Solution 10

Solid State Theory
Solutions Sheet 10.
Exercise 1.
FS 2014
Prof. Manfred Sigrist
Drude conductivity and reflectivity of metals and semiconductors
In exercise 3.1 we have used the semi-classical equations of motion in order to study Bloch oscillations
and damping effects. In this exercise we want to derive the Drude conductivity from the semi-classical
description and use it to study the reflectivity of metals and semiconductors.
(a) Consider a semi-classical electron subject to an oscillating electric field
E(t) = E 0 e−iωt
(1)
and model the scattering processes using the relaxation-time approximation with time constant τ .
Write the semi-classical equation of motion and solve it in Fourier space to derive the dynamic
Drude conductivity
ωp2
τ
,
(2)
σ(ω) =
4π 1 − iωτ
where the plasma frequency is defined as
ωp2 =
4πne2
.
m
(3)
(b) Use the expression for the Drude conductivity to obtain an expression for the reflectivity R(ω) of
a simple metal or semiconductor using the connection between σ(ω) and R(ω) given in Sec. 6.2.2.
of the lecture notes. To take into account the effect of the bound (or core) electrons, use as a
phenomenological ansatz for the dielectric function
(ω) = ∞ + Drude (ω) − 1 .
(4)
Here ∞ is assumed to be constant in the frequency range of interest, related to the fact that the
energy scale for exciting core electrons is much higher than the typical energy scales for the itinerant
electrons. Plot the reflectivity for the cases ∞ ∈ {1, 20} and τ ωp ∈ {2, 40, ∞}!
Usually, ∞ is much larger in semiconductors than in metals. Can you think about a possible
explanation for this behavior?
(c) In semiconductors, most electrons are not free to move through the crystal but are bound to the
ions. In order to find a better description of their behavior than just using ∞ , model this binding
by an additional harmonic potential of the form V = 21 mω02 r 2 and perform the same analysis as in
part (a). How does this influence the reflectivity discussed in part (b)?
Solution.
(a) The force acting on the electron is given by
F (t) = qE(t) −
p(t)
,
τ
(S.1)
where the charge is q = −e. The semi-classical equation of motion then reads
˙
p(t)
= F (t) = qE(t) −
p(t)
.
τ
(S.2)
Fourier transforming and multiplying with τ leads to
− iωτ p(ω) = qτ E 0 − p(ω)
1
(S.3)
with the solution
qτ
E0 .
1 − iωτ
(S.4)
nq 2 τ
E(ω)
m(1 − iωτ )
(S.5)
nq 2 τ
.
m(1 − iωτ )
(S.6)
p(ω) =
Using p = mv and j = nqv, we obtain
j(ω) =
and consequently
σ(ω) =
Inserting the plasma frequency (3), we obtain the formula for the Drude conductivity,
σ(ω) =
ωp2
τ
.
4π 1 − iωτ
(S.7)
(b) Using the relation between conductivity and dielectric function we obtain
(ω) = ∞ +
ωp2 τ 2
ωp2 τ /ω
4πi
σ(ω) = ∞ −
+
i
.
ω
1 + ω2τ 2
1 + ω2τ 2
Parametrizing the complex index of refraction by
p
N (ω) = (ω) = n(ω) + iκ(ω) ,
(S.8)
(S.9)
and using the abbreviations
ωp2 τ 2
,
1 + ω2τ 2
ωp2 τ /ω
i (ω) =
,
1 + ω2τ 2
r (ω) = ∞ −
(S.10a)
(S.10b)
we find that
1
n(ω) = √
2
1
κ(ω) = √
2
r
r (ω) +
r
q
2r (ω) + 2i (ω) ,
−r (ω) +
q
2r (ω) + 2i (ω) .
(S.11a)
(S.11b)
The reflectivity is then given by the standard formula
R(ω) =
(n(ω) − 1)2 + κ2 (ω)
.
(n(ω) + 1)2 + κ2 (ω)
(S.12)
Plots of the reflectivity for different values of ∞ and τ ωp are shown in Fig. 1.
The reason why ∞ is usually larger in semiconductors than in metals relates to the different
electron densities in these two classes of materials: In the formulae above, everything is
√
measured in terms of ωp ∝ nc , where nc is the density of charge carriers. For metals, nc ∼
1022 cm−3 , whereas for doped semiconductors, nc ∼ 1013 − 1018 cm−3 . Correspondingly,
~ωp ∼ 10eV for metals and ~ωp . 100meV for heavily doped semiconductors. The simple
solution to why ∞ is higher in semiconductors is that the ’value of ∞’ depends on the
energy range one is interested in, so that also ∞ varies, as it incorporates the response of
all excitations with energies higher than the energy range under investigation. For higher
frequency cutoff there is a smaller number of excitations above this cutoff, so that ∞
decreases with increasing ωp (as ωcutoff ∼ ωp ) and ultimately drops to unity for very large
cutoff.
2
1.0
0.8
RHΩL
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
ΩΩ p
1.0
0.8
RHΩL
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
ΩΩ p
ΤΩp =2
ΤΩp =40
ΤΩp =¥
Figure 1: Frequency dependence of the reflectivity for ∞ = 1 (top) and ∞ = 20 (bottom).
(c) The force acting on a particle in a semiconductor is now given by
p(t)
− mω02 r(t)
τ
(S.13)
˙
mr(t)
− mω02 r(t) ,
τ
(S.14)
iωm
r(ω) − mω02 r(ω) .
τ
(S.15)
F (t) = qE(t) −
leading to the equation of motion
m¨
r (t) = qE(t) −
or in Fourier space
− mω 2 r(ω) = qE 0 +
3
Solving for r(ω) leads to
r(ω) =
q
E0 .
m(ω02 − ω 2 − iω/τ )
(S.16)
Using v(ω) = −iωr(ω) and j = nqv, we obtain the conductivity
σ(ω) =
ωp2
nq 2 τ
1
τ
=
.
m 1 − iωτ (1 − (ω0 /ω)2 )
4π 1 − iωτ (1 − (ω0 /ω)2 )
(S.17)
For the reflectivity we still keep the parameter ∞ to describe the inner shells of the
semiconductor. We then need to adapt Eqs. (S.8) and (S.10) to
r = ∞ −
ωp2 τ 2 (1 − (ω0 /ω)2 )
1 + [ωτ (1 − (ω0 /ω)2 )]2
ωp2 τ /ω
i =
,
1 + [ωτ (1 − (ω0 /ω)2 )]2
,
(S.18a)
(S.18b)
while Eqs. (S.11) and (S.12) still hold.
The results are shown in Fig. 2. We note that the region of strong reflectivity shifts from
around ω = 0 to a narrow range around the frequency of the harmonic potential, ω0 .
This effect can only be observed for sufficiently clean systems, where τ ωp 1. If we
now considered multiple harmonic potentials for the different orbitals, this would lead to
multiple such resonance lines that could be observed in the reflectivity.
1.0
0.8
RHΩL
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
ΩΩ p
ΤΩp =2
ΤΩp =40
ΤΩp =¥
Figure 2: Frequency dependence of the reflectivity for a semiconductor with ∞ = 20 and
ω0 = 12 ωp .
4

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