Marder M.P. Condensed Matter Physics

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Vibrations of a Classical Lattice

343

Figure 13.2. The dispersion relation for a one-dimensional lattice with nearest-neighbor

interactions calculated in Eq. (13.6).

this dispersion relation is plotted in Figure 13.2.

There are two characteristic features of this simple solution that occur quite

generally:

For small values of k, ω is proportional to the absolute value of k. This

is no accident; it is nothing but the recovery of linear elasticity for waves

much longer than the interatomic spacing. The slope of the dispersion curve,

duj/dk = αχ/Χ/Μ = c near the origin is the sound speed. The branch of so-

lutions whose frequency vanishes as k vanishes results in general from the

symmetry requiring the energy of the crystal to remain unchanged when all

ions are displaced by an identical amount, as shown in Eq. (13.19). Very

long wavelength displacements of the ions are indistinguishable from uniform

translation on short length scales, and therefore they have low energy and fre-

quency. Excitations of this type are often referred to as Goldstone modes,

after Goldstone (1963); see Section

26.3.3.

The solution repeats periodically as a function of k , with period 2π/α. This

occurs because the phonon problem concerns waves moving in a medium that

is periodic with period a, and occurs for exactly the same reason that energies

of electrons are periodic functions of wave vector k. Just as for electrons,

the region

[—π/α,

π/α] is called the Brillouin zone. This symmetry will be

discussed in greater generality as Eq. (13.21).

One-Dimensional Chain with Basis. Now consider a one-dimensional chain,

with lattice parameter a, in which atoms of two different masses, M\ and Mi, al-

ternate. Assuming again that each atom interacts only with its nearest neighbors

(Figure 13.3), one has

344

Chapter 13. Phonons

Figure 13.3. Ions of alternating masses M\ and M

interacting with nearest neighbors.

M\Ü\ :

ül

■

X(u

-2u[ +U

l-\<

2 .

3C(,

.'+·

■

+ u\

(13.7a)

(13.7b)

-u

e\é

kÎa

= %{e

-2e

lka

Assuming

solution (13.8a)

of the form

ika

,kla

= %{e

lka

- 2e

+ e,)e

Jkla

(13.8b)

LÜ

M\+M

±\/M} +

2M\M

COS

ka + M\

M\M

Set the determinant of the

system (13.8) to zero, and

solve the resulting

polynomial for ω.

(13.9)

The two solutions of Eq. (13.9) are two branches of the phonon dispersion

relation and are depicted in Figure 13.5. One of the branches vanishes linearly

near k = 0 and is the acoustic branch, as it corresponds to ordinary sound. The

second branch is restricted to higher frequencies and is the optical branch, since in

solids these phonons are characteristically excited by light. For small k, the two

branches take the form

2(Mi+M

)

2X(Mi+M

)

M\M

ka, e\ = 1; £2=1 +ika/2,

(13.10a)

ei =M

; e

= -M\(\ + ika/2). (13.10b)

Equation (13.10) shows as illustrated in Figure 13.4, that for the acoustic mode,

atoms within the unit cell move essentially in unison, while for the optical mode,

atoms within the unit cell vibrate out of phase.

Vibrations of a Classical Lattice

345

Figure 13.4. Configurations of atoms in optical and acoustic modes. In the acoustic mode,

atoms within a unit cell move in concert, while in the optical mode they vibrate against one

another in opposite directions.

2.0

1.5

^ ,.0

■Ai

0.5

0.0

-7I-/2 7Γ/2

Figure 13.5. Vibrational frequencies of a chain with two alternating masses, as a function

of the wave number k of oscillation, calculated in Eq. (13.9). The solution is depicted by

measuring ω in units of

y/%/M\

and setting Mi = 3Aii.

346 Chapter 13. Phonons

13.2.2 Classical Vibrations in Three Dimensions

Crystalline lattices are not one-dimensional chains of ions connected to nearest

neighbors by springs, nor even three-dimensional lattices of masses connected by

springs. Nevertheless, viewing them this way is essentially correct when one re-

stricts attention to small deviations of ions from their equilibrium locations. When

ions move only a small amount from equilibrium, the restoring forces on them are

linear, and the techniques developed for the one-dimensional chain continue to

apply-

Here is why linear restoring forces arise so generally. Suppose one knows the

location in the ground state of a collection of ions, 1 . . . N, and let u

. . . u

describe the vector displacement of these ions from their equilibrium locations

. . . R

. When ions move, the energy of the crystal goes up; take the energy

functional to be

E{u\u

..u

)..

(13.11)

Instead of making guesses about the form of the energy functional, as in Section

11.8,

assume all the variables

are small, and develop Eq. (13.11) in a Taylor

expansion in powers of the variables u.

The energy becomes

ßC 1

£ = E

+ ^

TrT

Τ^

ί>Φ«/?

Α + · · · · a and /3 are Cartesian indices running over

^—' ßu 2 ^"~' basis vectors of three-dimensional space.

Ot aß

11'

(13.12)

The first term in Eq. (13.12) is the cohesive energy, calculated with such effort in

Chapter 11, but which for present purposes is a dull constant and will callously

be neglected. Because the lowest energy state is a minimum as a function of ion

locations, the linear term in the expansion in u

must vanish, and the first nonzero

contribution must be quadratic. That is,

aß

The 3x3 matrix Φ" comes immediately from Taylor expansion of the energy and

Φ'

/3 — . ,/ Φ" is always a symmetric matrix. (13.14)

öu'

du'ß

Periodic Boundary Conditions. The formal study of electron motion employed

the simplification of periodic boundary conditions. The same assumption also sim-

plifies calculations concerning lattice vibrations. Proceeding along the direction of

any of the three primitive vectors for the Bravais lattice, every physical quantity,

including the displacements u of the ions, is assumed to repeat after some number

of cells is cells is passed. All points in the equilibrium crystal are equivalent; the

Vibrations

of a Classical Lattice

347

Figure 13.6. Along the axis of a cubic crys-

tal,

there is a longitudinal mode in the direc-

tion of wave propagation, k, and two trans-

verse modes perpendicular to it.

crystal has no surfaces or boundaries. A first consequence of this simplifying as-

sumption is that

φ[[β

depends only upon R

—

because if one displaces just two

ions u

and

leaving all others in equilibrium locations, the resulting energy can

depend only upon their relative locations.

Equation of Motion. The dynamical equation for phonons follows from Eq. (13.13).

The force on ion / is —d£/du

, which leads to

MU = — y Φ U . Subscripts are being suppressed; each Φ" is (13.15)

,, a 3 x 3 matrix.

Much of the theory of phonons can be developed without reference to the par-

ticular form or value that the matrices Φ" take, so the question of how to calculate

them will be left until later. In fact, it would be hard to proceed in any other way,

because in order to conceive experimental probes of

one has to know what

their consequences will be. One should not be misled by the simplifications in the

starting examples;

will usually be nonzero when R

and R

are far from nearest

neighbors, particularly in insulators, where the ions carry charges interacting with

each other by Coulomb forces.

13.2.3 Normal Modes

Equation

( 13.15)

is solved by plane waves, but as they travel in a three-dimensional

crystal, one must keep track of information about their polarization. So take

have the form

Sr = ?g

—

"*". e is a unit vector that will describe the polar- (13.16)

ization of the vibration.

Substituting Eq. (13.16) into Eq. (13.15) gives

Μ^

? = ]ΓΦ

ν*-("')ε (13.17a)

_ _,

T./Si Si'\ ni Φ(ϊ) does not depend upon /

= Φ(£)β, With Φ(&) = 22

e Φ

because a11

Physical quantities 13.17b)

,/ depend only upon W

—

If it were not for the three-dimensional nature of the problem, Eq. (13.17)

would constitute an explicit solution; as things are, the solution is near at hand.

What remains is a matrix equation for the polarization vector e. The matrix Φ(&)

is real and symmetric, and therefore has three orthogonal eigenvectors for every k.

348

Chapter 13. Phonons

Let e^, v =

... 3 be the three unit vectors that diagonalize Φ(ϋ), and let Φ„ be

the corresponding eigenvalues. It follows immediately that

kv M

Roughly speaking, the three polarization vectors e^ comprise one longitudinal

mode for which e points along k, and two transverse modes for which e is perpen-

dicular to k (see Figure 13.6). This statement is not absolutely correct. It would

only be true if

one

had an isotropie crystal, invariant under all rotations, but no such

crystal exists. The closest that one can come is in a cubic crystal. When k points

along a crystalline axis, then by symmetry one polarization vector points along k,

and the two remaining polarization vectors correspond to degenerate energies, and

point along perpendicular axes.

There are some important symmetries that Φ" must always obey. The energy

of the crystal cannot change if all ions are simultaneously displaced by a single

vector. Therefore,

J2$

'=0.

(13.19)

^Φ(Ι = 0)=Ο (13.20)

Because of

the

assumption of periodic boundary conditions, the allowed values

of k are given exactly as in Eq. (6.7). Furthermore, notice that

Φ(% + Κ) = Φ(&), Look at Eq. (13.17), and notice that addition (13.21)

of any reciprocal lattice vector turns instantly

into a multiple of 2πί.

where K is any reciprocal lattice vector. Therefore, one may always take k to lie in

the first Brillouin zone. Lattice vibrations, just like electrons, are waves that travel

in the perfectly periodic potential described by ion locations R

. Exactly the same

k states can be used to classify the two sets of vibrations.

One difference between the electron problem and the phonon problem has to

do with the numbers of modes. There is no upper limit to the number of distinct

one-particle electronic states that can inhabit a lattice, and there is no limit to the

number of energy bands the states can fill. In contrast, a single Brillouin zone com-

pletely exhausts all the phonon states of a Bravais lattice. The reason for the

dif-

ference is that electron wave functions are defined everywhere in space, and values

of the wave function in between lattice sites are physically important. The phonon

states are completely described by their values at the lattice sites R, and any two

functions that are the same on these lattice sites are physically indistinguishable no

matter how they may wiggle in between.

13.2.4 Lattice with a Basis

Constructing a lattice with a basis in Section 13.2.1 led to a phonon spectrum

with more than one branch, including low-frequency acoustic modes and opti-

cal phonons with high frequencies at small wave vector. The same phenomenon

Vibrations

of a Classical Lattice 349

persists in three dimensions. Adding new atoms to a unit cell adds new degrees of

freedom to the lattice, and one must consider a correspondingly greater number of

normal modes to describe them. For example, in three dimensions with four atoms

per unit cell, one has 3x4 normal modes for every value of k.

The formal way to calculate these modes is to write

= -Σ

φων

/ν

(

l'n'

where the superscripts n and n' label the different atoms comprising the basis in

each unit cell. Proceeding on the path that led from Eq. (13.15) to Eq. (13.17), one

obtains

" = e"e

iu;

', (13.23)

Μ„ω

? = Σ Φ

ηη

'(k)?'. (13.24)

In the form Eq. (13.24), the practical mathematical problem that one has to solve

is difficult to make out. One way to describe the problem better is to define a new

index p that ranges over all degrees of freedom in the unit cell. With four atoms per

unit cell, p would range from

to 12. The first three values would correspond to

the x, y, and z coordinates of the first atom, the next three values would correspond

to the coordinates of the next atom, and so on. Using this notation,

= y^ Φρ

ι(%)<Ε

For

example, Φ

describes the force in the (13.25)

—' x direction on atom 1 when atom 2 moves in

P the y direction.

Example: Diamond Lattice. Consider atoms sitting on a diamond lattice (Figure

2.6),

interacting through central forces with their four nearest neighbors. The task

is to find the phonon dispersion relations. This calculation gives acceptable results

despite the fact that the diamond structure is never the ground state of particles

interacting with central forces, because only small deviations from equilibrium are

permitted.

To begin, it is necessary to record the potentials that result from central forces.

Suppose one has a collection of atoms sitting on a Bravais lattice R

with basis v",

take

= R

, and have particles interact with a potential of the form

U = - V^

4>nn'{\Ü

" — U

'"'

-R

'"']).

The subscript™'on fallows different breeds

2 ~^, of atoms to interact with different force laws.

lui n

(13.26)

Expanding to quadratic order in the small deviations u and using the fact that terms

linear in u must vanish, one finds that

~\ Σ

["''"-"''"

']*"''

'"'[«'"-3

(13.27)

Inl'n'

350

Chapter 13. Phonons

where the components of the 3x3 matrix

are

S" = T. Φηη·§ϊ\%-ΐ>ΐη-ϊ>ι>

' The subscripts on frange over

(13.28)

ra ">r—t\ n ;y, and z components.

^?[^W-^^(r)]

%^(r)}L

„„,. (13.29)

The condition that the crystal be in equilibrium when all if s vanish is

/ ~^nn'{

)V=Rl"-R

^' ™

S condition does not

demand that φ' van- (13.30)

~—f r

ish. Primes on φ mean derivatives.

Differentiating Eq. (13.27) to find the force on the atom at In gives

/„/ν

£ f"""«"^^, -«W<W)· ('3-31)

Deriving this relation is a slightly unpleasant two-line exercise in taking deriva-

tives of sums with large numbers of indices. It is best to write the full sum

,„ , ,. out explicitly and use relations like -2*- «'" = δτ,δα„δά

il im a fi

v J

In the present case, where there is only one potential function φ, and only the

four nearest neighbors at distance d are being considered, Eq. (13.29) really only

involves two numbers, which are

/1-/2 = (/>"(</)-V(d) (13.32)

and

Η = \φ\ά) (13.33)

^f'aß" = "^2~[/l —h]+S

ßf2 ^_ a„_

/„/

■

If «'" and «''"'are nearest (13.34)

l~ r—R"—R " neighbors, and zero otherwise.

So f depends only a little on details of the potential, and it mainly acquires its struc-

ture from the geometry of the lattice. Once the matrix f^L" has been computed,

Φ can be found from Eq. (13.31). Taking R

to be the origin, one needs to find Φ

only for / = 0, for n ranging over the basis (which in this case has two members,

the origin and | [111]) and for Ι'η' ranging over the nearest neighbors of each of

these two sites, a total of eight choices for nln'l''. Next one computes

Φ""'(k)

= ^

'"'

") There are not too many terms in the sum, be- (13.35)

,/ cause only nearest neighbors have nonzero Φ

"' "

and has four 3x3 matrices for the four combinations of

and n'. Finally, one forms

the 6 x 6 matrix Φ according to Eq. (13.25), and from its eigenvalues deduces the

vibrational frequencies of the phonons. The construction of the various matrices

can be performed relatively painlessly with the use of symbolic algebra, and the

final 6x6 matrix quickly diagonalized with any suitable numerical routine. Results

of this procedure, roughly appropriate for silicon, appear in Figure 13.7.

Vibrations

of a Quantum-Mechanical Lattice 351

Wave number

Figure 13.7. Calculation ofphonon frequencies from

Eq.

(13.35). Using/i/M= (2π)

·78

THz

and fijM = (2π)

·4

THz

the

results

give a

reasonable approximation

to the

phonon

structure of

silicon,

measured in

THz.

Results of more accurate computations together with

experimental data appear in Figure 13.18. Notation for Brillouin zone locations is given in

Figure 7.8.

13.3 Vibrations of a Quantum-Mechanical Lattice

The discussion of lattice vibrations has proceeded until now as if the world obeyed

classical mechanics. The reason for this negligence is that classical and quantum

mechanics find themselves in nearly complete agreement for harmonic oscillators.

The vibrational frequencies of a quantum-mechanical lattice are the same as its

classical counterpart. However, whereas in the classical case a mode may have

arbitrary amplitude, in the quantum mechanical case the modes are allowed only to

have discrete amplitudes. The energy of mode k, polarization v, is permitted only

to take values

/^> + i); (13.36)

where n > 0 is an integer. When n = 1, one says that a single phonon of wavenum-

ber k has been excited. Because an arbitrary number of phonons n may occupy a

given k state, one can interpret the excitations of a lattice as particles obeying Bose

statistics.

Despite the relatively minor formal differences between the results of the clas-

sical and quantum-mechanical analyses, there are several areas in which the phys-

ical influence of quantum mechanics is crucial. For example, at low temperatures,

the amplitude of lattice vibrations falls below the threshold of possibilities de-

scribed by Eq. (13.36), and specific heat differs substantially from classical ex-

352

Chapter 13. Phonons

pectations. For this and other reasons, a brief review of the quantum mechanical

theory now becomes necessary.

For a single harmonic oscillator described by the Hamiltonian

p2 i

•K= + -MUJ

(13.37)

2M 2

one defines raising and lowering operators

Μω

/ 1

/ R — i\l P This is the "raising" or "creation" operator, (13.38a)

V 2/z y 2hMüJ which when it acts upon the ground state of

the oscillator populates it with one excited state.

Μω - / 1

â= \ R-\-i\ P. And this is the "lowering" or "destruction" (13.38b)

V 2h y 2%Muj operator that drags the oscillator down the lad-

der of excited states each time it acts.

The Hamiltonian expressed in terms of these operators simplifies to

"K = Ηω ( ff a + |

= hxo ( fi + £

The number operator n = a'â measures the (13.39)

\ / V / degree of excitation of the oscillator, or equiv-

alently the number of phonons occupying it.

and the original particle position operator is

^v^

(

(13

40)

Second Quantization of Phonons. To treat motion of the atoms of a solid to

quadratic order, write the Hamiltonian

. ,.,

K=\] h | \_] " ^ Û

■

. . . The operators û

describe the deviation of an (13.41)

I *■*" ff/ ion from its equilibrium location R

, assumed

for the moment to belong to a monatomic Bra-

vais lattice.

In order to find all eigenvalues of Eq. (13.41), define the analog of

Eq.

(13.38),

1 v^ _;ï.s/^ r IMLÛI,, „, . / 1

ß7

Σ *-**%

■

W^û'

J—i—P

]

(13.42a)

/-^/ kv \\ Oft \\ OtiMi,^

VÜf^~ ~

kv L

V 2» "" ''^2tiMu^

1 "

4 =4=Σ

kv . Γ\1 ί-~ι

„ _ _ , e

ikä

'er

/Λ/ ^ kv

VNÎ

2h V

2hMusr

y ^-" y

"""^fci/

(13.42b)

A brief calculation (Problem 5) inverts these relations and expresses û

in terms

of the creation and annihilation operators as

= ^=Y far

'+ul

iMl

)

with Û

= J ^— er âr (13.43a)

kv ' *"

and