Kittel Charles - Introduction to Solid State Physics 8Th Edition - Solution Manual

CHAPTER 1 1. The vectors xˆ + yˆ + zˆ and − xˆ − yˆ + zˆ 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 θ = cos −1 1/ 3 = 90° + 19° 28' = 109° 28' .

2. The plane (100) is normal to the x axis. It intercepts the a' axis at 2a' and the c' axis at 2c' ; therefore the indices referred to the primitive axes are (101). Similarly, the plane (001) will have indices (011) when referred to primitive axes. 3. The central dot of the four is at distance

a

cos 60° a = a ctn 60° = cos 30° 3

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

⎛ a ⎞ ⎛c⎞ a =⎜ ⎟ +⎜ ⎟ , ⎝ 3 ⎠ ⎝2⎠ 2

or 2 2 1 2 a = c ; 3 4

c 8 = 1.633. a 3

1-1

CHAPTER 2

hkA is a plane defined by the points a1/h, a2/k, and a3 / A . (a) Two vectors that lie in the plane may be taken as a1/h – a2/k and a1 / h − a3 / A . But each of these vectors gives zero as its scalar product with G = ha1 + ka 2 + Aa3 , so that G must be perpendicular to the plane hkA . (b) If nˆ is the unit normal to the plane, the interplanar spacing is nˆ ⋅ a1/h . But nˆ = G / | G | , whence d(hkA) = G ⋅ a1 / h|G| = 2π / | G| . (c) For a simple cubic lattice G = (2π / a)(hxˆ + kyˆ + Azˆ ) , 1. The crystal plane with Miller indices

whence

1 G 2 h 2 + k 2 + A2 = = . d 2 4π 2 a2 1 3a 2 1 2. (a) Cell volume a1 ⋅ a 2 × a3 = − 3a 2 0

=

1 a 0 2 1 a 0 2 0 c

1 3 a 2 c. 2

xˆ (b) b1 = 2π

=

yˆ

4π 1 a 2 × a3 = − 3a 2 | a1 ⋅ a 2 × a3 | 3a c 2 0

zˆ

1 a 0 2 0 c

2π 1 ( xˆ + yˆ ), and similarly for b 2 , b3 . a 3 (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

VBZ = (2π)3

(a 2 × a 3 ) ⋅ (a 3 × a1 ) × (a1 × a 2 ) = (2π)3 / | (a1 ⋅ a 2 × a 3 ) | 3 | (a1 ⋅ a 2 × a 3 ) |

= (2π)3 / VC . 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

|F|2 =

1 − exp[−iM(a ⋅ ∆k)] 1 − exp[iM(a ⋅ ∆k)] ⋅ 1 − exp[−i(a ⋅ ∆k)] 1 − exp[i(a ⋅ ∆k)]

2 1 − cos M(a ⋅ ∆k) sin 12 M(a ⋅ ∆k) . = = 1 − cos(a ⋅ ∆k) sin 2 12 (a ⋅ ∆k)

(b) The first zero in sin

1 Mε occurs for ε = 2π/M. That this is the correct consideration follows from 2

1 1 1 sin M(πh + ε) = sin πMh cos Mε + cos sin Mε. πMh   

  

2 2 2 ± 1 zero, as Mh is an integer

5. S (v1 v 2 v 3 ) = f Σ e

−2πi(x j v1 +y j v 2 +z j v3 )

j

Referred to an fcc lattice, the basis of diamond is 000;

1 1 1 . Thus in the product 4 4 4

S(v1v 2 v3 ) = S(fcc lattice) × S (basis) , we take the lattice structure factor from (48), and for the basis

S (basis) = 1 + e

−i

1 π (v1 + v 2 + v3 ). 2

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.

6.

∞

f G = ∫ 4πr 2 (πa 0 Gr)−1 sin Gr exp ( −2r a 0 ) dr 3

0

= (4 G 3a 0 ) ∫ dx x sin x exp ( −2x Ga 0 ) 3

= (4 G 3a 0 ) (4 Ga 0 ) (1 + r G 2 a 0 ) 2 3

2

16 (4 + G 2 a 0 ) 2 . 2

The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G4 for

Ga 0 >> 1.

7. (a) The basis has one atom A at the origin and one atom B at

1 a. The single Laue equation 2

a ⋅ ∆k = 2π × (integer) 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) = fA + 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 1 a and the diffraction condition ( a ⋅ ∆k ) = 2π × (integer). 2 2

2-3

CHAPTER 3 1. E = (h/

2

2M) (2π λ ) 2 = (h/ 2 2M) (π L) 2 , with λ = 2L. 6

U(R) = 2Nε[9.114( σ R )12 − 12.253(σ R)6 ]. At equilibrium R 0 = 1.488σ6 , and U(R 0 ) = 2Nε( − 2.816).

2. bcc:

6

U(R) = 2Nε[12.132( σ R )12 − 14.454(σ R)6 ]. At equilibrium R 0 = 1.679σ6 , and U(R 0 ) = 2Nε( − 4.305). Thus the cohesive energy ratio bcc/fcc = 0.956, so that the fcc structure is

fcc:

more stable than the bcc.

3.

| U | = 8.60 Nε = (8.60) (6.02 × 1023 ) (50 × 10−16 ) = 25.9 × 109 erg mol = 2.59 kJ mol.

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

αe2 (1.75) (4.80 × 10−10 ) 2 = = 11.0 ×10−12 erg, R 3.66 × 10−8 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.

⎛ A αq 2 ⎞ U(R) = N ⎜ n − ⎟ ; α = 2 log 2 = Madelung const. R ⎠ ⎝R In equilibrium

⎛ nA αq 2 ⎞ ∂U nA n = N ⎜ − n +1 + 2 ⎟ = 0 ; R 0 = , ∂R αq 2 R0 ⎠ ⎝ R0 and

U(R 0 ) = −

Nαq 2 1 (1 − ). R0 n

3-1

b.

U(R 0 -R 0 δ) = U ( R 0 ) +

1 ∂2U 2 R 0 ( R 0δ ) + . . . , 2 2 ∂R

bearing in mind that in equilibrium (∂U ∂R) R

0

= 0.

⎛ n(n + 1)A 2αq 2 ⎞ ⎛ (n + 1) αq 2 2αq 2 ⎞ ⎛ ∂2U ⎞ − = N⎜ − ⎜ 2 ⎟ = N⎜ n+2 3 ⎟ 3 3 ⎟ R0 ⎠ R0 R0 ⎠ ⎝ ∂R ⎠R 0 ⎝ R0 ⎝ For a unit length 2NR0 = 1, whence 2 ⎛ ∂2U ⎞ αq 2 (n − 1) q 2 log 2 2 ∂ U . − = = (n 1) ; C R ⎜ 2⎟ = 0 4 2 ∂R 2 R 2R 0 R ⎝ ∂R ⎠R 0 0 0

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

x 2 e− x = 8.53 × 10−3. By trial and error we find x  9.2, or R0 = 3.00 Å. The actual KCl structure has R0 (exp) = 3.15 Å . For the imagined structure the cohesive energy is

U=

-αq 2 ⎛ p ⎞ U ⎜1⎟ , or 2 =-0.489 R0 ⎝ R0 ⎠ q

in units with R0 in Å. For the actual KCl structure, using the data of Table 7, we calculate

U = −0.495, q2

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),

3S11 = 2 (C11 − C12 ) + 1 (C11 + C12 ). 2

2

Thus Y = (C11 + C12 C11 − 2C12 ) (C11 + C12 ); further, also from (37), eyy = S21Xx, whence

σ = e yy e xx = S21 S11 = − C12 (C11 + C12 ).

9. For a longitudinal phonon with K || [111], u = v = w.

3-2

ω2ρ = [C11 + 2C44 + 2(C12 + C 44 )]K 2 3, or v = ω K = [(C11 + 2C12 + 4C44 3ρ )]1 2 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 e xx = −e yy =

1

2

e in (43). Then

U = 1 2 C11 ( 1 4 e2 + 1 4 e2 ) − 1 4 C12 e2 = 1 2 [ 1 2 (C11 − C12 )]e 2 ⎛ n(n + 1)A 2αq 2 ⎞ ⎛ (n + 1) αq 2 2αq 2 ⎞ ⎛ ∂2U ⎞ = N⎜ − = N⎜ − is the effective shear so that ⎜ n+2 3 ⎟ 3 3 ⎟ 2 ⎟ R0 ⎠ R0 R0 ⎠ ⎝ ∂R ⎠R 0 ⎝ R0 ⎝ constant. 12a. We rewrite the element aij = p – δij(λ + p – q) as aij = p – λ′ δij, where λ′ = λ + p – q, and δij 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

u (r, t) = u 0 ei[(K

3) (x + y + z) −ωt]

;

v(r, t) = v0 ei[. . . . .] ; w(r, t) = w 0 ei[. . . . .] , as the displacements for waves in the [111] direction. On substitution in (57) we obtain the desired equation. Then, by (a), one root is

ω2ρ = 2p + q = K 2 (C11 + 2C12 + 4C 44 ) / 3, and the other two roots (shear waves) are

ω2ρ = K 2 (C11 − C12 + C44 ) / 3. 13. Set u(r,t) = u0ei(K·r – t) and similarly for v and w. Then (57a) becomes 2

2

2

ω2ρu 0 = [C11K y + C44 (K y + K z )]u 0 + (C12 + C44 ) (K x K y v 0 + K x K z w 0 ) and similarly for (57b), (57c). The elements of the determinantal equation are

3-3

M11 = C11K x + C 44 (K y + K z ) − ω 2 ρ; 2

2

2

M12 = (C12 + C44 )K x K y ; M13 = (C12 + C44 )K x K z . and so on with appropriate permutations of the axes. The sum of the three roots of sum of the diagonal elements of the matrix, which is

ω2ρ

is equal to the

(C11 + 2C44)K2, where 2

2

2

K 2 = K x + K y + K z , whence v1 + v 2 + v3 = (C11 + 2C44 ) ρ , 2

2

2

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, C442, C443; thus one criterion of stability is C44 > 0. The next minor is C11 C44 3, or C11 > 0. Next: C443 (C112 – C122), 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

1 2 Mu S . The force 2

between atoms s and s+1 is –C(us – us+1); the potential energy associated with the stretching of this bond is

1 C(u s − u s+1 ) 2 , and we sum over all bonds to obtain the total potential energy. 2 b. The time average of

1 1 2 Mu S is Mω2 u 2 . In the potential energy we have 2 4

u s +1 = u cos[ωt − (s + 1)Ka] = u{cos(ωt − sKa) ⋅ cos Ka + sin (ωt − sKa) ⋅ sin Ka}. Then

u s − u s +1 = u {cos(ωt − sKa) ⋅ (1 − cos Ka) − sin (ωt − sKa) ⋅ sin Ka}.

We square and use the mean values over time:

< cos 2 > = < sin 2 > =

1 ; < cos sin > = 0. 2

Thus the square of u{} above is

1 2 u [1 − 2cos Ka + cos 2 Ka + sin 2 Ka] = u 2 (1 − cos Ka). 2 The potential energy per bond is cos Ka) this is equal to

1 2 Cu (1 − cos Ka), and by the dispersion relation ω2 = (2C/M) (1 – 2

1 Mω2 u 2 . Just as for a simple harmonic oscillator, the time average potential 4

energy is equal to the time-average kinetic energy. 2. We expand in a Taylor series

⎛ ∂2u ⎞ ⎛ ∂u ⎞ 1 u(s + p) = u(s) + pa ⎜ ⎟ + p 2 a 2 ⎜ 2 ⎟ + " ; ⎝ ∂x ⎠s 2 ⎝ ∂x ⎠s On substitution in the equation of motion (16a) we have

M

∂2u ∂2u 2 2 = ( Σ p a C ) , p p>0 ∂t 2 ∂x 2

which is of the form of the continuum elastic wave equation with

4-1

v 2 = M −1 Σ

p >0

p 2a 2Cp .

3. From Eq. (20) evaluated at K = π/a, the zone boundary, we have

−ω2 M1u = −2Cu ; −ω2 M 2 v = −2Cv . 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.

4.

ω2 =

sin pk 0 a 2 A Σ (1 − cos pKa) ; p >0 M pa

∂ω2 2A = Σ sin pk 0a sin pKa ∂K M p >0 1 (cos (k 0 − K) pa − cos (k 0 + K) pa) 2 When K = k0,

∂ω2 A = Σ (1 − cos 2k 0 pa) , ∂K M p>0 which in general will diverge because Σ 1 → ∞. p 5. By analogy with Eq. (18),

Md 2 u s dt 2 = C1 (vs − u s ) + C2 (vs −1 − u s ); Md 2 vs dt 2 = C1 (u s − vs ) + C2 (u s +1 − vs ), whence −ω2 Mu = C1 (v − u) + C2 (ve − iKa − u); −ω2 Mv = C1 (u − v) + C2 (ueiKa − v) , and (C1 + C2 ) − Mω2

−(C1 + C2 e − iKa )

−(C1 + C2 eiKa )

(C1 + C2 ) − Mω2

=0

For Ka = 0, ω2 = 0 and 2(C1 + C2 ) M. For Ka = π, ω2 = 2C1 M and 2C 2 M.

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

M  4 ×10−23 g and R  2 × 10−8 cm; thus ωD  (5 × 10−10 ) (3 × 10−46 )1 2  3 × 1013 s −1 (c) The maximum phonon wavevector is of the order of 108 cm–1. If we suppose that ω0 is

sodium

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 ω = (2 / M)[γ (1 − cos Ka) + Σ (−1) (2e / p a )(1 − cos pKa)] . 2

P

2

3 3

p>0

At the zone boundary ω2 = 0 if

1 + σ Σ (−1) P [1 − (−1) P ]p −3 = 0 , p>0

or if σ Σ[1 − ( −1) ]p

−3

= 1 . The summation is 2(1 + 3–3 + 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 0

is just that for log 2, whence the root is σ = 1/(2 log 2) = 0.7213.

4-3

CHAPTER 5 1.

1 ω = ωm | sin Ka|. We solve this for K to obtain 2 2 −1 K = (2/a) sin (ω / ωm ) , whence dK/dω = (2 / a)(ωm − ω2 ) −1/ 2 and, from (15), D(ω)

(a) The dispersion relation is

= (2L/πa)(ωm − ω2 ) −1/ 2 . This is singular at ω = ωm. (b) The volume of a sphere of radius K in 3 3/2 Fourier space is Ω = 4πK / 3 = (4π / 3)[(ω0 − ω) / A] , and the density of orbitals near ω0 is 2

D(ω)= (L/2π)3 | dΩ/dω |= (L/2π)3 (2π / A 3/2 )(ω0 − ω)1/ 2 , provided ω < ω0. It is apparent that D(ω) vanishes for ω above the minimum ω0. 2. The potential energy associated with the dilation is

1 1 1 B(∆V/V) 2 a 3 ≈ k BT . This is k BT and not 2 2 2

3 k BT , because the other degrees of freedom are to be associated with shear distortions of the lattice cell. 2 2 −47 −24 3 Thus < ( ∆V) > = 1.5 × 10 ;(∆V) rms = 4.7 × 10 cm ; and ( ∆V) rms / V = 0.125 . Now 3∆a/a ≈ ∆V/V , whence (∆a) rms / a = 0.04 . 3.

(a)

/ ρV) Σω−1 , < R 2 > = (h/2

where

from

(20)

for

a

Debye

spectrum

Σω−1

= ∫ dω D(ω)ω−1 = 3VωD / 4π3 v3 , whence < R 2 > = 3h/ ωD / 8π2 ρv3 . (b) In one dimension from −1 (15) we have D(ω) = L/πv , whence ∫ dω D(ω) ω diverges at the lower limit. The mean square 1 2 / strain in one dimension is < (∂R/∂x) 2 > = ΣK 2 u 0 = (h/2MNv) ΣK 2 2 2 / = (h/2MNv) (K D / 2) = h/ ωD / 4MNv3 . 2

2

4. (a) The motion is constrained to each layer and is therefore essentially two-dimensional. Consider one plane of area A. There is one allowed value of K per area (2π/L)2 in K space, or (L/2π)2 = A/4π2 allowed values of K per unit area of K space. The total number of modes with wavevector less than K is, with ω = vK,

N = (A/4π2 ) (πK 2 ) = Aω2 / 4πv 2 . The density of modes of each polarization type is D(ω) = dN/dω = Aω/2πv2. The thermal average phonon energy for the two polarization types is, for each layer,

U = 2∫

ωD

0

where ωD is defined by N =

ωD Aω =ω dω , 2 0 2πv exp(hω/τ) − 1

D(ω) n(ω,τ) =ω dω = 2∫

∫

ωD

D

D(ω) dω . In the regime =ωD >> τ , we have

U≅

2Aτ3 2πv 2 = 2

∫

∞

0

5-1

x2 dx. ex − 1

Thus the heat capacity C = k B ∂U/∂τ ∝ T . 2

(b) If the layers are weakly bound together, the system behaves as a linear structure with each plane as a vibrating unit. By induction from the results for 2 and 3 dimensions, we expect C ∝ T . But this only holds at extremely low temperatures such that τ h/ ω is just the term to the right of the summation symbol, so that B∆ = γU (T) . (c) By definition of γ, we have δω / ω = −γδV/V , or d log ω = −δ d log V . But θ ∝ ωD , whence d log θ = −γ d log V . ∂F/∂∆ = 0 becomes B∆ = γΣ

5-2

CHAPTER 6

h/ 2 2 1. The energy eigenvalues are ε k = k . The mean value over the volume of a sphere in k space is 2m

=

h/ 2 ∫ k 2 dk ⋅ k 2 3 h/ 2 2 3 = ⋅ k F = εF . 2m ∫ k 2 dk 5 2m 5

The total energy of N electrons is

3 U0 = N ⋅ εF . 5 2a. In general p = –∂U/∂V at constant entropy. At absolute zero all processes are at constant entropy (the 23

3 h/ 2 ⎛ 3π2 N ⎞ = N ⎜ ⎟ , whence 5 2m ⎝ V ⎠

3 Third Law), so that p = −dU 0 dV, where U 0 = Nε F 5 p=

2 U0 ⋅ . (b) Bulk modulus 3 V 2

dp 2 dU 0 ⎞ 2 U 0 ⎛ 2 ⎞ U 0 10 U 0 ⎛ 2 U0 B = −V . = V⎜− + +⎜ ⎟ = ⎟= ⋅ 2 dV 9 V ⎝ 3 V 3V dV ⎠ 3 V ⎝ 3 ⎠ V (c) For Li,

U0 3 = (4.7 × 1022 cm −3 ) (4.7 eV) (1.6 × 10−12 erg eV) V 5 = 2.1 × 1011 erg cm −3 = 2.1 ×1011 dyne cm −2 , whence B = 2.3 × 1011 dyne cm–2. By experiment (Table 3.3), B = 1.2 × 1011 dyne cm–2. 3. The number of electrons is, per unit volume, n =

∫

∞

0

dε D(ε) ⋅

1 e

(ε−µ ) τ

+1

, where D(ε) is the density

of orbitals. In two dimensions

m ∞ 1 dε (ε−µ) τ 2 ∫0 πh/ +1 e m = 2 (µ + τ log (1 + e −µ τ )), πh/

n=

where the definite integral is evaluated with the help of Dwight [569.1].

2 × 1033  1057 nucleons, and roughly an equal number of electrons. In a 4a. In the sun there are −24 1.7 × 10 white dwarf star of volume

6-1

4π (2 × 109 )3 ≈ 3 × 1028 cm3 3 the electron concentration is ≈

1057 ≈ 3 × 1028 cm −3 . Thus 28 3 × 10

h/ 2 1 1 (3π2 n)2 3 ≈ 10−27 ⋅1020 ≈ 10−7 ergs, or ≈ 3.104 eV. (b) The value of kF is not 2m 2 2 1/3 / F  hc / 3 √n. (c) A affected by relativity and is ≈ n , where n is the electron concentration. Thus ε F  hck εF =

change of radius to 10 km = 106 cm makes the volume ≈ 4 × 1018 cm3 and the concentration ≈ 3 × 1038 cm – . Thus ε F ≈ 10

3

−27

(3.1010 ) (1013 ) ≈ 2.10 −4 erg ≈ 108 eV. (The energy is relativistic.)

5. The number of moles per cm3 is 81 × 10–3/3 = 27 × 10–3, so that the concentration is 16 × 1021 atoms cm– . The mass of an atom of He3 is (3.017) (1.661) × 10–24 = 5.01 × 10–24 g. Thus

3

ε F  [(1.1 × 10−54 ) 10−23 ][(30)(16) × 1021 ]2 3 ≈ 7 × 10−16 erg, or TF ≈ 5K. 6. Let E, v vary as e–iwt. Then

v=−

eE m eτE 1 + iωτ =− ⋅ , −iω + (1 τ) m 1+ (ωτ)2

and the electric current density is

j = n( − e)v =

ne 2 τ 1 + iωτ ⋅ E. m 1+ (ωτ)2

7. (a) From the drift velocity equation

iωv x = (e m)E x + ωc v y ; iωv y = (e m)E y − ωc v x . We solve for vx, vy to find

(ωc − ω2 )v x = iω( e m )E x + ωc ( e m )E y ; 2

(ωc − ω2 )v y = iω( e m )E y + ωc ( e m )E x . 2

We neglect the terms in ωc2. Because j = n(–e)v = σE, the components of σ come out directly. (b) From the electromagnetic wave equation

c 2∇ 2 E = ε∂ 2 E ∂t 2 , we have, for solutions of the form ei(kz – ωt), the determinantal equation

ε xx ω2 − c 2 k 2 ε xy ω2 = 0. ε yx ω2 ε yy ω2 − c 2 k 2

6-2

Here ε xx = ε yy = 1 − ωP

ω2 and ε xy = −ε yx = i ωcωp ω3 . The determinantal equation gives the

2

2

dispersion relation. 8. The energy of interaction with the ion is

e∫

r0

0

( ρ r ) 4πr 2dr = −3e2

2r0 ,

where the electron charge density is –e(3/4πr03). (b) The electron self-energy is

ρ2 ∫

r0

0

(

)(

)

dr 4πr 3 3 4πr 2 r −1 = 3e 2 5r0 .

The average Fermi energy per electron is 3εF/5, from Problem 6.1; because N V = 3 4πr0 , the average 3

is 3 ( 9π 4 )

23

h/ 2 10mr0 . The sum of the Coulomb and kinetic contributions is 2

U=−

1.80 2.21 + 2 rs rs

which is a minimum at

1.80 4.42 = 3 , or rs = 4.42 1.80 = 2.45. 2 rs rs The binding energy at this value of rs is less than 1 Ry; therefore separated H atoms are more stable. 9. From the magnetoconductivity matrix we have

jy = σ yx E x =

(

For ωcτ >> 1, we have σ yx ≅ σ0 ωc τ = ne τ m 2

ωc τ

1 + ( ωc τ )

2

σ0 E x .

) ( mc eBτ ) = neB c .

10. For a monatomic metal sheet one atom in thickness, n ≈ 1/d3, so that

R sq ≈ mv F nd 2 e 2 ≈ mv F d e 2 . If the electron wavelength is d, then mv F d ≈ h/ by the de Broglie relation and

R sq ≈ h/ e 2 = 137 c in Gaussian units. Now

6-3

R sq ( ohms ) = 10−9 c 2 R sq ( gaussian ) ≈ ( 30 )(137 ) ohms ≈ 4.1kΩ .

6-4

CHAPTER 7 1a. The wavevector at the corner is longer than the wavevector at the midpoint of a side by the factor √2. As ε ∝ k2 for a free electron, the energy is higher by (√ 2)2 = 2. b. In three dimensions the energy at a corner is higher by (√ 3)2 than at the midpoint of a face. c. Unless the band gap at the midpoint of a face is larger than the kinetic energy difference between this point and a corner, the electrons will spill over into the second zone in preference to filling up the corner states in the first zone. Divalent elements under these conditions will be metals and not insulators. 2. ε = h/ k 2m , where the free electron wavevector k may be written as the sum of a vector K in the reduced zone and of a reciprocal lattice vector G. We are interested in K along the [111] direction: from 2

2

Chap. 2, K = (2 π / a) (1,1,1) u, with 0 < u <

1 , will lie in the reduced zone. 2

The

lattice

G´s

of

the

reciprocal

G = ( 2 π a ) [( h − k + A ) xˆ + ( h + k − A ) yˆ +

h, k, A

are

any

integers.

are

given

( −h + k + A ) zˆ ],

(

Then ε = h/

( 2π a ) [( u + h − k + A ) + ( u + h + k − A ) + ( u − h + k + A ) 2

2

2

2

by where

2

2m

)

]. We now

have to consider all combinations of indices h, k, A for which the term in brackets is smaller than 6[3(1/2)2] or 9/2. These indices are (000);

(

) (

)

(

)

( 1 1 1 ) ; ( 100 ) , ( 0 10 ) ,

and

( 00 1 ) ;

(100), (010), and

(001); (111); 1 10 , 10 1 , and 0 1 1 ; (110), (101), and (011). 3. (a) At k = 0 the determinantal equation is (P/Ka) sin Ka + cos Ka = 1. In the limit of small positive P this equation will have a solution only when Ka  1. Expand the sine and cosine to obtain in lowest order

1 2 The energy is ε= ( Ka ) . 2 h/ 2 K 2 2m  h/ 2 P ma 2 . (b) At k = π/a the determinantal equation is (P/Ka) sin Ka + cos Ka = –1. In the same limit this equation has solutions Ka = π + δ, where δ  1 . We expand to obtain 1 ( P π )( −δ ) + ⎛⎜ −1 + δ2 ⎞⎟ = −1, which has the solution δ = 0 and δ = 2P/π. The energy gap is 2 ⎠ ⎝ E g = h/ 2 2ma 2 ( 2πδ )  h/ 2 2ma 2 ( 4P ) . P

(

)

(

)

4. (a) There are two atoms in the basis, and we label them a and b. Then the crystal potential may be

⎛ ⎝

1 1 1 ⎞ a, y + a, z + a ⎟ and the Fourier transform has  4 4 4 ⎠ 1 i( G x + G y + G z ) a ⎞ ⎛ 4 ˆ then the exponential is = U1G ⎜ 1 + e ⎟ . If G = 2Ax,    ⎝ ⎠

written as U = U1 + U 2 = U1 ( r ) + U1 ⎜ x + components U G = U1G + U 2G 

1 i Aa

e2





= eiπ = −1, and U G = 2A = 0, so that this Fourier component vanishes. Note that the quantity in 

parentheses above is just the structure factor of the basis. (b) This follows directly from (44) with U set equal to zero. In a higher order approximation we would go back to Eq. (31) where any non-vanishing U G enters. 

7-1

5. Let k = K + iH ; λ ±1 =

2 ⎤ h/ 2 ⎡⎛ 1 ⎞ 2 G ⎢⎜ ⎟ ± iGH − H ⎥ . 2m ⎣⎢⎝ 2 ⎠ ⎦⎥

The secular equation (46) is now

λ1 − ε

U

U

λ −1 − ε

= 0,

h/ 2 ⎛ 1 ⎞ σ =ε− ⎜ G⎟ , 2m ⎝ 2 ⎠ 2

and for H > 1

the

magnetoconductivity tensor (6.64) reduces to

⎛ Qe−2 ne τe ⎜ −1 Qe σ ≈ m e ⎜⎜ ⎝ 0 2

−Qe−1 0 ⎞ ⎟ pe 2 τh Qe−2 0 ⎟ + mh 0 1 ⎟⎠

⎛ Q −h2 Q −h1 ⎜ −1 −2 ⎜ −Q h Q h ⎜ 0 0 ⎝

0⎞ ⎟ 0⎟ . 1 ⎟⎠

We can write nec Qe/B for ne τe / m e and pec Qh/B for pe τh / m h . The strong field limit for σyx 2

2

follows directly. The Hall field is obtained when we set

jy = 0 =

⎛ n ec ⎡ p ⎞ ⎤ + ⎢(n − p) E x + ⎜ ⎟ Ey ⎥ . H⎣ ⎝ Qe Q h ⎠ ⎦

The current density in the x direction is

jx =

⎤ ec ⎡⎛ n p ⎞ + ⎢⎜ ⎟ E x − (n − p) E y ⎥ ; B ⎣⎝ Q e Q h ⎠ ⎦

using the Hall field for the standard geometry, we have

ec ⎡⎛ n p ⎞ (n − p) 2 jx = ⎢⎜ + ⎟+ H ⎢⎝ Q e Q h ⎠ ⎛ n p + ⎜ ⎢⎣ ⎝ Qe Q h

8-2

⎤ ⎥ Ex . ⎞⎥ ⎟⎥ ⎠⎦

CHAPTER 9 1.

2a. −

−

π = 0.78 ×108 cm −1 a

π b

= 0.78 × 108 cm -1 −

π = 1.57 ×108 cm −1 a

9-1

π = 0.7 8 ×1 08 cm −1 a

b.

N = 2×

πk 2F

( 2π / k )

2

n = N/L2 = k F2 / 2π k F = 2πn 1 n = × 1016 els/cm 2 8 k F = 0.89 ×108 cm −1 c.

3a. In the hcp structure there is one atom whose z coordinate is 0 and one at the basis for G c =



2π zˆ is c 

1 c . The structure factor of 2

SGc (basis) = 1 + e− iπ = 1 − 1 = 0, 

so that by the same argument as in Problem 9.4 the corresponding component UG c of the crystal potential



is zero. b. But for U 2G c the structure factor is 

S2Gc (basis) = 1 + e−i2 π = 2. 

c. The two valence electrons can just fill the first BZ. All we need is an adequate energy gap at the zone boundary and for simple hex. there is no reason against a gap. d. In hcp there will be no gap (at least in lowest order) on the top and bottom faces of the BZ, by the argument of part a.

9-2

4.

dk e 5a. h/  = − v × B; dt c  / hGc T= evB 10−27 erg sec) (2 × 108 cm −1 ) (3 × 1010 cm s −1 )  (5 × 10−10 esu) (108 cm d −1 ) (103 gauss)  1.2 ×10−10 sec.

9-3

b. The electron moves in a direction normal to the Fermi surface -- more or less in a straight line if the Fermi surface is close to planar in the region of interest. The magnetic field puts a wiggle on the motion, but the field does not make the electron move in a helix, contrary to the behavior of a free electron. 6a.

Region I:

⎛ h/ 2 d 2 ⎞ − U 0 ⎟ ψ = εψ ⎜− 2 ⎝ 2m dx ⎠ h/ 2 k 2 ψ = A cos kx ; ε = − U 0 (*) 2m Region II:

h/ 2 d 2 ψ = εψ − 2m dx 2

ψ = B e

Boundary condition

− qx

h/ 2 q 2 ; ε=− 2m

(*)

1 dψ continuous. ψ dx k tan (ka / 2) = q ,

(**)

with k and q related to ε as above. b. The lazy way here is to show that the ε’s in the equations marked (*) above are equal when k and q are connected by (**), with ε = –0.45 as read off Fig. 20. This is indeed so. 7a. ∆ (

1 2πe , where S = πkF2, with kF = 0.75 × 108 cm–1 from Table 6.1, for potassium. Thus )= / S H hc

1 2 ∆( )   0.55 × 10−8 G −1. 2 H 137 k F e

9-4

b.

ωc R = vF ; R =

/ c v F mc hk = F eB eB

 0.5 × 10−3 cm π R 2  0.7 ×10−6 cm 2 .

H1 = (h/ / m) k ⋅ p . Then (18) is an eigenfunction of H 0 with   2 2 the eigenvalue ε n (0) + h/ k / 2m . In this representation the diagonal matrix element of H1 is equal to

8. Write (17) as H = H 0 + H1 , where

(h/ / m) ∫ dV u 0 (r) k ⋅ p U 0 (r). In a cubic crystal U 0 (r) will be even or odd with respect to the      inversion operation r → − r , but p is an odd operator. It follows that the diagonal matrix element    vanishes, and there is no first-order correction to the energy. The function U k (r) to first order in H1 is  

< j0 | H1 | n0 > U k (r) = U 0 (r) + Σ′ ,    j ε n (0) − ε j (0) and the energy to second order is

|< n0 | k ⋅ p | j0 >|2   / ε n (k) = ε n (0) + (hk) / 2m + (h/ / m) Σ′ .  ε n (0) − ε j (0) j 2

2

The effective mass ratio is the coefficient of h/ k / 2m , or 2

2

|< n0 | p | j0 >|2 m 2 ′ =1 + Σ .  m* m j ε n (0) − ε j (0) 9a.

∫ dV

w*(r − rn ) w (r − rm )

= N −1 Σ Σ eik'⋅rn e − ik⋅rm k

k'

∫ dVψ

* k'

(r)ψ k (r)

= N −1 Σ eik⋅(rn − rm ) k

where the summation is zero unless n = m, when it is equal to N. b. w(x − x n ) = N

−1 2

U 0 (x) Σ eik(x − x n ) . The summation is k

equal to

9-5

N 2

∫

i2πp (x − x n ) Na



iπ (x − x n ) a

−iπ (x − x n ) a

Σe P

e

i2πp (x − x n ) Na dp

−N 2

=

sin [ π( x − x n ) a ] , π (x − x n ) Na

w (x − x n ) = N1 2 u 0 (x)

sin [ π( x − x n ) a ] . π (x − x n ) a

=

e

−e i2π( x − x n ) Na

whence

10a. jy = σ0 (Q–1 Ex + sEy) = 0 in the Hall geometry, whence Ey = – Ex/sQ. b. We have jx = σ0 (Q–2 Ex – Q–1 Ey), and with our result for Ey it follows that

jx = σ 0 (Q −2 + s −1Q −2 ) E x , whence ρ = E x jx = (Q

2

σ0 )

s . s +1

9-6

CHAPTER 10 d2B 1 = B; this is the London equation. The proposed solution is seen directly to dx 2 λ 2 ⎛ 1 ⎞ satisfy this and to satisfy the boundary conditions B ⎜ ± δ ⎟ = Ba . (b) For δ < < λL, ⎝ 2 ⎠ 1a.

2

x 1⎛ x ⎞ cosh = 1 + ⎜ ⎟ + … 2 ⎝ λL ⎠ λ 1⎛ δ ⎞ δ cosh = 1+ ⎜ ⎟ +… 2λ 2 ⎝ 2λ ⎠ 2

(

)(

)

therefore B ( x ) = Ba − Ba 1 8λ 2 δ 2 − 4x 2 .

2a. From (4), dFS = −MdB a at T = 0. From Problem 1b, M(x) = −

(

)

(

)

1 1 ⋅ 2 Ba ⋅ δ2 − 4x 2 , 4π 8λ

whence FS ( x,Ba ) − FS ( 0 ) =

1 2 δ 2 − 4x 2 Ba . 2 64πλ

b. The average involves

∫ ( 1 2δ

0

1 3 4 δ3 δ − ⋅ δ − 4x dx 2 3 8 = 2 δ2 , = 1 1 3 δ δ 2 2 2

2

)

whence 1 2⎛ δ⎞ ∆F = Ba ⎜ ⎟ , for δ > 2 ; also, ω2 > λ c c λ c2 2

Thus the normal electrons play no role in the dispersion relation in the low frequency range. 4. The magnetic influence of the core may be described by adding the two-dimensional delta function Φ 0 δ ( ρ ) , where φ0 is the flux quantum. If the magnetic field is parallel to the z axis and div B = 0, then B − λ 2∇ 2 B = Φ 0 δ ( ρ ) ,

or

10-2

⎛ ∂ 2 B 1 ∂B ⎞ λ2 ⎜ 2 + ⎟ − B = − Φ 0δ ( ρ ) . ⎝ ∂ρ ρ ∂ρ ⎠

This equation has the solution B ( ρ ) = ( Φ 0 2πλ 2 ) K 0 ( ρ λ ) , where K0 is a hyperbolic Bessel function* infinite at the origin and zero at infinity:

(Φ ( ρ >> λ ) B ( ρ ) ( Φ ( ρ . 2 6mc

∞

−2r a 0 1 2 2 ⋅ π ⋅ = 3a 0 . 4 r dr e Here < r > = 3 ∫ πa 0 0 2

The numerical result follows on using N = 6.02 × 1023 mol–1. 2a. Eu++ has a half-filled f shell. Thus S = 7 × 1/2 = 7/2. The orbitals mL = 3, 2, 1, 0, –1, –2, –3 have one spin orientation filled, so that L = ΣmL = 0. Also J = L + S = 7/2. Hence the ground state is 8S7/2. b. Yb+++ has 13 electrons in the f shell, leaving one hole in the otherwise filled shell. Thus L = 3, S = 1/2, J = 7/2 -- we add S to L if the shell is more than half-filled. The ground state symbol is 2F7/2. c. Tb+++ has 8 f electrons, or one more than Eu++. Thus L = 3; S = 7/2 – 1/2 = 3; and J = 6. The ground state is 7F6. 3a. The relative occupancy probabilities are ______ e

− ( ∆+µB ) kT

( Here ∆

stand s for k B ∆ )

______ e −∆ kT ______ e

− ( ∆−µB ) kT

______ 1

The average magnetic moment is e

= µ

where Z = 1 + e

− ( ∆−µB ) kT

− ( ∆−µB ) kT

b. At high temperatures e −∆ kT → 1 and

11-1

−e Z

− ( ∆+µB ) kT

+ e −∆ kT + e

− ( ∆+µB ) kT

.

⎛ µB ⎞ ⎛ µB ⎞ + … ⎟ − ⎜1 − + …⎟ ⎜1 + kT ⎠ ⎝ kT ⎠ → µ ⋅ ⎝ 4 2 2 Nµ µB ; χ→ . = 2kT 2kT c. The energy levels as a function of field are:

If the field is applied to take the system from a to b we increase the entropy of the spin system from ≈ 0 to ≈ N log 2. If the magnetization is carried out constant total entropy, it is necessary that the lattice entropy be reduced, which means the temperature ↓ . 4a. Z = 1 + e −∆ T ; k B ∆e −∆ T k ∆ E= = ∆ TB −∆ T 1+ e e +1 ∆ ∆T e ⎛ ∂E ⎞ T2 = ∆ C=⎜ k . B ⎟ 2 ⎝ ∂T ⎠ ∆ e∆ T + 1

(

)

2

b. For ∆ T = − ∫ d∆ ( ∆ ∆ 0 ) tanh ∆ τ, 0

∆0

(

)

< C > = k B ∫ d∆ ∆ 2 ∆ 0 τ2 sech 2 ( ∆ τ ) 0

x0

= ( k B τ ∆ 0 ) ∫ dx x 2 sech 2 x , 0

where x ≡ ∆/τ. The integrand is dominated by contributions from 0 < ∆ < τ, because sech x decreases exponentially for large values of x. Thus ∞



( k Bτ ∆0 ) ∫ dx x 2 sech 2 x . 0

8.

< µ > eµB τ − e−µB τ 2 sinh x = = µ 1 + 2 cosh x 1 + 2 cosh x

11-4

CHAPTER 12 1. We have Sρ+δ = Sρ eik ⋅δ . Thus   

dSρ

(

x

)

⎛ 2JS ⎞ y 6 − ∑ ei k ⋅δ Sρ =⎜ ⎟ δ dt  ⎝ h/ ⎠ 

⎛ 2JS ⎞ ⎡ y 6 − 2 ( cos k x a + cos k y a + cos k z a ) ⎤⎦ Sρ ; =⎜ ⎟ ⎣  ⎝ h/ ⎠

dSρ

y

⎛ 2JS ⎞ ⎡ x 6 − 2 ( cos k x a + cos k y a + cos k z a ) ⎤⎦ Sρ . = −⎜ ⎟ ⎣ dt  ⎝ h/ ⎠ 

These equations have a solution with time-dependence ∼ exp(–iωt) if ω = ( 2JS h/ ) ( 6 − 2 cos k x a − 2 cos k y a − 2 cos k z a ) .

2. U = ∑ n k h/ ωk = h/ ∫ dω D( ω) ω < n ( ω) > . If ω = Ak2, then dω dk = 2Ak = 2 A ω1 2 , k 

and D( ω) =

4π ω 1 1 ω1 2 . = 8π3 A 2 Aω1 2 4π2 A 3/ 2

Then U=

h/ 1 dω ω3 2 βh/ ω . 32 ∫ 4π A e −1 2

At low temps,

∫ ( h/ β ) 

∞

1

52

∫ dx 0

x3 2 1 ⎛5⎞ ⎛5 ⎞ = Γ ⎟ ς ⎜ ;1⎟ 5 2 gamma ⎜ x e − 1 ( h/ β ) function ⎝ 2 ⎠ zeta ⎝ 2 ⎠ function

[See Dwight 860.39] U  0.45 ( k BT )

52

/ π2 A 3 2 h/ 3 2

/ ) . C = dU dT  0.113 k B ( k BT hA 32

12-1

3. M A T = C ( B − µM B − εM A ) ( B = applied field ) M BT = C ( B − εM B − µM A ) Non-trivial solution for B = 0 if µC T + εC = 0; TC = C ( µ − ε ) µC T + εC Now find χ = ( M A + M B ) B at T > TC : MT = 2CH − CM ( ε + µ ) ; χ =

2C T + C (µ + ε)

∴θ TC = ( µ + ε ) ( µ − ε ) .

4. The terms in U eA + U c + U K which involve exx are 1 2 2 C11e xx + C12 e xx ( e yy + e zz ) + B1α1 e xx . 2 Take ∂/∂exx: C11e xx + C12 ( e yy + e zz ) + B1α1 = 0, for minimum. 2

Further: C11e yy + C12 (e xx + e zz ) + B1α 2 = 0 . 2

C11e zz + C12 ( e xx + e yy ) + B1α 3 = 0 . 2

Solve this set of equations for exx: C − α 2 ( C11 + 2C12 ) e xx = B1 12 . ( C11 − C12 )( C11 + 2C12 ) 2

Similarly for eyy, ezz, and by identical method for exy, etc.

12-2

5a. U ( θ ) = K sin 2 θ − Ba M s cos θ  Kϕ2 − Ba M s

1 2 ϕ , for θ = π + ϕ 2

and expanding about small ϕ .

1 Ba M s . Thus at Ba = 2K/Ms the magnetization 2 reverses direction (we assume the magnetization reverses uniformly!). For minimum near ϕ = 0 we need K >

b. If we neglect the magnetic energy of the bidomain particle, the energies of the single and bidomain particles will be roughly equal when M s d 3 ≈ σ w d 2 ; or d c ≈ σ w M s . 2

2

For Co the wall energy will be higher than for iron roughly in the ratio of the (anisotropy 2 constant K1)1/2, or 10. Thus σ w ≈ 3 ergs cm . For Co, Ms = 1400 (at room °

temperature), so M s ≈ 2 ×106 erg cm . We have dc ≈ 3 2 ×106 ≈ 1.10−6 cm, or  100 A, as the critical size. The estimate is very rough (the wall thickness is dc; the mag. en. is handled crudely). 2

3

6. Use the units of Eq. (9), and expand tanh

Then (9) becomes m 

m m 1 m3 = − + " . [Dwight 657.3] t t 3 t3

m m3 − +"; t 3t 3 3 ( t 3 − t 2 )  m 2 ; m 2  3t 2 (1 − t ) ,

but 1 – t is proportional to Tc – T, so that m ∞ Tc − T for T just below Tc. 7. The maximum demagnetization field in a Néel wall is –4 πMs, and the maximum self1 energy density is ( 4πM s ) M s . In a wall of thickness Na, where a is the lattice constant, 2 2 the demagnetization contribution to the surface energy is σdemag ≈ 2πM s Na . The total

(

)

wall energy, exchange + demag, is σ w ≈ ( π2 JS2 Na 2 ) + 2πM s Na , by use of (56). The

minimum is at 12-3

2

∂σ w ∂N = 0 = −π2 JS2 N 2 a 2 + 2πM s a , or 2

12

⎛1 ⎞ 2 N = ⎜ π JS2 M s a 3 ⎟ , ⎝2 ⎠ and is given by σ w ≈ πM sS ( 2π J a )

12

≈ (10 ) (103 )(10−4 10−8 )

12

≈ 10 erg cm 2 ,

which is larger than (59) for iron. (According to Table 8.1 of the book by R. M. White and T. H. Geballe, the Bloch wall thickness in Permalloy is 16 times that in iron; this large value of δ favors the changeover to Néel walls in thin films.) 8. (a) Consider the resistance of the up and down spins separately. Magnetizations parallel: R↑↑ (up ) = σ p−1 ( L / A) + σ p−1 ( L / A) = 2σ p−1 ( L / A)

R↑↑ (down) = σ a−1 ( L / A) + σ a−1 ( L / A) = 2σ a−1 ( L / A) These resistances add in parallel: R↑↑ = R↑↑ (down) R↑↑ (up) /[ R↑↑ (down) + R↑↑ (up)] = 2( L / A) /(σ a + σ p ) Magnetizations antiparallel: R↑↓ (up ) = σ p−1 ( L / A) + σ a−1 ( L / A) R↑↓ (down) = σ a−1 ( L / A) + σ p−1 ( L / A) = R↑↓ (up)

These (equal) resistances add in parallel : R↑↓ = R↑↓ (up ) / 2 = ( L / A)(σ a−1 + σ p−1 ) / 2 The GMRR is then: GMRR = R↑↓ / R↑↑ − 1 = (σ a−1 + σ p−1 )(σ a + σ p ) / 4 − 1 = (σ a / σ p + σ p / σ a − 2) / 4 (b) For the ↑↓ magnetization configuration, an electron of a given spin direction must always go through a region where it is antiparallel to the magnetization. If σa → 0, then the conductance is blocked and the resistance R↑↓ is infinite.

12-4

CHAPTER 13 1. Consider a coil which when empty has resistance R0 and inductance L0. The impedance is Z0 = R0 – iωL0. When the coil is filled with material of permeability µ = 1 + 4πχ the impedance is Z = R 0 − iωL0 (1 + 4πχ ) = R 0 − iωL0 (1 + 4πχ′ + 4πiχ′′ ) , or Z = R 0 + 4πωχ′′L0 − iωL0 (1 + 4πχ′ ) .  

 

R

2a.

dF dFx xˆ + Fx  = dt dt 

L

dxˆ  +" . dt

⎛ dF ⎞ ⎛ = ⎜  ⎟ + ⎜ Fx ⎝ dt ⎠ R ⎝

dyˆ dxˆ  + Fy  + Fz dt dt

dzˆ ⎞ ⎟. dt ⎠

Now

(

)

dyˆ dxˆ dzˆ  = ( Ω × xˆ ) ;  = Ω × yˆ ;  = ( Ω × zˆ ) .       dt dt dt dxˆ Fx  + ⋅⋅⋅⋅ = Ω × F .   dt b.

dM  = γM × B   dt

;

⎛ dM ⎞ × M = γM × B . ⎜  ⎟ +Ω   ⎝ dt ⎠ R  

⎛ Ω⎞ ⎛ dM ⎞ ×⎜ B +  ⎟ . ⎜  ⎟ = γM  ⎝ γ⎠ ⎝ dt ⎠R c. With Ω = −γB0 zˆ we have   ⎛ dM ⎞ × B1xˆ , ⎜  ⎟ = γM   ⎝ dt ⎠R

so that M precesses about xˆ with a frequency ω = γB1. The time t1/2 to give t1/2ω = π is   t1/2 = π/γB1. d. The field B1 rotates in the xy plane with frequency Ω = γB0 . 

13-1

2

1 1 ⎛a⎞ 2 z z z z 3a. < Bi > = ⎜ ⎟ ∑ ∑ < I j I k > , where for I = we have < I j I k > = δ jk . Thus 2 4 ⎝N⎠ j k 2

a2 ⎛a⎞ 1 . < Bi > = ⎜ ⎟ ∑ δ jk = 4N ⎝ N ⎠ 4 jk 2

4

b.

⎛a⎞ 4 z z z z < Bi > = ⎜ ⎟ ∑ < I j I k IA I m > . jk m A ⎝N⎠

Now 1 [δ jk δkA δAm + δ jk δAm 16 +δ jA δkm + δ jm δ kA ], and

< I j I k IA I m >= z

z

z

z

4

4

2 ⎛a⎞ 1 ⎛ a ⎞ 3N [N + 3N 2 ] − ⎜ ⎟ . < Bi > = ⎜ ⎟ ⎝ N ⎠ 16 ⎝ N ⎠ 16 4

4. For small θ, we have U K − Kθ2 . Now the magnetic energy density U M = − BM cos θ − 1 − BM + BMθ2 , so that with proper choice of the zero of energy the anisotropy energy is 2 equivalent to a field BA = 2K M

along the z axis. This is valid for θ d

f = B cosh K ( z − d 2 ) for 0 < z < d . This solution assures that ϕ will be continuous across the boundaries if B = A/cosh(Kd/2). To arrange that the normal component of D is continuous, we need ε(ω) ∂ϕ/∂z continuous, or ε(ω) = – tanh(Kd/2).

14-3

CHAPTER 15 1a. The displacement under this force is

1 ∞ α ( ω) e − iωt dω. ∫ −∞ 2π

x (t) =

∫ α ( ω) e

With ω = ωR + iωI, the integral is

− iωR t ωI t

e

dω. This integral is zero for t < 0

because we may then complete a contour with a semicircle in the upper half-plane, over which semicircle the integral vanishes. The integral over the entire contour is zero because α(ω) is analytic in the upper half-plane. Therefore x(t) = 0 for t < 0. 1b. We want ∞

1 e − i ω t dω x (t) = ∫ ω02 − ω2 − iωρ , 2π −∞

(A)

which is called the retarded Green’s function of the problem. We can complete a contour integral by adding to x(t) the integral around an infinite semicircle in the upper halfplane. The complete contour integral vanishes because the integrand is analytic everywhere within the contour. But the integral over the infinite semicircle vanishes at t < 0, for then exp ⎡⎣ −i ( ωR + iωI ) ( − t ) ⎤⎦ = exp ( −ωI t ) exp ( iωR t ) , which → 0 as |ω| → ∞. Thus the integral in (A) must also vanish. For t > 0 we can evaluate x(t) by carrying out a Cauchy integral in the lower half-plane. The residues at the poles are

(

± 12 ω0 − 14 ρ2 2

)

1 2

(

2 exp ( − 12 ρt ) exp ⎡⎢ ∓ i ω0 − 14 ρ2 ⎣

)

1 2

t ⎤⎥ , ⎦

so that

(

x ( t ) = ω0 − 14 ρ2 2

)

1 2

(

exp ( − 12 ρt ) sin ω0 − 14 ρ2

2. In the limit ω → ∞ we have α′ ( ω) → −∑ f j ω2

from (9), while from (11a)

15-1

2

)

1 2

t.

α′ ( ω ) → −

∞

2 sα′′ ( s ) ds . πω2 ∫0

3. The reflected wave in vacuum may be written as − E y ( refl ) = Bz ( refl ) = A′ e

− i ( kx + ωt )

,

where the sign of Ey has been reversed relative to Bz in order that the direction of energy flux (Poynting vector) be reversed in the reflected wave from that in the incident wave. For the transmitted wave in the dielectric medium we find E y ( trans ) = ck Bz ( trans ) εω = ε −1 2 Bz ( trans ) = A"e (

i kx −ωt )

,

by use of the Maxwell equation c curl H = ε∂E/∂t and the dispersion relation εω2 = c2k2 for electromagnetic waves. The boundary conditions at the interface at x = 0 are that Ey should be continuous: Ey (inc) + Ey (refl) = Ey (trans), or A – A' = A''. Also Bz should be continuous, so that A + A' = ε1/2 A''. We solve for the ratio A'/A to obtain ε1/2 (A – A') = A + A', whence A' 1 − ε1 2 = , A ε1 2 + 1 and r≡

E ( refl ) E ( inc )

=−

A' ε1 2 − 1 n + ik − 1 . = = A ε1 2 + 1 n + ik + 1

The power reflectance is 2 ⎛ n − ik − 1 ⎞ ⎛ n + ik − 1 ⎞ ( n − 1) + K R ( ω) = r ∗ r = ⎜ . ⎟⎜ ⎟= 2 ⎝ n − ik + 1 ⎠ ⎝ n + ik + 1 ⎠ ( n + 1) + K 2 2

4. (a) From (11) we have σ " ( ω) = −

∞ σ ' (s ) 2ω P∫ 2 ds. π 0 s − ω2

In the limit ω → ∞ the denominator comes out of the integrand and we have

15-2

lim ω→∞

σ " ( ω) =

∞

2 σ ' ( s ) ds. πω ∫0

(b) A superconductor has infinite conductivity at zero frequency and zero conductivity at frequencies up to ωg at the energy gap. We can replace the lost portion of the integral (approximately σ'nωg) by a delta function σ'nωg δ(ω) in σ's(ω) at the origin. Then the KK relation above gives σ ''s ( ω) =

2 σ 'n ωg . πω

(c) At very high frequencies the drift velocity of the conduction electrons satisfies the free electron equation of motion mdv dt = − eE ; − iωmv = − eE ,

so that the current density is j = n ( −e ) v = − ine 2 E mω

and ωσ'' (ω) = ne2/m in this limit. Then use (a) to obtain the desired result. 5. From (11a) we have ε ' ( ω) − 1 =

2 ∞ δ ( s − ωg ) ωp 4πne 2 P∫ 2 ds . = 2 2 2 m s − ω ω − ω 0 g

6. n2 – K2 + 2inK = 1 + 4πiσ0/ω. For normal metals at room temperature σ0 ∼ 1017 – 1018 sec–1, so that in the infrared ω σ0 . Thus n 2 K 2 , so that R 1 − 2 n and n

( 2πσ0 ω) , whence

R 1−

( 2ω πσ0 ) . (The units of σ0 are sec–1 in CGS.)

7. The ground state of the line may be written ψ g = A1B1A 2 B2 … A N BN . Let the asterisk denote excited state; then if specific single atoms are excited the states are ∗ ∗ ϕ j = A1B1A 2 B2 … A j B j … A N BN ; θ j = A1B1A 2 B2 … A jB j … A N BN . The hamiltonian acts

thusly: H ϕ j = ε A ϕ j + T1θ j + T2θ j−1 ; H θ j = ε Bθ j + T1ϕ j + T2ϕ j + 1.

15-3

An eigenstate for a single excitation will be of the form ψ k = ∑ eijka ( αϕ j + βθ j ) . We j

form H ψ k = ∑ eijka [αε A ϕ j + αT1θ j + αT2 θ j−1 j

θ

+ βε B j + β T1ϕ j + β T2ϕ j + 1].

(

)

= ∑ eijka [ αε A + β T1 + e − ika β T2 ϕ j j

(

)

+ αT1 + βε B + eika αT2 θ j ] = Eψ k = ∑ eijka [αEϕ j + β Eθ j ].

This is satisfied if

( εA − Ε ) α + ( T1 + e−ika T2 ) β = 0; ( T1 + eika T2 ) α + ( εB − E ) β = 0. The eigenvalues are the roots of

εA − E

T1 + e − ika T2

T1 + eika T2 ε B − E

15-4

= 0.

CHAPTER 16 1.

e2 x 3 ⋅ = eE; ex = r 3 E = p; α = p E = r 3 = a H . r r

4π 4π 3 P = 0 inside a conducting sphere. Thus p = a P = a 3E 0 , and 3 3 α = p E0 = a 3 .

2. E i = E 0 −

3. Because the normal component of D is continuous across a boundary, Eair = εEdiel, where Eair = 4πQ/A, with Q the charge on the boundary. The potential drop between the 1⎞ ⎛ two plates is E air qd + E diel d = E air d ⎜ q + ⎟ . For a plate of area A, the capacitance is ε⎠ ⎝ C=

A . 1⎞ ⎛ 4πd ⎜ q + ⎟ ε⎠ ⎝

It is useful to define an effective dielectric constant by 1 1 = +q . εeff ε If ε = ∞, then εeff = 1/q. We cannot have a higher effective dielectric constant than 1/q. For q = 10–3, εeff = 103. 4. The potential drop between the plates is E1 d + E2 qd. The charge density Q D1 εE1 iσ = = = E2 , A 4π 4π ω by comparison of the way σ and ε enter the Maxwell equation for curl H. Thus E1 +

Q=

4πiσ ⎛ 4πiσ ⎞ E 2 ; V = E 2d ⎜ + q⎟ ; εω ⎝ εω ⎠

σAi Q E 2 ; and thus C ≡ = ω V

and εeff = (1 + q )

A , ⎛ 1 iω q ⎞ 4πd ⎜ − ⎟ ⎝ ε 4πσ ⎠

ε . 1 − ( iωεq 4πσ )

16-1

(CGS)

5a. E int = E 0 −

E int =

4π 4π E P = E0 − χ int . 3 3

E0 . 4π χ 1+ 3

b. P = χ E int =

χ E0 . 4π χ 1+ 3

6. E = 2P1/a3. P2 = αE = 2αP1/a3. This has solution p1 = p2

0 if 2α = a 3 ; α =

1 3 a . 2

7 (a). One condition is, from (43), γ ( TC − T0 ) − g 4 Ps + g 6 Ps = 0 . 2

4

The other condition is 1 1 1 2 4 6 γ ( Tc − T0 ) Ps − g 4 Ps + g 6 Ps = 0 . 2 4 6 Thus 1 1 2 4 g 4 Ps + g 6 Ps ; 2 3 2 1 3 g4 2 2 g 6 Ps = g 4 ; Ps = . 3 2 4 g6

− g 4 Ps + g 6 Ps = − 2

4

(b) From the first line of part (a), 2

2

2 3 g4 9 g4 3 g4 γ ( Tc − T0 ) = − = . 4 g6 16 g 6 16 g 6

8. In an electric field the equilibrium condition becomes − E + γ ( T − Tc ) P + g 4 P3 = 0 , where the term in g6 is neglected for a second-order transition. Now let P = Ps + ∆P . If we retain only linear terms in ∆P , then − E + γ ( T − Tc ) ∆P + g 4 3Ps ∆P = 0 , with use of 2

(40).

Further,

we

can

eliminate

Ps

2

∆P E = 1 2 γ ( Tc − T ) .

16-2

because

Ps = ( γ g 4 )( Tc − T ) . 2

Thus

9 a. ← a → i→

b.

cos

←i

i→

←i

i→

←i

π ( na ) a

← 2a → i i

i i

i i

Deforms to new stable structure of dimers, with lattice constant 2 × (former constant). c.

10. The induced dipole moment on the atom at the origin is p = αE, where the electric −3 field is that of all other dipoles: E = 2 a 3 ∑ p n = 4p a 3 ∑ n ; the sum is over

(

)

(

)(

)

positive integers. We assume all dipole moments equal to p. The self-consistency condition is that p = α(4p/a3) (Σn–3), which has the solution p = 0 unless α ≥ (a3/4) (1/Σn–3). The value of the summation is 1.202; it is the zeta function ζ(3).

16-3

CHAPTER 17 1. (a) The interference condition for a linear lattice is a cos θ = nλ. The values of θ that satisfy this condition each define a cone with axis parallel to the fiber axis and to the axis of the cylindrical film. Each cone intersects the film in a circle. When the film is flattened out, parallel lines result. (b) The intersection of a cone and a plane defines a conic section, here a hyperbola. (c) Let a, b be the primitive axes of a square lattice. The Laue equations (2.25) give a • ∆k = 2 πq; b • ∆k = 2 πr, where q, r are integers. Each equation defines a set of planes. The intersections of these planes gives a set of parallel lines, which play in diffraction from a two-dimensional structure the role played by reciprocal lattice points in diffraction from a three-dimensional structure. In the Ewald construction these lines intersect a sphere of radius k = 2 π/λ in a set of points. In two dimensions any wavelength (below some maximum) will give points; in three dimensions only special values of λ give points of intersection because one more Laue equation must be satisfied. The points correspond to the directions k' of the diffraction maxima. If the photographic plate is flat the diffraction pattern (2 dim.) will appear distorted.

Points near the direction of the incident beam are shown. (d) The lattice of surface atoms in the (110) surface of an fcc crystal is simple rectangular. The long side of the rectangle in crystal (real) space is a short side in the reciprocal lattice. This explains the 90° rotation between (21a) and (21b). 2.

∫

∞

0

With

the

trial

function

x

exp

(–ax),

the

normalization

integral

is

dx x 2 exp ( −2ax ) = 1 4a 3 . The kinetic energy operator applied to the trial function

gives

(

)

(

)(

)

− h/ 2 2m d 2 u dx 2 = − h/ 2 2m a 2 x − 2a exp ( −ax ) while Vu = eEx2 exp (–ax). The definite integrals that are needed have the form

∫

∞

0

(

)

dx x n exp( − ax) = n! a n+1 . The expectation value of the energy is < ε > = h/ 2 2m a 2

+ ( 3eE 2a ) , which has an extremum with respect to the range parameter a when

(

)

d < ε > da = h/ 2 2m 2a − 3eE 2a 2 = 0, or a 3 = 3eEm 2h/ 2 . The value of < ε > is a minimum at this value of a, so that

( = ( h/

)( ) + ( 3eE 2 ) ( 2h/ 2m ) ( 3eE 2 ) ( 2 + 2 ) ,

< ε > min = h/ 2 2m 3eEm h/ 2 2

13

23

23

17-1

−2 3

13

2

3eEm

)

13

where the last factor has the value 1.89 …. The Airy function is treated in Sec. 10.4 of the NBS Handbook of mathematical functions. 3. (a) D(ε ) =

dN dk d (πk 2 ) m m 2 = = 2A 2 2 dk dε (2π / L) dk = k π=

where A = L2. Note: There are two flaws in the answer m / πh 2 quoted in the text. First, the area A is missing, meaning the quoted answer is a density per unit area. This should not be a major issue. Second, the h should be replaced by = .

(b) N =

2 ⋅ πk F2 2 (2π / L)

=> ns = N / A = k F2 / 2π

L m where ns is the 2D sheet density. For a square sample, W=L, so: W ns e 2τ 2π m Rs = 2 2 and using =k F / m = vF : kF e τ h 1 2π= Rs = = 2 2 e kF A k F vF e τ

(c) Rs =

17-2

CHAPTER 18

1. Carbon nanotube band structure. (a) b i ⋅ a j = 2πδ ij

=>

b1 = (− 2aπ ,

2π 3a

b 2 = ( 2aπ ,

),

2π 3a

).

(b) The angle between K and b1is 30o ; A right triangle is formed in the first BZ with two sides of length K and b1/2. Now b1 = 43πa , so: K = (b1/2)/cos(30o)=

4π/3a .

(c) Quantization of k along x: kx(na)=2πj= kx =2π j/na. Assume n = 3i, where i is an integer. Then: kx = K(j/2i). For j = 2i, kx=K. Then ∆K = k y ˆj and there is a massless subband. (d) For n = 10, kx =2π j/10a =K(3j/20). The closest k comes to K is for j = 7, where ∆kx = K/20. Then:

ε 11 = 2=vF (4π / 3a) / 10 = 1.8 eV. The next closest is for j = 6, where ∆kx = K/10, twice the previous one. Therefore: ε22 = 2ε11 . (e) For the lowest subband: ∆k = ( K / 20) 2 + k y2 , so: 2

ε 2 = [(=K / 20vF )v F2 ]2 + (=k y v F ) 2 This is of the desired form, with m * = =K / 20v F . m * / m = =K / 20mvF = 0.12 . 2. Filling subbands

ε (nx , n y ) =

= 2π 2 (nx2 + n y2 ) 2mW 2

=> States are filled up to ε (2,2) =

= 2π 2 (8) 2mW 2

= 2 k12,1 = 2π 2 = (8 − 2) => 2m 2mW 2

k1,1 =

6π 2 2 6 => n1,1 = k1,1 = W π W

= 2 k 22,1 = 2π 2 = (8 − 5) => (2,1) subband: 2m 2mW 2

k 2,1 =

3π 2 2 3 => n2,1 = k 2,1 = π W W

(1,1) subband:

18-1

(2,1) subband: same. n=

2 6 4 3 + W W

= 5.9 x 108 /m.

3. Breit-Wigner form of a transmission resonance (a) cos(δϕ ) ≅ 1 − δϕ 2 / 2

; | ri |= 1− | ti |2 ≅ 1 − 12 | ti |2 − 18 | ti |4

The denominator of (29) is then: 1 + (1− | t1 |2 )(1− | t 2 |2 ) − 2(1 − 12 | t1 |2 − 18 | t1 |4 )(1 − 12 | t 2 |2 − 18 | t 2 |4 )(1 − 12 δϕ 2 )

= 14 (| t1 | 4 + | t 2 |4 ) + 12 | t1 | 2 | t 2 |2 +δϕ 2 = 14 (| t1 |2 + | t 2 |2 ) 2 + δϕ 2 4 | t1 |2 | t 2 |2 ℑ= . (| t1 |2 + | t 2 |2 ) 2 + 4δϕ 2 (b) δϕ = 2 Lδk and δk / δε = ∆k / ∆ε = (π / L) / ∆ε . Combining: δϕ = (2 L)(π / L)δε / ∆ε => δϕ / 2π = δε / ∆ε

(c) Combining: 4 | t1 |2 | t 2 |2 (∆ε / 2π ) 2 ℑ= (∆ε / 2π ) 2 (| t1 |2 + | t 2 |2 ) 2 + 4δε 2

which is (33).

4. Barriers in series and Ohm’s law (a) 1 1− | r1 |2 | r2 |2 1− | r1 |2 | r2 |2 − | t1 |2 | t 2 |2 1 − (1− | t1 |2 ) | r2 |2 − (1− | r1 |2 ) | t 2 |2 = = + = + 1 1 ℑ | t1 |2 | t 2 |2 | t1 |2 | t 2 |2 | t1 |2 | t 2 |2

= 1+

(b) σ 1D =

1 − (| r2 |2 + | t 2 |2 )+ | t1 |2 | r2 |2 + | r1 |2 | t 2 |2 | r2 |2 | r1 |2 = + + which gives (36). 1 | t1 |2 | t 2 |2 | t 2 |2 | t1 |2 n1D e 2τ 2k F e 2τ = , m πm

and

=k F = vF m

But: A B = vFτ B = 2vFτ

=>

=>

σ 1D =

σ 1D =

2vF e 2τ 2e 2 (2vFτ ) = h =π

2e 2 A B . h

5. Energies of a spherical quantum dot

(a) ∫ E ⋅ da = Qencl / εε o R+d

V=

∫ R

qdr 4πεε o r

E = q / 4πεε o r 2 Integrating from inner to outer shell:

=>

A

2

=

1 ⎞ q d ⎛1 ⎟= ⎜ − 4πεε o ⎝ R R + d ⎠ 4πεε o R( R + d ) q

18-2

C=

q R( R + d ) = 4πεε o V d

U=

(c) For d >> R , U

ε 0,0

=

e2

. Also ε 0, 0 =

4πεε o R

e2

e2 e2 d = . C 4πεε o R( R + d )

R2 A = εε o . d d

C ≅ 4πεε o

(b) For d

U

ε 0,0

=

e2

2m* R 2 4πεε o R = 2π 2 ⋅

e2 2m* R 2 2m* R 2 R = ⋅ = 4πεε o R = 2π 2 4πεε o = 2π 2 π 2 aB* ⋅

6. Thermal properties in 1D (a) D(ω ) =

2K 1 L = 2π / L v πv

ωD

∞

dωD(ω )=ω =L =L ⎛ k BT ⎞ ωdω U tot = ∫ ≅ = ⎟ ⎜ ∫ exp(=ω / k BT ) − 1 πv 0 exp(=ω / k BT ) − 1 πv ⎝ = ⎠ 0 Obtaining value from table of integrals: Lk B2T 2 π 2 π 2 Lk B2T 2 U tot = = 3hv =πv 6 2π 2 Lk B2T CV = ∂U tot / ∂T V == 3hv (b) The heat flow to the right out of reservoir 1 is given by: ∞

2∞

xdx

∫ exp( x) − 1 0

∞

DR (ω ) dω=ω =ℑ =ℑ ⎛ k BT1 ⎞ π 2 π 2 k B2T12 ωdω JR = ∫ = ℑ ⋅v⋅ ℑ= = ⎟ ⎜ L exp(=ω / k BT1 ) − 1 2π ∫0 exp(=ω / k BT1 ) − 1 2π ⎝ = ⎠ 6 6h 0 2

and similarly for JL. The difference is: JR − JL =

Let T1 = T + ∆T , T2 = T

π 2 k B2 ℑ

(T − T ) => (T − T ) ≈ 2T∆T

6h

2 1

2 1

=>

2 2

for small ∆T.

2 2

JR − JL =

18-3

π k ℑ 2

2 B

3h

∆T which gives (78).

CHAPTER 20 1. U = nEI. The number of ways to pick n from N is N! / (N − n) !n!. The number of ways to put n into N' = n'! / (N' − n) !n!.

⎛ ⎞ N! N′ ! Entropy S = k B ⎜ log + log . ⎟ ⎜ ⎟ ′ N − n !n! N − n !n! ( ) ( ) ⎝ ⎠ N!  N log N − ( N − n ) log ( N − n ) − n log n log ( N − n )!n! log

N′ !  N′ log N′ − ( N′ − n ) log ( N′ − n ) − n log n ( N′ − n )!n!

∂U ∂S ⎛ ∂F ⎞ −T = 0 in equilibrium; thus ⎜ ⎟ = ∂n ⎝ ∂n ⎠T ∂n N−n N′ − n ⎞ ⎛ E I = k BT ⎜ log + log ⎟ n n ⎠ ⎝ ( N − n )( N′ − n ) . For n