Patterson, Bailey - Solid State Physics Introduction to theory

84 2 Lattice Vibrations and Thermal Properties

2.3 Three-Dimensional Lattices

Up to now only one-dimensional lattice vibration problems have been considered. They have the very great advantage of requiring only simple notation. The prolixity of symbols is what makes the three-dimensional problems somewhat more cumbersome. Not too many new ideas come with the added dimensions, but numerous superscripts and subscripts do.

2.3.1 Direct and Reciprocal Lattices and Pertinent Relations (B)

Let (a1, a2, a3) be the primitive translation vectors of the lattice. All points defined by

Rl = l1a1 + l2a2 + l3a3 ,

(2.171)

where (l1, l2, l3,) are integers, define the direct lattice. This vector will often be written as simply l. Let (b1, b2, b3) be three vectors chosen so that

ai b j = δij .

(2.172)

Compare (2.172) to (1.38). The 2π could be inserted in (2.172) and left out of (2.173), which should be compared to (1.44). Except for notation, they are the same. There are two alternative ways of defining the reciprocal lattice. All points described by

Gn = 2π(n1b1 + n2b2 + n3b3 ) ,

(2.173)

where (n1, n2, n3) are integers, define the reciprocal lattice (we will sometimes use K for Gn type vectors). Cyclic boundary conditions are defined on a fundamental parallelepiped of volume

Vf.p.p. = N1a1 (N2a2 × N3a3 ) ,

(2.174)

where N1, N2, N3 are very large integers such that (N1)(N2)(N3) is of the order of Avogadro’s number.

With cyclic boundary conditions, all wave vectors q (generalizations of the old q) in one dimension are given by

q = 2π[(n1 / N1)b1 + (n2 / N2 )b2 + (n3 / N3 )b3 ] .

(2.175)

The q are said to be restricted to a fundamental range when the ni in (2.175) are restricted to the range

−Ni / 2 < ni < Ni / 2 .

(2.176)

We can always add a Gn type vector to a q vector and obtain an equivalent vector. When the q in a fundamental range are modified (if necessary) by this technique to give a complete set of q that are closer to the origin than any other lattice point,

2.3 Three-Dimensional Lattices	85

then the q are said to be in the first Brillouin zone. Any general vector in direct space is given by

r =η1a1 +η2a2 +η3a3 ,

(2.177)

where the ηi are arbitrary real numbers.

Several properties of the quantities defined by (2.171) to (2.177) can now be derived. These properties are results of what can be called crystal mathematics. They are useful for three-dimensional lattice vibrations, the motion of electrons in crystals, and any type of wave motion in a periodic medium. Since most of the results follow either from the one-dimensional calculations or from Fourier series or integrals, they will not be derived here but will be presented as problems (Problem 2.11). However, most of these results are of great importance and are constantly used.

The most important results are summarized below:

1.

1

∑

R

exp(iq R ) = ∑

δ

q,G

.

N1N2 N3

l

G

n

l

n

2.

1

∑

q

exp(iq Rl ) = δ R

,0

N1N2 N3

l

(summed over one Brillouin zone).

3.In the limit as Vf.p.p → ∞, one can replace

∑q by Vf.p.p3. ∫ d3q . (2π)

(2.178)

(2.179)

(2.180)

Whenever we speak of an integral over q space, we have such a limit in mind.

4.

Ωa

∫

exp(iq Rl )d3q = δ R ,0 ,

(2.181)

(2π)3

one Brillouin zone

l

where Ωa = a1 a2 × a3 is the volume of a unit cell.

5.

1

∫

exp[i(G 1

−Gl ) r]d3r = δ 1

.

(2.182)

Ωa Ωa

l

l ,l

6.

1

∫all q space exp[iq (r − r1)]d3q = δ (r − r1) ,

(2.183)

(2π)3

where δ(r − r1) is the Dirac delta function.

7. 1 ∫V exp[i(q − q1) r]d3r =δ (q − q1) . (2.184)

(2π)3 f.p.p.→∞

86 2 Lattice Vibrations and Thermal Properties

2.3.2Quantum-Mechanical Treatment and Classical Calculation of the Dispersion Relation (B)

This Section is similar to Sect. 2.2.6 on one-dimensional lattices but differs in three ways. It is three-dimensional. More than one atom per unit cell is allowed. Also, we indicate that so far as calculating the dispersion relation goes, we may as well stick to the notation of classical calculations. The use of Rl will be dropped in this section, and l will be used instead. It is better not to have subscripts of subscripts of…etc.

Z

b

l

Y

X

Fig. 2.8. Notation for three-dimensional lattices

In Fig. 2.8, l specifies the location of the unit cell and b specifies the location of the atoms in the unit cell (there may be several b for each unit cell).

The actual coordinates of the atoms will be dl,b and

xl,b = dl,b − (l + b)

(2.185)

will be the coordinates that specify the deviation of the position of an atom from equilibrium.

The potential energy function will be V(xl,b). In the equilibrium state, by definition,

( xl,bV )all xl,b =0 = 0 .

(2.186)

Expanding the potential energy in a Taylor series, and neglecting the anharmonic terms, we have

V (xl,b ) = V0	+	1	∑l,b,l1,b1(α,β)	xαlb J αβ1 1	x β1 1 .	(2.187)
	2			lbl b	l b

2.3 Three-Dimensional Lattices	87

In (2.187), xαl,b is the αth component of xl,b. V0 can be chosen to be zero, and this choice then fixes the zero of the potential energy. If plb is the momentum (operator) of the atom located at l + b with mass mb, the Hamiltonian can be written

	1				α=3	1			α	α
H =				∑l (all unit cells),α=1				plb plb
		2					m
				b (all atoms			b				(2.188)
				within a cell)	Jαβ				xα	xβ
+	1			∑α=3,β =3							.
	2			l,b,l1,b1,α=1,β	=1	lbl1b1			lb	l1b1

In (2.188), summing over α or β corresponds to summing over three Cartesian coordinates, and

					2
				∂
	J	αβ	=		V
		lbl1b1		∂xαlb	∂x	β1 1
						l b

. (2.189)

all xlb =0

The Hamiltonian simplifies much as in the one-dimensional case. We make a normal coordinate transformation or a Fourier analysis of the coordinate and momentum variables. The transformation is canonical, and so the new variables obey the same commutation relations as the old:

xl,b =	1	∑q X q1,be−iq l ,	(2.190)
	N
pl,b =	1	∑q Pq1,be+iq l ,	(2.191)
	N

where N = N1N2N3. Since xl,b and pl,b are Hermitian, we must have

X 1	= X 1†	,	(2.192)
−q,b	q,b

and

P1

= P1† .

−q,b

q,b

Substituting (2.190) and (2.191) into (2.188) gives

H =

1

∑l,b

1 1

∑q,q1 Pq1,b

P11

ei(q+q1) l

2

mb N

1

q ,b

e−i(q l+q1 l1) .

+

1

∑

1

1

∑

1

J αβ

X

1α

X 1β

2

l,b,l

,b ,α,β N

q,q

l,b,l1,b1

q,b

q1,b1

(2.193)

(2.194)

88 2 Lattice Vibrations and Thermal Properties

Using (2.178) on the first term of the right-hand side of (2.194) we can write

H =

1

∑q,b

1

Pq1,b Pq1,†b

2

1

mb

−iq1

αβ

+

∑

1

1

∑l

,l1

J

1

1 e

2N

q,q

,b,b

l,b,l

,b

α,β

(l−l1)	e	−i(q+q1)	1α	X	1β	(2.195)
		X q,b			1	1 .
					q ,b

The force between any two atoms in our perfect crystal cannot depend on the position of the atoms but only on the vector separation of the atoms. Therefore, we must have that

J αβ	1	1	= J αβ1 (l − l1) .
l,b,l		,b	b,b

Letting m = (l − l1), defining

Kbb1 (q) = ∑m J bb1 (m)e−iq m ,

and again using (2.178), we find that the Hamiltonian becomes

H = ∑q H q ,

where

H q =	1	∑b	1	Pq1,b Pq1†,b +	1	∑	1	Kαβ1	X q1α,b X 11β†1 .
	2				2
			mb				b,b	b,b	q ,b
						α,β

(2.196)

(2.197)

(2.198a)

(2.198b)

The transformation has used translational symmetry in decoupling terms in the Hamiltonian. The rest of the transformation depends on the crystal structure and is found by straightforward small vibration theory applied to each unit cell. If there are K particles per unit cell, then there are 3K normal modes described by (2.198). Let ωq,p, where p goes from 1 to 3K, represent the eigenfrequencies of the normal modes, and let eq,p,b be the components of the eigenvectors of the normal modes. The quantities eq,p,b allow us to calculate15 the magnitude and direction of vibration of the atom at b in the mode labeled by (q, p). The eigenvectors can be chosen to obey the usual orthogonality relation

∑	b	e*	e	1	= δ	1 .	(2.199)
		qpb		qp b		p,p

It is convenient to allow for the possibility that eqpb is complex due to the fact that all we readily know about Hq is that it is Hermitian. A Hermitian matrix can always be diagonalized by a unitary transformation. A real symmetric matrix can always be diagonalized by a real orthogonal transformation. It can be shown that

15The way to do this is explained later when we discuss the classical calculation of the dispersion relation.

2.3 Three-Dimensional Lattices	89

with only one atom per unit cell the polarization vectors eqpb are real. We can choose e−q,p,b = e*q,p,b in more general cases.

Once the eigenvectors are known, we can make a normal coordinate transformation and hence diagonalize the Hamiltonian [99]:

			X q11, p = ∑b mb eqpb X qb1 .								(2.200)
The momentum	P11 , which is canonically conjugate to (2.200), is
	q, p
			Pq11, p = ∑b (1/			mb )eq*pb Pqb1 .					(2.201)
Equations (2.200) and (2.201) can be inverted by use of the closure notation
			∑ p eα*		eβ		1 = δαβ	δbb1 .			(2.202)
			qpb qpb
Finally, define
	a	q, p	=1/ 2 ω	q, p	P11		− i (ω	q, p	/ 2	) X 11† ,	(2.203)
					q, p					q, p

and a similar expression for a†q,p. In the same manner as was done in the onedimensional case, we can show that

[aq, p , aq†, p ] = δqq1δ pp1 ,

(2.204)

and that the other commutators vanish. Therefore the as are boson annihilation operators, and the a† are boson creation operators. In this second quantization notation, the Hamiltonian reduces to a set of decoupled harmonic oscillators:

H = ∑q, p ωq, p (aq†, p aq, p +	1	) .	(2.205)
	2

By (2.205) we have seen that the Hamiltonian can be represented by 3NK decoupled harmonic oscillators. This decomposition has been shown to be formally possible within the context of quantum mechanics. However, the only thing that we do not know is the dispersion relationship that gives ω as a function of q for each p. The dispersion relation is the same in quantum mechanics and classical mechanics because the calculation is the same. Hence, we may as well stay with classical mechanics to calculate the dispersion relation (except for estimating the forces), as this will generally keep us in a simpler notation. In addition, we do not know what the potential V is and hence the J and K ((2.189), (2.197)) are unknown also.

This last fact emphasizes what we mean when we say we have obtained a formal solution to the lattice-vibration problem. In actual practice the calculation of the dispersion relation would be somewhat cruder than the above might lead one to suspect. We gave some references to actual calculations in the introduction to Sect. 2.2. One approach to the problem might be to imagine the various atoms

90 2 Lattice Vibrations and Thermal Properties

hooked together by springs. We would try to choose the spring constants so that the elastic constants, sound velocity, and the specific heat were given correctly. Perhaps not all the spring constants would be determined by this method. We might like to try to select the rest so that they gave a dispersion relation that agreed with the dispersion relation provided by neutron diffraction data (if available). The details of such a program would vary from solid to solid.

Let us briefly indicate how we would calculate the dispersion relation for a crystal lattice if we were interested in doing it for an actual case. We suppose we have some combination of model, experiment, and general principles so the

J αβ 1 1

l,b,l ,b

can be determined. We would start with the Hamiltonian (2.188) except that we would have in mind staying with classical mechanics:

H =

1

∑

α=3

1

( pα

)2 +

1

∑

α=3,β =3

J αβ

xα

x

β

.(2.206)

1

1

1

1 1

2

l,b,α=1

l,b

2

l,b,l

1

β =1

l,b,l

lb

mb

,b ,α=1,

,b

l b

We would use the known symmetry in J:

J αβ	= J αβ	, J αβ	= J αβ	.	(2.207)
l,b,l1 ,b1	l1 ,b1 ,l,b	l,b,l1 ,b1	(l −l1 ),b,b1

It is also possible to show by translational symmetry (similarly to the way (2.33) was derived) that

∑l1,b1 J	αβ 1	1 = 0 .	(2.208)
	l,b,l	,b

Other restrictions follow from the rotational symmetry of the crystal.16

The equations of motion of the lattice are readily obtained from the Hamiltonian in the usual way. They are

mb xαlb = −∑l1,b1,β	J	αβ 1	1	x	β1	1 .	(2.209)
		l,b,l	,b		l	,b

If we seek normal mode solutions of the form (whose real part corresponds to the physical solutions)17

xαl,b = 1 xαb e−iωt+q l ,	(2.210)
mb

16Maradudin et. al. [2.26].

17Note that this substitution assumes the results of Bloch’s theorem as discussed after (2.39).

2.3 Three-Dimensional Lattices	91

we find (using the periodicity of the lattice) that the equations of motion reduce to

		ω2 xαb = ∑b1,β M αβ			1		x β1 ,			(2.211)
			q,b,b				b
where
			M αβ
			q,b,b1
is called the dynamical matrix and is defined by
M αβ 1	=	1	∑(l−l1) J αβ			1	1 e−iq (l−l	1	) .	(2.212)
		mbm 1
q,b,b			(l−l				)b,b
		b
These equations have nontrivial solutions provided that
	det(M αβ 1		−ω2δαβδ			1 ) = 0 .				(2.213)
		q,b,b		b,b

If there are K atoms per unit cell, the determinant condition yields 3K values of ω2 for each q. These correspond to the 3K branches of the dispersion relation. There will always be three branches for which ω = 0 if q = 0. These branches are called the acoustic modes. Higher branches, if present, are called the optic modes.

Suppose we let the solutions of the determinantal condition be defined by ωp2(q), where p = 1 to 3K. Then we can define the polarization vectors by

ω 2p (q)eαq, p,b = ∑b1,β M αβ 1 eqβ, p,b .	(2.214)
q,b,b

It is seen that these polarization vectors are just the eigenvectors. In evaluating the determinantal equation, it will probably save time to make full use of the symmetry properties of J via M. The physical meaning of complex polarization vectors is obtained when they are substituted for xαb and then the resulting real part of xαl,b is calculated.

The central problem in lattice-vibration dynamics is to determine the dispersion relation. As we have seen, this is a purely classical calculation. Once the dispersion relation is known (and it never is fully known exactly – either from calculation or experiment), quantum mechanics can be used in the formalism already developed (see, for example, (2.205) and preceding equations).

2.3.3 The Debye Theory of Specific Heat (B)

In this Section an exact expression for the specific heat will be written down. This expression will then be approximated by something that can actually be evaluated. The method of approximation used is called the Debye approximation. Note that in three dimensions (unlike one dimension), the form of the dispersion relation and hence the density of states is not exactly known [2.11]. Since the Debye

92 2 Lattice Vibrations and Thermal Properties

model works so well, for many years after it was formulated nobody tried very hard to do better. Actually, it is always a surprise that the approximation does work well because the assumptions, on first glance, do not appear to be completely reasonable. Before Debye’s work, Einstein showed (see Problem 2.24) that a simple model in which each mode had the same frequency, led with quantum mechanics to a specific heat that vanished at absolute zero. However, the Einstein model predicted an exponential temperature decrease at low temperatures rather than the correct T 3 dependence.

The average number of phonons in mode (q, p) is

	nq, p	=		1		.	(2.215)
			exp(	ωq, p / kT ) −1

The average energy per mode is

ωq, p nq, p ,

so that the thermodynamic average energy is (neglecting a constant zero-point correction, cf. (2.77))

	U = ∑q, p	ωq, p		.	(2.216)
		exp( ωq, p / kT ) −1

The specific heat at constant volume is then given by

			∂U		1	∑q, p	( ωq, p )2 exp( ωq, p / kT )
	Cv	=		=				.	(2.217)
					kT 2		[exp( ωq, p / kT ) −1]2
			∂T v

Incidentally, when we say we are differentiating at constant volume it may not be in the least evident where there could be any volume dependence. However, the ωq,p may well depend on the volume. Since we are interested only in a crystal with a fixed volume, this effect is not relevant. The student may object that this is not realistic as there is a thermal expansion of the solids. It would not be consistent to include anything about thermal expansion here. Thermal expansion is due to the anharmonic terms in the potential and we are consistently neglecting these. Furthermore, the Debye theory works fairly well in its present form without refinements.

The Debye model is a model based on the exact expression (2.217) in which the sum is evaluated by replacing it by an integral in which there is a density of states. Let the total density of states D(ω) be represented by

D(ω) = ∑p D p (ω) ,

(2.218)

where Dp(ω) is the number of modes of type p per unit frequency at frequency ω. The Debye approximation consists in assuming that the lattice vibrates as if it were an elastic continuum. This should work at low temperatures because at low temperatures only long-wavelength (low q) acoustic modes should be important. At

2.3 Three-Dimensional Lattices	93

high temperatures the cutoff procedure that we will introduce for D(ω) will assure that we get the results of the classical equipartition theorem whether or not we use the elastic continuum model. We choose the cutoff frequency so that we have only 3NK (where N is the number of unit cells and K is the number of atoms per unit cell) distinct continuum frequencies corresponding to the 3NK normal modes. The details of choosing this cutoff frequency will be discussed in more detail shortly.

In a box with length Lx, width Ly, and height Lz, classical elastic isotropic continuum waves have frequencies given by

2

2

2

2

2

=π

c

2 k j

+

l j

+

m j

ω j

,

(2.219)

L2

L2

L2

x

y

z

where c is the velocity of the wave (it may differ for different types of waves), and (kj, lj and mj) are positive integers.

We can use the dispersion relation given by (2.219) to derive the density of states Dp(ω).18 For this purpose, it is convenient to define an ω space with base vectors

eˆ1

πc

ˆ

eˆ2

πc

ˆ

eˆ3

πc

ˆ

=

i ,

=

j ,

and

=

k .

(2.220)

Lx

Ly

Lz

Note that

ω2

= k 2eˆ2 + l 2eˆ2

+ m2eˆ2 .

(2.221)

j

j

1

j

2

j

3

Since the (ki, li, mi) are positive integers, for each state ωj, there is an associated cell in ω space with volume

eˆ	(eˆ	2	×eˆ	3	) =	(πc)3	.	(2.222)
1						Lx Ly Lz

The volume of the crystals is V = LxLyLz, so that the number of states per unit volume of ω space is V/(πc)3. If n is the number of states in a sphere of radius ω in ω space, then

	n =	1	4π	ω3	V	.
		8	3		(πc)3

The factor 1/8 enters because only positive kj, lj, and mj are allowed. Simplifying, we obtain

	n =	π	ω3	V	.	(2.223)
		6		(πc)3

18We will later introduce more general ways of deducing the density of states from the dispersion relation, see (2.258).