Marder M.P. Condensed Matter Physics

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Problems

653

(a) Evaluate the integral

P12

= (l\P

\2).

(b) Evaluate

for the 2P

—>

IS transition.

ofthe2P state?

(d) Show that in a box with N hydrogen atoms, the transition rate

A21

from one

of the six 2P states to one of the two IS states is given by

1 c

^2l(w)(/

-/l) = —

g{uj)-.

on n

(21.49)

(e) Set n = 1, and find

A21

by integrating

A21 (io)D

h(io)

over ui.

(f) Put

A21

into a final form involving e, m,

and c. The numerical answer is

6.268-10

s-'.

Gunn effect: The conduction band of gallium arsenide has the form shown

in Figure 19.8. In the central minimum, electrons have a small effective mass

(m* = 0.07m), leading to a mobility on the order of 7500 cm

^"'s"

How-

ever, once the electron acquires a k in the minimum on the right, its effec-

tive mass increases to around m* = 0.4m, and its mobility decreases to 200

-1

The result is that electron velocity as a function of electric field

E looks as shown in Figure 21.14.

When a sufficiently large electrical field (~ 3000 V/cm) is applied to GaAs,

the crystal begins to emit pulses, typically at frequencies in the 50 GHz range.

The physical origin of the pulses lies in the fact that as electrons become more

energetic, it is favorable for them to jump from the center valley at Y to the

left valley at L in Figure 19.8. However, when they do so, their mobility

decreases, and they slow down. This leads to a traffic jam of charge, which

travels through the sample as a localized pulse. The goal of this problem is to

calculate properties of these moving pulses.

Model electron velocity versus field by

v{E)

= {

H\E

for E <E

for E > E

(21.50)

Electric

field

Figure 21.14. Schematic view of velocity as a function of field for electrons in gallium

arsenide. Electron mobility decreases for

fields

greater than the critical E

654 Chapter

21.

Optical Properties of Semiconductors

with /ii >

H2,

as shown in Figure 21.14. Denote net charge density as a

function of length x along the sample by n(x); n(x) vanishes far from the

pulse. Denote current by

and take it to be of the form

j = nev(E)-eD—, (21.51)

where e is the electron charge and D is a diffusion constant.

(a) Write down Poisson's equation for the electric field, and an equation relating

the current j and the time rate of change of the charge density n. In order to

describe a pulse moving at velocity v*, assume that all spatially varying fields

are moving in steady state at velocity v*, and use this assumption to simplify

the relation between current and charge density.

(b) Look for a solution such that charge density vanishes for large negative x and

large positive x. Assume that for large negative x the electric field approaches

a low value E\ < E

, and for large positive x it approaches a larger value

Ei > E

. Find an expression for dn/dE, and integrate it to find charge as a

function of electric field.

p,\E\

H2E2

= v*. (21.52)

Adding to these equations the condition that charge density vanish when the

field reaches

E2,

find v* in terms of the mobilities p\, p2, and E

(d) Find the electric field E{x) for E < E

(e) As a pulse of this type moves across a sample, what happens to the potential

drop from one end of the sample to the other? Suppose that one demands that

the potential drop across the sample remain constant as the pulse moves in

its interior. Draw a sketch indicating how one could construct solutions that

would make it possible to satisfy this condition.

Frenkel exciton I: To model the Frenkel exciton, consider a variant of the

tight-binding model. At each site of a lattice there is an electron, which can

occupy one of two states, an atomic ground state and an atomic excited state.

Denote the Wannier states by w

—

R) and w

—

R); suppose they are real

functions and are localized near the atomic sites.

When the atoms are assembled into a crystal, the ground state is given to good

approximation by a Hartree-Fock state, consisting of an antisymmetrized

product for N electrons of w

(?i

—

R), the product being taken over all N sites

in the lattice. In second-quantized notation, the ground state of the system is

) = TTct |0). „

(21.53)

11 R,g c- creates an electron in the atomic ground

state at site R.

Problems

655

Suppose now that the only quantum states of importance are the ground state

and the excited states where one electron at one site moves from the ground

to the excited state; such a state is

i*>=4%i*°>-

(2L54)

(a) Choose the zero of energy so that (/J|Ä|$o) = 0- If the states in Eqs. (21.53)

and (21.54) are all that matter, then the Hamiltonian is given by

Ä = |$o><$o|Ä|$oX$o| +£ \R)(R\Ü\R')(R'\. (21.55)

RR'

Show that

is an eigenstate, and find the eigenvalue. Use the fact that

R'\!K\R}

de-

pends only upon R'.

(b) Now proceed to evaluate the matrix elements

(R\$C\R')

that determine this

eigenvalue. The Hamiltonian "K contains kinetic energy and Coulomb inter-

action terms. There are many contributions to (R\Ji\R); however, this matrix

element causes something proportional to the unit operator to be added to the

Hamiltonian, is equivalent to a constant shift in energy, and can be neglected.

Therefore, focus on evaluating

(R\$C\R')

for R ^ R'. Assuming that Coulomb

interaction leads to the largest interaction between sites, take

- e ^—^ 1 This is the Schrödinger representation; do not

JX = — y p; z^-r. confuse the independent variable r/ with the (21.57)

TTv

\n-n>\

wave function centered at

which is denoted

by \R).

Write the Hamiltonian (21.57) in second quantized notation, using Eq. (C.10).

(R\X\R')=e

[pz

J '1

dr\dr2

-R')w

-R)w

-R')w

-R)

-R)w

-R')w

{ri -R')w

-R)

(21.58)

(d) Argue from localization of the wave functions that the second term in

Eq. (21.58) can be neglected, and

(R\Û\R') ^e

f $^% w

(h -R')w

-R)w

{7, -R')w

-R).

\n-r

(21.59)

656 Chapter

21.

Optical Properties of Semiconductors

5. Frenkel exciton II:

(a) Continuing the study of the Frenkel exciton, let 5R = R

—

R', 5r\

=~r\—R'

and 6r

= r

—R. Show that

\ri-r

2{Sf

- 8r

)

■

SR (8r

- 8r



(b) Define the dipole moment

Show that

p = e dr w

(r)w

(r)r.

(5\<îf\S'\~

^P^

)

mR)

~ÔRÏ-

^RÎ-

-P-

V(p-V

OR'

2(87]

- 8r

)

■

(21.60)

(21.61)

(21.62)

(21.63)

integrals to find the final expression.

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22.

Optical Properties of Insulators

22.1 Introduction

Insulators do not allow charge to travel long distances at low frequencies, but that

does not mean that their response to electric fields is simple. Even the response of

insulators to static fields is complicated, expressed in terms of quantities that are

difficult to define, and hard to calculate precisely. As the frequency of incoming

radiation increases, insulators become difficult to distinguish from metals, because

radiation passes through both in similar ways.

Almost all optically transparent materials are insulators. Insulators have no free

electrons to interact with light, and phonons provide the main way for insulating

crystals to interact with radiation. If phonons provided the only mechanism, insu-

lating crystals would generally be transparent. However, very small concentrations

of impurities and vacancies create localized states that have a large effect on optical

properties. The study of these color centers is one of most important topics in the

optical properties of insulators, and it has had an impact on applications ranging

from photography to the laser.

22.2 Polarization

Polarization is Ill-Defined. As defined by Eq. (20.6) the polarization is not pre-

cisely determined. Changes in the polarization are given by the current

but the

actual value of the polarization at any given time is rather arbitrary. This uncer-

tainty is unavoidable, as indicated in Figure 22.1. Depending upon how charges

are grouped together, the polarization can have any value desired. Measurable

quantities are not affected by changes in the definition of the polarization, because

every change in definition of bulk polarization simultaneously implies changing

the amount of surface charge. However, Figure 22.1 actually understates the diffi-

culty of defining polarization, because charge distributions are not point-like, but

distributed continuously as in Figure 11.1. By changing the origin of the unit cell

continuously, the dipole moment of the cell can also be made to vary continuously.

So long as one consistently keeps track of polarization and surface charge, there

should not ultimately be any difficulties.

22.2.1 Ferroelectrics

The need to define polarization becomes especially acute for the pyroelectric and

ferroelectric materials. Pyroelectric solids develop an electric field upon heating

and cooling. Any free charge in the environment quickly rushes to mask it, but un-

til then the field is measurable. Heating and cooling produce electric fields because

659

Condensed Matter

Physics,

Second Edition

by Michael P. Marder

660

Chapter 22. Optical Properties of Insulators

Figure

22.1.

When charges are distributed throughout a solid, the polarization density de-

pends upon how they are grouped. In (A), positive and negative charges have been grouped

with their nearest neighbors. In (B), they have been grouped so that the polarization now

points in the opposite direction. In compensation, there is now a substantial surface charge.

they cause expansion and contraction of the crystal along various axes. Each unit

cell is electrically polarized, and as the dipole moment changes, an electric field

develops. The ferroelectric crystals have an additional property, which is that each

unit cell can choose between two symmetrical forms where the dipole moments

point in opposite directions. The crystal can be caused to flip between one and the

other by application of an external field. Usually, a ferroelectric has a transition

temperature T

above which the polarization vanishes, and the substance becomes

paraelectric. External electric fields easily cause ferroelectrics to flip between the

two equivalent stable states just below the transition temperature, whereas far be-

low the transition temperature, electric fields powerful enough to cause the polar-

ization to flip may not be available in the laboratory.

The explanation of ferroelectrics in terms of dipole moments in each unit cell

seems intuitively clear, but is thrown into question to the- extent that this dipole

moment cannot be defined. Because thousands of ferroelectric compounds have

been created, measured, and tabulated (for example, in Landolt and Bornstein (New

Series) vols. 16 and 28), the best definition of the polarization is in terms of the

actual operations performed in experiment.

Figure 22.2. The spontaneous electric polarization of a sample is measured by cooling it

below the ferroelectric transition temperature in an electric field, and then measuring the

total charge that flows between the ends.

Polarization

661

The experiments depend upon being able to access the paraelectric phase above

, where the polarization can safely be taken to be zero. As indicated in Figure

22.2,

the experiment begins with an insulating sample in the paraelectric phase,

with electrical contacts at the ends. The aim is to measure all the charge that flows

between ends of

the

sample, so the electrical contacts at the ends must be very good

and are often obtained with liquid electrolytes. An electric field is applied to the

sample, and the temperature is then lowered through the ferroelectric transition.

The applied field ensures that the sample forms a single domain of dipoles all

pointing in the same direction. Finally, the applied electric field can be turned

off. The total charge Q that has flowed from one end of the sample to the other

since the beginning of

the

experiment gives the total polarization through P = Q/A,

following Eq. (20.6). It is really the change in polarization between two states of

the solid that is being measured, and nothing else can precisely be defined.

22.2.2 Berry phase theory of polarization

With the advent of band structure calculations, considerable effort has been ex-

pended in learning how to calculate ferroelectric polarizations or, more precisely,

to find the difference in polarization between the two symmetrical states of the

ferroelectric.

The computations are based conceptually upon the experimental procedure of

starting in a paraelectric phase without polarization and adiabatically changing the

temperature T until spontaneous polarization appears. This process corresponds

exactly to the situation in which the geometric phases of Section 8.4.3 should arise,

as pointed out by Resta (1994) and Kingsmith and Vanderbilt (1993).

Suppose that at time t = 0, one has an insulating crystal without any bulk po-

larization. Suppose that a long time later, by slowly changing a parameter such as

temperature, a polarization appears. Denote the Wannier functions at the later time

by w

(r). The dipole moment per unit cell changes from 0 to

P = \ / dr 7

(7)\

. Sum only over occupied bands n. (22.1)

As discussed at the beginning of Section 8.4.4, Wannier functions in insulators

are localized, and the integral in Eq. (22.1) should converge. Indeed, the integral in

Eq. (22.1) was defined in Eq. (8.56), making it possible to rewrite in terms of the

Berry connection, Eq. (8.54), as

*=-££* = -££*«*•

(222)

" nk

A comparison of theory and experiment for some crystals appears in Table 22.1.

22.2.3 Clausius-Mossotti Relation

The Clausius-Mossotti relation is the classic link between atomic polarization and

dielectric constants. Suppose one takes a single unit cell out a solid, subjects it to

662 Chapter 22. Optical Properties of Insulators

Table 22.1. Comparison of theory and experiment for

spontaneous polarization P of selected perovskite crystals.

KNb0

PbZr

1/2

BaTi0

Theory (C/m

) 0.35

0.74

0.2-0.4

Experiment (C/m

) 0.37 075^ 0.27

Sources: (a) Resta et

al.

(1993), (b) Saghi-Szabö et

al.

(1999) (c)

Berlincourt and Krueger (1959) (d)Wahl et al. (2008)

a uniform electric field of strength Ë, and finds that it responds by developing a

dipole moment p = aE. The question is how a periodic solid tiled with such unit

cells will respond to an external electric field. The dipole moment in each unit

cell creates electric fields that affect all other unit cells, so the problem must be

solved self-consistently. It is tricky because the sums that arise are divergent and

must be handled carefully. In fact, the field created within a solid by a collection

of dipoles depends upon the way the sum is cut off far away; that is, it depends

upon the macroscopic shape of the sample. The field also depends upon details of

how dipoles are arranged at the atomic level. The influence of local arrangement

simplifies considerably in the case of a crystal of cubic symmetry, so the following

discussion will focus on this special case.

Consider a sphere of dielectric material. The reasons to focus upon a sphere

are as follows:

When a sphere of dielectric material is placed into a uniform external electric

field of strength

ÈQ,

classical electrostatics shows that the polarization P and

electric field E inside the sphere are uniform, as shown in Figure 22.3 and

related by

£ = £o-yP. (22.3)

Therefore, one should expect that all the unit cells in a sample of this shape

develop the same microscopic polarization p. In a sample of arbitrary shape,

electrostatics would predict that fields and dipole moments should vary macro-

scopically in a complicated way throughout the sample, making it difficult to

relate microscopic and macroscopic behavior.

In a cubic crystal, the unit cell at the center of a sphere sees fields with cubic

symmetry, a fact that can be used to resolve conclusively the value of divergent

sums.

Focus on the unit cell at the center of the sphere. The field

acting upon it

is due to two sources. First, this unit cell sees the external field

EQ.

Second, it sees

the field

*.=-£V^

= E3^-|

(22.4)

R^O