Solid State Physics 1. Vincent Casey

Transcription

1 Solid State Physics 1 Vincent Casey Autumn 2017

2 Contents 1 Crystal Mechanics Stress and Strain Tensors Physical Meaning Simplification of the Matrix Deformation Modes of Cubic Crystals Elastic Waves in Solids Elastic Properties Waves in Crystals Velocity Static and Dynamic Models Linear Lattice Models Linear 1D Chain Model Diatomic Linear Chain Model Specific Heat Capacity Specific Heat Capacities Historical Specific Heat: Classical Model Einstein Model Debye Model Key Features Debye Model Phonons Actual Phonon Density of States Phonon Interactions Anharmonicity and Thermal Expansion Normal and Umklapp Phonon Interactions Neutron Scattering by a Crystal Inelastic Neutron Spectroscopy i

3 CONTENTS SSP1 4 Drude Free Electron Theory Classical Theory and Conduction in Solids Free Electron Theory (Drude s Free Electron Theory) Case I: No electric field Case II: Constant uniform electric field Case III: Time dependent sinusoidal electric field Hall Effect Drude Model and Metal Reflectivity Drude model and plasma frequency of metals Plasma oscillations in metals Plasma oscillations in metals with scattering Band Theory of Solids Band Theory of Solids Introduction Bloch s Theorem The Nature of the Crystal Potential Field Solving Schrodinger s equation for the K-P 1D model Dispersion Relation for the Krönig-Penney Model Some consequences of the band theory of solids Band Filling Band Shapes of Real Semiconductors Effective Mass and Holes Filling of the energy bands with electrons Occupation of allowed states Properties of Semiconductors Carrier Density Fermi Level in Semiconductors Law of Mass Action Temperature Dependence of Carrier Density Current Transport and Current Density Equations Unipolar Devices High Field Effects Explanation of NDR based on band structure Negative Differential Resistance NDR and Instability Transfer Electron Devices TEDs Modes of Operation Accumulation Domain Mode V Casey ii

4 SSP1 CONTENTS 7 pn Junction Diode Introduction Fabrication Idealized Junction Formation Energy band diagrams Pn Junction Equilibrium Band Diagram Currents flowing at equilibrium Barrier potential V B Device Equations at equilibrium The Depletion Approximation Depletion Region Width Maximum Electric Field Junction Capacitance Diode Information from C J measurements Pn Junctions under external bias Reverse Bias Forward Bias The diode or rectifier equation V Casey

5 Chapter 1 Crystal Mechanics 1

6 1.1. STRESS AND STRAIN TENSORS SSP1 1.1 Stress and Strain Tensors This section is drawn largely from: The Physics and Chemistry of Solids, Stephen Elliot, Wiley 1998 A solid is stressed by applying external forces such that both the net force and the net torque are zero, i.e. static equilibrium prevails. Figure 1.1: (a) Zero net force and (b) zero net torque, applied to a cubic volume element of face area A. The first suffix indicates the direction in which the stress (force) acts and the second suffix indicates the direction of the normal to the face upon which the stress acts. Zero Net Force achieved by exerting forces equal in magnitude, but in opposite directions, perpendicularly to opposite faces, tending to compress/elongate the sample, 1.1(a). Zero Net Torque achieved by applying equal and opposite forces in a parallel fashion to opposite faces, tending to shear the sample, 1.1(b). Therefore, the cube does not accelerate or rotate. Two kinds of stress are identified: { Normal, σi j with i = j; Shear, σ i j with i j. A force F x applied in the x-direction to a plane with area A x whose normal is also in the x-direction produces a stress component: σ xx = F x A x Likewise, the same force applied to a plane whose normal is in the y-direction produces the stress component: σ xy = F x A y V Casey 2

7 SSP STRESS AND STRAIN TENSORS Since stresses on opposite faces of the cube are equal and opposite we only need to consider three faces. Thus, stress in general is described by a second-rank tensor σ with nine components σ i j (i, j = x,y,z), conveniently written as a 3 3 array: σ = σ xx σ xy σ xz σ yx σ yy σ yz σ zx σ zy σ zz However, only six of these stress components are independent because of the constraint of zero torque, which implies that: σ xy = σ yx ;σ yz = σ zy ;σ zx = σ xz Special cases of stress have most of the six independent stress components equal to zero. For instance for uniaxial stress in the x-direction, the only non-zero component is σ xx = σ. For hydrostatic stress, the non-zero components are the diagonal components σ xx = σ yy = σ zz = P For pure shear, in which the shape but not the volume of the solid is changed, is represented by say σ xy = σ yx = σ being the only non-zero components. A particular orientation of the spatial coordinates can always be found such that the general stress matrix has diagonal components only, i.e. σ = σ x σ y σ z the six independent variables defining the stress system now being the three principal stresses σ i (i = x,y,z) acting along the principle axes, and the three variables needed to determine the orientation of the principle axes with respect to the original coordinate system. A general stress matrix, transformed to the principal axis system can always be written as σ x σ y σ z = σ σ σ 0 + (σ x σ 0 ) (σ y σ 0 ) (σ z σ 0 ) 3 V Casey

8 1.1. STRESS AND STRAIN TENSORS SSP1 Figure 1.2: Uniform hydrostatic compressive stress and pure shear deformation (at constant volume). where the new stress component is given by σ 0 = (σ x + σ y + σ z )/3 The first term of the transformed stress matrix represents a purely hydrostatic term, with σ 0 = P = (σ x + σ y + σ z )/3, which causes a change in volume(dilatoric), but not of shape, of an elastically isotropic solid. The second term of the transformed stress matrix represents a pure shear (deviatoric), causing a change of shape, but not of volume, of a solid, since the sum of the diagonal components equals zero, i.e. x,y,z σ ii = 0 The application of an external force to a solid causes a deformation because different points in the material are displaced by different amounts. Consider a point P initially at r and a neighbouring point Q initially at r + u, displaced under the action of stress to P at (r + r) and Q at (r + r + u + u) respectively. For small relative displacements ( u << r ), the components of the relative displacement are given by u i = u i x x + u i y y + u i z z for i = x,y,z. The strain components e i j ε i j are then defined in terms of the dimensionless displacement gradients as and e i j = e ji = e ii = u i i (i = x,y,z) ( ui j + u ) j (i = x,y,z) i V Casey 4

9 SSP STRESS AND STRAIN TENSORS Figure 1.3: 2D representation of infinitesimal homogeneous elastic strain. Therefore, the strain components can, like the stress components, be written as a (3 3) matrix (symmetric) with again only six of the nine components being independent because of the requirement that the off-diagonal components obey the condition e i j = e ji in order to exclude rigid rotations. Special cases of strain include uniaxial strain in the x-direction with only one non-zero strain component being e xx = e; uniform dilation or compression, resulting from hydrostatic stress, with e i j = e (i = x,y,z); and pure shear with e i j = e ji = e (i = x, j = y) or other orthogonal pair. As with a general stress, a general strain can be separated into a dilationstrain component (volume change) and a pure strain (deviatoric-strain) component (shape change). For a general strain component e i j = e 0 3 δ i j + (e i j e 0 3 δ i j) where the Kronecker delta symbol has the properties: δ i j = 1, δ i j = 0, i = j i j Since there are only six independent stress and strain components, a convenient short-hand notation is to relabel the component indices as follows: 5 V Casey

10 1.1. STRESS AND STRAIN TENSORS SSP1 xx 1 yy 2 zz 3 yz 4 zx 5 xy 6 Hooke s law states that the stress and strain are directly proportional to each other. Thus, in terms of elastic-stiffness coefficients, C i j, a stress coefficient can be written as a function of a strain as σ i = 6 j=1 C i j e j The relationship between stress and strain coefficients can also be written in matrix form as σ 1 σ 2 σ 3 σ 4 σ 5 σ 6 = C 11 C 12 C 13 C 14 C 15 C 16 C 21 C 22 C 23 C 24 C 25 C 26 C 31 C 32 C 33 C 34 C 35 C 36 C 41 C 42 C 43 C 44 C 45 C 46 C 51 C 52 C 53 C 54 C 55 C 56 C 61 C 62 C 63 C 64 C 65 C 66 Alternatively, a strain coefficient can be expressed in terms of the elastic-compliance coefficients, S i j, as a function of the stress: 6 e i = S i j σ j j=1 where both elastic compliance and stiffness are quantities describing a material as an elastic continuum Physical Meaning Physical meaning of e i j e 1 e 2 e 3 e 4 e 5 e 6 e 11 = u x x e 22 = u y y e 33 = u z z V Casey 6

11 SSP STRESS AND STRAIN TENSORS Figure 1.4: Stress Strain Matrix Simple tensile strains along the x,y,z axes are positive for tension and negative for compression. Deformation in the xy plane of a rectangular thin film corresponds to: e 12 0;e 21 0 e 11 = e 22 = 0 e 3i = e i3 = 0;i = x,y,z With angles greatly exaggerated, this corresponds to: e 12 u x y e 21 u y x which equals the angles respectively. Figure 1.5: Pure Shear Possible combinations of e 12 and e 21 are: 7 V Casey

12 1.1. STRESS AND STRAIN TENSORS SSP1 Pure Shear e 12 = e 21 Pure Rotation e 12 = e 21 Simple Shear e 12 0;e 21 = 0 Figure 1.6: A pure rotation. e = Figure 1.7: Simple Shear 0 γ/2 0 γ/ e i j = ε i j + ω i j 0 γ/2 0 γ/ ε i j = 1/2(e i j + e ji ) ω i j = 1/2(e i j e ji ) e i j = e ji - symmetric with respect to i and j: measures shape change! V Casey 8

13 SSP STRESS AND STRAIN TENSORS Figure 1.8: Comparing pure strain and rotation. ω i j = ω ji - antisymmetric with respect to i and j: measures rotation. ω i j = 0(i = j) Therefore ω i j has three independent elements implying that three independent rotations are possible, one about each axis Simplification of the Matrix Of the 36 elastic stiffness (or compliance) coefficients in the most general case of triclinic crystals, 21 are independent and non zero as a result of the general condition C i j = C ji The presence of higher symmetry reduces the number of coefficients further. For cubic crystals, just three components are independent, C 11,C 12 and C 44 with C 11 = C 22 = C 33, C 12 = C 13 = C 23, C 44 = C 55 = C 66, and all other coefficients being zero. C 11 C 12 C C 12 C 11 C C 12 C 12 C C C C 44 These elastic stiffness constants are related to the corresponding elastic compliance coefficients via the relationships S 11 = C 11 +C 12 (C 11 C 12 )(C C 12 ) > 0, 9 V Casey

14 1.1. STRESS AND STRAIN TENSORS SSP1 S 12 = C 12 (C 11 C 12 )(C C 12 ) 0, S 44 = 1 C 44 as obtained from inversion of the stiffness matrix. In the case of elastically isotropic solids (amorphous solids), the number of independent elastic coefficients decreases to two since, in addition to the equalities applicable to cubic materials, the coefficients are further linearly related to each other by the equation C 11 = C C 44 Lamé Constants: For such isotropic materials, the two independent elastic stiffness coefficients are conventionally referred to as the Lamé constants: and λ C 12 with µ = C 44 and so C 11 (λ + 2µ) Uniaxial Stress: For uniaxial stress,young s modulus is defined as E = σ xx e xx E = 1 = (C 11 C 12 )(C C 12 ) (3λ + 2µ) = µ S 11 C 11 +C 12 (λ + µ) for the case of isotropic media. Under such stress loading, the sample also deforms in directions perpendicular to the stress direction. This behaviour is quantified by Poisson s ratio, defined as with ν = transverse strain normal strain = e yy e xx ν = S 12 C 12 = S 11 (C 11 +C 12 ) = λ 2(λ + µ) V Casey 10

15 SSP STRESS AND STRAIN TENSORS where 0 ν 0.5. Hydrostatic Stress: In the case of hydrostatic stress (pressure), the bulk modulus B (equal to the inverse of the compressibility, κ = B 1 relates the pressure to the dilation (fractional change in volume) e 0, B = P e 0 with B = 1 3(S S 12 ) = (C C 12 ) (3λ + 2µ) = 3 3 The bulk modulus is connected to microscopic quantities relating to the interatomic potential. Pure Shear: Finally, for the case of pure shear, the shear (or rigidity) modulus is defined as: shear stress G = shear strain = σ xy e xy giving for isotropic solids G = 1 2(S 11 S 12 ) = (C 11 C 12 ) C 44 = µ 2 Some Interrelationships: The various elastic moduli are inter-related; for example and G = B = E 2(1 + ν E 3(1 2ν) Deformation Modes of Cubic Crystals Each independent elastic constant is associated with a fundamental mode of deformation. For an isotropic material there are two modes: dilation and shear. The three independent constants in a cubic crystal require there to be three independent modes of deformation. 11 V Casey

16 1.1. STRESS AND STRAIN TENSORS SSP1 (a) Dilation by hydrostatic stress Under a hydrostatic pressure P: σ 1 = σ 2 = σ 3 = P and the dilation = e 1 + e 2 + e 3 is σ 4 = σ 5 = σ 6 = 0 = e 1 + e 2 + e 3 without rotation. Using these relationships we can show that the bulk modulus B is given by: B = p = 1 3 (C C 12 ) Figure 1.9: Two shear modes for a cubic crystal. Shear on a cube face parallel to a cube axis. For example on the (010) plane in the [001] direction: σ 4 = C 44 e 4 C 44 is a shear modulus, which by convention we name µ 0. Shear at 45 to a cube axis For example, on the (110) plane in the [1 10] direction. In this case the shear modulus is: µ 1 = 1 2 (C 11 C 12 ) V Casey 12

17 SSP STRESS AND STRAIN TENSORS This allows us define an anisotropy factor, µ 0 2C 44 = µ 1 (C 11 C 12 ) which of course will be one for an isotropic material. In terms of the Lamé constants: λ = C 12 ;µ = C 44 ;λ + 2µ = C 11 which explains why cubic crystals are not in general isotropic. W Al Fe Cu Na µ 0 µ Structure bcc fcc bcc fcc bcc 13 V Casey

18 1.1. STRESS AND STRAIN TENSORS SSP1 V Casey 14

19 Chapter 2 Elastic Waves in Solids 15

20 2.1. ELASTIC PROPERTIES SSP1 Introduction to Lattice Dynamics, Martin T. Dove, Cambridge topics in mineral physics and chemistry. 2.1 Elastic Properties Forces acting on atoms disturb them elastically from their equilibrium positions. A Taylor series expansion of the energy near the minimum (equilibrium position) yields: Figure 2.1: Equilibrium interatomic distance according to Madelung: energy minimum U 0 at R 0. U(R) = U 0 + U R (R R 0 ) U R0 2 R 2 (R R 0 ) R0 For small displacements, neglect terms of order 3 and higher. At equilibrium, U R so: where: U(R) = U U R 2 R0 (R R 0 ) 2 = U 0 + ku2 R0 2 k = 2 U R 2 V Casey 16 R0

21 SSP ELASTIC PROPERTIES and following Hooke s law in simplest form: u = R R 0 F = 2 U R 2 = ku The elastic properties are described by considering a crystal as a homogeneous continuum medium rather than a periodic array of atoms. In general, the problem is formulated in the following terms: Applied forces are described in terms of stress σ = F/A Displacements of atoms are described in terms of strain ε = δl/l = u x Elastic constants C relate stress σ and strain ε so that σ = Cε For an arbitrary 3D crystal the stress and the strain are tensors. For a local hydrostatic pressure due to force F(F x,f y,f z ): Stress components, σ i j,(i, j = 1,2,3) where x 1,y 2,z 3. For compressive stress, i = j, (σ 11,σ 22,σ 33 ) and for shear stress, i j, (σ 12,σ 21,σ 13,σ 31,σ 23,σ 32 ). The shear forces must come in pairs: σ i j = σ ji, i.e. no rotation/angular acceleration so the stress tensor is diagonal with six components. The displacement produced by the vibration is represented by the vector u = u x x,u y y,u z z. The strain tensor components are defined as: ε i j = u i x j Compressive strain: Shear strain: ε xx = u x x,ε yy = u y y,ε zz = u z z ε xy = u x y,ε yx = u y x,ε xz = u x z,ε zx = u z x,ε yz = u y z,ε zy = u z y, 17 V Casey

22 2.1. ELASTIC PROPERTIES SSP1 Since σ i j and σ ji are always applied together, we can define the shear strains symmetrically: ε i j = ε ji = 1 ( ui + u ) j 2 x j x i so the strain tensor is diagonal and has six components. σ xx C 11 C 12 C 13 C 14 C 15 C 16 σ yy C 21 C 22 C 23 C 24 C 25 C 26 σ zz σ yz = C 31 C 32 C 33 C 34 C 35 C 36 C 41 C 42 C 43 C 44 C 45 C 46 σ zx C 51 C 52 C 53 C 54 C 55 C 56 σ xy C 61 C 62 C 63 C 64 C 65 C 66 ε xx ε yy ε zz ε yz ε zx ε xy Figure 2.2: General matrix form of Hooke s Law highlighting compression, shear and mixed coefficients. V Casey 18

23 SSP WAVES IN CRYSTALS 2.2 Waves in Crystals We will now examine the vibrational behaviour of atoms in solids where λ a. The vibrations are thermally activated with a characteristic activation energy k B T. Firstly, vibrational excitations are collective modes: all atoms in the material take part in the vibrational mode. The influence of the translational periodicity of the structure of crystals has a dramatic effect on the vibrational behaviour when the wavelength of the vibrations becomes comparable to the size of the unit cell. When the vibration wavelength is much larger than the structural variation of the material, the solid may be considered as an elastic continuum (continuum approximation, see above). How can we visualise a vibration wave traveling through a crystal, where the space that vibrates is not continuous (like a string on a musical instrument) but is composed of discrete atoms? The answer is to think of the wave as representing displacements, u(x,t), of the atoms from their equilibrium position. Considering lattice vibrations three major approximations are made: atomic displacements are small, u << a, where a is the lattice constant forces acting on atoms are assumed to be harmonic/hookian, F = Cu adiabatic approximation is valid - electrons follow atoms so that the nature of the bond is not affected by vibrations Figure 2.3: Longitudinal vibration (force) on an element of an elastic continuum. The discreteness of the lattice must be taken into account where λ a. For long waves λ >> a, one may disregard the discrete/atomic nature of the solid and treat it as a continuous medium, i.e. a continuum. In this case vibrations propagate as ordinary acoustic waves, i.e. elastic waves, at the speed of sound in the material. m d2 u dt 2 = F (ρadx) d2 u = F(x + dx) F(x) dt2 19 V Casey

24 2.2. WAVES IN CRYSTALS SSP1 ρ 2 u t 2 = 1 F A x = σ xx x Assuming that the wave propagates along the [100] direction, and writing Hooke s law in the form σ xx = C 11 ε xx where σ is the stress and ε is the strain ε xx = u x x, we can rearrange the above into wave equation form: which has the general solution: 2 u t 2 = C 11 2 u x ρ x 2 (2.1) u(x,t) = u 0 e i(kx ωt) (2.2) where k is the wavevector (momentum vector) given by k = 2π λ and ω is the angular frequency ω = 2π f. The coefficient of the term on the right of the wave equation is 1/v 2 and so: v L = C 11 ρ = ω k This is the longitudinal sound velocity in the continuum. We can also have transverse vibrations in the medium. In this case, we use the shear coefficients and the shear stress and strain. Simple analysis gives: ρ 2 u t 2 = σ xy x, σ xy = C 44 ε xy and ε xy = u x. Figure 2.4: Transverse vibration (force) on an element of an elastic continuum. The transverse wave equation is: V Casey 20 2 u t 2 = C 44 2 u x ρ x 2 (2.3)

25 SSP WAVES IN CRYSTALS where the transverse sound velocity is now determined by the shear coefficient and the density: C 44 v T = ρ. There are two independent transverse modes: displacements along y and z. For k in the [100] direction in cubic crystals, by symmetry, the velocities of the two modes will be the same, i.e. degenerate. Normally C 11 > C 44 v L > v T Considering waves propagating in different directions in a cubic crystal. In general, the sound velocity will depend on a combination of elastic constants. v = where C e f f is an effective lattice constant. C e f f ρ Mode k [100] k [110] k [111] L C (C 11 +C C 44 ) 1 3 (C C C 44 ) T 1 C 44 C (C 11 C 12 +C 44 ) T 2 C (C 11 C 12 ) 1 3 (C 11 C 12 +C 44 ) Velocity The wave (or phase) velocity, defined as v = ω k = C 11 ρ gives a simple linear relationship between ω and k. Sound waves of differing frequency/wavelength propagate with the same velocity: the medium is said to be non-dispersive. In general, another velocity associated with a traveling wave can be defined, and this is important for waves traveling in dispersive media where the linearity between ω and k breaks down. The group velocity, defined as or, in general, v g = ω k, v g = k ω(k) 21 V Casey

26 2.2. WAVES IN CRYSTALS SSP1 is a measure of the velocity of a wave packet, composed of a group of plane waves, and having a narrow spread of frequencies about some mean value, ω. For acoustic waves with long wavelengths (k 0), i.e. in the elastic continuum limit, the phase and group velocities are equal. In a liquid, only longitudinal vibrations (modes) are supported (shear modulus is zero). The situation is more complicated in solids, where more than one elastic modulus is non-zero. As a consequence, both longitudinal and transverse acoustic modes exist even in isotropic solids, having in general different sound velocities. The situation is even more complicated for anisotropic crystals Static and Dynamic Models The static lattice model which is only concerned with the average positions of atoms and neglects their motions can explain a large number of material features such as: chemical properties; material hardness; shapes of crystals; optical properties; Bragg scattering of X-ray, electron and neutron beams; electronic structure as well as electrical properties. There are, however, a number of properties that cannot be explained by a static model. These include: thermal properties such as heat capacity; effects of temperature on the lattice, e.g. thermal expansion; the existence of phase transitions, including melting; transport properties, e.g. thermal conductivity, sound propagation; the existence of fluctuations, e.g. the temperature factor; certain electrical properties, e.g. superconductivity; dielectric phenomenon at low frequencies; V Casey 22

27 SSP LINEAR LATTICE MODELS interaction of radiation with matter, e.g. light and thermal neutrons. Are the atomic motions that are revealed by these factors random, or can we find a good description for the dynamics of the crystal lattice? The answer is that the motions are not random but are constrained and determined by the forces that atoms exert on each other. 2.3 Linear Lattice Models Linear 1D Chain Model all atoms identical (mass m) lattice spacing a Figure 2.5: A linear monatomic lattice: mass and spring model. For small vibrations, the force on any one atom is proportional to its displacement relative to all the other atoms. Choose atom s: F s = C p (u s+p u s ) p Here, p takes on both positive and negative values, C is the force constant which for a given direction now also depends on p, i.e. is large for p = 1, smaller for p = 2, etc.. F s = ma = m d2 u s dt 2 = C p (u s+p u s ) p We should expect harmonic vibrations represented by a traveling wave of amplitude u, where x s =sa: u s = ue i(kx s ωt) 23 V Casey

28 2.3. LINEAR LATTICE MODELS SSP1 and Substituting for u s and u s+p gives u s+p = ue i[k(s+p)a ωt]. Now C p = C p, therefore: mω 2 = C p (e ikpa 1). p mω 2 = C p (e ikpa + e ikpa 2) p>0 ω 2 = 1 m p>0c p [2cos(kpa) 2] ω 2 = m 2 C p [1 cos(kpa)] p>0 ( ) = m 4 C p cos 2 kpa 2 p>0 For simplicity set p = 1, i.e. nearest neighbour interaction only: ω 2 = 4C 1 m sin2 ka 2 4C ω = 1 m sin2 ka 2 [Note that C 1 is the effective spring constant for nearest neighbour interaction and is not necessarily the principal stress coefficient C 11 which we have used in our elastic theory.] ( ) ±π ω max = ω = 2 a C1 which is periodic in k with period 2π/a. It is clear from the dispersion relationship that lattice waves in a one dimensional lattice constitute a dispersive system for which the velocity varies with frequency and wavelength. Only for very small k, i.e. for very long wavelengths, does the velocity become a constant. The periodic form of ω(k) shown above suggests that some simplification in notation should be possible. Consider, for instance, a wave with wavenumber k compared to a wave with wave number k = k + n(2π / a). V Casey 24 m

29 SSP LINEAR LATTICE MODELS Figure 2.6: Dispersion curve for a linear monatomic lattice. u = u 0 e i(k x ωt) = u 0 e i[(k+n2π /a)x ωt] = ue i(n2π /a)x = ue i(n2π /a)sa = ue in2πs = u The displacement for k is therefore the same as for k. k consists of a wave of smaller wavelength than that corresponding to k, passing through all the atoms, but containing more oscillations than needed for the description. We can describe the displacement of the atoms in these vibrations most easily by looking at the limiting cases k = 0 and k = π a. The situation at k = 0 corresponds to an infinite wavelength; this means that all of the atoms of the lattice are displaced in the same direction from their rest position by the same displacement magnitude. For long wavelength vibrations neighbouring atoms are displaced by the same amount in the same direction. Since the long-wavelength longitudinal vibrations correspond to sound waves in the crystal, all of these vibrations with a similarly shaped dispersion curve, whether transverse or longitudinal vibrations are involved, are called acoustical branches of the vibration spectrum. [When k = 0, dω / dk = ω / k = velocity of sound]. The atomic displacements at k = π/a can be seen by substituting this value of k and x = sa into the general form of the solution for the displacement of a wave: 25 V Casey

30 2.3. LINEAR LATTICE MODELS SSP1 u s = ue i(kx ωt) + ue i(kx+ωt) = 2u( 1) s e iωt showing that neighbouring atoms are displaced by the same distance in opposite directions, giving rise to a minimum physically meaningful wavelength λ min = 2a. This is equivalent to the Bragg reflection condition nλ = 2d sinθ in its one dimensional equivalent form λ = a with n = 1 and d = a. Waves with k = π/a are unable to propagate through the crystal; this is consistent with the fact that the group velocity at k = π/a is equal to zero. At k = π/a, ω has a maximum and therefore dω / dk = 0. Since ω=0 when k = 0 and the latter corresponds to an infinite wavelength; i.e. represents one extreme of the spectrum and is therefore a good choice as the origin. There is no additional information available by extending the plot outside of the region π / a k π/a; we may place a physical interpretation on this maximum value of k = π/a - this represents the smallest wavelength that can propagate in the lattice, i.e. λ = 2a. Figure 2.7: Longitudinal and transverse modes supported by a linear monatomic lattice. In addition to longitudinal vibrations, the linear lattice supports transverse displacements leading to two independent sets (in mutually perpendicular planes) of vibrations that can propagate along the lattice. The forces acting in a transverse displacement are weaker than those in a longitudinal one. They give rise to a new branch of dispersive modes lying below the longitudinal branch. V Casey 26

31 SSP LINEAR LATTICE MODELS Diatomic Linear Chain Model If there are different kinds of atom in the structure, e.g. CsCl, then we must allow for different displacements of the different masses in, for instance, a diatomic linear chain. A similar analysis to the above leads to the following equations for nearest neighbour interaction: M 1 ü s = C 1 (v s+1 /2 + v s 1 /2 2u s ) M 2 v s+1 /2 = C 1 (u s+1 + u s 2v s+1 /2) If we assume that Figure 2.8: A linear diatomic lattice. i(ksa ω t) u s = ue v s+1 /2 = ve i[k(s+1 /2)a ω t) then we obtain the following dispersion relation: M 1 M 2 ω 4 2C 1 (M 1 + M 2 )ω 2 + 4C1 2 sin 2 ka 2 = 0 which when solved gives: ( ) [ ω 2 M1 + M 2 = C 1 ± C1 2 M 1 M 2 ( M1 + M 2 M 1 M 2 ) 2 4C2 1 M 1 M 2 sin 2 ka 2 ]1 / 2 or ( ω 2 ± = A ± A 2 Bsin 2 ka ) 1 / V Casey

32 2.3. LINEAR LATTICE MODELS SSP1 Figure 2.9: The two groups of vibrational modes supported by a linear diatomic lattice: acoustical modes and optical modes. There are now two solutions for ω 2, providing two distinct groups of vibrational modes. The first group, associated with ω 2, contains the acoustic modes dealt with above. The second group arises with ω 2 +, and contains the optical modes. These correspond to the movement of the different atom sorts in opposite directions. The name arises because in ionic crystals, they cause an electric polarization and can therefore be excited by light, which as a result is strongly absorbed. However, these optical modes occur in all crystals with two or more different atoms. V Casey 28

33 Chapter 3 Specific Heat Capacity 29

34 3.1. SPECIFIC HEAT CAPACITIES SSP1 3.1 Specific Heat Capacities Historical Classical: Dulong and Petit (1819, C v =3Nk, Correct at high temperature!) Einstein: Based on Planck s quantum hypothesis (1901; Quantised energy, Showed exponential dependence of C v ). Debye: Showed complete dependence (1912) Specific Heat: Classical Model The classical model for specific heats considered the atoms as being simple harmonic oscillators vibrating about a mean position in the lattice. Each atom could be simulated by three simple harmonic oscillators (SHOs) vibrating in mutually perpendicular directions. For a classical SHO: Average kinetic energy = 1 / 2 kt Average potential energy = 1 / 2 kt Total average energy per oscillator = kt Total average energy per atom = 3kT For N atoms the total average energy = U = 3NkT The specific heat capacity is C v = ( du dt = 3Nk = 3R )v Independent of T Classical treatment - Dulong and Petit - the specific heat capacity of a given number of atoms of a given solid is independent of T and is the same for all solids Einstein Model The above treatment of the vibrational behaviour of materials has been entirely classical. For a harmonic solid, the vibrational excitations are the collective independent normal modes, having frequencies ω determined by the dispersion relationship ω(k) with the allowed values of k set by the boundary conditions. In the classical limit, the energy of a given mode with frequency ω, determined by the wave amplitude, can take any value. V Casey 30

35 SSP SPECIFIC HEAT CAPACITIES Figure 3.1: Experimentally measured specific heat and model values. For the Boltzmann energy distribution: as T is raised P(E high ) increases and so the energy of the atomic vibrations becomes greater as we go from low to high T. Einstein produced a theory of heat capacity based upon Planck s quantum hypothesis (Planck s treatment of electromagnetic radiation). He assumed that: each atom is in an identical quantum harmonic oscillator potential well; each atom of the solid vibrates about its equilibrium position with an angular frequency ω; each atom has the same frequency and vibrates independently of other atoms. The quantum mechanical result (can be treated analytically), treating each normal mode as an independent harmonic oscillator with frequency ω. The energy is quantised and can only take values characterised by the quantum number n(k, p) for a particular branch p. A vibrational state of the whole crystal is thus specified by giving the excitation numbers n(k, p) for each of the 3N normal modes. Instead of describing the vibrational state of a crystal in terms of this number, it is more convenient and conventional to say, equivalently, that there are n(k, p) phonons (i.e. particle like entities representing the quantised elastic waves). Einstein:- replaced the classical SHO with a Quantum Harmonic Oscillator, QHO; where energy does not increase continuously but in discrete steps; and the corresponding quantum particle is the phonon! 31 V Casey

36 3.1. SPECIFIC HEAT CAPACITIES SSP1 Figure 3.2: The QHO potential and eigenstates. For a 1D QHO: E n = (n + 1 / 2 ) ω (3.1) n=0,1, The quantum mechanical expression for the energy implies that: the vibrational energy of a solid is non zero even when there are no phonons present the residual energy of a given mode, 1/2 ω, is the zero point energy The one dimensional energy partition function takes the form: Z 1D = β ω(n+1/2) e n 0 where β = 1/k B T. This may be rearranged to a more revealing form: Z 1D = e β ω 2 nβ ω e n 0 where the summation is an infinite geometric series and so: Therefore: Z 1D = e β ω 2 Z 1D = 1 1 e β ω e β ω 2 1 e β ω (3.2) V Casey 32

37 SSP SPECIFIC HEAT CAPACITIES Now we would like to use the energy partition function to get the energy expectation value. We have: Show this as an exercise! so with lnz taking the form: E = 1 Z Z β = ln Z β (3.3) lnz = β ω 2 β ln [1 e ω] where, E = ln Z β = [ ω 2 ] ωe β ω [ ] 1 = ω 1 e β ω e β ω 1 [ ] 1 E = ω 2 + n B(β ω) (3.4) n B (x) = 1 e x 1 is the Bose occupation factor. The mode ω is an excitation that is excited on average up to the n th B level! Alternatively, there is a boson orbital which is occupied by n B bosons (obey Bose- Einstein statistics). C = E T = T ( ωn B(β ω)) = ω [ ] 1 T e β ω 1 Show this. e β ω C 1D = k B (β ω) 2 (e β ω 1) 2 (3.5) e β ω C 3D = 3k B (β ω) 2 (e β ω 1) 2 (3.6) (a) High T β ω << 1 or ω << kt and e ω kt 1 + ω / kt C = 3k B (β ω) β ω (1 + β ω 1) 2 3k B 33 V Casey

38 3.1. SPECIFIC HEAT CAPACITIES SSP1 independent of ω. This is the classical limit because the energy steps, ω, are now small compared with the mean energy of the oscillator. (b) Intermediate T The full Einstein formula must be used. e β ω C 3D = 3k B (β ω) 2 (e β ω 1) 2 where θ E = ω / k, the so called Einstein temperature (this is an ad-hoc parameter!). Figure 3.3: Einstein model specific heat capacity general form. (c) Low T β ω >> 1 and e ω kt >> 1 C 3D 3k B (β ω) 2 1 (e β ω ) 3R( θe T ) 2 e θ E T where θ E = ω/k B. The system gets locked into the zero point energy state and the system specific heat C approaches zero exponentially as T 0. V Casey 34

39 SSP DEBYE MODEL Einstein model successfully accounts for high temperature specific heat, C, i.e. classical limit 3k B ; successfully predicts that C falls with decreasing T ; however, exponential decrease is not observed; if low frequencies are present, then ω will be small, much smaller than kt even at low temperatures; C will remain at 3k B T to much lower frequencies and the fall off is not as dramatic as predicted by the Einstein model; assumption of an average single frequency ω is too simplistic; need a spread of frequencies - a spread of states! 3.2 Debye Model Key Features C does not go to zero as T goes to zero as rapidly as predicted by the Einstein model. Reason - the overly restrictive requirement that all normal modes should have the same frequency. Long λ modes have lower frequency than short λ modes. Long λ modes are much harder to freeze out than short λ modes. Energy spacing for long λ modes are much closer together than short λ modes. Therefore C does not decrease as rapidly with T as suggested by the E model. The long λ modes are able to contribute to C even at low T Debye Model The frequencies of the normal modes are estimated by treating the solid as a continuous isotropic medium. This approach is reasonable because the only modes excited at low T are the long λ modes, i.e. λ > r 0. Assume a linear dispersion relationship: v = f λ = ω/k where k = 2 π/λ or ω(k) = v k. 35 V Casey

40 3.2. DEBYE MODEL SSP1 [ ] 1 E = 3 ω(k) k 2 + n B(β ω(k)) (3.7) Unlike photons, phonons have three polarizations: one longitudinal and two torsional (transverse). Hence the factor of 3! We also must sum over a range of k. Let L be the macroscopic dimension of the system. e ikr = e ik(r+l x) k = n 2π L x where n is integer. What are the possible values of k? For large L, 2π/L will be very small and the sum over all k can be replaced with an integral: k L 2π In 3D wrap crystal into hypertorous in 4D: let L x = L y = L z = L. Each k point will occupy a volume of k space given by: Therefore: (2π) 3 L 3 dk L3 k 3D (2π) 3 dk 3D The step or interval in k space is so small that we may integrate over k space. E = 3 L3 2π 3 [ ] 1 dk ω(k) 2 + n B(β ω(k)) (3.8) Each excitation mode has a boson frequency ω(k) which is occupied on average n B (β ω(k)) times. Consider k space as a spherical shell (recall medium is isotropic). Can consider the surface of this shell at arbitrary k and work out the volume of k space in the interval dk from the surface area: dk x dk y dk z 4π k 2 dk 0 E = 3 4πL3 2π 3 V Casey 36 0 ω 2 dω 1 [ ] 1 v 3 ( ω) 2 + n B(β ω)

41 SSP DEBYE MODEL Figure 3.4: 3D to 1D integral E = 0 [ ] 1 dω g(ω)( ω) 2 + n B(β ω) g(ω) = 3 4πL3 ω 2 2π 3 v [ 3 12 π ω = L 3 2 ] (2π) 3 v 3 [ 12 π ω 2 ] = N (2π) 3 v 3 n ( ) 9 ω 2 = N ω 3 d ω 3 d = 6π2 nv 3 = 6π 2 N V v3 ω d = ( 6π 2 N V ) 1/3 v (3.9) λ d 2πv ω d r 0 ( ) V 1/3 N 37 V Casey

42 3.2. DEBYE MODEL SSP1 E = 0 [ ] 1 dω g(ω)( ω) 2 + n B(β ω) g(ω) dω is the total number of oscillation modes between ω and ω + dω The density of allowed states or just density of states. Count the number of states in an interval and multiply by the expected energy per mode and integrate over all frequencies. Let x = β ω 9N E = ω 3 d 0 9N E = ω 3 d E = [ ] 1 dω ω n B(β ω) 0 9 N (β )4 ω 3 d dω ω 3 1 e β ω 1 0 dx ω 3 1 e x 1 dx ω e x 1 Can be written as a special case of the Riemann Zeta Function ζ(4) = π4 15 E = 9 N k4 B T 4 π 4 ( ω d ) 3 15 (3.10) C = 36 N kb 4 T 3 π 4 15 ( ω d ) 3 (3.11) C = N k B (k B T ) 3 12 π 4 ( ω d ) 3 5 C = N k B (k B T ) 3 12 π 4 ( ω d ) 3 5 (3.12) Shows T 3 dependence! Prefactor of the T 3 term can be calculated in terms of known quantities such as the velocity of sound in the crystal - i.e. is physical! Often we replace ω d by k B θ d where: θ d = ω d k B V Casey 38

43 SSP DEBYE MODEL C = N k B ( T θ d ) 3 12 π 4 5 (3.13) Material θ d (K) Diamond 1850 Berrilium 1000 Silicon 625 Copper 315 Silver 215 Lead 88 There is no upper limit on C as T increases. Should level off to 3 k B N at high temperature. Physically this means that there is no upper limit for k and the number of oscillation modes. Since ω = v k we need to find an upper limit for ω - a cut-off frequency - ω c. Debye imposed an upper limit by insisting that the total number of degrees of freedom, i.e. modes, should be limited to 3N, the number of atoms in the system. This neatly sets ω c. E = 3 N = ωc 0 ωc 0 dω g(ω) dω g(ω)( ω)n B (β ω) Low T implies large β. Therefore ω c has no impact - activity is all down near the zero point energy. n B (β ω) = 1 e β ω β ω 1 = k B T ω E = k B T ωc 0 dω g(ω) For high temperature all modes will be accessible and so the integral evaluates to 3 N and so: and, C = 3 N k B. E = 3 N k B T 3 N = ωc 0 ωc dω g(ω) = 9 N dω ω2 0 ω 3 d = 3 N ω3 c ω 3 d 39 V Casey

44 3.3. PHONONS SSP1 Therefore ω c = ω d The cut-off frequency is exactly the Debye frequency. There is a physical basis for ω d and so also for θ d! 3.3 Phonons Actual Phonon Density of States Neutron scattering data may be used to elucidate the phonon density of states and dispersion relation for a crystal. Multiple branches adds complexity to the actual distribution since these will superimpose on one another. Since the group velocity must disappear at some frequencies and it appears in the denominator of the density of states function, there will be singularities in the distribution - known as van Hove singularities. Figure 3.5: Actual density of states (upper curve) for aluminium compared to Debye model. The dashed curves show the individual transverse and longitudinal branches Phonon Interactions In pure defect free crystals (if such existed) phonons will pass through each other and reflect from the crystal surfaces without internal scattering or energy loss. In real crystals phonons interact, scatter and decay due to imperfections, impurities V Casey 40

45 SSP PHONONS and anharmonicity. It is useful to think of them as being grouped together to form particles, i.e. a group of waves with differing wavelengths and phases but having a reasonably well defined group velocity. At a given temperature we may expect a kinetic equilibrium to establish itself and so simple Kinetic Theory might yield some useful insights to the thermal properties of solids. Figure 3.6: As the crystal expands the equilibrium spacing shifts to the right reducing the symmetry of the potential profile, ie. less well approximated by an harmonic potential. Considering non-metals first, heat would have to be carried by phonons alone since there are no free electrons available. We know from every day experience that heat does not travel through insulators at the speed of sound! Therefore there must be some retarding mechanism operative to reduce the effective phonon velocity - phonon interactions. Experimentally heat is found to diffuse through insulating solids with a mean free path on the order of 10 nm, i.e. much less than the macroscopic dimensions of the solid under test. This is similar to values for ordinary gases and so provides encouragement to try out the kinetic theory with these solids. The thermal conductivity of an ideal gas is given by: K = 1 3 C v Λ where v is the RMS velocity, C is the specific heat capacity and Λ is the mean free path. At low temperature there are relatively few phonons excited and so phonon interaction is limited leading to large Λ s (mean free paths). The specific heat 41 V Casey

46 3.3. PHONONS SSP1 Figure 3.7: Kinetic theory of gases applied to phonons. capacity C will be zero at absolute zero C = 0, T = 0 and will increase as T 3 at low temperature. The thermal conductivity will do likewise at low temperature. At the other extreme of temperature, i.e. high temperature, there will be lots of phonons and so lots of interactions: Λ will decrease; C will have saturated at a constant value and so the thermal conductivity decreases roughly in proportion to the number density of phonons n which will in turn be proportional to the temperature. K Λ 1 n 1 T At intermediate temperatures the thermal conductivity will be set by the mean free path and the size of the sample Anharmonicity and Thermal Expansion At higher temperature more phonons are excited and the anharmonicity of the potential increases. The effects of anharmonicity are more pronounced at higher temperatures. Phonons transfer energy to the lattice. It gives rise to thermal expansion - equilibrium position shifts to the right. Not surprisingly, the thermal expansion coefficient has a similar temperature profile to specific heat capacity. V Casey 42

47 SSP PHONONS Normal and Umklapp Phonon Interactions We can use the Brillouin zone representation to examine simple phonon combination. For two phonons normal combination occurs where their vector sum is less than half a reciprocal lattice vector, i.e. < G/2, and so the limiting case is defined by q 1 = q 2 G/2. The resultant phonon will be within the first Brillouin zone, see Figure3.8, and this interaction is referred to as normal phonon scattering. Figure 3.8: Normal and Umklapp phonon scattering. If however, the phonons combine to give a resultant phonon q 3 > G/2 which is outside the first Brillouin zone, q 3 becomes oppositely directed to q 1 and q 2, see Fig.3.8. It is effectively reflected at the zone boundary in a process known as Umklapping. Therefore the crystal structure is both a source and a sink for crystal momentum in units of the reciprocal lattice vector G. The crystal acts as a momentum buffer enabling both energy and momentum balance to be achieved Neutron Scattering by a Crystal Consider a neutron that is incident upon a crystal. The neutron interacts strongly with only the atomic nuclei in the crystal. It may pass right through the crystal absorbing and emitting phonons. The phonon occupation number in the crystal may change due to this interaction. Conservation of energy requires that the change in the energy of the neutron is equal to the energy of the phonons it absorbed less the energy of the phonons it emitted. Therefore the change in energy of the neutron 43 V Casey

48 3.3. PHONONS SSP1 contains information about the phonon frequencies. However, a second conservation law is required to disentangle this information from the scattering data. The second law is known as the conservation of crystal momentum (symmetries of the Hamiltonian imply conservation laws). If we define the crystal momentum of a phonon to be times its wavevector, then the change in neutron momentum is just the negative of the chnage in total phonon crystal momentum, to within an additive reciprocal lattice vector. This crystal momentum of a phonon is not, however, accompanied by any real momentum of the ionic system. Because there are two conservation laws, it turns out to be possible to extract the explicit forms of the ω(k) from the neutron scattering data in a simple way. Zero-Phonon Scattering The crystal initial and final states are identical. Energy conservation implies that the energy of the neutron is unchanged - the scattering is elastic. Crystal momentum conservation implies that the neutron momentum can only change by G where G is a reciprocal lattice vector. If we write the incident and scattered momenta as: then these restrictions become: p = q, p = q, q = q, q = q + G These are the Laue conditions that the incident and scattered X-ray wave vectors must satisfy in order for elastically scattered X-rays to produce a Bragg peak. Since a neutron with momentum p = q can be viewed as a plane wave with wave vector q, this is not surprising. Elastically scattered neutrons, which do not create or destroy phonons, are found only in directions that satisfy the Bragg condition, and give precisely the same structural information about the crystal as elastic X-ray scattering. One-Phonon Scattering One phonon scattering provides the most accessible spectroscopic information. In the case of the more important absorption mechanism, conservation of energy and momentum implies that E = E + ω(k) p = p + k + G V Casey 44

49 SSP PHONONS where k is the wave vector for a particular absorbed phonon in a specific branch. In the case of emission E = E ω(k) p = p k + G where the phonon has been emitted into a branch with wave vector k. Figure 3.9: Neutron incident on a lattice and event diagram for one-phonon interaction. In either case the crystal momentum law may be used to represent k in terms of the neutron momentum transfer, p p. Furthermore, the additive reciprocal lattice vector that appears in this relation can be ignored, when the resulting expression for k is substituted into the energy conservation law, since each ω(k) is a periodic function in the reciprocal lattice: ω(k ± G) = ω(k). As a result the two conservation laws yield one equation or: p 2 ( = p2 p ) p + ω 2M n 2M n p 2 ( = p2 p p ) ω 2M n 2M n In a given experiment, the incident neutron momentum and energy are usually specified. Thus for a given phonon dispersion relation ω(k) the only unknowns are the three components of the final neutron momentum p. Generally, a single equation relating the three components of a vector p will, if it has any solutions 45 V Casey

50 3.3. PHONONS SSP1 at all, specify a surface in three dimensional ω(p ) space. If we only examine neutrons emerging in definite direction we will specify the direction of p, and can therefore expect to find solutions at only a single point on the surface. If we select a general direction we will see neutrons scattered by one-phonon processes only at a few discrete values of p, and correspondingly only at a few discrete energies E = p 2 /2M n. Knowing the energy and the direction in which the scattered neutron emerges, we can construct p p and E E, and can therefore conclude that the crystal has a normal mode with frequency (E E)/ and wave vector ±(p p)/. A point in the crystal s phonon spectrum has been established. By varying the controllable parameters such as incident energy, orientation of the crystal and direction of detection, a large number of such points may be collected in order to map out the entire phonon spectrum (assuming it is possible to distinguish the phonons scattered in one-phonon processes from all other processes) Inelastic Neutron Spectroscopy The neutron matter interaction is weak. Typical parameters are: m f p 1 cm λ m E 0.08 ev kt 0.025eV Neutrons do have an intrinsic magnetic moment, i.e. spin, and so will interact with atoms and ions in a crystal with magnetic moments. Neutrons are neutral sub-atomic particles with no electrical charge. Because of this, these unassuming particles are non-destructive and can penetrate into matter much deeper than charged particles such as electrons. In addition, because of their spin, neutrons can be used to probe magnetism on an atomic scale. There are two main methods of producing neutrons for materials research. One is by splitting uranium atoms in a nuclear fission reactor. The other, called spallation, involves firing high-energy protons into a metal target, such as mercury or tungsten, to induce a nuclear reaction that produces neutron beams. ILL is the most intense reactor neutron source in the world. ISIS is the most productive spallation neutron source in the world. Neutron sources play a crucial role in research across the scientific spectrum, from nuclear and elementary particle physics, chemistry and materials science to engineering and life sciences. Neutrons have unique advantages as a probe of atomic-level properties: V Casey 46

51 SSP PHONONS Figure 3.10: European Spallation Source (ESS) is due to complete in Lund, Swedan by 2020, Figure 3.11: Three axis neutron spectrometer. 47 V Casey

52 3.3. PHONONS SSP1 The process of neutron scattering is non-destructive, so that delicate or valuable samples can be studied. Neutrons are penetrating, so that they can look deep inside engineering samples to study, for example, welds. Neutrons with energies in the range of atomic motions have wavelengths of the order of the distances between atoms making them very good at studying both where atoms are and how they are moving. Neutrons are good at seeing light atoms, such as hydrogen, in the presence of heavier ones. Neutrons are good at distinguishing neighbouring elements in the periodic table. Different isotopes of the same element scatter neutrons differently. For example, extra information can be gained by swapping hydrogen atoms with their deuterium isotopes in part of a sample. Neutrons have a magnetic moment, meaning that they can be used to study the magnetic properties of materials. V Casey 48

53 Chapter 4 Drude Free Electron Theory 49

54 4.1. CLASSICAL THEORY AND CONDUCTION IN SOLIDS SSP1 4.1 Classical Theory and Conduction in Solids Texts Simon The Oxford Solid State Basics Myers Introductory Solid State Physics Kittel Introduction to Solid State Physics Ashcroft & Mermin Solid State Physics Free Electron Theory (Drude s Free Electron Theory) Figure 4.1: Inorganic crystalline solids. Before 1900 it was known that most conductive materials obeyed Ohm s Law I = V /R. In 1897 J. J. Thompson discovers the electron as the smallest charge carrying constituent of matter with a charge equal to -q q = C In 1900 Paul Drude formulated a theory for conduction in metals using the electron concept. The theory assumed: 1. Metals have a large density of free electrons. V Casey 50

55 SSP CLASSICAL THEORY AND CONDUCTION IN SOLIDS 2. The electrons move according to Newton s laws until they scatter from ions, defects,etc. 3. After a scattering event the momentum of the electron is completely random (i.e. has no relation to its momentum before scattering - memory loss ) Figure 4.2: Electron scattering without and with an electric field. Thermal velocity: v th = 3kT m 0 (4.1) Average thermal energy is 3 2 kt and kinetic energy is 1 2 mv2 th! Free electrons obey Maxwell-Boltzmann statistics and Newton s Laws. Parameters: n free electron density (number per unit volume) E electric field -q electron charge m 0 electron rest mass Let τ be the scattering time and 1/τ be the scattering rate. This means that the probability of scattering in a small time interval dt is: dt/τ. The probability of not scattering in time dt is then: ( 1 dt ) τ. Let p(t) be the average electron momentum at time t, then we have: ( p(t + dt) = 1 dt ) ( ( ) p(t) q τ dt E(t)dt) + (0) (4.2) τ Scattering - average momentum after scattering is zero - dt τ (0) term! 51 V Casey

56 4.1. CLASSICAL THEORY AND CONDUCTION IN SOLIDS SSP1 No scattering - motion follows Newton s laws - first term on the right above! ( 1 dt ) ( p(t) q τ E(t)dt) Therefore: d p(t) = q dt E(t) p(t) (4.3) τ showing the interplay between the external force and the internal frictional damping type term. (Eq of motion, free electron: F = m 0 a = qe.) Three scenarios to deal with: 1. No electric field. 2. Constant uniform electric field. 3. Time dependent sinusoidal electric field Case I: No electric field d p(t) dt = p(t) τ p(t) = p(0) e t τ Steady state solution: p(t) = 0. Random scattering averages momentum to zero. If you impart momentum to the electrons, on average, they will relax back to zero momentum exponentially with a time constant τ Case II: Constant uniform electric field The exponential term becomes negligible with time leaving only the field term p(t) = qτ E. The electron drift velocity is defined as: where v d = p m 0 = qτ m 0 E = µ n E µ n = qτ m 0 (4.4) is the electron mobility (drift velocity per unit electric field, units: cm 2 /V s). The electron current density J (units: Amps/cm 2 ) is: J = n q v d = nqµ n E = σ E V Casey 52

57 SSP CLASSICAL THEORY AND CONDUCTION IN SOLIDS where n is the free electron density (units: cm 3 ) (dc conductivity, units: Siemens/cm) σ 0 = nqµ n = nq2 τ m 0 (4.5) Case III: Time dependent sinusoidal electric field d p(t) = q dt E(t) p(t) τ There is no steady state solution in this case. Assume the electric field, average momentum and currents are all sinusoidal with phasors given as follows: [ E(t) = Re E(ω)e iωt] p(t) = Re [ p(ω)e iωt] [ J(t) = Re J(ω)e iωt] d p(t) dt v d (ω) = p(ω) m 0 The electron current density is: = q E(t) p(t) τ iω p(ω) = q E(ω) p(ω) τ p(ω) = qτ 1 iωτ E(ω) qτ m 0 = 1 iωτ E(ω) J(ω) = n( q) v d (ω) = σ(ω) E(ω) nq 2 τ m σ(ω) = 0 1 iωτ = σ 0 1 iωτ - a very important result! Result is applied to propagation of em radiation in metals. (4.6) Gives right result at dc ω = V Casey

58 4.1. CLASSICAL THEORY AND CONDUCTION IN SOLIDS SSP1 Ignores magnetic field effects (B only adds a term which is v/c smaller than the electric field term). Must assume that the electric field does not vary significantly locally, i.e. the wavelength of the em wave is much greater than the mean free path of the electron λ >> l. The electric field has impact only since last collision. Holds for visible light since wavelengths are 10 3 to 10 4 Å. Drude theory accounts for Ohm s Law Considering unit volume of the metal, which contains n free electrons, each of charge q, then the total charge will be nq. The total charge crossing unit area per unit time will be nqv d. This is the definition of current density and so: J = nq v d = nqµ n E J = σ E where σ = nqµ n is the sample conductivity which is reciprocally related to the resistivity: σ = 1 ρ By simple geometric considerations it is easy to show that the current density leads to: I = A ρl V R = ρl A i.e. Ohm s law, which is one of the successes of the Drude theory. Since lattice scattering is expected to increase with increasing temperature the theory also explains why many metals have positive temperature coefficients of resistance, i.e. resistance increases with temperature. The current density equation is known as the microscopic version of Ohm s Law. Note also that σ = nq2 τ m 0 using the expression for drift velocity in the current density equation. V Casey 54

59 SSP CLASSICAL THEORY AND CONDUCTION IN SOLIDS Matthiessen s Rule According to the Drude theory of conduction we would expect the resistivity of a metal to become smaller and smaller as the temperature is reduced reaching a value of zero at absolute zero, due to reduced lattice vibration. However, real crystals always contain defects such as vacancies, dislocations and impurities. These will scatter electrons also and contribute to a background resistivity which would remain even if absolute zero could be reached. We can attribute a characteristic time τ i the relaxation time to the electron inter-collision time for each type of impurity and defect. Considering impurities only, the number of collisions for the i th impurity is 1/τ i and the total number of collisions for impurities is 1/τ i. If τ L is the relaxation time for lattice scattering, then the total number of scatters per second is 1 τ T = 1 τ i + 1 τ L (4.7) where τ T is the overall relaxation time. However, from (??) and using (??), and so (4.7) becomes 1 τ = nq2 ρ m 0 (4.8) ρ T = ρ i + ρ L (4.9) after canceling the common factor, nq 2 /m. Calling ρ i, ρ 0 gives ρ T = ρ 0 + ρ L (4.10) where ρ 0 is the residual resistivity which is independent of temperature and roughly proportional to the amount of impurities present. ρ L is the lattice scattering resistivity which approaches zero at zero Kelvin and rises approximately linearly (basis of platinum resistance thermometer) above a characteristic temperature for each metal (the Debye Temperature). Clearly, we can include additional terms to take dislocations, defects, etc., into account as appropriate. Equation (4.10) is known as Matthiessen s rule. It is important because it leads to the idea that, in an ideally pure metal, the resistivity is due solely to the thermal vibrations of the lattice. The purity of a metal may be expressed in terms of the ratio of its resistivity at room temperature (293 or 300K) and its resistivity at liquid helium temperature (4.2K), i.e. the residual resistivity. However, it must be used with care since only impurities in solid solution scatter electrons well: precipitates make only a small contribution. 55 V Casey

60 4.1. CLASSICAL THEORY AND CONDUCTION IN SOLIDS SSP1 Mean Free Path The average distance traveled by an electron between collisions is: λ = τ v (4.11) where v is the average velocity of electrons made up of thermal and drift velocities. Normally v th /v d > 10 6 and so λ τv th (4.12) Considering copper at room temperature (300K), σ = Sm 1, n = cm 3, τ = s, λ = 2.3nm. The atomic spacing in copper is about 0.2nm. Therefore, using classical theory, the electron appears to travel ten times the average distance between atoms before colliding with one. However, the experimentally determined value for the mean free path in copper at room temperature is about 53nm much larger than the classically predicted value Figure 4.3: Charge particles in magnetic fields: cyclotron/betatron/mass spectrometer Hall Effect A transverse voltage develops across a conductor that is carrying current whilst in a magnetic field. The Lorentz force on the carriers is given by the cross product of the current density and the magnetic field. d p(t) dt + p(t) τ V Casey 56 ( ) = q E + p B m 0

61 SSP CLASSICAL THEORY AND CONDUCTION IN SOLIDS Figure 4.4: Schematic of Hall effect experimental arrangement. In steady state, the current is time independent, i.e. constant, and hence: ( ) p(t) = q E + p B τ m 0 p x τ = qe x qp yb z m o p y τ = qe y + qp xb z m o p z τ = qe z Multiply both across by nqτ/m o noting that ω c = qb/m (the cyclotron frequency) and simplifying: σ o E x = j x + ω c τ j y σ o E y = ω c τ j x + j y ( Ex E y ) = 1 σ 0 ( 1 ωc τ ω c τ 1 For the Hall effect j y = 0 implying that: E y = ω cτ σ 0 j x )( jx j y ) R H = E y = ω cτ = 1 j x B z σ 0 B z nq 57 V Casey

62 4.1. CLASSICAL THEORY AND CONDUCTION IN SOLIDS SSP1 Figure 4.5: Hall probe. Figure 4.6: The Rev counter/pulse Counter. V Casey 58

63 SSP DRUDE MODEL AND METAL REFLECTIVITY Table 4.1: Hall Coefficients for some metals. Metal R H nq Na Cu Ag -0.8 Pd Pt Cd +0.5 W +1.2 Be +5.0 Hall voltage, V H = w E y may be measured directly by placing contact electrodes on the sides of the specimen: V H = R H I B t = 1 IB nq t (4.13) where I, B and sample thickness t are easily determined experimentally. Therefore R H and/or n can be calculated. Show how t rather than w enters the above relationships! Importance: Carrier type by sign of Hall Coefficient 1/nq, +1/pq. Carrier density. Not in agreement with experiment in relation to: Positive Hall coefficients found in some metals. Magnetoresistance The magnetoresistance is zero according to Drude since: j x (B) = σ 0 E x = j x. However, metals do display magnetoresistive effects. 59 V Casey

64 4.2. DRUDE MODEL AND METAL REFLECTIVITY SSP1 Figure 4.7: Incident, transmitted and reflected electromagnetic waves - air to metal. 4.2 Drude Model and Metal Reflectivity When electromagnetic waves are incident on an air-metal interface there is a reflected wave: The reflection coefficient is: What is ε(ω) for metals? From Maxwell s equation: Ampere s law: Γ = E R E I = ε0 ε(ω) ε0 + ε(ω) Phasor form: H ( r,t) = J( r,t) + ε 0 E( r,t) t H ( r,t) = J( r) iωε 0 E( r) = σ(ω) E( r) iωε 0 E( r) = iωε e f f (ω) E( r) where ( ε e f f (ω) = ε i σ(ω) ) ωε 0 is the effective (complex) dielectric constant for metals. The metal reflection coefficient becomes: (4.14) Γ = E R ε0 ε e f f (ω) = E I ε0 + ε e f f (ω) V Casey 60

65 SSP DRUDE MODEL AND METAL REFLECTIVITY Using the Drude expression for frequency dependent conductivity, Eqn.4.6, the frequency dependence of the reflection coefficient of metals can be explained adequately all the way from radio frequencies (RF) to optical frequencies Drude model and plasma frequency of metals For small frequencies (ωτ) << 1): σ(ω) σ 0 = nq2 τ m 0 ( ε e f f (ω) ε i σ ) 0 ωε 0 For large frequencies (ωτ) >> 1) (collision-less plasma regime): σ(ω) σ 0 iωτ = i nq2 m 0 ω ε e f f (ω) ε 0 ( 1 ω2 p ω 2 where the plasma frequency is: nq ω p = 2 (4.15) ε 0 m 0 Electrons behave like a collision-less plasma. For ω p > ω >> 1 τ, ε real, negative; decay - no propagation! When ε is positive (ω < ω p ), oscillatory; radiation propagates; transparent Plasma oscillations in metals The electric field generated by the charge separation: The force on the electrons is: E = nqu ε 0 F = qe = nq2 u ε 0 ) 61 V Casey

66 4.2. DRUDE MODEL AND METAL REFLECTIVITY SSP1 Figure 4.8: The electric field generated by charge movement from one region (positive) to another region of extent u. As a result of this force the electron displacement u will obey Newton s second law: F = m 0 d 2 u(t) dt 2 d2 u(t) dt 2 A second order system with solution: = qe = nq2 u(t) ε 0 = ω 2 pu(t) u(t) = Acos(ω p t) + Bsin(ω p t) Plasma oscillations are charge density oscillations Plasma oscillations in metals with scattering From Eqn.(4.3)we know in the presence of scattering: d p(t) dt = q E(t) p(t) τ d 2 u(t) m 0 dt 2 = q E(t) m 0 du(t) τ dt E(t) = nqu(t) ε 0 Combining these yields the differential equation: or d 2 u(t) dt 2 = ω 2 pu(t) 1 τ du(t) dt V Casey 62

67 SSP DRUDE MODEL AND METAL REFLECTIVITY d 2 u(t) dt du(t) + ω 2 τ dt pu(t) = 0 A second order system with damping! Case I: Underdamped ω p > 1 2τ u(t) = e γt [Acos(Ωt) + Bsin(Ωt)] Damped plasma oscillations. Case II: Overdamped γ = 1 2τ Ω = ω 2 p γ 2 ω p < 1 2τ u(t) = Ae γ 1t + Be γ 2t γ 1 = 1 2τ + 1 4τ 2 ω2 p No oscillations. γ 2 = 1 2τ 1 4τ 2 ω2 p 63 V Casey

68 4.2. DRUDE MODEL AND METAL REFLECTIVITY SSP1 V Casey 64

69 Chapter 5 Band Theory of Solids 65

70 5.1. BAND THEORY OF SOLIDS SSP1 5.1 Band Theory of Solids Application of Schrodinger s Equation to a one dimensional crystal Introduction In the free electron model of a metal the effects of the positive ions, i.e. the lattice, are neglected. We saw earlier that all energies were allowed in the free electron case. The free electron theory (U = 0 or constant) in the form of the Drude theory was reasonably successful in explaining many metallic properties. However, it was not possible to account for the differences between metals, insulators and semiconductors using this theory. In addition, the specific heat capacity of solids was found to be much smaller than that predicted by the free electron theory. Figure 5.1: A realistic periodic potential for a 1D crystal.( Teeth shaped curves correspond to potential along a line of ions; continuous curve corresponds to potential along a line between planes of atoms; broken dashed curves are for isolated ions). In order to account for the differences between electronic materials, the influence of the crystal lattice on the electrons must be taken into account when applying Schrödinger s equation to the crystal. The results, even for a very simple model of the lattice potential, are surprising. We find that a whole new theory of solids emerges, the Band Theory of Solids. In this theory, entities such as bandgaps, positive and negative electron masses, positive and negative electrons, i.e. holes and electrons, emerge as natural consequences of the fact that the crystal lattice gives rise to a periodic potential energy profile within the crystal. V Casey 66

71 SSP BAND THEORY OF SOLIDS Bloch s Theorem For a 1D (monatomic) crystal the potential repeats itself with spatial period given by the lattice constant a U(x + a) = U(x) Bloch s theorem tells us that for such a potential the solutions to the Schrödinger equation can be taken to satisfy the condition: Ψ(x + a) = e ika Ψ(x) for some constant K. Let D be the displacement operator :D f (x) = f (x + a) For a periodic potential such as U(x), D commutes with the Hamiltonian operator: [D,H] = 0 This means we are free to choose eigenfunctions of H that are simultaneous eigenfunctions of D: DΨ = λψ or, Ψ(x + a) = λψ(x) Now since this relationship holds for all of x, λ cannot be zero. It is complex and so can be expressed in exponential form: λ = e ika for some constant K. It follows from the above and using the Born-von Karmen boundary condition that: Ψ(x + Na) = e ikna Ψ(x) so e ikna = 1 or NKa = n2π, and hence e ikna Ψ(x) = Ψ(x) K = n2π,(n = 0,±1,±2,...) Na and K is real. The important outcome of Bloch s theorem is that we need only solve the Schrödinger equation within a single cell (0 x a). We may then easily generate all other (infinite number?) of solutions. Note that since K is real, this means that: Ψ(x + a) 2 = Ψ(x) 2 and so while Ψ(x) is not periodic its expectation value is. 67 V Casey

72 5.1. BAND THEORY OF SOLIDS SSP The Nature of the Crystal Potential Field X-ray diffraction has shown that crystals consist of well defined periodic arrays of atoms. Let us consider a one dimensional crystal with atoms which consist of positively charged cores surrounded by loosely bound outer electrons (the outer electrons are loosely bound due to the screening effects of the inner electron shells). Since the atoms are in a periodic arrangement, it is only reasonable to assume that the potential field arising from the atom cores will also be periodic. The outer electrons, i.e. the conduction electrons, move in this periodic potential. A realistic potential profile, based upon Coulomb s potential, i.e. U(x) α 1/x, might look like that shown in the Fig However, it is extremely difficult to solve Schrodinger s equation for a realistic periodic potential. Krönig and Penney suggested a simplified model consisting of a 1D array of square well potentials of width a, separated by potential barriers of height U 0 and width b. It is assumed that for any electron, everything else in the crystal can be represented by this effective potential. This one electron is then considered to be representative of all other electrons in the system. The value of this model is that Schrödinger s equation may be solved individually for both regions. By establishing continuity equations at the boundaries and using Bloch s theorem, the form of overall solutions are arrived at. Figure 5.2: Square Well Potential representation of the lattice potential Solving Schrodinger s equation for the K-P 1D model 2 d 2 Ψ(x) 2m dx 2 + (E U(x))Ψ(x) = 0 (5.1) d 2 Ψ(x) dx 2 V Casey 68 2m (U(x) E)Ψ(x) = 0 (5.2) 2

73 SSP BAND THEORY OF SOLIDS Figure 5.3: Krönig Penney Model of a 1-D crystal. Region I 0 x a, U(x) = 0 Solution: where α 2 = 2mE 2 d 2 Ψ(x) dx 2 + 2mE Ψ(x) = 0 (5.3) 2 Ψ(x) = Ae iαx + Be iαx (5.4)... an energy term Region II a x l, U(x) = U 0 Solution d 2 Ψ(x) dx 2 2m 2 (U 0 E)Ψ(x) = 0 (5.5) Ψ(x) = Ce γx + De γx (5.6) γ 2 = 2m 2 (U 0 E) (5.7) Now since the barrier is finite, there must be some probability of penetration by the electrons. Therefore, both the wave function Ψ and its first derivative Ψ must be continuous at points such as x = 0, a, l, etc.. This allows us establish relationships between the constants A,B,C and D. Continuity at, x = 0 69 V Casey

74 5.1. BAND THEORY OF SOLIDS SSP1 Ψ 1 (0) = Ψ 2 (0) A + B = C + D (5.8) and at x = a Ψ 1 (0) = Ψ 2 (0) iα[a B] = γ[c D] (5.9) Ψ 1 (a) = Ψ 2 (a) Ae iαa + B iαa = Ce γa + De γa (5.10) and at x = l, using Bloch s theorem Ψ 1 (a) = Ψ 2 (a) iα[ae iαa B iαa ] = γ[ce γa De γa ] (5.11) Ψ 1 (l) = Ψ 1 (0)e ikl = Ψ 2 (l) (A + B)e ikl = Ce γl + De γl (5.12) Ψ 1 (l) = Ψ 1 (0)eikl = Ψ 2 (l) iα(a B)e ikl = γ[ce γl De γl ] (5.13) Equations have a solution only if the determinant of the coefficients of A,B,C and D vanish or if, α 2 γ 2 Sinh(γb) Sinh(αa) +Cosh(γb) Cosh(αa) = Cos(kl) (5.14) 2αγ for E < U 0. For E > U o, γ becomes purely imaginary. Equation 5.14 does not change significantly if we replace γ by iγ: γ 2 α 2 Sin(γb) Sin(αa) +Cos(γb) Cos(αa) = Cos(kl) (5.15) 2αγ (Cosh x = Cos ix; Sinh x = 1/i Sin ix ) We could plot the left hand side of the last two equations versus E/U 0, (see Bar-Lev p69). However, to obtain a more convenient form Kröonig and Penney considered the case where the potential barrier becomes a delta function, that is, the case where U 0 is infinitely large, over an infinitesimal distance b, but the product U 0 b remains finite and the same, i.e. as U 0,b 0, U 0 b = constant. Now γ 2 U 0 and also goes to infinity as U 0. Therefore Lim(γ 2 α 2 ) U0 = γ 2 What happens the product γb as U 0 goes to infinity? V Casey 70

75 SSP BAND THEORY OF SOLIDS b becomes infinitesmal as U 0 becomes infinite; however, since γ is only proportional to U 0 it does not go to infinity as fast as b goes to zero so, the product γb goes to zero as U 0 goes to infinity. Therefore: Also, since l = (a + b). Equation 5.15 then reduces to γb 0 as U 0 Sin(γb) γb Cos(γb) 1 as U 0 as b 0 a l or γ 2 b sin(αl) + cos(αl) = cos(kl) (5.16) 2α γ 2 ab Sin(αl) + cos(αl) = cos(kl) (5.17) 2 (αl) If we define ( γ 2 ) Uo ab P = Lim 2 b 0 which is a measure of the barrier strength or stopping power, then or P Sin(αl) αl +Cos(αl) = Cos(kl) (5.18) PSinc(αl) +Cos(αl) = Cos(kl) (5.19) This is the dispersion relationship for electrons in a periodic 1D crystal. 71 V Casey

76 5.1. BAND THEORY OF SOLIDS SSP theta Figure 5.4: Dispersion curve for K-P model with a barrier stopping power of P= Figure 5.5: Dispersion curve for K-P models with increasing barrier stopping power, 0,1,4,10. V Casey 72

77 SSP BAND THEORY OF SOLIDS Dispersion Relation for the Krönig-Penney Model When Schrodinger s equation is applied to the K-P model as outlined above, solutions are only possible which satisfy the following dispersion relationship between energy (alpha term) and momentum (k term): P Sin α l α l where l = a + b, the lattice spacing; + Cos α l = Cos kl α 2 = 2mE 2 P = mabu o 2 Clearly the right hand side of this relationship has limits of +1 and 1. This imposes bounds upon the left hand side of the equation. In other words, this relationship allows one determine whether a given α value is allowed or forbidden. The best way to proceed is to plot the left hand side of the equation as a function of α. The parameter P is a measure of the stopping power of the barrier; it involves the product U o b. P has the value 3.11 for the simple crystal K-P potential shown earlier, figure The plot of the LHS of the dispersion relation for a range of values of P is shown in figure The bounds imposed by the RHS of the relation are also shown. There are regions of αl space and hence energy which are allowed and other regions which are forbidden. A clearer picture emerges when energy E is plotted versus wave vector k as shown in figure5.6. This latter plot is obtained directly from the plot of the dispersion relation. Consider, for instance, the case where P = 3.11, figure As αl increases from zero, the LHS of the function is out of bounds. It comes into bounds at α 1 l, and hence energy E 1, where Cos(kl) = +1. This will occur for kl = 0, 2π, etc., (k = n 2π/l). Let us take k = 0 at this point, for convenience. The function remains within bounds until α 2 l (energy E 2 ), at which point Cos(kl) = 1. Here kl = π or k = π/l and the first allowed energy band is defined. As αl increases further, the function goes out of bounds again. The next allowed zone starts at α 2 l (energy E 2 ), where Cos(kl) is still equal to 1. Thus, in terms of the E versus k diagram, there is a discontinuity in energy at k = π/l; an energy gap occurs. The top of the second allowed band occurs where Cos(kl) is again equal to +1 (here k = 2π/l). The process may be repeated to give as many allowed and forbidden bands as required. The function is symmetric in k space so it is really only necessary to plot for positive k values and then map the curves into negative k space. 73 V Casey

78 5.1. BAND THEORY OF SOLIDS SSP1 Since the dispersion relation will have the same value if Cos(kl) is replaced by Cos(k + n2π/l)l, the energy curves are unchanged if shifted in k space by n2π/l. We can therefore represent all the information in the region π/l << k << +π/l. This is known as the reduced zone scheme, figure5.7 Figure 5.6: Extended zone scheme Figure 5.7: Reduced zone scheme V Casey 74

79 SSP SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS 5.2 Some consequences of the band theory of solids Band Filling Aufbau Principle: Electrons fill the energy bands starting with the lowest available level and working upwards through the band! Conductor (Metal) Conductivity occurs if: -the upper conduction band is partially filled, -a full band overlaps an empty band. Figure 5.8: Conductor Insulator: Each band is either full or empty and no overlap occurs. Also the bandgap is large, i.e. E g > 3eV. Semiconductor: Full valence band. Small energy gap, i.e. E g 1eV Band Shapes of Real Semiconductors The interatomic distance and potential will depend upon direction within the crystal. This implies that the shape of the E versus k diagrams will depend upon the direction of k, and so one may expect much more complicated band shapes than Figure 5.9: Insulator 75 V Casey

80 5.2. SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS SSP1 Figure 5.10: Semiconductor those obtained earlier for the simple 1D case. There will in general be a number of minima (depending upon the k direction) in the conduction band. The valence band maximum is always at k = 0. Arising from these considerations, two general categories of semiconductor may be defined. These are: -direct bandgap semiconductors, and -indirect bandgap semiconductors. Figure 5.11: Direct bandgap semiconductor Direct Bandgap Semiconductors: GaAs; InSb; InP; CdS Indirect Bandgap Semiconductors: Si; Ge; GaP; AlSb. For direct band gap semiconductors such as GaAs, the lowest conduction band minimum occurs at k = 0. For indirect bandgap semiconductors such as Si and Ge, the lowest conduction band minimum occurs near the zone edge. An electron V Casey 76

81 SSP SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS Figure 5.12: Indirect bandgap semiconductor transition between the bands in direct bandgap semiconductors is essentially a single event involving only an energy change. However, in the case of an indirect bandgap semiconductor, an electron transition between, for instance, the top of the valence band and the minimum of the conduction band involves both an energy change step, E, and a momentum change, k, which of course is less likely to occur Effective Mass and Holes What happens when an electron is accelerated in a lattice with a periodic potential? p = h λ = k F = ma = qφ a = qφ m a = dv g dt v g = dω dk = 1 de dk a = dv g dt = 1 d 2 E dk dk 2 dt 77 V Casey

82 5.2. SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS SSP1 Figure 5.13: Group velocity and effective mass for the valence band of a semiconductor de = qφdx = qφv g dt = qφ de dk dt dk dt = qe a = qφ d 2 E 2 dk 2 m = 2 d 2 E/ dk 2 (5.20) V Casey 78

83 SSP SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS Figure 5.14: Parabolic approximation for the valence and conduction band edges When an electron is accelerated in a crystal, its response is no longer determined by the electron rest mass. Instead the electron behaves as though it has an effective mass given by 5.20 above. The effective mass of the electron is determined by the curvature of the E versus k diagram. This is illustrated in the diagram below. Near the bottom of the band, the effective mass is positive and here electrons behave as normal except that the mass is no longer constant or equal to the rest mass. Near the top of a given band we see that the mass of the electron may be negative. In simple terms, this means that the carriers here move in the opposite direction to the applied force. The simplest interpretation of this is to assume that we are now dealing with positive electrons or holes as they are almost universally referred to Filling of the energy bands with electrons. Parabolic Approximation For semiconductor work we are interested mainly in the top of the valence band and the bottom of the conduction band where the E versus k relationship is parabolic. m is therefore constant. Energies in these regions may be approximated by: E = E C + 2 k 2 2m n E = E V 2 k 2 2m p (5.21) (5.22) 79 V Casey

84 5.2. SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS SSP1 Density of States in Semiconductors Once the band-structure has been determined, the next task is to determine the number of modes or states in the band or more importantly, the density of states per unit energy within a given band. Once we have this we can then use appropriate statistics to determine how the bands are filled. Since we are interested in the electronic properties of solids, we will be interested in working out how the bands are filled with electrons. Our interest will be confined to semiconductors with either an almost empty conduction band (n-type) or an almost full valence band (p-type). Therefore we need only consider the bottom of the conduction band or the top of the valence band and so we may use the parabolic approximation to simplify the analysis. Consider a box shaped crystal of sides L x,l y,l z with interatomic spacings l x,l y,l z respectively. There are N x = L x /l x, N y = L y /l y, and N z = L z /l z corresponding values of k x,k y,k z. A single level, therefore, requires a section of Brillouin zone in the k x direction of width k x = 2π / lx N x = 2π L x (5.23) and the volume inkspace taken by a single level is k x k y k z, since two states with opposing spin can have the same energy, a single state requires only half this volume. Therefore a single state volume is: 1 2 k x k y k z = (2π)3 2L x L y L z The number of allowed states N in k space in a spherical shell of radius k and thickness dk is: N = 4πk 2 dk (2π) 3 / 2L x L y L z = The density of states per unit volume is: dn = ( ) k 2 dk π ( ) k 2 L x L y L z dk π For the bottom of the conduction band, E = E C + 2 k 2 2m n, implying: and also, k 2 = 2m n 2 (E E C) (5.24) V Casey 80

85 SSP SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS Figure 5.15: Density of states for parabolic valence and conduction bands dk = m n 2 k de Therefore, the density of states in the conduction band is, dn = 4π h 3 (2m n) 3 /2 (E E C ) 1 /2 de = N C (E)dE where the density of states function is given by N C (E) = 4π h 3 (2m n) 3 /2 (E E C ) 1 /2 (5.25) Similarly for the valence band, the density of states function has the form, dn = 4π h 3 (2m p) 3 /2 (E E V ) 1 /2 de = N V (E)dE N V (E) = 4π h 3 (2m p) 3 /2 (E E V ) 1 /2 (5.26) Occupation of allowed states. For collections of particles, e.g. atoms, molecules, electrons, a statistical treatment which describes the average rather than detailed properties of a typical component of the complete assembly of particles is most useful since the behaviour of the group of particles can then be deduced directly. The type of statistics used depends on - 81 V Casey

86 5.2. SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS SSP1 Figure 5.16: Maxwell-Boltzmann distribution for a classical ensemble of particles the type of particle present (neutral, charged, mass, etc..) the possible interactions between them. Classical particles obey Maxwell-Boltzmann statistics where there is no restriction on the energy of the particles, e.g. ideal gas, dn = N F MB (E) de where N is the number of neutral molecules per cubic meter. F MB is the distribution function and gives the fraction of the total number of molecules per unit volume in the energy range de. In the high energy region, F MB (E) e E k B T Particles which obey the exclusion principle, e.g. electrons, interact quantum mechanically in such a way that the occupancy of a particular state is restricted by the Pauli exclusion principle. For such particles, Fermi-Dirac statistics apply. The Fermi-Dirac distribution function has the form, 1 F FD = 1 + e ( E EF k B T ) (5.27) This gives for any ensemble obeying the exclusion principle, the probability that a particular stateeis occupied. For high energy states, i.e. E >> E F, F FD =FMB. At high energies, the number of electrons distributed over many available states is small and there are many more energy levels than electrons to occupy them. Under these conditions, there is little chance of two or more electrons occupying the same state and whether V Casey 82

87 SSP SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS Figure 5.17: Fermi-Dirac distribution function for electrons at temperatures 2, 50 and 100 K the exclusion principle is included in the statistics or not becomes irrelevant to the form of the distribution. At very low temperatures there is a small but finite probability that electrons will occupy available states for which E > E F but the probability rapidly decreases with increasing energy. As the temperature is increased, this tail of the probability function becomes more pronounced and the probability of occupancy of higher energy states is correspondingly increased. Note that the probability that an electron occupies the Fermi Energy Level, E F, (referred to simply as the Fermi level) is always 1 / 2 independent of the actual temperature. Also, the probability function is symmetric about the Fermi level Properties of Semiconductors Revision Intrinsic Semiconductors Extrinsic Semiconductors Doping Carrier Density A Conduction Band [ N(E)dE = N C (E)F FD (E)dE = 4π h 3 (2m n) 3 /2 (E E C ) 1 /2 The density of electrons in the whole conduction band is, e E E F k B T ] de 83 V Casey

88 5.2. SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS SSP1 n = ETop E C N C (E)F FD (E)dE (5.28) Since E F is located at least a few k B T below E C, we can make the following approximations, We can replace F FD (E) by F MB (E) since (E E F /k B T ) >> 1 Recall F MB (E) goes to zero as E goes to infinity. Therefore, we can replace the top limit of the integral by without changing the result. Therefore, Now n = 4π h 3 (2m n) 3 /2 (E E C ) 1 /2 e E C ( ) E EF k B T ( ) ( ) ( ) e E EF k B T = e E EC k B T.e EC E F k B T de n = 4π h 3 (2m n) 3 /2 e Let x = (E-E C )/k B T ; also de=k B Tdx Substituting, ( ) EC E F k B T E C (E E C ) 1 /2 e ( (E E C ) 1 /2 = kt 1 /2 E EC k B T n = 4π h 3 (2m nk B T ) 1 2 e 0 ( ) EC E F k B T x 1 π /2 dx = 2 0 )1 / 2 ( ) E EC k B T x 1 2 e x dx de n = 2 ( ) h 3 (2πm nk B T ) 3 /2 e EC E F k B T ( ) n = N C e EC E F k B T (5.29) (5.30) where N C is a constant and is known as the Effective Density of States in the conduction band. V Casey 84

89 SSP SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS B Valence Band Similarly for holes in the valence band, p = 2 ( h 3 (2πm pk B T ) 3 /2 e EF E V k B T ) (5.31) ( p = N V e EF E V k B T ) (5.32) where N V is a constant and is known as the Effective Density of States in the valence band. At 300K Si GaAs N C 2.8x10 19 cm 3 4.7x10 17 cm 3 N V 1.04 x cm 3 7.0x10 18 cm Fermi Level in Semiconductors Intrinsic semiconductors In intrinsic semiconductors, n = p = n i where n i is the intrinsic carrier density. N C e ( ) EC E Fi k B T ( ) EFi = N V e E V k B T E N C E V 2E Fi ( C m )3 / 2 = e k B T p = N V m n ( m p E Fi = E C + E V 2 + 3k BT 4 Ln m n ) (5.33) In both Si and Ge,m n m p. Therefore, E Fi is practically mid-way between E C and E V, i.e. at E G / Law of Mass Action np = N C N V e ( ) ( ) EC E Fi EFi k B T e E V k B T ( ) np = N C N V e EG k B T = n 2 i 85 V Casey

90 5.2. SOME CONSEQUENCES OF THE BAND THEORY OF SOLIDS SSP1 ( ) np = AT 3 e EG k B T (5.34) where A is a constant. The electron-hole product in semiconductors in thermal equilibrium is a function of T only since m n and m p and E G are relatively independent of T. This is a general result holding for both intrinsic and extrinsic semiconductors and is known as the law of mass action; the electron-hole product is a constant at a given temperature. n i = 1.45 x cm 3 (Si) n i = 1.79 x 10 6 cm 3 (GaAs) Fermi Level in Extrinsic Semiconductors Doping Figure 5.18: Symbols used with extrinsic semiconductors n-type Addition of Gr V donor Impurity level just below E C! p-type Addition of Gr IV acceptor Impurity level just above E V! N D Density of donor atoms N D + Density of ionised donors N A Density of acceptor atoms N A + Density of ionised acceptors nfree electron density pfree hole density n i Intrinsic electron/hole carrier density. Extrinsic Semiconductor n p Charge Neutrality must prevail p + N + D = n + N A (5.35) n p = N ( ) C e EC 2E F +E V k B T N V V Casey 86