
∂T
= C
v
, the specific heat of the solid. 770
2.10 Phonon Collision Rate 771
The collision rate τ
−1
of phonons depends on 772
1. Anharmonic effects which cause phonon–phonon scattering 773
2. Defects and impurities which can scatter phonons and 774
3. The surfaces of the crystal which can also scatter phonons 775
Only the phonon–phonon collisions are very sensitive to temperature, since 776
the phonon density available to scatter one phonon varies with temperature. 777
For a perfect infinite crystal, defects, impurities, and surfaces can be ignored. 778
Phonon-phonon scattering can degrade the thermal current, but at very 779
low temperature, where only low frequency (ω ω
D
or k k
D
) phonons 780
are excited, most phonon–phonon scattering conserves crystal momentum. 781
By this we mean that in the real scattering processes shown in Fig. 2.19, no 782
reciprocal lattice vector K is needed in the conservation of crystal momentum, 783
and Fig. 2.19a would contain a delta function δ(k
1
+ k
2
− k
3
), Fig. 2.19b a 784
δ(k
1
− k
2
− k
3
), and Fig. 2.19c a δ(k
1
+ k
2
− k
3
− k
4
). This occurs because 785
each k-value is very small compared to the smallest reciprocal lattice vector 786
K. These scattering processes are called Nprocesses(for normal scattering 787
processes), and they do not degrade the thermal current. 788
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
2.11 Phonon Gas 73
¢
¢
¢
Fig. 2.19. Phonon–phonon scattering (a) Scattering of two phonons into one
phonon, (b) Scattering of one phonon into two phonons, (c) Scattering of two pho-
nons into two phonons
T
κ
Fig. 2.20. Temperature dependence of the thermal conductivity of an insulator
At high temperatures phonons with k values close to a reciprocal lattice 789
vector K will be thermally excited. In this case, the sum of k
1
and k
2
in 790
Fig. 2.19a might be outside the first Brillouin zone so that k
3
= k
1
+ k
2
−K. 791
It turns out that these processes, U-processes (for Umklapp processes) do 792
degrade the thermal current. At high temperatures it is found that τ is 793
proportional to temperature to the −n power, where 1 ≤ n ≤ 2. The high tem- 794
perature specific heat is the constant Dulong–Petit value, so that according 795
to (2.163) κ
T
∝ T
−n
at high temperature. 796
At low temperature, only U-processes limit the thermal conductivity (or 797
contribute to the thermal resistivity). But few phonons with k ≈ k
D
are 798
present at low temperature. A rough estimate would give e
−¯hω
D
/Θ
for the 799
probability of U-scattering at low temperature. Therefore, τ
U
, the scattering 800
time for U-processes is proportional to e
Θ
D
/Θ
. Since the low temperature 801
specific heat varies as T
3
, (2.163) would predict κ
T
∝ T
3
e
T
D
/T
for the thermal 802
conductivity at low temperature. The result for the temperature dependence 803
of thermal conductivity of an insulator is sketched in Fig. 2.20. 804
2.11 Phonon Gas 805
Landau introduced the concept of thinking of elementary excitations as par- 806
ticles. He suggested that it was possible to have a gas of phonons in a 807
crystal whose properties were analogous to those of a classical gas. Both the 808
atoms or molecules of a classical gas and the phonons in a crystal undergo 809
collisions. For the former, the collisions are molecule–molecule collisions or 810
molecule–wall of container collisions. For the latter they are phonon–phonon, 811
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
74 2 Lattice Vibrations
phonon–imperfection or phonon–surface collisions. Energy is conserved in 812
these collisions. Momentum is conserved in molecule–molecule collisions in 813
a classical gas and in N-process phonon–phonon collisions in a phonon gas.Of 814
course, the number of particles is conserved in the molecule–molecule collisions 815
of a classical gas, but phonons can be created or annihilated in phonon–phonon 816
collisions, so their number is not a conserved quantity. 817
The sound waves of a classical gas are oscillations of the particle density. 818
They occur if ωτ 1, so that thermal equilibrium is established very quickly 819
compared to the period of the sound wave. They also require that momentum 820
be conserved in the collision process. 821
Landau
4
called normal sound waves in a gas first sound.Heproposedan 822
oscillation of the phonon density in a phonon gas that named second sound. 823
This oscillation of the phonon density (or energy density) occurred in a crystal 824
if ωτ
N
1 (as in first sound) but ωτ
U
1 so that crystal momentum is 825
conserved. Second sound has been observed in He
4
and in a few crystals. 826
4
L. Landau, J. Phys. U.S.S.R. 5, 71 (1941).
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
2.11 Phonon Gas 75
Problems 827
2.1. Consider a three-dimensional Einsteinmodelinwhicheachdegreeof828
freedom of each atom has a vibrational frequency ω
0
. 829
(a) Evaluate G(ω), the number of modes per unit volume whose frequency 830
is less than ω. 831
(b) Evaluate g(ω)=
dG(ω)
dω
. 832
(c) Make a rough sketch of both G(ω)andg(ω) as a function of ω. 833
2.2. For a one-dimensional lattice a phonon of wave number k has frequency 834
ω
k
= ω
0
sin
|k|a
2
for a nearest neighbor coupling model. Now approximate this 835
model by a Debye model with ω = s|k|. 836
(a) Determine the value of s, the sound speed, and k
D
,theDebye837
wave vector. 838
(b) Sketch ω as a function of k for each model over the entire Brillouin 839
zone. 840
(c) Evaluate g(ω) for each model and make a sketch of g(ω)vs.ω for each. 841
2.3. Consider a diatomic linear chain. Evaluate u
q
/v
q
for the acoustic and 842
optical modes at q =0andatq =
π
2a
. 843
2.4. Consider a linear chain with two atoms per unit cell (each of mass M ) 844
located at 0 and δ,whereδ<
a
2
, a being the primitive translation vector. Let 845
C
1
be the force constant between nearest neighbors and C
2
the force constant 846
between next nearest neighbors. Determine ω
±
(k =0)andω
±
(k =
π
a
).
d
847
2.5. Show that the normal mode density (for samll ω)inad-dimensional 848
harmonic crystal varies as ω
d−1
. Use this result to determine the temperature 849
dependence of the specific heat. 850
2.6. In a linear chain with nearest neighbor interactions ω
k
= ω
0
sin
|k|a
2
. Show 851
that g(ω)
2
πa
1
√
ω
2
0
−ω
2
. 852
2.7. For a certain three-dimensional simple cubic lattice the phonon spectrum 853
is independent of polarization λ and is given by 854
ω(k
x
,k
y
,k
z
)=ω
0
sin
2
k
x
a
2
+sin
2
k
y
a
2
+sin
2
k
z
a
2
1/2
.
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
76 2 Lattice Vibrations
(a) Sketch a graph of ω vs. k for 855
1. k
y
= k
z
=0and0≤ k
x
≤
π
a
(i.e., along Γ → X), 856
2. k
z
=0andk
x
= k
y
=
k
√
2
for 0 ≤ k ≤
√
2π
a
(i.e., along Γ → K), 857
3. k
x
= k
y
= k
z
=
k
√
3
for 0 ≤ k ≤
√
3π
a
(i.e., along Γ → L). 858
(b) Draw the ω vs. k curve for the Debye approximation to these dispersion 859
curves as dashes lines on the diagram used in part (a). 860
(c) What are the critical points of this phonon spectrum? How many are 861
there? 862
(d) Make a rough sketch of the Debye density of states g(ω). How will the 863
actual density of states differ from the Debye approximation? 864
(e) Using this example, discuss the shortcomings and the successes of the 865
Debye model in predicting the thermodynamic properties (like specific 866
heat) of solids. 867
2.8. For a two-dimensional crystal a simple Debye model takes ω = sk for the 868
longitudinal and the single transverse modes for all allowed k values up to the 869
Debye wave number k
D
. 870
(a) Determine k
D
as a function of
N
L
2
,whereN is the number of atoms 871
and L
2
is the area of the crystal. 872
(b) Determine g(ω), the density of normal modes per unit area. 873
(c) Find the expression for the internal energy at a temperature T as 874
an integral over the density of states times an appropriate function of 875
frequency and temperature. 876
(d) From the result of part (c) determine the specific heat c
v
. 877
(e) Evaluate c
v
for k
B
T ¯hω
D
=¯hsk
D
. 878
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
2.11 Phonon Gas 77
Summary 879
In this chapter, we discussed the vibrations of the atoms in solids. Quantum 880
mechanical treatment of lattice dynamics and dispersion curves of the normal 881
modes are described. 882
The Hamiltonian of a linear chain is written, in the harmonic approx- 883
imation, as H =
i
P
2
i
2M
+
1
2
i,j
c
ij
u
i
u
j
, where P
i
is the momentum and 884
u
i
= R
i
− R
0
i
is the deviation of the ith atom from its equilibrium position. 885
A general dispersion relation of the normal modes is Mω
2
q
=
N
l=1
c(l)e
iqla
. 886
The normal coordinates are given by 887
q
k
= N
−1/2
n
u
n
e
−ikna
; p
k
= N
−1/2
n
P
n
e
+ikna
.
The inverse of q
k
and p
k
are u
n
=N
−1/2
k
q
k
e
ikna
; P
n
= N
−1/2
k
p
k
e
−ikna
. 888
The quantum mechanical Hamiltonian is given by H =
k
H
k
,where 889
H
k
=
ˆp
k
ˆp
†
k
2M
+
1
2
Mω
2
k
ˆq
k
ˆq
†
k
.
The dynamical variables q
k
and p
k
are replaced by quantum mechanical oper- 890
ators ˆq
k
and ˆp
k
which satisfy the commutation relation [p
k
,q
k
]=−i¯hδ
k,k
. It 891
is convenient to rewrite ˆq
k
and ˆp
k
in terms of the operators a
k
and a
†
k
,which 892
are defined by 893
q
k
=
¯h
2Mω
k
1/2
a
k
+ a
†
−k
; p
k
= ı
¯hM ω
k
2
1/2
a
†
k
− a
−k
.
The a
k
’s and a
†
k
’s satisfy
a
k
,a
†
k
−
= δ
k,k
and [a
k
,a
k
]
−
=
a
†
k
,a
†
k
−
=0. 894
The displacement of the nth atom and its momentum can be written 895
u
n
=
k
¯h
2MNω
k
1/2
e
ikna
a
k
+ a
†
−k
,
P
n
=
k
ı
¯hω
k
M
2N
1/2
e
−ikna
a
†
k
− a
−k
.
The Hamiltonian of the linear chain of atoms can be written
896
H =
k
¯hω
k
a
†
k
a
k
+
1
2
,
and its eigenfunctions and eigenvalues are
897
|n
1
,n
2
,...,n
N
>=
a
†
k
1
n
1
√
n
1
!
···
a
†
k
N
n
N
√
n
N
!
|0 >
and E
n
1
,n
2
,...,n
N
=
i
¯hω
k
i
(n
i
+
1
2
). 898