Introduction to Solid State Physics, 8th Edition Charles Kittel CHAPTER 1: CRYSTAL STRUCTURE. Periodic Array of Atoms. Fundamental Types of Lattices. Index System for Crystal Planes. Simple Crystal Structures. Direct Imaging of Atomic Structure. Nonideal Crystal Structures. Crystal Structure Data. CHAPTER 2: WAVE DIFFRACTION AND THE RECIPROCAL LATTICE. Diffraction of Waves by Crystals. Scattered Wave Amplitude. Brillouin Zones. Fourier Analysis of the Basis. CHAPTER 3: CRYSTAL BINDING AND ELASTIC CONSTANTS. Crystals of Inert Gases. Ionic Crystals. Covalent Crystals. Metals. Hydrogen Bonds. Atomic Radii. Analysis of Elastic Strains. Elastic Compliance and Stiffness Constants. Elastic Waves in Cubic Crystals. CHAPTER 4: PHONONS I. CRYSTAL VIBRATIONS. Vibrations of Crystals with Monatomic Basis. Two Atoms per Primitive Basis. Quantization of Elastic Waves. Phonon Momentum. Inelastic Scattering by Phonons. CHAPTER 5: PHONONS II. THERMAL PROPERTIES. Phonon Heat Capacity. Anharmonic Crystal Interactions. Thermal Conductivity. CHAPTER 6: FREE ELECTRON FERMI GAS. Energy Levels in One Dimension. Effect of Temperature on the Fermi-Dirac Distribution. Free Electron Gas in Three Dimensions. Heat Capacity of the Electron Gas. Electrical Conductivity and Ohm’s Law. Motion in Magnetic Fields. Thermal Conductivity of Metals. CHAPTER 7: ENERGY BANDS. Nearly Free Electron Model. Bloch Functions. Kronig-Penney Model. Wave Equation of Electron in a Periodic Potential. Number of Orbitals in a Band. CHAPTER 8: SEMICONDUCTOR CRYSTALS. Band Gap. Equations of Motion. Intrinsic Carrier Concentration. Impurity Conductivity. Thermoelectric Effects. Semimetals. Superlattices. CHAPTER 9: FERMI SURFACES AND METALS. Construction of Fermi Surfaces. Electron Orbits, Hole Orbits, and Open Orbits. Calculation of Energy Bands. Experimental Methods in Fermi Surface Studies. CHAPTER 10: SUPERCONDUCTIVITY. Experimental Survey. Theoretical Survey. High-Temperature Superconductors. CHAPTER 11: DIAMAGNETISM AND PARAMAGNETISM. Langevin Diamagnetism Equation. Quantum Theory of Diamagnetism of Mononuclear Systems. Paramagnetism. Quantum Theory of Paramagnetism. Cooling by Isentropic Demagnetization. Paramagnetic Susceptibility of Conduction Electrons. CHAPTER 12: FERROMAGNETISM AND ANTIFERROMAGNETISM. Ferromagnetic Order. Magnons. Neutron Magnetic Scattering. Ferrimagnetic Order. Antiferromagnetic Order. Ferromagnetic Domains. Single Domain Particles. CHAPTER 13: MAGNETIC RESONANCE. Nuclear Magnetic Resonance. Line Width. Hyperfine Splitting. Nuclear Quadrupole Resonance. Ferromagnetic Resonance. Antiferromagnetic Resonance. Electron Paramagnetic Resonance. Principle of Maser Action. CHAPTER 14: PLASMONS, POLARITONS, AND POLARONS. Dielectric Function of the Electron Gas. Plasmons. Electrostatic Screening. Polaritons. Electron-Electron Interaction. Electron-Phonon Interaction: Polarons. Peierls Instability of Linear Metals. CHAPTER 15: OPTICAL PROCESSES AND EXCITONS. Optical Reflectance. Excitons. Raman Effects in Crystals. Energy Loss of Fast Particles in a Solid. CHAPTER 16: DIELECTRICS AND FERROELECTRICS. Macroscopic Electric Field. Local Electric Field at an Atom. Dielectric Constant and Polarizability. Structural Phase Transitions. Ferroelectric Crystals. Displacive Transitions. CHAPTER 17: SURFACE AND INTERFACE PHYSICS. Surface Crystallography. Surface Electronic Structure. Magnetoresistance in a Two-Dimensional Channel. p-n Junctions. Heterostructures. Semiconductor Lasers. Light-Emitting Diodes. CHAPTER 18: NANOSTRUCTURES. Imaging Techniques for Nanostructures. Electronic Structure of 1D Systems. Electrical Transport in 1D. Electronic Structure of 0D Systems. Electrical Transport in 0D. Vibrational and Thermal Properties of Nanostructures. CHAPTER 19: NONCRYSTALLINE SOLIDS. Diffraction Pattern. Glasses. Amorphous Ferromagnets. Amorphous Semiconductors. Low Energy Excitations in Amorphous Solids. Fiber Optics. CHAPTER 20: POINT DEFECTS. Lattice Vacancies. Diffusion. Color Centers. CHAPTER 21: DISLOCATIONS. Shear Strength of Single Crystals. Dislocations. Strength of Alloys. Dislocations and Crystal Growth. Hardness of Materials. CHAPTER 22: ALLOYS. General Consideration. Substitutional Solid Solutions – Hume-Rotherby Rules. Order-Disorder Transformation. Phase Diagrams. Transition Metal Alloys. Kondo Effect. CHAPTER 1 1. The vectors ˆ ˆ ˆ + + x y z and ˆ ˆ ˆ − − + x y z are in the directions of two body diagonals of a cube. If θ is the angle between them, their scalar product gives cos θ = –1/3, whence . 1 cos 1/3 90 19 28' 109 28' − θ = = °+ ° = ° 2. The plane (100) is normal to the x axis. It intercepts the a' axis at and the c' axis at ; therefore the indices referred to the primitive axes are (101). Similarly, the plane (001) will have indices (011) when referred to primitive axes. 2a' 2c' 3. The central dot of the four is at distance cos60 a ctn 60 cos30 3 a a ° = ° = ° from each of the other three dots, as projected onto the basal plane. If the (unprojected) dots are at the center of spheres in contact, then 2 2 2 a c a , 2 3 ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or 2 2 2 1 c 8 a c ; 1.633. 3 4 a 3 = = 1-1 CHAPTER 2 1. The crystal plane with Miller indices hk is a plane defined by the points a � 1/h, a2/k, and . (a) Two vectors that lie in the plane may be taken as a 3 / � a 1/h – a2/k and 1 3 / h / − � a a . But each of these vectors gives zero as its scalar product with 1 2 h k 3 = + + � G a a a , so that G must be perpendicular to the plane . (b) If is the unit normal to the plane, the interplanar spacing is hk� ˆn 1 ˆ /h ⋅ n a . But , whence . (c) For a simple cubic lattice ˆ / | | = n G G 1 d(hk ) G / h| | 2 / | G| = ⋅ = π � a G ˆ ˆ ˆ (2 / a)(h k ) = π + + � G x y z , whence 2 2 2 2 2 2 2 1 G h k . d 4 a + + = = π � 1 2 3 1 1 3a a 0 2 2 1 1 2. (a) Cell volume 3a a 0 2 2 0 0 ⋅ × = − a a a c 2 1 3 a c. 2 = 2 3 1 2 1 2 3 2 3 ˆ ˆ 4 1 1 (b) 2 3a a 0 | | 2 2 3a c 0 0 2 1 ˆ ˆ ( ), and similarly for , . a 3 × π = π = − ⋅ × π = + x ˆ c y z a a b a a a x y b b (c) Six vectors in the reciprocal lattice are shown as solid lines. The broken lines are the perpendicular bisectors at the midpoints. The inscribed hexagon forms the first Brillouin Zone. 3. By definition of the primitive reciprocal lattice vectors 3 3 2 3 3 1 1 2 1 2 3 3 1 2 3 3 C (a a ) (a a ) (a a ) ) (2 ) / | (a a a ) | | (a a a ) | / V . BZ V (2 (2 ) × ⋅ × × × = π ⋅ × ⋅ × = π = π For the vector identity, see G. A. Korn and T. M. Korn, Mathematical handbook for scientists and engineers, McGraw-Hill, 1961, p. 147. 4. (a) This follows by forming 2-1 2 2 1 2 2 1 2 1 exp[ iM(a k)] 1 exp[iM(a k)] |F| 1 exp[ i(a k)] 1 exp[i(a k)] sin M(a k) 1 cosM(a k) . 1 cos(a k) sin (a k) − − ⋅ ∆ − ⋅ ∆ = ⋅ − − ⋅ ∆ − ⋅ ∆ ⋅ ∆ − ⋅ ∆ = = − ⋅ ∆ ⋅ ∆ (b) The first zero in 1 sin M 2 ε occurs for ε = 2π/M. That this is the correct consideration follows from 1 zero, as Mh is an integer 1 1 sin M( h ) sin Mh cos M cos Mh sin M . 2 2 ± π + ε = π ε + π ε ����� ����� 1 2 5. j 1 j 2 j 3 2 i(x v +y v +z v ) 1 2 3 S (v v v ) f e j − π = Σ Referred to an fcc lattice, the basis of diamond is 1 1 1 000; . 4 4 4 Thus in the product 1 2 3 S(v v v ) S(fcc lattice) S (basis) = × , we take the lattice structure factor from (48), and for the basis 1 2 3 1 i (v v v ). 2 S (basis) 1 e − π + + = + Now S(fcc) = 0 only if all indices are even or all indices are odd. If all indices are even the structure factor of the basis vanishes unless v1 + v2 + v3 = 4n, where n is an integer. For example, for the reflection (222) we have S(basis) = 1 + e–i3π = 0, and this reflection is forbidden. 3 2 1 G 0 0 3 3 0 0 3 2 3 2 2 0 0 0 2 2 2 0 6. f 4 r ( a Gr) sin Gr exp ( 2r a ) dr (4 G a ) dx x sin x exp ( 2x Ga ) (4 G a ) (4 Ga ) (1 r G a ) 16 (4 G a ) . ∞ − = π π − = − = + + ∫ ∫ 0 The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G4 for 0 Ga 1. >> 7. (a) The basis has one atom A at the origin and one atom 1 B at a. 2 The single Laue equation defines a set of parallel planes in Fourier space. Intersections with a sphere are a set of circles, so that the diffracted beams lie on a set of cones. (b) S(n) = f 2 (integer) ⋅∆ π× a k = A + fB e–iπn. For n odd, S = fA – 2-2 fB; for n even, S = fA + fB. (c) If fA = fB the atoms diffract identically, as if the primitive translation vector were 1 a 2 and the diffraction condition 1 ( ) 2 (integer). 2 ⋅∆ = π × a k 2-3 CHAPTER 3 1. 2 2 2 2 E (h 2M) (2 ) (h 2M) ( L) , with 2L / / = π λ = π λ . = 2. bcc: 12 6 U(R) 2N [9.114( R ) 12.253( R) ]. = ε σ − σ At equilibrium and 6 6 0 R 1.488 = σ , 0 U(R ) 2N ( 2.816). = ε − fcc: 12 6 U(R) 2N [12.132( R ) 14.454( R) ]. = ε σ − σ At equilibrium and Thus the cohesive energy ratio bcc/fcc = 0.956, so that the fcc structure is more stable than the bcc. 6 6 0 R 1.679 = σ , 0 U(R ) 2N ( 4.305). = ε − 23 16 9 3. | U | 8.60 N (8.60)(6.02 10 ) (50 10 ) 25.9 10 erg mol 2.59 kJ mol. − = ε = × × = × = This will be decreased significantly by quantum corrections, so that it is quite reasonable to find the same melting points for H2 and Ne. 4. We have Na → Na+ + e – 5.14 eV; Na + e → Na– + 0.78 eV. The Madelung energy in the NaCl structure, with Na+ at the Na+ sites and Na– at the Cl– sites, is 2 10 2 12 8 e (1.75) (4.80 10 ) 11.0 10 erg, R 3.66 10 − − − α × = = × × or 6.89 eV. Here R is taken as the value for metallic Na. The total cohesive energy of a Na+ Na– pair in the hypothetical crystal is 2.52 eV referred to two separated Na atoms, or 1.26 eV per atom. This is larger than the observed cohesive energy 1.13 eV of the metal. We have neglected the repulsive energy of the Na+ Na– structure, and this must be significant in reducing the cohesion of the hypothetical crystal. 5a. 2 n A q U(R) N ; 2 log 2 Madelung const. R R ⎛ ⎞ α = − α = = ⎜ ⎟ ⎝ ⎠ In equilibrium 2 n 0 2 n 1 2 0 0 U nA q n N 0 ; R R R R + ⎛ ⎞ ∂ α = − + = = ⎜ ⎟ ∂ α ⎝ ⎠ A , q and 2 0 0 N q 1 U(R ) (1 ). R n α = − − 3-1 b. ( ) ( ) 2 2 0 0 0 0 0 2 1 U U(R -R ) U R R R ..., 2 R ∂ δ = + δ + ∂ bearing in mind that in equilibrium R0 ( U R) 0. ∂ ∂ = 2 2 n 2 3 3 3 2 0 0 0 0 U n(n 1)A 2 q (n 1) q 2 N N R R R R R R 2 + ⎛ ⎞ ⎛ ⎛ ⎞ ∂ + α + α = − = − ⎜ ⎟ ⎜ ⎜ ⎟ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ 2 0 q ⎞ α ⎟ ⎠ For a unit length 2NR0 = 1, whence 0 2 2 2 2 2 0 4 2 2 2 0 0 R 0 R U q U (n 1)q log 2 (n 1) ; C R R R 2R R ⎛ ⎞ ∂ α ∂ − = − = = ⎜ ⎟ ∂ ∂ ⎝ ⎠ . 6. For KCl, λ = 0.34 × 10–8 ergs and ρ = 0.326 × 10–8Å. For the imagined modification of KCl with the ZnS structure, z = 4 and α = 1.638. Then from Eq. (23) with x ≡ R0/ρ we have 2 x 3 x e 8.53 10 . − − = × By trial and error we find or R x 9.2 � , 0 = 3.00 Å. The actual KCl structure has R0 (exp) = 3.15 Å . For the imagined structure the cohesive energy is 2 2 0 0 -αq p U U= 1- , or =-0.489 R R q ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ in units with R0 in Å. For the actual KCl structure, using the data of Table 7, we calculate 2 U 0.495, q = − units as above. This is about 0.1% lower than calculated for the cubic ZnS structure. It is noteworthy that the difference is so slight. 7. The Madelung energy of Ba+ O– is –αe2/R0 per ion pair, or –14.61 × 10–12 erg = –9.12 eV, as compared with –4(9.12) = –36.48 eV for Ba++ O--. To form Ba+ and O– from Ba and O requires 5.19 – 1.5 = 3.7 eV; to form Ba++ and O-- requires 5.19 + 9.96 – 1.5 + 9.0 = 22.65 eV. Thus at the specified value of R0 the binding of Ba+ O– is 5.42 eV and the binding of Ba++ O-- is 13.83 eV; the latter is indeed the stable form. 8. From (37) we have eXX = S11XX, because all other stress components are zero. By (51), 11 11 12 11 12 3S 2 (C C ) 1 (C C ). = − + + Thus 2 2 11 12 11 12 11 12 Y (C C C 2C ) (C C ); = + − + further, also from (37), eyy = S21Xx, whence yy 21 11 12 11 12 xx e e S S C (C C ) σ = = = − + . 9. For a longitudinal phonon with K || [111], u = v = w. 3-2 2 2 11 44 12 44 1 2 11 12 44 [C 2C 2(C C )]K 3, or v K [(C 2C 4C 3 )] ρ ω ρ = + + + = ω = + + This dispersion relation follows from (57a). 10. We take u = – w; v = 0. This displacement is ⊥ to the [111] direction. Shear waves are degenerate in this direction. Use (57a). 11. Let 12 xx yy e e = − = e in (43). Then 2 2 1 1 1 1 2 4 4 4 11 12 2 1 1 2 2 11 12 U C ( e e ) C e [ (C C )]e = + − = − 2 so that 2 2 n 2 3 3 3 2 0 0 0 0 U n(n 1)A 2 q (n 1) q 2 N N R R R R R R 2 + ⎛ ⎞ ⎛ ⎛ ⎞ ∂ + α + α = − = − ⎜ ⎟ ⎜ ⎜ ⎟ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ 2 0 q ⎞ α ⎟ ⎠ is the effective shear constant. 12a. We rewrite the element aij = p – δ ij(λ + p – q) as aij = p – λ′ δ ij, where λ′ = λ + p – q, and δ ij is the Kronecker delta function. With λ′ the matrix is in the “standard” form. The root λ′ = Rp gives λ = (R – 1)p + q, and the R – 1 roots λ′ = 0 give λ = q – p. b. Set i[(K 3) (x y z) t] 0 i[. . . . .] 0 i[. . . . .] 0 u (r, t) u e ; v(r,t) v e ; w(r,t) w e , + + −ω = = = as the displacements for waves in the [111] direction. On substitution in (57) we obtain the desired equation. Then, by (a), one root is 2 2 11 12 44 2p q K (C 2C 4C )/3, ω ρ = + = + + and the other two roots (shear waves) are 2 2 11 12 44 K (C C C )/3. ω ρ = − + 13. Set u(r,t) = u0ei(K·r – t) and similarly for v and w. Then (57a) becomes 2 2 2 2 0 11 y 44 y z 12 44 x y 0 x z 0 u [C K C (K K )]u (C C ) (K K v K K w ) ω ρ = + + + + + 0 and similarly for (57b), (57c). The elements of the determinantal equation are 3-3 2 2 2 2 11 11 x 44 y z 12 12 44 x y 13 12 44 x z M C K C (K K ) M (C C )K K ; M (C C )K K . ω ρ; = + + − = + = + and so on with appropriate permutations of the axes. The sum of the three roots of 2 ω ρ is equal to the sum of the diagonal elements of the matrix, which is (C11 + 2C44)K2, where 2 2 2 2 x y z 2 2 2 1 2 3 11 44 K K K K , whence v v v (C 2C ) , ρ = + + + + = + for the sum of the (velocities)2 of the 3 elastic modes in any direction of K. 14. The criterion for stability of a cubic crystal is that all the principal minors of the quadratic form be positive. The matrix is: C11 C12 C12 C12 C11 C12 C12 C12 C11 C44 C44 C44 The principal minors are the minors along the diagonal. The first three minors from the bottom are C44, C44 2, C44 3; thus one criterion of stability is C44 > 0. The next minor is C11 C44 3, or C11 > 0. Next: C44 3 (C11 2 – C12 2), whence |C12| < C11. Finally, (C11 + 2C12) (C11 – C12)2 > 0, so that C11 + 2C12 > 0 for stability. 3-4 CHAPTER 4 1a. The kinetic energy is the sum of the individual kinetic energies each of the form 2 S 1 Mu . 2 The force between atoms s and s+1 is –C(us – us+1); the potential energy associated with the stretching of this bond is 2 s 1 1 C(u u ) 2 s+ − , and we sum over all bonds to obtain the total potential energy. b. The time average of 2 2 2 S 1 1 Mu is M u . 2 4 ω In the potential energy we have s 1 u u cos[ t (s 1)Ka] u{cos( t sKa) cos Ka sin ( t sKa) sin Ka}. + = ω − + = ω − ⋅ + ω − ⋅ s s 1 Then u u u {cos( t sKa) (1 cos Ka) sin ( t sKa) sin Ka}. + − = ω − ⋅ − − ω − ⋅ We square and use the mean values over time: 2 2 1 cos sin ; cos sin 0. 2 < > = < > = < > = Thus the square of u{} above is 2 2 2 2 1 u [1 2cos Ka cos Ka sin Ka] u (1 cos Ka). 2 − + + = − The potential energy per bond is 2 1 Cu (1 cos Ka), 2 − and by the dispersion relation ω2 = (2C/M) (1 – cos Ka) 2 2 1 this is equal to M u . 4 ω Just as for a simple harmonic oscillator, the time average potential energy is equal to the time-average kinetic energy. 2. We expand in a Taylor series 2 2 2 2 s s u 1 u u(s p) u(s) pa p a ; x 2 x ⎛ ⎞ ∂ ∂ ⎛ ⎞ + = + + + ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ � On substitution in the equation of motion (16a) we have 2 2 2 2 p 2 2 p 0 u u M ( p a C ) t x > ∂ ∂ = Σ ∂ ∂ , which is of the form of the continuum elastic wave equation with 4-1 2 1 2 2 p p 0 v M p a C − > = Σ . 3. From Eq. (20) evaluated at K = π/a, the zone boundary, we have 2 1 2 2 M u 2Cu ; M v 2Cv . −ω = − −ω = − Thus the two lattices are decoupled from one another; each moves independently. At ω2 = 2C/M2 the motion is in the lattice described by the displacement v; at ω2 = 2C/M1 the u lattice moves. 2 0 2 0 0 0 p 0 p 0 sin pk a 2 4. A (1 cos pKa) ; M pa 2A sin pk a sin pKa K M 1 (cos (k K) pa cos (k K) pa) 2 > > ω = Σ − ∂ω = Σ ∂ − − + When K = k0, 2 0 p 0 A (1 cos 2k pa) , K M > ∂ω = Σ − ∂ which in general will diverge because p 1 . Σ → ∞ 5. By analogy with Eq. (18), 2 2 s 1 s s 2 s 1 s 2 2 s 1 s s 2 s 1 s 2 iKa 1 2 2 iKa 1 2 Md u dt C (v u ) C (v u ); Md v dt C (u v ) C (u v ), whence Mu C (v u) C (ve u); Mv C (u v) C (ue v) , and − + − = − + − = − + − −ω = − + − −ω = − + − 2 iK 1 2 1 2 iKa 2 1 2 1 2 (C C ) M (C C e ) 0 (C C e ) (C C ) M − + − ω − + a = − + + − ω 2 1 2 2 1 2 For Ka 0, 0 and 2(C C ) M. For Ka , 2C M and 2C M. = ω = + = π ω = 6. (a) The Coulomb force on an ion displaced a distance r from the center of a sphere of static or rigid conduction electron sea is – e2 n(r)/r2, where the number of electrons within a sphere of radius r is (3/4 πR3) (4πr3/3). Thus the force is –e2r/R2, and the 4-2 force constant is e2/R3. The oscillation frequency ωD is (force constant/mass)1/2, or (e2/MR3)1/2. (b) For sodium and thus 23 M 4 10 g − × � 8 R 2 10 cm; − × � 10 46 1 2 D (5 10 ) (3 10 ) − − ω × × � (c) The maximum phonon wavevector is of the order of 10 13 1 3 10 s− × � 8 cm–1. If we suppose that ω0 is associated with this maximum wavevector, the velocity defined by ω0/Kmax ≈ 3 × 105 cm s–1, generally a reasonable order of magnitude. 7. The result (a) is the force of a dipole ep up on a dipole e0 u0 at a distance pa. Eq. (16a) becomes 2 P 2 3 3 p>0 (2/ M)[ (1 cosKa) ( 1) (2e / p a )(1 cos pKa)] . ω = γ − + Σ − − At the zone boundary ω2 = 0 if P P 3 p>0 1 ( 1) [1 ( 1) ]p− + σ Σ − − − = 0 , or if . The summation is 2(1 + 3 p 3 [1 ( 1) ]p 1 − σ Σ − − = –3 + 5–3 + …) = 2.104 and this, by the properties of the zeta function, is also 7 ζ (3)/4. The sign of the square of the speed of sound in the limit Ka is given by the sign of 1 << p 3 2 p>0 1 2 ( 1) p p , − = σ Σ − which is zero when 1 – 2–1 + 3–1 – 4–1 + … = 1/2σ. The series is just that for log 2, whence the root is σ = 1/(2 log 2) = 0.7213. 4-3