Advanced Solid State Physics

(Axel Boer) #1

10.4 Diffusive and Ballistic Transport


We now try to get a uniform picture of the transport e.g. if the frequency is so high that there is some
scattering of electrons but also ballistic transport. In this frequency regime metals can be described by
a dielectric constant which sounds weird, because normally only insulators have a dielectric constant.
But for this particular frequency range the electric field changes faster than the electrons scatter and
therefore the electrons move out of phase just like they do in an insulator (polarization current). Hence
a dielectric constant is also useful to describe a metal in this frequency range.
In general the polarization and the electric displacement are connected to the electric field via a
matrixχ:


Pi(ω,k) =


j

ε 0 χij(ω,k)Ej(ω,k) (106)

Di(ω,k) =


j

ε 0 εij(ω,k)Ej(ω,k) (107)

The polarization and the electric displacement are both linear to the electric field which is linear
response theory. The susceptibility (χ = εP 0 E) and the dielectric constant (ε = εD 0 E) can be thought
of as generalized susceptibilities as above (see chapter 10.1) and therefore must satisfy the Kramers-
Kronig relations.
The displacement is related to the polarization like


D = ε 0 E+P,

with this equation, eqn. (106) and (107) the dielectric constant can be written as


ε(ω,k) = 1 +χ(ω,k).

Often this relation is written


ε(ω) = 1 +χ(ω).

This is the low - frequency (wavelength longer than the lattice constant) approximation which is
important e.g. for optics.
In an insulator in a changing electric field there are no free electrons to produce a current but there
is a displacement current which is the derivation of the displacement by time.


JD =

∂D

∂t

= ε 0 ε

∂E

∂t

(108)

For a sinusoidal electric fieldE(ω) = exp(iωt), the displacement current is


JD = iωε 0 ε(ω)E(ω), (109)

hence the displacement current and the electric field are 90°out of phase. The displacement current
and Ohm’s law (J = σE) put into Faraday’s law (∇×H = J−∂∂tD) gives


∇×H = σ(ω)E(ω)−iωε 0 ε(ω)E(ω) (110)
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