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 n̂ 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
Mε.
πMh sin
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 << 1 is
p −3 2
given by the sign of 1 = 2σ Σ ( −1) p p , which is zero when 1 – 2–1 + 3–1 – 4–1 + … = 1/2σ. The series
p
p>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 << =ωD ≈ =vN layer / L , where Nlayer/L is the number of
layers per unit length.
1 1 x
1
= (e + 1) /(e x − 1) = coth (x/2) , where
2 2
2
− x/2
− sx
− x/2
−x
x = h/ ω/k BT . The partition function Z = e Σ e = e /(1 − e ) = [2sinh (x/2)]−1 and the
5. (a) From the Planck distribution
<n>+
free energy is F = kBT log Z = kBT log[2 sinh(x/2)]. (b) With ω(∆) = ω(0) (1 – γ∆), the condition
1
h/ ω coth (h/ ω/2k BT) on direct differentiation. The energy
2
< n > 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
<ε> =
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 << G we have, with
⎛
h/ 2 ⎞ ⎛
h/ 2 ⎞
2
iG
iG
σ
+
H
⋅
σ
−
H
⎜
⎟⎜
⎟ = U1 ;
2m ⎠ ⎝
2m ⎠
⎝
2
⎛ h/ 2
⎞
2
GH ⎟ = U1 ;
σ −⎜
2m
⎝
⎠
2
h/ 2 2 U1 − σ
..
.
H = 2
h/
2m
2
G
2m
2
.
6. U(x,y) = – U[ei (2π/a) (x+y) + other sign combinations of ± x ± y]. The potential energy contains the four
reciprocal lattice vectors (2 π/a) (±1; ±1). At the zone corner the wave function ei(π/a) (x+y) is mixed with e–i
(π/a) (x+y)
. The central equations are
π π⎤
π⎤
⎡ π
; ⎥ − UC ⎢ − ; − ⎥ = 0 ;
a⎦
⎣a a⎦
⎣ a
π
π
π π
( λ − ε ) C ⎡⎢− ; − ⎤⎥ − UC ⎡⎢ ; ⎤⎥ = 0 ,
a⎦
⎣ a
⎣a a ⎦
( λ − ε ) C ⎡⎢
(
where λ = 2 h/
2
)
2m ( π a ) . The gap is 2U.
2
7-2
CHAPTER 8
1a.
E d = 13.60 eV ×
b.
r = aH × ε ×
m* 1
× 6.3 × 10−4 eV
m ε2
m
6 × 10−6 cm
m*
c. Overlap will be significant at a concentration
N=
1
≈ 1015 atoms cm −3
3
r
4π
3
2a. From Eq. (53), n (n 0 N d )
1/ 2
e − Ed / 2k BT , in an approximation not too good for the present example.
⎛ m*k B T ⎞
n0 ≡ 2 ⎜
2 ⎟
⎝ 2πh/ ⎠
3/ 2
≈ 4 ×1013 cm −3 ;
Ed
1.45 ; e −1.45 0.23 .
2k BT
n 0.46 ×1013 electrons cm −3 .
1
b.
RH = −
−1.3 × 10−14 CGS units
nec
3. The electron contribution to the transverse current is
⎛µ B
⎞
jy (e) neµ e ⎜ e E x + E y ⎟ ;
⎝ c
⎠
⎛ −µ n B
⎞
Ex + Ey ⎟ .
⎝ c
⎠
for the holes jy (h) neµ h ⎜
Here we have used
ωce τe =
µe B
for electrons;
c
ωch τh =
µh B
for holes.
c
The total transverse (y-direction) current is
0 = (neµ e − peµ h )(B/c)E x + (neµ e + peµ h )E y ,
2
2
(*)
and to the same order the total current in the x-direction is
jx = (peµ h + neµ e )E x .
Because (*) gives
8-1
pµ h − nµe 1
⋅ ,
pµ h + nµe c
2
Ey = Ex B
2
we have for the Hall constant
RH =
1 pµ h − nµ e
⋅
.
ec (pµ h + nµe ) 2
2
Ey
jx B
=
2
/ x / m t ; v y = hk
/ y / m t ; v z = hk
/ z /m A . The equation of motion
4. The velocity components are v x = hk
h/ dk/dt = − (e/c) v × B . Let B lie parallel to the kx axis; then
dk x / dt = 0; dk y / dt = −ωA k z ; ωA ≡ eB/mA c; dk z / dt = ωt k y ; ωt ≡ eB/m t c . We differentiate
in
k
space
is
with respect to time to obtain d k y / dt = −ωA dk z / dt ; on substitution for dkz/dt we have
2
2
d 2 k y / dt 2 + ωA ωt k y = 0 , the equation of motion of a simple harmonic oscillator of natural frequency
ω0 = (ωA ωt )1/ 2 = eB/(m A m t )1/ 2 c.
5.
Define
Qe ≡ eBτe / m e c; Q h = eBτh / m h c . In the strong field limit
Q >> 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 << λ.
96π
⎝λ⎠
2
c. Let us set
1 2 ⎛δ⎞
1 2
Baf ⎜ ⎟ =
Bac ,
96π
⎝ λ ⎠ 8π
2
10-1
where Baf is the critical field for the film and Bac is the bulk critical field. Then
Baf = 12 ⋅
3a. (CGS) curl H =
λ
Bac .
δ
4πj 1 ∂E 4π ⎛
c
⎞ 1 ∂E
+
=
A⎟+
.
⎜ σ0 E −
2
c c ∂t
c ⎝
4πλ
⎠ c ∂t
curl curl H = −∇ 2 H =
4π ⎛
c
⎞ 1 ∂ curl E
B⎟ +
.
⎜ σ0 curl E −
2
∂t
c ⎝
4πλ ⎠ c
Now in CGS in nonmagnetic material B and H are identical. We use this and we use the
Maxwell equation
curl E = −
1 ∂B
c ∂t
to obtain
1 1 ∂ 2 B 4πσ0 ∂B
∇ B− 2 B 2 2 − 2
= 0.
λ
c ∂t
c ∂t
2
If B ∼ ei ( k ⋅ r − ωt ) , then
−k 2 −
1 ω2 4πiσ0 ω
+
+
= 0.
λ 2 c2
c2
Q.E.D.
1 ω
ω2
1
4πσ0 ω
.
b. 2 = p2 >> 2 ; also, ω2 << 4πσ 0ω and 2 >>
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 << λ ) B ( ρ )
0
2πλ 2 ) n ( λ ρ ) ;
0
2πλ 2 ) ( πλ 2ρ ) exp ( − ρ λ ) .
12
The total flux is the flux quantum:
∞
∞
0
0
2π ∫ dρ ρ B ( ρ ) = Φ 0 ∫ dx x K 0 ( x ) = Φ 0 .
5. It is a standard result of mechanics that E = −grad ϕ − c −1∂ A ∂t. If grad ϕ = 0, when
(
)
we differentiate the London equation we obtain ∂j ∂t = c 2 4πλ L E. Now j = nqv and
∂j ∂t = nq ∂v ∂t = ( nq m ) E.
2
Compare
the
two
2
equations
for
∂j/∂t
to
find
c 2 4πλ L = nq 2 m.
2
*
Handbook of mathematical functions, U.S. National Bureau of Standards AMS 55, sec.
9.6.
6. Let x be the coordinate in the plane of the junction and normal to B, with
− w 2 ≤ x ≤ w 2. The flux through a rectangle of width 2x and thickness T is 2xTB =
φ (x). The current through two elements at x and –x, each of width dx is
/ ⎤⎦ dx = ( J 0 w ) cos ( 2xTeB hc
/ ) dx ,
dJ = ( J 0 w ) cos ⎡⎣ eΦ ( x ) hc
and the total current is
J = ( J0 w )
/ )
sin ( wTBe hc
w 2
/ ) dx = J
∫ cos ( xTe B hc
/ )
( wTBe hc
0
.
0
7a. For a sphere H ( inside ) = Ba − 4πM 3; for the Meissner effect H ( inside ) = −4πM,
whence Ba = −8πM 3.
b. The external field due to the sphere is that of a dipole of moment µ = MV, when V is
the volume. In the equatorial plane at the surface of the sphere the field of the sphere is
−µ a 3 = −4πM 3 = Ba 2. The total field in this position is 3Ba/2.
10-3
10-4
CHAPTER 11
1. From Eq. (10),
χ = −N
e2
< r2 > .
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> = 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> → 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 << 1, e
∆T
1 ⎛∆⎞
→ 1 and C → k B ⎜ ⎟ .
4 ⎝T⎠
5a. If the concentration in the spin-up band is N+ = 1/2 N (1 + ζ), the kinetic energy of all
the electrons in that band is
3 + h/ 2
N
3π2 N +
5
2m
(
)
23
= E 0 (1 + ζ )
53
,
and the magnetic energy is –N+ µ B = – 1/2 N(1 + ζ) µ B.
{
b. Now E tot = E 0 (1 + ζ )
53
+ (1 − ζ )
53
} − NζµB;
11-2
{
}
∂E tot 5
23
23
= E 0 (1 + ζ ) − (1 − ζ ) − NµB = 0
∂ζ
3
4
ζ
3
∴ζ =
9NµB 3µB
=
20E 0
2ε F
3Nµ 2
M = Nµζ =
B . Q.E.D.
2ε F
6a. The number of pairs of electrons with parallel spin up is
( )
1 +
N
2
2
1
2
= N 2 (1 + ζ ) ,
8
so that the exchange energy among the up spins is
1
2
− VN 2 (1 + ζ ) ;
8
and among the down spins the exchange energy is
1
2
− VN 2 (1 − ζ ) .
8
{
b. Using these results and those from Prob. 5 we have E tot = E 0 (1 + ζ )
(
)
1
− VN 2 1 + ζ 2 2 − NζµB. Thus (for ζ << 1)
8
∂E tot
∂ζ
20
1
E 0 ζ − VN 2 ζ − NµB = 0 ;
9
2
NµB
ζ=
20
1
E 0 − VN 2
9
2
NµB
=
2Nε F 1
− VN 2
3
2
and
3Nµ 2
M = Nµζ =
B
3
2ε F − VN
2
11-3
53
+ (1 − ζ )
53
}
c. For B = 0 and ζ = 0.
∂ 2 E tot
∂ζ 2
20
1
40 E 0 4 ε F
E 0 − VN 2 < 0 if V >
=
9
2
9 N2 3 N
7a. The Boltzmann factor gives directly, with τ = kBT
U = −∆
e∆ τ − e−∆ τ
= −∆ tanh ∆ τ ;
e∆ τ + e−∆ τ
C = k B dU dτ = k B ( ∆ k BT ) sech 2 ( ∆ k BT ) ,
2
because d tanh x/dx = sech2x.
b. The probability P(∆) d∆ that the upper energy level lies between ∆ and ∆ + d∆,
referred to the midpoint as the zero of energy, is P(∆) d∆ = (d∆) / ∆0. Thus, from (a),
∆0
< U >= − ∫ 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
∞
<C>
( 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
x̂ + 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 x̂ 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 << 1 . For a sphere the demagnetizing field is parallel
to M and exerts no torque on the spin system. Thus B0 + BA is the effective field.
5. We may rewrite (48) with appropriate changes in M, and with Banisotropy = 0. Thus
+
(
+
+
)
−iωM A = −iγ A M A λ M B + M B λ M A ;
+
(
+
)
+
− i ωM B = iγ B M B λ M A + M A λ M B .
The secular equation is
γAλ MB − ω
−γ Bλ M B
γ Aλ MA
γ Bλ M A − ω
or
13-2
=0,
ω2 − ω ( γ A λ M B − γ Bλ M A ) = 0 .
One root is ω0 = 0; this is the uniform mode. The other root is
ω0 = λ ( γ A M B − γ B M A ) = 0;
this is the exchange mode.
13-3
CHAPTER 14
∂ϕ
= kA sin kx e kz , and at the boundary this is equal to Exi. The normal
∂x
component of D at the boundary, but outside the medium, is ε(ω)kA cos kx, where for a
plasma ε(ω) = 1 – ωp2/ω2. The boundary condition is –kA cos kx = ε(ω)kA cos kx, or
ε(ω) = –1, or ωp2 = 2ω2. This frequency ω = ωp 2 is that of a surface plasmon.
1. E x0 = −
2. A solution below the interface is of the form ϕ(−) = A cos kx e kz , and above the
interface ϕ(+) = A cos kx e − kz , just as for Prob. (1). The condition that the normal
component of D be continuous across the interface reduces to ε1(ω) = –ε2(ω), or
1−
ω2p1
ω
2
= −1 +
2
ωp2
ω
2
, so that ω2 =
1 2
(ωp1 + ω2p2 ) .
2
3. (a) The equation of motion of the electrons is
−ω2 x e = −(e/me )E x + iωωe ye ; − ω2 ye = −(e/me )E y − iωωe x e . For the holes,
−ω2 x h = (e/m h )E x + iωωh y h ; − ω2 y h = (e/m h )E y − iωωh x h .
The result follows on forming ξe = xe + iye and ξh = xh + iyh. (b) Expand
−1
(ωe + ω) −1 ωe (1 − ω / ωe ) and (ωh − ω) ωh −1 (1 + ω / ωh ) . In this approximation
−1
−1
(ξh − ξe ) / E + (c/B)(ωh + ωe ) = (c2 / eB2 ) (m h + me ) .
4. From the solution to Problem 3 we have P + = pe 2 E + / m h ωh ω , where we have dropped
a term in ω2 in comparison with ωhω. The dielectric constant
ε(ω) = 1 + 4πP + / E + 4πpe 2 / m h ωh ω , and the dispersion relation ε(ω)ω2 = c2k2 becomes
4πpe2ω/(eB/c) = c2k2. Numerically, ω ≈ [(103 )(3 × 1010 ) /(10)(3 × 1022 )(5 × 10−10 )] ≈ 0.2 s −1 .
It is true that ωτ will be <<1 for any reasonable relaxation time, but ωc τ > 1 can be
shown to be the applicable criterion for helicon resonance.
5. md 2r/dt 2 = − mω2r = −eE = 4πeP/3 = −4πne 2r/3 . Thus ωο2 = 4πne 2 3m.
6. md 2r dt 2 = −mω2r = −(e c) ( v × Bzˆ ) − mωο r, where ωo2 = 4πne2/3m, from the
2
solution to A. Thus, with ωc ≡ eB/mc,
−ω2 x = iωωc y − ωo x ;
2
−ω2 y = −iωωc x − ωo y .
2
14-1
Form ξ ≡ x + iy; then −ω2ξ − ωωc ξ + ωο ξ = 0, or ω2 + ωωc − ωο = 0, a quadratic
2
2
equation for ω.
7. Eq. (53) becomes c 2 K 2 E = ω2 [ε (∞)E + 4πP], where P is the ionic contribution to the
polarization. Then (55) becomes
ω2 ε(∞) − c 2 K 2 4πω2
= 0,
2
ω2
Nq 2 M
− ωT
or
2
2
ω4 ε 2 (∞) − ω2 ⎡⎣c 2 K 2 + ε (ω)ωΤ + 4π Nq 2 M ⎤⎦ + c 2 K 2ωT = 0.
One root at K = 0 is ω2 = ωT + 4πNq 2 ε (∞) M. For the root at low ω and K we neglect
terms in ω4 and in ω2K2. Then
2
ω2 = c 2 K 2 ωT [ε (∞) ωΤ + 4πNq 2 M]
2
2
= c 2 K 2 [ε (∞) + 4πNq 2 MωT ] = c 2 K 2 ε (0) ,
2
where ε (0) is given by (58) with ω = 0.
8(a). σ = ne 2 T m = (ωp 4π)T = 0.73 × 1015 s −1 = 800 (Ω cm) −1
2
(b) ωp = 4πne 2 m*; m* = 4πne2 ωp = 4.2 × 10−27 g; m* m = 4.7.
2
2
9. The kinetic energy of a Fermi gas of N electrons in volume V is
U = N( 3 5) ( h/ 2 2m ) ( 3π2 N V ) 2 3 . Then dU/dV = – (2/3)U/V and d2U/dV2 = (10/9)U/V2.
The bulk modulus B = Vd 2 U dV 2 = (10 9) U V = (10 9) (3 5) n ( mv F 2 ) = nmv F 3.
The velocity of sound v = (B ρ)1 2 , where the density ρ = n (m + M) nM, whence
v ( m 3M )1 2 v F .
2
2
10. The response is given, with ρ = 1/τ, by
(
)
m d 2 x dt 2 + ρdx dt + ωp x = F ( t ) .
2
The conductivity σ does not enter this equation directly, although it may be written as σ
= ωp2τ/4π. For order of magnitude,
14-2
(
)(
)
σ = 1 10−6 9 × 1011 1018 s −1 ;
(
ρ = 1 τ = v F A 1.6 × 108
(
ωp = 4πne2 m
)
12
) ( 4 × 10 ) 0.4 × 10
−6
14
(
10 × 1023 × 23 × 10−20 10−27
s −1 ;
)
12
1.5 × 1016 s −1 .
The homogeneous equation has a solution of the form x ( t > 0 ) = Ae −λt sin ( ωt + φ ) ,
12
2
2
where ω = ⎡ωp + ( ρ 2 ) ⎤ and λ = ρ/2. To this we add the particular solution x =
⎣
⎦
–e/mω and find A and φ to satisfy the initial conditions x(0) = 0 and x ( 0 ) = 0.
11. The Laplacian ∇ 2 ϕ = 0, whence
d 2f
− K 2f = 0 .
2
dz
This has solutions
f = AeKz for z < 0
f = Ae
− K ( z −d )
for z > 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 << R ,
U=
and therefore
= 2π 2
2 m* R 2
=>
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 << N, N′ ,
= k BT log
n2
E I = k BT log NN′ n 2 ; n = NN′ e − EI 2k BT .
(
)
2. From (2), n = Ne–Ev/kBT. For sodium, N = 2.5 × 1022 atoms/cm3. Thus
(
)(
)
n 2.5 × 1022 4 × 10−18 ≈ 105 per cm3 .
13.6 3
⋅ eV. From Table 10.3 we have ε (∞) = 2.25 for NaCl, whence
ε2 4
h/ ω = 2.0 eV. The observed value quoted in Table 18.2 is 2.7 eV.
3a. h/ ω =
b. From Vol. I of Atomic energy levels, Circular 467 of the Nat. Bur. of Standards, 3p-3s
= 16960 cm–1 = 2.1 eV.
20-1
CHAPTER 21
1. In fcc the nearest neighbors to a given atom lie in [110] and equivalent directions -thus these directions are lines of closest packing. In bcc the nearest neighbors lie in [111]
and equivalent directions. There is a lively interest in the channeling of fast ions along
open channel directions in crystals.
2a.
b.
3. When a dislocation moves a distance L through the crystal, the crystal advances by a
distance b. The work σL2b may be expressed as
Thus F = σb is the force/unit length on
the dislocation.
21-1
CHAPTER 22
1. Disordered state: S given by Eq. (2.48) with <f> as averaged over 3 Cu and 1 Au atom.
Ordered state:
− iπ k + A
−i h +A
− iπ h + k
S ( hkA ) = f Au + f Cu (e ( ) + e ( ) + e ( ) ) Consider the following reflections: (x ≡
refl. present)
Indices
100
110
111
200
210
211
220
221
222
Disordered
no
no
x
x
no
no
x
no
x
Ordered
x
x
x
x
x
x
x
x
x
2. C = ∂E/∂T; now use E from Eq. (5): E = E0 + 2NUP2. Thus Cconfig = 4NUP(T) ∂P/∂T,
and we recall that U is negative, as otherwise an ordered state does not occur.
From Fig. 7b,
in this region a treatment based on P alone gives
Cconfig = 0. We should look at the short range order!
22-1