### Solution 11

```Solid State Theory
Solutions Sheet 11.
Exercise 1.
FS 2014
Prof. Manfred Sigrist
Relaxation time approximation
In this exercise we will show that the so-called single-relaxation-time approximation,
Z
∂f (k)
f (k) − f0 (k)
dd k 0
W (k, k0 )[f (k) − f (k0 )] −→ −
=−
,
d
∂t
(2π)
τ
coll
(1)
is a true solution to the Boltzmann equation under certain conditions.
We consider a spatially homogeneous two-dimensional metal with an isotropic Fermi surface (εk =
~2 k 2 /2m) at zero temperature. The impurity scattering responsible for a finite resistivity is described
by a delta potential in real space,
Vimp (r) = V0 δ(r).
(2)
The system is subject to a homogeneous and time-independent electric field along the x-axis.
a) Show that the transition rates W (k, k0 ) for the impurity potential (2) are constant and non-zero
only for scattering events conserving the energy of the incoming state.
b) Write down the static Boltzmann transport equation for this setup in the form
“drift-term” = “collision-integral”
(3)
and take advantage of the zero-temperature limit and the symmetries of the system to eliminate all
but angular variables.
c) In a case with only angular dependence, it turns out to be useful to expand the drift term ∇k f ·(eE)
and δf = f − f0 in Fourier modes
X
X
δf =
fl eilϕ ,
∇k f · (eE) =
dl eilϕ .
(4)
l
l
Rewrite the Boltzmann equation as a set of algebraic equations for the coefficients in the expansion
(4)
X
dm =
Lm,n fn .
(5)
n
d) What are the eigenvalues of the so-called collision operator Lm,n and what is their meaning? How
can one interpret vanishing eigenvalues?
e) Find a solution to equation (5) and compare δf to the single-relaxation-time approximation, equation (1).
Solution.
a) The transition rates are computed by using Fermi’s golden rule, relating the probability
per unit of time for an initial state labeled by k to scatter to some final state labeled by
k0 due to a collision with an impurity (cf. section 6.3 of the lecture notes).
W (k, k0 ) =
2π Nimp 0
|hk |Vimp |ki|2 δ(εk − εk0 ),
~
1
(S.1)
where Nimp denotes the total number of impurities in the system. The matrix elements
hk0 |V |ki are computed using the explicit form of the eigenstates for the free electron
Hamiltonian, i.e.
Z
0
1
0
V (k − k ) =
d2 r ei(k−k )·r V0 δ(r)
Ω
1
(S.2)
= V0 .
Ω
Where we used ψk (r) =
rates are given as
√1 eik·r ,
Ω
with Ω being the area of our system. Thus, the transition
W (k, k0 ) =
2πnimp 2
V0 δ(εk − εk0 ),
| Ω~{z }
(S.3)
W0
with nimp = Nimp /Ω giving the density of impurities. Due to the isotropy of the energy
εk ∼ k 2 they do not depend on k or k0 as long as the energy is conserved in the collision.
b) In the static limit, the left hand side of the Boltzmann equation
∂f (r, k, t)
dr
eE
+ ∇r f (r, k, t) ·
+ ∇k f (r, k, t) ·
∂t
dt
~
(S.4)
reduces with f (r, k, t) = f (k) to the third term proportional to the driving external field.
We restrict our considerations to linear order in E since we are interested in linear response
to the driving force. In consequence, we must replace ∇k f with ∇k f0 because δf is already
of order O(E).
This term represents the so-called drift term. The right hand side of the Boltzmann
equation is the collision integral, which for impurity scattering acquires the form (cf. Eq.
(6.22) in the lecture notes)
Z
∂f (k)
Ω
=−
d2 kW (k, k0 )(f (k) − f (k0 ))
2
∂t
(2π)
coll
Z
Ω
0
0
2
0 ) f0 (k) + δf (k) − f0 (k ) − δf (k )
=−
d
kW
δ(ε
−
ε
0
k
k
(2π)2
Z
Ω
m
=−
dθ dk kW0 2 δ(k − k 0 ) f0 (k) + δf (k) − f0 (k0 ) − δf (k0 )
2
(2π)
~ k
Z
Ω m
=−
dθW0 [δf (θ) − δf (θ0 )].
(S.5)
(2π)2 ~2
Taking (S.4) and (S.5) together we arrive at
Z
∼
eE
∇k f0 ·
= − dθ0 W 0 [δf (θ) − δf (θ0 )],
~
(S.6)
∼
where W 0 = mΩW0 /(2π~)2 .
X
dl eilθ = −
Z
∼
dθ0 W 0
l
!
X
l
2
fl eilθ −
X
l
ilθ0
fl e
,
(S.7)
which we project onto the m-th Fourier mode by multiplying on both sides with exp(−imθ)
and integrating over θ
!
Z
Z
Z
X
X
X
∼
0
dθe−imθ
dl eilθ = − dθe−imθ dθ0 W 0
fl eilθ −
fl eilθ .
(S.8)
l
l
l
This equation can be easily evaluated by using the relation
Z
dθeinθ = 2πδn,0 ,
(S.9)
and leaves us with
∼
2πdm = −W 0
XZ
dθdθ
0
e
−imθ ilθ
e
−imθ ilθ0
−e
e
fl
l
∼
= −(2π)2 W 0
X
(δm,l − δm,0 δl,0 )fl .
(S.10)
l
With the expression (S.10) we have reduced the problem of finding a solution to the
Boltzmann equation to solving a system of linear equations. In general, this system of
equations may be infinitely large and quite often only manageable numerically. In our
case, relating the equation (S.10) to the matrix equation on the exercise sheet, we find
that the matrix L has the following diagonal form


..
.




1




.
.


.




1
∼ 

.

0
(S.11)
Lm,n = −2π W 0 



1




..


.




1


..
.
d) We have reduced the problem to a simple matrix equation dl = Ll,m fm , where summation
over repeated indices is implied. In our case, since the scattering rates are constant, and in
particular independent of the scattering angle, the matrix turns out to be diagonal. The
eigenvectors of L represent the Fourier decomposition of eigenfunctions of the collision
operator and the eigenvalues represent inverse relaxation times. A vanishing eigenvalue
represents and infinite relaxation time, so that the corresponding eigenfunction defines
a conserved quantity. Any quantity that cannot be changed due to collisions will be
preserved. In our case, we have only one conserved quantity left, because the energy, that
can also not be changed within collisions has explicitly been projected out. This conserved
quantity, as can be seen from (S.11), corresponds to the zeroth Fourier component. This
constant shift of δf corresponds to a change in the particle number of the system and can
of course not be changed during collisions.
e) Since we know the exact form of the drift term, we can easily find its decomposition in
Fourier modes. The drift-term to linear order in E is defined as
∇k f0 ·
3
(eE)
~
(S.12)
which at zero temperature reduces to
~
e~kE
k · (eE) = −δ(εk − µ)
cos θ,
(S.13)
m
m
where θ is the angle between the k and the electric field, which points into the x-axis.
−δ(εk − µ)
In the above analysis, we have integrated out the energy variables, so we can drop the
delta function in (S.13) and express the Fourier decomposition as

 − e~kE , l = ±1
X
e~kE iθ
−iθ
ilθ
2m
.
(S.14)
(e + e ) ⇒ dl =
dl e = −

2m
0
, else
l
Because the matrix L is diagonal, it is straightforward to find a solution to the Boltzmann
equation,

e~kE 1


, l = ±1
∼
2m 2π W
.
(S.15)
fl =
0


0
, else
Finally, by comparing the coefficients from the expansion of the drift term in (S.14) to
those of the relaxation time approximation
∂f
δf
=− ,
(S.16)
∂t coll
τ
with the coeficients for δf from (S.15) we can identify the value of the relaxation time to
be
1
τ=
(S.17)
∼ .
2π W 0
Exercise 2.
Residual Resistivity of Copper
Recapitulate Section 6.3.1 and try to find an explanation for the data in Table 1. What is the major
reason for the increase of the resisitivity?
Hint: In analogy to doped semiconductors, assume that an impurity atom will adjust itself corresponding
to its neighbourhood. This means that it rejects all electrons from its outer shells in order to match the
occupancy of its surrounding atoms. Hence the impurity is left with an effective nuclear charge Z. The
explanation of the above table lies in this fact.
Solution.
In the lecture notes it is shown that
1
Z2
∝ N (F )nimp 4
τ
kTF
such that
ρ∝
and ρ ∝
N (F ) nimp 2
Z .
4
n
kTF
1
nτ
(S.18)
(S.19)
The first fraction only depends on the Fermi surface which is hardly affected by the impurities
and the second fraction does not change much because the doping increases the number of free
charge carriers only slightly (very small impurity density). Therefore we can conclude that
ρ ∝ Z 2.
4
(S.20)
As it was mentioned in the hint we assume that the optimal electronic configuration of the
impurity atom valence shell is identical with the valence shell of the surrounding atoms.
In this spirit we define the effective charge of the impurity as the number of electrons which
have to be removed from the impurity in order to have the same valence shell configuration as
the surrounding atoms, because those electrons are fully delocalized and can therefore no longer
screen the nucleus of the impurity atom (apart from Thomas-Fermi screening).
A look at the periodic table gives us the electronic configuration, e.g. copper has the valence
band configuration 4s1 . By calculating the difference of the valence electrons between copper
and the impurity we find the following table:
Impurity
Be
Mg
B
Al
In
Si
Ge
Sn
As
Sb
Resistivity
ρ/(10−8 Ωm)
0.64
0.6
1.4
1.2
1.2
3.2
3.7
2.8
6.5
5.4
Electronic configuration
Eff. charge Z
2s2
3s2
2s2 2p1
3s2 3p1
5s2 5p1
3s2 3p2
4s2 4p2
5s2 5p2
4s2 4p3
5s2 5p3
1
1
2
2
2
3
3
3
4
4
Table 1: Electronic configuration and the effective charge of the impurities.
Indeed, by comparing the values of the residual resistivity with the electronic configuration,
we find a good correlation in agreement with scaling (S.20). Note that this agreement is only
qualitative, not quantitative, as impurities with identical effective charges do not lead to precisely
the same residual resistivity.
5
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