Patterson, Bailey - Solid State Physics Introduction to theory

24 1 Crystal Binding and Structure

2.Monoclinic Symmetry. For each unit cell, α = γ = π/2, β ≠ α, a ≠ b, b ≠ c, and a ≠ c. The two Bravais lattices are shown in Fig. 1.12.

		a
		b
α	γ
	β	c
	(a)

a

b

c

(b)

Fig. 1.12. (a) The simple monoclinic cell, and (b) the base-centered monoclinic cell

3.Orthorhombic Symmetry. For each unit cell, α = β = γ = π/2, a ≠ b, b ≠ c, and a ≠ c. The four Bravais lattices are shown in Fig. 1.13.

	a
		b
α	γ
	β	c
	(a)	(b)	(c)

(d)

Fig. 1.13. (a) The simple orthorhombic cell, (b) the base-centered orthorhombic cell, (c) the body-centered orthorhombic cell, and (d) the face-centered orthorhombic cell

4.Tetragonal Symmetry. For each unit cell, α = β = γ = π/2 and a = b ≠ c. The two unit cells are shown in Fig. 1.14.

	a
		b
α	γ
	β	c
	(a)	(b)

Fig. 1.14. (a) The simple tetragonal cell, and (b) the body-centered tetragonal cell

1.2 Group Theory and Crystallography 25

5.Trigonal Symmetry. For each unit cell, α = β = γ ≠ π/2, < 2π/3 and a = b = c. There is only one Bravais lattice, whose unit cell is shown in Fig. 1.15.

Fig. 1.15. Trigonal unit cell

6.Hexagonal Symmetry. For each unit cell, α = β = π/2, γ = 2π/3, a = b, and a ≠ c. There is only one Bravais lattice, whose unit cell is shown in Fig. 1.16.

a

b

α β

γ c

Fig. 1.16. Hexagonal unit cell

7.Cubic Symmetry. For each unit cell, α = β = γ = π/2 and a = b = c. The unit cells for the three Bravais lattices are shown in Fig. 1.17.

	a
		b
α	γ
	β	c
	(a)	(b)

(c)

Fig. 1.17. (a) The simple cubic cell, (b) the body-centered cubic cell, and (c) the facecentered cubic cell

26 1 Crystal Binding and Structure

1.2.6 Schoenflies and International Notation for Point Groups (A)

There are only 32 point group symmetries that are consistent with translational symmetry. In this Section a descriptive list of the point groups will be given, but first a certain amount of notation is necessary.

The international (sometimes called Hermann–Mauguin) notation will be defined first. The Schoenflies notation will be defined in terms of the international notation. This will be done in a table listing the various groups that are compatible with the crystal systems (see Table 1.3).

An f-fold axis of rotational symmetry will be specified by f. Also, f will stand for the group of f-fold rotations. For example, 2 means a two-fold axis of symmetry (previously called C2), and it can also mean the group of two-fold rotations. f will denote a rotation inversion axis. For example, 2 means that the crystal is brought back into itself by a rotation of πfollowed by an inversion. f/m means a rotation axis with a perpendicular mirror plane. f 2 means a rotation axis with a perpendicular two-fold axis (or axes). fm means a rotation axis with a parallel mirror plane (or planes) (m = 2 ). f 2 means a rotation inversion axis with a perpendicular two-fold axis (or axes). f m means that the mirror plane m (or planes) is parallel to the rotation inversion axis. A rotation axis with a mirror plane normal and mirror planes parallel is denoted by f/mm or (f/m)m. Larger groups are compounded out of these smaller groups in a fairly obvious way. Note that 32 point groups are listed.

A very useful pictorial way of thinking about point group symmetries is by the use of stereograms (or stereographic projections). Stereograms provide a way of representing the three-dimensional symmetry of the crystal in two dimensions. To construct a stereographic projection, a lattice point (or any other point about which one wishes to examine the point group symmetry) is surrounded by a sphere. Symmetry axes extending from the center of the sphere intersect the sphere at points. These points are joined to the south pole (for points above the equator) by straight lines. Where the straight lines intersect a plane through the equator, a geometrical symbol may be placed to indicate the symmetry of the appropriate symmetry axis. The stereogram is to be considered as viewed by someone at the north pole. Symmetry points below the equator can be characterized by turning the process upside down. Additional diagrams to show how typical points are mapped by the point group are often given with the stereogram. The idea is illustrated in Fig. 1.18. Wood [98] and Brown [49] have stereograms of the 32 point groups. Rather than going into great detail in describing stereograms, let us look at a stereogram for our old friend D3 (or in the international notation 32).

The principal three-fold axis is represented by the triangle in the center of Fig. 1.19b. The two-fold symmetry axes perpendicular to the three-fold axis are represented by the dark ovals at the ends of the line through the center of the circle.

In Fig. 1.19a, the dot represents a point above the plane of the paper and the open circle represents a point below the plane of the paper. Starting from any given point, it is possible to get to any other point by using the appropriate symmetry operations. D3 has no reflection planes. Reflection planes are represented by dark lines. If there had been a reflection plane in the plane of the paper, then the outer boundary of the circle in Fig. 1.19b would have been dark.

1.2 Group Theory and Crystallography 27

Table 1.3. Schoenflies1 and international2 symbols for point groups, and permissible point groups for each crystal system

Crystal system	International symbol										Schoenflies symbol
Triclinic	1										C1
		1									Ci
Monoclinic	2										C2
		m									C1h
		(2/m)									C2h
Orthorhombic	222										D2
		2mm									C2v
		(2/m)(2/m)(2/m)									D2h
	4										C4
											S4
		4
Tetragonal		(4/m)									C4h
	422										D4
		4mm									C4v
											D2d
		42m
		(4/m)(2/m)(2/m)									D4h
		3									C3
	¯										C3i
Trigonal	3
	32										D3
		3m									C3v
											D3d
		3(2/ m)
		6									C6
	¯										C3h
	6
Hexagonal		(6/m)									C6h
	622										D6
		6mm									C6v
				m2							D3h
		6
		(6/m)(2/m)(2/m)									D6h
	23										T
											Th
Cubic	(2/ m)3
	432										O
											Td
		43m
	(4/ m										Oh
								)(3)(2/ m)

1A. Schoenflies, Krystallsysteme und Krystallstruktur, Leipzig, 1891.

2C. Hermann, Z. Krist., 76, 559 (1931); C. Mauguin, Z. Krist., 76, 542 (1931).

28 1 Crystal Binding and Structure

At this stage it might be logical to go ahead with lists, descriptions, and names of the 230 space groups. This will not be done for the simple reason that it would be much too confusing in a short time and would require most of the book otherwise. For details, Buerger [1.5] can always be consulted. A large part of the theory of solids can be carried out without reference to any particular symmetry type. For the rest, a research worker is usually working with one crystal and hence one space group and facts about that group are best learned when they are needed (unless one wants to specialize in crystal structure).

N

Plane	Lattice
through		Symmetry
equator	point	axis

S

Fig. 1.18. Illustration of the way a stereogram is constructed

(a)

(b)

Fig. 1.19. Stereogram for D3

1.2.7 Some Typical Crystal Structures (B)

The Sodium Chloride Structure. The sodium chloride structure, shown in Fig. 1.20, is one of the simplest and most familiar. In addition to NaCl, PbS and MgO are examples of crystals that have the NaCl arrangement. The space lattice is fcc (face-centered cubic). Each ion (Na+ or Cl−) is surrounded by six nearestneighbor ions of the opposite sign. We can think of the basis of the space lattice as being a NaCl molecule.

Table 1.4. Packing fractions (PF) and coordination numbers (CN)

	Crystal Structure						PF	CN
	fcc		2π		= 0.74			12
		6
	bcc		3π					8
		8			= 0.68
	sc			π		= 0.52		6
		6
	diamond		3π		= 0.34			4
		16

1.2 Group Theory and Crystallography 29

+ ion	- ion	(0, 0, 0) (a/4, a/4, a/4)
Fig. 1.20. The sodium chloride structure		Fig. 1.21. The diamond structure

The Diamond Structure. The crystal structure of diamond is somewhat more complicated to draw than that of NaCl. The diamond structure has a space lattice that is fcc. There is a basis of two atoms associated with each point of the fcc lattice. If the lower left-hand side of Fig. 1.21 is a point of the fcc lattice, then the basis places atoms at this point [labeled (0, 0, 0)] and at (a/4, a/4, a/4). By placing bases at each point in the fcc lattice in this way, Fig. 1.21 is obtained. The characteristic feature of the diamond structure is that each atom has four nearest neighbors or each atom has tetrahedral bonding. Carbon (in the form of diamond), silicon, and germanium are examples of crystals that have the diamond structure. We compare sc, fcc, bcc, and diamond structures in Table 1.4.

The packing fraction is the fraction of space filled by spheres on each lattice point that are as large as they can be so as to touch but not overlap. The coordination number is the number of nearest neighbors to each lattice point.

The Cesium Chloride Structure. The cesium chloride structure, shown in Fig. 1.22, is one of the simplest structures to draw. Each atom has eight nearest neighbors. Besides CsCl, CuZn (β-brass) and AlNi have the CsCl structure. The Bravais lattice is simple cubic (sc) with a basis of (0,0,0) and (a/2)(1,1,1). If all the atoms were identical this would be a body-centered cubic (bcc) unit cell.

The Perovskite Structure. Perovskite is calcium titanate. Perhaps the most familiar crystal with the perovskite structure is barium titanate, BaTiO3. Its structure is shown in Fig. 1.23. This crystal is ferroelectric. It can be described with a sc lattice with basis vectors of (0,0,0), (a/2)(0,1,1), (a/2)(1,0,1), (a/2)(1,1,0), and (a/2)(1,1,1).

30 1 Crystal Binding and Structure

Cs+

Cs+

Cs

Cs+

Cl–

Cs+

Cs+

Cs+

Cs+

Fig. 1.22. The cesium chloride structure

Ba2+

O2–

	Ba2+	Ba2+
	O2–
	Ba2+	O2–
	Ti4+

O2–

Ba2+

O2–

Ba2+

Ba2+

O2–

Ba2+

Fig. 1.23. The barium titanate (BaTiO3) structure

Crystal Structure Determination (B)

How do we know that these are the structures of actual crystals? The best way is by the use of diffraction methods (X-ray, electron, or neutron). See Sect. 1.2.9 for more details about X-ray diffraction. Briefly, X-rays, neutrons and electrons can all be diffracted from a crystal lattice. In each case, the wavelength of the diffracted entity must be comparable to the spacing of the lattice planes. For X-rays to have a wavelength of order Angstroms, the energy needs to be of order keV, neutrons need to have energy of order fractions of an eV (thermal neutrons), and electrons should have energy of order eV. Because they carry a magnetic moment and hence interact magnetically, neutrons are particularly useful for determining magnetic structure.10 Neutrons also interact by the nuclear interaction, rather than with electrons, so they are used to located hydrogen atoms (which in a solid have few or no electrons around them to scatter X-rays). We are concerned here with elastic scattering. Inelastic scattering of neutrons can be used to study lattice vibrations (see the end of Sect. 4.3.1). Since electrons interact very strongly with other electrons their diffraction is mainly useful to elucidate surface structure.11

Ultrabright X-rays: Synchrotron radiation from a storage ring provides a major increase in X-ray intensity. X-ray fluorescence can be used to study bonds on the surface because of the high intensity.

10For example, Shull and Smart in 1949 used elastic neutron diffraction to directly demonstrate the existence of two magnetic sublattices on an antiferromagnet.

11Diffraction of electrons was originally demonstrated by Davisson and Germer in an experiment clearly showing the wave nature of electrons.

1.2 Group Theory and Crystallography 31

1.2.8 Miller Indices (B)

In a Bravais lattice we often need to describe a plane or a set of planes, or a direction or a set of directions. The Miller indices are a notation for doing this. They are also convenient in X-ray work.

To describe a plane:

1.Find the intercepts of the plane on the three axes defined by the basis vectors (a1, a2, a3).

2.Step 1 gives three numbers. Take the reciprocal of the three numbers.

3.Divide the reciprocals by their greatest common divisor (which yields a set of integers). The resulting set of three numbers (h, k, l) is called the Miller indices for the plane. {h, k, l} means all planes equivalent (by symmetry) to (h, k, l).

To find the Miller indices for a direction:

1.Find any vector in the desired direction.

2.Express this vector in terms of the basis (a1, a2, a3).

3.Divide the coefficients of (a1, a2, a3) by their greatest common divisor. The resulting set of three integers [h, k, l] defines a direction. h, k, l means all vectors equivalent to [h, k, l]. Negative signs in any of the numbers are indicated by placing a bar over the number (thus h ).

1.2.9 Bragg and von Laue Diffraction (AB)12

By discussing crystal diffraction, we accomplish two things: (1) We make clear how we know actual crystal structures exist, and (2) We introduce the concept of the reciprocal lattice, which will be used throughout the book

The simplest approach to Bragg diffraction is illustrated in Fig. 1.24. We assume specular reflection with angle of incidence equal to angle of reflection. We also assume the radiation is elastically scattered so that incident and reflected waves have the same wavelength.

For constructive interference we must have the path difference between reflected rays equal to an integral (n) number of wavelengths (λ). Using Fig. 1.24, the condition for diffraction peaks is then

nλ = 2d sinθ ,

(1.23)

which is the famous Bragg law. Note that peaks in the diffraction only occur if λ is less than 2d, and we will only resolve the peaks if λ and d are comparable.

12A particularly clear discussion of these topics is found in Brown and Forsyth [1.4]. See also Kittel [1.13, Chaps. 2 and 19].

32 1 Crystal Binding and Structure

Incident Reflected

θθ

θ θd θ Lattice planes

d sin θ

Fig. 1.24. Bragg diffraction

The Bragg approach gives a simple approach to X-ray diffraction. However, it is not easily generalized to include the effects of a basis of atoms, of the distribution of electrons, and of temperature. For that we need the von Laue approach.

We will begin our discussion in a fairly general way. X-rays are electromagnetic waves and so are governed by the Maxwell equations. In SI and with no charges or currents (i.e. neglecting the interaction of the X-rays with the electron distribution except for scattering), we have for the electric field E and the magnetic field H (with the magnetic induction B = μ0H)

	E = 0 , × H = ε0	∂E	, × E = −	∂B	, B = 0.
		∂t		∂t

Taking the curl of the third equation, using B = μ0H and using the first and second of the Maxwell equations we find the usual wave equation:

	2 E =	1	∂2 E	,	(1.24)
		c2	∂t2

where c = (μ0ε0)–1/2 is the speed of light. There is also a similar wave equation for the magnetic field. For simplicity we will focus on the electric field for this discussion. We assume plane-wave X-rays are incident on an atom and are scattered as shown in Fig. 1.25.

In Fig. 1.25 we use the center of the atom as the origin and rs locates the electron that scatters the X-ray. As mentioned earlier, we will first specialize to the case of the lattice of point scatterers, but the present setup is useful for generalizations.

		|kf| = |ki| = k	r	Observation
		kf		point
Point of scattering			R = rs + r
	ki
Source		rs θ′
point	ri

Fig. 1.25. Plane-wave scattering

1.2 Group Theory and Crystallography 33

The solution of the wave equation for the incident plane wave is

Ei (r) = E0 exp[i(ki ri −ωt)] ,

(1.25)

where E0 is the amplitude and ω = kc. If the wave equation is written in spherical coordinates, one can find a solution for the spherically scattered wave (retaining only dominant terms far from the scattering location)

E	s	= K E(r ) eikr			,	(1.26)
		1	s	r

where K1 is a constant, with the scattered wave having the same frequency and wavelength as the incident wave. Spherically scattered waves are important ones since the wavelength being scattered is much greater than the size of the atom. Also, we assume the source and observation points are very far from the point of scattering. From the diagram r = R − rs, so by squaring, taking the square root, and using that rs /R << 1 (i.e. far from the scattering center), we have

				r
	r = R 1		−	s		cosθ			′ ,	(1.27)
				R
	from which since krs cosθ kf · rs;
	kr kR − k f						rs .			(1.28)
	Therefore
		eikR			i(k		−k		) r
	Es = K1E0			e		i		f	s e−iωt ,	(1.29)
		R

where we have used (1.28), (1.26), and (1.25) and also assumed r−1 R−1 to sufficient accuracy. Note that (ki − kf) · rs, as we will see, can be viewed as the phase difference between the wave scattered from the origin and that scattered from rs in the approximation we are using. Thus, the scattering intensity is proportional to |P|2 (given by (1.32)) that, as we will see, could have been written down immediately. Thus, we can write the scattered wave as

		Esc = FP ,							(1.30)
	where the magnitude of F2 is proportional to the incident intensity E0 and
		F		=		K1E0			(1.31)
								,
						R
	P = ∑s e−i k rs ,								(1.32)
	summed over all scatterers, and
		k = k f − ki .							(1.33)