Patterson, Bailey - Solid State Physics Introduction to theory

64 2 Lattice Vibrations and Thermal Properties

2.2.4 Classical Diatomic Lattices: Optic and Acoustic Modes (B)

So far we have considered only linear lattices in which all atoms are identical. There exist, of course, crystals that have more than one type of atom. In this Section we will discuss the case of a linear lattice with two types of atoms in alternating positions. We will consider only the harmonic approximation with nearestneighbor interactions. By symmetry, the force between each pair of atoms is described by the same spring constant. In the diatomic linear lattice we can think of each unit cell as containing two atoms of differing mass. It is characteristic of crystals with two atoms per unit cell that two types of mode occur. One of these modes is called the acoustic mode. In an acoustic mode, we think of adjacent atoms as vibrating almost in phase. The other mode is called the optic mode. In an optic mode, we think of adjacent atoms as vibrating out of phase. As we shall show, these descriptions of optic and acoustic modes are valid only in the longwavelength limit. In three dimensions we would also have to distinguish between longitudinal and transverse modes. Except for special crystallographic directions, these modes would not have the simple physical interpretation that their names suggest. The longitudinal mode is, usually, the mode with highest frequency for each wave vector in the three optic modes and also in the three acoustic modes.

A picture of the diatomic linear lattice is shown in Fig. 2.4. Atoms of mass m are at x = (2n + 1)a for n = 0, ±1, ±2,..., and atoms of mass M are at x = 2na for n = 0, ± 1,... The displacements from equilibrium of the atoms of mass m are labeled d mn and the displacements from equilibrium of the atoms of mass M are labeled d Mn . The spring constant is k.

From Newton’s laws10

mdnm = k(dnM+1 − dnm ) + k(dnM − dnm ) ,	(2.82a)
and
MdnM = k(dnm − dnM ) + k(dnm−1 − dnM ) .	(2.82b)
It is convenient to define K1 = k/m and K2 = k/M. Then (2.82) can be written
dnm = −K1(2dnm − dnM − dnM+1)	(2.83a)
and
dnM = −K2 (dnM − dnm − dnm−1) .	(2.83b)

10When we discuss lattice vibrations in three dimensions we give a more general technique for handling the case of two atoms per unit cell. Using the dynamical matrix defined in that section (or its one-dimensional analog), it is a worthwhile exercise to obtain (2.87a) and (2.87b).

2.2 One-Dimensional Lattices (B)	65

d nM–1	dnm–1	dnM	dnm	dnM+1
M	m	M	m	M
k		k	k	k
…				…
		a
x = 2(n – 1)a	(2n – 1)a	2na	(2n + 1)a	2(n + 1)a
	Fig. 2.4. The diatomic linear lattice
Consistent with previous work, normal mode solutions of the form
	dnm = Aexp[i(qxnm −ωt)] ,			(2.84a)
and
	dnM = B exp[i(qxnM −ωt)]			(2.84b)

will be sought. Substituting (2.84) into (2.83) and finding the coordinates of the atoms (xn) from Fig. 2.4, we have

−ω2 Aexp{i[q(2n +1)a −ωt]} = −K1(2Aexp{i[q(2n +1)a −ωt]}

−B exp{i[q(2na) −ωt]}

					− B exp{i[q(n +1)2a −ωt]})				,
	−ω2B exp{i[q(2na) −ωt]} = −K2 (2B exp{i[q(2na) −ωt]}
					− Aexp{i[q(2n +1)a −ωt]}
					− Aexp{i[q(2n −1)a −ωt]})
	or
	ω2 A = K (2A − Be−iqa − Be+iqa ) ,								(2.85a)
				1
	and
	ω2B = K2 (2B − Ae−iqa − Ae+iqa ) .								(2.85b)
	Equations (2.85) can be written in the form
		ω2		− 2K1	2K1 cos qa			A	(2.86)
				cos qa	ω2 − 2K			= 0 .
	2K		2			2		B

Equation (2.86) has nontrivial solutions only if the determinant of the coefficient matrix is zero. This yields the two roots

ω12 = (K1 + K2 ) − (K1 + K2 )2 − 4K1K2 sin 2 qa ,

(2.87a)

66 2 Lattice Vibrations and Thermal Properties

and

ω22 = (K1 + K2 ) + (K1 + K2 )2 − 4K1K2 sin 2 qa .

(2.87b)

In (2.87) the symbol √ means the positive square root. In figuring the positive square root, we assume m < M or K1 > K2. As q → 0, we find from (2.87) that

ω1 = 0 and ω2 = 2(K1 + K2 ) . As q → (π/2a) we find from (2.87) that

ω1 = 2K2 and ω2 = 2K1 .

Plots of (2.87) look similar to Fig. 2.5. In Fig. 2.5, ω1 is called the acoustic mode and ω2 is called the optic mode. The reason for naming ω1 and ω2 in this manner will be given later. The first Brillouin zone has −π/2a ≤ q ≤ π/2a. This is only half the size that we had in the monatomic case. The reason for this is readily apparent. In the diatomic case (with the same total number of atoms as in the monatomic case) there are two modes for every q in the first Brillouin zone, whereas in the monatomic case there is only one. For a fixed number of atoms and a fixed number of dimensions, the number of modes is constant.

In fact it can be shown that the diatomic case reduces to the monatomic case when m = M. In this case K1 = K2 = k/m and

ω12 = 2k / m − (2k / m) cos qa = (2k / m)(1− cos qa), ω22 = 2k / m + (2k / m) cos qa = (2k / m)(1+ cos qa).

But note that cos qa for −π/2 < qa < 0 is the same as −cos qa for π/2 < qa < π, so that we can just as well combine ω12 and ω22 to give

ω = (2k / m)(1 − cos qa) = (4k / m)sin 2 (qa / 2)

for −π < qa < π. This is the same as the dispersion relation that we found for the linear lattice.

The reason for the names optic and acoustic modes becomes clear if we examine the motions for small qa. We can write (2.87a) as

ω1	2K1K2	qa	(2.88)
	(K1 + K2 )

for small qa. Substituting (2.88) into (ω2 − 2K1)A + 2K1 cos (qa)B = 0, we find

B

2K

K

2

q2a

2 (K

1

+ K

2

) − 2K

1

qa→0

= −

1

→ +1.

(2.89)

A

2K1 cos qa

Therefore in the long-wavelength limit of the ω1 mode, adjacent atoms vibrate in phase. This means that the mode is an acoustic mode.

		2.2 One-Dimensional Lattices (B) 67
	ω
	2(K1 + K2 )
ω2	2K1	Optic mode
ω1	2K2
		Acoustic mode
–π/2a		π/2a	q

Fig. 2.5. The dispersion relation for the optic and acoustic modes of a diatomic linear lattice

Fig. 2.6. (a) Optic and (b) acoustic modes for qa very small (the long-wavelength limit)

It is instructive to examine the ω1 solution (for small qa) still further:

	ω1	=	2K1K2		qa =	2k 2 /(mM )	qa =	ka	q . (2.90)
			(K1	+ K2 )		k / m + k / M		(m + M ) / 2a

For (2.90), ω1/q = dω/dq, the phase and group velocities are the same, and so there is no dispersion. This is just what we would expect in the long-wavelength limit.

Let us examine the ω2 modes in the qa → 0 limit. It is clear that

ω2	2(K	+ K	2	) +	2K1K2		q2a2	as qa → 0.	(2.91)
2	1				(K1	+ K2 )

68 2 Lattice Vibrations and Thermal Properties

Substituting (2.91) into (ω2 − 2K1)A + 2K1 cos (qa)B = 0 and letting qa = 0, we have

2K2 A + 2K1B = 0 ,

or

mA + MB = 0 .

(2.92)

Equation (2.92) corresponds to the center of mass of adjacent atoms being fixed. Thus in the long-wavelength limit, the atoms in the ω2 mode vibrate with a phase difference of π. Thus the ω2 mode is the optic mode. Suppose we shine electromagnetic radiation of visible frequencies on the crystal. The wavelength of this radiation is much greater than the lattice spacing. Thus, due to the opposite charges on adjacent atoms in a polar crystal (which we assume), the electromagnetic wave would tend to push adjacent atoms in opposite directions just as they move in the long-wavelength limit of a (transverse) optic mode. Hence the electromagnetic waves would interact strongly with the optic modes. Thus we see where the name optic mode came from. The long-wavelength limits of optic and acoustic modes are sketched in Fig. 2.6

In the small qa limit for optic modes by (2.91),

ω2 = 2k(1/ m +1/ M ) .

(2.93)

Electromagnetic waves in ionic crystals are very strongly absorbed at this frequency. Very close to this frequency, there is a frequency called the restrahl frequency where there is a maximum reflection of electromagnetic waves [93].

A curious thing happens in the q → π/2a limit. In this limit there is essentially no distinction between optic and acoustic modes. For acoustic modes as q → π/2a, from (2.86),

(ω2 − 2K1)A = −2K1B cos qa ,

or as qa → π/2,

A	= K1	cos qa	= 0 ,
B
		K1 − K2

so that only M moves. In the same limit ω2 → (2K1)1/2, so by (2.86) 2K2 (cos qa)A + (2K1 − 2K2 )B = 0 ,

or

B	= 2K2	cos qa	= 0 ,
A
		K2 − K1

so that only m moves. The two modes are sketched in Fig. 2.7. Table 2.2 collects some one-dimensional results.

2.2 One-Dimensional Lattices (B)	69

Fig. 2.7. (a) Optic and (b) acoustic modes in the limit qa → π/2

Table 2.2. One-dimensional dispersion relations and density of states†

Model Dispersion relation Density of states

Monatomic

ω = ω0

sin

qa

D(ω)

1

ω02 −ω2

2

Small q

Diatomic

[M > m,

μ = Mm/(M + m)]

ω2

1

1

4

– acoustic

−

−

sin

2 qa

D(ω) constant

2

μ

μ

Mm

2

1

1

4

– optical

ω

+

−

sin

2

D(ω) | q |

−1

μ

μ

2

Mm

qa

† q = wave vector, ω = frequency, a = distance between atoms.

2.2.5 Classical Lattice with Defects (B)

Most of the material in this Section was derived by Lord Rayleigh many years ago. However, we use more modern techniques (Green’s functions). The calculation will be done in one dimension, but the technique can be generalized to three dimensions. Much of the present formulation is due to A. A. Maradudin and coworkers.11

11 See [2.39].

70 2 Lattice Vibrations and Thermal Properties

The modern study of the vibration of a crystal lattice with defects was begun by Lifshitz in about 1942 [2.25] and Schaefer [2.29] has shown experimentally that local modes do exist. Schaefer examined the infrared absorption of H− ions (impurities) in KCl.

Point defects can cause the appearance of localized states. Here we consider lattice vibrations and later (in Sect. 3.2.4) electronic states. Strong as well as weak perturbations can lead to interesting effects. For example, we discuss deep electronic effects in Sect. 11.2. In general, the localized states may be outside the bands and have discrete energies or inside a band with transiently bound resonant levels.

In this Section the word defect is used in a rather specialized manner. The only defects considered will be substitutional atoms with different masses from those of the atoms of the host crystal.

We define an operator p such that (compare (2.36))

pun = ω2 Mun +γ (un+1 − 2un + un+1) ,

(2.94)

where un is the amplitude of vibration of atom n, with mass M and frequency ω. For a perfect lattice (in the harmonic nearest-neighbor approximation with γ = Mωc2 /4 = spring constant),

pun = 0 .

This result readily follows from the material in Sect. 2.2.2. If the crystal has one or more defects, the equations describing the resulting vibrations can always be written in the form

pun = ∑k dnk uk .

(2.95)

For example, if there is a defect atom of mass M 1 at n = 0 and if the spring constants are not changed, then

dnk = (M − M 1)ω 2δn0δk0 .

(2.96)

Equation (2.95) will be solved with the aid of Green’s functions. Green’s functions (Gmn) for this problem are defined by

pGmn = δmn .

(2.97)

To motivate the introduction of the Gmn, it is useful to prove that a solution to (2.95) is given by

un = ∑l,k Gnl dlk uk .

(2.98)

Since p operates on index n in pun, we have

pu

n

= ∑

pG

nl

d

lk

u

k

= ∑

δ

nl

d

lk

u

k

= ∑

k

d

nk

u

k

,

l,k

l,k

and hence (2.98) is a formal solution of (2.95).

2.2 One-Dimensional Lattices (B) 71

The next step is to find an explicit expression for the Gmn. By the arguments of Sect. 2.2.2, we know that (we are supposing that there are N atoms, where N is an even number)

	δmn =	1	∑sN=−01exp	2πis	(m − n) .	(2.99)
		N		N

Since Gmn is determined by the lattice, and since periodic boundary conditions are being used, it should be possible to make a Fourier analysis of Gmn:

	Gmn =	1	∑sN=−01 gs exp	2πis	(m − n) .
		N		N

From the definition of p, we can write

		s			s
	p exp 2πi		(m −n)	= ω 2 M exp 2πi		(m −n)
		N			N

+γ	exp	2πi	s	(m − n −1)	− 2 exp 2πi	s	(m −
			N			N

		s
	+exp 2πi		(m − n +1) .
		N

(2.100)

	(2.101)
n)

To prove that we can find solutions of the form (2.100), we need only substitute (2.100) and (2.99) into (2.97). We obtain

1

N −1

2

s

s

∑s=0 gs ω

M exp 2πi

(m

− n)

+γ exp 2πi

(m − n −1)

N

N

N

s

s

−

2 exp 2πi

(m

− n)

+ exp 2πi

(m − n +1)

(2.102)

N

N

1

N −1

s

=

∑s=0 exp

2πi

(m

− n) .

N

N

Operating on both sides of the resulting equation with

∑m−n exp

2πi

−

(m

− n)s′

,

N

we find

∑s g s {ω2 Mδss′ − 2γδss′[1 − cos(2πs / N )]} = ∑s δss′ .

(2.103)

Thus a G of the form (2.100) has been found provided that

gs =

1

=

1

.

(2.104)

Mω2 − 2γ (1 − cos 2πs / N )

Mω2 − 4γ sin 2

(πs / N )

72 2 Lattice Vibrations and Thermal Properties

By (2.100), Gmn is a function only of m − n, and, further by Problem 2.4, Gmn is a function only of |m − n|. Thus it is convenient to define

Gmn = Gl ,

(2.105)

where l = |m − n| ≥ 0.

It is possible to find a more convenient expression for G. First, define

				Mω2
	cosφ =1 −						.		(2.106)
					2γ
	Then for a perfect lattice
	0 < ω2 ≤ ωc2 =					4γ			,
						M
	so
	1 ≥1 −	Mω	2		≥ −1.				(2.107)
		2γ

Thus when φ is real in (2.106), ω2 is restricted to the range defined by (2.107). With this definition, we can prove that a general expression for the Gn is12

			1		Nφ
G	n	=		cot		cos nφ + sin | n | φ .	(2.108)
			2γ sinφ		2

The problem of a mass defect in a linear chain can now be solved. We define the relative change in mass e by

e = (M − M 1) / M ,

(2.109)

with the defect mass M 1 assumed to be less than M for the most interesting case. Using (2.96) and (2.98), we have

un = Gn Meω2u0 .

(2.110)

Setting n = 0 in (2.110), using (2.108) and (2.106), we have (assuming u0 ≠ 0, this limits us to modes that are not antisymmetric)

1		= 2		γ sinφ		= eMω2		= 2eγ (1 − cosφ) ,
	Gn		cot(Nφ /		2)
or
				sinφ			= e(1 − cosφ) ,
				cot(Nφ / 2)

12For the derivation of (2.108), see the article by Maradudin op cit (and references cited therein).

φ(e) φ(0) + ∂φ

∂e e=0

φ(0) = 2πs / N .

2.2 One-Dimensional Lattices (B) 73

	or
	tan	Nφ	= e tan	φ .	(2.111)
		2
				2

We would like to solve for ω2 as a function of e. This can be found from φ as a function of e by use of (2.111). For small e, we have

e . (2.112)

From (2.111),

(2.113)

Differentiating (2.111), we find

d

Nφ

d

φ

tan

=

e tan

,

de

2

de

2

or

N

sec2

Nφ

∂φ

= tan φ

+

e

sec2 φ ∂φ

,

2

2

∂e

2

2

2 ∂e

or

∂φ

=

tanφ 2

.

(2.114)

∂e

e=0

(N 2) sec2 (Nφ

2)

e=0

Combining (2.112), (2.113), and (2.114), we find

φ

2πs

+

2e

tan πs

.

(2.115)

N

N

N

Therefore, for small e, we can write

2πs

2e

πs

cosφ cos

+

tan

N

N

N

2πs

2e

πs

2πs

2e

tan

πs

= cos

cos

tan

−sin

sin

N

N

N

N

N

N

(2.116)

cos

2πs

−

2e

tan πs sin

2πs

N

N

N

N

= cos

2πs

−

4e

sin2 πs .

N

N

N