A Modern Course in the Quantum Theory of Solids Capitulo 1

7/25/2019 A Modern Course in the Quantum Theory of Solids Capitulo 1

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Chapter 1

Lattice Dynamics

When the structure and cohesion of solids are studied, we assume that the

atoms or ions in solids stay at their respective equilibrium positions. This

is sufficient for the purpose of studying their structural and binding prop-

erties. However, when we pursue to understand many other properties of

solids, such as their thermodynamic properties, the picture of static atoms

or ions in solids becomes inadequate and their dynamics must be taken into

consideration. As a matter of fact, atoms or ions in solids never stay persis-

tently at their equilibrium positions at finite temperatures. Instead, they

move back and forth (that is, they vibrate or oscillate) constantly about

their equilibrium positions. This kind of motion is referred to as lattice vi-

brationsand the entire subject related to lattice vibrations is called lattice

dynamics or crystal dynamics.

Lattice dynamics can be said to be the oldest branch of solid state

physics. To be convincing, we now trace some of the early important de-

velopments in lattice dynamics. In 1907, Einstein1 published his work onthe lattice specific heat, entitled Plancks theory of radiation and the the-

ory of specific heat (the birth of the Einstein model on the lattice specific

heat). In 1912, Born and von Karman2 published their work on lattice

vibrations, entitled On vibrations in space lattices(the birth of the formal

theory of lattice dynamics), and Debye3 published his work on the lattice

specific heat, entitledOn the theory of specific heat(the birth of the Debye

model on the lattice specific heat). There are many other early landmark

developments.

1A. Einstein, Annalen der Physik 22, 180 (1907).2M. Born and Th. von Karman, Physikalische Zeitschrift 13, 297 (1912); ibid. 14, 15

(1913).3P. Debye, Annalen der Physik (Leipzig) 39, 789 (1912).

1


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2 A Modern Course in Quantum Theory of Solids

In 1950s, Brockhouse4

pioneered the measurement of the spectrum oflattice vibrations (the dispersion relations of normal modes of lattice vi-

brations) of a solid through inelastic neutron scattering experiments. The

development of this experimental technique provided a great impetus to

the study of lattice dynamics in various types of solids.

Lattice vibrations are very important because they play vital roles in

many physical properties of solids. Lattice vibrations can scatter electrons

in a metal and thus yield resistance to the motion of electrons, which leads

to the increase in the resistivity of the metal. Lattice vibrations can take

heat from or give heat to the environment and thus affect the heat capacity

of a solid. A certain kind of lattice vibrations interact with photons and

thus have an impact on the optical properties of a solid. The interaction

of lattice vibrations with conduction electrons in a metal can even change

the ground state of the electrons in a fundamental way and render them to

be superconducting. The consequences of lattice vibrations on the physical

properties of solids are so many that one can hardly give an exhausted list

in a limited space.

Because of their paramount importance, a thorough study of latticevibrations is undoubtedly necessary. Surprised or not, lattice vibrations

call for both classical and quantum theories for their complete descriptions.

The necessity of a quantum theory of lattice vibrations is clearly testified

by the inability of classical theory of lattice vibrations to produce the cor-

rect temperature dependence of the specific heat of a solid as observed in

experiments. Lattice vibrations must be studied in three stages before their

effects on physical properties can be fully unveiled. In the first stage, vari-

ous vibrational modes are obtained through solving the classical equationsof motion of atoms or ions. In the second stage, vibrational modes are

quantized according to the canonical quantization rules. This is the first

time that lattice vibrations are quantized. The second stage acts only as a

transition. For a better understanding of their properties and for the conve-

nience of their applications, lattice vibrations are quantized for the second

time in the third stage. With the second quantization, the consequences of

lattice vibrations on physical properties of solids can be fully investigated.

Regardless of the type of bonding, any solid can be taken as composed

of electrons and nuclei. These two kinds of particles are intimately coupled

4B. N. Brockhouse and A. T. Stewart, Physical Review 100, 756 (1955).


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Lattice Dynamics 3

together. The Hamiltonian of a solid is then given byH= He+ Hn+ Hen,

He= i

2

2m2i +

1

2

1

40

i=j

e2

|ri rj | ,

Hn = I

2

2MI2I+

1

2

1

40

I=J

ZIZJe2

|RIRJ| ,

Hen=

1

4

0iI

ZIe2

|ri RI|,

(1.1)

wherem is the mass of an electron, MI and ZIe are the mass and charge

of nucleusI, ri and RIdenote the positions of the ith electron and theIth

nucleus, respectively, He is the Hamiltonian of the subsystem of electrons,

Hnis the Hamiltonian of the subsystem of nuclei, and Henis the interaction

Hamiltonian between electrons and nuclei.

The Hamiltonian in Eq. (1.1) is referred to as the fundamental Hamil-

tonian of a solid in the sense that all the properties of the solid can be

computed if the eigenvalues and eigenstates ofHcould be found exactly.Unfortunately, it is not in sight at all that any one could accomplish that.

Therefore, we have no choice but make some approximations to be able

to proceed to understand any physical properties of a solid. Because

the electrons and ions are coupled together, the separation of the elec-

tronic and nuclear motions would be of great help. This is provided by

the BornOppenheimer approximationthat is also known as the adiabatic

approximation.

This chapter is organized as follows. The BornOppenheimer approxi-mation is first introduced in Sec. 1.1 so that we can concentrate only on the

motion of nuclei (or atoms or ions) thereafter. We then attempt to develop

the classical theory of lattice vibrations as gently as possible, with the full

classical theory established in the end.

In Sec. 1.2, we introduce the harmonic approximation and derive the

harmonic lattice potential energy for a three-dimensional crystal with a

multi-atom basis. We then proceed to find the normal modes of lattice

vibrations of a solid under the harmonic approximation. Attention should

be paid to the way we solve the classical equations of motion of atoms: We

expand the displacement of an atom in terms of its Fourier components (i.e.,

make a Fourier transformation of the displacement of atoms with respect

to positions of primitive cells and time) so that the differential equations

are converted into algebraic equations.


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In finding the normal modes of lattice vibrations in a crystal, we followthese steps:

(1) Establish the classical equations of motion of atoms using Newtons

second law with the force acting on an atom derived from the harmonic

lattice potential energy.

(2) Fourier transform the displacements of atoms and convert the differen-

tial equations into algebraic equations.

(3) Find the allowed values of wave vector.

(4) Solve the resultant algebraic equations for the frequencies of normalmodes.

(5) Introduce the normal coordinates and polarization vectors for normal

modes.

(6) Solve for the polarization vectors.

(7) Discuss the properties of polarization vectors.

(8) Derive an expression for the displacements of atoms.

(9) Derive the Hamiltonian of the crystal under study.

The results obtained in the last two steps will be used in the quantization

of lattice vibrations.

1.1 BornOppenheimer Approximation

The characteristic speed of an electron in a solid is 106 m/s, while that of a

nucleus is 105 m/s. Thus, the electrons in a solid move much faster than the

nuclei. This is because the electrons are much lighter than the nuclei, m 103MI, while the momenta they acquire through various interactions are

comparable. Because of the much higher mobility of the electrons than that

of the nuclei, when the configuration of the nuclei changes, the electrons

can respond instantaneously and thus remain essentially in the electronic

ground state. We can thus assume that the nuclei remain at their stationary

positions when the ground state of the electronic subsystem is solved. The

full potential energy of the nuclei is then obtained by taking the electronic

contributions into account and used subsequently to solve for the motion of

the nuclei. The disentanglement of the motion of the electrons and nuclei

in such a manner is known as the BornOppenheimer approximationorthe

adiabatic approximation.

We now describe the BornOppenheimer approximation in more details.

For brevity in notations, we first introduce collective notations for the co-


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Lattice Dynamics 5

ordinates of the electrons and nuclei. Letr= {r1, r2, , rNe} collectivelydenote the electronic coordinates and R ={R1,R2, ,RN} the nuclearcoordinates. Here Neis the number of electrons and Nthe number of nuclei.

Let (r, R) = (r; R)(R) be the wave function of the solid with (r; R)

and(R) the electronic and nuclear wave functions, respectively. The semi-

colon in(r; R) indicates that R is taken as a parameter when the motion

of the electronic subsystem is solved. We start with the eigenequation of

the Hamiltonian H= H(r, R) (i.e., the stationary Schrodinger equation of

the solid)

H(r, R)(r, R) =E(r, R), (1.2)

whereEis the eigenvalue ofH(r, R). To proceed, we rearrange the above

equation as

1

(r; R)

He(r) + Hen(r, R)

(r; R)

= 1

(r; R)(R) E Hn(R)

(r; R)(R)

. (1.3)

To emphasize the coordinate dependence, we have explicitly displayed

proper coordinate variables in the Hamiltonians. Because of the entan-

glement of the variables r and R, the two sides of Eq. (1.3) can not both

equal either a constant or a function of only r or R. However, since the

nuclei can be taken as remaining at their stationary positions when the

ground state of the electronic subsystem is solved, the eigenequation for

the electronic states

He(r) + Hen(r, R)(r; R) = E(R)(r; R) (1.4)can be solved with the nuclei remaining in the configuration R, where E(R)is the electronic eigenenergy in the nuclear configuration R. Inserting the

above equation into Eq. (1.3), we obtainHn(R) + E(R)

(r; R)(R)

= E(r; R)(R). (1.5)

The motion of the nuclei is thus separated from that of the electrons. This

is a great step forward since we now have a recipe to solve the eigenequation

of the Hamiltonian of the solid albeit it is done approximately. If the vari-

ables in Eq. (1.3) had been separated exactly, we would have had an exact

solution to the problem. In a sense, the BornOppenheimer approximation

is equivalent to solving Eq. (1.2) with the separation of variables. Hence,

the impreciseness in the BornOppenheimer approximation is caused by

the forceful application of the separation of variables to Eq. (1.2).


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The electronic energyE(R) is referred to as the adiabatic contributionto the potential energy of the nuclei. The other terms related to electronicwave functions are referred to as the non-adiabatic contribution and they

can be inferred from Eq. (1.5). Multiplying both sides of Eq. (1.5) by

(r; R) from left and then integrating over r (i.e., over r1, r2, , rNe),we have

I

2

2MI2I+ (R)

(R) =E(R), (1.6)

where (R) is the nuclear potential energy and is given by

(R) =1

2

1

40

I=J

ZIZJe2

|RIRJ| + E(R) + na(R). (1.7)

The term na(R) in (R) is the non-adiabatic contribution and is given by

na(R) = I

2

MI

dr (r; R)I(r; R)

I

I

2

2MI

dr (r; R)2I(r; R) (1.8)

with

dr =

dr1dr2 drNe . Due to the afore-mentioned slow motionof the nuclei in comparison with the electrons, the non-adiabatic contribu-

tion is usually small and can be taken into account perturbatively. The

non-adiabatic contribution is very important to some physical properties

of a solid since it describes the interaction between electrons and lattice

vibrations.

When only the electronic states are of concern, Eq. (1.4) can be solved

for a set of fixed nuclear positions (i.e., a fixed nuclear configuration). In

consideration of the large masses and slow motion of the nuclei, their mo-tion is often solved using classical mechanics. In such a case, a potential

energy surface can be mapped out by solving Eq. (1.4) for different nuclear

configurations and then used in classical computations for the motion of

nuclei.

In practice, the nuclei in Eq. (1.2) are often replaced with ions or ion

cores since the core electrons play a much less role than valence electrons

in determining the properties of a solid.

1.2 Lattice Potential Energy and Harmonic Approximation

From the above discussions, we see that the nuclear potential energy, re-

ferred to as the lattice potential energy hereafter, can be obtained only


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Lattice Dynamics 7

after the electronic motion has been solved for all the nuclear configura-tions. Thus, it seems that the understanding of the lattice dynamics of a

crystal is impossible without the knowledge of the electronic states. How-

ever, the lattice potential energy can also be obtained empirically with the

input from experiments. Such an approach is called the pseudopotential

method. In any event, the lattice potential energy is assumed to be known

from now on. In this sense, our treatment is of phenomenological nature.

With the kinetic energy expressed in terms of momenta of atoms, the lattice

Hamiltonian is given by

H=i

p2i2mi

+ (r1, r2, , rN), (1.9)

where we have used the lowercase letter i to label an atom, the lowercase

letter m to denote its mass, the bold lowercase letter p to denote its mo-

mentum, and the bold lowercase letter r to denote its position. The bold

capital letter Ris now reserved for the lattice vectors. Also, for brevity we

will generally refer to atoms as the constituents of a crystal in this section

even though they may be ions. But, ions will be used when an ionic crys-tal is explicitly referred to. For pairwise interactions between atoms, the

lattice potential (r1, r2, , rN) can be written as

(r1, r2, , rN) = 12

Ni=j=1

(ri rj), (1.10)

where(ri rj) is the interaction energy between atomsiandj.For the given lattice potential energy of a crystal, the problem we face

is what to do with it to develop a theory for the lattice dynamics of thecrystal. To accomplish this, we make good use of the fact that atoms move

only in the close vicinities of their equilibrium positions (that is, the am-

plitudes of their vibrations are small). In the first step, we Taylor-expand

the lattice potential energy in terms of the displacements of atoms from

their equilibrium positions and keep only up to the second-order terms in

the expansion. This practice is known asthe harmonic approximation. The

lattice Hamiltonian in the harmonic approximation is referred to as a har-

monic Hamiltonian. The crystal with a harmonic Hamiltonian is referred

to as a harmonic crystal. The terms of orders higher than the second

order are referred to as anharmonic terms. Ordinarily, the contributions

from the anharmonic terms are negligibly small and can be safely ignored.

However, for crystals with extraordinary properties, such as ferroelectric

crystals and crystals that can undergo structural phase transformations,


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the anharmonic terms become important. For such crystals, the effects ofthe cubic and quartic anharmonic terms are often considered.

We now consider a three-dimensional crystal with a multi-atom basis.

To be able to keep track of the algebras comfortably, we first describe clearly

how the atoms are labeled and their positions denoted.

As generally done, we label each primitive cell by the lattice site on

which the primitive cell sits. Thus, the ith primitive cell locates on the

ith lattice site and its position vector is given by Ri that is the lattice

vector of the ith lattice site. Because of the presence of bases of atoms in

a crystal, the number of atoms in each primitive cell is greater than one.The atoms within each primitive cell are indexed by positive integers, with

Greek letters (,, ) often used for the variables of indices. For ap-atombasis, we have= 1, 2, ,p. To refer to an atom within a primitive cell,we can say the th atom within the ith primitive cell. The position of

an atom within a primitive cell is given in the local Cartesian coordinate

system associated with the primitive cell with the origin at the tip of the

position vector of the primitive cell, denoted by d for theth atom. Thus,

the equilibrium position of the th atom within the ith primitive cell in acrystal is given by Ri+ d.

Shown in Fig. 1.1 is a simple cubic crystal with a two-atom basis. The

CsCl crystal has such a structure. Because atom 1 in a primitive cell locates

at the tip of the position vector of the primitive cell, its position vector is

zero in the local Cartesian coordinate system associated with the primitive

cell and is thus not shown in the figure.

With the displacement of an atom from its equilibrium position taken

into account, the instantaneous position ri of theth atom within theithprimitive cell is given by

ri=Ri+ d+ ui. (1.11)

In components, the above equation reads

ri, =Ri+ d+ ui,. (1.12)

The lattice potential energy in Eq. (1.10) is now expressed as

=1

2

i=j

(Ri+ d Rj d+ ui uj) (1.13)

for a crystal with a multi-atom basis. Taylor-expanding(Ri+ dRj d+uiuj) in terms ofuiujaboutRi+dRjdand keeping


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Lattice Dynamics 9

(a)

O

x

y

z

i

(b)

O

x

y

z

Ri

ri1

ui1

d2

ri2

ui2

(c)

Fig. 1.1 Lattice vibrations in a simple cubic crystal with a multi-atom basis. (a) Staticlattice. To indicate that the atoms of the second kind locate at the centers of the cubes(the primitive cells), the body diagonals of one cube are drawn. (b) Dynamic lattice.The ith primitive cell is shaded and marked by i close to its rear lower-left corner thatis chosen as the origin of the local Cartesian coordinate system associated with theprimitive cell. The global Cartesian coordinate system is also shown. (c) Description ofatomic positions. Shown are the position Ri of the ith primitive cell, the position d2

of the second atom within the primitive cell, the displacements u

i1 and u

i2, and theinstantaneous positions ri1 and ri2 of the two atoms within the primitive cell. Notethat the position vector d1 of the first atom is zero within the local Cartesian coordinatesystem associated with the primitive cell and is not shown.

only terms up to the second order, we have

(Ri+ d Rj d+ ui uj) (Ri+dRjd)+

(Ri+dRjd)(ui, uj,)

+12

(ui, uj,)(Ri+dRjd)(ui, uj,), (1.14)

where and denote the first- and second-order partial derivatives of

with respect to components of a lattice vector

(Ri+ d Rj d) = (Ri+ d Rj d)Ri

,

(Ri+ d

Rj

d) =

2(Ri+ d Rj d)RiRi

.

(1.15)

The harmonic lattice potential energy is then given by

harm = 0+1

4

i=j

(ui,uj,)(Ri+dRjd)(ui,uj,),

(1.16)


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where0=

1

2

i=j

(Ri+ d Rj d) (1.17)

is the total cohesive (or lattice) energy. The first-order term vanishes be-

cause of the equilibrium conditioni (i=j)

(Ri+ d Rj d) = 0.

We can remove the constraint i =jon the dummy summation variablesin Eq. (1.16) upon noticing the presence of the factors (ui, uj,) and(ui,uj,). However, to avoid the possible divergence in (Ri+dRjd), we set it to be identically zero for i =j, which is permissiblebecause the term with i= j did not appear in Eq. (1.16). We can then

rearrange harm as follows

harm = 0+1

4

i, j

(ui, uj,)(Ri+dRjd)(ui uj,)

= 0+1

2

ij

,

(Ri+dRjd)(ui, ui,ui, uj, )

= 0+1

2

ij

,

j

(Ri+ d Rj d )

ij

(Ri+ d Rj d)

ui, uj,

= 0+

1

2ij

,

ui, D, (Ri Rj)uj, , (1.18)

where we have introduced matrixD(RiRj) whose (, )th element isgiven by

D, (Ri Rj) =j

(Ri+ d Rj d )

ij

(Ri+ d Rj d). (1.19)Note that the row of D is indexed by the combination of and and

the column by the combination of and . Thus, for a three-dimensional

crystal with ap-atom basis,Dis a 3p3p matrix. Note also that the depen-dence on the positions of atoms within a primitive cell has been transferred

into the subscripts. The Fourier transform of D(Ri Rj) with respect


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Lattice Dynamics 11

to Ri Rj is a very important quantity from which the dynamics of thelattice vibrations in the crystal can be inferred.Because of the presence of the basis, not all elements of D(Ri Rj)

are even functions ofRi Rj . However, it still has several other usefulproperties.

(1) As indicated by its argument,D,(RiRj) depends on Ri and Rjonly in the form ofRiRj.

(2) From its definition in Eq. (1.19), it is seen that D,(Rj Ri) =D,(RiRj). It also holds that D,(RjRi) =D,(RiRj).

(3) The summation ofD, (Ri Rj) overi or j vanishesj

D, (Ri Rj) = 0. (1.20)

1.3 Normal Modes of a Three-Dimensional Crystal with a

Multi-Atom Basis

Having discussed the harmonic lattice potential energy of a crystal, we nowturn to solving the problem of lattice vibrations for the crystal by finding

the normal modes of its lattice vibrations using classical mechanics. We

begin with setting up the classical equations of motion for atoms in the

crystal.

1.3.1 Equations of motion of atoms

Differentiating the harmonic lattice potential energy for a three-dimensional

crystal with a multi-atom basis in Eq. (1.18) with respect to ui, , weobtain theth component of the force exerting on the th atom within the

ith primitive cell due to all other atoms in the crystal

Fi, = harm

ui,

= 12

ui,

ij

,

ui, D, (RiRj)uj,

= 1

2ij

,

D

,

(RiRj)uj, ii

+ ui, D, (RiRj)ij

= j

D, (Ri Rj)uj, . (1.21)


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It follows from Newtons second law mui, =Fi, thatmui, =

j

D, (Ri Rj)uj, , (1.22)

wherem is the mass of the th atom in a primitive cell. Note that there

is an equation of the above form for each atom in the crystal and for each

coordinate component of the displacement of each atom. Thus, we have

3N p equations in three dimensions. The solutions to the above equations

are to be found by expressingui, in terms of its Fourier components

ui, (t) =k

1N m

Q(k, )ei(kRit) (1.23)

with N the total number of primitive cells in the crystal. The fact that

ui,(t) takes only on real values leads to the property that Q(k, ) =

Q(k, ) for the Fourier coefficient Q(k, ) .

1.3.2 Allowed values of wave vector k

The allowed values ofk are to be found from the Born-von Karman bound-ary condition that, for a crystal with a multi-atom basis, is stated as follows

ui1+N1, i2i3, ,(t) =ui1, i2+N2, i3, ,(t) =ui1i2, i3+N3, ,(t)

=ui1i2i3, ,(t), (1.24)

whereN1,N2, andN3are respectively the numbers of primitive cells along

basis vectors a1, a2, and a3.

The above equation indicates that the crystals wraps itself up in all

three directions of a1, a2, and a3. Inserting Eq. (1.23) into Eq. (1.24)

yields

eiN1ka1 = eiN2ka2 = eiN3ka3 = 1. (1.25)

Since the above equations do not change ifk is changed by any reciprocal

lattice vector K, we restrict k to be within the first Brillouin zone with

the understanding that the wave vectors differing only by reciprocal lattice

vectors are all equivalent. To infer the allowed values ofk from the above

equation, we express k as

k= x1b1+ x2b2+ x3b3,where 0 |x1|,|x2|,|x3| 1 and b1, b2, and b3 are the primitive vectorsof the reciprocal lattice. Upon making use of the orthonormality relation

between bi and aj, bi aj = 2ij, we haveei2N1x1 = ei2N2x2 = ei2N3x3 = 1


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Lattice Dynamics 13

from which it follows thatx1 =n1/N1, n1= 0,1,2, ,(N1/2 1), N1/2,x2 =n2/N2, n2= 0,1,2, ,(N2/2 1), N2/2,x3 =n3/N3, n3= 0,1,2, ,(N3/2 1), N3/2.

The allowed values ofk are then given by

k= (n1/N1)b1+ (n2/N2)b2+ (n3/N3)b3 (1.26)

with ni = 0,1,2, ,(Ni/2 1), Ni/2 for i = 1, 2,3. From thevalue ranges ofn1,n2, andn3, we see that the total number of the allowed

values of k is N = N1N2N3, the total number of primitive cells in the

monatomic crystal. This statement is applicable to any crystal regardless

of its dimensionality and no matter whether or not it has a basis.

1.3.3 Allowed values of frequency

We now find the allowed values of frequency. Substituting Eq. (1.23) intoEq. (1.22) yields

m 2Q(k, )N

k

ei(kRit)

= k

j

D, (Ri Rj) Q (k, )N m

ei(kRjt),

2

Q(k, ) =

1mmj D, (RiRj)eik(RiRj)

Q (k, ),

D, (k) 2

Q(k, ) = 0, (1.27)

where D, (k) is the dynamical matrix for a three-dimensional crystal

with a multi-atom basis and is given by

D, (k) = 1

m

mj

D, (Ri Rj)eik(RiRj). (1.28)

To be able to infer some conclusions without solving the above equations

explicitly, we must acquire the necessary knowledge on the properties of the

dynamical matrix. We now show that it is a Hermitian matrix. Taking the

Hermitian conjugation of D(k) and making use of D,(Rj Ri) =


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D, (Ri Rj), we haveD, (k) =

1mm

j

D,(Ri Rj)eik(RiRj)

= 1mm

j

D, (RjRi)eik(RiRj )

= 1mm

j

D, (Ri Rj)eik(RiRj), (1.29)

that is,D, (k) = D, (k) or D

(k) = D(k). (1.30)

To arrive at the final result on the third line in Eq. (1.29), we have set

RjRi Ri Rj .We now go back to the equations in Eq. (1.27). First of all, they imply

thatthe squares of the frequencies of the normal modes are the eigenvalues

of the dynamical matrix. The Hermitian property of the dynamical matrix

guarantees that all the solutions of2 are real. They are also nonnegative

for stable crystals. Since these equations are homogeneous linear equations

for Q (k, )s, the secular equation for the determination of frequencies

follows from the sufficient and necessary condition for the existence of non-

trivial solutions

det |D, (k) 2 | = 0. (1.31)The above equation is an algebraic equation of order 3pwithp the number

of atoms in the multi-atom basis. Thus, it has 3p different solutions for

2

and 6p different solutions for at each wave vector k if no degeneracyoccurs. Hence, there are 3p branches of normal modes. The Latin letter s

is used as the branch variable. The frequency in branch s will be denoted

byks. Since there are Ndifferent allowed values ofk, there are in total

3pNnormal modes of lattice vibrations in a crystal with a p-atom basis.

Note that degeneracy may occur in some regions of the first Brillouin zone.

Taking into account the fact that there are in total 6pallowed values of

, we can express the coefficient Q(k, ) in the expansion ofui, (t) in

Eq. (1.23) as follows

Q(k, ) =

3ps=1

Q(k, ks)ks+ Q(k, ks),ks

. (1.32)

We now find out how many branches among the 3pbranches are acous-

tical branches and how many are optical branches. For this purpose, we


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Lattice Dynamics 15

study the dynamical matrix at k = 0. Setting k to zero in Eq. (1.28) andmaking use of Eq. (1.19), we obtain

D, (0) = 1mm

j

D, (Ri Rj). (1.33)

Setting k = 0 in Eq. (1.31), we obtain the equation for determining fre-

quencies at k= 0

det |D, (0) 2| = 0. (1.34)

Noticing that the above equation is just the eigenequation for D(0), wesee that the zero eigenvalues of D(0) correspond to acoustical branches.

Therefore, the number of acoustical branches is given by the dimension

of D(0) less its rank. The rank of D(0) can be found by the Gaussian

elimination method in linear algebra. With the rows and columns ofD(0)

indexed in the order = 1x,1y, 1z, 2x,2y, 2z, , px, py, pz, matrixD(0) takes on the following form

j

D11, xx

m1

j

D11, xy

m1

j

D11, xz

m1

j

D1p,xxm1mp

j

D1p,xym1mp

j

D1p,xzm1mp

j

D21, xxm2m1

j

D21, xym2m1

j

D21, xzm2m1

j

D2p,xxm2mp

j

D2p,xym2mp

j

D2p,xzm2mp

.

.....

.

... . .

.

.....

.

..j

Dp1, xxmpm1

j

Dp1, xympm1

j

Dp1, xzmpm1

j

Dpp, xx

mp

j

Dpp,xy

mp

j

Dpp, xz

mp

If we multiply the th column for = 1, 2, , p 1 by m/mp,respectively, and then add the results to the pth column, we obtain thefollowing result on the th row in the pth column

1mmp

j

D, (Ri Rj) = 0

for=x, y, z, where we have made use of the property ofD,(RiRj)in Eq. (1.20). Therefore, the last three columns ofD(0) have been brought

to zero through the elementary column operations to matrix D(0). After

this, no additional columns can be brought to zero because of the absence

ofDp, (Ri Rj) for=x, y, z in matrix D(0). Therefore, the rank ofD(0) is 3p 3. This implies that three acoustical branches are present in athree-dimensional crystal with ap-atom basis and that the remaining3p3branches are optical branches. This conclusion is verified in Fig. 1.2 by

the experimental results of inelastic neutron scattering on an NaCl crystal


16/42


that has a two-ion basis. Figure 1.2 shows that NaCl has three acousticalbranches (one longitudinal and two transverse acoustical branches, that

is, one LA branch and two TA branches) and three optical branches (one

longitudinal and two transverse optical branches, that is, one LO branch

and two TO branches). Note that the two transverse acoustical branches

are degenerate along the directions (, 0,0) and (, , ) and so are the two

transverse optical normal modes.

0 1 0 0.5(,0,0) (,,0) (,,)

0

10

20

30 LO

2TO

LA

2TAh

(meV)

LO

TO1

TO2

LATA1

TA2

LO

2TO

LA

2TA

Fig. 1.2 Dispersion relations of the normal modes in an NaCl crystal at 80 K. Thesymbols denote experimental data of inelastic neutron scattering by Raunio et. al. [G.Raunio, L. Almqvist, and R. Stedman, Physical Review 178, 1496 (1969)]. The linesrepresent cubic-spline interpolations of the experimental data.

From the above results we can infer that, for a three-dimensional

monatomic crystal, there are three branches of normal modes and they areall acoustical branches. If a monatomic crystal is taken as a crystal with a

one-atom basis for the purpose of counting branches of normal modes, we

see that the above conclusion is also applicable to a monatomic crystal.

Taking into account the facts that a one-dimensional crystal of inert

gas atoms has only one branch of acoustical normal modes and that a one-

dimensional ionic crystal has one branch of acoustical normal modes and

one branch of optical normal modes, we can draw a general conclusion that

there is (are) d acoustical branch(es) andd(p

1) optical branch(es) in a

d-dimensional crystal with ap-atom basis.

The above conclusion can be also stated in terms of the numbers of

acoustical and optical normal modes. In a d-dimensional crystal of size

of N primitive cells with a p-atom basis, there are dN acoustical normal

modes andd(p 1)Noptical normal modes.


17/42

Lattice Dynamics 17

1.3.4 Polarization vectorsHaving discussed the frequencies of all normal modes, we now study

the solutions for the Fourier coefficients Q(k, ) in Eq. (1.23). Since

Q(s)(k, ks) for normal mode ks can only be determined within a multi-

plicative factor from Eq. (1.27), we set

Q(s)(k, ks) =Q(k, ks)(s) (k) (1.35)

and demand that the vector (s) (k) = (

(s)1 (k),

(s)2 (k),

(s)3 (k)) be nor-

malized. Inserting the above expression into Eq. (1.27) and specializingEq. (1.27) for branch s, we obtain the equations for (s) (k)s

D, (k) 2ks

(s) (k) = 0. (1.36)

The vector (s) (k) is referred to as the polarization vector of normal

mode ks on atom . The polarization vectors possess the following prop-

erties

(s)

(k) =

(s)

(k), (1.37)

(s)

(k)(s) (k) =ss , (1.38)

s

(s)

(k)(s) (k) = . (1.39)

Equation (1.37) indicates that the effect of taking the complex conjugation

of a polarization vector is equivalent to taking the inversion of its wave-

vector variablek in k-space. Eq. (1.38) gives usthe orthonormality relation

of polarization vectors. Eq. (1.39) gives us the completeness relation ofpolarization vectors.

1.3.5 Displacements of atoms

We can derive an expression for the displacement of an atom in a three-

dimensional crystal with a multi-atom basis from Eqs. (1.23), (1.32),

and (1.35). We have

uj, (t) = 1

N mks

qks(t)(s) (k)eikRj , (1.40)

whereqks(t)s arethe generalized coordinatesof normal modes, referred to

as the normal coordinates, and are given by

qks(t) =Q(k, ks)eikst + Q(k, ks)eikst. (1.41)


18/42


Using the above expression and Q

(k, ks) =Q(k, ks), we can verifythatqks(t) possesses the following propertyqks(t) =qks(t). (1.42)

1.3.6 Hamiltonian of a crystal with a multi-atom basis

To derive the Hamiltonian for a monatomic crystal, we first express its ki-

netic and interaction potential energies in terms of the above-introduced

normal coordinates. Making use of the expression of the displacement

uj, (t) of an atom in Eq. (1.40), we have for the kinetic energy

T =j

1

2mu

2j, (t) =

1

2N

kkss

j

qks(t)qks (t)(s) (k)

(s) (k

)ei(k+k)Rj

=1

2

kss

qks(t)qks (t)(s) (k)

(s) (k) =

1

2

ks

qks(t)qks(t), (1.43)

where we have made use of the orthonormality relation of the polarization

vectors given in Eq. (1.38). With the constant term omitted and in terms of

the normal coordinates, the harmonic lattice potential energy in Eq. (1.18)is given by

harm =1

2

ks

2ksq

ks(t)qks(t). (1.44)

The Lagrangian of the crystal is then given by

L= T harm = 12

ks

qks(t)qks(t) 1

2

ks

2ksqks(t)qks(t). (1.45)

To obtain the Hamiltonian of the crystal, we must first find the momentum

conjugate to qks(t). Differentiating L with respect to qks(t), we have

pks(t) = L

qks(t)=

1

2

qks(t)

ks

qks (t)qks (t)

=1

2

ks

qks (t)kkss+ qks (t)kkss

= qks(t) = qks(t). (1.46)

The Hamiltonian of the crystal then follows from the above Lagrangian in

the standard wayH=

ks

pks(t)qks(t) L

=1

2

ks

pks(t)pks(t) +1

2

ks

2ksqks(t)qks(t). (1.47)


19/42

Lattice Dynamics 19

Since the Hamiltonian has been expressed as a sum of the Hamiltoniansof 3pNindependent harmonic oscillators, we have hitherto solved the prob-

lem of lattice vibrations of a three-dimensional crystal with a multi-atom

basis at the level of classical mechanics. In order to see how good the results

obtained so far, we now compute the contribution of lattice vibrations to

the specific heat (the lattice specific heat) using the above results.

1.4 Classical Theory of the Lattice Specific Heat

For the computation of the lattice specific heat, we first evaluate the internal

energy u per unit volume of the crystal. The internal energy is given by

the sum of the energies of individual harmonic oscillators weighted by the

Boltzmann factor eH/Zwith = 1/kBTthe inverse of temperature and

Z =

ksdqksdpks eH the canonical partition function. The internal

energy per unit volume is then given by

u= 1

ZV

ks

dqksdpks HeH

= 1ZV

ks

dqksdpks eH = 1

V

ln Z

. (1.48)

Our problem then reduces to the evaluation ofZ. Making use of Eq. (1.47),

we have

Z=

ks

dqksdpks eks(pkspks+

2

ksq

ksqks)/2

=ks

dqksdpks e

(pkspks+

2

ksqksqks)/2

.

The above maneuvers have reduced a 6pN-fold integral into a product

of 2-fold integrals. Note that, because of the relations qks = qks and

pks =pks, q

ks and p

ks are not independent variables. For our goal is the

evaluation of the lattice specific heat, we do not even need to evaluate the

2-fold integral explicitly. What we need to do is to extract the temperature-

dependence ofZfrom the above equation. This can be easily accomplished

by making a change of variables

qks =

qks, pks =

pks.

We then have

Z= 1

3pN

ks

dq

ksdpks e

(p ksp

ks+2

ksq

ksq

ks)/2 =

A

3pN,


20/42


where the temperature-independent value of the 2-fold integral has beendenoted by A. The internal energy per unit volume is then given by

u= 1

V 3pN kBT = 3pnkBT (1.49)

from which the lattice specific heat per unit volume follows

cv = u

T = 3pnkB, (1.50)

wherepis the number of atoms in a primitive cell and n= N/Vis the num-

ber of primitive cells per unit volume. For a three-dimensional monatomiccrystal without a multi-atom basis, p= 1. We then havecv = 3nkB. The

result in Eq. (1.50) is the well-known DulongPetit law for the lattice spe-

cific heat of solids. Expressing it in joules per kelvin per mole, we have

cv = 3R with R the gas constant, R = 8.314 JK1 mol1. Expressing itin calories per kelvin per mole, we have cv = 3R6 calK1 mol1 withR 1.986 calK1 mol1.

Unfortunately, the result in Eq. (1.50) is in consistency with the exper-

iment only in the high-temperature limit. While the above result indicatesthat cv is a constant at all temperatures, the experiment reveals that cvtends to zero essentially in the cubic power ofT as Tgoes to zero. There-

fore, classical theory of lattice vibrations is insufficient in explaining the

temperature dependence of the lattice specific heat. To resolve this incon-

sistency, we now quantize the lattice vibrations.

1.5 Quantization of Lattice Vibrations

We now quantize the normal modes of lattice vibrations derived in the

classical theory to develop a quantum theory for lattice dynamics. The

quantization process consists of two steps. In the first step, the canonical

quantization scheme is utilized to quantize the normal coordinates and

momenta of normal modes. In the second step, the combinations of the

quantum operators of the normal coordinates and momenta of the normal

modes give rise to new operators, the annihilation and creation operators

of phonons, with phonons being quanta of lattice vibrations. An important

quantity to obtain in the quantization process is the expression of the atomic

displacement in terms of the annihilation and creation operators of phonons.

This expression is referred to as the quantum field operatorof the atomic

displacement since it describes the field of the atomic displacement in terms

of quantum operators.


21/42

Lattice Dynamics 21

The first quantization is achieved by replacing the classical normal co-ordinatesqks in Eq. (1.41) and the corresponding momenta pks of normal

modes by operators qks and pks that are required to satisfy the commuta-

tion relationsqks, p

ks

= kk ss ,

qks, qks

=

pks, pks

= 0. (1.51)

Note that qks and pks have the following properties

qks = qks, p

ks = pks. (1.52)

In the framework of the first quantization, the atomic displacements

and the Hamiltonian of the crystal corresponding to Eqs. (1.40) and (1.47),respectively, are given by

uj, = 1

N m

ks

qks(s) (k)e

ikRj , (1.53)

H=1

2

ks

pkspks+

1

2

ks

2ksq

ksqks. (1.54)

In the second quantization, we introduce the following annihilation and

creation operators of phonons

aks =

ks2

1/2qks+ i

kspks

,

aks =

ks2

1/2qks i

kspks

.

(1.55)

The operators aks and aks satisfy the following commutation relations

aks, aks

= kk ss ,

aks, aks

=

aks, a

ks

= 0.(1.56)

Inverting the expressions in Eq. (1.55), we can express qks and pks asfollows

qks =

2ks

1/2aks+ a

ks

,

pks = i

ks2

1/2aks aks

.

(1.57)

In terms of the annihilation and creation operators of phonons, the

quantum field operator of the atomic displacements and the Hamiltonian

of the crystal are given by

uj, =ks

2N mks

1/2(s) (k)

aks+ a

ks

eikRj , (1.58)

H=ks

ks

aksaks+ 1/2

. (1.59)


22/42


The eigenvalues and eigenstates of the crystal Hamiltonian Hare givenby

En =ks

(nks+ 1/2)ks,

|n =ks

|nks =ks

1nks!

aks

nks |0,n= {nks | ks}, nks = 0, 1,2, .

(1.60)

The time dependence of aks and aks

can be derived through the Heisen-

berg equation of motion. It is found that

aks(t) = eikstaks, a

ks(t) = e

ikstaks. (1.61)

Inserting the above expressions for the time dependence of aks and aks

into Eq. (1.58), we obtain the time-dependent quantum field operator of

the atomic displacements

uj, (t) =ks

2N mks

1/2(s) (k)

eikstaks+ e

ikstaks

eikRj

=ks

2N mks

1/2(s) (k)e

i(kRjkst)aks

+ (s)

(k)ei(kRjkst)aks

.

(1.62)

1.5.1 Statistics for phonons

In treating finite-temperature problems related to phonons, the statisticsfor phonons is an indispensable piece of instrument. Since phonons are

bosons of spin zero, they obey the BoseEinstein statistics. However, we

can directly compute the thermal distribution using the eigenvalues of the

crystal Hamiltonian in Eq. (1.60). According to Boltzmann, the crystal

takes the eigenstate|n as its state with the probability of eEn/kBT/Zwith Z=

ne

En/kBT the canonical partition function. We now evaluate

the average number of phonons in the single-phonon state |ks, nks, whichis given by

nks = 1Z

n

nkseEn/kBT

= 1

Z

{nks}

nksks

e(nks+1/2)ks/kBT.


23/42

Lattice Dynamics 23

Making use of standard algebraic manipulations in statistical mechanics,we have

nks = 1Z

nks=0

nkse(nks+1/2)ks/kBT

nks=0e(nks+1/2)ks/kBT

{n

ks}

ks

e(nks+1/2)ks/kBT

=

nks=0

nkse(nks+1/2)ks/kBT

nks=0e(nks+1/2)ks/kBT

= 1

eks/kBT 1 .

That is

nks = 1

eks/kBT 1 (1.63)which is just what the BoseEinstein statistics gives for phonons. Note

that, because phonons in a crystal are constantly annihilated and created,

their number is not conserved and their chemical potential is zero.

1.6 Phonon Density of States

In phonon-related problems, we often need to perform the summation ofthe form

ks F(ks) over the phonon wave vector k and branch s, where

F(ks) depends on k only through the phonon dispersion relation ks.

Such a sum can be converted into an integral over phonon frequencies for

the benefit of reducing a three-dimensional integral (that results from con-

verting the sum over kinto an integral over k) to a one-dimensional integral.

This is especially useful in numerical computations. The conversion into

an integral over phonon frequencies can be easily implemented by using the

property of the Dirac -function: d ( ks) = 1. Inserting this

magic one into the summation

ks F(ks), we have

1

V

ks

F(ks) = 1

V

ks

d ( ks)F(ks)

=

d F()

1

V

ks

( ks)

=

d g()F(), (1.64)

where g() is the phonon density of states, with g()d the number ofphonon states per unit volume in the frequency range from to + d,

and is given by

g() = 1

V

ks

( ks) =s

dk

(2)3( ks). (1.65)


24/42


The phonon density of states in branch s is given bygs() =

1

V

k

( ks) =

dk

(2)3( ks). (1.66)

g() is then the summation ofgs() over phonon branches. We can also ex-

press the phonon density of states in terms of an integral over the constant-

frequency surface. To obtain this expression, we write dk = ddk with

d the area element on the constant-frequency surface S, ks = , and

express( ks) in terms of the componentk ofk perpendicular to theconstant-frequency surface S

( ks) = (k k 0)kks ,wherekksis the derivative in the direction of the normal of the constant-

frequency surface S. We then have

gs() =

S

d

dk(2)3

(k k 0)kks = 1

(2)3

S

dkks . (1.67)This alternative expression of the phonon density of states can differenti-

ate the importance in the contributions of various normal modes to thedensity of states and disclose the singularities in the dispersion relations.

Ifkks = 0 at some particular wave vector k0, this expression indicates

that the vicinity aroundk0makes an important contribution to the phonon

density of states since the integrand diverges at k0. This leads to a peak

in the phonon density of states at the corresponding frequency. Such a

frequency is known as a van Hove singularity. The wave vectors that con-

tribute to van Hove singularities are referred to as critical pointsof the first

Brillouin zone.

1.7 Lattice Specific Heat of Solids

Since lattice vibrations (phonons) contribute to a variety of physical proper-

ties of solids, the results obtained in the previous section find their extensive

applications in these properties. Here we concentrate on the phonon con-

tribution to the specific heat (the lattice specific heat) of solids since the

specific heat of solids is one of the few problems that first gave us hints on

the inaccuracy of the classical theory in its description of the microscopic

world.

We will first derive a general expression for the lattice specific heat using

the eigenvalues of the crystal Hamiltonian in Eq. (1.60). We will then study

the Debye and Einstein models for the lattice specific heat.


25/42

Lattice Dynamics 25

1.7.1 General expression of the lattice specific heatWe follow the standard approach in thermodynamics for the computation

of the lattice specific heat. We first derive the internal energy of the crystal.

The lattice specific heat is then computed from the internal energy. The

thermal distribution function in Eq. (1.63) gives us the average phonon

number in the single-phonon state|ks. Since each phonon in the single-phonon state|ks carries an energy ofks, the internal energy uper unitvolume of the crystal is given by

u= 1V

ks

nks ks = 1V

ks

kseks/kBT 1

=s

dk

(2)3ks

eks/kBT 1 , (1.68)

where the k-integration is over the first Brillouin zone of the crystal. The

lattice specific heat per unit volume cv is then given by

cv = u

T =

T s

dk

(2)3ks

eks/kBT

1

. (1.69)

The above equation is referred to as the general expression for the lat-

tice specific heat. To compute the lattice specific heat using the above

expression, we must know the dispersion relations of the normal modes

(the phonon dispersion relations). From our previous experience, we know

that it is a great challenge to compute the phonon dispersion relations for

real crystals. In any event, if the phonon dispersion relations are known,

Eq. (1.69) can be then utilized to compute the lattice specific heat per unit

volume. However, even without knowing the explicit phonon dispersion re-

lations, we can still evaluate approximately the lattice specific heat in the

high- and low-temperature limits.

1.7.2 High-temperature limit

In the high-temperature limit, ks/kBT 1. We can then expand theBoseEinstein distribution function in Eq. (1.69) as follows

1

eks/kBT 1= 1

ks/kBT+ (ks/kBT)2/2! + (ks/kBT)3/3! + =

kBT

ks 1

2+

ks12kBT

+ .The second term is a constant and does not contribute to the lattice

specific heat. The contributions from the third and other higher-order


26/42


terms are much smaller than the contribution from the first term becauseks/kBT 1 and they are quantum corrections to the result from thefirst term alone. Retaining only the first term in the above expansion, we

have

limT

cv T

s

dk

(2)3 ks kBT

ks

=kBs

dk

(2)3 = 3pkB/vc = 3pnkB, (1.70)

where n = 1/vc = N/V is the number of primitive cells per unit volume

and p the number of atoms in a primitive cell. For a three-dimensional

monatomic crystal without a multi-atom basis, we have cv = 3nkB. The

result in Eq. (1.70) is just the DulongPetit law and it indicates that the

DulongPetit law is valid only at high temperatures.

1.7.3 Low-temperature limit

At low temperatures, the probabilities for the normal modes of high fre-

quencies to be occupied by phonons are extremely small. Thus, we can takeonly the normal modes of low frequencies into account in the computation

of the lattice specific heat at low temperatures. Since the optical normal

modes are of high frequencies in comparison with the acoustical phonons,

their contributions are neglected. For the acoustical normal modes, only

those of low frequencies make substantial contributions. From the compu-

tations of the dispersion relations of the normal modes in the last chapter,

we know that the dispersion relation for acoustical normal modes of low

frequencies can be well approximated by a linear dependence on the wavenumber,ks cs(k)k, wherecs(k) is the speed of sound that depends onlyon the direction ofk (denoted by k) and does not on the magnitude ofk.

Because ecs(k)k/kBT becomes even smaller for large values ofk, the error

introduced by extending the k-integration in Eq. (1.69) from over the first

Brillouin zone to over the entire reciprocal space is negligibly small at low

temperatures. We thus extend the region of the k-integration in Eq. (1.69)

to the entire reciprocal space. With the above-introduced simplifications,

the lattice specific heat of a solid at low temperatures is given bylimT0

cv 122

T

s

d

k

4

0

dk cs(k)k

3

ecs(k)k/kBT 1

= 6

2

kBT

c

3kB

0

dx x3

ex 1 ,


27/42

Lattice Dynamics 27

where1

c3 =

1

3

s

d

k

4

1

c3s(k)(1.71)

is the average of the inverse of the cubed speeds of sound of the normal

modes of the three acoustical branches. The remaining integral can be

performed by first multiplying the numerator and denominator by ex and

then expanding 1/(1 ex) as a Taylor series

limT0

cv 62

kBT

c

3

kB

0

dx x3ex

1 ex

= 6

2

kBT

c

3kB

n=1

0

dx x3enx

= 6

2

kBT

c

3kB

n=1

6

n4 =

22

5

kBT

c

3kB, (1.72)

where we have made use ofn=11/n

4

= 90/4

. This is a remarkableresult! It implies that the lattice specific heat tends to zero cubically as

the temperature goes to zero, in excellent agreement with the experiment.

The problem of the lattice specific heat at low temperatures has thus been

solved with the quantization of lattice vibrations! The experimental data

of the specific heat of diamond at low temperatures are given in Fig. 1.3

together with a fit to cv =AT3.

0.0

0.01

0.02

0.03

0 20 40 60 80

cv

(calK-1mol-1)

T [K]

Fig. 1.3 Low-temperature specific heat of diamond. The open circles represent theexperimental data [W. DeSorbo, Journal of Chemical Physics 21, 876 (1953)]. The solidline is a fit to cv =AT3 with A= 4.774 108 calK4 mol1.


28/42


Since a diamond crystal is an insulator (a semiconductor with a largeband gap), its low-temperature specific heat consists of only the lattice spe-

cific heat. From Fig. 1.3, it is seen that the specific heat of diamond at low

temperatures indeed follows the T3-power law and that the experimental

data is well fitted to cv =AT3 with A= 4.774 108 calK4 mol1.

1.8 Debye Model

To evaluate the contribution of the lattice vibrations to the specific heatof a solid, Debye put forward his model for lattice vibrations in 1912. In

the Debye model, only three acoustical branches are used to describe all

the lattice vibrations in a solid. It is assumed that all the normal modes

in the three acoustical branches have the same speed of sound c, that the

dispersion relation is linear in the wave number k, =ck, and that there

exists an upper limit (calledthe Debye wave vectorand denoted by kD) for

the wave number. The maximum wave number kD is determined through

demanding that the number of acoustical normal modes in the model beequal to the actual number of acoustical normal modes in the crystal. The

quantity D = ckD is called the Debye frequency. In the Debye model,

the contribution of optical normal modes to the specific heat is taken into

account through high-frequency acoustical normal modes.

For the convenience of finding expressions forkDand D, we first assume

that they are known and derive the phonon density of states in the Debye

model. From Eq. (1.65), we have in the Debye model

gD() = 3

V

k

( ck) = 322

kD0

dk k2( ck)

= 32

22c3() (D ), (1.73)

where (x) is the step function, (x) = 1 for x > 0, = 0 for x < 0. Note

that the phonon density of states in the Debye model is quadratic in for

0< D (this is a characteristic of the Debye model) and that it is zero

for D.

We now find expressions for kD and D. For a three-dimensional

monatomic crystal without a multi-atom basis, the total number of acousti-

cal normal modes is 3N withNthe number of primitive cells. The number

of acoustical normal modes in the Debye model is given byVd gD().


29/42

Lattice Dynamics 29

We thus have

3N=V

d gD() = 3V

22c3

d 2() (D )

= 3V

22c3

D0

d 2 = 3DV

22c3 (1.74)

from which it follows that

D=

62

n1/3

c, kD =

62

n1/3

, (1.75)

wheren= N/Vis the number of primitive cells per unit volume.

We now compute the lattice specific heat within the Debye model. From

the general expression for the lattice specific heat in Eq. (1.69), we have

cDv = 3

22

T

kD0

dk ck3

eck/kBT 1

= 9nkB T

TD3

T D/T0

dx x

3

ex 1

, (1.76)

where D = ckD/kB = D/kB is the Debye temperature. The Debye

temperature D has since been used to characterize crystals. For a plot of

cDv versus temperature, see Fig. 1.6. In general, Ddepends on temperature

[see below for a more detailed discussion]. Unfortunately, the specific heat in

Eq. (1.76) can not be given in a closed form at an intermediate temperature.

However, the closed forms can be approximately obtained at high and low

temperatures.

1.8.1 High-temperature limit

In the high-temperature limit, since D/T 1, the values of the integra-tion variablexare very small in the entire integration interval. We can then

expand the exponential function ex in the denominator of the integrand in

Eq. (1.76) as a Taylor series and retain only the first two terms. We then

have

cDv = 9nkB

T

T

D

3T

D/T0

dx x2

= 3nkB. (1.77)

We have thus recovered the DulongPetit law in the high-temperature limit.


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1.8.2 Low-temperature limitSince D/T 1 in this limit, the upper limit of the integral in Eq. (1.76)can be extended to infinity. We then have

cDv = 9nkB

T

T

D

3T

0

dx x3

ex 1

=124

5

T

D

3nkB 234

T

D

3nkB, (1.78)

where the result for the integral in Eq. (1.72) has been used. Hence, the

lattice specific heat at low temperatures also follows the T3-power law in

the Debye model. The success of the Debye model at low temperatures lies

at the physical fact that only low-frequency acoustical single-phonon states

are occupied with appreciable probabilities at low temperatures.

1.8.3 Debye temperature

As mentioned in the above, the Debye temperature has been used to char-acterize a solid. As a matter of fact, it is one of the most important charac-

teristics of a solid. It reflects the density, structural stability, and bonding

strength of the solid. Structure defects in a solid can be also identified

through the variation in its Debye temperature. The Debye temperature

is also the characteristic energy scale of phonons in the solid and used in

comparison of energy scales with other elementary excitations. The mag-

nitudes of the Debye temperature vary widely among solids: It can be as

large as over 2, 000 K, such as in diamond, and as small as below 40 K,such as in cesium. The typical value of the Debye temperature D can be

taken as several hundred Kelvins. Since the Fermi temperature F of the

electron gas in a metal is typically several ten thousand Kelvins, the ratio

D/Fis typically of the order of 102. Thus, the energy scale of phonons

in a metal is very small compared to that of electrons. This fact will be

extensively exploited in the study of the electronphonon interaction.

The Debye temperature of a solid can be inferred from several different

physical quantities of the solid, such as the entropy, the specific heat, the

speed of sound, the elastic constants, and etc.

The Debye temperature of a solid in general varies with temperature.

The variation is large in some solids and small in others. The tempera-

ture dependence of the Debye temperature of a perfect crystalline solid is

chiefly caused by the electronphonon interaction and the anharmonicity


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Lattice Dynamics 31

in lattice vibrations. The temperature dependence of the Debye tempera-ture in diamond is shown in Fig. 1.4 from which it is seen that the Debye

temperature in diamond is high and that its variation is large. The Debye

temperature peaks at about 60 K with a peak value of about 2, 250 K. It is

about 1, 850 K at 25 K and 1, 870 K at 300 K. The large Debye temperature

in diamond leads to a small lattice specific heat in diamond as shown in

Fig. 1.3.

1900

2000

2100

2200

2300

100 200 300

D

[K]

T[K]

Fig. 1.4 Debye temperature as a function of temperature in diamond [W. DeSorbo,Journal of Chemical Physics 21, 876 (1953)].

The Debye temperatures of alkali metals are small compared to the De-

bye temperature of diamond, with lithium having the largest Debye temper-

ature (about 375 K) among the alkali metals and its Debye temperature not

varying appreciably with temperature. The Debye temperatures of the re-

maining alkali metals, sodium, potassium, rubidium, and cesium, are shown

in Fig. 1.5 as functions of temperature. It is seen the Debye temperatures

of these alkali metals do not vary much with temperature, either.

1.9 Einstein Model

Performing the derivative with respect to temperature T in the general

expression of the lattice specific heat in Eq. (1.69), we obtain

cv =kB

V

ks

(ks/2kBT)2

sinh2(ks/2kBT)=

kBV

ks

E(ks/2kBT), (1.79)


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50

100

150

1 10 100 200 300

D

[K]

T [K]

Fig. 1.5 Debye temperature as a function of temperature for alkali metals Na, K, Rb,and Cs (from top to bottom) from T= 1 to 300 K [D. L. Martin Physical Review 139,150 (1965)].

where we have converted the integration over k into a summation over k

and introduced an auxiliary functionE(x) given by

E(x) = x2

sinh2(x). (1.80)

The functionE(x) is calledthe Einstein function. For the convenient eval-

uation of the contribution of the optical phonons to the lattice specific

heat, Einstein treated the optical normal modes as independent harmonic

oscillators and assumed that they have the identical frequencyE. This is

the well-known Einstein model for the lattice specific heat. Note that the

acoustical phonons are not taken into account in the Einstein model. Be-cause the optical phonons all have nonzero frequencies, the lattice specific

heat given by the Einstein model has an incorrect temperature dependence

at low temperatures.

The lattice specific heat per unit volume in the Einstein model is simply

given by

cEv =poptN kB

V E(E/2kBT) =poptnkB

(E/T)2eE/T

eE/T

1

2 , (1.81)

where E = E/kB is the Einstein temperature and popt is the number

of optical branches. Note that, as T 0, cEv poptnkB(E/T)2eE/T.Although cEv 0 as T 0, the temperature dependence is incorrect asmentioned above with the reason given there. AsT , cEvpoptnkB.If the number of optical branches is equal to three, cEv at high temperatures


33/42

Lattice Dynamics 33

agrees with that given by the DulongPetit law and with that given by theDebye model.

The temperature dependence of the lattice specific heats predicted

by the Debye and Einstein models are plotted in Fig. 1.6. For the lat-

tice specific heat given by the Debye model, cv/3nkB is plotted, whereas

cEv /poptnkB is plotted for the Einstein model.

0.0

0.5

1.0

0.0 0.5 1.0 1.5

cv

D3nkB

,cv

Epoptn

kB

T D , T E

Debye

Einstein

Fig. 1.6 Lattice specific heats predicted in the Debye and Einstein models as functionsof reduced temperature T /D orT /E. The solid line is for c

Dv and the dashed line for

cEv .

From Fig. 1.6, it is seen that the lattice specific heats from the Debye

and Einstein models both tends to zero as temperature goes to zero and

approach the result given by the DulongPetit law at high temperatures.Overall,cEv is smaller than c

Dv. Note that c

Ev goes to zero much faster than

cDv does. This is due to the erroneous behavior ofcEv at low temperatures

mentioned in the above.

1.10 Effect of Thermal Expansion on Phonon Frequencies

When temperature varies, a crystal expands or shrinks, which leads to the

variation in phonon frequencies. This is the subject we investigate in this

section. We start from the general description of the thermal expansion.

The dimensionlessGruneisen parameter(named after Eduard Gruneisen),

(T), is used to describe the thermal expansion. The Gruneisen parameter


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is defined by

(T) =BT

cv, (1.82)

where is the volume thermal expansion coefficient, = ln V/T, BTthe isothermal bulk modulus, BT = VP/V = P/ln V with P thepressure, and cv the specific heat per unit volume. To derive an explicit

expression for (T), we need to compute the pressure P that is given by

P =

F/VT

in terms of the Helmholtz free energy F,F = kBTln Z,where the canonical partition function Zis given by

Z=n

eEn =ks

nks

eks(nks+1/2) =ks

2

sinh(ks/2kBT). (1.83)

Here the eigenvalues of the crystal Hamiltonian in Eq. (1.60) have been

used. The Gruneisen parameter is then given by

(T) = 1

cv

T

F

V

T

V

= kBcv

T

T

ln Z

V

T

V

=

ks

ksE(ks/2kBT)

ks

E(ks/2kBT), (1.84)

where the Einstein function E(x) is given in Eq. (1.80) and ks isthe mode

Gruneisen parameter for normal mode ks and is given by

ks = ln ksln V

. (1.85)

Note that Eq. (1.84) implies that (T) is a weighted average of the mode

Gruneisen parameters with the weight for normal mode ksgiven by the nor-

malized Einstein function: E(ks/2kBT) divided by

ks E(ks/2kBT).

For a one-dimensional crystal, the mode Gruneisen parameter is given

by

ks = ln ksln L

(1.86)

with L the length of the one-dimensional crystal. Take a one-dimensional

crystal of inert gas atoms of mass mas an example. The phonon dispersion

relation is given byk = (4K/m)1/2| sin(ka/2)| for such a one-dimensional

crystal, whereKis the force constant. We have

k = ln kln a

= (ka/2) cot(ka/2). (1.87)


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Lattice Dynamics 35

Note that the mode Gruneisen parameter is negative for all normal modes.The Gruneisen parameter is then given by

(T) =

k

(ka/2) cot(ka/2)E(k/2kBT)

k

E(k/2kBT). (1.88)

The values of (T) at a number of temperatures are evaluated numer-

ically with the results plotted in Fig. 1.7 as a function of T / with

= (4K/m)1/2/kB.

-1.0

-0.9

-0.8

-0.7

0.0 0.5 1.0

T

Fig. 1.7 Plot of the Gruneisen parameter of a one-dimensional crystal of inert gas atomsas a function of the reduced temperature T /.

From Fig. 1.7, it is seen that the Gruneisen parameter for this one-

dimensional crystal is negative at all temperatures, which implies that the

normal mode frequencies decrease as the crystal expands.

1.11 Specific Heat of a Metal

The specific heat of a pure metallic crystal consists of the electronic and

lattice specific heats. At low temperatures, the electronic specific heat takeson the form cev = T with the electronic specific heat coefficientand the

lattice specific heat takes on the formcLv =AT3 [cf. Eqs. (1.72) and (1.78)].

Thus, the specific heat of a pure metal is given by

cv =cev+cLv = T+ AT

3. (1.89)


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0.0

0.4

0.8

0 0.5 1.0 1.5

cv

[mcalK-1mol-1]

T [K]

(a)

0.3

0.4

0.5

0.6

0.0 0.5 1.0 1.5 2.0 2.5cv

T[mcalK-2mol-

1]

T2 [K2 ]

(b)

Fig. 1.8 Low-temperature specific heat of sodium. (a) Specific heatcv as a function oftemperature T. The open circles represent the experimental data [D. L. Martin, PhysicalReview 124, 438 (1961)]. The solid line is a linear least-squares fit of the experimentaldata to cv = T+ AT3. (b) Specific heat divided by temperature, cv/T, as a functionof T2. The open circles represent the same experimental data as in (a) but now cv /Tis plotted as a function ofT2. The solid straight line is a linear least-squares fit of theexperimental data to cv/T =+ AT2.

To see how well the form of the low-temperature specific heat in Eq. (1.89)is obeyed in a real simple metal, we show the low-temperature specific heat

of sodium in Fig. 1.8.

In Fig. 1.8(a), the experimental data of the low-temperature specific

heat of sodium are shown. From the linear least-squares fit of the ex-

perimental data to the expression in Eq. (1.89), it is seen that the low-

temperature specific heat of sodium does follow the law prescribed in

Eq. (1.89). Although the coefficients and A can be determined from

the above linear least-squares fit of the experimental data, it has be-

come a custom that the coefficients are determined by plotting the ex-perimental data in the manner of cv/T versus T2. Such a plot is shown

in Fig. 1.8(b) together with the linear least-squares fit of the experimen-

tal data to cv/T = +AT2. In such a plot, the intercept on the ver-

tical axis yields the value for and the slope of the straight line gives

the value for A. It has been found that 0.335 mcalK2mol1 andA 0.107 mcalK4mol1.

The above example demonstrates that the low-temperature specific heat

of a simple metal follows the law given in Eq. (1.89). However, deviationsfrom this law are observed in metallic compounds, especially in metallic

compounds that contain d or felements, such as heavy fermion systems5.

5G. R. Stewart, Reviews of Modern Physics 56, 755 (1984).


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PROBLEMS 37

Problems

1-1 Consider a linear chain in which alternate ions have masses m1 and

m2 and only nearest neighbors interact through a spring of force

constant K. Find the dispersion relations for the normal modes.

Discuss the limiting cases for m1 m2 andm1=m2.1-2 A commonly-seen example of a simple one-dimensional crystal is a

line of point masses, each of which has two nearest neighbors: One

is distance d away and the other distance (a

d) away (d

a)

in equilibrium. The two neighboring point masses are connected bysprings, with the force constants of the springs between the near-

distanced point masses and between the far-distanced point masses

given by K andG(K > G), respectively.

(1) Write down the harmonic crystal potential energy in terms of the

displacements of point masses from their equilibrium positions.

(2) Set up the classical equations of motion for the point masses.

(3) Solve for the frequencies and polarization vectors of the normal

modes of the lattice vibrations from the classical equations of

motion.

1-3 Consider a one-dimensional crystal of atoms of mass m. Only the

interactions up to the next nearest neighbors are taken into account

and are modeled by springs with the force constant for the nearest-

neighbor interaction given by K and that for the next-nearest-

neighbor interaction given by G.

(1) Compute the dispersion relation of the normal modes.(2) Find the condition onGso that the dispersion curve peaks inside

the first Brillouin zone.

(3) Find the expressions for the group and phase velocities and eval-

uate them at the peak position of the dispersion curve under the

condition found in (2).

1-4 Consider a linear chain of atoms of mass m with the nearest neigh-

boring atoms connected by springs of force constant K. In addition,

the motion of each atom is damped, with the damping forceujexerted on the jth atom, where uj is the displacement of the jth

atom from its equilibrium position. Assume that (mK)1/2.(1) Write down the equations of motion of atoms with the damping

taken into account.


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(2) Find the dispersion relation k(3) Find the relaxation time of the normal modes.

1-5 Consider a linear chain of polarizable molecules with the nearest-

neighbor equilibrium distance a. All the molecules are fixed to their

positions. However, each molecule has an internal degree of freedom

that obeys the equation of motion2p/t2 = 20p + E 20, wherepis the electric dipole moment of the molecule (assumed to be parallel

to the chain), E the local electric field due to all other molecules,

and the polarizability. Find the dispersion relation(k) for smallamplitude polarization waves (optical normal modes). Discuss the

dependence of(0) on.

1-6 A triatomic linear chain consists of three different types of atoms

of masses m1, m2, and m3, respectively. As usual, it is assumed

that only nearest-neighboring atoms interact and the interactions are

modeled as being mediated through springs of force constants be-

tween atoms of types 1 and 2, between atoms of types 2 and 3, and

between atoms of types 3 and 1. Derive an equation that deter-

mines the frequencies of normal modes and describe the properties of

solutions.

kx

ky

/a

/a

/a

/a

M

Z

X

Fig. 1.9 First Brillouin zone of the two-dimensional square lattice. Three high-symmetry points, , X, and M, and three high-symmetry lines, , , and Z, areshown.

1-7 Consider a two-dimensional crystal with a square Bravais lattice

whose first Brillouin zone is given in Fig. 1.9. With only interactions

between the nearest and next nearest neighbors taken into account,


39/42

PROBLEMS 39

the harmonic lattice potential energy of the crystal is given by =

1

2a2Kij

( Ri Rj)(uiuj)

2+

1

4a2G(ij)

( Ri Rj)(uiuj)

2,

whereij indicates summation over the nearest neighbors and (ij)summation over the next nearest neighbors. Here a is the lattice

constant, Ris are Bravais lattice vectors, and uis are deviations of

atoms from their equilibrium positions.

(1) Construct the dynamical matrix.

(2) Find the frequencies and polarization vectors of normal modesalong the lines , , and Z, respectively.

(3) Plot the dispersion relations along these three high-symmetry

lines.

1-8 The lattice dynamics of a simple cubic crystal of lattice constant a

and atom mass M is studied here with only interactions between

nearest-neighboring atoms taken into account. The interactions are

modeled as being mediated through springs of force constant .

(1) Write down the potential energy of the crystal and construct the

dynamical matrix.

(2) Solve for the dispersion relations.

(3) Plot the dispersion relations along [100] and label branches prop-

erly. Give the physical reason for the zero-frequency normal-

mode branches.

(4) Indicate the pattern of displacements of eight atoms in the con-

ventional cell for the mode at k = (/a, 0, /a) with displace-

ments alongex.1-9 Consider a three-dimensional monatomic Bravais lattice in which each

ion of mass M interacts only with its nearest neighbors with the

interaction potential energy given by(rirj) =K(|rirj |d)2/2,wheredis the equilibrium spacing between the atoms andKthe force

constant of the spring connecting the atoms.

(1) Show that the frequencies of the three normal modes at each

wave vector kare given bys(k) = [s(k)/M]1/2, wheres(k)s

are the eigenvalues of the 3 3 matrixD = 2KR=0sin2(k R/2)RR with, =x, y, z.

(2) Now apply the above result to a monatomic FCC crystal. Show

that, if k is in the [100] direction, k = (k, 0, 0), then the fre-

quency of the longitudinal acoustical normal mode is given by


40/42


L = 2(2K/M)1/2

sin(ka/4) and the frequency of the two de-generate transverse acoustical normal modes is given by T =

2(K/M)1/2 sin(ka/4). Also consider the cases for k in the [110]

and [111] directions

1-10 A three-dimensional crystal has a two-atom basis. The masses of the

two atoms in the basis arem1andm2, respectively. Let v1andv2 be

their velocities. Show that, for an optical normal mode at the center

of the first Brillouin zone (k= 0), m1v1+ m2v2= 0.

1-11 The quantum field operator of atomic displacements for a three-dimensional crystal with a multi-atom basis is given in Eq. (1.62).

We now derive the quantum field operator of atomic momenta. In

analogy with the definition of momentum in classical mechanics, let

Pj, (t) =muj, (t)/t.

(1) Write down the explicit expression ofPj, (t) in terms of oper-

ators aks and aks.

(2) Show that [uj, (t), P, (t)] = ij . Therefore, Pj, (t)

is the momentum field operator conjugate to the displacementfield operator uj, (t).

(3) Show that uj, (t) and Pj, (t) are Hermitian operators.

1-12 In this problem, the Hamiltonian for a three-dimensional crystal with

a multi-atom basis will be derived.

(1) Specializing the above-obtained expression for Pj, (t) to the

time-independent case and making use of the resultant ex-

pression, express the kinetic energy of the crystal, T =j, P

j, Pj, /2min terms of operators aks and a

ks.

(2) Using the expression of uj, , express the harmonic lattice po-

tential energy of the crystal, harm = (1/2)j

, u

j,

D, (RjR)u, in terms of operators aks and aks.(3) Derive the Hamiltonian for a three-dimensional crystal with a

multi-atom basis.

1-13 Take phonons in a crystal as if they move in a box of volume V andstudy the thermodynamics of this gas of phonons.

(1) Evaluate the canonical partition function of phonons, Z =ne

En/kBT with En an eigenvalue of the crystal Hamiltonian,

and the Helmholtz free energy F = kBTln Z.


41/42

PROBLEMS 41

(2) From the thermodynamic relation S =F/TV, computethe entropyS.

(3) Express S in terms of the thermal average of the occupation

number nB(ks) nks of the single-phonon state|ks.

1-14 Consider a one-dimensional crystal of inert gas atoms. Let L be the

length of the crystal and N the number of atoms. Let a be the

lattice constant. (i) Evaluate the phonon density of states for this

crystal. (ii) Derive an integral expression for the lattice specific heat

of the crystal. (iii) Evaluate the lattice specific heat in the high- andlow-temperature limits.

1-15 Reconsider the above problem within the Debye model. (i) Find the

phonon density of states within the Debye model. (ii) Determine the

Debye frequency. iii. Find a general expression for the lattice specific

heat. (iv) Evaluate the lattice specific heat in the high- and low-

temperature limits. (v) Compare the exact and Debye results for the

lattice specific heat by plotting them together from zero temperature

to the Debye temperature.1-16 Consider the relative size of the electronic and lattice contributions

to the specific heat of a metal using the Sommerfeld theory for the

electrons and the Debye model for the phonons. (i) Find an expres-

sion for the ratiocev/cLv . (ii) Determine the temperature T

at which

cev =cLv . (iii) Give an estimate on the order of magnitude for T

in

alkali metals.

1-17 A number of values of the specific heat of potassium at low temper-

atures are given in Table 1.1. (i) Plot Cv versus T andCv/T versus

T2. (ii) Perform a linear least-squares fit of the experimental data

forCv/T toCv/T =+ AT2 and determineandA. (iii) Estimate

the Debye temperature of potassium at low temperatures.

Table 1.1 Low-temperature specific heat of potassium in mJ K1mol1 [W. H. Lienand N. E. Phillips, Physical Review 133, A1370 (1964)]. The temperatureT is in K.

T Cv T Cv T Cv T Cv0.260 4 0.585 2 0.288 5 0.657 8 0.364 4 0.885 8 0.451 5 1.177 00.278 1 0.630 6 0.289 4 0.665 7 0.373 4 0.918 0 0.457 8 1.208 0

0.295 3 0.678 6 0.306 7 0.710 4 0.393 5 0.973 3 0.483 5 1.302 00.250 1 0.559 2 0.327 0 0.768 7 0.399 4 1.003 0 0.496 9 1.353 00.265 0 0.596 9 0.337 9 0.796 2 0.423 1 1.021 0 0.543 5 1.551 00.269 8 0.606 6 0.347 8 0.836 2 0.427 4 1.102 0 0.594 4 1.786 0


42/42


1-18 Assume that the dispersion relation of an optical phonon branch in asolid takes on the form(k) =0 Ak2 neark= 0, where0 andAare positive constants. Find the phonon densities of states for < 0and > 0, respectively.

1-19 The dispersion relation of the acoustical branch in a one-dimensional

crystal with a two-atom basis of identical atoms of mass m is given

by

a(k) = 1

m1/2K+ G

K2 + G2 + 2KG cos(ka)1/2

1/2

,

whereKandG are the force constants of the springs connecting the

two nearest neighbors of an atom. Evaluate the phonon density of

states for this acoustical branch.

1-20 At finite temperatures, phonons in a crystal are constantly created

and annihilated so that their number, nph =

ks nks with nks the

number of phonons in single-phonon state|ks, fluctuates greatly.(i) Evaluate the thermal average number of phonons per unit vol-

umenph

=Z1n npheEn/kBT =Z1ksn nkseEn/kBT with

Z =ne

En/kBT and En the eigenvalues of the harmonic lat-

tice Hamiltonian in a three-dimensional crystal with a multi-atom

basis. (ii) Find the expressions fornph in the high- and low-temperature limits. (iii) Evaluate the variance of the number of

phonons, var(nph) = n2ph nph2.1-21 Reexamine the Gruneisen parameter(T) of a one-dimensional crys-

tal of inert gas atoms with the phonon dispersion relation given by

k = (4K/m)1/2

|sin(ka/2)

|. (i) Show algebraically that (0) =

1

atT= 0. (ii) Show analytically that (T ) = ln 2 asT .(iii) Evaluate numerically (T) for T / from 0 to 1.5 and plot the

results. Here = (4K/m)1/2/kB.

Documents

A Modern Course in the Quantum Theory of Solids Capitulo 1