Patterson, Bailey - Solid State Physics Introduction to theory

34 1 Crystal Binding and Structure

P can be called the (relative) scattering amplitude.

	γ′		kf
					γ′ = π/2 − θ
	γ			k
		θ
					γ = π/2 + θ
			θ
ki					kf
		θ
	θ
		rs		θ

d

d

ki θ

Fig. 1.26. Schematic for simpler discussion of scattering

It is useful to follow up on the comment made above and give a simpler discussion of scattering. Looking at Fig. 1.26, we see the path difference between the two beams is 2d = 2 rs sin θ. So the phase difference is

ϕ = 4λπ rs sinθ = 2krs sinθ ,

since |kf| = |ki| = k. Note also
	π			π			= 2krs sinθ ,
k rs = krs cos		2	−θ	− cos	2	+θ

which is the phase difference. We obtain for a continuous distribution of scatterers

P = ∫ exp(−i k rs )ρ(rs )dV ,

(1.34)

where we have assumed each scatterer scatters proportionally to its density.

We assume now the general case of a lattice with a basis of atoms, each atom with a distribution of electrons. The lattice points are located at

Rpmn = pa1 + ma2 + na3 ,

(1.35)

where p, m and n are integers and a1, a2, a3 are the fundamental translation vectors of the lattice. For each Rpmn there will be a basis at

R j = a ja1 + b ja2 + c ja3 ,

(1.36)

where j = 1 to q for q atoms per unit cell and aj, bj, cj are numbers that are generally not integers. Starting at Rj we can assume the electrons are located at rs so the electron locations are specified by

r = Rpmn + R j + rs ,

(1.37)

1.2 Group Theory and Crystallography 35

r rs

Rj

Rpmn

Fig. 1.27. Vector diagram of electron positions for X-ray scattering

as shown in Fig. 1.27. Relative to Rj then the electron’s position is

rs = r − Rpmn − R j .

If we let ρj (r) be the density of electrons of atom j then the total density of electrons is

ρ(r) = ∑ pmn ∑qj =1 ρ j (r − R j − Rpmn ) .	(1.38)
By a generalization of (1.34) we can write the scattering amplitude as
P = ∑pmn ∑ j ∫ ρ j (r − R j − Rpmn )e−i k r dV.	(1.39)

Making a dummy change of integration variable and using (1.37) (dropping s on rs) we write

P = ∑ pmn e	−i	k Rpmn	∑ j e	−i	k R j	∫ ρ j (r)e	i k r
								dV .

For N 3 unit cells the lattice factor separates out and we will show below that

3	k
∑pmn exp(−i k Rpmn ) = N δGhkl		,
where as defined below, the G are reciprocal lattice vectors. So we find
P = N 3δGhklk Shkl ,		(1.40)
where Shkl is the structure factor defined by
Shkl = ∑ j e−iGhkl Rj f jhkl	,	(1.41)
and fj is the atomic form factor defined by
f jhkl = ∫ ρ j (r)e−iGhkl r dV .		(1.42)

Since nuclei do not interact appreciably with X-rays, ρj (r) is only determined by the density of electrons as we have assumed. Equation (1.42) can be further simplified for ρj (r) representing a spherical distribution of electrons and can be worked out if its functional form is known, such as ρj (r) = (constant) exp(–λr).

36 1 Crystal Binding and Structure

This is the general case. Let us work out the special case of a lattice of point scatterers where fj = 1 and Rj = 0. For this case, as in a three-dimension diffraction grating (crystal lattice), it is useful to introduce the concept of a reciprocal lattice. This concept will be used throughout the book in many different contexts. The basis vectors bj for the reciprocal lattice are defined by the set of equations

ai b j = δij ,

(1.43)

where i, j → 1 to 3 and δij is the Kronecker delta. The reciprocal lattice is then defined by

Ghkl = 2π(hb1 + kb2 + lb3) ,							(1.44)
where h, k, l are integers.13 As an aside, we mention that we can show that
b	=	1	a	2	× a	3	(1.45)
1		Ω

plus cyclic changes where Ω = a1 (a2 × a3) is the volume of a unit cell in direct space. It is then easy to show that the volume of a unit cell in reciprocal space is

Ω	= b (b	× b ) =	1	.	(1.46)
RL	1 2	3	Ω
The vectors b1, b2, and b3 span three-dimensional space, so					k can be expanded in
terms of them,
k = 2π(hb1 + kb2 + lb3) ,					(1.47)
where now h, k, l are not necessarily integers. Due to (1.43) we can write
Rpmn	k = 2π( ph + mk + ln) ,				(1.48)

with p, m, n still being integers. Using (1.32) with rs = Rpmn, (1.48), and assuming a lattice of N3 atoms, the structure factor can be written:

P = ∑Np=−01 e−i2πph ∑mN=−10 e−i2πmk ∑nN=−01 e−i2πnl .

This can be evaluated by the law of geometric progressions. We find:

| P |2

2

2

2

= sin

πhN sin

πkN sin

πlN .

sin

2

πh

sin

2

πk

sin

2

πl

(1.49)

(1.50)

For a real lattice N is very large, so we assume N → ∞ and then if h, k, l are not integers |P| is negligible. If they are integers, each factor is N2 so

| P |2 = N 6δhintegers,k,l .

(1.51)

13Alternatively, as is often done, we could include a 2π in (1.43) and remove the multiplicative factor on the right-hand side of (1.44).

1.2 Group Theory and Crystallography 37

Thus for a lattice of point ions then, the diffraction peaks occur for

k = k f − ki = Ghkl = 2π(hb1 + kb2 + lb3) ,

(1.52)

where h, k, and l are now integers.

ki		kf
θ	θ		k = G′
	θ

Fig. 1.28. Wave vector–reciprocal lattice relation for diffraction peaks

Thus the X-ray diffraction peaks directly determine the reciprocal lattice that in turn determines the direct lattice. For diffraction peaks (1.51) is valid. Let

Ghkl = nG′h′k′l′, where now h′, k′, l′ are Miller indices and G′h′k′l′ is the shortest vector in the direction of Ghkl. Ghkl is perpendicular to (h, k, l) plane, and we show in Problem 1.10 that the distance between adjacent such planes is

dhkl =	2π	.				(1.53)
	Gh′′k′l′
Thus
| G |= 2k sinθ = n | G′			|= n	2π	,	(1.54)
	h′k′l′			dhkl
so since k = 2π/λ,
nλ = 2dhkl sinθ ,						(1.55)

which is Bragg’s equation.

So far our discussion has assumed a rigid fixed lattice. The effect of temperature on the lattice can be described by the Debye–Waller factor. We state some results but do not derive them as they involve lattice-vibration concepts discussed in Chap. 2.14 The results for intensity are:

I = IT =0e−2W ,

(1.56)

where D(T) = e–2W, and W is known as the Debye–Waller factor. If K = k – k′, where |k| = |k′| are the incident and scattered wave vectors of the X-rays, and if

14 See, e.g., Ghatak and Kothari [1.9].

38 1 Crystal Binding and Structure

e(q, j) is the polarization vector of the phonons (see Chap. 2) in the mode j with wave vector q, then one can show14, 15 that the Debye–Waller factor is

	2W =	2	∑q, j	K e(q, j)	coth	ω j (q)	,	(1.57)
		2MN		ω j (q)		2kT

where N is the number of atoms, M is their mass and ωj(q) is the frequency of vibration of phonons in mode j, wave vector q. One can further show that in the Debye approximation (again discussed in Chap. 2): At low temperature (T << θD)

2W =		3		2K 2
							= constant ,		(1.58)
	4M kθD
and at high temperature (T >> θD)
2W =			3			T		K 2 T ,	(1.59)
		M
				θD θD

where θD is the Debye Temperature defined from the cutoff frequency in the Debye approximation (see Sect. 2.3.3). The effect of temperature is to reduce intensity but not broaden lines. Even at T = 0 the Debye–Waller factor is not unity so there is always some “diffuse” scattering, in addition to the diffraction.

As an example of the use of the structure factor, we represent the bcc lattice as a sc lattice with a basis. Let the simple cubic unit cell have side a. Consider a basis at R0 = (0,0,0)a, R1 = (1,1,1)a/2. The structure factor is

S	hkl	= f	0	+ f e−i2π(h+k +l)a / 2 = f		0	+ f (−1)h+k +l .	(1.60)
				1			1
Suppose also the atoms at R0 and R1 are identical, then f0 = f1 = f so
				Shkl = f (1+ (−)h+k +l ),				(1.61)
				= 0	if h + k +l is odd,
				= 2 f	if h + k +l is even.

The nonvanishing structure factor ends up giving results identical to a bcc lattice.

Problems

1.1.Show by construction that stacked regular pentagons do not fill all twodimensional space. What do you conclude from this? Give an example of a geometrical figure that when stacked will fill all two-dimensional space.

1.2.Find the Madelung constant for a one-dimensional lattice of alternating, equally spaced positive and negative charged ions.

15 See Maradudin et al [1.16]

1.2 Group Theory and Crystallography 39

1.3.Use the Evjen counting scheme [1.19] to evaluate approximately the Madelung constant for crystals with the NaCl structure.

1.4.Show that the set of all rational numbers (without zero) forms a group under the operation of multiplication. Show that the set of all rational numbers (with zero) forms a group under the operation of addition.

1.5.Construct the group multiplication table of D4 (the group of three dimensional rotations that map a square into itself).

1.6.Show that the set of elements (1, –1, i, –i) forms a group when combined under the operation of multiplication of complex numbers. Find a geometric group that is isomorphic to this group. Find a subgroup of this group. Is the whole group cyclic? Is the subgroup cyclic? Is the whole group abelian?

1.7.Construct the stereograms for the point groups 4(C4) and 4mm(C4v). Explain how all elements of each group are represented in the stereogram (see Table 1.3).

1.8.Draw a bcc (body-centered cubic) crystal and draw in three crystal planes that are neither parallel nor perpendicular. Name these planes by the use of Miller indices. Write down the Miller indices of three directions, which are neither parallel nor perpendicular. Draw in these directions with arrows.

1.9.Argue that electrons should have energy of order electron volts to be diffracted by a crystal lattice.

1.10.Consider lattice planes specified by Miller indices (h, k, l) with lattice spacing determined by d(h,k,l). Show that the reciprocal lattice vectors G(h,k,l) are orthogonal to the lattice plane (h,k,l) and if G(h,k,l) is the shortest such reciprocal lattice vector then

d(h, k,l) = 2π . G(h, k,l)

1.11.Suppose a one-dimensional crystal has atoms located at nb and αmb where n and m are integers and α is an irrational number. Show that sharp Bragg peaks are still obtained.

1.12.Find the Bragg peaks for a grating with a modulated spacing. Assume the grating has a spacing

dn = nb +εb sin(2πknb) ,

where ε is small and kb is irrational. Carry your results to first order in ε and assume that all scattered waves have the same geometry. You can use the geometry shown in the figure of this problem. The phase φn of scattered wave n at angle θ is

ϕn = 2λπ dn sinθ ,

40 1 Crystal Binding and Structure

where λ is the wavelength. The scattered intensity is proportional to the square of the scattered amplitude, which in turn is proportional to

E ≡ ∑0N exp(iϕn )

for N+1 scattered wavelets of equal amplitude.

θ

d2

d2sinθ

1.13.Find all Bragg angles less than 50 degrees for diffraction of X-rays with wavelength 1.5 angstroms from the (100) planes in potassium. Use a conventional unit cell with structure factor.

2 Lattice Vibrations and Thermal Properties

Chapter 1 was concerned with the binding forces in crystals and with the manner in which atoms were arranged. Chapter 1 defined, in effect, the universe with which we will be concerned. We now begin discussing the elements of this universe with which we interact. Perhaps the most interesting of these elements are the internal energy excitation modes of the crystals. The quanta of these modes are the “particles” of the solid. This chapter is primarily devoted to a particular type of internal mode – the lattice vibrations.

The lattice introduced in Chap. 1, as we already mentioned, is not a static structure. At any finite temperature there will be thermal vibrations. Even at absolute zero, according to quantum mechanics, there will be zero-point vibrations. As we will discuss, these lattice vibrations can be described in terms of normal modes describing the collective vibration of atoms. The quanta of these normal modes are called phonons.

The phonons are important in their own right as, e.g., they contribute both to the specific heat and the thermal conduction of the crystal, and they are also important because of their interaction with other energy excitations. For example, the phonons scatter electrons and hence cause electrical resistivity. Scattering of phonons, by whatever mode, in general also limits thermal conductivity. In addition, phonon–phonon interactions are related to thermal expansion. Interactions are the subject of Chap. 4.

We should also mention that the study of phonons will introduce us to wave propagation in periodic structures, allowed energy bands of elementary excitations propagating in a crystal, and the concept of Brillouin zones that will be defined later in this chapter.

There are actually two main reservoirs that can store energy in a solid. Besides the phonons or lattice vibrations, there are the electrons. Generally, we start out by discussing these two independently, but this is an approximation. This approximation is reasonably clear-cut in insulators, but in metals it is much harder to justify. Its intellectual framework goes by the name of the Born–Oppenheimer approximation. This approximation paves the way for a systematic study of solids in which the electron–phonon interactions can later be put in, often by perturbation theory. In this chapter we will discuss a wide variety of lattice vibrations in one and three dimensions. In three dimensions we will also discuss the vibration problem in the elastic continuum approximation. Related topics will follow: in Chap. 3 electrons moving in a static lattice will be considered, and in Chap. 4 electron– phonon interactions (and other topics).

42 2 Lattice Vibrations and Thermal Properties

2.1 The Born–Oppenheimer Approximation (A)

The most fundamental problem in solid-state physics is to solve the many-particle Schrödinger wave equation,

H ψ = i

∂ψ

,

(2.1)

c ∂t

where Hc is the crystal Hamiltonian defined by (2.3). In a sense, this equation is the “Theory of Everything” for solid-state physics. However, because of the many-body problem, solutions can only be obtained after numerous approximations. As mentioned in Chap. 1, P. W. Anderson has reminded us, “more is different!” There are usually emergent properties at higher levels of complexity [2.1]. In general, the wave function ψ is a function of all electronic and nuclear coordinates and of the time t. That is,

ψ =ψ(ri , Rl ,t) ,

(2.2)

where the ri are the electronic coordinates and the Rl are the nuclear coordinates. The Hamiltonian Hc of the crystal is

2

i2

2

l2

1

∑i,′

e2

H c = −∑i

− ∑l

+

2

j

2m

2M

l

4πε

0

| r

− r

j

|

e2Zl

i

(2.3)

1

e2Zl Zl'

′

− ∑i,l

+

2

∑l,l′

.

4πε

0

| r

− R

|

4πε

0

|

R

− R

|

i

l

l

l'

In (2.3), m is the electronic mass, Ml is the mass of the nucleus located at Rl, Zl is the atomic number of the nucleus at Rl, and e has the magnitude of the electronic charge. The sums over i and j run over all electrons.1 The prime on the third term on the right-hand side of (2.3) means the terms i = j are omitted. The sums over l and l′ run over all nuclear coordinates and the prime on the sum over l and l′ means that the l = l′ terms are omitted. The various terms all have a physical interpretation. The first term is the operator representing the kinetic energy of the electrons. The second term is the operator representing the kinetic energy of the nuclei. The third term is the Coulomb potential energy of interaction between the electrons. The fourth term is the Coulomb potential energy of interaction between the electrons and the nuclei. The fifth term is the Coulomb potential energy of interaction between the nuclei.

1Had we chosen the sum to run over only the outer electrons associated with each atom, then we would have to replace the last term in (2.3) by an ion–ion interaction term. This term could have three and higher body interactions as well as two-body forces. Such a procedure would be appropriate [51, p. 3] for the practical discussion of lattice vibrations. However, we shall consider only two-body forces.

2.1 The Born–Oppenheimer Approximation (A) 43

In (2.3) internal magnetic interactions are left out because of their assumed smallness. This corresponds to neglecting relativistic effects. In solid-state physics, it is seldom necessary to assign a structure to the nucleus. It is never necessary (or possible) to assign a structure to the electron. Thus in (2.3) both electrons and nuclei are treated as point charges. Sometimes it will be necessary to allow for the fact that the nucleus can have nonzero spin, but this is only when much smaller energy differences are being considered than are of interest now. Because of statistics, as will be evident later, it is usually necessary to keep in mind that the electron is a spin 1/2 particle. For the moment, it is necessary to realize only that the wave function of (2.2) is a function of the spin degrees of freedom as well as of the space degrees of freedom. If we prefer, we can think of ri in the wave function as symbolically labeling all the coordinates of the electron. That is, ri gives both the position and the spin. However, i2 is just the ordinary spatial Laplacian.

For purposes of shortening the notation it is convenient to let TE be the kinetic energy of the electrons, TN be the kinetic energy of the nuclei, and U be the total Coulomb energy of interaction of the nuclei and the electrons. Then (2.3) becomes

H c = TE +U +TN .	(2.4)
It is also convenient to define
H 0 = TE +U .	(2.5)

Nuclei have large masses and hence in general (cf. the classical equipartition theorem) they have small kinetic energies. Thus in the expression Hc = H0 + TN, it makes some sense to regard TN as a perturbation on H0. However, for metals, where the electrons have no energy gap between their ground and excited states, it is by no means clear that TN should be regarded as a small perturbation on H0. At any rate, one can proceed to make expansions just as if a perturbation sequence would converge.

Let M0 be a mean nuclear mass and define

		m		1/ 4
K =	M0			.
If we define
H L = −∑l		M 0					2	l2	,	(2.6)
						2m
		Ml
then
T = K 4H						L	.			(2.7)
N
The total Hamiltonian then has the form
H c = H 0 + K 4H L ,										(2.8)