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Patterson, Bailey - Solid State Physics Introduction to theory

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528 9 Dielectrics and Ferroelectrics

P. Aigrain in 1960. Since their frequency depends on the Hall coefficient, they have been used to measure it in solids. Their dispersion relation shows that lower frequencies have lower velocities. When high-frequency helicons are observed in the ionosphere, they are called whistlers (because of the way their signal sounds when converted to audio).

For electrons (charge −e) in E and B fields with drift velocity v, relaxation time τ, and effective mass m, we have

d

 

1

 

 

m

 

+

τ

v = −e(E + v × B).

(9.84)

 

dt

 

 

 

Assuming B = Bkˆ and low frequencies so ωτ << 1, we can neglect the time derivatives and so

vx = − eτmEx ωcτvy ,

v

y

= −

eτEy

+ω τv

x

,

(9.85)

 

 

 

 

 

m

c

 

 

vz

= −

eτEz

,

 

 

 

 

 

 

 

 

 

 

 

 

 

m

 

 

 

 

where ωc = eB/m is the cyclotron frequency. Letting, σ0 = m/ne2τ, where n is the number of charges per unit volume, and the Hall coefficient RH = −1/ne, we can write (noting j = −nev, j = v/RH):

vx = σ0 RH (Ex + Bvy ) ,

(9.86)

vy = σ0 RH (Ey Bvx ) .

(9.87)

Neglecting the displacement current, from Maxwell’s equations we have:

 

× B = μ0 j,

 

× E = −

B

.

 

 

 

 

 

 

t

 

Assuming ·E = 0 (overall neutrality), these give

 

2 E = μ0

j

.

(9.88)

 

 

 

t

 

If solutions of the form E = E0exp[i(kx ωt)] and v = v0exp[i(kx ωt)] are sought, we require:

k 2 E = −iωμ0

 

v

,

 

 

 

 

 

 

 

RH

Ex = i

ωμ0

 

vx

 

,

 

k 2

 

RH

 

 

 

 

 

 

 

9.4 Dielectric Screening and Plasma Oscillations (B)

529

 

 

 

 

 

 

 

Ey = i

 

ωμ0

vy

 

 

.

 

 

 

 

 

 

 

 

 

k 2

 

 

RH

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Thus

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

iσ

ωμ

0

 

 

 

σ

 

R

 

 

 

 

Bv

 

= 0,

1

0

 

 

v

x

0

H

 

y

 

 

 

 

 

k 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

σ

 

 

R

 

Bv

 

+

 

iσ

 

 

ωμ

0

 

 

= 0.

0

H

x

1

0

 

 

 

v

y

 

 

 

 

 

 

 

 

 

 

k 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Assuming large conductivity, σ0ωμ0/k2 >> 1, and large B, we find:

ω =

k 2

 

RH

 

B =

k

2

 

B,

 

 

 

μ0

 

 

μ0ne

or the phase velocity is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

v p =

 

ω

=

ωB

 

,

 

 

 

 

k

 

 

μ0ne

 

 

(9.89)

(9.90)

(9.91)

independent of m. Note the group velocity is just twice the phase velocity. Since the plasma frequency ωp is (ne2/0)1/2, we can write also

v p = c

ωωc

.

(9.92)

ω2p

 

 

 

Typically vp is of the order of sound velocities.

9.4.2 Alfvén Waves (EE)

Alfvén waves occur in a material with two kinds of charge carriers (say electrons and holes). As for helicon waves, we assume a large magnetic field with electromagnetic radiation propagating along the field. Alfvén waves have been observed in Bi, a semimetal at 4 K. The basic assumptions and equations are:

1.× B = μ0 j, neglecting displacement current.

2.× E = −∂B/∂t, Faraday’s law.

3.ρv· = j × B, where v is the fluid velocity, and the force per unit volume is dominated by magnetic forces.

4.E = −(v × B), from the generalized Ohm’s law j/σ = E + v × B with infinite conductivity.

5.B = Bx ˆi + By ˆj , where Bx = B0 and is constant while By = B1(t).

530 9 Dielectrics and Ferroelectrics

6.Only the jx, Ex, and vy components need be considered (vy is the velocity of the plasma in the y direction and oscillates with time).

7.v· = ∂v/∂t, as we neglect (v· )v by assuming small hydrodynamic motion. Also we assume the density ρ is constant in time.

Combining (1), (3), and (7) we have

 

 

 

 

vy

 

 

 

 

 

 

 

 

 

 

 

 

 

B

 

 

μ

0

ρ

 

 

=[( × B) × B]

y

 

1

B .

t

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

By (4)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ez = −B0vy ,

 

 

 

 

 

 

 

so

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

μ0 ρ

Ez

= B

 

B1 .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B

t

 

0

 

 

x

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

By (2)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ez

 

= −

B1

,

 

 

 

 

 

 

 

 

 

 

 

 

 

x

 

t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

so

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2E

 

 

 

 

B2

2B

 

 

 

 

B2

 

2E

 

t2

z

= −

 

0

 

 

 

 

1

= +

 

 

0

 

 

 

z .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

μ0 ρ xt

 

 

 

μ0ρ x2

This is the equation of a wave with velocity

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

vA =

B

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

μ0 ρ

 

 

 

 

 

 

 

 

(9.93)

(9.94)

(9.95)

(9.96)

the Alfvén velocity. For electrons and holes of equal number density n and effective masses me and mh,

vA =

B

,

(9.97)

μ0n(me + mh )

 

 

 

Notice that vA = (B2/μ0ρ)1/2 is the velocity in a string of tension B2/μ0 and density ρ. In some sense, the media behaves as if the charges and magnetic flux lines move together.

A unified treatment of helicon and Alfvén waves can be found in Elliot and Gibson [9.5] and Platzman and Wolff [9.15]. Alfvén waves are also discussed in space physics, e.g. in connection with the solar wind.

9.5 Free-Electron Screening

531

 

 

9.5 Free-Electron Screening

9.5.1 Introduction (B)

If you place one charge in the midst of other charges, they will redistribute themselves in such a way as to “damp out” the long-range effects of the original charge. This long-range damping is an aspect of screening. Its origin resides in the Coulomb interactions of charges. This phenomenon was originally treated classically by the Debye–Huckel theory. A semiclassical form is called the Thomas– Fermi Approximation, which also assumes a free-electron gas. Neither the Debye– Huckel Theory nor the Thomas–Fermi model treats screening accurately at small distances. To do this, it is necessary to use the Lindhard theory.

We begin with the linearized Thomas–Fermi and Debye–Huckel methods and show how to use them to calculate the screening due to a single charged impurity. Perhaps the best way to derive this material is through the dielectric function and derive the Lindhard expression for it for a free-electron gas. The Lindhard expression for ε(ω→0, q) for small q then gives us the Thomas–Fermi expression. Generalization of the dielectric function to band electrons can also be made. The Lindhard approach follows in Sect. 9.5.3.

9.5.2 The Thomas–Fermi and Debye–Huckel Methods (A, EE)

We assume an electron gas with a uniform background charge (jellium). We assume a point charge of charge Ze (e > 0) is placed in the jellium. This will produce a potential φ(r), which we assume to be weak and to vary slowly over a distance of order 1/kF where kF is the wave vector of the electrons whose energy equals the Fermi energy. For distances close to the impurity, where the potential is neither weak nor slowly varying our results will not be a very good approximation. Consistent with the slowly varying potential approximation, we assume it is valid to think of the electron energy as a function of position.

Ek =

2k 2

eϕ(r) ,

(9.98)

2m

 

 

 

where is Planck’s constant (divided by 2π), k is the wave vector, and m is the electronic effective mass.

In order to exhibit the effects of screening, we need to solve for the potential φ. We assume the static dielectric constant is ε and ρ is the charge density. Poisson’s

equation is

 

2ϕ = ερ ,

(9.99)

532 9 Dielectrics and Ferroelectrics

where the charge density is

ρ = eZδ (r) + n0e ne ,

(9.100)

where eZδ(r) is the charge density of the added charge. For the spin 1/2 electrons obeying Fermi–Dirac statistics, the number density (assuming local spatial equilibrium) is

 

n =

 

 

1

 

dk

,

 

 

 

(9.101)

 

exp[β(Ek μ)] +1 4π3

 

 

 

 

 

 

 

where β = 1/kBT and kB is the Boltzmann constant. When φ = 0, then n = n0, so

n0

= n0 (μ) =

 

 

1

 

 

 

 

dk

.

(9.102)

 

exp[β((

 

 

 

 

 

 

 

 

2k 2 / 2m) μ)] +1 4π3

 

Note by (9.98) and (9.102), we also have

 

 

 

 

 

 

 

 

 

n = n0[μ + eϕ(r)] .

 

 

 

 

 

(9.103)

This means the charge density can be written

 

 

 

 

 

 

 

ρ = eZδ(r) + ρind (r) ,

 

 

 

 

(9.104)

where

 

 

 

 

 

 

 

 

 

 

 

 

ρind (r) = −e[n (μ + eϕ(r)) n (μ)] .

(9.105)

 

 

0

 

 

0

 

 

 

 

 

We limit ourselves to weak potentials. We can then expand n0 in powers of φ and obtain:

ρind (r) = −e2

n0

ϕ(r) .

(9.106)

μ

 

 

 

 

 

 

The Poisson equation then becomes

 

 

 

 

2ϕ = − 1

Zeδ(r) e2

n0

ϕ(r) .

(9.107)

μ

ε

 

 

 

 

A convenient way to solve this equation is by the use of Fourier transforms. The Fourier transform of the potential can be written

ϕ(q) = ϕ(r) exp(iq r)dr ,

(9.108)

9.5 Free-Electron Screening

533

 

 

with inverse

ϕ(r) =

1

ϕ(q) exp(iq r)dq ,

 

(2π)3

 

and the Dirac delta function can be represented by

 

δ (r) =

 

 

1

 

exp(iq r)dq .

 

(2π)3

 

 

 

 

 

 

 

 

Taking the Fourier transform of (9.107), we have

 

 

2

 

1

 

 

 

 

2 n0

 

q

ϕ(q) =

 

 

Ze e

 

 

 

ϕ(q) .

ε

 

 

 

μ

 

 

 

 

 

 

 

 

 

 

 

Defining the screening parameter as

 

 

 

 

 

 

 

 

 

 

 

kS2

 

e2

n

 

 

 

 

=

 

 

 

0

,

 

 

 

 

ε

 

 

 

 

 

 

 

 

 

μ

 

 

we find from (9.111) that

(9.109)

(9.110)

(9.111)

(9.112)

ϕ(q) =

Ze

1

.

ε

q2 + kS2

 

 

Then, using (9.109), we find from (9.113) that

ϕ(r) = 4Zeπεr exp(kS r) .

Equations (9.112) and (9.114) are the basic equations for screening. For the classical nondegenerate case, we have from (9.102)

(9.113)

(9.114)

n0 (μ) = exp(βμ)exp(β 2k 2 / 2m)

dk

,

(9.115)

4π3

 

 

 

 

 

 

 

so that by (9.112)

 

 

 

 

 

 

 

kS2

 

e2

n

 

 

 

 

=

 

0

,

 

 

(9.116)

ε

kBT

 

 

 

 

 

 

 

 

we get the classical Debye–Huckel result. For the degenerate case, it is convenient to rewrite (9.102) as

n0( μ ) = D( E ) f ( E )dE ,

(9.117)

534 9 Dielectrics and Ferroelectrics

so

 

 

 

 

n0

= D( E )

f

dE ,

(9.118)

μ

μ

 

 

 

where D(E) is the density of states per unit volume and f(E) is the Fermi function

f (E) =

 

1

.

(9.119)

exp[β(E μ)] +1

since

 

 

 

f (E)

δ (E μ) ,

(9.120)

μ

 

 

 

at low temperatures when compared with the Fermi temperature; so we have

nμ0 D(μ) .

Since the free-electron density of states per unit volume is

D(E) =

1

 

 

2m 3/ 2

E ,

2π

 

 

 

 

 

 

 

2

 

 

2

 

 

and the Fermi energy at absolute zero is

 

 

 

 

 

 

μ =

 

2

(3π 2n )2 / 3

,

 

 

 

 

 

 

2m

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

where n0 = N/V, we find

 

 

 

 

 

 

 

 

 

 

D(μ) =

 

3n0

 

,

 

 

 

2μ

 

 

 

 

 

 

 

 

 

(9.121)

(9.122)

(9.123)

(9.124)

which by (9.121) and (9.112) gives the linearized Thomas–Fermi approximation. If we further use

μ =

 

3

k

B

T

 

,

 

(9.125)

2

 

 

 

 

 

F

 

 

 

we find

 

 

 

 

 

 

 

 

 

 

 

k 2

 

e2

 

 

n

 

 

 

 

=

 

 

 

 

 

0

 

 

,

(9.126)

 

ε

k

T

 

 

S

 

 

F

 

 

 

 

 

 

 

 

B

 

9.5 Free-Electron Screening

535

 

 

which looks just like the Debye–Huckel result except T is replaced by the Fermi temperature TF. In general, by (9.112), (9.118), (9.119), and (9.122), we have for free-electrons,

 

2

 

e2n

 

F

(η)

 

 

 

k

=

 

 

 

0

 

1/ 2

 

,

 

(9.127)

εk

 

T

F

(η)

 

 

S

 

B

 

 

 

 

 

 

 

 

 

F

1/ 2

 

 

 

 

where η = μ/kBT and

 

 

 

 

 

 

 

 

 

 

 

 

F1/ 2

(η) =

 

 

xdx

 

,

(9.128)

 

 

 

 

0

 

exp(x η) +1

 

 

is the Fermi integral. Typical screening lengths 1/kS for good metals are of order 1 Å, whereas for typical semiconductors 60 Å is more appropriate. For η << −1, F′1/2(η)/F1/2(η) ≈ 1, which corresponds to the classical Debye–Huckel theory, and for η >> 1, F′1/2(η)/F1/2(η) = 3/(2η) is the Thomas–Fermi result.

9.5.3 The Lindhard Theory of Screening (A)

Here we do a more general discussion that is self-consistent.5 We start with the idea of an external potential that determines a set of electronic states. Electronic states in turn give rise to a charge density from which a potential can be determined. We wish to show how we can determine a charge density and a potential in a self-consistent way by using the concept of a frequencyand wave-vector- dependent dielectric constant.

The specific problem we wish to solve is that of the self-consistent response to an applied field. We will assume small applied fields and linear responses. The electronic response to the applied field is called screening, and it arises from the interaction of the electrons with each other and with the external field. Only

screening by a free-electron gas will be considered.

Let a charge ρext be placed in jellium, and let it produce a potential φext (by itself). Let φ be the potential caused by the extra charge, the free-electrons, and the uniform background charge (i.e. extra charge plus jellium). We also let ρ be the corresponding charge density. Then

 

 

ext

 

2ϕext = −

 

ρ

,

(9.129)

 

ε

2ϕ = −

ρ

.

(9.130)

ε

The induced charge density ρind is then defined by

 

ρind = ρ ρext .

(9.131)

5This topic is also treated in Ziman JM [25, Chap. 5], and Grosso and Paravicini [55 p245ff].

536 9 Dielectrics and Ferroelectrics

We Fourier analyze the equations in both the space and time domains:

q2ϕext (q,ω) = ρext (q,ω) ,

ε

q2ϕ(q,ω) = ρ(q,ω) ,

ε

ρ(q,ω) = ρext (q,ω) + ρind (q,ω) .

Subtracting (9.132a) from (9.132b) and using (9.132c) yields:

εq2[ϕ(q,ω) ϕext (q,ω)] = ρind (q,ω) .

We have assumed weak field and linear responses, so we write

ρind (q,ω) = g(q,ω)ϕ(q,ω) ,

which defines g(q, ω). Thus, (9.133) and (9.134) give this as

εq2[ϕ(q,ω) ϕext (q,ω)] = g(q,ω)ϕ(q,ω) .

Thus,

ϕ(q,ω) = ϕext (q,ω) ,

ε(q,ω)

where

ε(q,ω) =1g(q,ω) .

εq2

(9.132a)

(9.132b)

(9.132c)

(9.133)

(9.134)

(9.135)

(9.136)

(9.137)

To proceed further, we need to calculate ε(q, ω) directly. In the process of doing this, we will verify the correctness of the linear response assumption. We write the Schrödinger equation as

H 0 k = Ek k .

(9.138)

We assume an external perturbation of the form

δV (r,t) = [V exp(i(q r +ωt)) +V exp(i(q r +ωt))]exp(αt) . (9.139)

The factor exp(αt) has been introduced so that the perturbation vanishes as t = −∞, or in other words, as the perturbation is slowly turned on. V is assumed real. Let

H = H 0 +δV .

(9.140)

9.5 Free-Electron Screening

537

 

 

We then seek an approximate solution of the time-dependent Schrödinger wave equation

 

 

 

 

 

 

 

= i

ψ

.

 

 

(9.141)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

t

 

 

 

We seek solutions of the form

 

 

 

 

 

 

 

 

 

 

 

 

ψ = kCk(t) exp(iEkt /

) k.

(9.142)

Substituting,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k(H 0 +δV )Ck(t) exp(iEkt /

) k

(9.143)

 

 

 

 

 

= i

kCk(t) exp(iEkt /

) k.

 

 

 

 

 

 

 

 

 

 

 

 

t

 

 

 

 

 

 

 

Using (9.138) to cancel two terms in (9.143), we have

 

 

kδVCk(t) exp(iEkt /

) k

= i

kCk(t) exp(iEkt / ) k. (9.144)

Using

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k′′ k

= δkk′′ ,

 

 

(9.145)

 

 

 

 

 

k′′ δV k

= k′′ δV k′ ± q δ k′′±q ,

(9.146)

 

 

 

 

 

 

 

 

 

 

 

k

 

 

Ck′′(t) =

 

1

Ck′′+q exp(iEk′′+qt /

) k′′ δV k′′ + q exp(iEk′′t /

)

 

i

 

 

 

 

 

 

 

 

 

(9.147)

+

 

1

Ck′′−q exp(iEk′′−qt /

) k′′ δV k′′ − q exp(iEk′′t /

 

) .

 

 

i

 

 

 

 

 

 

 

 

 

 

Using (9.139), we have

 

 

 

 

 

 

 

 

Ck′′(t) =

 

1

Ck′′+q exp(i(Ek′′+q Ek′′)t /

)V exp(iωt) exp(αt)

 

i

 

 

 

 

 

 

 

 

 

 

 

(9.148)

 

 

 

1

 

 

 

 

 

 

 

 

+

 

Ck′′−q exp(i(Ek′′−q Ek′′)t /

)V exp(iωt) exp(αt) .

 

 

 

 

 

i

 

 

 

 

 

 

 

 

 

 

We assume a weak perturbation, and we begin in the state k with probability f0(k), so we have

Ck′′(t) = f0 (k)δk′′,k + λCk(1′′) (t) .

(9.149)

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