SOLID STATE PHYSICS I (PHS- 316)

LECTURER: NGWA Engelbert AFUOTI. ( ngwaengelbert@yahoo.com)

ENSAB

CONTENTS

1. Introduction…………………………………………………………...22.1 Classes of solids………………………………………………………42.2 Crystal structures……………………………………………………..52.2.1 Translational symmetry………………………………………………52.2.2 Bravais lattice…………………………………………………………6

3 Some Group Theory, Theory of Symmetry, Schoenflies Notations, Non-Cubic Crystallographic Point Group…………………………… 8

3.1 Definition and Simple Properties of Groups………………………….83.2 Unitary Transformation………………………………………………103.3 Symmetry Transformation……………………………………………10

3.3.1 Point Group………………………………………………………… . 11 4. Periodic Function of Reciprocal Lattice, Miller Indices……………...23 4.1 Recesciprocal lattice…………………………………………………..26 4.2 Miller indices………………………………………………………… 26 5. Lattice Dynamics…………………………………………………….. 28 5.1 Vibrational Modes of a Monatomic Lattice………………………….. 28 5.1.1 Linear Monatomic Chain…………………………………………….. 28 5.1.2 Density of State………………………………………………………. 37 5.2 Vibrational Modes of a Diatomic Linear Chain……………………… 38 5.3 Vibrational Modes in Three-Dimensional Crystal…………………… 45

5.4 Normal Vibration of a Three-Dimensional Crystal………………….. 52 5.5 Second Quantization of the Phonon Field…………………………… 57 6. Electronic State in Ideal Crystal…………………………………….. 62

6.1 The zone theory……………………………………………………… 62

6.1.1 Bloch theorem………………………………………………………. 64

6.2 The Approximation of the Nearly Free Electron

Model (N.F.E)……………………………………………… ……… 70

6.3 The Tight-Binding Approximation (T.B.A)………………............... 75

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1. INTRODUCTION

Solid state Physics is that branch of physics that deals with the structure and properties of solids. It is the largest branch of condensed matter physics, which study rigid matter, or solids, through methods such as quantum mechanics, crystallography, electromagnetism and metallurgy. Solid state physics considers how the large-scale properties of solid materials result from their atomic-scale properties. Solid-state physics thus forms the theoretical basis of materials science, as well as having direct applications, for example in the technology of transistors and semiconductors. Solids consist of atoms or molecules packed closely together-frequently in a very orderly way. Most of the physical properties of solids depend upon the way in which the atoms or molecules are packed, or arranged; that is, on the internal structure of the solid. For this reason the study of the structure of solids, called crystallography, is of basic importance to all other investigations in solid state physics. The study of the attractive forces between atoms or molecules in solids is another basic aspect; it has helped greatly in the understanding of many of the properties of solids. For example, since the attractive forces are electrical in nature, they play a very important role in determining such electrical properties of solids as conductivity. These forces also determine to a large extent the optical and thermal properties of solids, such as refractive power and the ability to conduct heat. Solids are rarely, if ever, pure. (A pure substance consists of one chemicalelement or compound only; impurities are traces of other elements mixed in with the main element or compound.) The study of the influence of impurities upon the properties of solids is an important aspect of solid-state physics. For example, devices such as the solar battery and the transistor were made possible only after scientists had learned how to control impurities in silicon and similar substances.

Frequently an atom or molecule is unexpectedly absent in the packing arrangement of a solid. The gap left in the structure is called a vacancy. Vacancies may profoundly alter the mechanical properties of materials. For example, a piece of cold-worked metal has many more vacancies than a comparable piece of annealed metal, and is usually much harder and tougher. Thus the study of vacancies by solid-state physicists is of considerable interest to metallurgists.

The atoms or molecules at the surface of a solid are not packed in the same way as those in the interior; unlike interior atoms or molecules, they are not completely surrounded by neighboring atoms or molecules. Thus the surface behavior differs substantially from that of the rest of the solid. The study of solid surfaces and of films (that is, very thin solids consisting mostly of surface) has led to the use in electronics of thin films with unusual electrical properties and to ways of increasing the resistance of metal surfaces to corrosion.

The crystal lattice can vibrate. These vibrations are found to be quantised, the quantised vibrational modes are known as phonons. Phonons play a major role in

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many of the physical properties of solids, such as the transmission of sound. In insulating solids, phonons are also the primary mechanism by which heat conduction takes place. Phonons are also necessary for understanding the lattice heat capacity of a solid, as in the Einstein model and the later Debye model.

One of the principal aims of solid-state physicists is to figure out how to exploit properties of solids for something useful. For example, detailed study of band-structure led to development of PN-junction, which led to the invention of the transistor, of which there are now several varieties (bipolar-junction transistors, junction FETs, Metal-Oxide Semiconducting FETs, Modulation-Doped FETs, Pseudomorphic High-Electron Mobility Transistors, etc). Study of band structure also led to the ability to create a population inversion in group III-V material for the lasing process to be possible for the laser diode. Peltier coolers, often used on CDs and microprocessors, have been invented by studying phonon interactions of solids.

Over the past years this branch of physics has undergone a considerable evolution with current research fields like:

- Quantum electronics

- Semiconductor device fabrication in the semiconductor industry

- Quasicrystals - Spin glass - Superconductivity and many others.

In these lectures we exclude the so-called amorphous solids (e.g. glass, some times also called a supper cooled liquid) from consideration. We also exclude from consideration systems known as liquid crystals. Concerned in these lectures is to present a systematic study of the principles underlying solid state physics. The general notion of crystalline structures, properties of solid such as – mechanical, electrical, thermal, optical …, are therefore not left out. Concerned is also on the microscopic description of the physical phenomena in solids. Quantum mechanics is the basic tool for the adequate description of phenomena at the atomic level. Thus a complete solid state theory must be quantum mechanically examined. Physical phenomena are examined using models; for example, a simple picture of a solid can be seen as an ensemble of point masses coupled to each other by springs. Point masses may be made to oscillate by electromagnetic radiations, sound and so on. The study based on this model result in

- The frequency spectrum of a mechanical oscillator ( we examine the acoustic and optical branches of the spectrum)

- Understanding the variation of specific heat with temperature and so on.

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2.1 Classes of solids

Solids can be classified according to their properties (more profoundly by their chemical binding) and their structures (symmetries). Based on their chemical binding, there are basically four types of solid which we briefly present below with the aim to give mere reasons why atoms combined to form solids:

- Molecular crystals - Examples of molecular crystals are nitrogen (N2) and rare gas crystals such as argon (Ar). Molecular crystals consist of chemically inert molecules (neutral molecules which have little or no affinity for adding or sharing additional electrons). They are also referred too as chemically saturated units. The interaction between chemically saturated units is described by the Van der Waals forces. Quantum mechanics describes these forces as being due to correlations in the fluctuating distributions of charges in the chemically saturated units. The appearance of virtual excited states causes transitory dipole moments to appear on adjacent atoms, and if these dipole moments have the right directions, then the atoms can be attracted to one another. The Van der Waals forces are weak, short range forces, and hence molecular crystals are characterised by low melting and boiling point. The forces in molecular crystals are almost central forces (central forces act along the line joining the atoms), and they make efficient use of their binding in closed packed crystal structures. We should not forget to mention that there is also a repulsive force which keeps the lattice from collapsing.

- Ionic crystals – Examples of ionic crystals are sodium chloride ( NaCl) and lithium fluoride ( LiF). Ionic crystals also consist of chemically saturated units (the ions which form their basic unit are in rare gas configurations). The ionic bond is due mostly to the electrostatic attraction between oppositely charged ions. However there is the core repulsion between the ions due to an overlapping of electron clouds (as constrained by the Pauli principle) to prevent the lattice from collapsing. Since the Coulomb forces of attraction are strong, long range, nearly two body central forces, ionic crystal are characterised by close packing and rather tight binding. These crystals also show good ionic conductivity at high temperatures, good cleavage, and strong infrared absorption.

- Metallic crystals – Examples of metals are sodium ( ) and copper (Cu). A metal such as sodium is viewed as being composed of positive ion cores ( ) immersed in a “sea” of free conduction electrons which comes from the removal of the 3s electron from the atomic sodium Na. One reason for the binding in metals is lowering of the kinetic energy of the “free” electrons relative to their energy in the atomic 3s state. In a metallic crystal the valence electrons are free (within the constrained of the Pauli principle) to wander throughout the crystal causing them to have a smoother wave function and less kinetic energy. Lowering of the kinetic energy implies binding. Due to the important role of free electrons in

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binding, metals are good electrical and thermal conductors. A more quantum mechanical description of metallic bonded crystals is given by the Wigner-Seitz theory.

- Valence crystals – An example of valence crystal is carbon in diamond form. Some insight into a covalent bond can be gained by considering them as being caused by sharing electrons between atoms of unfilled shells. Sharing electrons can lower the energy because the electrons can get into lower energy states without violating the Pauli principle. Covalent crystals are characterised by hardness, poor cleavage, poor electronic and ionic conductivity. The forces in covalent bonds can be thought of as short range, two-body, but not central forces. The covalent bond is very directional, and the crystals tend to be loosely packed. The quantum mechanical description of this bonding is given by the Heitler-London theory.

We shall present a systematic study of bonding in crystals in our lectures in solid physics II.

2.2 Crystal structures

Many properties of materials are affected by their crystal structure. This structure can be investigated using a range of crystallographic techniques, including X-ray crystallography, neutron diffraction and electron diffraction. The chemical element position on the periodic table uniquely determines the crystal properties. Each crystal structure differs from the other in atomic structure. The regularity and symmetry of the given crystal structure makes it homogeneous and isotropic. For a homogeneous structure, for all interior point, there exists a point with identical properties to that of the first point located at a certain distance from it. If the properties of a crystal structure are the same for all directions, then we say the structure is isotropic. Below we give the definition of crystal structure terms and some facts about crystal structure. 2.2.1 Translational symmetry

A crystal has periodic properties due to the regular array of atoms on it. The symmetry of the space periodic field is that of the crystal lattice. The basis of the space periodicity is defined by three fundamental period , and such that 2.2.1.1 Where are integers and any set of leaves the crystal invariable. The space periodic field  ; 2.2.1.2

Where is a radius vector , vector is the principal translation or lattice vector. It is also called crystal axis, lattice constant and or radius vector of the

lattice site.

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The characteristic feature of a crystal structure is the translational invariant (2.2.1.2). For the most complete description of the translational invariant, we select the shortest , that are not coplanar for which (2.2.1.2) is true. A crystal structure that satisfies (2.2.1.2) possesses translational symmetry. Hence a crystal is said to be translationally symmetric or periodic if there exist three linearly indepent vectors , and such that a translation by brings one back to an equivalent point in the crystal. The set of operations that elements differ by the selection on is a translational group. The geometrical array of points generated by a translation is a lattice. Each point in the array is the lattice site. If one place at each point on the lattice sites a collection or basis of atoms, the resulting structure is called a crystal structure. Construct fundamental periods from the centre of the given atom. The volume of the parallelepiped with sides is called an elementary cell or unit cell. A primitive cell is an elementary cell with corners coinciding with lattice points i.e. a cell with lattice sites only at the corners. For simplicity a primitive cell is one with the shortest translation vectors . It is clear that a crystal is made up of identical primitive cells. An elementary cell is simple if it contains only one atom and otherwise compound. There is also a type of translation operation that relates objects within a unit cell so that the same objects are found at coordinates that are half multiples of unit cell distances along two or three of the axes. These operation are for example responsible for the face and body-centered lattices found in three dimensions as seen in Bravais lattices which we see in the next section . The ensemble of all symmetry elements (operations) in a crystal structure is a space group. A symmetry operation lives the crystal environment invariable or that operation which brings the crystal back onto itself. (2.2.1.2) satisfies this requirement. In addition to translation operation, there are also rotation, reflection, and inversion operations. A symmetry operation with no translation is a point operation. Examples are rotation about an axis, reflection about a plane and inversion about a point. Point operation applies to object and translation to lattice.

2.2.2 Bravais lattice A crystal lattice has an infinite set of translational symmetry. Let and be two fundamental periods:

, 2. 2.2.1

Also

2.2.2.2

where is the symmetric inverse to . As and are integers, then

. Construct on any lattice sites an elementary cell (parallelepiped) using the fundamental period (2.2.2.2), some atoms are found at the vertices of the elementary cell. The vertices overlap relative to the parallel translation by a fundamental period. The totality of such equilibrium sites that overlap with each other relative to the fundamental period forms the Bravais lattice of the

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crystal. It may be seen that for every lattice, we may find three primitive translation such that any motion from one position in the crystal to another with exactly the same physical environment, is given by

2.2.2.3

The quantities are three smallest non zero lattice vectors. The vectors given by equation (2.2.2.3) define a set of points; that is known as the Bravais lattice of the crystal. The unit cell is the smallest volume for which translation from the cell interior point through the vector enables all other points in the crystal to be reached. The unit cell then contains just one Bravais lattice point and we can construct a unit cell about each lattice point. The fundamental concept in the description of any crystal structure is the Bravais lattice. Crystals have only 14 different possible parallelepiped network of points (i.e. the fourteen unit cell represents the only possible ways space can be filled without gaps, that is consonant with the restriction of translation and rotation symmetry ). These are the 14 Bravais lattices corresponding to the 7 crystal system ( symmetries). The list of the 7 crystal system and the 14 Bravais lattices is provided at the end of the next chapter.

3 SOME GROUP THEORY THEORY OF SYMMETRY SCHOENFLIES NOTATION NON-CUBIC CRYSTALLOGRAPHIC POINT GROUP

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To discuss the symmetry properties of solids, one needs an appropriate formalism. The most concise formalism for this is an abstract mathematical theory called group theory. Group theory was developed long before it found application in physics. With the development of Quantum Mechanics in the first quarter of the twentieth century, it was realised that this abstract mathematical theory can be of considerable use in analysing the eigen states of energy of a physical system. However, we shall be interested at this stage it use in describing crystal symmetry. To use the crystal symmetry operation as examples of groups, we start by introducing some group theory notations and defining various terms which helps in organizing in one’s mind the various sorts of symmetries that are presented to us by crystals.

3.1 Definition and Simple Properties of Groups

A group is a collection of elements with a rule to combine two elements given. This process of combination is called group multiplication which must satisfy the following properties:

- Closure. If g and h are any two elements of the group , then is an element of

- Associative law. If g, h, x are arbitrary elements of , then

- Existence of the identity element. There is an element, e in G such that , where g is any element in G, e is called the identity element.

- Existence of inverse. For every element g there exist another element (denoted by ) in G, such that . is called the inverse of the element g.

From the above axioms we can prove that and . By definition . Now multiply the relation from the left by e, we obtain

or . Thus . Similarly, multiply by g from the right, one obtains or . Thus In general group multiplication is not commutative (i.e. ). A commutative group is known as an Abelian group (i.e. ). For a finite group, the number n of elements is called the order of the group.We now consider an example of a group to fix our ideas. Example 1. The four numbers 1, , -1, where form a group under ordinary multiplication. If we multiply any two elements the result is one of the four elements. Thus closure is satisfied. The multiplication is associative. The element 1 is obviously the identity element. Also, and , showing that is the inverse of the element and verse versa and -1 is it own inverse. If we denote the element by , the elements of the group can be specified as

, , , . Such a group in which all the element are powers of a single element is called a cyclic group. For a cyclic group of order n, with generator ,

. This group constitute a particular case of the abelian group.

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A subgroup of is a subset of elements of which form a group under the same rule of group multiplication as in . For every group there are two trivial sub groups: (i) the identity element, e, by itself forms a sub group, (ii) also the whole group can be considered a sub group of itself. A subgroup of is called a proper subgroup if . In example 1, the numbers (1.-1) forms a subgroup of the original group. If , , are some basis elements of a group such that ; . It can be seen that . In , is conjugate relative to and in

, is conjugate relative to . It therefore follows that and are conjugate relative to each other. If , and

, 3.1.1

then the set of elements and form a class with the basis . This follows that two elements and form a class if there exist an element in the group such that (3.1.3) is satisfied. Suppose is an operator denoting the change of the coordinate system, then if we must have

or ie

Equation (3.1.3) constitutes a general definition: a set of members of a group that are obtained by similarity transformation of the form ( ) with the operator also being a member of the group constitute a class. It is easy to show that the elements of a class are conjugate elements.

It is obvious that , then it follows that, ,

or . We have use the fact that

. Since and, then is the definition of

the conjugate. It follows that and are conjugate to one another. In example 1, let us consider the elements conjugate to : , Thus is conjugate only to itself and forms a class of one element. The same is true for all the other elements. This is a general characteristic of all abelian groups. Also in every group the identity element forms a class by it self. This is

because the only element conjugate to e is e:

3.2 Unitary Transformation Let’s introduce the notion of the vector space. For this we construct the system of basis 3.2.1 having the property . The system in (3.2.1) is called basis of the vector space. The vector of a point in space in define as

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3.2.2 This is the element of vector space. The set of all unitary transformation in an n- dimensional vector space forms a group. This group is isomorphic to the group of all unitary matrices which forms a group under unitary transformation. Two groups and are said to be isomorphic to each other if there is a one to one correspondence between their elements i.e. . 3.3 Symmetry Transformation

We talk of a symmetry operation of a function as the operation of rotation, reflection and translation of the function which leaves the system of coordinate unchanged. Thus a symmetry transformation overlaps a body onto itself. Suppose we have a function ; . Then the operation displaces the function by a. The conversion is called the active point of view. For the passive point of view, we imagine the function remaining fixed and we rotate or translate the coordinate system. We take the active point of view in these lectures. The set of all symmetry transformations which leaves a physical system invariant forms a group. Group multiplication in this case is the application of two transformations successively. If each transformation leaves the system invariant, the product will also leave the system invariant. The product is thus a symmetry transformation. The successive application of three transformations obeys associativity. The identity element is the transformation that leaves the system undisturbed. Finally, for every symmetry transformation there is the inverse

operation. (for example, the inverse of the operation “rotation by in the anti-

clockwise direction about the z-axis” will be “rotation by about the same axis

in the clockwise direction”). The inverse operation will leave the physical quantity invariant and is thus a symmetry operation. A crystalline solid may also contain symmetry operations which are not group product of its rotation, inversion, and translational symmetry elements. There are two possible types of symmetry of this type: - Screw axis symmetry. A screw axis is the combination of a rotation about an axis with a displacement parallel to the axis. - Glide plane symmetry. A glide plane is a reflection with a displacement parallel to the glide plane axis.

3.3.1 Point Group

A complete set of symmetry operations in a crystal is it space group. Point groups are collections of crystal symmetry operations which form a group and also leave one point fixed. The point group controls the external morphology of a crystal. There are 32 such crystallographic point groups with 7 crystal system which is specified by the types of symmetry of the 14 Bravais lattices. In general associating bases of atoms to the 14 Bravais lattices gives a total of 230 periodic

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patterns ie there are 230 space groups. Of these space groups, 73 are simple group product of point groups and translational groups. These are the so called symmorphic space groups. The rest of the space groups have screw or glide symmetries. Point groups denoted in the Scheonflies notation are:

. Where the denotation C- cyclic, D- dihedral, S- spiegel (mirror), h, v and d are horizontal, vertical and diagonal respectively.

3.3.2. group

The rotation through an integral multiple about some axis is called an n-fold

rotation axis denoted by ; . correspond to rotation and . The structure is mapped back onto itself in this case. Any real crystal exhibits both translational and rotational symmetry. The mere fact the crystal must have translational symmetry places restriction on the type of rotational symmetry that one can have. This may be stated in the theorem; a crystal can have only one-, two-, three-, four- and six-fold axis of symmetry. The proof of this may be facilitated by the geometrical construction below:

Fig. 3.1 In fig. 3.1 is a vector drawn to a lattice point (one of the points defined by

), and is another lattice point. is chosen so as to be the closest lattice point

to in the direction of one of the primitive translations; thus is the minimum separation distance between lattice points in that direction. The coordinate system is chosen so that the y-axis is parallel to . It will be assumed that a line parallel to the x-axis and passing through the lattice point defined by is an n-fold axis of symmetry. Since all lattice points are equivalent, there must be

a similar axis through the tip of . If , then a counter-clockwise rotation

of about by produces a new lattice vector . Similaly a clockwise

x

y

z

ap

a a

rR

1RrR1

rR

R 1R

a

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rotation by the same angle of about produces a new lattice point . From

fig 3.1, is parallel to the y-axis and . Further,

. Therefore or

. This equation can be satisfied only for or . This is the result that was to be proved.

It may be shown that if there exists then obviously , they lie on vertical planes; ie and that we see further. Do the operation twice, thrice, …, k – times then we have the following:

Twice

Thrice

………. ………. ……..k-times

If n is a multiple of k then . If k=n , then ; The structure

overlaps or is brought back onto itself. e denotes an identity element under multiplication.

are rotation through k= 0, 1, …, n-1 about an axis through

the centre and perpendicular to the structure plane. Elements of the given

group are : , k= 0, 1, …, n-1 or

The order of is n, generated by the element . Generated implies

each element of the given group may be obtain by multiplication of

by itself an appropriate number of times. The group is abelian as

; has classes. It should be noted that

each has only one symmetry element of the n-fold rotation axis; has no symmetry element.

3.3.2.1 REFLECTION

A reflection maps every point onto it mirror image about a plane. Denote a reflection about a plane; a two-fold reflection about that plane is:

3.3.2.2 ROTATION-REFLECTION

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The operation followed by in a plane perpendicular to that rotation axis is an n-fold rotation–reflection (the group order is the number element in that set ).

is the operation of reflection about the plane perpendicular to the given axis.

AB- rotation axis, -planes

- solid angle between plane

and M- observable point

M B Fig. 3.2

denotes rotation about AB: points A and B are invariant under : . Consider fig 3.3

Fig. 3.3

and denote mirror reflection about planes and respectively. is the mirror reflection of relative to . Rotation through the solid angle ; 3.3.2.2.1 Thus, the product of two reflections about an axis perpendicular to two planes and is equivalent to twice the angle between the two planes . (3.3.2.2.1) is a reflection from the plane to . Rotation is in the reverse direction to

. Multiply (3.3.2.2.1) by :

; but , 3.3.2.2.2

Thus the product of rotation and reflection about a plane passing through the rotational axis is equivalent to the reflection in another plane intersecting with the first through half the angle of rotation. Denote inversion operator by : 3.3.2.2.3

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A

Multiple (3.3.2.2.3) by then we have or . Also . The group has two elements: i.e. e and and denoted as . Thus, the

group has only a center of symmetry.

3.3.2.3 group It is the group of all axis of symmetry of a regular prism . has the subset of

elements corresponding to the set of planes .

The group has 2n elements. The order of the group is 2n. The group is not abelian. E.g. :1) n-odd, rotation through and fall into one and the same class. Thus

the set of elements . may be taken as rotation in the reverse

direction to . Subtract from n , unity due to unit element; then (n-1)- even.

Thus the set of elements and fall into two classes . Among

rotations n-1 are pair-wise conjugate. They form classes each with two

elements , . form a class. All two-fold

axis form one class. Hence the number of classes q: .

Here all axes are equivalent.

2) n- even, we have and = constitute a class. The

rest of the n-2 rotations pair-wise conjugate gives again classes of two

elements each. It should be noted that . Thus the classes ,

, ; . Thus the number of classes with two

elements is . It should be seen that the number of classes q:

.

Consider the group . Examine the equilateral triangle in fig 3.4

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Fig. 3.4

group contains group with elements . Consider rotation

through rad about , , planes. These are two-fold rotations that correspond to two-fold axes and Examine also in fig 3.5

Fig. 3.5

In addition to the four-fold rotations axis ,the group contains the two-fold axis perpendicular to the four-fold axis plus additional two-fold axis ad required by the existence of the four-fold axis. The order of the group is 8 with the elements

.

3.3.2.4 T. Group (regular tetrahedron group) The system of axis of the T. group is the symmetry axis of a regular tetrahedron. The three-fold axis passes through each vertices of the tetrahedron. Three-fold

axis corresponds to two rotations and . There are two-fold axes at the

opposite faces of a tetrahedron. The two-fold axis correspond to one rotation

The above rotation correspond to an identical element . The two-fold axes are mutually perpendicular. The three-fold axis forms with each other the tetrahedron angle . The T group has: an ide - an identity element ;

- three rotation ;

- four rotation of and

each;

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Thus the group has N elements: . The order of the group is

thus 12 . The T group as seen has four classes: , , , ;

here indicates the axis. The number of classes q = 4 .

3.3.2.5 O. Group ( octahedral group)

It is a cubic group. It is a group of all symmetry operation of a cub in which

there are three four-fold axis , , through the centre of opposite axis ;

four three-fold axis and through opposite vertices ; six two-fold axis

through the middle of opposite sides. The elements of the group are: ,

, , , . The class number is, and

number of elements .

GROUPS OF THE SECOND KIND

3.3.2.6 Group is the simplest group of the second kind. It is a cyclic group. Thus the entire group is obtained from the elements of the first kind. is group rotation about a 2n-fold rotation-reflection axis. The group has 2n elements:

.

The operation is equivalent to a rotation-reflection about a horizontal plane:

, as two reflections is equal to an identity element. Consider

: . Thus the group has two elements and I and denoted as

Consequently the group has 2n elements.

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Consider ; then: , then for , the

group has an inversion I. Denote such a group by . The

is abelian; As each element form a class in the group, then the class number is .

3.3.2.7 Group

In addition to the n-fold axis of , it contains a single mirror plane perpendicular to the axis. The has 2n element; i.e. n-rotation:

and n rotation-reflection operations:

. Here . As each element in the group forms

a class there are classes. The group is Abelian.If then the group has a centre of symmetry:

The simplest group has two element and denoted as ; ,

. 3.3.2.8 Group

In addition to the n-fold rotation axis of the group has a mirror plane that contains the axis of rotation plus as many additional vertical mirror planes as the existence of the n-fold axis requires. The has 2n elements; i.e. n-rotation

about the n-fold axis: and n-reflection in the vertical planes:

. The group is not Abelian. The group is isomorphic to the

group:

, :

, :

The distribution of element of in terms of their classes depend on even and odd n.

1) n-even. , Constitutes a class each and the rest of n-2

rotation which are pair-wise conjugate gives again classes of two

elements each. The classes are: , , and

classes.

2) n-odd. All symmetry planes are equivalent. Reflections in them belong to one and the same class . Amongst rotation about the axis , n-operation are

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pair-wise conjugate with each other. They form classes each with

two elements: . The set form also a class.

Thus classes.

3.3.2.9 Group

This is the most symmetric of the groups. It contains all elements of plus a mirror plane perpendicular to the n-fold axis of . In addition to the n-fold

elements in : , , the group has n-reflections

and n rotation-reflections operations: . Thus the group has 4n

elements; the order of the group is 4n.The reflection commutes with the rest of the elements of the group ; , where .

* If n-even, then an inversion of an element in the group: Thus from the above , the number of classes in is doubled compared to the

group

1) n-odd ;

2) n-even ;

However the classes in coincides with those of (rotation about the axis) and the rest obtained due to multiplication by . If n-odd, then reflection in the vertical plane belongs to the same class and if n-even, then they form two

classes. The reflection-rotation operation and are pair-wise conjugate with each other.

2.3.2.10 Group

contains elements of plus planes containing the n-fold axis that bisect the angles between the two-fold axis. Elements of this group are:

- The group with the 2n-fold rotation–reflection about the horizontal

axis with elements: ; it has 2n elements.

- The reflection in diagonal plane with elements: ; it has

n elements- The reflection about the two-fold axis with elements: ; it

has n elements Thus the group has 4n elements. The number of classes is independent of the parity of n.

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Considering group, then for even-n we have a two-fold rotation axis and for odd-n , an inversion I. It follows that the number of classes in is

.The group is not Abelian. The elements of the group are:

elements

elements

elements

The total number of element is 4n. The element and coincide. It is seen

that for even-n we have and for odd-n, we have I. Thus for class number, we have The group is a group of all symmetry elements of a figure that is made up of two regular n-fold prism place on each other such that the side of one passes through the centre of the other.

3.3.2.11 Group

The group is an n-fold symmetric group of a regular tetrahedron including all improper operations. By improper operation any operation that takes a right handed object onto a left handed one. The group has 3-fold axes with elements

and giving a total N = 8 elements, and three 2-fold axes with element

giving a total of N =3 elements, and then an identity element . It has the

mirror planes: , making a total of six and three of the elements:

and , as making a total of six element. Thus in the group, 24 elements are found:

Element Number of elements, 8

3

16

, ( ) 6

The classes are: , , , , . It follows that the number of

classes is . The group is not abelian.

3.3.2.12 Group Adding an inversion to yields . Thus

The group has N = 24 elements; 12 from group and 12 due to inversion. It has classes; 4 from group and 4 due to inversion.

19

3.3.2.13 GroupThe group is an n-fold symmetric group of a cube (thus the O group) including also improper operations that the horizontal reflection ( ) admit;

It is obtained by adding an inversion to the O. group. The O. group has:

Element Number of elements, , 9

, 8

6

1Total=24

Considering the fact that , the group has N = 24 elements; 10 classes as compare to 5 classes of the O. group.

CRYSTALLOGRAPHIC POINT GROUP

SYSTEM NOTATION BRAVAIS LATTICE

CLASSES ELEMENTS

TRICLINIC 1 1, 2

MONOCLINIC 2 2, 2, 4

RHOMBIC 0R ORTHOROMBIC

4 4, 4, 8

RHOMBOHEDRAL OR TRIGONAL

2 3, 6, 6, 6, 12

TETRAGONAL 1 4,4,8,8,8,8,12

HEXAGONAL 1 6,6,12,12,12,12,24

CUBIC 3 12,24,24,24,48

The list of the 7 crystal system and the 14 Bravais lattices

20

PROBLEMS

1. Consider the symmetry group of transformation of an equilateral triangle (i.e. the dihedral group ).

a) Write down the symmetry operations and work out the group multiplication table.

b) Identify all the subgroups c) Identify the classes.

2. Show that an element of a group constitutes a class by itself, if and only if it commutes with all the elements of .

3. Show that the set of elements which are the inverses of the elements of a class of a group also forms a class.

4. Consider a class of a group. Each element of is use to generate a cyclic group. Prove that the order is the same for each element of the class.

5. Prove that a crystal can have only one-, two-, three-, four-, and six-fold axis of symmetry.

21

4. PERIODIC FUNCTION OF RECIPROCAL LATTICE MILLER INDICES

4.1 RECIPROCAL LATTICE

We introduce a second Bravais lattice namely the reciprocal lattice with the primitive translation that is shown further in the lecture. We select the Cartesian system of coordinate , , for a primitive cell. The vectors , 2, 3 are

dependent on the crystal symmetry. The vectors and are in the same

direction as in figure 4.1.1.

22

Fig. 4.1.1

For the example of the use of the reciprocal lattice, we consider the microscopic physical space value (said ) in the crystal to be periodic. Periodicity implies that

4.1.1

When we wish to deal with crystals of finite size, the question on boundary condition usually arises: suppose the crystal is very large, and then we expect the precise form of this not to affect the physical description of properties over the broad crystal. We call this the cyclic or periodic boundary condition. Here the value of any wave function is the same at equivalent points on opposite sites of the crystal. This periodicity is represented mathematically by (4.4.1) or

Let N be the number of primitive cells counting from the direction of each primitive translation , = 1, 2, 3. Suppose this brings us to a point not only physically identical but mathematically identical to the original point. Thus we

may have . This follows that so that every eigen value

of is root of unity of the form . As the translation operator

commutes amongst themselves and with the Hamiltonian , then for all vectors it may be found that

We expand (4.1.1) in Fourier series:

4.1.2

Here , 2, 3 are some integers. Let`s do the transformations , = 1, 2, 3 4.1.3 Where is the transformation matrix, then

4.1.4 is the desired expansion using the reciprocal lattice: here ,

. In order to find we use property (4.1.1):

In order that it should be identical to (4.1.4), then or

23

4.1.5 where is some integer and:

4.1.6

  ;  ; 4.1.7

, , are not coplanar vectors. It’s a necessary and sufficient condition to define vector . From (4.1.7) it is obvious that has the dimensions of inverse length. Thus it is a reciprocal vector to vector and is a direct vector, then:

 ;  ;

Here, is the volume of elementary cell of the direct lattice and

, 2, 3 is the vector of the reciprocal lattice with primitive translation

4.1.9 It is obvious that 4.1.10

Take scale product of (4.1.8) by either vector , or then

4.1.11

where is the volume of the elementary cell of the reciprocal lattice. We may obtain (4.1.11) in another manner. Let’s represent the expansion (4.1.2) in the form

 

where is the volume of the unit cell of the direct lattice. Suppose and be

any two vectors. Consider first the Fourier representation of the Dirac delta function:

4.1.12

This is qualitatively obvious as completely localized functions must have the Fourier component. Consider equation (4.1.1`), then we may be able to reduce equation (4.1.12) to sum of integration over the unit cell namely

This follows that

4.1.13

Here the generalized Kronical symbol is defined 4.1.14

24

Equation (4.1.13) implies the result ( ). When use in 4.1.4, suppose and are two distinct direct lattice vectors; then the analogous equation

4.1.15

where is the volume of the unit cell of the reciprocal lattice which is called the Brillouin zone (BZ). It is the basic region inside which the dispersion relation of electron waves is discussed. Let’s consider two expansions that are formally true:

4.1.16and

4.1.17

From equation (4.1.12), we may write:

4.1.18

This yield .The reciprocal lattice has all symmetries as the direct lattice. In reciprocal space is set of reciprocal lattice. Its sites are specified by the radius vector related to the translation vector of the direct lattice through (4.1.6). In real space, each site may be surrounded by a polyhedron with properties: All points within it are near to the given site than any other one. Such a construction is the Wigner-Seitz cell. The first BZ is the analogue in reciprocal space. Dispersion laws, calculations in the band theory are carried out in the the BZ .

4.2 MILLER INDICES

Differentiate atomic planes in crystal. This may be done using Miller indices. The coordinate of three atoms lying on the plane defined the position and orientation of crystal plane. Select any atomic plane with three atoms as in fig. 4.2.1

25

Fig. 4.2.1

Here , , are coordinates where the given plane intersects with the main coordinate axes (in unit of the lattice constant). Examine the ratio

; ; 4.2.1

Here , , are called Miller indices denoted as . , , are whole numbers. If the plane intersects the given coordinate axes at infinity, then zero is the corresponding Miller indices. The selection of the indices implies a separate plane or a family of parallel planes. The Miller indices is negative if the plane intersect the given coordinate axes in the domain where it is negative. The minus sign is put over the number not in front. For example we write and not

.

Problems1. Draw the plane of the faces of a cubic crystal with the following Miller Indices: (100), (010), (001), .2. Show that a space lattice of a face centered cubic can be body centered cubic and vice-versa in a reciprocal space.3. Show that in a cubic system, the angles between normal to the planes

and is defined by the formula

26

5. LATTICE DYNAMICS

5.1 VIBRATIONAL MODES OF A MONOATOMIC LATTICE5.1.1 LINEAR MONOATOMIC CHAIN

The atoms in a solid are always in constant thermal induced motion. Models in which we discuss this motion are required. We may conveniently illustrate many of the critical concepts in one dimension. This heading examines in details the dynamics of a linear chain of atoms. For the sake of convenience, we assume only one atom per unit cell for the given crystalline solid. Here we consider only classical motion of the chain with no reference to quantization:1-We assume atomic arrangement with minimum energy to be a periodic array of atoms. In order to avoid surface effects, we consider the familiar periodic boundary condition: The crystal with it N lattice vector is part of an infinite Bravais lattice; then the potential energy of the system is expanded in powers of atomic displacements from the equilibrium position. The harmonic approximation is used in the expansion of the potential energy where orders higher than the quadratic are ignored. The approximation is good for mere solids at room temperature. This can be further defined using the perturbation method. The harmonic approximation leads to a classical equation of motion that is conveniently solvable by introducing special linear combination of atomic displacement called the normal coordinates. Express the potential and kinetic energy of the system in terms of the normal coordinate. This contain no cross terms. In it, we have only quadratic terms. This reduces the equation of motion into a set of independent harmonic oscillator equations that are easily solvable. The solutions of the system are called normal modes.2-Despite large size and apparent complexity of the system, the ideal crystal perfect periodicity enables us to define required normal modes frequency of the

27

system. This we assume periodic boundary conditions. Generally, crystal motion is examined as a linear superposition of normal modes. These normal modes correspond to the solution of traveling plane waves. Apply some of the properties of these solutions to the case of a one dimensional chain. Examine properties of a finite chain of N atoms with interatomic distances

(lattice constant) as in fig 5.1.1

I

II

Fig. 5.1.1To find normal modes, we impose further, boundary conditions at the end of the finite chain. Neighboring atoms are coupled by a spring with force constant . In fig. 5.1.1 I, we have equilibrium position, and in 5.1.1 II, the atoms are displaced out of their equilibrium position with displacement of the atom . For the above system each atom has a degree of freedom, and for the entire system, N degrees of freedom. The above model is well described by a linear elementary Bravais lattice cell where the positions of the atom are determined by

the translation vector where is an integer showing the equilibrium

position of the atom in the given chain. The atoms are linked among themselves in the chain by coupling force such that the excitations are propagated in the chain in the form of waves in compression, and all the atoms are displaced from their equilibrium position. We assumed that the coupling between near neighbors is much stronger than that between distance neighbors. If we consider the transmission of plane waves along the ox-axis of the chain, then the displacement where is the position and , the time moment of the wave. Considering the linear array of atoms with the interatomic spacing , then the equation can only be described at each atomic site, and have no value for intermediate values of , and as a consequence, the given equation is made more precised in the form , where is considered as the displacement of the atomic site at the time moment, . If we assume that the displacement of the atoms from their equilibrium position is small, then the displacement function (the potential energy) is small such

that we expand the function in atomic displacement

28

5.1.1The subscript zero means that the derivatives are evaluated at equilibrium. The ground state configuration is the equilibrium value and thus may be the reference point (origin). We neglect as it will not affect our answer. As the potential has a minimum at the equilibrium position; then it follows that

 . This is because, in the equilibrium configuration, there is no net force

acting on the nucleus. The dynamic problem which (5.1.1) gives rise to is only solvable in closed form if the anharmonic terms are neglected. For small oscillations their effect is presumably much smaller than the harmornic terms. The cubic and higher order terms are responsible for certain effects which completely vanishes if they are left out. Whether or not one can neglect them depends on what one wants to describe. We need anharmonic terms to explain thermal expansion, a small correction (linear in temperature) to the specific heat of an insulator at high temperatures, and the thermal resistivity of insulators at high temperatures. The effect of the anharmonic terms is to introduce interactions between the various normal modes of the lattice vibration. This shall not be our point of concern in this heading, so we neglect their effects. At equilibrium, is minimum, thus the remaining term in (5.1.1) is quadratic in displacement and describes the harmonic oscillations of the atomic lattice:

5.1.2

Such an approximation is called harmonic approximation. Here and corresponds to the displacement of two neighboring atoms. Further to avoid surface effects, we take the periodic boundary conditions: the crystal with N lattice vector is part of an infinite crystal that is made of crystal vibrating with identically atoms. In order that the chain should have the property of translational invariance, it is necessary that

5.1.3

where is a tensor. The kinetic energy of the lattice is

.

Let’s find the equation of motion of the system. For that we write the Lagrangian of the system as follows

29

and the equation of motion

Considering the fact that

, then

5.2.4

The solution of this equation then proceeds in the usual way. In order to avoid surface effect, we consider the familiar boundary conditions. The translational symmetry of the lattice guarantee’s that the tensor depends only on the difference . For crystals with more than one atom per primitive cell will require separate equations for each of the atoms per primitive cell. Equation (5.1.4) is a large number of couple differential equations. It has no direct solution. If we suppose that the wave length , then we have a continuous distribution of the given wave as is the waves in a string. We consider interaction constant representing coupling between pairs of atoms. We assume coupling between near neighbors to be large as to that between distant neighbors. Hence if (5.1.4) represents point particles coupled by springs, then we expect the spring constant to be larger for nearer neighbors. Hence we are concerned with short range interactions. Let’s find the normal modes of vibration, i.e. type of movement such that all the atoms oscillates with time with the same procession according to the law . We find the solution of (5.1.4) in the form of progressive wave 5.1.5Then considering (5.1.4)

5.1.6

The tensor P has the property of translational symmetry. If we do the translation and , then the difference is invariant relative to this

translation and consequently and should describe a given defined state.

Thus the quantity and are connected through the relation

and 5.1.7This follows that is a complex value function and may be represented , where has the dimension of inverse length and and implies that

; 5.1.8

where is lattice vector and has the sense of wave vector. Considering (5.1.5) and (5.1.8), it follows that 5.1.9

30

Here , where is the wave length of the wave. If we substitute (5.5.9)

into (5.1.6), then it follows that

5.1.10

if we let then

5.1.11

(5.1.11) is a dispersion relation. In the simplification, the sine function counsel’s out as they are odd function with respect to and the cosine function remains as they are even functions. (5.1.11) shows that the frequency of the the distribution of the wave in the crystal for different are different. This phenomenon is called dispersion. It follows that , which implies the system has a certain centre of inversion. Considering (5.1.11), the frequency of vibration of the crystal does not depend on . This implies that all atoms in the chain vibrate with the same frequency.

Let’s examine how changes. Let then

5.1.12

from where it follows that  . So and are physically equivalent quantities. This of course is true since the crystal is translationally invariant. Let’s find the domain of change of . Let’s find the domain of change of . For the

minimum value, it follows that  ; since then it follows

that . This range of is called the first Brillouin zone (BZ); which is

defined in physics as the zone for different . The centre of the BZ corresponds to , where is an infinitesimal value of . This implies, we may expand

in the sense that

5.1.13 We consider (5.1.11) and we take the limits , then it follows that

. This expresses the fact that the forces on any atom are the same if each atom is displaced from the equilibrium position by the same amount. Also since is an even function of and hence , thus for sufficiently small , or , 5.1.14Equation (5.1.14) is a dispersion expression for waves propagating without

dissipation; i.e. their group velocity = equals their phase velocity

; =

31

This is the type of vibrations in a continuum. The small approximation is a low frequency or long wavelength approximation; hence the discrete nature of the lattice is unimportant. The significance of the phase velocity should be questioned if we consider that many may be use to describe the same lattice wave. If we consider a periodic medium, say a crystal in contact with a homogeneous region, then the phase velocity is not a uniquely defined concept; which follows that

, here is the minimum wave length of the distribution of the

wave in the chain. This follows that the entire chain is made of two sublattice where all the even atoms vibrate out of phase with respect to the odd ones. The wavelength less than cannot propagate down the linear chain lattice. We know that the wave motion are periodic occurring time and space. If the adjacent atoms are always out of phase and for the adjacent atoms should be in phase. In this case, is not a wave motion but a translation of the entire chain. Suppose we have a chain of N atoms as in fig. 5.12

1 2 3 ….. N

We enclose this chain so that the distance between the first and the is a. The enclosed chain has N intervals. In order that the translational symmetry of the crystal should not be disturbed, they should satisfy the condition 5.1.15 Condition (5.1.15) also known as the Born Von Karman condition implies that the numbers concerns one and the same atom. This implies that or

, and

5.1.16

This implies that is quantified. This follows that is quantified and is found only in expression of the type for which . Then there is no

change in adding multiple to the wave vector . Considering the inequality

and (5.1.16), then we have

5.1.17

Thus the allowable quasi parameter takes countable even values and N the number of elementary cells in the chain. It is seen that the same cell of the atomic

displacement may be represented by any one of the vectors . It should

be noted that is the vector of the reciprocal lattice for the one dimensional

crystal. The redundancy is often expressed as periodicity in reciprocal space of

32

certain physical quantities is a consequence of the periodicity in real space of the physical system. It should be noted values of magnitude greater than gives no new solution to the equation of motion and thus gives no new information and should be of no interest. Consider a particular case, let from (5.1.8), we consider three values , , that correspond to  ; respectively. If we consider the

fact that then

then 5.1.18

Considering (5.1.4) and (5.1.18) then we have

5.1.19

or One the right hand side of (5.1.19`) of the square bracket, we may see a displacement to the left and to the right of the given atom. Equation (5.1.19 `) resembles the force in classical mechanics . From the general law of dispersion (5.1.11) we have: and considering (5.1.18)

5.1.20

and or

5.1.21

The (+) or (-) sign in equation (5.1.21) represents waves traveling either to the right or to the left. The negative sign in (5.1.21) also correspond to the domain of negative . The frequency and wave vector may not independently be chosen. Specifying one, the others are determined from the dispersion relation (5.1.21) for the chain. From the dispersion relation (5.1.21), it follows that, what ever the choice of the wave vector or wave length , the angular frequency may never be greater than unity. This is contrary to an ideal string that will propagate the wave in any frequency. We may see from (5.1.21) that the vibration of the atoms is independent of . This implies that all the atoms in the chain oscillate with one frequency.

33

We may see that in fig 5.13 within the interval the dispersion curve

flattens at the edge of the BZ. This flattening is thought of as reflected waves

(waves in the opposite direction) that gradually increases until at and the

amplitude is equal to the incident component. It may be seen that the graph is symmetric between and . This implies

that waves to the left and right are identical. There is a repetition at each  ;

where is an integer. This identical set of atomic

displacement may be represented by any of the set of the wave vector

. Note that at long wave length the frequency is linear in wave number . At large wave number, the dispersion curve approaches the BZ and phase horizontal.

Consider N atoms and the periodic boundary condition; it follows that we have N allowed values of as the given wave. Considering N atoms, we have N degrees of freedom and thus as a consequence, we may say all the modes are found. Considering (5.1.21) it can be seen that, we have only one branch to the vibration spectrum. Formula (5.1.21) is the dispersion law and its graph is shown in fig 5.13. From the figure it is seen that the maximum frequency that the wave may have in the lattice is . This frequency corresponds to the value

. The values outside the given limit do not give us any new

information.

In the case of long wave, i.e. for small; , in this case (5.1.21)

takes the form

5.1.22

q

q

dq

d

a

a

0

Fig. 5.1.3

34

which is a linear dependence of . The phase velocity (wave velocity) takes the form

5.1.23

and the group velocity

5.1.24

This implies that . That should have been expected as waves of long wave length will not be sensitive to the discreteness of the structure. In other words a large number of the atoms will participate like on a homogeneous line. Thus long waves correspond to the propagation of sound wave. There is the puzzling result

that the group velocity goes to zero at . This confirms the fact that in

the chain, the modes of vibration characterized stationary waves. If we consider

that then we expand in Taylor series:

.

If we consider only the first two terms of the expansion considering that powers higher than are infinitesimally small then,

This shows formerly that the group velocity of the wave packet is . As we

have seen this velocity is zero at the boundary of the BZ . This is energy

associated with the wave packet and located within it. The group velocity is the

velocity with which energy is carried and that velocity goes to zero for . If

we consider , the velocity of an acoustic wave where E is the Young’s

modulus and the density, then

and .

Thus the phase velocity

hence 5.1.26

and

5.1.27

It is seen that for small (see fig 5.1.4) we have

35

0

Fig. 5.1.4

5.1.2 DENSITY OF STATE

Consider restricted to the first BZ for convenience. The notion of normal modes with chosen outside the first BZ is physically indistinguishable from the one association with within the first zone with being the reciprocal lattice vector. We evaluate how many modes have frequency in the range

. Defined the number of modes and call the distribution function the density of mode. The density of mode refers to the number of

modes d£ per unit frequency interval : . In the quantum picture the

given mode correspond to the different possible states of phonons. The modal frequency corresponds to an allowed phonon state of energy . This follows that the density of modes may be referred too as density of state of phonons. We already know that the condition for the allowed wave vector for the

wave in a chain of N atoms is and then and from (5.1.1) we

have:

5.1.2.1

and or

; 5.1.2.2

It should be noted that, is finite at the point of frequency . The divergence at is not physically unrealistic. Since it is an integral, this singularity points no problem. It may be shown that the total number of modes is N;

36

5.2 VIBRATIONAL MODES OF A DIATOMIC LINEAR CHAIN

We generalize discussion by considering the propagation of lattice vibration in crystal with more than one atom per unit cell. We may do this such that the mass of nearest neighbors differs from the mass of next nearest neighbor. The manner of sodium chloride structure makes a simpler generalization by modifying masses at alternate points as in fig 5.2.1.

Fig. 5.2.1The propagation of waves remains one dimensional. Our analyses are carried out in analogy with those in heading (5.1.1). In the case of two atoms per primitive cell, there arises a new phenomenon. In the dispersion law, there arises two branches called the acoustic and optical branch. The optical mode appears in the infra-red frequencies and can be excited by infra-red radiations. Consider for convenience a linear Bravais lattice for which the elementary cell of parameter has two atoms. Such a system has 2N degrees of freedom. In fig 5.1.1 is a linear chain with two types of atoms arranged alternately. A distance separate two neighboring atoms. Suppose we have light and heavy atoms with masses and respectively (i.e. ). We consider the force of interaction between atoms to be short range. This implies that the atoms interact only with their neighbors and as a result the equation of motion takes the form:

5.2.1

Where is a force constant related with the elasticity. If we take into consideration the fact that the interaction of the atoms with different masses has different amplitudes, then we find the solution of equation (5.2.1) in the form of progressive wave in the crystal: ,

5.2.2

It can be seen in and that we assume the same and for the motion of two types of atoms. This is from the fact that, we assume a simple wave motion for which both types of atoms participate. It should be noted that though the

37

displacement of both types of atoms are different, they both contribute to one and the same wave disturbance. If we substitute (5.2.2) into (5.2.1), then we have:

5.2.3

This system of homogeneous equation has a solution only if their determinant is equal to zero:

= 0 5.2.4

The determinant gives immediately the dispersion law:

5.2.5

Let , , 5.2.6

Then

5.2.7

The negative values of have no physical sense and as consequence, we are interested only on the positive values.If we consider the case of (5.1.3), then each value of correspond to each value of (univalent dependence of on ) and for the case of (5.2.7), each value of

corresponds to two values of ( bivalent dependence of on as a result of sign) and as consequence, we have two modes of vibration of the type (5.2.2)

which repeat for every . It should be expected as unit cell in this case has a size

of and the first zone runs from to . We have N values of in the

given interval and thus, for the two complete branches, are required 2N normal modes corresponding to 2N degrees of freedom of 2N atoms that are confined to a line. Examine what differentiate the two branches in (5.2.7):

38

5.2.8

called acoustic branch and

5.2.9

called optical branch. The splitting into acoustic and optic branches occurs for any crystal with two atoms per primitive cell. It occurs also for the transverse as well as longitudinal modes. For a general crystal with two atoms per primitive cell, there are six modes. Our structure chosen for treatment is selected for mathematical convenience. The typical crystal with two atoms per primitive cell is NaCl structure. Let’s

investigate the two branches. Examine the Brillouin’s zone , if we

consider the fact that and are even functions , then we limit ourselves to

the domain .

For = 0, we have (0)=0, (0)= 5.2.10

For = , we have

( )= , ( )= 5.2.11

For an infinitely increasing wave length ( and ), we expand (5.2.8) and (5.2.9) in MacLaurin’s series limiting ourselves only to the first term of the series:

5.2.12

and for the optical

then

5.2.13

39

It is seen that the dispersion of the optical branch is weak at the neighbourhood of = 0. Relations (5.2.12) and (5.2.13) are shown in figures 5.2 2. It is seen that the

optical branch decreases in a parabolic form.

forbiden zone

0

Fig. 5.2.2 Consider the group velocity for the acoustic and optical branches

 ;

For = , if then . It follows that ,

if (i.e. ) then ;

Thus for ,

* For = 0, considering (5.2.12), we have and from (5.2.13),

we have . Examine the case where (see figure

5.2.2).It can be seen that the transition from the compound to a simple lattice will take the lattice constant twice less than the same, it follows that the Brillouin’s

zone should be increases twice for the compound lattice or

40

and for , it follows and we have figure 5.2.3 and fig

5.2.4. If in (5.2.6) and (5.2.7) we let , then

5.2.14

and

5.2.15

Fig. 5.2.3

0

0

Fig.5.2.4

In the previous heading (see 5.1.2) we had and if we

substitute in place of the quantity , then we have . It

follows that in the compound lattice, the acoustic branch satisfies the limiting

case. If from the part of the branch ( ), we try to do the change of , =

, g = 1 then for , the optical branch is the prolongation of

41

the acoustic branch ‘(see figure 5.2.4). Then, we may see that the cosine solution should be the duplication of the sine solution .This implies that, each point on the cosine curve represents already a mode. This is an extended zone representation .We examine the physical sense of the difference between the acoustic and optical mode of vibration of atoms in the chain. In this connection, we compare the amplitude of vibration and the phase of this vibration of neighbouring atoms of each of the branches. It should be noted from (5.2.2)

, from where we have

5.2.16 We examine the system of equation 5.2.3 from where we have

5.2.17It follows that the amplitude are complex values. However, the physical sense

belongs to the real part and at the point = 0 and = (limiting vibration) they

are real. Examine the case = 0, that is the case of long waves length , in this case , we have

5.2.18

It implies that for the case of long wave vibration, the displacement of light and heavy atoms coincide. The elementary cell vibrates as a whole. The vibrations of neighbouring atoms are in phase. This follows that, both type of atoms move together in a long wave length; pressure wave. We thus have a vibration of a density and as the vibration of a density is sound vibration hence, the name acoustic vibration. Consider (5.2.7), (5.2.6), and (5.2.9), then for = 0, we have

This implies that one vibration lacks the other in the cell i.e. the atoms vibrate in opposite direction ( in opposite phase). The two atoms in the cells are out of phase by . The centre of mass of each cell that has two types of atom is at rest; we have 5.2.20Dividing (5.2.20) by :

42

=0

we have the displacement of the centre of mass. The heavy and light atoms have zero displacement for long wave length. If there is a charge difference of the two atoms, this causes a polarization wave with wave vector . Such a system may either absorbs or emits light from where the name optical vibration. For mere solids, this frequency is similar to that of infrared light and thus the name optical vibration. Thus, optical modes occur at infra-red frequencies. They can be excited with infra-red radiations.

Consider the case of the limiting value for the Brillouin zone

5.2.21

This follows that in the limit of short acoustic wave, the light atoms are at rest and the heavy ones displace and as consequence the density changes as the atoms are heavy. Let’s consider the case of the optical branch,

5.2.22

It is known from (5.2.11) that , . Let’s remove the

indeterminacy in (5.2.22). Suppose =  :

5.2.23

.

We expand this taking that is very small then : . This implies

that the heavy atoms are at rest and the light atoms are vibrating.

5.3 VIBRATIONAL MODES IN THREE- DIMENSIONNALCRYSTAL

We consider the lattice dynamics in the crystal base on the harmonic interaction potential. The force constant on the nearest neighbour interaction is considered stronger. We consider a crystal in which every atom undergoes small oscillation around their equilibrium position. Considering a harmonic potential, we find solution to the equations of motion of atoms in the form of plane wave (sinusoidal or cosinusoidal). Sinusoidal or cosinusoidal time dependant is determined by the

43

phonon frequency that is a function of the phonon wave vector . We consider an elementary cell with s atoms that have masses , . The crystal with

volume V= has elementary cells with being the volume of an elementary cell has a total of Ns atoms. Each atom in the crystal has three degrees of freedom, then for Ns atoms, 3Ns degrees of freedom. Consider a crystal to be a parallelepiped with edges of length that are parallel to the

corresponding lattice vector. Taking as basis vectors , we denote the

equilibrium position by  :

5.3.1

where are integers. The displacement of the atom from the

equilibrium position is denoted by so that the position vector of the

atom is : where has cartesian components , .

We proceed classically then the total kinetic energy may be written:

5.3.3

where is the mass of the atom. Since the vibration of the lattice involves, small execution from the equilibrium position, then we may expand the many body potential in powers of atomic displacement  :

Here, the subscript zero implies that the derivatives are evaluated at the equilibrium configuration. The ground state potential energy , may always be said equal to zero and there by makes a reference point. This will not affect our answer. We assume that the electrons in the crystal always have time to adjust it selves to the configuration with lowest energy even during crystal vibration. In fact at the equilibrium configuration, there is no net force on the nuclei. If we limit ourselves to the harmonic approximation in which terms in the expansion of the order higher than quadratic are ignored, then W becomes:

5.3.5

where 5.3.6

The quantity 5.3.6 depends on the atomic arrangement of the crystal and the interatomic potential. It is seen that the harmonic potential W of the 3-dimensional crystal is written in terms of the displacement vector of each atom from its equilibrium position as in (5.3.5). The sum externs over all pairs of atoms located at n and n’. At low temperatures ( ), it is not too unreasonable to go by the harmonic approximation. The separation between atoms is of the order of one Armstrong

44

and at some temperature; vibration has amplitude of the order of 0.1 Armstrong. Experiments will show our idealisation to be false. In our case, the general approach is to make idealisation then find corrections to this approach to give better results. It should be noted that the harmonic approximation is a good approximation; a classical mechanical theorem is used to have exact solution to the equation of motion. This can be done by first finding special linear combinations called normal coordinates of the atomic displacement coordinates. The potential and the kinetic energy of the system contain no cross terms if express in the normal coordinates, they contain only quadratic terms. We then find equations of motion that reduce to set of independent harmonic oscillation equations that are easily solvable. The solutions are called normal modes of the system. Write the Lagrangian L of the system; , and then the equation of motion becomes:

or 5.3.7

This gives a system of 3Ns equations. These equations are true for each Cartesian component of displacement . This shows that (5.3.7) forms a large number of coupled differential equations and thus may not be solved directly.Consider the properties of (5.3.6):

1) . It is invariant relative to all indices .This is as a result of

the independence of the order of differential equation in (5.3.6).

2) . The expression in (5.3.5) is invariant under translation

by the lattice vector (5.3.1). 3) We suppose that at the initial moment, the crystal is at the temperature T=0. This follows there is no vibration and consequently, the displacement is a constant. If all atoms in the crystal have the same displacement say from the

equality: then the entire system is simply displaced without interval distortions. Then the right hand side of (5.3.7) will have the same value as when the displacement vanishes:

5.3.8

from where is followed that

5.3.9

which is the condition that (5.3.8) vanishes for arbitrary . Equation (5.3.9) expresses the fact that the forces on any atom is zero if each atom is displaced from the equilibrium position by the same amount say . Expression (5.3.5) is invariant relative to the translation by the lattice vector (5.3.1). Then it satisfies Bloch’s theorem:

5.3.10

45

where is the phonon frequency associated with the wave vector . Substitute this in (5.3.7)

. 5.3.11

Let’s call

5.3.12

the dynamic matrix. Thus the phonon properties can be determined from the dynamic matrix . If we move from the sum with respect to n’ to the sum with

respect to , then we see that the right hand side (5.3.11) is independent of n. We expect that as the left hand side is independent of n, then it follows that:

  5.3.13

Introducing the Kroneker dalta symbol, the left hand side of (5.3.13) can be written in the form

then

5.3.14

It is a system of homogeneous equation for the quantity . This system has non trivial solution if the determinant is zero. This is a condition for which the phonon frequency may be obtained. Thus

5.3.15

It is a determinant of the type 3s 3s with 3s independent eigen values and there are (3s-3) optical modes. The number of equations in (5.3.14) is obtained from the number of possible polarization direction (three of them). That is two transverse and one longitudinal times the number of atoms in the unit cell. Considering the fact that the production is a factor N smaller than the number of equations in (5.3.7), then the solution of (5.3.14) is quite possible. In principle, the system should be solvable for all possible value of q for which we have N of them. Practically, it is sufficient to solve for a reasonable number of q. Other solution are obtained from interpolation, thus the equation of the determinant (5.3.15) gives the phonon spectral of the crystal. This gives us information about the rule of electron-electron, electron-ion and electron-phonon interaction. It explains phonon characteristic responsible for atomic properties.

PROPERTIES OF THE DYNAMIC MATRIX If we consider (5.3.12), we see that the dynamic matrix is hermitian. That is it takes real values 5.3.16

The characteristic equation has a real spectrum. This implies that is real. Physically, it is obvious that may not be less than zero as considering (5.3.10),

46

there appears the term which implies the instability of the lattice. If we solve

(5.3.15), then we have 3s different values: , =1,2,

…,3s ; is a 3-dim vector and thus, the surface is a 4-dimensional hyper

surface. If we do the change then it is easy to see that 

5.3.17

and if we substitute in (5.3.15), we see that the determinant changes rows for columns and vice-versa and thus

5.3.18

Since , it implies

5.3.19

If in (5.3.11), we let then

5.3.20

Let

5.3.21

where is the vector of the reciprocal lattice. If we substitute (5.3.21) into

(5.3.10), then:

Here is an integer and =1. The minimum of in order to have

the minimum value of is the inverse value of . If

or or then we either have or or

from where it follows that , =1,2,3. Similarly, , =1,2,3 . This

follows that

5.3.22

It is obvious that with the result, there is no appearance of the new wave and thus , =1,2,3 5.3.23 which is the domain of physically non differentiable value of q. For the cubic

crystal, we have , , where is the

component of q for the axis. Thus, the physical non differentiable domain of

is the Brillouin zone (BZ); for the dispersion law in place of the graph of ,

we have . There are points in the BZ where the dynamic matrix is

real. That is when disappears, that is when:

1)

2)

47

3) If in the lattice, there is a centre of inversion, then the sum with respect to n’ may be divided into two for which at the end the tends to a cosine function that is real.We examine the case when q = 0, this is the limiting case for long wave vectors. If we consider (5.3.11), then:

5.3.24

For , then the right hand side of (5.3.24) is equal to zero. This is a

particular case of the one-dimensional lattice for q = 0 for which . If we consider (5.3.9), then the right hand side of (5.3.24) is equal to zero and for that 5.3.25

There exists three directions which independent of the symmetry of the

crystal. This is because q = 0 corresponds to identical displacement (see 5.3.25) of all atoms of the crystal and cannot cause any change. Hence,

5.3.26

This is in accordance with Bohm-Staver for longitudinal wave at q = 0. Not withstanding, there is more than one type of atom in the case where atoms occupy physically equivalent site. The different type of atoms may vibrate out of phase with one another. Thus (5.3.26) may not hold for all modes. However it must hold for three modes; these three modes are called acoustic modes. This is due to the fact that at low q, there are frequencies of the order of those of sound. The remaining mode that equation (5.3.26) does not hold has frequencies in the infrared region of the spectrum. They are called optical modes. From (5.3.25), it follows that the amplitude is independent of the type of atom and from (5.3.10), we have ; which follows that the entire elementary cell vibrate as a total whole with different phases. If we assume , then such a wave does not fill the discreteness of the structure in an elastic medium. If the crystal is isotropic, then for the given waves, we have one longitudinal and two transverse waves. In an anisotropic crystal, it is not possible to see longitudinal and transverse wave. A crystal as a whole is anisotropic. We examine a cubic crystal for longitudinal and transverse waves simply for convenience. We know that

=1,2,…..,3s and for the case of =1,2,3, we have and for the case of

(3s-3), we have (3s-3 optical modes). Then from (5.3.24), we have:

5.3.27

As is symmetric with respect to indices n and n’ then from (5.3.27),

and as then

5.3.28

48

If we multiply (5.3.28) by a factor considering then

5.3.29

This gives us a character of the vibration for (3s-3) cases, in particular, for the elementary cell, the centre of mass remains at rest .Such a vibration is called optical. The centre of gravity is constant. From , for , we

expand in the series:

5.3.30 

We may bring the sum , to the diagonal form that leads us

to . All the acoustic frequencies are equal to zero for . Here, we have

degeneracy. The dependence for , implies

5.3.31

that for which we have the acoustic vibration. In the crystal of type, we have s = 2 and =1 6. This follows that there exists 3 branches which are

obviously acoustic; see figure 5.3.1. The inferior branches are small for small and linearly tend to zero which we call acoustic and the rest of the (3s-3) branches are optical and have branches with longitudinal and transverse vibrations. The velocity of propagation of longitudinal waves is greater than those of transverse wave as the frequencies of vibration of longitudinal waves are greater than those of transverse waves ( ). Here L and T stand for longitudinal and transverse modes respectively.

49

0

Fig. 5.3.1

CYCLIC BOUNDARY CONDITION FOR THREE-DIMENSIONNAL CASES (BORN-VON KARMAN CYCLIC CONDITION)

Let’s be given a cell as illustrated in figure 5.3.2:

Fig. 5.3.2

We have the cyclic condition if for , =1,2,3 then in (5.3.10) i.e. in

, the displacement is invariant since

5.3.32

where and whole numbers from where

5.3.33

It can be shown that

5.3.34

where is a vector of reciprocal lattice and is now quasi discrete. It can be

shown that the variation of q may be bounded by the limits of the BZ (Wigner-Seitz cell ) and may be found from

50

5.3.35

and from here, considering (5.3.34), we have

5.3.36

where takes G/2 values , the number of elementary cells in the

principal domain of the crystal . If then is equal to

values and is a whole number. We know that the volume of the BZ is equal to

. We may calculate the number of different vibrations per unit volume of

the given crystal which is equal to the density of vibration; thus

where V is the volume of the given crystal.

5.4. NORMAL VIBRATION OF A THREE DIMENSIONAL CRYSTAL

The kinetic energy of the lattice is

5.4.1

and the potential energy is

5.4.2

It is convenient for us to transform to normal modes. This implies that instead of the system of vibrating atoms, we deal with the set of wave (modes). The amplitude of these are called normal coordinates. It may be seen further that, they are dependant on the wave vector  :

5.4.3

where are plane waves in the crystal and is the amplitude of the

wave with polarization . If

5.4.4

then is complex value and we can show that the expression on the right

hand side of (5.4.3) is real:

5.4.5

51

We introduce the following normalization condition that we prove:

5.4.6

where m represents the mass of all atoms in the system ; that is . Let’s

define

5.4.7

If as implies , which is the number of the lattice vector .

If then and are physically equivalent. If we consider the

lattice vector , then we can do the transformation in (5.4.7). The crystal is a closure of the entire space then is invariant relative to such a transformation and thus

and

Since is a vector of the direct lattice and since , then it follows that only = 0, thus

. 5.4.8

The quantity is called lattice sum. Here is a lattice vector of a reciprocal

lattice. It is obvious that . Let’s get equation for the amplitude of

the wave i.e.

5.4.9a

. 5.4.9b

Let’s multiply (5.4.9a) and (5.4.9b) by and respectively and we take

the sum with respect to the indices k and after which we subtract the second from the first and we have :

5.4.10

We do the change of indices and if and there is no

degeneracy, then the different branches do not intersect and thus and the

right hand side of (5.4.10) is zero. If , then and the right hand side

of (5.4.10) may be equal to zero. Then to what should be the sum

for . If we examine a particular case, then the lattice is simple for k=1:

52

If the amplitude is taken in such a way that there are a unit polarization vector i.e.

, then it follows that

5.4.11

But from definition . If we substitute (5.4.3) into (5.4.1),

then we have:

=

Thus

5.4.12

Let’s examine the expression for the potential energy:

=

=

From (5.4.9), then,

and from (5.4.11) then

and considering (5.4.8) , then we have

5.4.13

The total energy of vibration of the crystal is:

5.4.14

53

This expression is the energy of the system of independent harmonic oscillation with frequencies . In order to represent the energy of the system in the form of independent harmonic oscillator, we have to express the complex valued function

through real normal coordinate such that the condition (5.4.4) i.e.

should be automatically satisfied. Let’s be given the Lagrangian :

then the equation of motion for the normal coordinate is obtained from

from where we have

, 5.4.15

Vector has Ns values and the system (5.4.15) has 3Ns equations. The normal

coordinate are complex values. We moved from complex valued normal coordinate to real valued normal coordinate. The wave in the crystal should be a moving wave and not a standing wave and thus, the real normal coordinate should be canonical. Thus, the canonical transformation:

5.4.16

There is difficulties here; considering (5.4.4), then we have .

As then we have equations and not 3Ns as expected. Thus

we select such that 

5.4.17

Let

5.4.18

then :

5.4.19

Thus from (5.4.17), (5.4.18), (5.4.19), we have:

5.4.20

We show that are normal coordinate. For this, we classify the dependence

of on time . Let’s derivate with respect to time:

54

It is necessary to obtain

5.4.21

from where:

5.4.22

It follows from here that

, 5.4.23

for a given and j, from (5.4.3) considering (5.4.20), we have

and from

,

which shows that in the crystal we have a moving wave. If now, we substitute (5.4.20) into (5.4.14); then it follows that

Then from condition (5.4.22), we have:

=

.

As the Brillouin zone has a centre of inversion, then

5.4.24

And from here considering (5.4.18), for , we have:

5.4.25

Here are normal coordinates. Let’s find the Hamilton function of the crystal

vibration. The general momentum is given by:

5.4.26

Thus

5.4.27

5.5 SECOND QUANTIZATION OF THE PHONON FIELD

55

For higher temperatures T in a crystal, the kinetic energy is must higher than the potential energy. In this case, we may renormalize the potential energy. If the temperatures are lower, then the potential energy is much higher than the kinetic energy. When we examine the interaction of the electron with the crystal lattice vibrations then the above notions are very convenient to take note of. In the coordinate representation, the Hamiltonian of the system has the form

5.5.1

and

5.5.2

where in the coordinate representation, we have

5.5.3

Let’s solve the Schrödinger equation 5.5.4 We find the eigen value E and the eigen states of the operator (5.5.2). The eigen states describe the small vibrations of the crystal lattice and E is the eigen value of the energy. As the Hamiltonian (5.5.2) represents the sum of independent identical summands then the solution of (5.5.4) may be found in the form 5.5.5

where is a wave function of single oscillator for which (5.5.4) becomes

5.5.6

and

5.5.7

where

The equation (5.5.6) for the squarable and integrable function represents a standard problem in quantum mechanics for a harmonic oscillator .The solution of this problem is well known. Let’s introduce the dimensionless quantity:

5.5.8

then

5.5.9

where is the Hermite polynomial:

56

5.5.10

We know that:

=

Thus

5.5.11

Here we have a dimensionless value .

Dirac was the first person who introduced that in the case of non interacting electrons, it is convenient to use the method of the second field quantization to describe the particles. This follows that, the field is represented as a quasi classical particles. This method is used to investigate the field with variable number of particles. Introduce the second quantization for harmonic oscillator:

5.5.12

Thus

or 5.5.13

In the same manner, we find

5.5.14

We examine the rule of commutation of these operators. Hence, we act them on the wave function of the oscillator. From the property of the Hermite polynomial; it follows that:

5.5.15

Then from here and (5.5.12), we have:

57

5.5.16

It follows from here that when acts on the state of the given

oscillator, it reduces (increases) the phonon number by one quantum number of oscillator. The energy of the oscillator is:

The state of the oscillator for which =0 is a state with no phonons. The

expression gives the excitation energy of the system. This energy depends on N and increases with an increase in N. The excitation energy has a form of the total energy of an ideal gas. The particle of this gas i.e. the quanta of the energy of excitation, are called phonons. From Dirac’s notation:

5.5.17

Here ( ) is called annihalation (creation) operator. It follows that

5.5.18

Thus from (5.5.18), we have: 

5.5.19

or

5.5.19’

also

 ;  ;

The commutation in (5.5.19) is true for base particles. In the quantum state, there may exist an arbitrary number of bosons (phonons). The wave function is symmetric relative to the change of position of two phonons. Let’s find the Hamiltonian of the vibrating crystal in the representation of the second field quantization. Considering (5.5.2) and (5.5.11), the dimensionless Hamiltonian of the system has the form

5.5.21

Where

5.5.23

We evaluate , thus:

58

5.5.24

Thus the eigen states of (5.5.24) are those of since this operator has a sense

of and does not change the state then it is referred to as the phonon

operator. The value describes the ground state ( zero point) called

vacuum state energy with no phonon i.e. . This energy implies that at the temperature of zero Kelvin, the atoms in the crystal are in state of oscillatory motion. The additional energy is a considerable quantity, it shows also that the localization of the atoms in their exact position of equilibrium leads to the indeterminacy of their velocity according to Heisenberg relation. The excited state to just one phonon: 5.5.25and if

5.5.26

and 5.5.27then it follows that;

5.5.28

This is the energy of the normal vibration of the oscillators. It is an approximate solution (from the expansion of the potential energy). Phonons as quanta of elementary excitation are fictitious transformation. Phonon may be called quanta of sound vibration. Phonons in the process of interaction with others may disappear and new ones arise. The question is on the mean value of the phonon number per normal vibration. Plank solved this problem and this number is equal to:

Bohr-Einstein distribution.

In the formula , the quantity is the phonon number. It should be noted that at any finite temperature, a crystal is completely full of phonons. They contribute significantly to the warming (heating) of the crystal only at low temperatures or high intensity. If optical branch are excited, the resultant is optical phonon. If the temperature is raised phonon are created order wise they are destroyed. There is the point of conservation of energy in the process of creation and annihilation as energy flow into or out of the crystal through heat conduction as expected.

59

6. ELECTRONIC STATE IN IDEAL CRYSTALS

6.1 THE ZONE THEORY

All solids are made of atoms i.e. they represent a totality of nuclei and electrons. In crystalline solids, the nuclei of atoms are situated at sites of crystalline lattice and has a space periodicity .The stationary state of the particles in the crystal is described by the Schrödinger equation

6.1.1

where is the Hamiltonian of the totality of the particles, i.e. the Hamiltonian of the solid is the eigen function and E the eigen value. Let  , be

the radius vector of the electron and  , be the radius vector of the

nuclei. Let be the mass of the nuclei of the family and that of the electron. The kinetic energy of the system is that due to the motion of the electrons and that due the motion of the nuclei

6.1.2

and the potential energy of the totality is that due to the interaction between the electrons and the nuclei and amongst electrons

6.1.3

Thus the Hamiltonian

6.1.4

60

The movement of the electrons in the field of the nuclei at rest

6.1.5

Thus in this case we neglect thermal motion and

6.1.6

The wave function (6.1.4) depends on the coordinate of all particles:

To solve the equation (6.1.1) till the present moment is very difficult and we move to certain approximations. We suppose that which follows that the nuclei perform their motion about a certain equilibrium position while those of the electron are progressive and their velocities are much higher than those of the nuclei. In this case we may suppose that the nuclei are practically at rest compare with the motion of the electron. This is called the adiabatic approximation or the Born-Oppenheimer approximation. In this case we may consider the radius vector of the nuclei as not variable but constant. In this connection the Schrödinger equation becomes more and more simple. If nuclei are at rest then their kinetic energy is zero and their interactions become a constant;

6.1.7

This may be neglected considering choice of the origin of the energy. Considering adiabatic approximation, the variables are not separable and thus, we sole equation (7.1.1) approximately. First approximation stem from the fact, the principal role is played by the kinetic energy and the interactions are neglected. Thus

6.1.8

Here the variables are separable

7.1.9

where is the wave function for the free particle

61

6.1.10

and

6.1.11

(7.1.10) is not true wave function (It does not respect the Pauli exclusion Principle). It should be noted that the physical situation of the atomic system that contains two or more electrons is always such that, the corresponding wave function should be anti-symmetric with respect to the interchange of the position of the electrons. The wave function is dependent on the spatial and spin coordinate of the electrons and the nuclei and the time. According to the Pauli exclusion principle it is only two electrons with opposite spins that can be found in the same quantum state. The total wave function of the system which satisfy this principle should be anti-symmetric i.e. the wave function changes sign for the interchange of the position of any two electrons. The function does not satisfy this condition. The anti-symmetric wave function is written in the form of Slater determinant;

6.1.12

The factor assumes the normalization of

If we denote by the potential energy of an electron in the crystal, then it follows to solve the Schrödinger equation:

6.1.13

In a crystal the arrangement of atoms in space are strictly periodic and as a consequence, the total potential of the crystal should posses a three-dimensional periodicity.

7.1.1 BLOCH THEOREM

Considering (7.1.13), then the potential acting on an electron in the crystal obeys the property of periodicity

6.1.1.1

62

where . Consider the translational vector

6.1.1.2

This translation is called translational symmetry. If for that translation the Hamiltonian is invariant i.e.

6.1.1.3

It can be shown that all translational symmetry of a given system form a cyclic

group. This follows from group theory. If is true (7.1.1.3) then it follows that

commutes with  : . We can also

proceed as follows: let then :

from here if follows that

this follows the commutation of and . It follows from the theory of

commutation that if two operators commute then they have one and the same

eigen function i.e. operator and have one and the same eigen function.

Let us find the form of . If we consider (7.1.1.2), then

Thus the explicit form of the translational operator ;

6.1.1.4

This operator commutes with the Hamiltonian. Let us find the eigen value of this operator. This can be obtain from the eigen value equation

6.1.1.5

Where is the eigen value of then

63

7.1.1.6

and as the translation does not change the normalization of the wave function of the electronic system, then we have

6.1.1.7

Let show again that and commute. The operator  ;

where and operator of the kinetic and potential energy respectively. Consider , then select an arbitrary wave function  :

Thus

6.1.1.8

Hence and commute. It should be noted that . Consider

the case of the kinetic energy operator:

If we consider (6.1.1.7) and ( ) then commutes with

. Consequently is a constant of motion and

Suppose we are given two translation operators and then from (6.1.1.6)

we have . From here it follows that

. It means that our eigen value takes the form

6.1.1.9

This is the Bloch’s theorem. If we consider the eigen value (6.1.1.9), then

64

In relation (6.1.1.9), represents the wave vector characteristic of the quantum state of an electron in the crystal . Its physical sense is the number of wave length in the interval ;

6.1.1.10

The relation 6.1.1.10 characterizes the properties of the translation of the wave function. This is an indication that when an electron is displaced in a periodic field, the eigen function of the Hamiltonian satisfy the condition of translation i.e in 6.1.1.10. In this wise we may write and and we show that . The establishment of this relation is a crucial problem in solid state physics. For any wave function satisfying the Schrödinger equation with the periodic potential, there always exists such a vector that the translation with respect to the lattice vector result in the product (plane waves)

on having the periodicity of the Bravais lattice.

Theorem 6.1.1.10 also satisfies plane waves i.e. which is also solution of the Schrödinger equation with potential equals to zero. Bloch gave some specific form to the wave function in 6.1.1.9 and in particular

6.1.1.12

This is the plane modulated wave. It is a Bloch wave and is the Bloch’s

amplitude. There exist some supplementary condition to . If we carry the translation (6.1.1.10), then we have:

6.1.1.13

From here it follows that

6.1.1.14

From here follows the condition for the domain of periodicity:

As and are scalarly multiplied, then and belong to one and the same

eigen value. It follows from here that, quasi wave vector of an electron in a crystal has the same properties as the wave vector . We have seen this earlier in our previous heading; this corresponds to the Brillouin zone

.

6.1.1.15

65

Here, , =1,2,3 are the primitive vectors. For the finite crystal to have a periodic property, it is necessary to introduce the cyclic boundary condition,

6.1.1.15Here, the wave vector most be real. Apply Bloch’s theorem to the boundary condition (6.1.1.15) then we have

from here, it requires that: and or

6.1.1.17 Here

is an integer. It follows that

is quasi discrete. If we

substitute (6.1.1.17) into (6.1.1.15) then we have

6.1.1.18 It follows from here that takes values; then from (6.1.1.17)

6.1.1.19 Let’s examine the order of the Bloch’s amplitude; we normalize the wave

function to unity where is the volume of elementary cell and

, the volume of the crystal lattice, Thus

The last integral is taken with respect to the first elementary cell and .

Though, it is difficult to give the exact solution to (6.1.1.13). It is necessary to have an idea to the energy spectrum. Let’s substitute (6.1.1.11) in (6.1.1.13), then we have:

6.1.1.20

from here, we may find the property for ; let’s take the complex conjugate of (6.1.1.20):

6.1.1.21 Suppose in (6.1.1.21), we do the change ,

6.1.1.22

66

and 6.1.1.23 Condition (6.1.1.22) shows that the energy E is an even function of . Considering the property of periodicity then may be represented in the Fourier series and also

6.1.1.24

6.1.1.25Here and are vector of the reciprocal lattice. and are the Fourier

coefficients since. is a periodic function, then it follows that . Let’s substitute (6.1.1.25) and (6.1.1.24) into (6.1.1.20) then we have

and

6.1.1.26 The solution (6.1.1.26) is based on the fact that all terms are function of . As is an arbitrary vector, then it follows from (6.1.1.26) that

6.1.1.27

This is a system of homogeneous equations relative to the coefficient . In order that (6.1.1.27) should be solvable, it is necessary that it determinant should be equal to zero. The diagonal elements are:

And the non-diagonal elements is unknown. This follows that the equation

cannot be solvable but we can have a representation of energy spectrum. The determinant has N roots: which are the eigen values. It

should again be noted that i.e. energy has no centre of inversion

and then we may have the sketch of the energy spectrum.

We take for convenience the one dim case.

E

67

Forbiden zone

0

Fig. 6.1.1.1

The correct form of is unknown fig 6.1.1.1 is drown from intuition as

.

6.2 THE APPROXIMATION OF THE NEARLY FREE ELECTRON

MODEL (N.F.E)

In the N.F.E model, a crystal is considered as a special lattice of ions immersed in an electron gas. The perturbation in the eigen value problem is the departure of the true potential from zero. If we consider a vacuum, then for a zero potential, an

electron wave function is a plane wave normalize in the entire space .

In the crystal, the normalization integral is conveniently represented as a sum of the contribution from all the lattice site, and as the contributions are the same, then they may be taken out of the sum and what is left is the total number of the sites. On introducing the volume of the Weigner Seitz cell where is the volume of the given crystal it is found that the plane wave may just as well be normalized to the Weigner Seitz cell. Let’s solve 6.1.1.27:

6.2.1

The periodic potential is assume to be small and is a perturbation; from (6.2.1) we have

Considering the perturbation theory, we have: from ,

it follows that

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6.2.2

This shows that we have a free wave in the crystal. If we substitute (6.2.2) in (6.2.1’) for

then

6.2.3

for , we have ; for , , , and

6.2.4

and also

6.2.5

for the first order perturbation theory, if we substitute

in (6.2.1’)

then,

if , then

and as then . If , then

and

6.2.6

We find the energy for the 2nd order perturbation, if we consider and we take , then

6.2.7

from here, we have

This

is the energy for the 2nd order perturbation. Thus the energy for the system

6.2.8

However, it is entirely very possible that if we take the correct reciprocal lattice , we may always find such that

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6.2.9

This is the condition of discontinuity of the energy and

6.2.10

Let’s find out the geometrical physical sense of (6.2.9). This shows that and

are orthogonal. Consider fig 6.2.1 P

Fig. 6.2.1 The plane pp’ is perpendicular to the vector . All the vectors in the plane pp’

will either be perpendicular to or . All vectors on pp’ may be obtained

through the sum of and on condition that the head of should be on pp’ plane. Thus all the vectors with their heads on the pp’ plane satisfy expression (6.2.9). This is the equation of the plane perpendicular to . The totality of the

planes passing through the middle of which are perpendicular to them

determine all the vectors satisfying the condition or

. The equation (6.2.9) may be used to construct the Brillouin zone of

all crystalline solids from their reciprocal lattice. We may see from above that the given perturbation theory is not physical as

appear to be greater than . This is connected with the destruction of the N.F.E. .As there exist that satisfies (6.2.10) then there exist a plane on which there is a scattering of the free moving electrons. This correspond to the forbidden zones in the energy band diagram. Consider now an amended variant of the perturbation theory; i.e. we consider

for the direction of not satisfying (6.2.9). Let’s consider (6.2.1’)If , then

6.2.11

If , then

6.2.12

For the solution of the above equation, the determinant of the system should be equal to zero:

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From where we have

6.2.13 or

6.2.14 If we consider (6.2.9) and (6.2.10) then it follows that:

6.2.15

or 6.2.16

The two solutions in (6.2.16) correspond to the minimum and maximum

; and are separated by a forbidden gap of which

6.2.17

The Brillouin zone is the zone for space and the energy-zone is the zone for which

6.2.18

and it is an equation in -space. The energy is discontinuous in planes that are intersecting. We show that between

and , there exist a forbidden zone. Suppose

6.2.19

where is parallel to . If we substitute (6.2.19) in (6.2.16), then we have:

and as is arbitrary, then

6.2.20

Equation (6.2.13) is the resonance condition and follows that . Consider

6.2.21

and from (6.2.21), we have

6.2.22

If we substitute this in (6.2.14), then we have;

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Although we are constructing the theory of at the neighborhood of which

is again smaller than , we have:

6.2.23

And as the 2nd summand at the right hand side of (6.2.23) is greater than zero, then it follows that lower part of the energy is less than and upper part, greater than . From here, it is obvious that between

and , we have a

forbidden zone. For the exact resonance, ; the energies and

coincide and we have a degeneracy (there is a coincidence at zero

approximation). It follows the degeneracy of two known perturbed states. At the point of resonance, we have a mixture of two wave functions. If we consider the

one dimensional case, in place of , we put , then in place of , we have

. This follows that:

6.2.24

This follows that

since , then follows fig 6.2.1

forbidden zone

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second allowable zone

first allowable zone

0

Fig. 6.2.1

7.2 THE TIGHT-BINDING APPROXIMATION (TBA)

In the previous headings, emphases were put on the plane wave part of the Bloch function neglecting the atomic part. We may take an alternative approach with emphasis of the atomic function at the neighborhood of the nuclei. The T.B.A is connected with the overlap of the atomic wave function enough to require some correction to the picture of the isolated atoms. It should be noted that this correction is not much as to render atomic descriptions completely irrelevant. The T.B.A is much suitable for electrons that are tightly bound to their atoms as for the case of the band built for inner electrons or d-electrons whose orbits may have relatively small radii. When developing the tight-Binding-approximation we should consider that at the neighborhood of each lattice site, a complete Hamiltonian of the crystal should be approximated with the Hamiltonian of the single atom located at a lattice site; it is also assume that, the boundaries of the atomic Hamiltonian are well localized. Let’s consider the valence electron for which the periodic potential strongly changes at the neighborhood of the nucleus. If we consider the model of the Tight-Binding electron, then it follows that in the first approximation, the electron is tightly bound to the atomic nucleus and the interaction with the rest of the atom is examined as a perturbation. Considering the theory of the tight Binding approximation, we may take the wave function of the zero approximation to be

. This is the atomic wave function. This implies that the wave function of the electron interact only with the nucleus where it is seated. In the wave function and are coordinate of the electron and nucleus respectively. This wave function may also be written as: 6.3.1 and satisfy the Schrödinger equation:

6.3.2

6.3.3

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In a crystal, an electron with energy may be localized in any atom. Let

be the number of elementary cells. It should follows from above that different states correspond to the first energy ; this follows, quantum degeneracy; it follows in such a case to construct the eigen function of the zero approximation applying quantum mechanics and in particular

6.3.4

If is the wave function of the zero approximation in the crystal, then it may

be represented in the form and has a property of periodicity. It follows that,

6.3.5 Here,

6.3.6

Then considering (6.3.5), we have

This follows that (6.3.5) is a periodic function with respect to coordinate thus the selection of is correct. It therefore follows that (6.3.4) has the form

6.3.7

This is the zero approximation correction to the wave function since the function is normalized to unity then we may find the normalization factor for the wave function  :

6.3.3 We may reduce the double sum to a single one through a change of variables:

  , , thus

The integrant is the product of the function and and is

dependent not only on the position of the nuclear but also on the distance that separate them. Suppose

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* If then * If the distance between two nuclei is large, then the corresponding wave function does not overlap and will practically be equals to zero that is

Thus for only small values of . Supposes is fixed then and

have the same values. Hence, the summation relative to will be equivalent to summation relative to :

It should be noted that the quasi wave vector is such that . It will be

quite enough to use the first order perturbation theory. It follows that we may use the wave function (6.3.7) and from where follows the Schrödinger equation:

6.3.9

Where is an intrinsic periodic potential of an electron. Consider all

interactions, then we substitute (6.3.4) into (6.3.9) and we have

6.3.10which is the first order of the perturbation theory. If we have the wave function in the zero approximation of the perturbation theory, then we have:

6.3.11

Let’s take the difference (6.3.10) and (6.3.11), then we have:

6.3.12Multiply this relation by and integrate the resultant with respect to the configuration coordinate. Considering the fact that: from here, we have:

or

6.3.14 The quantity is the exchange energy which result from the fact that any electron may be found about any nucleus with a certain probability; this peculiarity takes into consideration the fact that for the formation of we have the wave function of two atoms at the distance of . This

implies that two nuclei at the separation of can exchange electrons. This exchange is done with the help of the field of all the nuclei excluding that of the

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two nuclei in question. The exchange is also done with the help of the component of the periodic self- consistent field of the crystal. It should be noted that

is only considerable for small distances. If , then is large. That is to

say, the exchange is only for close neighboring atoms. It should be noted that is considerable at the neighborhood of two neighboring nuclei. Consider (6.3.14), then

Considering (6.3.6), then

6.3.15Considering (6.3.15), we may already obtain some information about the spectrum. gives a shift of the energy level upwards or downwards. When the atoms are brought closer together, there arises an interaction that leads to the splitting of the energy level. Consider a crystal that is made up of atoms that are distant from each other. Let’s examine fig. 6.3.1.

0 r U

V Fig. 6.3.1 Let’s investigate

At the neighborhood of zero, and and already for the neighboring sites . Thus, in order to evaluate , it is necessary to consider the neighboring sites; thus

6.3.16

It follows that for a cubic crystal, we have:

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6.3.17

Let’s examine the bottom of the permissible zone, here and

then

This is evaluated at the center of the Brillouin zone. Let’s examine the ends of the

first Brillouin zone, here ; this follows that

This should be the ceiling of the zone. Then . Considering the value of and , we may see that the width of the forbidden zone is defined by the parameter A. The value of this is for the cubic crystal. Let’s examine the center of the Brillouin zone. We examine the value of the relation (6.3.17) as

6.3.18

This follows that at the neighborhood of the bottom of the zone the dispersion is quadratic in q and

where 6.3.11

is the effective mass of the hole. Let’s examine the neighborhood of the ceiling.

Consider , from (6.3.17), we have

as ;

At the ceiling again, we have the quadratic dispersion relation and

6.3.20

which is the effective mass of the electron.

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PROBLEMS1. In the general case, in the problem an the chemical bonding in molecules and solid state , it is necessary to solve the Schrödinger equation for system of N nuclei and n electrons amongst which act Coulomb forces. Let’s neglect relativistic effect that may arise due to the dependence of the spins of terms. Write this equation using the denotation below

Coordinates Mass ChargeNucleus

Electronsm -e

2. In order to describe the motion of electrons in a system, consisting of N nuclei and n electrons, it is necessary to make a very good approximation. Here, we assume that the nuclei are at rest because their mass is greater than that of the electron. Write the Schrödinger equation for a system of N fixed nuclei and n electrons. Use the atomic units, that is assume the electronic charge e, electronic

mass m and the Planck constant to be the equal to unity.

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3. Not withstanding that the eigen value in the Schrödinger equation for the

electron include the inter-nuclei repulsion it is often called the

electronic energy. In it, is found as a parameter, the inter-nuclei distance, it plays the role of the potential energy in the Schrödinger equation for the motion in the field of the nuclei. If we assume that the center of mass of diatomic molecule is fixed, then the Schrödinger equation describes the motion of two of its nuclei A and B and has the form

where

is the reduce mass and . Find and E considering that the potential

has the form

. Such a potential correspond to a force acting

between atoms and proportional to the displacement of the inter-nuclei distance R from its equilibrium value (here k is a force constant). For convenience, let its potential energy be given by its value . Hint: The problem is similar to that of a particle of mass in a centre potential field . Consequently, in spherical system of coordinate , the wave function has the form

.

ADMINISTRATIVE

These lectures are basically the lectures taught by Prof. LUKONG Cornelius FAI in the University of Dschang. All errors associated with typing and failures to transmit information are mind. I shall be happy to welcome questions and critics in and out of class or through e-mail.The grading policy of this course is as follows:o Class participation, 10 % o Homework, 10 %o Quiz, 10 %o Final exams 70 %

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