Quinn J.J., Yi K.-S. Solid State Physics: Principles and Modern Applications

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Uncorrected Proof

BookID 160928 ChapID 08 Proof# 1 - 29/07/09

8.12 Surface Waves 241

Fig. 8.21. Dielectric function ε(ω)ofapolarcrystal

Since ε

, α

,andα are all positive, this equation has a solution only in 436

the region where ε(ω) < 0. Recall that ε(ω)versusω looksasshownin437

Fig. 8.21. ε(ω) is negative if ω

<ω<ω

. The dispersion relation, (8.106), is 438

written by 439

ε(ω)

+ ε(ω)

∞

(ω

− ω

)

(ω

− ω

)+ε

∞

(ω

− ω

)

. (8.107)

The denominator can be written as

440

D ≡ (ε

+ ε

∞

)ω

− (ε

+ ε

∞

)=(ε

+ ε

∞

)(ω

− ω

), (8.108)

where the surface phonon frequency ω

is given by 441

+ ε

∞

+ ε

∞

. (8.109)

It is easy to show that ω

<ω

. The dispersion relation can be 442

written 443

∞

(ω

− ω

)

(ε

+ ε

∞

)(ω

− ω

)

. (8.110)

444

Figure 8.22 shows the right-hand side of (8.110) as a function of frequency. 445

Since surface modes can occur only where q

> 0andε(ω) < 0, we see that 446

the surface mode is restricted to the frequency region ω

<ω<ω

.Itis 447

not diﬃcult to see that as cq

→∞, the surface polariton approaches the 448

frequency ω

. It is also apparent that at ω = ω

, c

= ε

.Thisgivesthe 449

dispersion curve sketched in Fig. 8.23. 450

Uncorrected Proof

BookID 160928 ChapID 08 Proof# 1 - 29/07/09

242 8 Dielectric Properties of Solids

Fig. 8.22. Dispersion curves of phonon–polariton modes

Fig. 8.23. Dispersion relation of surface phonon–polariton modes. Also shown are

the bulk mode ω

and ω

−

which can occur outside the region ω

<ω<ω

Uncorrected Proof

BookID 160928 ChapID 08 Proof# 1 - 29/07/09

8.12 Surface Waves 243

Problems 451

8.1. Use the eigenstates of the hydrogen atom to evaluate its atomic polariz- 452

ability α. 453

8.2. If atomic hydrogen formed a cubic lattice, what would its static dielectric 454

constant be? 455

8.3. Evaluate the reﬂectivity for an S-polarized and a P-polarized electro- 456

magnetic wave incident at an angle θ from vacuum on a material of dielectric 457

function ε(ω) as illustrated in the ﬁgure below. 458

One can take E =(E

, 0, 0)e

iωt−iq·r

and E =(0,E

iωt−iq·r

as the S- 459

and P-polarized electric ﬁelds, respectively. Remember that q · E =0and 460

q =(0,q

). 461

8.4. A degenerate polar semiconductor contains n

free electrons per unit 462

volume in the conduction band. Its dielectric function ε(ω)isgivenby 463

ε(ω)=ε

∞

− ω

−

where ω

and ω

are the LO and TO phonon frequencies, and ω

4πn

. 464

1. Show that ε(ω) can be written as 465

ε(ω)=ε

∞

(ω

− ω

−

)(ω

− ω

)

(ω

− ω

)

and determine ω

−

and ω

. 466

Uncorrected Proof

BookID 160928 ChapID 08 Proof# 1 - 29/07/09

244 8 Dielectric Properties of Solids

2. Make a sketch of ε(ω)versusω; be sure to indicate the locations of ω

, ω

, 467

−

, ω

, ε

,andε

∞

. 468

3. Determine the dispersion relation of the longitudinal and transverse modes, 469

i.e. ω as a function of q. 470

4. In which regions of frequency are the transverse waves unable to propagate? 471

5. Consider a vacuum–degenerate polar semiconductor interface. Use the 472

results obtained in the text to determine the dispersion relations of the 473

surface modes. 474

6. Make a sketch of ω versus q

is parallel to the interface) for these surface 475

modes and for the bulk modes which have q

=0. 476

Uncorrected Proof

BookID 160928 ChapID 08 Proof# 1 - 29/07/09

8.12 Surface Waves 245

Summary 477

In this chapter, we studied dielectric properties of solids in the presence of 478

an external electromagnetic disturbance. We ﬁrst reviewed elementary elec- 479

tricity and magnetism, and introduced concept of local ﬁeld inside a solid. 480

Then dispersion relations of self-sustaining collective modes and reﬂectivity 481

of a solid are studied for various situations. Finally, the collective modes local- 482

ized near the surface of a solid are also described and dispersion relations of 483

surface plasmon-polariton and surface phonon-polariton modes are discussed 484

explicitly. 485

When an external electromagnetic disturbance is introduced into a solid, 486

it will produce induced charge density and induced current density. These 487

induced densities produce induced electric and magnetic ﬁelds. The local ﬁeld 488

(r) at the position of an atom in a solid is given by 489

= E

+ E

where E

, E

are, respectively, the external ﬁeld, depolarization ﬁeld 490

(= −λP), Lorentz ﬁeld (=

4πP

), and the ﬁeld due to the dipoles within the 491

Lorentz sphere





i∈L.S.

3(p

·r

−r



. The local ﬁeld at the center of a

492

sphere of cubic crystal is simply given by 493

sphere

= E

−

4π

P +

4π

P +0=E

The induced dipole moment of an atom is given by p = αE

.Thepolariza- 494

tion P is given, for a cubic crystal, by P =

Nα

1−

4πNα

E ≡ χE, where N is the 495

number of atoms per unit volume and χ is the electrical susceptibility. The 496

electrical susceptibility and the dielectric function (ε =1+4πχ) of the solid 497

are 498

χ =

Nα

1 −

4πNα

; ε =1+

4πNα

1 −

4πNα

The relation between the macroscopic dielectric function ε and the atomic

499

polarizability α is called the Clausius–Mossotti relation: 500

ε − 1

ε +2

4πNα

The total polarizability of the atoms or ions within a unit cell can usually

501

be separated into three parts: 502

1. Electronic polarizability α

: The displacement of the electrons relative to 503

the nucleus 504

2. Ionic polarizability α

: The displacement of an ion itself with respect to its 505

equilibrium position 506

3. Dipolar polarizability α

dipole

: The orientation of any permanent dipoles by 507

the electric ﬁeld in the presence of thermal disorder. 508

Uncorrected Proof

BookID 160928 ChapID 08 Proof# 1 - 29/07/09

246 8 Dielectric Properties of Solids

In the presence of a ﬁeld E, the average dipole moment per unit volume is 509

given by ¯p

= p L





, where L(ξ) is the Langevin function. The dipolar

510

polarizability α

dipole

shows strong temperature dependence. The electronic 511

polarizability α

and the ionic polarizability α

ion

are almost independent of 512

temperature. 513

In a metal, the conduction electrons are free and the dielectric function 514

becomes 515

ε(ω)=1−

4πNe

− iω/τ

=1−

− iω/τ

In an ionic crystal, we have

516

ε(ω)=ε

∞



− ω



Here, ω

and ω

are the TO and LO phonon frequencies, respectively. We 517

note that ω

>ω

since ε

>ε

∞

in general. 518

For the propagation of light in a material characterized by ε(ω), the 519

external sources j

and ρ

vanishes. Therefore, we have 520

∇×E = −

iω

B; ∇×B =

iωε(ω)

The two Maxwell equations for ∇×E and ∇×B can be combined to give a

521

wave equation: 522



ε(ω) − q



E + q (q ·E)=0.

For an inﬁnite homogeneous medium of dielectric function ε(ω), a general

523

dispersion relation of the self-sustaining waves is written as 524

ε(ω)



ε(ω) − q



=0.

The two transverse modes and one longitudinal mode are characterized,

525

respectively, by 526

ε(ω)

; ε(ω)=0.

For the interface (z = 0) of two diﬀerent media of dielectric functions ε

527

and ε

, the boundary conditions give us the general dispersion of the surface 528

wave: 529

=0 or

ε(ω)

=0.

where

530

−

and α =

−

ε(ω).

Uncorrected Proof

BookID 160928 ChapID 09 Proof# 1 - 29/07/09

9 1

Magnetism in Solids 2

9.1 Review of Some Electromagnetism 3

9.1.1 Magnetic Moment and Torque 4

We begin with a brief review of some elementary electromagnetism. A current 5

distribution j(r) produces a magnetic dipole moment at the origin that is 6

given by 7

m =



r × j(r)d

r. (9.1)

If j(r) is composed from particles of charge q

at positions r

moving with 8

velocity v

, j(r)=



δ(r − r

), and the magnetic moment m is 9

m =



× v

. (9.2)

For a single particle of charge q moving in a circle of radius r

at velocity v

, 11

we have 12

m =

(9.3)

and m is perpendicular to the plane of the circle. The current i in the loop is

given by q divided by t =

2πr

, the time to complete one circuit. Thus 14

i =

2πr

. (9.4)

We can use this in our expression for m to get

m =

. (9.5)

Here, a = πr

is the area of the loop. We can write m =

if we associate 16

vector character with the a of the loop. 17

Uncorrected Proof

BookID 160928 ChapID 09 Proof# 1 - 29/07/09

248 9 Magnetism in Solids

Fig. 9.1. A magnetic moment m due to a current loop of radius r

located in a

magnetic ﬁeld B

If a magnetic ﬁeld were imposed on a magnetic moment, the magnetic 18

moment would experience a torque. To show this we begin with the Lorentz 19

force 20

F =

× B. (9.6)

For a charge dq the force dF is given by

dF =

× B =

ds × B. (9.7)

Here, ds is an inﬁnitesimal element of path length (see, for example, Fig. 9.1).

The torque τ is given by

r × dF. 23

τ =



r × dF =



r × ds × B. (9.8)

But

r × ds = a, and hence we have

τ =

a × B = m × B. (9.9)

9.1.2 Vector Potential of a Magnetic Dipole 25

If a magnetic dipole m is located at the origin, it produces a vector potential 26

at position r given by 27

A(r)=

m × r

. (9.10)

Of course the magnetic ﬁeld B(r)=∇×A(r). If we have a magnetization

(magnetic dipole moment per unit volume), then A(r)isgivenby 29

A(r)=





M(r



) × (r − r



)

|r − r



. (9.11)

Uncorrected Proof

BookID 160928 ChapID 09 Proof# 1 - 29/07/09

9.1 Review of Some Electromagnetism 249

As we did with the electric polarization P(r), we can transform this equation 30

into two parts. 31

A(r)=





∇



× M(r



)

|r − r



M(r



) ×



|r − r



, (9.12)

where

n is a unit vector outward normal to the surface S. The volume integra- 33

tion is carried out over the volume V of the magnetized material. The surface 34

integral is carried out over the surface bounding the magnetized object. Since 35

A(r) is related to a current density by 36

A(r)=





j(r



)

|r − r



, (9.13)

the vector potential produced by a magnetization is equivalent to volume

distribution of current 38

(r)=c∇×M(r) (9.14)

and a surface distribution of current

(r)=cM(r) × ˆn. (9.15)

Recall that if E = 0, Maxwell’s equation for ∇×B is

∇×B =

4π

+ j

4π

+4π∇×M. (9.16)

Deﬁning H = B − 4πM gives

∇×H =

4π

(9.17)

which shows that H is that part of the ﬁeld arising from the free current 43

density j

. As we stated before the two Maxwell equations 44

∇·E =4π (ρ

+ ρ

ind

) , (9.18)

and

∇×B =

E +

4π

+ j

ind

)+4π∇×M (9.19)

are sometimes written in terms of D and H.

∇·D =4πρ

(where D = E +4πP and ∇·P = −ρ

pol

with bound charge density ρ

pol

)and 47

∇×H =

D +

4π

Uncorrected Proof

BookID 160928 ChapID 09 Proof# 1 - 29/07/09

250 9 Magnetism in Solids

9.2 Magnetic Moment of an Atom 48

9.2.1 Orbital Magnetic Moment 49

Let us consider the nucleus to be ﬁxed and evaluate the orbital contribution 50

of the electron currents to the magnetic moment of an atom. 51

m =



× v

. (9.20)

Since q

= − e for every electron, and every electron has mass m

,wecan 52

write 53

m = −



×m

= −



total angular momentum

of the electrons



. (9.21)

We know



× m

is quantized and equal to ¯hL,where|L| =0, 1, 2,... 54

and L

=0, ±1, ± 2,...,±L.Thus,wehave 55

m = −

e¯h

L = −μ

L. (9.22)

Here, μ

e¯h

=0.927 ×10

−20

(ergs/gauss) or 5.8 ×10

−2

(meV/T) is called 56

the Bohr magneton. The Bohr magneton corresponds to the magnetic 57

moment of a 1s electron in H. 58

9.2.2 Spin Magnetic Moment 59

In addition to orbital angular momentum ¯hL,eachelectroninanatomhasan 60

intrinsic spin angular momentum ¯hs, giving a total spin angular momentum 61

¯hS where 62

S =



. (9.23)

The z-component of spin is s

= ±

, and the spin contribution to the magnetic 63

moment is ∓μ

. Thus, for each electron, there is a contribution −2μ

s to 64

the magnetic moment of an atom. If we sum over all spins, the total spin 65

contribution to the magnetic moment is 66

= −2μ



= −2μ

S. (9.24)

Note that the factor of 2 appearing in this expression is not exact. It is actually

given by g =2(1+

2π

− 2.973

+ ···)  2 × 1.0011454. However, in our 68

discussion here we will take the g-factor as 2. 69