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118

5 Metals

which are perpendicular to q. The matrix element is indeed zero if k and q are parallel. Combining all terms yields for the complex conductivity

ˆ

= ωm 8 2kF

2

 

ω

qvF

 

 

2

 

 

 

 

 

 

 

 

 

 

 

+

 

+

 

 

 

 

 

 

 

 

 

 

σ (q, ω)

 

iN e2

3

 

 

q

 

 

 

3

+ i

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

ω+i

 

 

 

 

 

 

 

16q

 

 

2kF

 

 

 

 

2

 

 

 

 

q

 

 

1

 

 

 

 

qvF

 

 

 

 

 

 

q

 

ω+i

 

 

 

 

 

 

3kF

1

 

 

q

 

ω + i

 

 

 

 

 

Ln

 

2kF

 

qvF

+ 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

qvF

 

 

 

 

 

 

16q

 

 

 

 

qvF

 

 

 

 

 

 

 

 

2kF

1

 

 

2kF +

 

 

2

2

 

q

 

 

ω+i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

ω+i

+ 1

 

 

 

 

 

 

3kF

 

1

 

 

 

q

 

ω + i

 

 

 

 

 

Ln

 

 

2kF

+

qvF

 

 

. (5.3.6)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In the limit of small relaxation rate (τ → ∞) and using Eq. (5.2.23) for the expression for the logarithm Ln, these equations reduce to

 

 

 

4 mqvF

 

 

 

3π

N e2

 

σ (q, ω)

 

N e2

3π k

F

1

 

 

 

 

 

=

 

ωm

16q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

2

 

 

 

 

2

2

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

+

 

 

< 1

 

qvF

 

 

 

 

2kF

 

qvF

2kF

 

 

 

 

2

2

 

 

ω

 

q

 

 

 

 

 

ω

 

 

 

 

q

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

< 1 <

 

qvF

2kF

qvF

2kF

qvF

 

ω

 

 

q

 

 

ω

 

 

q

 

ω

 

 

 

 

 

 

 

 

 

ω

 

 

q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

> 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2kF

 

 

 

 

 

 

 

 

 

 

 

 

 

qvF

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

σ2(q, ω) =

ωm

8 2kF

2

+ 3

qvF

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+ 1

 

 

 

 

 

 

 

 

 

 

N e2

 

3

 

 

q

 

 

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

q

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

2kF

qvF

2

 

 

 

 

ω

 

 

16q

 

 

 

 

2kF

 

qvF

 

 

 

 

3kF

1

 

 

 

 

q

 

 

 

ω

 

 

 

 

 

 

ln

 

2kF

qvF

+ 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

q

 

 

ω

 

 

 

 

 

3kF

 

 

 

 

 

 

q

 

 

 

ω

 

 

 

 

 

 

 

 

 

+

 

 

+

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2kF

 

qvF

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ln

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

ω

 

 

 

 

16q

2kF

+ qvF

 

 

 

 

 

 

+

 

1

 

 

 

 

 

 

 

2kF

 

qvF

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+ 2qkF

(5.3.7a)

. (5.3.7b)

Both the real and imaginary parts of the conductivity are plotted in Fig. 5.12 as functions of frequency ω and wavevector q. Of course, it is straightforward to calculate the dielectric constant ˆ(q, ω) and Lindhard response function χˆ (q, ω).

The overall qualitative behavior of the conductivity components can easily be inferred from Fig. 5.11. First, absorption can occur only when electron–hole excitations are possible; outside this region σ1(q, ω) = 0. On the left hand side of Fig. 5.11, in this dissipationless limit, for small q a power expansion gives

σ2(q, ω) =

N e2

1 +

q2v2

+

 

3q4v2

(5.3.8)

ωm

 

5ω2

16kF2ω2 + · · · .

 

 

 

 

F

 

 

F

 

5.3 Transverse response for arbitrary q values

119

Fig. 5.12. (a) Real and (b) imaginary parts of the frequency and wavevector dependent conductivity σˆ (q, ω) calculated after Eqs (5.3.7) in the limit τ → ∞. The conductivity is plotted in arbitrary units, the frequency is normalized to the Fermi frequency ωF = vFkF/2,

and the wavevector is normalized to the Fermi wavevector kF. There are no excitations for

ω < qvF 2q2vF/ kF.

120

5 Metals

The region qωvF + 2qkF < 1, also indicated in Fig. 5.11, is called the quasi-static limit; for small q values this indicates the region where, due to the low frequency of the fluctuations, screening is important. For small q values, series expansion gives in the τ → ∞ limit

σ1(q, ω) =

3 N e2π

1

 

ω2

q2

 

 

(5.3.9)

 

 

 

 

 

,

4 qvFm

q2vF2

4kF2

where we have retained an extra term compared with the Boltzmann equation (5.2.22).

5.4 Longitudinal response

So far the discussion in this chapter has focused on the transverse response, i.e. the response of the medium to a vector potential A(q, ω). Now we discuss the response to static (but later also time dependent) charges. First, the low frequency limit (ω 0) is analyzed; this leads to the well known Thomas–Fermi screening of the electron gas. Next, the response as obtained from the selfconsistent field approximation (Section 4.3.1) is discussed in detail; this gives the response functions appropriate for arbitrary q values. Finally, we examine the collective and single-particle excitations of an electron gas at zero temperature. Although we confine ourselves to the framework of the classical electrodynamics, and first quantization is used, the problem can also be treated in the corresponding second quantized formalism [Hau94, Mah90].

5.4.1 Thomas–Fermi approximation: the static limit for q < kF

If we place an additional charge e into a metal at position r = 0, the conduction electrons will be rearranged due to Coulomb repulsion; this effect, called screening, can be simply estimated by the so-called Thomas–Fermi method.

Starting from Poisson’s equation (2.1.7) in Coulomb gauge

2 ind(r) = −4π δρ(r) = −4π [ρ(r) ρ0] = 4π e[N (r) N0] , (5.4.1)

where δρ(r) = ρ(r) ρ0 is the deviation from the uniform charge density, δ N (r) = N (r) N0 is the deviation from the uniform particle density, and (r) is the electrostatic potential at position r. We assume that (r) is a slowly varying function of distance r, and that the potential energy of the field V (r) = −e (r) is simply added to the kinetic energy of all the electrons (the so-called rigid band approximation); i.e.

 

 

h2

 

/

3

E(r) =

EF e ind(r) =

2¯m

3π 2 N (r) 2

 

 

 

5.4 Longitudinal response

121

 

h2

 

 

/

 

h2k2

 

EF =

2¯m

3π 2 N0

2

 

3 =

¯2mF

.

(5.4.2)

The carrier density can be expanded around the Fermi energy EF in terms of (r)

 

1

 

 

2m

 

3/2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

N (r) =

 

 

 

 

 

 

 

 

[EF e ind(r)]3/2

 

 

3π

2

h2

 

 

 

 

 

 

 

¯

 

 

 

 

3/2

 

 

 

 

 

 

3e ind(r)

 

 

1

 

 

2m

 

3/2

1

+ · · ·

 

 

 

 

 

 

 

 

 

EF

 

 

 

 

 

 

 

 

3π

2

h2

2

E

F

 

 

 

 

 

 

 

¯

 

 

 

e ind(r)

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(5.4.3)

=

N0

 

N0

 

 

 

 

 

 

,

 

 

 

 

 

 

 

 

 

2

 

 

EF

 

 

 

 

 

 

 

 

 

 

 

 

where the unperturbed carrier density is (see Eq. (5.4.2))

 

 

 

 

 

 

 

N0 =

 

1

 

2mEF/h¯ 2 3

/

2 .

 

 

 

 

 

(5.4.4)

 

 

 

 

 

 

 

 

 

 

 

 

 

3π 2

 

 

 

 

 

 

The Poisson equation is then written as

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6π N0e2

 

 

 

 

 

 

 

 

 

 

 

 

 

2 ind(r) = 4π eδ N (r) = −

 

 

 

 

 

 

ind(r) = −λ2 ind(r) ;

(5.4.5)

 

 

 

 

EF

 

where the Thomas–Fermi screening parameter λ

 

 

 

 

 

 

 

 

 

 

 

 

 

6

π N

2

 

 

 

1/2

 

 

 

 

 

 

 

 

 

 

1

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

/

 

 

λ =

 

0e

 

 

=

 

4π D(EF)e2

 

 

 

(5.4.6)

 

 

F

 

 

 

 

 

 

 

 

 

 

 

E

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

is closely related to the density of states at the Fermi level

 

 

 

 

 

 

D(EF) =

 

3N

=

mkF

.

 

 

 

 

 

 

(5.4.7)

 

 

 

2EF

π 2h2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¯

 

 

 

 

 

 

 

 

 

 

Interestingly, the screening does not depend on the mass of the particles or other transport properties because it is calculated in the static limit. Note that without assuming a degenerate free-electron gas we can write more generally

λclass = 4π e2

N0

 

1/2

,

 

 

 

 

E

 

 

which in the classical (non-degenerate) limit leads to the Debye–Huckel¨ screening

λDH = 4π N0e2/kBT 1/2

since ∂E/∂ N0 = 3kBT /2N0. For an isotropic metal the solution

 

ind

(r)

 

exp{−λr}

(5.4.8)

r

 

 

 

is called the screened Coulomb or Yukava potential. It is appropriate only close to the impurity; at distances larger than 1/2kF the functional form is fundamentally

122

5 Metals

different [Kit63]. The screening length 1is about 1 A˚ for a typical metal, i.e. slightly smaller than the lattice constant, suggesting that the extra charge is screened almost entirely within one unit cell.

The Fourier transform of Eq. (5.4.8) can be inverted to give the screened potential

 

 

 

 

(q, 0) =

4π e

 

;

 

 

 

(5.4.9)

 

 

 

 

 

 

 

 

 

 

 

q2 + λ2

 

 

 

and, using Eq. (3.1.31),

 

 

 

 

 

 

 

 

 

 

1(q, 0) = 1

4π δρ(q, 0)

= 1

+

λ2

= 1 +

6π N0e2

(5.4.10)

 

 

 

 

 

 

 

.

q2

(q, 0)

q2

EFq2

For q 0, the dielectric constant diverges; at distances far from the origin the electron states are not perturbed.

5.4.2 Solution of the Boltzmann equation: the small q limit

The Thomas–Fermi approximation is appropriate for static screening (ω = 0). If we are interested in the frequency dependence, we have to utilize the Boltzmann equation (5.2.7) for the case of a longitudinal field; from there we obtain the longitudinal current, in a fashion similar to Eq. (5.2.13). Assuming quasi-free electrons or a spherical Fermi surface, after some algebra we arrive at the following expression for the longitudinal dielectric constant:

(q, ω)

 

1

 

3ωp2

 

1

 

ω + i

Ln

 

qvF ω i

 

,

(5.4.11)

=

+ q2vF2

+

 

 

ˆ

 

 

2qvF

qvF

ω

i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

in analogy

to what is

derived for the

transverse

dielectric

constant using

Eq. (5.2.19). If q is small, utilizing the Taylor expansion in q yields the expression

 

 

ωp2

1 +

3q2vF2

 

ˆ(q, ω) = 1

 

 

+ · · ·

(5.4.12)

+ i/τ )2

5+ i/τ )2

for qvF < |ω + i|.

In the limit 1

ω, this expression is identical to

Eq. (5.2.21) obtained for the transverse response and resembles the Drude formula (5.1.9) as the q = 0 case. This is not surprising, for at q = 0 the distinction between longitudinal and transverse response becomes obsolete. For finite q, the expression tends to the static limit

ˆ

=

 

 

3ωp2

 

 

ω2

+

 

π ω

 

 

 

+ q2vF2

q2vF2

 

2qvF

 

(q, ω)

 

1

 

 

1

 

 

 

i

 

.

(5.4.13)

As the plasma frequency ωp2 = 4π N e2/m, the first term is identical to that derived within the framework of the Thomas–Fermi approximation.

5.4 Longitudinal response

123

5.4.3 Response functions for arbitrary q values

The semiclassical calculation of the q dependent conductivity is limited to small q values since it does not include the scattering across the Fermi surface; the appearance of vF in appropriate expressions merely reflects through the quantum statistics the existence of the Fermi surface. If we intend to take into account these processes, we must consider the time evolution of the density matrix as in Section 4.3.1. A more detailed discussion of the problems can be found in various textbooks [Cal91, Mah90, Pin66, Zim72].

The response functions to be evaluated are

 

 

4π

 

 

 

 

 

iω

 

ˆ(q, ω) = 1

 

χˆ (q, ω)

and

σˆ (q, ω) =

 

χˆ (q, ω)

(5.4.14)

q2

q2

with the Lindhard function

 

 

 

 

 

 

 

 

 

χˆ (q, ω) =

 

2e2

 

 

dk

 

 

fk0+q fk0

 

(5.4.15)

(2π )3

E(k + q) E(k) h+ i/τ )

 

 

 

 

 

 

 

 

 

 

 

¯

 

 

 

 

as derived in Eq. (4.3.21). Following a traditional treatment of the problem we evaluate the dielectric constant, but we also give the appropriate expressions for the conductivity. We will evaluate these expressions in three dimensions: the discussion of the one-dimensional and two-dimensional response functions can be found in Appendix F.

For T = 0, the Fermi–Dirac distribution fk0 = 1 for k < kF and 0 for k > kF, which allows the integral to be solved analytically,3 and we find:

 

 

2

 

 

 

+ 2q

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

q

 

ω+i

 

1

ˆ

= −

 

 

2 E

F)

2kF

qvF

 

 

 

 

q

ω+i

 

χ (q, ω)

 

e D(

1

 

kF

1

 

q

 

 

ω +

1

 

Ln

 

2kF

qvF

+ 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

qvF

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ω+i

 

 

2kF

 

 

 

+

 

 

 

+

 

 

 

 

 

2

 

 

 

q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2q

2kF

 

qvF

 

 

 

 

q

 

+

ω+i

1

 

 

 

 

 

 

 

 

2kF

 

 

 

 

 

 

 

kF

1

 

q

 

 

 

 

 

Ln

 

qvF

1

 

 

, (5.4.16)

 

 

 

 

 

ω + 1

 

 

 

2kF

+

qvF

+

 

 

where D(EF) is the density of states of both spin directions at the Fermi level,

3 Following the method given in footnote 2 on p. 111, the steps are as follows:

 

dk

 

 

 

 

 

 

 

 

(k kF)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2vF/2kF (vF/ kF)(k · q) + i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ω q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kF

k dk Ln

ω q2vF/2kF

qvFk/ kF

 

i

 

 

 

 

 

 

 

 

 

 

= π

0

 

 

ω

+

 

 

 

 

 

 

 

 

 

 

 

 

q2vF/2kF

qvFk/ kF

i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

q2vF/2kF)

 

 

 

q2vF2 + q2vF/2kF)2

Ln

 

ω q2vF/2kF

qvF

i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

+

 

 

 

 

 

ω q2vF/2kF

qvF

+ i

 

 

2q

 

 

qv

/ k2

 

 

 

 

 

 

2q2v2

/ k2

 

 

 

 

 

 

 

π k2

 

 

 

F

 

F

 

 

 

 

 

 

 

 

F

F

q 2

 

 

 

 

 

+

+

 

 

 

 

 

 

 

ω

 

 

q

 

 

 

1

1

ω

 

 

 

 

q2vF/2kF

i/τ ) qvF

 

 

=

 

F

 

 

 

 

+

 

 

 

+

 

 

Ln

 

+ +

.

 

 

2q

qvF

2kF

2

qvF

2kF

q2vF/2kF

i/τ ) qvF

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

The first part is solved in a similar way by using k + q instead of k.

124

5 Metals

the well known Lindhard response function. Again, Ln denotes the principal part of the complex logarithm. The frequency and wavevector dependence of the real and imaginary parts of the dielectric response function χˆ (q, ω) are calculated by utilizing Eq. (3.2.7). In the 10 limit,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2kF qvF

2

 

 

 

 

 

 

 

q

 

 

 

ω

+

1

 

 

 

= −

2

 

 

+ 2q

 

 

 

 

 

 

 

2kF

 

 

qvF

 

 

 

 

 

e2 D( F)

 

 

kF

 

 

 

 

 

 

 

 

 

q

 

 

 

 

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

2kF

qvF

1

 

 

χ1(q, ω)

 

 

 

 

E

 

1

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ln

 

 

q

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

q

 

 

 

 

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kF

 

 

 

 

q

 

 

 

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2k

 

 

 

 

 

 

qv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(5.4.17a)

 

 

 

 

 

2q

2kF

qvF

 

 

 

2kF

 

qvF

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ln

 

 

 

 

+

 

 

 

 

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

 

 

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2π ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

π kF

 

 

 

 

 

 

 

q

 

 

ω

 

2

 

 

 

 

 

 

 

 

+

 

 

 

 

 

< 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

qvF

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2kF

 

 

 

 

qvF

 

 

 

 

 

 

 

 

 

 

 

 

 

e D(EF)

 

q

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

χ

(q, ω)

 

 

 

 

 

 

2kF

qvF

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

 

ω

 

 

 

 

 

 

 

 

q

 

 

ω

 

 

 

 

= −

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

< 1<

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

| 2kF qvF |

2kF

+ qvF

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

| 2qkF qωvF | > 1 .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(5.4.17b)

Both functions are displayed in Fig. 5.13, where the frequency axis is normalized to vFkF and the wavevector axis to kF. It is obvious that both for q → ∞ and ω → ∞ the response function χ1(q, ω) 0, expressing the fact that the variations of the charge distribution cannot follow a fast temporal or rapid spatial variation of the potential.

In the static limit (Thomas–Fermi approximation, ω 0), the real part of the Lindhard dielectric response function of Eq. (5.4.17a) is reduced to

 

e2 D(

)

 

 

k

 

1 q

 

 

 

 

 

q 2kF

 

 

χ1(q, 0) = −

EF

 

1 +

F

 

 

 

 

ln

 

 

+

 

 

(5.4.18)

2

 

q

4 kF

q

2kF

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

for a spherical Fermi surface. For small q values

 

 

 

 

 

 

 

 

 

 

 

χ1(q, 0) =

δρ(q, 0)

=

λ2

=

3 N0e2

,

 

(5.4.19)

 

 

 

 

 

 

 

 

 

 

 

 

(q, 0)

 

4π

2

 

EF

 

 

 

which is independent of q. For finite q values, χ1(q) decreases with increasing q and the derivative has a logarithmic singularity at q = 2kF. The functional form is displayed in Fig. 5.14 along with the results found for one and two dimensions. The singularity is well understood and can be discussed by referring to Figs 5.10 and 5.11. By increasing q towards kF, the number of states which contribute to the integral (5.4.15), fk0+q fk0, progressively increases, until q is equal to 2kF, and the sum does not change when q increases beyond this value. When q 2kF there is a vanishingly small number of states, for which E(k + q) E(k) (these states giving, however, a large contribution to the integral), and a combination of these

5.4 Longitudinal response

125

Fig. 5.13. Wavevector and frequency dependence of the Lindhard response function χˆ (q, ω) of a free-electron gas at T = 0 in three dimensions calculated by Eqs (5.4.17).

(a) The real part χ1(q, ω) and (b) the imaginary part χ2(q, ω) are both in arbitrary units.

126

5 Metals

Lindhard function χ1(q) / χ1(q = 0)

0

1D

2D

3D

2kF Wavevector q

Fig. 5.14. The static Lindhard dielectric response function χ1 as a function of wavevector q at T = 0 in one, two, and three dimensions (1D, 2D, and 3D, respectively). Here χ1(q)/χ1(0) is plotted for normalization purposes. At the wavevector 2kF the Lindhard function χ1(q) diverges in one dimension, has a cusp in two dimensions, and shows a singularity in the derivative in three dimensions.

two effects leads to a decreasing χ1(q, ω) near q 2kF and to a logarithmical singularity in the derivative. For one dimension, near q = 2kF there is a large number of states with E(k + q) E(k), and this in turn leads to a singularity in the response function in one dimension also shown in Fig. 5.14. As expected, the two-dimensional case lies between these two situations, and there is a cusp at q = 2kF.

All this is related to what is in general referred to as the Kohn anomaly [Koh59], the strong modification of the phonon spectrum, brought about (through screening) by the logarithmic singularity at 2kF. The effect is enhanced in low dimensions due to the more dramatic changes at 2kF. In addition to these effects, various broken symmetry states, with a wavevector dependent oscillating charge or spin density may also develop for electron–electron interactions of sufficient strength; these are discussed in Chapter 7.

5.4 Longitudinal response

127

The singularity of the response function at q = 2kF leads also to a characteristic spatial dependence of the screening charge around a charged impurity. While in the Thomas–Fermi approximation the screening charge decays exponentially with distance far from the impurity, the selfconsistent field approximations yield the well known Friedel oscillations with a period 2kF, which in three dimensions reads

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ρ =

 

A

cos{2kF + φ}

 

;

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(5.4.20)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

here both A and φ depend on the scattering potential [Gru77].

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

After having evaluated χˆ (q, ω) it is straightforward to write down the following

expression for the dielectric constant:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

q

 

 

ω+i

 

 

 

ˆ

 

=

 

+ q

2vF2 2

 

+ 4q

2kF

 

ω

qvF

 

 

 

 

 

 

 

 

 

 

 

 

q

 

ω+i

 

1

(q, ω)

 

1

 

 

3ωp

1

 

 

 

 

 

kF

 

 

1

 

 

 

 

 

q

 

 

 

 

+ 1

 

 

 

 

Ln

 

2kF

 

qvF

+ 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

qvF

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2kF

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

q

 

 

 

ω+i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4q

 

 

2kF

 

 

 

 

 

qvF

 

 

 

 

 

 

 

 

 

q

+

ω+i

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2kF

 

 

 

 

 

 

 

 

 

 

kF

 

1

 

 

q

 

 

 

 

ω

 

+ 1

 

 

 

 

 

 

 

 

qvF

 

 

1

 

 

 

, (5.4.21)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ln 2kF

+

 

qvF

 

 

+

 

 

 

which for negligible damping, 10, reduces to

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

4q

 

2kF qvF

2

 

 

 

q

 

 

 

 

 

ω

 

1

 

 

 

 

 

= + q

2 +

 

 

 

2kF

 

 

 

 

qvF

 

 

 

 

1(q, ω)

 

1

 

 

λ2

1

 

 

kF

 

 

 

 

 

1

 

 

 

 

q

 

 

 

 

 

ω

 

 

 

 

 

 

ln

 

2kF

 

qvF

+ 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

q

 

 

 

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kF

 

 

 

 

 

 

 

 

 

 

 

q

 

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2k

 

 

 

 

 

qv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

, (5.4.22a)

 

 

 

 

4q

 

 

2kF

qvF

 

 

 

 

2kF

+

 

qvF

1

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ln

 

 

 

 

 

F

 

 

 

F

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

 

 

 

ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where λ2 = 3ωp2/vF2 is the Thomas–Fermi screening parameter introduced above. In the same limit the imaginary part becomes

 

 

 

 

3π ωp2ω

 

 

 

q

 

ω

 

2

 

 

q

+

 

ω

< 1

q

 

 

ω

 

 

 

 

3π ωp2kF

 

 

 

 

 

 

 

q

 

ω

 

 

 

 

 

 

 

2q3vF3

 

 

 

 

 

 

 

 

 

 

2kF

 

qvF

 

 

 

 

 

2

(q, ω)

=

 

1

 

 

 

 

 

 

 

 

 

 

 

< 1 <

 

 

 

 

(5.4.22b)

4q3v2

2kF

 

qvF

 

 

 

2kF

 

 

qvF

 

2kF

 

 

qvF

 

 

 

F

 

 

 

 

 

 

|

 

 

 

− |

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

| 2qkF qωvF | > 1 .

 

 

 

 

In the static limit ω 0, the imaginary part disappears and the expression for(q, ω) from Eq. (5.4.10) is recovered, as expected. In Fig. 5.15 these components are shown as a function of frequency and wavevector. For small q values we can expand the energy as

Ek+q = Ek +

∂Ek

q

+

1

 

2Ek

q2

,

(5.4.23)

 

 

 

k

2 k2