Kittel Charles Introduction To Solid State Physics 8Th Edition Solution Manual

C HAPTER 1

1. The vectors ˆ x ++ y ˆ z ˆ and ˆ −−+ xy ˆ z ˆ 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= 1 cos 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

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 ⎛ a ⎞ ⎛⎞ c

or

a = c;

2 2 1 2 c8

3 4 a3

C HAPTER 2

1. The crystal plane with Miller indices hk is a plane defined by the points a 1 /h, a 2 /k, and a 3 / . (a) Two vectors that lie in the plane may be taken as a 1 /h – a 2 /k and a 1 /h − a 3 / . But each of these vectors gives zero as its scalar product with G = h a 1 + k a 2 + a 3 , so that G must be perpendicular to the plane hk . (b) If is the unit normal to the plane, the interplanar spacing is ˆn na ˆ ⋅ 1 /h . But n ˆ = G /| G | , whence d(hk ) G =⋅ a 1 / h| | 2 / | G| G =π . (c) For a simple cubic lattice G =π (2 / a)(h x ˆ + k y ˆ + z ˆ ) ,

whence

1 1 3a a0 2 2

2. (a) Cell volume aa 1 ⋅× 2 a 3 =−

x ˆ + y ˆ ), and similarly for b 2 , b 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

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

|F| =

− 1 exp[ i(a − ⋅∆ k)]

− 1 exp[i(a ⋅∆ k)]

− 1 cos(a ⋅∆ k)

sin 2 (a ⋅∆ k)

(b) The first zero in sin M ε occurs for ε = 2π/M. That this is the correct consideration follows from

sin M( h π+ε= ) sin π Mh cos M ε+ cos Mh sin M . π ε

zero,

as Mh is an integer

Referred to an fcc lattice, the basis of diamond is 000; . Thus in the product

S(v v v ) S(fcc lattice) S (basis) 12 3 = × ,

we take the lattice structure factor from (48), and for the basis

− 1 2 i (v π 1 ++ v 2 v ). 3

S (basis) 1 e =+

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 v 1 +v 2 +v 3 = 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.

2 3 − 6. 1 f

π 4 r ( a Gr) sin Gr exp ( 2r a ) dr π ∫ −

The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G 4 for

Ga 0 >> 1.

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

a ⋅∆ k= 2 π× (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) = f A +f B e –iπn . For n odd, S = f A – a ⋅∆ k= 2 π× (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) = f A +f B e –iπn . For n odd, S = f A –

were

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

C HAPTER 3

2 2 2 1. 2 E = (h 2M) (2 / πλ= ) (h 2M) ( / π L) , with λ = 2L .

12 6 6 2. bcc: 6 U(R) = 2N [9.114( ε σ R) − 12.253( σ R) ]. At equilibrium R

0 = 1.488 σ, and U(R ) 0 = 2N ( 2.816). ε−

12 6 6 fcc: 6 U(R) = 2N [12.132( ε σ R) − 14.454( σ R) ]. At equilibrium R

0 = 1.679 σ, and U(R ) 0 = 2N ( 4.305). ε− Thus the cohesive energy ratio bcc/fcc = 0.956, so that the fcc structure is

more stable than the bcc.

3. |U| =

8.60 N ε

23 − 16 = 9 (8.60) (6.02 × 10 ) (50 × 10 ) = 25.9 × 10 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 H 2 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

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.

U(R) = N ⎜ n − ⎟ ; α= 2 log 2 = Madelung const.

In equilibrium

∂ 2 U ⎛ nA α q ⎞

N ⎜ − n1 + + 2 ⎟ = 0;R 0 =

and

Nq 2 α

U(R ) 0 =−

b. 2 U(R -R )

0 0 δ= UR () 0 +

2 R 0 ( R 0 δ+ ) ...,

2R ∂

bearing in mind that in equilibrium ( U R) ∂∂ R = 0.

2 2 2 ⎛ 2 ∂ U ⎞ ⎛ n(n 1)A + 2q α ⎞ ⎛ (n 1) q +α 2 α q⎞ ⎜ 2 ⎟ = N ⎜

n2 +

For a unit length 2NR 0 = 1, whence

6. For KCl, λ = 0.34 × 10 –8 ergs and ρ = 0.326 × 10 –8 Å. For the imagined modification of KCl with the ZnS structure, z = 4 and α = 1.638. Then from Eq. (23) with x ≡ R 0 / ρ we have

− xe 3 = 8.53 × 10 .

By trial and error we find x 9.2, or R 0 = 3.00 Å. The actual KCl structure has R 0 (exp) = 3.15 Å . For the imagined structure the cohesive energy is

⎟ , or =-0.489 2

U in units with R 0 in Å. For the actual KCl structure, using the data of Table 7, we calculate 2 =− 0.495, q

units as above. This is about 0.1% lower than calculated for the cubic ZnS structure. It is noteworthy that the difference is so slight.

7. The Madelung energy of Ba + O – is – αe 2 /R per ion pair, or –14.61 × 10 –12 0 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 R 0 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 e XX =S 11 X X , because all other stress components are zero. By (51),

3S 11 = 2 (C 11 − 12 C ) 1 (C + 11 + 12 C ).

2 Thus 2 Y = (C

11 + CC 12 11 − 2C 12 ) (C 11 + 12 C );

further, also from (37), e yy =S 21 X x ,

whence σ= e yy e xx = S 21 S 11 =− C 12 (C 11 + C) 12 .

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

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).

1 xx

11. Let e =−

e yy =e 2 in (43). Then

so that ⎜ 2 ⎟ = N ⎜

3 ⎟ is the effective shear

n2 +

constant.

12a. We rewrite the element a ij =p– δ ij ( λ + p – q) as a

ij =p– λ′ ij , where λ′ = λ + p – q, and ij is the Kronecker delta function. With λ′ the matrix is in the “standard” form. The root λ′ = Rp gives λ = (R – 1)p + q, and the R – 1 roots λ′ = 0 give λ = q – p.

v(r, t) i[. . . . .] = v

w(r, t) i[. . . . .] = w

as the displacements for waves in the [111] direction. On substitution in (57) we obtain the desired equation. Then, by (a), one root is

2 ωρ= 2 2p q += K (C

11 + 2C 12 + 4C ) / 3, 44

and the other two roots (shear waves) are

2 ωρ= 2 K (C

11 − C 12 + 44 C ) / 3.

13. Set u(r,t) = u 0 e i(K · r – t) and similarly for v and w. Then (57a) becomes

2 2 2 ωρ= 2 u

0 [C K 11 y + 44 C (K y + K )]u z 0

(C 12 + 44 C ) (K K v x y 0 + KKw) x z 0

and similarly for (57b), (57c). The elements of the determinantal equation are

11 = CK 11 x + 44 C (K y + K) z − ω ρ; M 12 = (C 12 + 44 C )K K ; x y

M 13 = (C 12 + 44 C )K K . x z

and so on with appropriate permutations of the axes. The sum of the three roots of 2 ωρ

is equal to the sum of the diagonal elements of the matrix, which is

(C 2 11 + 2C 44 )K , where

+ K y + K , whence z

v 1 + v 2 + v 3 = (C 11 + 2C ) 44 ρ ,

for the sum of the (velocities) 2 of the 3 elastic modes in any direction of K.

14. The criterion for stability of a cubic crystal is that all the principal minors of the quadratic form be positive. The matrix is:

C 11 C 12 C 12 C 12 C 11 C 12 C 12 C 12 C 11 C 44 C 44

C 44

The principal minors are the minors along the diagonal. The first three minors from the bottom are C 44 , C 44 2 ,C 3 44 ; thus one criterion of stability is C

3 44 C > 0. The next minor is

2 12 ), whence |C 12 |<C 11 . Finally, (C 11 + 2C 12 ) (C 11 –C 12 ) > 0, so that C 11 + 2C 12 > 0 for stability.

11 C 44 , or C 11 > 0. Next: C 3 2 44 2 (C 11 –C

C HAPTER 4

1a. The kinetic energy is the sum of the individual kinetic energies each of the form

Mu . S The force

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

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

b. The time average of

Mu is M S ω u. In the potential energy we have

u s1 + = u cos[ t (s 1)Ka] u{cos( t sKa) cos Ka ω−+ = ω− ⋅ +

sin ( t sKa) sin Ka}. ω− ⋅

Then u s − u s1 + = u {cos( t sKa) (1 cos Ka) ω− ⋅− −

sin ( t sKa) sin Ka}. ω− ⋅

We square and use the mean values over time:

cos >=< sin >= ; < cos sin >= 0.

Thus the square of u{} above is

u [1 2cos Ka cos Ka sin Ka] u (1 cos Ka). − + + = −

The potential energy per bond is 2 Cu (1 cos Ka), − and by the dispersion relation ω = (2C/M) (1 –

cos Ka) this is equal to M ω u. Just as for a simple harmonic oscillator, the time average potential

energy is equal to the time-average kinetic energy.

2. We expand in a Taylor series

22 ⎛ ∂ u ⎞

u(s p) + = u(s) pa + ⎜ ⎟ + pa ⎜ 2 ⎟ + ; ⎝ ∂ x ⎠ s 2 ⎝ ∂ x ⎠ s

On substitution in the equation of motion (16a) we have

22 ∂ u

M 2 =Σ ( paC)

p0 >

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

2 − 1 v 22 = M Σ paC

p0 >

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

Thus the two lattices are decoupled from one another; each moves independently. At ω 2 = 2C/M 2 the motion is in the lattice described by the displacement v; at ω 2 = 2C/M 1 the u lattice moves.

4. ω= A Σ

2 2 sin pk a

0 (1 cos pKa) ; −

(cos (k 0 − K) pa cos (k − 0 + K) pa)

When K = k 0 ,

which in general will diverge because Σ→∞ 1 .

5. By analogy with Eq. (18),

= 1 C (u s − v ) C (u s + 2 s1 + − v ), whence s 2 − iKa

For Ka =πω= , 2C M and 2C M. 1 2

6. (a) The Coulomb force on an ion displaced a 2 distance r from the center of a sphere of static or rigid conduction electron sea is – e 2 n(r)/r , where the

3 3 2 number of electrons within a sphere of radius r is (3/4 πR 2 ) (4πr /3). Thus the force is –e r/R , and the

2 3 force constant is e 1/2 /R . The oscillation frequency ω D is (force constant/mass) , or (e /MR ) . (b) For

1/2

sodium 46 1 2 M 4 10 × g and R × 2 10 cm; thus ω

D (5 10 × ) (3 10 × )

− 23 − 8 − 10 −

0 is 5 associated with this maximum wavevector, the velocity defined by ω –1

8 3 10 s –1 × (c) The maximum phonon wavevector is of the order of 10 cm . If we suppose that ω

13 − 1

0 /K max ≈ 3 × 10 cm s , generally a reasonable order of magnitude.

7. The result (a) is the force of a dipole e p u p on a dipole e 0 u 0 at a distance pa. Eq. (16a)

2 becomes 33 ω= (2 / M)[ (1 cos Ka) γ− +Σ− ( 1) (2e / p a )(1 cos pKa)] . −

p>0

At the zone boundary ω 2 = 0 if

− 1 3 +σΣ−

p>0 ( 1) [1 ( 1) ]p −− =0,

or if –3 σΣ−− [1 ( 1) ]p = 1 . The summation is 2(1 + 3 +5 + …) = 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

− 3 –3

given by the sign of –1 12 =σΣ− ( 1) p p , which is zero when 1 – 2 +3 –4 + … = 1/2σ. The series

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

C HAPTER 5

1. (a) The dispersion relation is ω=ω m | sin Ka|. We solve this for K to obtain

− 1 2 2 − K 1/ 2 = (2/a) sin ( / ωω

m ) , whence dK/d ω= (2 / a)( ω−ω m )

and, from (15), D( ) ω

2 2 − = 1/ 2 (2L/ a)( πω−ω

) . This is singular at ω = ω m . (b) The volume of a sphere of radius K in

3 Fourier space is 3/2 Ω=π 4 K / 3 (4 / 3)[( =π ω−ω

0 ) / A] , and the density of orbitals near ω 0 is

D( )= (L/2 ) | d /d | (L/2 ) (2 / A )( 1/ 2 ω π Ωω= π π ω−ω

0 ) , provided ω < ω 0 . It is apparent that

D(ω) vanishes for ω above the minimum ω 0 .

2. The potential energy associated with the dilation is

B( V/V) a ∆ ≈ kT B . This is kT B and not

3 kT B , because the other degrees of freedom are to be associated with shear distortions of the lattice cell.

2 − 47 − 24 Thus 3 <∆ ( 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) 1 < R >= (h/2 V) / ρ Σω , where from (20) for a Debye spectrum Σω

D /4v π , whence < R >=ω 3h / D /8 πρ v . (b) In one dimension from − (15) we have 1 D( ) ω=π L/ v , whence ∫ω d D( ) ωω diverges at the lower limit. The mean square

− 1 2 3 3 2 2 2 =∫ω 3 d D( ) ωω= 3V ω

2 1 2 2 strain in one dimension is <∂∂ ( R/ x) >=Σ Ku 0 = (h/2MNv) K / Σ

2 2 = 3 (h/2MNv) (K /

D / 2) =ω h / D / 4MNv .

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/4π 2 allowed values of K per unit area of K space. The total number of modes with wavevector less than K is, with ω =

vK,

2 2 2 N 2 = (A/4 ) ( K ) π π =ω A /4v. π

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

U = 2 D( ) n( , ) ω ωτωω= d 2 d, ω ∫

0 ∫ 0 π 2 2 v exp(h / ) 1 ωτ−

where ω D is defined by N = ∫ D D( ) ω dω . In the regime ω >> τ D , we have

3 2A 2 τ ∞ x

2 2 dx.

2v

Thus the heat capacity 2 C = k

B ∂ ∂τ ∝ U/ T .

(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 << ω ≈ D vN layer /L , where N layer /L is the number of

layers per unit length.

5. (a) From the Planck distribution <>+= n (e + 1) /(e −= 1) coth (x/2) , where

B . The partition function Ze = Σ e = e /(1 e ) [2sinh (x/2)] − = and the

free energy is F = k B T log Z = k B T log[2 sinh(x/2)]. (b) With ω(∆) = ω(0) (1 – γ∆), the condition

∂ ∂∆ = 1 F/

0 becomes B ∆ = γΣ / ω h coth (h /2k T) / ω B on direct differentiation. The energy

<>ω n h/ is just the term to the right of the summation symbol, so that B ∆=γ U (T) . (c) By definition of γ, we have δω ω = −γδ /V / V , or d log ω = −δ d log V . But θ∝ω D , whence

d log θ = −γ d log V .

C HAPTER 6

1. The energy eigenvalues are 2 ε=

k. The mean value over the volume of a sphere in k space is

The total energy of N electrons is

U 0 =⋅ε N F .

2a. In general p = –∂U/∂V at constant entropy. At absolute zero all processes are at constant entropy (the

2 2 3 23 3 h / ⎛ 3N π ⎞

Third Law), so that p =− dU dV 0 , where U 0 = N ε F = N

⎜ ⎟ , whence

5 5 2m ⎝ V ⎠

2 U p =⋅ 0 . (b) Bulk modulus

U 0 3 22 − 3 − = 12 (4.7 × 10 cm ) (4.7 eV) (1.6 10 × erg eV) V 5 11 − 3 11 − = 2 × 2.1 10 erg cm = × 2.1 10 dyne cm ,

11 whence B = 2.3 × 10 –2 dyne cm . By experiment (Table 3.3), B = 1.2 × 10 dyne cm .

3. The number of electrons is, per unit volume, n = ∫

0 ε d D( ) ε⋅ ( ε−µ τ ) , where D(ε) is the density e + 1

of orbitals. In two dimensions

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

4a. In the sun there are 57

− 24 10 nucleons, and roughly an equal number of electrons. In a

white dwarf star of volume

93 28 (2 10 ) 3 × ≈× 3 10 cm

28 − the electron concentration is 3 ≈

28 ≈× 3 10 cm . Thus

3.10 eV. (b) The value of k F is not 2m

F (3 n) π ≈ 10 ⋅ 10 ≈

10 ergs, or ≈

affected by relativity and is ≈ n 3 , where n is the electron concentration. Thus ε

F hck / F hc / √n. (c) A

6 18 3 38 change of radius to 10 km = 10 – cm makes the volume ≈ 4 × 10 cm and the concentration ≈ 3 × 10 cm

3 − 27 10 13 − 4 . Thus 8 ε≈

F 10 (3.10 ) (10 ) ≈

2.10 erg 10 eV. ≈ (The energy is relativistic.)

5. The number of moles per cm 3 is 81 × 10 –3 /3 = 27 × 10 –3 , so that the concentration is 16 × 10 21 atoms cm –

. The mass of an atom of He –24 is (3.017) (1.661) × 10 = 5.01 × 10 g. Thus

F [(1.1 10 × ) 10 ][(30)(16) 10 ] × ≈× 7 10 erg, or T F ≈ 5K.

6. Let E, v vary as e –iwt . Then

and the electric current density is

7. (a) From the drift velocity equation

iv ω= x (e m)E x +ω c v;iv y ω= y (e m)E y −ω c v. x

We solve for v x ,v y to find

We neglect the terms in ω 2 c . Because j = n(–e)v = σE, the components of σ come out directly. (b) From the electromagnetic wave equation

2 2 2 c 2 ∇ E = ε∂ Et ∂ ,

we have, for solutions of the form e i(kz – ωt) , the determinantal equation

ck εω xy

xx

εω yx

εω− yy ck

2 2 2 Here 3 ε=ε=−ω

1 P ω and ε = −ε = ω ω xy yx i c p ω . The determinantal equation gives the

xx yy

dispersion relation.

8. The energy of interaction with the ion is

r 4 r dr π =− 3e 2 ∫ r,

r 0 2 e 2 () ρ

where the electron charge density is –e(3/4πr 3

0 ). (b) The electron self-energy is

2 r 0 3 2 − 1 ρ 2 ∫ dr 4 r 3 4 r r

3e 5r . 0

The average Fermi energy per electron is 3ε 3 F /5, from Problem 6.1; because NV34r = π 0 , the average

23 2 is 2 394 ( π ) /

h 10mr 0 . The sum of the Coulomb and kinetic contributions is

U =−

which is a minimum at

The binding energy at this value of r s is less than 1 Ry; therefore separated H atoms are more stable.

9. From the magnetoconductivity matrix we have

ωτ c

j y =σ yx E x =

1 +ωτ () c

c τ >> 1, we have σ≅σωτ= yx 0 c ( ne τ m mc eB ) ( τ= ) neB c .

For ω 2

10. For a monatomic metal sheet one atom in thickness, n ≈ 1/d 3 , so that

22 R 2

sq ≈ mv nd e F ≈ mv d e F .

If the electron wavelength is d, then mv d F ≈ h/ by the de Broglie relation and

sq ≈ he / = 137 c

in Gaussian units. Now in Gaussian units. Now

≈ 4.1k Ω .

C HAPTER 7

1a. The wavevector at the corner is longer than the wavevector at the midpoint of a side by the factor √2. As ε ∝ k 2 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 2. 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

Chap. 2, K = (2 π / a) (1,1,1) u, with 0u << , will lie in the reduced zone.

The G´s of the reciprocal lattice are given by

G = ( 2 π a[hk )( −+ )( x ˆ + hk +− ) y ˆ + ( −++ hk ) ˆ z ], where

( / h 2m )

h, k, 2 are any integers. Then ε=

( 2a[uhk π )( +−+ )( + uhk ++− )( + uhk −++ ) ]. We now

have to consider all combinations of indices

h, k, for which the term in brackets is smaller than

( 1 1 1 ; 100 , 0 10 , )()() and () 00 1 ; (100), (010), and (001); (111); ()() 1 10 , 10 1 , and ( 011; ) (110), (101), and (011).

6[3(1/2) 2 ] or 9/2. These indices are (000);

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

() Ka . The energy is ε=

2 2 2 / 2 h K 2m / h P ma . (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 2 ( ⎞ P π −δ + − + δ = − )( ) ⎜ 1 ⎟ 1, which has the solution δ = 0 and δ = 2P/π. The energy gap is

g = ( / h 2ma ) ( 2 πδ ) ( / h 2ma ) () 4P .

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

written as U = U 1 + U 2 = Ur 1 () + Ux 1 ⎜ + a, y + a, z + a ⎟ and the Fourier transform has

components U G = U 1G + U 2G = U 1G 1e +

iG x

⎜ 4 ⎟ . If G = 2Ax, ˆ then the exponential is

e 2 i Aa

i = π e = −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.

5. Let k =+ Ki; H λ= ± 1 ⎢ ⎜ G ⎟ ± iG H − H ⎥ .

2m ⎢ ⎣ ⎝ 2 ⎠

The secular equation (46) is now

H << G we have, with σ=ε−

6. U(x,y) = – U[e i (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 e –i is mixed with e (π/a) (x+y) . The central equations are

i(π/a) (x+y)

λ−ε ) C ⎢ ; − UC − ; − = 0 ;

λ−ε ) C ⎢ − ; − ⎥ − UC ;

2 where 2 λ= (

2 h 2m / ) () π a. The gap is 2U.

C HAPTER 8

1a. E d =

13.60 eV × × 2 × 6.3 10 eV

c. Overlap will be significant at a concentration

N = 4 π 3 ≈ 10 atoms c m

1/ 2 2a. From Eq. (53), − d E / 2k T n (n N ) e B

0 d , in an approximation not too good for the present example.

E d − 1.45

13 − n 3 × 0.46 10 electrons cm .

b. R H =−

− × 1.3 10 CGS units

nec

3. The electron contribution to the transverse current is

j (e) ne y µ e ⎜

⎛ −µ n B ⎞

for the holes j (h) ne y µ h ⎜

Here we have used

ωτ= for electrons; ωτ= h ce e B ch h for holes.

The total transverse (y-direction) current is

2 = 2 0 (ne µ−µ

e pe h )(B/c)E x + (ne µ+µ e pe h )E , y

and to the same order the total current in the x-direction is

j x = (pe µ+µ h ne )E . e x

Because (*) gives Because (*) gives

4. The velocity components are v x = hk / m ; v / x t y = hk / m ; v / y t z = hk /m / z . The equation of motion in k space is / h dk/dt =− (e/c) v B × . Let B lie parallel to the k x axis; then dk / dt x = 0; dk / dt y = −ω k; z ω≡ eB/m c; dk / dt z =ω t k; y ω≡ t eB/m c t . We differentiate

2 with respect to time to obtain 2 d k / dt

= −ω dk / dt z ; on substitution for dk z /dt we have 2 d k / dt 2

+ωω t k y = 0 , the equation of motion of a simple harmonic oscillator of natural frequency

t ) = eB/(m m ) t c.

5. Define Q e ≡ eB / m c; Q τ e e h = eB τ h /m h c . In the strong field limit Q >> 1 the

magnetoconductivity tensor (6.64) reduces to

2 ⎛ Q h Q h 0 ⎞ ne τ e ⎜ − 1 − 2 ⎟ pe τ h ⎜ − 1 − 2 ⎟

2 We can write nec Q 2

e /B for ne τ e /m e and pec Q h /B for pe τ h /m h . The strong field limit for σ yx

follows directly. The Hall field is obtained when we set

ec ⎡

j y == 0 ⎢ (n p) E − x + ⎜ + ⎟ E y ⎥ .

The current density in the x direction is

ec ⎡ ⎛ n

j x = ⎢ ⎜ + ⎟ E x −− (n p) E y ⎥ ;

using the Hall field for the standard geometry, we have

ec 2 ⎡ ⎛ n p ⎞ (n p) − ⎤ j x = ⎢ ⎜ +

H ⎢ ⎝ Q e Q h ⎠ ⎟⎛ n

C HAPTER 9

8 - = 1 × 0.78 10 cm

8 − −= 1 × 1.57 10 cm

b.

π 2 k N F =× 2

( 2 2/k π )

2 n 2 = N/L = k/2

F π k F = 2n π

1 16 2 n =× 10 els/cm

F = × 0.89 10 cm

c.

3a. In the hcp structure there is one atom whose z coordinate is 0 and one at c . The structure factor of

2 π the basis for G c =

z ˆ is c

−π S (basis) 1 e i

G c =+

=−= 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

−π S i2

2G c (basis) 1 e =+ = 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.

T = evB − 27 8 − 1 10 − 10 1 erg sec) (2 10 cm ) (3 10 cm s ) × ×

− 10 8 − 1 (5 10 3 × esu) (10 cm d ) (10 gauss) − × 10 1.2 10 sec.

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:

Region II:

Boundary condition

ψ continuous.

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.

1 2 π e 2 8 –1

7a. ∆ () = , where S = πk F , with k F = 0.75 × 10 cm from Table 6.1, for potassium. Thus H hc / S

2 0.55 10 × G.

H 137 k e F

8. Write (17) as H = H 0 + H 1 , where H 1 = (h / m) k p / ⋅ . Then (18) is an eigenfunction of H 0 with

2 the eigenvalue 2 ε n (0) h k / 2m + / . In this representation the diagonal matrix element of H 1 is equal to

(h / m) dV u (r) k p U (r). / 0 ⋅ 0 In a cubic crystal U (r) 0 ∫ will be even or odd with respect to the

inversion operation r → −r , but is an odd operator. It follows that the diagonal matrix element p vanishes, and there is no first-order correction to the energy. The function U (r) k to first order in H 1 is

and the energy to second order is

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

ε n (0) −ε j (0)

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

∫ dV w*(r r ) w (r r ) − n − m

= * N ΣΣ e e ∫ dV ψ

where the summation is zero unless n = m, when it is equal to N.

b. − 12 ik(x x ) w(x x ) − −

= N U (x) e

0 n Σ . The summation is

equal to

10a. j –1 y =σ 0 (Q E x + sE y ) = 0 in the Hall geometry, whence E y =–E x /sQ.

b. We have j –1 x =σ 0 (Q E x –Q E y ), and with our result for E y it follows that

–2

− 2 − 1 − j 2 x = σ 0 (Q + sQ)E x ,

whence ρ= E x j x = (Q σ 0 )

s1 +

C HAPTER 10

dB 2 1 1a. 2 = 2 B; this is the London equation. The proposed solution is seen directly to

dx λ

satisfy this and to satisfy the boundary conditions B ⎜ ±δ= ⎟ a B . (b) For δ < < λ L ,

x 2 1 ⎛ x ⎞ cosh =+ 1 ⎜ ⎟ + …

δ 2 1 ⎛ δ ⎞ cosh =+ 1 ⎜ ⎟ + …

2 2 therefore 2 Bx () =

B a − B18 a ( λ )( δ− 4x . )

2a. From (4), dF S =− MB d a at T = 0. From Problem 1b,

Mx () =− ⋅ 2 B a ⋅δ− ( 4x , )

whence

S ( F x,B a ) − F0 S () =

2 ( δ− 4x B . ) a

b. The average involves

∫ 0 ( δ− 4x dx δ−⋅ )

whence

2 ⎛⎞ δ ∆= F B a ⎜⎟ , for δ << λ .

c. Let us set

B af ⎜⎟ ⎛⎞= B ac ,

B af = 12 ⋅ B. ac δ

4j1E π ∂ 4 π ⎛

c ⎞ 1E ∂

3a. (CGS) curl H =

c ⎞ 1 ∂ curl E

curl curl H = −∇ H = ⎜ σ 0 curl E −

4 πλ ⎠ c ∂ t

Now in CGS in nonmagnetic material B and H are identical. We use this and we use the Maxwell equation

1B ∂

curl E =−

ct ∂

to obtain

2 1 1 ∂ B 4 πσ 0 ∂ B

∇− B 2 B 2 2 − 2 = 0.

If i Be ∼ ( kr ⋅−ω t, ) then

2 1 ω 4i πσω 0

−− k 2 + 2 +

2 = 0 . Q.E.D.

b. 2 = 2 >> 2 ; also, ω << πσ ω 4 0 and 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 Φδρ 0 () , where φ 0 is the flux quantum. If the magnetic field is parallel to

the z axis and div B = 0, then

2 B 2 −λ∇ B =Φδ

or

2 ⎛ ∂ B 1B ∂ ⎞

⎟ − = −Φ δ B 0 () ρ .

⎝ ∂ρ

ρ ∂ρ ⎠

This equation has the solution 2 B () ρ=Φ (

0 2 πλ ) K 0 ( ρ λ where K ) , 0 is a hyperbolic

Bessel function * infinite at the origin and zero at infinity:

( 2 ρ << λ )()

B ρ ( Φ 0 2 πλ ) n ( λρ ) ;

( ρ >> λ )() B ρ ( Φ 0 2 πλ ) ( πλ ρ 2 ) exp ( −ρ λ . )

2 12

The total flux is the flux quantum:

2 πρρ ∫ d B () ρ=Φ 0 ∫ dx x K x 0 () =Φ 0 .

− 5. It is a standard result of mechanics that 1 E =− grad ϕ− c ∂ A ∂ t. If grad ϕ = 0, when

( c4 πλ L ) E. Now j = nqv and

2 we differentiate the London equation we obtain 2 ∂∂= jt

( nq m E. ) Compare the two equations for ∂j/∂t to find

* 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 − w2 ≤≤ 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

dJ = ( J w cos e 0 ) ⎡ ⎣ Φ () x hc dx / ⎦ ⎤ = ( J w cos 2xTeB hc dx , 0 )( / )

and the total current is

w2

sin wTBe hc ( / )

J = ( Jw 0 ) ∫ cos xTe B hc dx ( / ) = J 0 .

0 ( wTBe hc / )

7a. For a sphere ( H inside ) = B a −π 4 M 3; for the Meissner effect ( H inside ) =−π 4 M,

whence B a =−π 8M 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 3 a =−π 4M3B =

a 2. The total field in this position is 3B a /2.

C HAPTER 11

1. From Eq. (10),

e 2 χ=− 2 N

6mc

Here < r >=

3 ⋅π 4 ∫ r dr e ⋅ = 3a . 0

2 1 2 − 2r a 0 2

23 The numerical result follows on using N = 6.02 × 10 –1 mol .

2a. Eu ++ has a half-filled f shell. Thus S = 7 × 1/2 = 7/2. The orbitals m L = 3, 2, 1, 0, –1, –2, –3 have one spin orientation filled, so that L = Σm L = 0. Also J = L + S = 7/2. Hence

the ground state is 8 S

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 2 F

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 7 F

3a. The relative occupancy probabilities are

− ∆+µ ( ) ______ e B kT (

Here ∆ stand for k s B ∆ )

−∆ ______ e kT

− ∆−µ ( ) B kT

______ e ______ 1

The average magnetic moment is

− ∆+µ ( ) where B kT

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

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 . ↓

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

2 + 23 N 53 (

3N π ) = E1 0 ( +ζ ) ,

5 2m

and the magnetic energy is –N + µ B = – 1/2 N(1 + ζ) µ B.

tot = E 0 { ( 1 + ζ ) +−ζ ( 1 ) } − ζµ N B ;

53 b. Now 53 E

= E 0 { ( 1 +ζ ) −−ζ ( 1 ) −µ NB0 =

∂ E tot 5 23 23

B. Q.E.D.

2 ε F 6a. The number of pairs of electrons with parallel spin up is

= N1 ( +ζ, )

so that the exchange energy among the up spins is

VN 1 ( +ζ ) ;

and among the down spins the exchange energy is

VN 1 ( −ζ ) .

tot = E 0 { ( 1 +ζ ) +−ζ ( 1 ) }

53 b. Using these results and those from Prob. 5 we have 53 E

VN 1 ( +ζ ) 2N − ζµ Thus (for ζ << 1) B.

∂ E tot 20 1 2

E 0 ζ− VN ζ−µ= NB 0;

NB µ

E 0 − VN

9 2 NB µ

= 2N ε F 1 2 − VN

and

3N 2 µ

M = µζ = N

2 ε− F VN

7a. The Boltzmann factor gives directly, with τ = k B T

e −∆ τ − e

U = −∆ ∆τ −∆ τ = −∆ tanh ∆τ ;

2 = 2 C k dU d

B τ= k B ( ∆ k T sech B ) ( ∆ kT, B )

because d tanh x/dx = sech 2 x.

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),

< >= − U d ∆∆∆ ( tanh ∫ ∆τ 0 ) ,

2 2 <>= 2 C k

B d ∫ ∆∆∆τ ( 0 ) sech () ∆τ

k B τ∆ 0 ) dx x sech x, ∫

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

2 <> 2 C (

k B τ∆ 0 ) dx x sech x . ∫

µτ B −µ τ <µ> B e − e 2 sinh x

µ + 1 2 cosh x

+ 1 2 cosh x

C HAPTER 12

ik 1. We have ⋅δ S ρ+δ = Se ρ . Thus

dS x

⎛ 2JS ⎞

dt ρ ⎝ h / ⎠ ( δ )

⎟⎣ − 6 2 cos k a cos k a cos k a S ; ( x + y + z ) ⎤ ρ

dS y

⎛ 2JS =− x ⎜

⎞⎡ − 6 2 cos k a cos k a cos k a S . ( x + y + z ) ⎤ ρ

⎝ h / ⎠ ⎟⎣

dt

These equations have a solution with time-dependence ∼ exp(–iωt) if

ω= ( 2JS h 6 2 cos k a 2 cos k a 2 cos k a / ) ( − x − y − z ) .

2 2. 12 U =∑ nh

/ ω= k hd / ∫ ωωω< D () n () ω> . If ω = Ak , then d ω dk = 2Ak = 2A ω ,

and

D () ω= 3 12 = 2 3/ 2 .

8 π A 2A ω

Then

4A π ∫

32 1 U = 2 32 d ωω βω h / .

At low temps,

1 32 x

52 ∫ dx x = 52 gamma Γ ⎜⎟ ς ⎜ ;1 ⎟

() h / β 0 e − 1 () h / β function ⎝⎠ 2 zeta ⎝ 2 function ⎠

[See Dwight 860.39]

52 2 32 U 32 (

0.45 k T B ) / π A h/

= 32 C dU dT 0.113 k

B ( k T hA B / ) .

MT A = CB ( −µ M B −ε M A )( = B applied field )

MT B = CB ( −ε M B −µ M A )

Non-trivial solution for B = 0 if

T +ε C µ C

= 0; T C = C ( µ− ε )

µ C T +ε C

Now find χ= ( M A + M B ) B at T > T C :

2C

MT = 2CH CM − ( ε+µχ= ) ; TC + ( µ+ε )

∴θ T C =µ+ε ( )( µ−ε ) .

4. The terms in U e + U c + U K which involve e xx are

Ce 11 xx + Ce 12 xx ( e yy + e zz ) +α B 1 1 e. xx

Take ∂/∂e xx :

11 xx + C 12 ( e yy + e zz ) +α= B 1 1 0, for minimum.

Ce 11 zz + C 12 ( e xx + e yy ) +α= B 1 3 0.

Solve this set of equations for e xx :

12 −α 2 ( C 11 + 2C 12 )

e xx = B 1 .

( C 11 − C 12 )( C 11 + 2C 12 )

Similarly for e yy ,e zz , and by identical method for e xy , etc.

5a.

U 2 () θ=

K sin θ− a B M cos s θ

K ϕ− BM a s ϕ , for θ=π+ϕ

and expanding about small ϕ .

For minimum near ϕ= 0 we need K > BM a s . Thus at B a = 2K/M s the magnetization

reverses direction (we assume the magnetization reverses uniformly!).

b. If we neglect the magnetic energy of the bidomain particle, the energies of the single and bidomain particles will be roughly equal when

2 3 2 Md 2

≈σ w d ; or d c ≈σ w M. s

For Co the wall energy will be higher than for iron roughly in the ratio of the (anisotropy

constant K 2

1 ) , or 10. Thus σ≈ w 3 ergs cm . For Co, M s = 1400 (at room

temperature), so M s ≈× 2 10 erg cm . We have dc 3 2 10 ≈ × ≈

1.10 cm, or 100 A, as the critical size. The estimate is very rough (the wall thickness is d c ; the mag. en. is

handled crudely).

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

m 3 m 1m tanh = − 3 + . [Dwight 657.3]

t 3t

Then (9) becomes m

t 3t

( − t ) m;m 3t 1 t , ( − )

3 2 2 2 3t 2

but 1 – t is proportional to T c – T, so that m ∞ T c − for T just below T T c .

7. The maximum demagnetization field in a Néel wall is –4 πM s , and the maximum self-

energy density is ( 4MM π s ) s . In a wall of thickness Na, where a is the lattice constant,

2 the demagnetization contribution to the surface energy is 2 σ demag ≈π 2 M Na . s The total

( JS Na ) +π ( 2 M Na , s ) by use of (56). The

2 2 2 wall energy, exchange + demag, is 2 σ≈π

minimum is at minimum is at

MS2Ja s ( π ) ≈ () 10 10 ( )( 10 10 ) ≈ 10 erg cm ,

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:

↑↑ ( up ) = σ p ( L / A ) + σ p ( L / A ) = 2 σ p ( L / A )

↑↑ ( down ) = σ a ( L / A ) + σ a ( L / A ) = 2 σ a ( 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: − 1 − 1 R

↑↓ ( up ) = σ p ( L / A ) + σ a ( L / A )

↑↓ ( down ) = σ a ( L / A ) + σ p ( L / A ) = R ↑↓ ( up )

These (equal) resistances add in parallel : − 1 − 1 R

↑↓ = R ↑↓ ( up ) / 2 = ( L / A )( σ a + σ p ) / 2

The GMRR is then: − 1 − GMRR 1 = R

↑↓ / R ↑↑ − 1 = ( σ a + σ p )( σ a + σ p ) / 4 − 1

(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 σ a → 0, then the conductance is blocked and the resistance R ↑↓ is infinite.

C HAPTER 13

1. Consider a coil which when empty has resistance R 0 and inductance L 0 . The impedance is Z 0 =R 0 – iωL 0 . When the coil is filled with material of permeability

µ = + πχ 14 the impedance is Z = R 0 −ω iL14 0 ( + πχ = ) R 0 −ω iL14 0 ( + πχ + π χ ′ 4i ′′ ) , or

Z = R 0 + πωχ 4 ′′ L 0 −ω iL14 0 ( + πχ ′ ) .

dz = Ω× ˆ (

x; ˆ ) = Ω× y; ˆ = Ω× ( z. ˆ )

dy ˆ

dt ( ) dt

⎟ ⎞=γ×+ M ⎜ B ⎟ .

⎝ dt ⎠ R

c. With Ω = −γ Bz 0 ˆ we have

so that M precesses about with a frequency ω = γB ˆx 1 . The time t 1/2 to give t 1/2 ω = π is t 1/2 = π/γB 1 .

d. The field B 1 rotates in the xy plane with frequency Ω=γ B. 0

1 3a. < B i >= ⎜⎟ ∑∑< I j k I > , where for I = we have < II > = δ . Thus N j k

IIII j k m >= [ δδδ+δ jk k m jk δ m

4. For small θ, we have 2 U K −θ K . Now the magnetic energy density U M =− BM cos θ−

1 − 2 BM + BM θ , so that with proper choice of the zero of energy the anisotropy energy is

2 equivalent to a field

B A = 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 B 0 +B A is the effective field.

5. We may rewrite (48) with appropriate changes in M, and with B anisotropy = 0. Thus

A =−γ i A ( M A λ M B + M B λ M A ) ;

−ω + iM

−ω iM B =γ i B ( M B λ M A + M A λ M B ) .

The secular equation is

γλ A M B −ω

γλ A M A

= 0, −γ λ B M B γλ B M A −ω

or

ω−ωγλ 2 (

A M B −γλ B M A ) =0.

One root is ω 0 = 0; this is the uniform mode. The other root is

ω=λγ 0 ( A M B −γ B M A ) = 0;

this is the exchange mode.

C HAPTER 14

1. kz E x0 =− = kA sin k xe , and at the boundary this is equal to E xi . The normal ∂ x

component of D at the boundary, but outside the medium, is ε(ω)kA cos kx, where for a

2 plasma ε(ω) = 1 – ω 2 p /ω . The boundary condition is –kA cos kx = ε(ω)kA cos kx, or

2 ε(ω) = –1, or ω 2 p = 2ω . This frequency ω=ω p 2 is that of a surface plasmon.

2. A solution below the interface is of the form kz ϕ−= () A cos kx e , and above the − interface kz ϕ+= () A cos kx e , just as for Prob. (1). The condition that the normal

component of D be continuous across the interface reduces to ε 1 (ω) = –ε 2 (ω), or

ω p2

p1

1 − 2 =−+ 1 2 , so that ω= ( ω+ω p1 p2 ).

3. (a) The equation of motion of the electrons is

2 −ω 2 x

e =− (e/m )E e x + ωω i e y; e −ω y e =− (e/m )E e y − ωω i e x e . For the holes,

2 −ω 2 x

h = (e/m )E h x + ωω i h y; h −ω y h = (e/m )E h y − ωω i h x. h

The result follows on forming ξ e =x e + iy e and ξ h =x h + iy h . (b) Expand − 1 − 1 − ( 1 ω+ω

e ) ω e (1 −ωω / e ) and ( ω−ω h ) ω h (1 +ωω / h ) . In this approximation

h e )/E (c/B)( ω+ω h e ) (c / eB ) (m = h + m e ) .

2 4. From the solution to Problem 3 we have + P = pe E / m

h ω ω , where we have dropped h

a term in ω 2 in comparison with ω

h ω. The dielectric constant

2 2 εω=+π 2 ()14P/E π 4 pe / m

h ωω h , and the dispersion relation ε(ω)ω =c k becomes

2 2 2 3 10 22 − 10 − 4πpe 1 ω/(eB/c) = c k . Numerically, ω≈ [(10 )(3 10 ) /(10)(3 10 )(5 10 × × × )] 0.2 s ≈ . It is true that ωτ will be <<1 for any reasonable relaxation time, but ω c τ > 1 can be

shown to be the applicable criterion for helicon resonance.

2 2 2 2 2 5. 2 md /dt r =−ω=− m r e E =π 4 e /3 P =−π 4 ne /3 r . Thus ω=π

4 ne 3m.

2 2 2 2 2 6. 2 md r dt =−ω=− m r (e c) ( v × B)m z ˆ −ω ο r , where ω o = 4πne /3m, from the

solution to A. Thus, with ω c ≡ eB/mc,

2 −ω = ωω − ω 2 x i

o x;

2 2 −ω = − ωω − ω y i c x o y .

2 2 2 Form ξ ≡ x + iy; then 2 −ω ξ − ωω ξ + ω ξ = 0,

or 0, ω + ωω − ω = c ο a quadratic equation for ω.

2 2 7. Eq. (53) becomes 2 cKE = ω ε(∞) + π [ E 4 P], where P is the ionic contribution to the polarization. Then (55) becomes

⎣ cK + ε(ω)ω + π Τ 4 Nq M ⎤ ⎦ + cK ω= T 0.

2 2 One root at K = 0 is 2 ω=ω+π

4 Nq ε(∞) M. For the root at low ω and K we neglect

4 2 terms in ω 2 and in ω K . Then

2 2 2 2 2 ω= 2 cK ω

[ ε(∞)ω + π Τ 4 Nq M]

= cK[ ε(∞) + π 4 Nq M ω T ]cK = ε(0) ,

where ε (0) is given by (58) with ω = 0.

15 − 1 − 8(a). 1 σ= ne m =ω (

0.73 10 s × = 800 ( cm) Ω

2 2 2 2 − (b) 27 ω=π

4 ne m*; m* 4 ne =π ω= p

4.2 10 × g; m* m = 4.7.

9. The kinetic energy of a Fermi gas of N electrons in volume V is

2 2 23 2 2 U 2 = N( 3 5) ( h 2m ) ( 3 N V ) / π . Then dU/dV = – (2/3)U/V and d U/dV = (10/9)U/V .

2 2 2 The bulk modulus 2 = B Vd U dV = (10 9) U V = (10 9) (3 5) n ( mv

F 2) = nmv F 3. The velocity of sound 12 v = (B ) , ρ where the density ρ= n (m M) + nM, whence

v 12 ( m 3M ) v.

10. The response is given, with ρ = 1/τ, by

+ρ dx dt +ω p x ) = Ft. ()

2 2 m d x dt 2

The conductivity σ does not enter this equation directly, although it may be written as σ =ω 2

p τ/4π. For order of magnitude,

1 10 )( × 9 10 ) 10 s ;

8 − 6 14 − ρ=τ= 1 1 v

F ( × 1.6 10 )( × 4 10 ) × 0.4 10 s ;

( 4 ne m )( 10 10 × × × 23 10 10 )

16 − × 1 1.5 10 s .

−λ The homogeneous equation has a solution of the form t xt ( >

0 ) = Ae sin ( ω+φ t ) ,

⎡ 2 2 where 12 ω=ω+ρ p () 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 x0 () = 0.

11. The Laplacian 2 ∇ϕ= 0, whence

This has solutions

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).

C HAPTER 15

1a. The displacement under this force is

−ω xt it () =

e d. ω

−ω i R t ω I With ω = ω t

R + iω I , the integral is ∫ αω () e e d ω . 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

xt () =

−ω 1 it e d ω

(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 half- plane. 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 ⎡ −ω+ω− i ( R i I ) () t ⎤ = exp ( ⎣ −ω ⎦ I t exp i ) ( ω R 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

±ω−ρ 1 1 1 2 1 ( 0 4 ) exp ( −ρ 2 t exp )

ω−ρ 0 4 ) t,

so that

xt () =ω−ρ 1 ( 0 4 ) exp 1 ( 1 −ρ 2 t sin ) ( ω−ρ 0 4 ) t.

2. In the limit ω → ∞ we have

from (9), while from (11a)

αω→− ′ () 2 s α ′ ′ () s ds

3. The reflected wave in vacuum may be written as

− ( i kx + t ω − )

y () E refl = z () B refl = ′ A e

where the sign of E y has been reversed relative to B z 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 B trans z ( ) εω

z ( B trans ) = A"e

− 12 ( i kx −ω t )

2 2 by use of the Maxwell equation c curl H = ε∂E/∂t and the dispersion relation εω 2 =c k for electromagnetic waves.

The boundary conditions at the interface at x = 0 are that E y should be continuous: E y (inc) + E y (refl) = E y (trans), or A – A' = A''. Also B z should be continuous, so that A + A'

=ε 1/2 A''. We solve for the ratio A'/A to obtain ε (A – A') = A + A', whence

() E refl

A' 12 ε− 1 n ik 1 +−

The power reflectance is

2 ⎛ 2 n ik 1 −− ⎞⎛ n ik 1 +− ⎞ ( n1 − ) + K R () ω=∗= rr ⎜

⎝ n ik 1 −+ ⎠⎝ n ik 1 ++

⎠ 2 ( n1 + ) + K

4. (a) From (11) we have

2 ∞ ω σ 's ()

σω=− " () P ∫ 2 2 ds.

π 0 s −ω

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

σ ' () s ds.

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

'' s () ω= σω ' n g .

(c) At very high frequencies the drift velocity of the conduction electrons satisfies the free electron equation of motion

mdv dt =− eE ; −ω i mv =− eE ,

so that the current density is

jn 2 = () −

ev =− ine E m ω

and ωσ'' (ω) = ne 2 /m in this limit. Then use (a) to obtain the desired result.

5. From (11a) we have

π 2 4 ne δ−ω

εω−= ' () 1 P

∫ 2 2 ds = 2 2 .

0 s −ω

ω−ω g

17 6. n 18 –K + 2inK = 1 + 4πiσ

0 /ω. For normal metals at room temperature σ 0 ∼ 10 – 10

2 sec 2 , so that in the infrared ω σ Thus

n K , so that R12n − and

n –1 ( 2πσ ω , 0 ) whence R1 − ( 2 ω πσ . 0 ) (The units of σ 0 are sec in CGS.)

7. The ground state of the line may be written ψ= g ABAB 1 1 2 2 … AB. N N Let the asterisk denote excited state; then if specific single atoms are excited the states are

ABAB 1 1 2 2 … AB j j … AB; N N θ= j ABAB 1 1 2 2 … AB j j … AB. N N The hamiltonian acts thusly:

H ϕ=εϕ+θ+θ j A j T 1j T 2 j1 − ;

H θ=εθ+ϕ+ϕ+ j B j T 1 j T 2 j 1.

An eigenstate for a single excitation will be of the form ijka ψ=

∑ e ( αϕ + βθ j j ) . We

form

H ijka ψ=

∑ e[ αε ϕ + α θ + α θ A j T 1j T 2 j1 −

+βε θ

B j +βϕ+βϕ+ T 1 j T 2 j 1] .

ijka

− ika

= ∑ e[ ( αε + β + A T 1 e β T 2 ) ϕ j

+ α + βε + ika (

=ψ= ijka E

∑ e[E αϕ+βθ j E j ].

This is satisfied if

− ( ika ε−Εα+

A ) ( T 1 + e T 2 ) β= 0 ;

T 1 + eT 2 ) α+ε− ( B E ) β= 0.

ika

The eigenvalues are the roots of

− ε− ika

= 0.

ika

T 1 + eT 2 ε− B E

C HAPTER 16

3 3 1. 3 ⋅= eE; ex = rE = p; α= pE == r a

2. E i = E 0 − P = 0 inside a conducting sphere. Thus p = aP = E , and a 0

α= 3 pE

0 =. a

3. Because the normal component of D is continuous across a boundary, E air = εE diel , where E air = 4πQ/A, with Q the charge on the boundary. The potential drop between the

two plates is E air qd E + diel d = E air dq ⎜ + ⎟ . For a plate of area A, the capacitance is

⎛ 1 ⎞ 4dq π ⎜ + ⎟

It is useful to define an effective dielectric constant by

1 =+ 1 q. ε eff ε

If ε = ∞, then ε eff = 1/q. We cannot have a higher effective dielectric constant than 1/q.

For q = 10 3 ,ε eff = 10 .

4. The potential drop between the plates is E 1 d+E 2 qd. The charge density

= E, 2 (CGS)

by comparison of the way σ and ε enter the Maxwell equation for curl H. Thus

2 E ; and thus C ≡ =

and ε=+ eff ( 1q

)( 1 − ωε i q4 πσ )

6. E = 2P 1 /a .P 2 = αE = 2αP 1 /a . This has solution p 1 =p 2 0 if 2 α= a; α= a .

7 (a). One condition is, from (43),

T C − T 0 ) − gP 4 s + gP 6s =. 0

The other condition is

T c − TP 0 ) s − gP 4 s + gP 6s = 0.

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

4g 6 16 g 6 16 g 6

8. In an electric field the equilibrium condition becomes 3 −+γ E ( TTPgP − c ) + 4 = 0 ,

where the term in g 6 is neglected for a second-order transition. Now let P =+∆ P s P . If

we retain only linear terms in 2 ∆ P , then −+γ E (

TT − c ) ∆+ P g 3P 4 s ∆ = , with use of P 0

2 (40). Further, we can eliminate 2 P

because P s =γ ( g 4 )( T c − T ) . Thus ∆ PE12T = γ ( c − T ) .

9 a. ←→ a cos () na

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 3 − field is that of all other dipoles: 3 E = ()

2a ∑ p n = ( 4p a ) () ∑ n ; the sum is over

positive integers. We assume all dipole moments equal to p. The self-consistency

condition is that p = α(4p/a 3 ) (Σn ), which has the solution p = 0 unless α ≥ (a /4) (1/Σn –3 ). The value of the summation is 1.202; it is the zeta function ζ(3).

C HAPTER 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. With the trial function x exp (–ax), the normalization integral is

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

( / h 2m d u dx ) =− ( / h 2m a x 2a exp )( − ) ( − ax )

while Vu = eEx 2 exp (–ax). The definite integrals that are needed have the form

0 dx x exp( ax) = n! a − . The expectation value of the energy is <ε>= ( / h 2m a ) + ( 3eE 2a , ) which has an extremum with respect to the range parameter a when

n+1

( / h 2m 2a 3eE 2a ) − = 0, or a = 3eEm 2h/ . The value of <ε> is a

2 2 3 d 2 <ε> da =

minimum at this value of a, so that

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

min

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

where A = L 2 .