Kittel Solid State Physics 3.4 Solution

Solutions provided here is combined from 6th and 8th edition of Charles Kittel, Introduction to Solid State Physics exercises, so it is better to take in account problem's name instead of number - in different editions a set of problems can be slightly differ for a same chapter. When it is convenient CGS system of units will be used instead of SI.

Contents 1.1 Tetrahedral bonds

1.2 Indices of planes

1.3 Hcp structure

2.1 Interplanar separation

2.2 Hexagonal space lattice

2.3 Volume of Brillouin zone

2.4 Width of diffraction maximum

2.5 Structure factor of the diamond

2.6 Form factor of atomic hydrogen

2.7 Diatomic line

3.1 Quantum solid

3.2 Cohesive energy of bcc and fcc neon

3.3 Solid molecular hydrogen

3.4 Possibility of ionic crystals R+ R−

3.5 Linear ionic crystal

3.6 Cubic ZnS structure

3.7 Divalent ionic crystals

3.8 Young's module and Poisson's ratio

3.9 Longitudinal wave velocity

3.10 Transverse wave velocity

3.11 Effective shear constant

3.12 Determinantal approach

3.13 General propagation direction

3.14 Stability criteria

4.1 Vibrations of square lattice

4.2 Monatomic linear lattice

4.3 Continuum wave equation

4.4 Basis of two unlike atoms

4.5 Kohn anomaly

4.6 Diatomic chain

4.7 Atomic vibrations in metal

4.8 Soft phonon modes*

5.1 Singularity in density of states

5.2 Rms thermal dilation of crystal cell

5.3 Zero point lattice displacement and strain

5.4 Heat capacity of layer lattice

5.5 Gr¨ uneisen constant*

6.1 Kinetic energy of electron gas

6.2 Pressure and bulk modulus of an electron gas

6.3 Chemical potential in two dimensions

6.4 Fermi gases in astrophysics

6.5 Liquid He3

6.6 Frequency dependence of the electrical conductivity

6.9 Static magnetoconductivity tensor

6.10 Maximum surface resistance

7.1 Square lattice, free electron energies

7.2 Free electron energies in reduced zone

7.3 Kronig-Penney model

Problem 1.1: Tetrahedral bonds Back to contents

Problem specification The angles between the tetrahedral bonds of diamond are the same as the angles between the body diagonals of a cube as in Fig. 12. Use elementary vector analysis to find the value of the angle.

Solution Elementary vectors for lattice of diamond type(FCC) is: a~1 = 12 a(−1; 1; 1) a~2 = 21 a(1; −1; 1) a~3 = 12 a(1; 1; −1) To find angle between vectors, it is necessary to refer to vector scalar product: a~1 ∗ a~2 = |a~1 | ∗ |a~2 | ∗ cosφ From here angle can be determined as: cosφ =

a~1 ∗ a~2 |a~1 | ∗ |a~2 |

Scalar product of two vectors is: a~1 ∗ a~2 =

a a a2 a2 (−1; 1; 1) ∗ (1; −1; 1) = (−1 − 1 + 1) = − 2 2 4 4

Product of vectors modules is: |a~1 | ∗ |a~2 | = (

ap 3a2 (3))2 = 2 4

Determining angle: 2

cosφ =

−a a~1 ∗ a~2 1 = 3a42 = − |a~1 | ∗ |a~2 | 3 4

1 φ = arccos(− ) = 109.27 3

Problem 1.2: Indices of planes Back to contents

Problem specification Consider the planes with indices (100) and (001); the lattice is FCC, and the indices refer to the conventinal cubic cell. What are the indices of these planes when referred to the primitive axes of Fig. 13?

Solution

To determine plane indices, it is necessary to prolongate axes to intersection with planes. Points of intersection are shown on figure. For plane (100) coordinates of intersection will be (2;0;2), which results in Miller index (101). For plane (001) intersection plane is (0;2;2), thus Miller index is (011)

Problem 1.3: Hcp structure Back to contents

Problem specification q Show that the c/a ratio for an ideal hexagonal close-packed structure is 83 = 1.633. If c/a significantly larger than this value, the crystal structure may be thought of as composed of planes of closely packed atoms, the planes being loosely stacked.

Solution

Figure shows hexagonal close packed structure. For first, lets drop a perpendicular from second layer of atoms into base level. Next geometrical step will be drawing intersection of medianas.

Finding h (height of equilateral triangle): r h=

1 a2 − ( a)2 = a 2

3 4

Using property of equilateral triangle: r 1 1 3 a d= h= ∗a = √ 3 3 4 2 3 5

Now determine cathetus b:

r b=

1 + ( a)2 = 2

1 2 1 a + a2 = 4 12

4a2 a =√ 12 3

Half-width of structure height is being calculated as: r r r 2 a c 2 2 = a2 − = a2 = a 2 3 3 3 Thus, c/a ratio is: r r c 2 8 =2 = a 3 3

Problem 2.1: Interplanar separation Back to contents

Problem specification Consider a plane hkl in a crystal lattice. a) Prove that the reciprocal lattice vector G = hb1 + kb2 + lb3 is perpendicular to this plane. 2π b) Prove that the distance between two adjacent planes of the lattice is d(hkl) = |G| c) Show for a simple cubic lattice that d2 =

a2 h2 +k2 +l2

Solution

a) Image shows arbitrary (hkl) plane intersects axes at ( h1 ; k1 ; 1l ) Vector G has coordinates (h; k; l). To prove this part, it is necessary to prove that vectors, which are constructed plane (hkl) are perpendicular to G. It it necessary to recall scalar product between G and this vectors. Taking as an example one pair of vectors, which are determine (hkl) plane: x~1 = ( h1 ; 0; − 1l ) x~2 = (0; − h1 ; k1 ) Using scalar product property: ~ · x~1 = (h; k; l) · ( 1 ; 0; − 1 ) = 1 + 0 − 1 = 0 G h l Same for the second vector: ~ · x~2 = (h; k; l) · (0; − 1 ; 1 ) = 0 − 1 + 1 = 0 G h k ~ is perpendicular to plane (hkl). As dot product between these vectors is 0, then vector G

b) Lets take a normal vector to plane (hkl). Shortest distance to the plane (hkl) will be: d=

1 a~1 · ~n h

Normal vector can be determined as:

~ G |G|

~n = Finding distance: d=

~ 1 2π G 2πh = ~a · = ~ ~ h |G| h|G| |G|

c) Direct lattice vectors for simple cubic lattice are: a~1 = a(1; 0; 0) a~2 = a(0; 1; 0) a~3 = a(0; 0; 1) Reciprocal lattice vectors easy to find: b~1 =

2π a (1; 0; 0)

b~2 =

2π a (0; 1; 0)

b~3 =

2π a (0; 0; 1)

~ is: Vector G ~ = hb~1 + k b~2 + lb~3 G Writing condition for distance d: d=

2π ~ |G|

2π 2 h2 ( 2π a )

2 k 2 ( 2π a )

2 l2 ( 2π a )

2π a =q =√ 2 + k 2 + l2 2π 2 2 h 2 2 ( a ) (h + k + l )

Finally: d2 =

a2 h2 + k 2 + l2

Problem 2.2: Hexagonal space lattice Back to contents

Problem specification The primitive translation vectors of the hexagonal space lattice may be taken as: √

a~1 = (a

3 x 2 )~ √

a~2 = −(a

+ ( a2 )~y

3 x 2 )~

+ ( a2 )~y

a~3 = c~z √

a) Show that the volume of the primitive cell is ( 23 )a2 c. b) Show that the primitive translations of the reciprocal lattice are: √ )~ x + ( 2π b~1 = ( a2π y a )~ 3 √ )~ x + ( 2π y b~2 = −( a2π a )~ 3

z b~3 = ( 2π c )~ so the lattice is its own reciprocal, but with a rotation of axes. c) Describe and sketch the first Brillouin zone of the hexagonal space lattice.

Solution a) Volume of the primitive cell is calculated as: √ a 3 a 0 2 2 √ √ √ √ √ 2 2 2 √ a 3 a a 3 a c 3 a c 3 a c 3 V = |a~1 · a~2 × a~3 | = − a 3 a 0 = ∗ ∗c+( ∗a ∗ c) = + = 2 2 2 2 2 2 4 4 2 0 0 c b) Reciprocal lattice vectors can be found from primitive translation vectors as: a~3 b~1 = 2π a~1a~·2a~×2 × a~3 a~1 b~2 = 2π a~1a~·3a~×2 × a~3 a~2 b~3 = 2π a~1a~·1a~×2 × a~3

Applying formula for first vector b~1 :

b~1 = 2π

a~2 × a~3 = a~1 · a~2 × a~3

2π

√ a2 c 3 2

x ˆ √ ∗ − a 3 2 0

yˆ a 2

zˆ √ 4π ac ac 3 2π 2π √ = x ˆ + yˆ) = √ x yˆ ( ˆ+ 0 2c 3 2 2 a a a 3 c 9

For second vector:

b~2 = 2π

a~3 × a~1 = a~1 · a~2 × a~3

2π

√ a2 c 3 2

x ˆ ∗ 0 √ a 3 2

yˆ 0 a 2

zˆ √ 4π ac ac 3 2π 2π √ = (− x ˆ + yˆ) = − √ x ˆ+ yˆ c 2c 3 2 2 a a a 3 0

Third vector:

b~3 = 2π

a~1 × a~2 = a~1 · a~2 × a~3

2π

√ a2 c 3 2

x ˆ √ ∗ a 3 2 √ − a 3 2

yˆ a 2 a 2

zˆ √ √ √ 4π a a 3 a a 3 4π a2 3 2π √ ( ˆ + 0 ∗ yˆ + ( ∗ + ∗ )ˆ z) = )ˆ z= zˆ 0 = 2 √ (0 ∗ x 2 2 2 2 2 2 c a c 3 a c 3 0

Figure shows redrawn hexagonal close packed structure in reciprocal space in 2D. Filled by red color area is first Brillouin zone.

Problem 2.3: Volume of Brillouin zone Back to contents

Problem specification 3

Show that the volume of the first Brillouin zone is (2π) Vc , where Vc is the volume of a crystal primitive cell. Hint: The volume of a Brillouin zone is equal to the volume of the primitive parallelepiped in Fourier space. Recall the vector identity (~c × ~a) × (~a × ~b) = (~c · ~a × ~b)~a

Solution Volume of the first Brillouin zone is: V = |b~1 · b~2 × b~3 | Volume of the direct lattice: Vc = |a~1 · a~2 × a~3 | Reciprocal lattice vectors relation to primitive translation vectors: a~3 b~1 = 2π a~1a~·2a~×2 × a~3 a~1 b~2 = 2π a~1a~·3a~×2 × a~3 a~2 b~3 = 2π a~1a~·1a~×2 × a~3

Then volume can be found: (a~2 × a~3 ) · (a~3 × a~1 ) × (a~1 × a~2 ) V = |b~1 · b~2 × b~3 | = (2π)3 |a~1 · a~2 × a~3 |3 Applying proposed identity: (a~3 × a~1 ) × (a~1 × a~2 ) = (a~3 · a~1 × a~2 )a~1 Then volume can be calculated as: V = (2π)3

(a~2 × a~3 ) · (a~3 · a~1 × a~2 )a~1 (a~2 × a~3 ) · (a~1 · a~2 × a~3 )a~1 = (2π)3 3 |a~1 · a~2 × a~3 | |a~1 · a~2 × a~3 |3

It is possible to eliminate in numerator and denominator the same values: V = (2π)3

(a~2 × a~3 ) · a~1 (2π)3 1 3 = = (2π) |a~1 · a~2 × a~3 |2 |a~1 · a~2 × a~3 | Vc

Problem 2.4: Width of diffraction maximum Back to contents

Problem specification We suppose that in a linear crystal there are identical point scattering centers at every lattice Ppoint ρm = ma, where m is an integer. By analogy with (20) the total radiation amplitude will be proportional to F = exp(−ima · ∆k). The sum over M lattice points is

F =

1 − exp(−iM (a · ∆k)) 1 − exp(−i(a · ∆k))

by the use of the series

PM −1 m=0

xm =

1 − xM 1−x

a) The scattered intensity is proportional to |F |2 . Show that 1 sin2 M (~a · ∆k) 2 |F |2 ≡ F ∗ F = 1 sin2 (~a · ∆k) 2 b) We know that a diffraction maximum appears when ~a · ∆k = 2πh, where h is an integer. We change ∆k slightly and define in ~a · ∆k = 2πh + such that gives the position of the first zero in sin2 21 M (~a · ∆k). Show that = 2π M , so 1 that the width of the diffraction maximum is proportional to M and can be extremely narrow for macroscopic values of M. The same result holds true for a three-dimensional crystal.

Solution a) To find |F |2 it is necessary to multiply initial F by a conjugate value: 1 − e−iM (a·∆k) 1 − eiM (a·∆k) · = 1 − e−i(a·∆k) 1 − ei(a·∆k) 1 − eiM (~a·∆k) − e−iM (~a·∆k) + e−iM (~a·∆k) · eiM (~a·∆k) = = 1 − ei(~a·∆k) − e−i(~a·∆k) + e−i(~a·∆k) · ei(~a·∆k) 2 − eiM (~a·∆k) − e−iM (~a·∆k) 1 − eiM (~a·∆k) − e−iM (~a·∆k) + 1 = = 1 − ei(~a·∆k) − e−i(~a·∆k) + 1 2 − ei(~a·∆k) − e−i(~a·∆k) |F |2 ≡ F ∗ F =

Using trigonometric equation: eix + e−ix 2 Taking out −2 out of numerator and denominator in order to use described above relation: cos(x) =

iM (~ a·∆k)

|F |2 =

−iM (~ a·∆k)

+e 2 − 2( e ) 2 − eiM (~a·∆k) − e−iM (~a·∆k) 2 − 2cos(M (~a · ∆k)) 1 − cos(M (~a · ∆k)) 2 = = = i(~ a ·∆k) −i(~ a ·∆k) i(~ a ·∆k) −i(~ a ·∆k) 2 − 2cos(~a · ∆k) 1 − cos(~a · ∆k) 2−e −e e +e 2 − 2( ) 2

This trigonometric identities can be useful: 12

1 − 2cos(2x) 2 x 1 − cos(x) sin2 ( ) = 2 2

sin2 (x) =

Finally, for |F |2 : 1 sin2 ( M (~a · ∆k)) 1 − cos(M (~a · ∆k)) 2 = |F | = 1 1 − cos(~a · ∆k) sin2 ( (~a · ∆k)) 2 2

b) Writing initial conditions: 1 1 sin( M (~a · ∆k)) = sin( M (2πh + )) 2 2 Further using trigonometric identity: sin(α + β) = sin α · cos β + cos β · sin α Applying identity: 1 1 1 1 M M 1 ) + cos(πhM ) · sin( ) sin( M (2πh + )) = sin( M · 2πh) · cos( M ) + cos( M · 2πh) · sin( M ) = sin(πhM ) · cos( 2 2 2 2 2 2 2 Equating expression to 0: M M ) + cos(πhM ) · sin( )=0 2 2 From here it is clear that first summand is zero, as sin(πhM ) = 0, so it is necessary to investigate second summand: sin(πhM ) · cos(

M )=0 2 M M sin( ) = 0; ⇒ = πn 2 2 2πn = M cos(πhM ) · sin(

For the given conditions, n = 1: =

2π M

Problem 2.5: Structure factor of the diamond Back to contents

Problem specification The crystal structure of diamond is described in Chapter 1. The basis consists of eight atoms if the cell is taken as the conventional cube. a) Find the structure factor S of this basis. b) Find the zeros of S and show that the allowed reflections of the diamond structure satisfy v1 + v2 + v3 = 4n, where all indices are even and n is any integer, or else all indices are odd (Fig. 19). (Notice that h,k,l may be written for v1 , v2 , v3 and this is often done.)

Solution Diamond has a face-centered cubic lattice with basis, consisting of: r~1 = a(0; 0; 0); r~2 = a( 41 ; 14 ; 14 );

a) Structure factor of diamond lattice is: ~ ~ ~ Sbasis = f e−i·G·r~1 + e−i·G·r~2 = f 1 + e−i·G·r~2 ~ and r~2 : Keep in mind that expressions for G ~ = v1 b~1 + v2 b~2 + v3 b~3 ; G r~2 =

1 1 1 a~1 + a~2 + a~3 ; 4 4 4

Then, for second summand of structure factor: ~

e−i·G·r~2 = e−i·2π( 4 v1 + 4 v2 + 4 v3 ) = e−i· 2 (v1 +v2 +v3 ) Structure factor of basis is:

π Sbasis = f 1 + e−i· 2 (v1 +v2 +v3 )

b) Now it is possible to analyze obtained structure factor for diamond. From previous expression it is possible to determine next:     0, v1 + v2 + v3 = 2n     Sbasis = (1 ± i)f, v1 + v2 + v3 = 2n + 1        2f, v1 + v2 + v3 = 4n

Problem 2.6: Form factor of atomic hydrogen Back to contents

Problem specification −2r

e a0 πa30

For the hydrogen atom in its ground state, the number density is n(r) = Show that the form factor is fG =

16 (4+G2 a20 )2

, where a0 is the Bohr radius.

Solution Form factor is determined as: Z ∞ Z ∞ Z ∞ Z ∞ −2r −2r −2r e a0 2 sin Gr 4π 4 sin Gr a0 e e a0 r sin Grdr fj = 4π n(r)r2 dr = 4π r dr = r sin Grdr = 3 3 3 Gr πa0 Gr πa0 G 0 a0 G 0 0 0 Now it is possible to introduce sin function by exponentials: sin Gr =

eiGr − e−iGr 2i

Replacing sin function by above formula: Z ∞ Z ∞ Z ∞ −2r eiGr − e−iGr −2r −2r −2r 2 2 4 iGr −iGr a0 a0 e rdr = 3 e (e a0 +iGr − e a0 −iGr )rdr (e −e )rdr = 3 fj = 3 a0 G 0 2i ia0 G 0 ia0 G 0 General solution for integral is known(for a > 0, n = 0, 1, 2...): Z ∞ xn e−ax dx = 0

For example, one part will be look as follows: Z ∞ Z 2r e− a0 +iGr rdr = 0

∞

n! +1

Second integral will be looking as follows: Z ∞ Z − a2r −iGr rdr = e 0 0

∞

( a20

1! − iG)2

( a20

1! + iG)2

e−r( a0 −iG) rdr =

e−r( a0 +iG) rdr =

Applying to initial integral:  fj =

2 ia30 G

∞

−2r a0 +iGr

−e

−2r a0 −iGr

)rdr =



2   ia30 G

Trying to make a ratios difference under a common divisor:   2 2  2 2 4 + iG − − iG a0 2  a0 2  a20 +  fj = 3  = 3  2 ia0 G ia0 G 4 2 + G 2 a

1 2 a0

− iG

4iG a0

2 −

fj =

2   ia30 G



8iG a0

16   a40



To obtain final equation, multiplying ratio for a40 :  fj =



1  16   2  = 4  2  a 4 0 + G2 + G2 a2

4 a20

 1

4 a20

+ G2

 2  =

2 a0

− G2 − a42 − 0 2 4 2 + G 2 a

Simplifying ratio: 

16 2

(4 + G2 a20 )

 2  + iG

4iG a0

 − G2  

Problem 2.7: Diatomic line Back to contents

Problem specification Consider a line of atoms ABAB...AB, with an A-B bond length of 21 a. The form factors are fA , fB , respectively. The incident beam of x-rays is perpendicular to the line of atoms. a) Show that the interference condition is nλ = a cos θ, where θ is the angle between the diffracted beam and the line of atoms. b) Show that the intensity of the diffracted beam is proportional to |fA − fB |2 for n odd, and to |fA + fB |2 for n even. c) Explain what happens if fA = fB .

Solution

a) From geometrical interrelation: ~ = |K~f | · sin α = |K~f | · sin(90◦ − θ) = |K~f |(sin 90◦ cos θ − cos 90◦ sin θ) = |K~f | · cos θ G In same time, from property of reciprocal lattice for 1D: G = nb = n ·

2π a

Now it is possible to combine two results: n·

2π = |K~f | · cos θ a

Considering conservation energy law: ~ i | = |K~f | = k = |K

2π λ

Finally: n·

2π 2π = · cos θ a λ nλ = a · cos θ

b) Intensity of the diffracted beam is: I ∝ |SG |2 Structure factor of diatomic chain: 1

SG = fA · e−i2π·0 + fB · e−i2π· 2 n = fA + fB · e−iπn

Structure factor is: SG =

    fA − fB ,

n is odd

   fA + fB ,

n is even

According to relation with structure factor intensity depends as:     |fA − fB |2 , n is odd I∝    |fA + fB |2 , n is even c) Given condition fA = fB means that these atoms are identical, as they have same form factors. For odd numbers of n, resulting intensity will be equal to 0, as fA − fB = 0. In case of even numbers, resulting intensity will be equal to 4f .

Problem 3.1: Quantum solid Back to contents

Problem specification In a quantum solid the dominant repulsive energy is the zero-point energy of the atoms. Consider a crude one-dimensional model of crystalline He4 with each He atom confined to a line segment of length L. In the ground state the wave function within each segment is taken as a half wavelength of a free particle. Find the zero-point kinetic energy per particle.

Solution Kinetic energy for system will be described as follows: E=

~2 k 2 2m

Wave vector ~k is equal: k=

2π λ

By initial conditions it is clear that: λ = 2L Kinetic energy will be equal then: E=

~2 k 2 ~2 = 2m 2m

2π 2L

2 =

~2 π 2 2mL2

Problem 3.2: Cohesive energy of bcc and fcc neon Back to contents

Problem specification Using the Lennard-Jones potential, calculate the ratio of the cohesive energies of neon in the bcc and fcc structures (Ans. 0.956). The lattice sums for the bcc structures are: X = p−12 = 9.11418; ij j

= p−6 ij = 12.2533;

Solution For FCC structure: X

= p−12 = 12.13188; ij

= p−6 ij = 14.45392;

Total potential energy of the crystal is:   X σ 12 X σ 6 1  − U = · 4N  2 p R p R ij ij j j General idea of solution is to calculate potential energies of BCC and FCC structures at equilibrium position, and then find their ratio. First, lets calculate for BCC:   X σ 12 X σ 6  U = 2N  − pij R pij R j j Finding values at equilibrium position, for this it is necessary to take a derivative of energy by R and equate it to 0: dU =0 dR dU σ6 σ 12 = −2N 12 · 9.11418 13 − 6 · 12.2533 7 = 0 dR R R 12 · 9.11418

σ6 σ 12 − 6 · 12.2533 =0 R13 R7

109.3706

σ 12 σ6 = 73.5198 7 13 R R

σ6 = 0.672 R6 Calculating same ratio for FCC: 12 · 12.13188

σ6 σ 12 − 6 · 14.45392 =0 R13 R7

145.58256

σ 12 σ6 = 86.72352 7 13 R R

σ6 = 0.59569 R6 Now it is possible to calculate ratio P

j pij R Ubcc bcc =P 12 Uf cc σ j

pij R

f cc

−

Ubcc Uf cc : σ pij R σ pij R

6 bcc

9.11418 · (0.672)2 − 12.2533 · 0.672 4.1158 − 8.2342 = 12.13188 · (0.59569)2 − 14.45392 · 0.59569 4.3049 − 8.61

f cc

Finally, ratio of cohesive energies for neon is: Ubcc 4.1184 = = 0.9566 Uf cc 4.3051

Problem 3.3: Solid molecular hydrogen Back to contents

Problem specification ˚. For H2 one finds from measurements on the gas that the Lennard-Jones parameters are = 50 × 10−16 erg and σ = 2.96A Find the cohesive energy in kJ per mole of H2 ; do the calculation for an fcc structure. Treat each H2 molecule as a sphere. kJ , much less that we calculated, so that quantum corrections must The observed value of the cohesive energy is 0.751 mol be very important.

Solution For an FCC structure it is known that: X

= p−12 = 12.13188; ij

= p−6 ij = 14.45392;

General algorithm to solve this problem is next: 1. Find a minimum of the cohesive energy with values corresponding to FCC lattice. σ 6 ) . 2. Find ratio ( R

3. Insert found ratio to the expression of cohesive energy U(R) and finally evaluate it. Lets start from first part: dU =0 dR dU σ 12 σ6 = −2N 12 · 12.13188 13 − 6 · 14.45392 7 = 0 dR R R 12 · 12.13188

σ 12 σ6 − 6 · 14.45392 7 = 0 13 R R

145.58256

σ 12 σ6 = 86.72352 R13 R7

σ 6 The ( R ) ratio is finally:

σ6 = 0.59569 R6 Inserting found values into expression for U(R). (Take into account that value of was converted to Joules units):   X σ 12 X σ 6  = 2·6.022×1023 ·50×10−16 ·10−7 12.13188 · (0.59569)2 − 14.45392 · 0.59569 U (R) = 2N  − p R p R ij ij j j Finally: U (R) = 602.2 · [4.3049 − 8.61] = 602.2 × (−4.3051) = −2592.53

J mol

= −2.59253

kJ mol

Problem 3.4: Possibility of ionic crystals R+ R− Back to contents

Problem specification Imagine a crystal that exploits for binding the coulomb attraction of the positive and negative ions of the same atom or molecule R. This is believed to occur with certain organic molecules, but it is not found when R is a single atom. Use the data in Tables 5 and 6 to evaluate the stability of such a form of Na in the NaCl structure relative to normal metallic sodium. Evaluate the energy at the observed interatomic distance in metallic sodium, and use 0.78 eV as the electron affinity of Na.

Solution From tables it is possible to get next parameters: • One electron ionization energy, ENa atom ionization = 5.14 eV. • Cohesive energy of a single Na atom is 1.113 eV. • Nearest-neighbor distance for metallic Na is 3.66 ˚ A. • Electron affinity energy, ENa electron affinity = 0.78 eV. • Madelung constant α for NaCl structure is 1.747565. Expression for potential energy of ionic crystal consisting of 2N ions is (CGS): ρ N αq 2 1− U =− R0 R0 As ρ is significantly less than R0 , it is possible to neglect second term in parenthesis and obtain for our case(N = 1): UN a+ N a − = −

αq 2 40.314 · 10−20 40.314 · 10−20 1.747565 · (4.803 × 10−10 )2 =− =− = −11.014 · 10−12 erg =− −8 −8 R0 3.66 × 10 3.66 × 10 3.66 × 10−8

Note that R0 here is taken for metallic Na, as it required from problem specification. It is more comfortable to convert energy to electron-volts, to compare energies more easily: UN a+ N a− = −11.014 · 10−12 erg = −11.014 × 10−12 · 6.2415 × 1011 eV = −68.743 × 10−1 eV = −6.87 eV Cohesive energy of two ions N a+ N a− can be found as: Ecohesive = UN a+ N a− − ENa atom ionization + ENa electron affinity = 6.87 − 5.14 + 0.78 = 2.51 eV For a single ion, cohesive energy will be 1.255 eV, which is larger than value for metallic Na 1.113 eV.

Problem 3.5: Linear ionic crystal Back to contents

Problem specification Consider a line of 2N ions of alternating charge ±q with a repulsive potential energy a) Show that at the equilibrium separation(CGS): 2N q 2 ln 2 1 U (R0 ) = − 1− R0 n

A Rn

between nearest neighbors.

b) Let the crystal be compressed so that R0 → R0 (1 − δ). Show that the work done in compressing a unsit length of the crystal has the leading term 21 Cδ 2 , where(CGS): C=

(n − 1)q 2 ln 2 R0

q . Note: we should not expect to obtain this result from the expression for To obtain the results in SI, replace q 2 by 4π 0 U (R0 ), but we must use the complete expression for U(R).

Solution a) For one-dimensional case, the Madelung constant α is equal to 2 ln 2. General expression for the potential energy is: αq 2 −R ρ U (R) = N zλe − R In this equation first part represents repulsive potential and second part is responsible for the attractive potential. In our problem specification we need to use expression RAn as a repulsive potential. So, expression for the potential energy will be look like: A αq 2 U (R) = N − Rn R As it is necessary to obtain expression at equilibrium separation, it is necessary to apply it's condition: dU =0 dR nA αq 2 N − n+1 + 2 = 0 R0 R0 αq 2 nA = n+1 R02 R0 Multiplying both sides for R0 :

αq 2 nA = n R0 R0

From here: R0n =

nAR0 αq 2

Inserting this expression for initial expression for U(R): " # 2 A αq 2 A αq 2 Aαq 2 αq 2 αq αq 2 N αq 2 1 U (R0 ) = N − = N nAR0 − =N − =N − = −1 R0n R0 R0 nAR0 R0 nR0 R0 R0 n αq 2 23

Finally inserting value of Madelung constant into last expression: 2N q 2 ln 2 1 2N q 2 ln 2 1 U (R0 ) = −1 =− 1− R0 n R0 n b) Idea to prove statement for this part is to use expression for U(R) as was mentioned in note to problem, and expand expression near U (R0 → R0 (1 − δ)) in form: 0 1 00 U (R)R0 →R0 (1−δ) = U (R0 (1 − δ)) + U (R0 (1 − δ)) · R0 (1 − δ) + U · (R0 (1 − δ))2 + ... 2

Taking into account that fact

dU dR

= 0 at equilibrium, so calculating second derivative: n(n + 1)A 2αq 2 d2 U =N − dR2 Rn+2 R03

From part a) it is known: R0n =

nAR0 αq 2

Inserting into our expression to continue: " # d2 U n(n + 1)A 2αq 2 (n + 1)αq 2 2αq 2 N αq 2 = N = N − = (n − 1) − nAR0 2 dR2 R03 R03 R03 R03 αq 2 R0 From above expression of Taylor series it is clear that: C=

1 1 (n − 1)q 2 ln 2 d2 U 2N ln 2q 2 (n − 1) · R02 2 = · R02 = 3 2N dR 2N R0 R0

Problem 3.6: Cubic ZnS structure Back to contents

Problem specification Using λ and ρ from Table 7 and the Madelung constants given in the text, calculate the cohesive energy of KCl in the cubic ZnS structure described in Chapter 1. Compare with the value calculated for KCl in the NaCl structure.

Solution For first, finding an energy at state of the equilibrium separation: U (R)

zλ ρ

· e−

R0 ρ

−

dU dR i 2

αq R02

zλ − ρ e

R0 ρ

zλe−

R0 ρ

R02 zλe−

R0 ρ

R02 e−

R0 ρ

h R0 N zλe− ρ −

αq 2 R02

ραq 2 R02

αq 2 R

= ραq 2 =

ραq 2 zλ

General idea to solve equation is to divide left part on ρ2 : αq 2 R02 − Rρ0 e = 2 ρ zλρ In this way it is possible to obtain: x2 e−x =

αq 2 zλρ

Values of constants are(taken from table data): αZnS

1.6381

0.326˚ A

zλNaCl structure

2.05 · 10−8 erg

As KCl in NaCl structure has z equal to 6, in ZnS structure there is 4 neighboring ions, then zλ for these corrections will be equal to: 2.05 · 10−8 erg · 4 zλZnS structure = = 1.367 · 10−8 erg 6 Returning to energy expression: x2 e−x =

αq 2 1.6381 · (4.803 · 10−10 )2 37.789 · 10−20 = = = 0.0087474 zλρ 1.367 · 10−8 · 0.326 · 10−8 0.432 · 10−16

This transcendent equation is need to be solved by using numerical methods. Simple program using Python language is listed below: 25

import math a s m e p s i l o n = 1 e −10 left = 0 r i g h t = 20 steps = 0

# # # #

Define accuracy L e f t boundary o f x R i g h t boundary o f x I t e r a t i o n counter

def f ( x ) : return m.pow( x , 2 ) ∗ m. exp(−x ) − 0 . 0 0 8 7 4 7 4 # D e f i n e f u n c t i o n h e r e while ( r i g h t − l e f t ) > f l o a t ( e p s i l o n ) : center = ( l e f t + right ) / float (2) val = f ( center ) ∗ f ( right ) i f ( val < 0): l e f t = float ( center ) else : right = float ( center ) steps = steps + 1 print " R e s u l t = " , c e n t e r , "by" , s t e p s , " i t e r a t i o n s " By result of these calculations we have: x = 9.17111692812 From our notation it is clear that: x= From here R0 can be evaluated:

R0 = 9.17111692812 ρ

R0 = 9.17111692812 · 0.326 · 10−8 cm = 2, 989˚ A

At equilibrium separation it is possible to calculate energy between pair of ions: N αq 2 0.326 · 10−8 ρ 1.6381 · (4.803 · 10−10 )2 Utot = − 1 − 1− =− R0 R0 2.989 · 10−8 2.989 · 10−8 Finally for KCl in Zns structure: Utot = −

37.789 · 10−20 · 0, 89 = 11.252 · 10−12 erg 2.989 · 10−8

Now calculte for KCl in NaCl structure: Utot = − −

1.747565 · (4.803 · 10−10 )2 3.147 · 10−8

0.326 · 10−8 1− = 11.483 · 10−12 erg 3.147 · 10−8

From this results it is seems that cohesive energies at equilibrium separation for KCl at different lattice configurations NaCl and ZnS are very similar.

Problem 3.7: Divalent ionic crystals Back to contents

Problem specification Barium oxide has the NaCl structure. Estimate the cohesive energies per molecule of the hypothetical crystals Ba+ O− and Ba++ O−− referred to separated neutral atoms. The observed nearest-neighbor internuclear distance is R0 = 2.76˚ A; the first and second ionization potentials of Ba are 5.19 and 9.96 eV; and the electron affinities of the first and second electrons added to the neutral oxygen atom are 1.5 and -9.0 eV. The first electron affinity of the neutral oxygen atom is the energy released in the reaction O + e → O− . The second electron affinity is the energy released in the reaction O− + e → O−− . Which valence state do you predict will occur? Assume R0 is the same for both forms, and neglect the repulsive energy.

Solution Cohesive energy at equilibrium separation can be determined by expression: ρ N αq 2 1− Utot = − R0 R0 Neglecting repulsive energy means that term

ρ can be removed thus giving: R0 Utot = −

N αq 2 R0

Next it is necessary to calculate cohesive energies for barium oxide of two valence states and compare cohesive energies. Value of N is equal to one pair of ions, thus N = 1. 1) Ba+ O− Utot = −

40.314 · 10−20 N αq 2 1.747565 · (4.803 · 10−10 )2 =− = −14.607 · 10−12 erg =− −8 R0 2.76 · 10 2.76 · 10−8

More useful to convert into electron-volts and take a module of energy: Utot = 14.607 · 10−12 erg · 6.2415 · 1011 = 91.167 · 10−1 eV = 9.116 eV Calculating energy to form specified ionic crystal from neutral atoms of barium and oxygen: Ecohesive = 9.116 eV − 5.19 eV + 1.5 eV = 5.426 eV Doing the same with valence = 2: 2) Ba++ O−− Utot = −

N αq 2 1.747565 · (2 · 4.803 · 10−10 )2 =− = −58.428 · 10−12 erg = −36.467 eV R0 2.76 · 10−8

Final energy expression is: Ecohesive = 36.467 eV − 5.19 eV − 9.96 eV + 1.5 eV − 9.0 eV = 13.817 eV From these calculations it is obvious that divalent barium oxide has higher cohesive energy, thus ionic crystal of Ba++ O−− is more stable than single-valence BaO.

Problem 3.8: Young's module and Poisson's ratio Back to contents

Problem specification A cubic crystal is subject to tension in the [100] direction. Find expressions in terms of the elastic stiffnesses for Young's modulus and Poisson's ratio as defined in Fig. 21.

Solution From problem specification we know that object is subject to tension in direction [100], which means that from 6 components of the stress we will have only one non-zero component linked with X direction: Xx 6= 0 As Yy , Zz , Yz , Zx , Xy are equal to 0 (no stress in these directions), then it is clear from (37) and (38) that present strain is: exx = S11 Xx Young's modulus is defined as a stress/strain ratio. Using definition and previous expression to obtain: E=

Xx Xx 1 = = exx S11 Xx S11

Now to determine Young's modulus it is necessary to obtain expression for S11 . It is possible, if we use (51): C11 − C12

C11 + 2C12

1 S11 − S12 1 S11 + 2S12

Then we can obtain from these relations: 1 , ⇒ C11 − C12 1 = , ⇒ C11 + 2C12

S11 − S12 = S11 + 2S12

1 C11 − C12 1 S11 = − 2(C11 + 2C12 ) 2

S12 = S11 − S12

Equating two expressions for S12 to determine S11 : S11 −

1 1 S11 = − C11 − C12 2(C11 + 2C12 ) 2 28

Gathering S11 in one part of expression: S11 +

S11 1 1 = + 2 C11 − C12 2(C11 + 2C12 )

Processing: 3S11 2

3S11

S11

1 1 + C11 − C12 2(C11 + 2C12 ) 2 1 + C11 − C12 C11 + 2C12 2 (C11 + 2C12 ) + (C11 − C12 ) (C11 − C12 ) (C11 + 2C12 ) 3C11 + 3C12 (C11 − C12 ) (C11 + 2C12 ) C11 + C12 (C11 − C12 ) (C11 + 2C12 ) C11 + C12 2 +C C 2 C11 11 12 − 2C12

Now it is easy to find Young's modulus: E= By definition, Poisson's ratio is equal to

2 1 C 2 + C11 C12 − 2C12 Xx = = 11 exx S11 C11 + C12

strain in Y direction : strain in X direction ν=

eyy S21 Xx S21 = = exx S11 Xx S11

Instead of S21 it is possible to substitute S12 (S21 = S12 by symmetry) for a cubic crystal: ν=

S12 S11

Determining S12 : S12 S12 S12 S12

1 C11 − C12 1 C11 + C12 − = (C11 − C12 ) (C11 + 2C12 ) C11 − C12 C11 + C12 − (C11 + 2C12 ) = (C11 − C12 ) (C11 + 2C12 ) C12 = − (C11 − C12 ) (C11 + 2C12 ) = S11 −

Returning back to the Poisson's ratio: eyy S12 ν= = = exx S11

C12 C12 (C11 − C12 ) (C11 + 2C12 ) =− C11 + C12 C11 + C12 (C11 − C12 ) (C11 + 2C12 )

−

Problem 3.9: Longitudinal wave velocity Back to contents

Problem specification v u1 u (C11 + 2C12 + 4C44 ) t3 Show that the velocity of a longitudinal wave in the [111] direction of a cubic crystal is given by vx = ρ iK(x + y + z) √ 3 Hint: For such a wave u = v = w. Let u = u0 e e−iωt , and use Eq. (57a).

Solution Using hint, let consider wave propagating as: iK(x + y + z) √ 3 u = u0 e e−iωt Equation (57a) is: ρ

∂2u ∂2u = C11 2 + C44 2 ∂t ∂x

∂2u ∂2u + 2 ∂y 2 ∂z

+ (C12 + C44 )

∂2v ∂2w + ∂x∂y ∂x∂z

Calculating derivatives assuming that u = v = w (as exponent in differential is exponent, same exponential part will be canceled): K2 K2 K2 K2 K2 − − −ω 2 ρ = C11 − + C44 − + (C12 + C44 ) − 3 3 3 3 3 2 2 2 2K K 2K −ω 2 ρ = C11 − + C44 − + (C12 + C44 ) − 3 3 3 2 2 2 2 K 2K 2K 2K ω 2 ρ = C11 + C44 + C12 + C44 3 3 3 3 3ω 2 ρ = ω2

K 2 C11 + 2K 2 C12 + 4K 2 C44 K 2 (C11 + 2C12 + 4C44 ) 3ρ s K 2 (C11 + 2C12 + 4C44 ) 3ρ

Velocity of wave propagation: s ω v= = K

K 2 (C11 + 2C12 + 4C44 ) 3ρ K

s =

C11 + 2C12 + 4C44 3ρ

Problem 3.10: Transverse wave velocity Back to contents

Problem specification v u1 u (C11 − C12 + C44 ) t3 Show that the velocity of transverse waves in the [111] direction of a cubic crystal is given by vx = ρ Hint: See Problem 9.

Solution For transversal wave it is essential that one wave function should be equal to zero and other two are opposite. Then solution can go by few ways: 1)v

= −w

u =

0 and using expression (57b)

2)w v

= −u =

0 and using expression (57c)

Expressions for these waves are: iK(x + y + z) √ 3 w = w0 e e−iωt iK(x + y + z) √ 3 u = u0 e e−iωt Lets demonstrate solution for variant 2), using expression (57c): 2 2 ∂2v ∂2ω ∂ u ∂2ω ∂ ω ∂2ω + ρ 2 = C11 2 + C44 + + (C12 + C44 ) ∂t ∂z ∂x2 ∂y 2 ∂x∂z ∂y∂z Calculating derivatives taking into account different signs of functions: 2 K2 K2 K K2 + C44 − − + (C12 + C44 ) = C11 − 3 3 3 3 2 2 2 K 2K K −ω 2 ρ = C11 − + C44 − + (C12 + C44 ) 3 3 3 2 2 2 2 K 2K K K ω 2 ρ = C11 + C44 − C12 − C44 3 3 3 3 −ω 2 ρ

3ω 2 ρ ω2 ω

= K 2 C11 − K 2 C12 + K 2 C44 K 2 (C11 − C12 + C44 ) 3ρ s 2 K (C11 − C12 + C44 ) = 3ρ

Expression for velocity of wave propagation: s ω v= = K

K 2 (C11 − C12 + C44 ) 3ρ K

s =

C11 − C12 + C44 3ρ

Problem 3.11: Effective shear constant Back to contents

Problem specification Show that the shear constant 21 (C11 − C12 ) in a cubic crystal is defined by setting exx = −eyy = 21 e and all other strains equal to zero, as in Fig. 22. Hint: Consider the energy density (43); look for a C 0 such that U = 12 C 0 e2

Solution Lets write expression for U (43): U=

1 1 C11 (e2xx + e2yy + e2zz ) + C44 (e2yz + e2zx + e2xy ) + C12 (eyy ezz + ezz exx + exx eyy ) 2 2

From problem specification it is known that we have only deformations exx , eyy , so working with expression for energy in this case: 1 1 U = C11 (e2xx + e2yy + e2zz ) + C44 (e2yz + e2zx + e2xy ) + C12 (eyy ezz + ezz exx + exx eyy ) 2 2 1 U = C11 (e2xx + e2yy ) + C12 (exx eyy ) 2 Taking into account initial conditions: 1 1 2 1 2 1 2 1 1 2 1 2 1 2 1 1 U = C11 ( e + e ) + C12 (− e ) = C11 · e − e C12 = e C11 − C12 2 4 4 4 2 2 4 2 2 2 Finally, to get expression in requested form U = 12 C 0 e2 : 1 1 1 1 1 U = e2 C11 − C12 = (C11 − C12 ) e2 2 2 2 2 2

Problem 3.12: Determinantal approach Back to contents

Problem specification It is known that an R-dimensional square matrix with all elements equal to unity has roots R and 0, with the R occurring once and the zero occurring R-1 times. If all elements have the value p, then the roots are Rp and 0. a) Show that if the diagonal elements are q and all other elements are p, then there is one root equal to (R − 1)p + q and R − 1 roots equal to q − p. b) Show from the elastic equation (57) for a wave in the [111] direction of a cubic crystal that the determinantal equation which gives ω 2 as a function of K is    q − ω2 ρ    p    p

q − ω2 ρ

   =0   

1 2 1 K (C11 + 2C44 ) and p = K 2 (C12 + C44 ). This expresses condition that three linear homogeneous algebraic 3 3 equations for the three displacement components u, v, w have a solution. Use the result of part (a) to find the three roots of ω 2 ; check with the results given for Problems 9 and 10. where q =

Solution a) Lets start from initial matrix: 



 q p p       p q p        p p q It is possible to make a diagonal matrix from initial by subtracting p from each row:    q−p    0    0

q−p

      

Now we have a diagonal matrix, so it is possible to use matrix properties described in problem specification: λ − (q − p)

λ − (q − p)

= Rp

Here used also a common rule to obtain characteristic matrix equation, from which it is easy to determine roots: λ

q−p

Rp + (q − p) = p(R − 1) + q

b) For proposed conditions, waves propagation are: u = u0 ei(Kx x+Ky y+Kz z−ωt) v

= v0 ei(Kx x+Ky y+Kz z−ωt)

= w0 ei(Kx x+Ky y+Kz z−ωt)

Vectors projections obey rule: Kx2 + Ky2 + Kz2

Kx = Ky = Kz

K √ 3

Thus, plane waves can be written as: K (x+y+z−ωt) i√ 3

u =

u0 e

v0 e

w0 e

K i√ (x+y+z−ωt) 3

K (x+y+z−ωt) i√ 3

Equations (57) are: ∂2u ρ 2 ∂t ∂2v ρ 2 ∂t ∂2w ρ 2 ∂t

= = =

2 2 ∂ v ∂2w ∂2u ∂ u ∂2u C11 2 + C44 + 2 + (C12 + C44 ) + ∂x ∂y 2 ∂z ∂x∂y ∂x∂z ∂2w ∂2v ∂2u ∂2v ∂2v + C11 2 + C44 + + (C + C ) 12 44 ∂y ∂x2 ∂z 2 ∂x∂y ∂y∂z 2 2 2 2 ∂ w ∂ w ∂ w ∂ u ∂2v C11 2 + C44 + + + (C12 + C44 ) ∂z ∂x2 ∂y 2 ∂x∂z ∂y∂z

Inserting expressions for waves into equations (57) consequently: 2

−ω ρu

−ω 2 ρv

−ω 2 ρw

2K 2 K2 K2 K2 C11 u − C44 u + (C12 + C44 ) − v− w − 3 3 3 3 2 2 2 2 K 2K K K − C11 v − C44 v + (C12 + C44 ) − u− w 3 3 3 3 K2 2K 2 K2 K2 − C11 w − C44 w + (C12 + C44 ) − u− v 3 3 3 3

K2 2K 2 K2 K2 C11 u + C44 u + v(C12 + C44 ) + w(C12 + C44 ) 3 3 3 3 K2 2K 2 K2 K2 −ω 2 ρv + C11 v + C44 v + u(C12 + C44 ) + w(C12 + C44 ) 3 3 3 3 K2 2K 2 K2 K2 −ω 2 ρw + C11 w + C44 w + u(C12 + C44 ) + v(C12 + C44 ) 3 3 3 3 −ω 2 ρu +

1 1 2 1 2 u −ω 2 ρ + K 2 (C11 + 2C44 ) + v K (C12 + C44 ) + w K (C12 + C44 ) 3 3 3 1 2 1 1 2 2 2 u K (C12 + C44 ) + v −ω ρ + K (C11 + 2C44 ) + w K (C12 + C44 ) 3 3 3 1 2 1 2 1 u K (C12 + C44 ) + v K (C12 + C44 ) + w −ω 2 ρ + K 2 (C11 + 2C44 ) 3 3 3

It is possible to define now determinantal equation:  1 1 2 1 2 K (C12 + C44 ) K (C12 + C44 )  −ω 2 ρ + K 2 (C11 + 2C44 ) 3 3 3   1 2 1 1 2  K (C12 + C44 ) −ω 2 ρ + K 2 (C11 + 2C44 ) K (C12 + C44 )  3 3 3   1 2 1 1 2 K (C12 + C44 ) K (C12 + C44 ) −ω 2 ρ + K 2 (C11 + 2C44 ) 3 3 3 If we replace by q =

1 2 1 K (C11 + 2C44 ) and p = K 2 (C12 + C44 ): 3 3   −ω 2 ρ + q    p    p

    =0   



−ω 2 ρ + q

   =0   

To find roots it is possible to use expressions for q and p: λ=q−p=

1 2 1 1 K (C11 + 2C44 ) − K 2 (C12 + C44 ) = K 2 (C11 − C12 − C44 ) 3 3 3

This was a trivial root of characteristic matrix equation which describe transverse wave propagation in [111] direction. For longitudinal waves(R=3): λ = p(R − 1) + q = 2p + q =

1 1 1 2 K (2C12 + 2C44 ) + K 2 (C11 + 2C44 ) = K 2 (C11 + 2C12 + 4C44 ) 3 3 3

Obtained result fully agree results from previous problems.

Problem 3.13: General propagation direction Back to contents

Problem specification a) By substitution in (57) find the determinantal equation which expresses the condition that the displacement R(r) = [u0 x ˆ + v0 yˆ + ω0 zˆ]ei(K·r−ωt) be a solution of the elastic wave equations in a cubic crystal. b) The sum of the roots of a determinantal equation is equal to the sum of the diagonal elements aii . Show from part (a) that the sum of the squares of the three elastic wave velocities in any direction in a cubic crystal is equal to (C11 + 2C44 )/ρ. Recall that vs2 = ω 2 /K 2 .

Solution For case of general propagation direction: Kx 6= Ky 6= Kz Plane waves are described as: u = u0 ei(Kx x+Ky y+Kz z−ωu t) v

= v0 ei(Kx x+Ky y+Kz z−ωv t)

= w0 ei(Kx x+Ky y+Kz z−ωw t)

Equations (57) are: ∂2u ∂t2 ∂2v ρ 2 ∂t ∂2w ρ 2 ∂t ρ

= = =

2 2 ∂2u ∂ u ∂2u ∂ v ∂2w + C + + + (C + C ) 44 12 44 ∂x2 ∂y 2 ∂z 2 ∂x∂y ∂x∂z 2 2 2 ∂ v ∂2v ∂2w ∂ v ∂ u C11 2 + C44 + 2 + (C12 + C44 ) + ∂y ∂x2 ∂z ∂x∂y ∂y∂z ∂2u ∂2v ∂2w ∂2w ∂2w C11 2 + C44 + + (C + C ) + 12 44 ∂z ∂x2 ∂y 2 ∂x∂z ∂y∂z C11

Calculating: −ωu2 ρu

= −C11 Kx2 u + C44 (−Ky2 − Kz2 )u + (C12 + C44 )(−Kx Ky v − Kx Kz w)

−ωv2 ρv

= −C11 Ky2 v + C44 (−Kx2 − Kz2 )v + (C12 + C44 )(−Kx Ky u − Ky Kz w)

2 −ωw ρw

= −C11 Kz2 w + C44 (−Kx2 − Ky2 )w + (C12 + C44 )(−Kx Kz u − Ky Kz v)

= −ωu2 ρu + C11 Kx2 u + C44 (Ky2 + Kz2 )u + (C12 + C44 )(Kx Ky v + Kx Kz w)

= −ωv2 ρv + C11 Ky2 v + C44 (Kx2 + Kz2 )v + (C12 + C44 )(Kx Ky u + Ky Kz w)

2 = −ωw ρw + C11 Kz2 w + C44 (Kx2 + Ky2 )w + (C12 + C44 )(Kx Kz u + Ky Kz v)

= u(−ωu2 ρ + C11 Kx2 + C44 (Ky2 + Kz2 )) + v(Kx Ky (C12 + C44 )) + w(Kx Kz (C12 + C44 ))

= u(Kx Ky (C12 + C44 )) + v(−ωv2 ρ + C11 Ky2 + C44 (Kx2 + Kz2 )) + w(Ky Kz (C12 + C44 ))

2 ρw + C11 Kz2 w + C44 (Kx2 + Ky2 )) = u(Kx Kz (C12 + C44 )) + v(Ky Kz (C12 + C44 )) + w(−ωw

Getting determinantal equation: 



Kx Ky (C12 + C44 ) Kx Kz (C12 + C44 )  C11 Kx2 + C44 (Ky2 + Kz2 ) − ωu2 ρ    Kx Ky (C12 + C44 ) C11 Ky2 + C44 (Kx2 + Kz2 ) − ωv2 ρ Ky Kz (C12 + C44 )    2 ρ Kx Kz (C12 + C44 ) Ky Kz (C12 + C44 ) C11 Kz2 + C44 (Kx2 + Ky2 ) − ωw

   =0   

Sum of determinantal equation root is equal to sum of diagonal elements: 2 )=0 C11 Kx2 + C44 (Ky2 + Kz2 ) + C11 Ky2 + C44 (Kx2 + Kz2 ) + C11 Kz2 + C44 (Kx2 + Ky2 ) − ρ(ωu2 + ωv2 + ωw 2 ) = C11 (Kx2 + Ky2 + Kz2 ) + 2C44 (Kx2 + Ky2 + Kz2 ) ρ(ωu2 + ωv2 + ωw

Next identities are valid: Kx2 + Ky2 + Kz2

= K2

2 ωu2 + ωv2 + ωw

= ω2

2 vu2 + vv2 + vw

= vs2

Obtaining: ω 2 ρ = K 2 (C11 + 2C44 ) Recalling identity vs2 = ω 2 /K 2 :

K 2 (C11 + 2C44 ) ω2 C11 + 2C44 ρ vs2 = 2 = = K K2 ρ

Problem 3.14: Stability criteria Back to contents

Problem specification The criterion that a cubic crystal with one atom in the primitive cell be stable against small homogeneous is that the energy density (43) be positive for all combinations of strain components. What restrictions are thereby imposed on the elastic stiffness constants? (In mathematical language the problem is to find the conditions that a real symmetric quadratic form should be positive definite. The solution is given in books on algebra; see also Korn and Korn, Mathematical Handbook, McGraw-Hill, 1961, Sec.13.5-6.). 2 2 Answer : C44 > 0, C11 > 0, C11 − C12 > 0 and C11 + 2C12 > 0. For an example of the instability which results when C11 ∼ = C12 , see L.R. Testardi Phys. Rev. Letters 15, 250(1965).

Solution Elastic stiffness matrix for a cubic crystal is: 



 C11    C  12    C  12    0     0    0

C12

C11

C12

C11

C44

                  

To prove that a real symmetric quadratic form should be positive definite, it is necessary to prove that by given coefficients all minors of elastic stiffness matrix will be positive. It is seen that only independent minors are on main diagonal of matrix, because of it's configuration. So, starting from first minor: C11 > 0 This statement is obvious, as here only one component and it is necessary to be positive. Next minor:    C11   C12

C12  >0  C11

From here it is seen that: 2 2 2 2 C11 − C12 > 0 → C11 > C12

Third minor:





 C11    C  12   C12

C12 C11 C12

C12    C12  >0   C11

Calculating: 3 3 3 2 2 2 C11 + C12 + C12 − C11 C12 − C11 C12 − C11 C12

> 0

(C11 − C12 )2 (C11 + 2C12 ) > 0 39

From here: C11 + 2C12 > 0 Further it is seen that last will be a multiplication of C11 on C44 , as C11 > 0, then C44 must be also greater than 0. Summarize conditions: C11

2 C11

2 C12

C11 + 2C12 > C44

0 0

Problem 4.1: Vibrations of square lattice Back to contents

Problem specification We consider transverse vibrations of a planar square lattice of rows and columns of identical atoms, and let ul,m denote the displacement normal to the plane of the lattice of the atom in the lth column and mth row (Fig.13). The mass of each atom is M. Assume force constants such that the equation of motion is

d2 ul,m = C[(ul+1,m + ul−1,m − 2ul,m ) + (ul,m+1 + ul,m−1 − 2ul,m )] dt2

a) Assume solutions of the form ul,m = u(0)ei(lKx a+mKy a−ωt) where a is the spacing between nearest-neighbor atoms. Show that the equation of motion is satisfied if ω 2 M = 2C(2 − cos Kx a − cos Ky a) This is the dispersion relation for the problem. b) Show that the region of K space for which independent solutions exist may be taken as a square of side 2π/a. This is the first Brillouin zone of the square lattice. Sketch ω versus K for K = Kx with Ky = 0, and for Kx = Ky .

Solution a) For this part assuming solution in next form: ul,m = u(0)ei(lKx a+mKy a−ωt) Inserting into initial expression: −M ω 2 ul,m = C[(ul+1,m + ul−1,m − 2ul,m ) + (ul,m+1 + ul,m−1 − 2ul,m )] Now it is possible to establish next relations: ul+1,m

ul,m · eiKx a

ul−1,m

ul,m · e−iKx a

ul,m+1

ul,m · eiKy a

ul,m−1

ul,m · e−iKx a 41

Lets rewrite expression using this properties: −M ω 2 ul,m −M ω 2 cos x

= Cul,m (eiKx a + e−iKx a − 2 + eiKy a + e−iKy a − 2) = C(eiKx a + e−iKx a + eiKy a + e−iKy a − 4) =

eix +e−ix 2

⇒

−M ω 2

= C(2 cos Kx a + 2 cos Ky a − 4)

−M ω 2

2C(cos Kx a + cos Ky a − 2)

M ω2

2C(2 − cos Kx a − cos Ky a)

π π π π b) Independent solutions exist only within first Brillouin zone, which has range [− , ] and range r = + . Described a a a a 2D lattice is cubic lattice, so in directions m and l lattice constant is same and equal to a. Area of reciprocal lattice is will be multiplication of ranges of first Brillouin zone. It is easy to see that it will be: S =r·r =

2π 2π · a a

2π a

For case when K = Kx and Ky = 0 plot will be looking as on Figure 4.

For case when Kx = Ky it is possible to obtain next expression for law of motion: M ω2

2C(2 − 2 cos Kx a)

ω2

8C sin2 12 Kx M

So only difference from previous case will be that curve will be slightly lower than previous line as it obvious from expression.

Problem 4.2: Monatomic linear lattice Back to contents

Problem specification Consider a longitudinal wave us = u cos(ωt − sKa) which propagates in a monatomic linear lattice of atoms of mass M, spacing a, and nearest-neighbor interaction C. a) Show that the total energy of the wave is E = 12 M

s (dus /dt)

+ 21 C

s (us

− us+1 )2

where s runs over all atoms. b) By substitution of us in this expression, show that the time-average total energy per atom is 1 2 2 4Mω u

+ 21 C(1 − cos Ka)u2 = 12 M ω 2 u2

where in the last step we have used the dispersion relation (9) for this problem.

Solution a) Classical representation of energy is: E = Ekinetic + Epotential Kinetic energy of moving with velocity v body of mass m is: Ek =

1 mv 2 2

If we take as m the mass of one particle M and sum over all atoms, it is obvious: 1 X 2 1 X M v = M (dus /dt)2 2 2 s s

Ek =

Potential energy of monatomic chain can be described well in harmonic oscillator approximation(in case of longitudinal wave): F

−kx

−

dEp

−F dx = kxdx

dEp dx

1 2 kx 2 Here k is force constant C, and displacement x can be represented as strain between plane us and us+1 : Ep

Ep =

= k

Rx 0

xdx =

1 X 2 1 X C x = C (us − us+1 )2 2 2 s s

Thus, total energy can be written as: E = Ek + Ep =

1 X 1 X M (dus /dt)2 + C (us − us+1 )2 2 2 s s

b) To demonstrate this part, for first need to insert us into expression, find kinetic energy, describe potential energy and finally sum these energies to find total energy. Lets proceed with kinetic energy: Ek =

1 X 1 1 M (dus /dt)2 = M (−ωu cos(ωt − sKa))2 = M ω 2 u2 (cos(ωt − sKa))2 2 2 2 s

Time-average value of cos2 x = 12 , so kinetic energy is: Ek =

1 1 1 1 M ω 2 u2 (cos(ωt − sKa))2 = M ω 2 u2 · = M ω 2 u2 2 2 2 4

Now it is necessary to work with potential energy. Displacements are: us

u cos(ωt − sKa)

us+1

u cos[(ωt − sKa) − ka]

Lets define: α

ωt − sKa

= u cos(α)

So, displacements can be restated as:

us+1

= u cos(α − β)

Expanding parenthesis gives: (us − us+1 )2

= u2 cos2 α − 2u2 cos α cos(α − β) + u2 cos2 (α − β)

cos(α − β)

(us − us+1 )2

= u2 [cos2 α − 2 cos α cos(α − β) + cos2 (α − β)] =

cos α cos β − sin α sin β

= u2 [cos2 α − 2 cos α(cos α cos β − sin α sin β) + (cos α cos β − sin α sin β)2 ] = = u2 [cos2 α − 2 cos2 α cos β + 2 cos α sin α sin β + (cos2 α cos2 β − 2 cos α cos β sin α sin β + sin2 α sin2 β)] = To make expression simpler, it is necessary to take into account time-average values of functions: hcos2 αi

= hsin2 αi =

hsin α cos αi

(us − us+1 )2

= u2 [ 21 − cos β +

1 2

0 1 2

cos2 β +

1 2

sin2 β] =

= u2 [ 21 − cos β + 12 (cos2 β + sin2 β)] = u2 [1 − cos β] So, potential energy will be expressed as: Ep =

1 X 1 1 C (us − us+1 )2 = Cu2 (1 − cosβ) = Cu2 (1 − cosKa) 2 2 2 s 44

Total energy is a sum of kinetic and potential energy: E = Ek + Ep =

1 1 M ω 2 u2 + Cu2 (1 − cosKa) 4 2

Now it is possible to use expression (7) as it more convenient than (9) as we have cos function: ω2

1 2 2 Cu (1

− cosKa) →

1 2 2 4Mω u

2C M (1

− cos Ka)

1 2 2 4Mω u

According to this total energy is: =

+ 41 M ω 2 u2 = 12 M ω 2 u2

Final expression then: 1 1 1 M ω 2 u2 + Cu2 (1 − cos Ka) = M ω 2 u2 4 2 2

Problem 4.3: Continuum wave equation Back to contents

Problem specification Show that for long wavelengths the equation of motion (2) reduces to the continuum elastic wave equation 2 ∂2u 2∂ u = v ∂t2 ∂x2

where v is the velocity of sound.

Solution Equation of motion (2) is: d2 us = C(us+1 + us−1 − 2us ) dt2 In left side we have already a time derivative of us , so need to rewrite right part. In continuum approximation we state that distance a between planes is very small, so it is possible to expand right part into Taylor series near planes boundary: M

∂u a+ ∂x ∂u = u(s) − a+ ∂x

us+1

us+a = u(s) +

us−1

us−a

1 ∂2u 2 a + ... 2 ∂x2 2 1∂ u 2 a − ... 2 ∂x2

Inserting values into expression neglecting terms higher than 2 order, and in same time change time differential to partial, because in this case us will depend on few variables: ∂2u ∂t2 ∂2u M 2 ∂t ∂2u M 2 ∂x ∂2u ∂x2

= C(us+1 + us−1 − 2u(s)) ∂u 1 ∂2u 2 ∂u 1 ∂2u 2 = C u(s) + a+ a + u(s) − a + a − 2u(s) ∂x 2 ∂x2 ∂x 2 ∂x2 2 ∂ u 2 = C a ∂x2 Ca2 ∂ 2 u = M ∂x2

From other side, it is possible to establish: v

dω dk r 4C (1 − cos Ka) M r r 4C Ca2 dω 1 = cos Ka · a = cos Ka dk M 2 M

In continuum approximation planes situated so close, so Ka → 0, cos Ka → 1: r Ca2 v = M 2 Ca v2 = M So, finally we replace

Ca2 by a v 2 to obtain a classical wave equation in form: M 2 ∂2u 2∂ u = v ∂x2 ∂x2

Problem 4.4: Basis of two unlike atoms Back to contents

Problem specification For the problem treated by (18) to (26), find the amplitude ratios u/v for the two branches at Kmax = π/a. Show that at this value of K the two lattices act as if decoupled: one lattice remains at rest while the other lattice moves.

Solution Equations of motion in basis of two types of atoms are: d2 us dt2 d 2 vs M2 2 dt

= C(vs + vs+1 − 2us ) = C(us+1 + us − 2vs )

Solution in form of: us

ueisKa e−iωt

veisKa e−iωt

Substituting functions into motion equations obtaining:

For this problem K =

−M1 ω 2 u =

C(v · eisKa e−iωt + v · eisKa e−iωt e−ika − 2u · eisKa e−iωt )

−M2 ω 2 v

C(u · eisKa e−iωt eika + u · eisKa e−iωt − 2v · eisKa e−iωt )

−M1 ω 2 u =

Cv(1 + e−ika ) − 2Cu

−M2 ω 2 v

Cu(eika + 1) − 2Cv

π , so: a −M1 ω 2 u = Cv(1 − 1) − 2Cu −M2 ω 2 v

= Cu(−1 + 1) − 2Cv

M1 ω 2 u =

2Cu

M2 ω 2 v

2Cv

Solutions in this case will take form: ω2

ω2

2C M1 2C M2

Unlike case when K = 0, we see that at boundary of first Brillouin zone these motions are independent each other.

Problem 4.5: Kohn anomaly Back to contents

Problem specification We suppose that the interplanar force constant Cp between planes s and s + p is of the form Cp = A

sin pk0 a , pa

where A and k0 are constants and p runs over all integers. Such a form is expected in metals. Use this and Eq. (16a) to find an expression for ω 2 and also for ∂ω 2 /∂K. Prove that ∂ω 2 /∂K is infinite when K = k0 . Thus a plot of ω 2 versus K or of ω versus K has a vertical tangent at k0 : there is a kink at k0 in the phonon dispersion relation ω(K).

Solution Equation (16a) is: ω2 =

2 X Cp (1 − cos pKa) M p>0

Inserting Cp into (16a) gives: ω2 =

2 X sin pk0 a A (1 − cos pKa) M p>0 pa

Taking derivative of ω 2 by ∂K: ∂ω 2 2A P sin pk0 a = sin pKa · pa p>0 ∂K M pa ∂ω 2 2A P = p>0 sin pk0 a · sin pKa ∂K M Using trigonometric identity it is possible to restate expression: sin α · sin β ∂ω 2 ∂K ∂ω 2 ∂K For case K = k0 :

= = =

1 [cos(α − β) − cos(α + β)] 2 2A P 1 p>0 [cos(pk0 a − pKa) − cos(pk0 a + pKa)] M 2 A P cos(pa(k0 − K)) − cos(pa(k0 + K)) M p>0

∂ω 2 A X A X 2 = (1 − cos 2paK) = sin 2paK ∂K M p>0 2M p>0

It is seen that ratio will be infinite, as series diverge.

Problem 4.6: Diatomic chain Back to contents

Problem specification Consider the normal modes of a linear chain in which the force constants between nearest-neighbor atoms are alternately C and 10C. Let the masses be equal, and let the nearest-neighbor separation will be a/2. Find ω(K) at K = 0 and K = π/a. Sketch in the dispersion relation by eye. This problem simulates a crystal of diatomic molecules such as H2 .

Solution Lets restate equations of motion for case when different atomic planes have different force constants: 2

m ddt2u 2

m ddt2v

C1 (vs − us ) + C2 (vs−1 − us )

C2 (us+1 − vs ) + C1 (us − vs )

Solution will be in form of plane waves: us

= ueisKa e−iωt

= veisKa e−iωt

us+1

= ueisKa e−iωt eiKa

vs−1

= veisKa e−iωt e−iKa

10C1 = 10C

Inserting into motions expressions: −ω 2 m · ueisKa e−iωt

= eisKa e−iωt (C1 (v − u) + C2 (ve−iKa − u))

−ω 2 m · veisKa e−iωt

= eisKa e−iωt (C2 (ueiKa − v) + C1 (u − v))

−ω 2 mu = C1 (v − u) + C2 (ve−iKa − u) −ω 2 mv

= C2 (ueiKa − v) + C1 (u − v)

−ω 2 mu = C1 v − C1 u + C2 e−iKa v − C2 u −ω 2 mv

= C2 eiKa u − C2 v + C1 u − C1 v

= C1 u + C2 u − ω 2 mu − C2 e−iKa v − C1 v

= −C2 eiKa u − C1 u + C2 v + C1 v − ω 2 mv

= u(C1 + C2 − ω 2 m) + v(−C2 e−iKa − C1 )

u(−C2 eiKa − C1 ) + v(C2 + C1 − ω 2 m)

Now it is possible to write a determinantal equation:   C1 + C2 − ω 2 m   −C2 e−iKa − C1

 −C2 e−iKa − C1  =0  C1 + C2 − ω 2 m

For K = 0: C1 + C2 − ω 2 m − C2 − C1

ω2

(C1 + C2 − ω 2 m)2 − (−C2 − C1 )2

121C 2 − 22Cω 2 m + ω 4 m2 − 121C 2

ω 4 m2

22Cω 2 m

ω2

22C m

and second:

For K =

π a:

C1 + C2 − ω 2 m − C2 + C1

11C − ω 2 m − 9C

ω2

2C m

(C1 + C2 − ω 2 m)2 − (C2 − C1 )2

121C 2 − 22Cω 2 m + ω 4 m2 − 81C 2

ω 4 m2 − 22Cω 2 m + 40C 2

and second:

2C m 20C ω2 = m To draw schematically dispersion relation it is necessary to take in account that fact that: ω1

K=0 K=

π a

ω1 = 0 r ω1 =

K=0 K=

π a

2C m

22C r m 20C ω1 = m ω1 =

Problem 4.7: Atomic vibrations in metal Back to contents

Problem specification Consider point ions of mass M and charge e immersed in a uniform sea of conduction electrons. The ions are imagined to be in stable equilibrium when at regular lattice points. If one ion is displaced a small distance r from its equilibrium position, the restoring force is largely due to the electric charge within the sphere of radius r centered at the equilibrium 3 , which defines R. position. Take the number density of ions (or of conduction electrons) as 3 4πR p 2 3 a) Show that the frequency of a single ion set into oscillation is ω = e /M R . b) Estimate the value of this frequency for sodium, roughly. c) From (a), (b), and some common sense, estimate the order of magnitude of the velocity of sound in the metal.

Solution a) When ion is displaced from equilibrium position, electrostatic force is returning ion to initial position(in CGS): F =−

e2 3 4πr3 e2 r · · =− 3 2 3 r 4πR 3 R

In same time: d2 r dt2 d2 r e2 M 2 + 3r dt R M

k ω

= −

e2 r R3

e2 R3 r

k = M

e2 M R3

b) For sodium atom (according to Table 4): R

3.659 · 10−10 m = 3.659 · 10−8 cm

22.98(g/mol) = 3.8153 · 10−23 g 6.023 · 1023

(4.803 · 10−10 statC)2 = 23.069 · 10−20 statC 2

Calculating ω: s ω=

23.069 · 10−20 = 3.8153 · 10−23 · (3.659 · 10−8 )3

23.069 = 1013 3.8153 · 48.9877 · 10−27

23.069 = 1.234 · 1013 Hz 18.6902

c) Suppose we have a maximum displacement of ion from its equilibrium position order of R. Then velocity of sound can be evaluated as: ω v = k 2π 2π k = = λ R ωR 3.659 · 10−8 · 1.234 · 1014 v = = = 0.718 · 106 cm/s = 7.18 · 103 m/s 2π 6.28

Problem 4.8: Soft phonon modes* Back to contents

Problem specification Consider a line of ions of equal mass but alternating in charge, with ep = e(−1)p as the charge of the pth ion. The interatomic potential is the sum of two contributions: (1) a short-range interaction of force constant C1R = γ that acts between nearest neighbors only, and (2) a coulomb interaction between all ions. a) Show that the contribution of the coulomb interaction to the atomic force constants is CpC = 2(−1)p

e2 , where p3 a3

a is the equilibrium nearest-neighbor distance. b) From (16a) show that the dispersion relation may be written as P∞ ω2 = sin2 12 Ka + σ p=1 (−1)p (1 − cos pKa)p−3 2 ω0 2

e where ω 2 ≡ 4γ M and σ = γa3 . c) Show that ω 2 is negative (unstable mode) at the zone boundary Ka = π if σ > (2 ln 2)−1 = 0.721. Notice that the phonon spectrum is not that of a diatomic lattice because the interaction of any ion with its neighbors is the same as that of any other ion.

Solution

a) Line of ions with alternating charge can be considered as sum of electrical dipoles, which electric field magnitude can be described as(in case when searching E on the line far from current ion) in CGS: E

2p R3

ql = e(−1)p l

= pa

= E·e=

2(−1)p e · l · e 2(−1)p e2 = l (pa)3 p3 a3

As expression for force is known, it is possible to write a force constant: CpC =

2(−1)p e2 p3 a3

b) Equation (16a): ω2 =

2 X Cp (1 − cos pKa) M p>0 54

For this case, force constant will be consist of two parts - near neighbor potential C1R and coulomb interaction between all ions CpC : Cp = C1R + CpC Inserting already known expressions for these potentials into (16a) obtaining: ω2 =

∞ 2 X e2 (γ + 2(−1)p 3 3 )(1 − cos pKa) M p=1 p a

Here series changed range from p > 0 to p = 1 as for this problem we have p as a natural counter. ∞ 2 2 2 X p e 2 p e ω = γ − γ cos pKa + 2(−1) 3 3 − 2(−1) 3 3 cos pKa M p=1 p a p a Here need to recall to physical meaning of force constant C1R = γ. This coefficient is affecting only nearest neighbors, so in expression γ − γ cos pKa, counter p can be taken as 1, thus: 2 2 P∞ 2 p e ω = γ(1 − cos Ka) + 2(−1) 3 3 (1 − cos pKa) M p=1 p a 2 P 2 ∞ p e ω2 = (1 − cos Ka) + 2(−1) (1 − cos pKa) γM p=1 γp3 a3 2 2 P∞ 2 1 p e 2 (1 − cos pKa) ω = 2 sin 2 Ka + 2(−1) γM p=1 γp3 a3 2 4 P∞ 2 1 p e sin Ka + (−1) (1 − cos pKa) ω2 = 2 γM p=1 γp3 a3 P∞ ω 2 = ω02 p=1 sin2 21 Ka + σ(−1)p (1 − cos pKa)p−3 ω2

= ω02 · sin2 12 Ka + ω02

P∞

p=1

σ(−1)p (1 − cos pKa)p−3

Dividing by ω02 to obtain final expression: ∞ X ω2 2 1 = sin Ka + σ (−1)p (1 − cos pKa)p−3 ω02 2 p=1

c) Using condition Ka = π and assuming that fact ω 2 = 0 near zone boundary:

Calculation of series sum gives:

P∞ π + σ p=1 (−1)p (1 − cos pπ)p−3 2 P∞ 1 + σ p=1 (−1)p (1 − cos pπ)p−3

cos pπ

(−1) + 1 + (−1) + ... = (−1)p

1+σ

∞ X

sin2

P∞

p=1 (−1)

(1 − (−1)p )p−3

(−1)p (1 − (−1)p )p−3 u −2.1036

p=1

Inserting series sum into expression and changing equality to inequality as it necessary to obtain value of σ when frequency ω 2 will be negative: 1 + σ(−2.1036) < −2.1036σ 2.1036σ σ 55

< −1 > 1 > 0.4753755

To prove last part lets start from dispersion relation at zone boundary (Ka → 0): ∞ X 1 sin2 Ka + σ (−1)p (1 − cos pKa)p−3 = 0 2 p=1

In this approximation remember that sin of angle is proportional to angle, and second sin2 21 x = ( 21 Ka)2 + σ 1 2 2 4K a

P∞

2 1 p −3 p=1 (−1) (2 sin 2 pKa)p

+ 2σ

1 2 2 4K a

(sin2 12 pKa)p−3

p=1 (−1)

p 1 2 2 2 −3 ( 4 p K a )p

P∞

(−1)p 1 2 2 p (4K a )

(−1)p p

P∞

+ 2σ

p=1 (−1)

P∞

1 2 2 4K a

+ 2σ

p=1

1 + 2σ Calculating sum of series:

∞ X (−1)p p=1

P∞

p=1

= − ln 2 u −0.693

Returning to expression with result: 1 − 2σ ln 2

2 ln 2σ

1 u 0.721 2 ln 2

Thus in specified region of σ lattice is unstable.

1−cos x : 2

Problem 5.1: Singularity in density of states Back to contents

Problem specification a) From the dispersion relation derived in Chapter 4 for a monatomic linear lattice of N atoms with nearest neighbor interactions, show that the density of modes is D(ω) =

2N 1 ·p 2 − ω2 π ωm

where ωm is the maximum frequency. b) Suppose that an optical phonon branch has the form ω(K) = ω0 − AK 2 , near K = 0 in three dimensions. Show that 3 L 2π √ ω0 − ω for ω < ω0 and D(ω) = 0 for ω > ω0 . Here the density of modes is discontinuous. D(ω) = 2π A3/2

Solution Dispersion relation for monatomic linear lattice is: r ω=

4C 1 | sin Ka| M 2

Maximum value ω takes at it not difficult to see when sin is maximum at Ka = π, so lets determine ωm as maximum value: r r π 4C 4C · | sin | = ωm = M 2 M So, initial ω can be rewritten as: 1 ω = ωm sin Ka 2 Lets resolve this equation relatively K: 1 = ωm sin Ka 2 ω 1 = sin Ka 2 ωm ω 1 arcsin Ka = ωm 2 2 ω K = arcsin a ωm ω

Searching density of states: D(ω) dK dω

L dK π dω 0 2 ω 2 1 1 2 1 2 ω 2 1 s p m = arcsin = s · = = = p 2 2 2 2 a ωm a ωm aωm ω 2 − ω 2 aωm ωm − ω a ωm − ω 2 ω m 1− 2 ωm ωm =

Parameter L is a length of a line, and can be determined as number of particles N multiplied by inter-particle separation distance a: L=N ·a Finally have: D(ω) =

L dK Na 2 1 2N 1 p p = = 2 2 2 π dω π a ωm − ω π ωm − ω 2 57

b) Density of states in three dimensions: dN = dω 3 L 4πK 3 N = 2π 3 q 1 K = A (ω0 − ω) q 3 1 3 3 4π 3/2 (ω − ω) 0 A L 4π (ω0 − ω) L = N = 3/2 2π 3 2π 3A 3 3 dN L L 3p 4π 2π p = = D(ω) = · (ω − ω) · (−1) (ω0 − ω) 0 3/2 2π dω 2 2π 3A A3/2 D(ω)

It is seen that density of states is discontinuous as all values ω > ω0 are deprecated.

Problem 5.2: Rms thermal dilation of crystal cell Back to contents

Problem specification a) Estimate for 300 K the root mean square thermal dilation ∆V /V for a primitive cell of sodium. Take the bulk modulus as 7 × 1010 erg cm−3 . Note that the Debye temperature 158 K is less than 300 K, so that the thermal energy is of the order of kb T . b) Use this result to estimate the root mean square thermal fluctuation ∆a/a of the lattice parameter.

Solution a) Dilation expression is (35): ∆V = ∆ax + ∆ay + ∆az = 3∆a V Linkage of potential energy with dilation can be described as (53): δ=

1 2 Bδ 2

1 In same time, thermal energy is proportional to kT for each degree of freedom. Here we have only one degree of freedom, 2 and taking into account that thermal energy should be related to specified volume of N a we obtain: 1 kT 2V kT V

1 2 Bδ 2

= Bδ 2 kT VB

δ2

= a3 , where a is lattice parameter for Na

4.225˚ A = 4.225 · 10−8 cm

All necessary constants are present to obtain dilation:

b) To obtain

δ2

kT 1.38 · 10−16 · 300 414 · 10−16 414 · 10−2 = = = = 0.007842 VB (4.225 · 10−8 )3 · 7 · 1010 75.418 · 7 · 10−14 527.926 √ ∆V = 0.007842 = 0.0885 = 8.85% V

∆a ratio it is necessary to recall that: a V ∆V V ∆a a

= a3 ≡ =

3∆a a 1 ∆V · = 8.85/3 = 2.95% 3 V

Problem 5.3: Zero point lattice displacement and strain Back to contents

Problem specification 2 3~ωD , 8π 2 ρv 3 ~ P −1 where v is the velocity of sound. Start from the result (4.29) summed over independent lattice modes: hR2 i = ω . 2ρV 1 We have included a factor of to go from square amplitude to square displacement. 2 P −1 b) Show that ω and hR2 i diverge for a one-dimensional lattice, but that the mean square strain is finite. Con2 ∂R 2 ~ωD L 1P 2 2 sider h( K u0 as the mean square strain, and show that it is equal to for a line of N atoms each ) i= ∂x 2 4πM N v 3 2 of mass M, counting longitudinal modes only. The divergence of R is not significant for any physical measurement.

a) In the Debye approximation, show that the mean square displacement of an atom at absolute zero is hR2 i =

Solution a) In Debye approximation velocity of sound for all modes is taken as a constant. Lets replace summation integral: Z ωD X −1 ω = D(ω)ω −1 dω

ω −1 by an

Density of states are described by next expression(note that density of states counted for 3 modes as they equal in Debye approximation): 3V ω 2 D(ω) = 2π 2 v 3 Now it is possible to calculate sum: Z ωD Z ωD 2 2 X 3V ω 2 −1 3V 3V ωD 3V ωD ω −1 = ω dω = ωdω = · = 2π 2 v 3 2π 2 v 3 0 2π 2 v 3 2 4π 2 v 3 0 Now inserting sum value into expression for mean square value of displacement: hR2 i =

2 2 ~ 3V ωD 3~ωD ~ X −1 ω = · = 2ρV 2ρV 4π 2 v 3 8π 2 ρv 3

b) To prove that sum is diverges it is necessary to calculate integral in one-dimensional approximation: Z ωD Z ωD Z ωD X L dK −1 L dω L ω −1 = D(ω)ω −1 dω = ω dω = = ln(ωD − ∞) = −∞ π dω πv ω πv 0 0 0

So, this sum diverges, as was necessary to show. Now calculating mean square strain (insert (4.29) instead of u20 ): h(

∂R 2 ) i = ∂x K2

D(ω)2D

∂R 2 ) i = ∂x = = = = ρV

1 4(0 + 12 )~ R ωD 2 1P 2 2 K D(ω)dω K u0 = · 0 2 2 ρV ω ω2 v2 L dK L = π dω πv ~ R ωD 2 L K dω ρV ω 0 πv ~ R ωD ω 2 L dω ρV ω 0 v 2 πv ~L R ωD ωdω πρV v 3 0 2 ~L ωD πρV v 3 2 2 ~ωD L 2πρV v 3

= MN

∂R 2 ) i = ∂x

2 ~ωD L 2πM N v 3

Note: to obtain result from problem statement it possible is necessary to have a multiplication coefficient 2 instead of 4 in expression 4.29.

Problem 5.4: Heat capacity of layer lattice Back to contents

Problem specification a) Consider a dielectric crystal made up of layers of atoms, with rigid coupling between layers so that the motion of the atoms is restricted to the plane of the layer. Show that the phonon heat capacity in the Debye approximation in the low temperature limit is proportional to T 2 . b) Suppose instead, as in many layer structures, that adjacent layers are very weakly bound to each other. What form would you expect the phonon heat capacity to approach at extremely low temperatures?

Solution a) In current approximation it is obvious that atoms are limited in motion by a 2D surface, so for the specified problem it is necessary to obtain expression for U and see how it depends on temperature. So, lets express U : Z ωD ~ω dω U= D(ω) · ~ω τ e −1 0 To evaluate U it is necessary to obtain density of states in two dimensions first. For this it is necessary to remember general idea - density of states is amount of states in system per size unit: (size of whole system) × (size of elementary unit of system) 2 L N2D = · πK 2 2π 2 2 0 2 L ω L 2πω L2 ω Sω dN = ·π = · = = D(ω) = dω 2π v2 2π v2 2πv 2 2πv 2 N

Here square of length L was replaced by area S. Now it is possible to proceed to calculation of U (multiplier 2 is because for plane we have two possible polarization modes): U

R ωD 0

D(ω) ·

~ω e

~ω τ

−1 ~ω

dω =

R ωD Sω dω = · 0 2πv 2 e ~ω τ − 1 S~ R ωD ω2 dω = ~ω πv 2 0 e τ − 1

Introducing replace of variable from ω to x: x = ω

dω

~ω ~ω = τ kT xkT ~ kT dx ~

Proceeding further with calculation of U : U

= =

S~ k 2 T 2 kT R xD x2 dx = · 2 · πv 2 ~ ~ 0 ex − 1 2 2 R 2 SkT k T x x · 2 0D x dx 2 πv ~ e −1

So, our expression for U depends on T 2 so does the heat capacity. b) For case of weakly bound layers it is still expected heat capacity to be proportional to T 2 , only difference will be in numerical constants as for this case we will have 3 modes instead of 2 in pure plane approximation. 62

Problem 5.5: Gr¨ uneisen constant* Back to contents

Problem specification a) Show that the free energy of a phonon mode of frequency ω is kB T ln[2 sinh(~ω/2kB T )]. It is necessary to retain the zero-point energy 12 ~ω to obtain this result. b) If ∆ is the fractional volume change, then the free energy of the crystal may be written as P F (∆, T ) = 12 B∆2 + kB T ln[2 sinh(~ωK /2kB T )], where B is the bulk modulus. Assume that the volume dependence of ωK is δω/ω = −γ∆, where γ is known as the Gr¨ uneisen constant. If γ taken as independent of the mode K, show that F is a minimum with respect to ∆ P1 ~ω coth(~ω/2k when B∆ = γ B T ), and show that this may be written in terms of the thermal energy density as 2 ∆ = γU (T )/B. c) Show that on the Debye model γ = −∂ ln θ/∂ ln V . Note: Many approximations are involved in this theory: the result (a) is valid only if ω is independent of temperature; γ may be quite different for different modes.

Solution a) General expression for free energy is can be expressed as: F = −kT ln Z where Z - partition function. Partition function can be expressed as: Z=

e− kT

Energy of specified microstate En can be described by equation(4.27) when a specified mode is occupied by n phonons: 1 ~ω En = n + 2 Working with expression for Z: =

P∞

x =

~ω kT P∞

Z P∞

−nx n=0 e

Z cschx 1 x csch 2 2

− kT = n=0 e

P∞

− n=0 e

n=0

e− 2 · e−nx = e− 2

~ω(n+ 1 ) 2 kT

P∞

n=0

P∞

n=0

~ω

e− 2kT · e−

n~ω kT

e−nx

ex 1 ex = = x x −x e −1 e (1 − e ) 1 − e−x 1 1 x x = e− 2 · = csch 1 − e−x 2 2 1 = sinh x 1 x −1 = x = 2 sinh 2 2 sinh 2 =

Finally we get:

2 sinh

x −1 2

= −kT ln Z = −kT · ln

2 sinh

x −1 x ~ω = kT ln(2 sinh ) = kT ln 2 sinh 2 2 2kT 63

b) Lets write expression for F : F (∆, T ) =

X 1 B∆2 + kB T ln[2 sinh(~ωK /2kB T )] 2

We assume that ωK can be expressed as difference between initial frequency and final: ωK = ω0 − ωγ∆ To find a minimal value of F respect to ∆, it is necessary to calculate the first derivative: ∂F ∂∆ ∂F ∂∆

= =

B∆

0 1 ~(ω0 − ωγ∆) ~γω · 2 cosh B∆ + kT · − ~(ω0 − ωγ∆) 2kT 2kT 2 sinh 2kT ~(ω0 − ωγ∆) 2 cosh P ~γω 2kT · − B∆ + ~(ω0 − ωγ∆) 2 2 sinh 2kT ~(ω0 − ωγ∆) cosh P ~ω 2kT B∆ − γ · ~(ω0 − ωγ∆) 2 sinh 2kT P1 ~(ω0 − ωγ∆) B∆ − γ ~ω coth 2 2kT P1 ~(ω0 − ωγ∆) ~ω coth γ 2 2kT P

For next part lets take: U (T )

P1 ~ω coth 2

B∆

γU (T )

∆

γU (T ) B

~(ω0 − ωγ∆) 2kT

c) Lets have next relations: ∂ω ω

−γ∆

∆

∂V V

∂ω ω

−γ

∂ ln ω

−γ∂ ln V

−

∂V V

∂ ln ω ∂ ln V

In Debye approximation: ~ωD θ = kT T So, ω is proportional to the Debye temperature θ thus finally: γ=−

∂ ln θ ∂ ln V

Problem 6.1: Kinetic energy of electron gas Back to contents

Problem specification Show that the kinetic energy of a three-dimensional gas of N free electrons at 0 K is U0 =

3 N F 5

Solution Fermi energy for electrons in 3D is expressed by next relation: ~2 F = 2m

3π 2 N V

To find energy it is necessary to integrate over all occupied states: U0

= = = = =

2 R N ~2 3π 2 N 3 dN = dN = F 0 0 2m V 2 ~2 3π 2 3 R N 2 N 3 dN = 0 2m V 2 5 ~2 3π 2 3 3N 3 · = 2m V 5 2 ~2 3π 2 N 3 3 · N= 2m V 5 3 N F 5

Problem 6.2: Pressure and bulk modulus of an electron gas Back to contents

Problem specification a) Derive a relation connecting the pressure and volume of an electron gas at 0 K. Hint: use the result of Problem 1 and the relation between F and electron concentration. The result may be written as p = 32 (U0 /V ). b) Show that the bulk modulus B = −V (∂p/∂V ) of an electron gas at 0 K is B = 5p/3 = 10U0 /9V . c) Estimate for potassium, using Table 1, the value of the electron gas contribution to B.

Solution a) Using expression for pressure from thermodynamics(it is correct for case when T is 0 K): P

∂U0 ∂V

= = = =

−

∂U0 ∂V

2/3 3 ~2 3π 2 N 3 N F = N · 5 5 2m V −1/3 3 2 ~2 3π 2 N 3π 2 N N· · · − = 5 3 2m V V2 −1/3 3/3 2 ~2 3π 2 N 3π 2 N 1 3 N· · · − · = 5 3 2m V V V 2/3 3 2 ~2 3π 2 N 1 − N· · = 5 3 2m V V 2 U0 − 3 V

2 U0 3 V

b) Having expression for P it is possible to calculate bulk modulus B: ∂P −V ∂V 2 ∂U0 2U0 − = 2 3V ∂V 3V 2 2 U0 2U0 · − − = 3V 3 V 3V 2 4U0 6U0 − 2− = 9V 9V 2 10U0 − 9V2 ∂P 10U0 −V = −V − = ∂V 9V 2 10U0 9V

∂P ∂V

= = = =

= =

c) For potassium next parameters are necessary to obtain from Table 1: N

1.40 · 1022 cm−3

2.12 eV

Now it is possible to proceed with calculations: B

= =

10 U0 10 3 2 = · N F = N F = 9 V 9 5 3 2 · 1.40 · 1022 · 2.12 5.936 · 1022 −3 = = 1.979 · 1022 eV/cm 3 3

To compare this value with data of Table 3 Chapter 3 it is necessary to convert it to

1 eV

1.602 · 10−12 erg = 1.602 · 10−12 dyn · cm

1.979 · 1022 eV/cm

−3

· 1.602 · 10−12 dyn · cm = 3.17 · 1010

For K in Table 3 Chapter 3 value of B is 0.032 · 1012

dyn : cm2

dyn dyn = 0.0317 · 1012 2 cm2 cm

dyn , which is very close to obtained value. cm2

Problem 6.3: Chemical potential in two dimensions Back to contents

Problem specification Show that the chemical potential of a Fermi gas in two dimensions is given by:   πn~2   µ(T ) = kB T ln e mkB T − 1 for n electrons per unit area. Note: the density of orbitals of a free electron gas is two dimensions is independent of energy: D() = m/π~2 , per unit area of specimen.

Solution To find chemical potential, it is necessary to use density of states formula of electron gas: Z ∞ n= D()f ()d 0

Where D is known, f () is Fermi-Dirac distribution. As density of states of a free electron gas in 2D is not dependent on energy, it is possible to take it out from integral to obtain: n = nπ~2 m

1 m R∞ d −µ 0 2 π~ e kT + 1 R∞ d 0

−µ kT

−u Divide integral by expression e kT to obtain: nπ~2 m

R ∞ e−

−µ kT

−µ e− kT

This integral can be calculated by introduction of change:

d e−

x −µ kT

nπ~2 m nπ~2 m nπ~2 mkT nπ~2 mkT nπ~2 mkT

= e−

−µ kT

1 dx kT 1 R ∞ dx − kT 0 1 + x 1 − ln[x + 1]|∞ 0 kT h −µ i − ln e− kT + 1 |∞ 0 h ∞−µ i h 0−µ i − ln e− kT + 1 + ln e− kT + 1

µ ln e kT + 1

= − = = =

Taking exponentials in both parts to obtain: nπ~2

e mkT µ

e kT

= e kT + 1 nπ~2

= e mkT − 1 68

Taking logarithm of both parts to obtain finally: µ kT

nπ~2 ln e mkT − 1 nπ~2 µ = kT ln e mkT − 1 =

Problem 6.4: Fermi gases in astrophysics Back to contents

Problem specification a) Given M} = 2 × 1033 g for the mass of the Sun, estimate the number of electrons in the Sun. In a white dwarf star this number of electrons may be ionized and contained in a sphere radius 2 × 109 cm; find the Fermi energy of the electrons in electron volts. b) The energy of an electron in the relativistic limit >> mc2 is related to the wavevector as ≈ pc = ~kc. Show that the Fermi energy in this limit is F ≈ ~c(N/V )1/3 , roughly. c) If the above number of electrons were contained within a pulsar of radius 10 km, show that the Fermi energy would be ≈ 108 eV. This value explains why pulsars are believed to be composed largely of neutrons rather than of protons and electrons, for the energy release in the reaction n → p + e− is only 0.8 × 106 eV, which is not large enough to enable many electrons to form a Fermi sea. The neutron decay proceeds only until the electron concentration builds up enough to create a Fermi level of 0.8 × 106 eV, at which point the neutron, proton, and electron concentrations are in equilibrium.

Solution To take roughly, lets suppose that Sun is fully consists of hydrogen, which has mass of one mole 2 g. Each atom has one electron for the case of hydrogen. In one mole there is number of Avogadro atoms of hydrogen, so: Ne

2 · 1033 M} = = 0.166 · 1056 electrons M H 2 NA 2 · 6.02 · 10−23

Expression for Fermi energy: 2 2/3 ~2 3π N F = 2me V 4 3 4 V = πR = · 3.14 · (2 · 109 )3 = 4.188 · 8 · 1027 = 33.504 · 1027 cm3 3 3 2/3 2/3 (1.054 · 10−27 )2 3 · 9.869 · 0.166 · 1056 1.11 · 10−54 4.914 · 1056 F = = = 2 · 9.1 · 10−28 33.504 · 1027 18.2 · 10−28 33.504 · 1027 2/3 = 0.06 · 10−26 14.6 · 1027 = 0.06 · 10−26 · 5.97 · 1018 = 0.3582 · 10−8 erg = 2184.52 eV b) Expression for wave vector k: 1/3 3π 2 N k= V Inserting into expression of Fermi energy of electron in relativistic limit to obtain: 2 1/3 1/3 3π N N F = ~kc = ~c ≈ ~c V V

c) Using formula from b) and result from a) to obtain value: 2 1/3 3π N F = ~c V 4 V = π(10000)3 = 4.188 · 1015 m3 3 1/3 1/3 3 · 9.869 · 0.166 · 1056 4.914 · 1056 −25 F = 1.986 · 10−25 J · m = 1.986 · 10 J · m = 4.188 · 1015 4.188 · 1015 1/3 = 1.986 · 10−25 1.173 · 1041 J = 1.986 · 10−25 · 4.895 · 1013 J = =

9.721 · 10−12 J = 0.6 · 108 eV

Problem 6.5: Liquid He3 Back to contents

Problem specification The atom He3 has spin 21 and is a fermion. The density of liquid He3 is 0.081 g cm−3 near absolute zero. Calculate the Fermi energy F and the Fermi temperature TF .

Solution Fermi energy is determined as: F =

~2 2m

3π 2 N V

2/3

From initial conditions we know only density, but not the electron concentration. Necessary parameters to find is also a mass of one helium-3 atom, volume we will take equal to 1 cm3 . Molecular mass of He3 is 3 g/mol, so: N

~ = F

= =

g 23 0.081 cm mol−1 num ρNA 3 · 6.022 · 10 = = 0.162 · 1023 g M 3 mol cm3 g 3.016 mol M −23 g = 1 = 0.5001 · 10 NA 6.022 · 1023 mol

1.054 · 10−34 J · s = 1.054 · 10−27 erg · s 2/3 2/3 3 · π 2 · 0.162 · 1023 num (1.054 · 10−27 erg · s)2 1.11 · 10−54 cm3 · = · 3 · 9.869 · 0.162 · 1023 = −23 3 −23 2 · 0.5001 · 10 g 1 cm 1.0002 · 10 1.109 · 10−31 · (479.6334 · 1021 )2/3 = 1.109 · 10−31 · 61.273 · 1014 = 67.951 · 10−17 erg = 6.7951 · 10−16 erg

Now lets find Fermi temperature. As known, it can be found as: TF

F 6.7951 · 10−16 erg = = 4.924 K kB 1.38 · 10−16 erg K−1

Problem 6.6: Frequency dependence of the electrical conductivity Back to contents

Problem specification Use the equation m

dv v + dt τ

= −eE for the electron drift velocity v to show that the conductivity at frequency ω is σ(ω) = σ(0)

1 + iωτ , 1 + (ωτ )2

where σ(0) = ne2 τ /m.

Solution Here it is necessary to have a final expression dependent of frequency, so lets initial wave will be in exponential form: v

= v0 e−iωt

E0 e−iωt

Inserting into motion equation to obtain:

dv v m + dt τ −iωt v e 0 −iωt m −iωv0 e + τ mv −imωv + τ m − imω v τ v

= −eE = −eE0 e−iωt = −eE = −eE eE = −m − imω τ

Knowing that j = σE: nqv v −

σE ne σ

= σE → nev = −σE = −

σE ne

eE = −m − imω τ ne2 ne2 τ ne2 τ 1 = m = = = σ(0) m − imωτ m(1 − iωτ ) 1 − iωτ − imω τ

Finally multiply ratio by 1 + iωτ to obtain final result: σ(ω) = σ(0)

1 + iωτ 1 + (ωτ )2

So as we see from result, at high frequencies conductivity is decreases, gives the reason for skin-effect.

Problem 6.9: Static magnetoconductivity tensor Back to contents

Problem specification For the drift velocity theory of (51), show that the static current density can be written in matrix form as      jx      σ0 j  =  y  1 + (ωc τ )2     jz

 1   ω τ  c   0

−ωc τ 1 0

 Ex       E  0   y     Ez 1 + (ωc τ )2 0

In the high magnetic field limit of ωc τ 1, show that σyx = nec/B = −σxy In this limit σxx = 0, to order 1/ωc τ . The quantity σyx is called the Hall conductivity.

Solution a) Writing out expression (51):

d 1 B m + vx = −e Ex + vy dt τ c 1 B d + vy = −e Ey − vx m dt τ c 1 d + vz = −eEz m dt τ dv From problem specification, current is static, so = 0, and equation system can be rewritten as: dt mvx B = −e Ex + vy τ c B mvy = −e Ey − vx τ c mvz = −eEz τ Expressing projections of velocities: B eτ eτ Ex + vy = − Ex − ωc τ vy vx = − m c m eτ B eτ vy = − Ey − vx = − Ey + ωc τ vx m c m eτ vz = − Ez m Inserting vy into first motion equation (respectively vx ): B eτ mvx = −e Ex + − Ey + ωc τ vx τ c m eBτ mvx eB = − ωc τ vx − e Ex − Ey τ c mc mvx eB + ωc τ vx = −e (Ex − ωc τ Ey ) τ c m eB vx + ωc τ = −e (Ex − ωc τ Ey ) τ c 73

(1)

Dividing last expression by m and multiplying by τ to make it closer to desired: vx

1 + ωc2 τ τ

vx 1 + ωc2 τ 2

e (Ex − ωc τ Ey ) m eτ = − (Ex − ωc τ Ey ) m eτ = − (Ex − ωc τ Ey ) m (1 + ωc2 τ 2 )

= −

Knowing that current density is j = −nev and σ0 =

−

ne2 τ obtaining: m

= −

jx ne

= −

jx ne

eτ (Ex − ωc τ Ey ) m (1 + ωc2 τ 2 ) ne2 τ (Ex − ωc τ Ey ) m (1 + ωc2 τ 2 ) σ0 (Ex − ωc τ Ey ) 1 + ωc2 τ 2

It is easy to show expressions for jy and jz : jy

σ0 (Ey + ωc τ Ex ) 1 + ωc2 τ 2 For vz lets take motion equation and express velocity from j:

−

jz ne jz

eτ Ez m 2 ne τ Ez = σ0 Ez m

= − =

Now all expressions for current density are fit static magnetoconductivity tensor. b) From static magnetoconductivity tensor it is seen that: σyx =

σ0 · ωc τ 1 + (ωc τ )2

In high magnetic field limit ωc τ 1, so denominator can be expressed as:

σyx

ne2 τ σ0 ωc τ σ0 nec = = = m = eB (ωc τ )2 ωc τ B τ mc

From the same tensor it is obvious that: σxy = −σyx = −

nec B

Problem 6.10: Maximum surface resistance Back to contents

Problem specification Consider a square sheet of side L, thickness d, and electrical resistivity ρ. The resistance measured between opposite edges of the sheet is called the surface resistance: Rsq = ρL/Ld = ρ/d, which is independent of the area L2 of the sheet. (Rsq is called the resistance per square and is expressed in ohms per square, because ρ/d has dimensions of ohms.) If we express ρ by (44), then Rsq = m/nde2 τ . Suppose now that the minimum value of the collision time is determined by scattering from the surfaces of the sheet, so that τ ≈ d/vF , where vF is the Fermi velocity. Thus the maximum surface resistivity is Rsq ≈ mvF /nd2 e2 . Show for a monatomic metal sheet one atom in thickness that Rsq ≈ ~/e2 = 4.1kΩ, where 1kΩ is 103 ohms.

Solution Maximum surface resistivity is defined as: Rsq ≈

mvf nd2 e2

Problem is solved by making assumption that: n = Rsq

Rsq

1 d3 mvf d e2 ~ ~ = λ d ~ ~ 1.054 · 10−34 1.054 · 10−34 dd = = = = 0.41 · 104 Ω = 4.1kΩ e2 e2 (1.602 · 10−19 )2 2.566 · 10−38

mvF =

Problem 7.1: Square lattice, free electron energies Back to contents

Problem specification a) Show for a simple square lattice (two dimensions) that the kinetic energy of a free electron at a corner of the first zone is higher that of an electron at midpoint of a side face of the zone by a factor 2. b) What is the corresponding factor for a simple cubic lattice (three dimensions)? c) What bearing might be result of (b) have on the conductivity of divalent metals?

Solution

a) Lets take a look at the graphical representation of first Brillouin zone in 2D and place on the graph points corresponding to problem specification. First point is situated in corner of a zone, so it has a coordinates ( πa ; πa ) - here positive corner is just a suitable choice. Let k~1 will be connecting center of the first zone with it's corner. Next point is situated on the middle of the side, so it has coordinates ( πa ; 0) and vector connecting this point with the center of the first zone will be called k2 . Now, energy of a free electron can be expressed as: =

~2 k 2 2m

(2)

Where k corresponds to module of wave vector for specified electron state, so: k12

k12

1 2

π2 π2 π2 + 2 =2 2 2 a a a π2 π2 2 |k2 | = 2 + 0 = 2 a a k12 =2 k22

|k1 |2 =

b) It is easy to see from previous point that for a 3D situation of SC lattice, we will have a factor 3 as k1 will have 3 components, and k2 still have only one component. c) Divalent metals - metals with valence of 2, so they have not filled valence band. With the change described in (b) it is seen that difference in kinetic energy between corner of first Brillouin zone and middle of the face is increased. This means that energy of electrons will be higher so the conductivity of divalent metal will be a slightly higher.

Problem 7.2: Free electron energies in reduced zone Back to contents

Problem specification Consider the free electron energy bands of an fcc crystal lattice in the approximation of an empty lattice, but in the reduced zone scheme in which all k's are transformed to lie in the first Brillouin zone. Plot roughly in the [111] direction the energies of all bands up to six times the lowest band energy at the zone boundary at k = (2π/a)( 12 , 21 , 12 ). Let this be the unit of energy. This problem shows why band edges need not necessarily be at the zone center. Several of the degeneracies (band crossings) will be removed when account is taken of the crystal potential.

Solution

Problem 7.3: Kronig-Penney model Back to contents

Problem specification (a) For the delta-function potential and with P 1, find at k = 0 the energy of the lowest energy band. (b) For the same problem find the band gap at k = π/a.

Solution a) Energy can be found from relation: ~2 k 2 2m Final expression for Kronig-Penney model can be taken as initial point for this problem: =

P sin Ka + cos Ka = cos ka Ka In this limit: P

→

0, sin Ka → Ka

So, having now: P sin Ka + cos Ka = 1 Ka P Ka + cos Ka = 1 Ka P

1 − cos Ka

K 2 a2 2 2P 2 K = a2 ~2 · 2P ~2 K 2 ~2 P a2 = = = 2m 2m ma2 b) To find the energy gap we should consider two energies levels and find Eg = E1 − E2 . For k = π/a: P

P sin Ka + cos Ka = −1 Ka Now lets consider magnitude of Ka is slightly exceeding zone boundary π on some small δ → 0, so Ka = π + δ. This value of Ka will represent a state with higher energy (E2 ), and for ground relative state we will take Ka = π; difference

between this states will give us a band gap value. Now lets try to find a value of δ: P sin(π + δ) + cos(π + δ) π+δ P (sin π cos δ + cos π sin δ) + (cos π cos δ + sin π sin δ) π+δ Pδ − − cos δ π Pδ π Pδ π P π δ

−1

1 − cos δ

= = =

Now it is possible to find energy gap: E1

∝

E2 Eg

~2 K 2 ~2 π 2 = 2m 2ma2 π+δ 2 ~2 ( ) ~2 (π 2 + 2πδ + δ 2 ) ~2 K 2 a = = 2m 2m 2ma2 P, P 1 → δ 2 → 0

~2 (π 2 + 2πδ) ~2 π 2 ~2 πδ = + 2ma2 2ma2 ma2 ~2 π 2 ~2 πδ ~2 π 2 = E2 − E1 = + − = 2ma2 ma2 2ma2 2 2P 2 2 ~ π π ~ πδ 2~ P = = = ma2 ma2 ma2 =

δ2 2 δ 2 2P π

Kittel Solid State Physics 3.4 Solution

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Kittel Solid State Physics 3.4 Solution

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