PHZ6426: Fall 2013
Problem set # 6: Electron Transport. Phonons.
due Monday, 12/02 at the time of the class
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Office hours: TR 3 pm-4 pm
Please help your instructor by doing your work neatly. Every (algebraic) final result
must be supplemented by a check of units. Without such a check, no more
than 75% of the credit will be given even for an otherwise correct solution.
...
P1 [20 points ]Hall conductivity of Bloch electrons
In class, we derived a general formula for the Hall conductivity in the weak-field regime (ωc τ 1)
Z
dD k
1
1
∂f0
3 2B
σxy = 2e τ
v vy
− vx
−
,
D x
c
myx
myy
∂ε
(2π)
where D is the dimensionality,
va =
1 ∂εk
,
~ ∂kα
1 ∂ 2 εk
1
= 2
,
mαβ
~ ∂kα ∂kβ
and f0 is the equilibrium Fermi function.
a) Show that
σxy = 0
at T = 0 for a half-filled tight-binding band corresponding to a 2D square lattice with dispersion
εk = −2t [cos (kx a) + cos (ky a)] .
[10 points]
Solution
Since the off-diagonal term in the effective mass tensor is absent, we have at T = 0,
Z
I
dD k 2 1
vx2
3 2
σxy = 2e3 τ 2 B
v
δ
(ε
−
ε
)
=
2e
τ
B
dA
.
F
k
D xm
|vF | myy
yy
(2π)
(1)
where the integral goes over the Fermi surface, |vF | is the magnitude of the velocity evaluated on the Fermi
surface, and
1
1 ∂ 2 εk
2ta2
= 2
= 2 cos (ky a)
2
myy
~ ∂ky
~
1 ∂εk
2ta
vx =
=
sin (kx a)
~ ∂kx
~
The Fermi surface at half-filling in 2D is shown in Fig. 1 (red). Consider segment I on which ky = π/a−kx .
The contribution of the segment to the integral in (1)
Z π/a
h π
i
sin2 kx a
σIxy = 2e3 τ 2 B
dkx
cos
− kx a
|vF |
a
0
Z π/a
2
sin kx a
= −2e3 τ 2 B
dkx
cos(kx a).
|vF |
0
π
IV
I
−π
π
III
II
−π
FIG. 1: The Brillouin zone (black) and Fermi surface at half-filling (red) for a 2D tight-binding model on a square lattice.
Wavenumbers are measured in units of 1/a.
Since cosine is odd in the interval (0, π/a) , while the rest of the integrand is even, the integral vanishes.
Same is true for the remaining segments II − IV.
b) Find σxy at T = 0 for electrons (εF > 0) and holes (εF < 0) in graphene, using the Dirac model for
the spectrum εk = ±~vF k. [10 points] [Beware: although the spectrum is isotropic, the off-diagonal
components of the effective mass tensor are not equal to zero!]
Solution
Since the Dirac spectrum is isotropic, it is more convenient to keep the 2D integral over k in its original
form. Choosing the electron branch of the spectrum, ε = ~vF k and putting T to zero,
Z
d2 k
1
1
3 2
σxy = 2e τ B
− vx
δ (~vF k − εF ) .
2 vx vy m
myy
xy
(2π)
Derivatives
1 ∂εk
kx
= vF
~ ∂kx
k
1 ∂εk
ky
=
= vF
~ ∂ky
k
vx =
vy
1
1 ∂ 2 εk
vF kx ky
= 2
=−
mxy
~ ∂kx ∂ky
~k k 2
!
ky2
1
vF
=
1− 2
myy
~k
k
vx
1
1
vy
− vx
mxy
myy
kx
= vF
k
= −
σxy = −2e3 τ 2 B
Z
ky vF kx ky
kx vF
−vF
− vF
2
k ~k k
k ~k
vF3 kx2
vF3
=
−
cos2 φ.
~k 3
~k
dkk vF3
δ (~vF k − εF )
2
(2π) ~k
Z
0
2π
dφ cos2 φ
ky2
1− 2
k
!!
= −2πe3 τ 2 B
= −
Z
vF3
dkk vF3
1
3 2
2 ~k δ (~vF k − εF ) = −2πe τ B
2 ~2 v
F
(2π)
(2π)
1 3 2 2
e τ BvF .
2π~2
Using the result of the Problem 1 for the cyclotron mass in graphene mc = εF /vF2 and the expression
for the conductivity σ0 = e2 νF vF2 τ /2 with νF = εF /π~2 vF2 at B = 0, we see that σxy = −ωc τ σ0 with
ωc = eB/mc (SI units).
For εF < 0, one has to choose the hole branch of the spectrum εk = −~vF k. Since the integrand is
proportional to the third power of εk , the sign of the σxy changes.
P2 [25 points ] Consider an electron gas described by the Maxwell-Boltzmann statistics in the presence of an elastic scattering
mechanism with an energy dependent relaxation time τ () = A/s . Using the relaxation-time approximation
for the Boltzmann equation, show that the Wiedemann-Franz ratio is given by
κ
=
Tσ
2
5
kB
,
−s
2
e
(2)
where κ and σ are the thermal and charge conductivities, correspondingly.1
Solution
Suppose that we apply both the electric field and the temperature gradient. Then the Boltzmann equation in
the relaxation time approximation reads
v · ∇f − e (v · E)
∂f0
f − f0
=−
.
∂ε
τ (ε)
I assume here that τ may depend on the electron energy–this will be important for obtaining the numerical
factor in Eq. (2). In the presence of the temperature gradient, the distribution function in the gradient term is
an equilibrium function of the combination (ε − µ (r))/T (r)
ε − µ (r)
f = f0
.
kB T (r)
Note that the chemical potential in general depends on the temperature and, thus, on r. For small gradients,
∇f = −
ε−µ 0
∇µ 0
f0 ∇T −
f
2
kB T
kB T 0
where T = const and µ = const everywhere but inside the gradient trims. Using
∂f0
1 0
=
f ,
∂ε
kB T 0
we re-write ∇f as
∇f = −
ε−µ
∂f0
∂f0
∇T
− ∇µ
,
T
∂ε
∂ε
so that the Boltzmann equation becomes
−
ε−µ
∂f0
∂f0
f − f0
v·∇T
− ev · E0
=−
T
∂ε
∂ε
τ (ε)
where E0 = E + ∇µ/e is the effective electric field. Actually, measurement of the electric current involves a
contact between two metals, so that the gradient of the chemical potential must always be added to the electric
field. Then the electrical conductivity measures the response to an effective field E0 rather than to E. With this
in mind, we relabel E0 → E. The non-equilibrium part of the distribution function is given by
ε−µ
∂f0
∂f0
f − f0 = τ (ε)
v·∇T
+ ev · E
T
∂ε
∂ε
The electrical current
d3 k
Z
j = −2e
= 2e
2
Z
(2π)
d3 k
3 v (f
− f0 )
∂f0
3 v (v · E) − ∂ε
(2π)
d3 k
ε−µ
3 v (v·∇T )
T
(2π)
Z
+ 2e
∂f0
−
∂ε
The term, proportional to ∇T , describes the thermoelectric effect. Assuming that E||∇T ||ẑ, using isotropy of
the energy spectrum, and switching from integration over k to that over ε, we obtain for the electrical current
along ẑ
Z
Z
e2
∂f0
e
ε−µ
∂f0
2
2
j =
dεν (ε) v (ε) τ (ε) −
E+
dεν (ε) v (ε)
τ (ε) −
∂z T
3
∂ε
3
T
∂ε
≡ LEE E + LET ∂z T
(3)
where ν (ε) is the density of states and v (ε) is the electron velocity. The thermal current is defined as
Z
d3 k
jT = 2
3 v (ε − µ) (f − f0 ) ,
(2π)
which, after the same manipulations as for the charge current, is reduced to
Z
Z
2
1
(ε − µ)
∂f0
e
∂f0
dεν (ε) v 2 (ε)
τ (ε) −
∂z T −
dεν (ε) v 2 (ε) (ε − µ) τ (ε) −
E
3
T
∂ε
3
∂ε
≡ −LT T ∂z T − T LET E.
jT = −
The electrical conductivity is measured under the conditions of constant temperature, which yields
Z
e2
∂f0
σ = LEE =
dεν (ε) v 2 (ε) τ (ε) −
.
3
∂ε
The thermal conductivity is measured in an open circuit when no electric current flows through the sample.
Equating j to zero in Eq. (3), we find the electric field generated to compensate the charge flow produced by
the temperature gradient
E=−
LET
∂z T.
LEE
Substituting this relation into the thermal current and using the definition of the thermal conductivity jT =
−κ∂z T, we obtain
κ = LT T − T
L2ET
.
LEE
The Lorentz ratio
L≡
κ
1 LT T LEE − T L2ET
= 2
σT
e T
L2EE
(4)
0
Let’s focus on the numerator in Eq. (4). Defining temporarily A (ε) ≡ ν (ε) v 2 (ε) τ (ε) − ∂f
, we have
∂ε
T LT T LEE −
T L2ET
Z
Z
h
i
2
dε0 A (ε) A (ε0 ) (ε − µ) − (ε − µ) (ε0 − µ)
Z
Z
Z
Z
=
dε dε0 A (ε) A (ε0 ) (ε − µ) (ε − ε0 ) = dε dε0 A (ε) A (ε0 ) [ε (ε − ε0 ) − µ (ε − ε0 )]
=
dε
Z
=
Z
dε
0
0
0
dε A (ε) A (ε ) ε (ε − ε ) =
Z
dεA (ε) ε
2
Z
0
0
dε A (ε ) −
Z
2
dεεA (ε) .
(5)
The integral of the µ (ε − ε0 ) term vanishes because of symmetry. An important consequence is that the chemical
potential drops out and is thus not needed to be known explicitly (this is important for the non-degenerate case
where µ depends on T.) Note that the cancellation of the chemical potential was only possible because we kept
the thermoelectric term.
Substituting (5) back into (4), we obtain
1
L= 2
e T
"R
dεA (ε) ε2
R
−
dεA (ε)
R
2 #
dεεA (ε)
R
dεA (ε)
(6)
Now we specify the form of the density of states ν (ε) = Aε1/2 and the velocity v (ε) = A0 ε1/2 (free 3D
electrons), xassume that τ (ε) = B/εs , and focus on the non-degenerate case, when f0 = Ce−ε/kB T (and
−ε/kB T
0
/kB T ). Constants A, A0 , B, and C are not important as the cancel out from the ratios of
− ∂f
∂ε = Ce
integrals in (6). Rescaling x = ε/kB T and canceling common factors of T , we re-write L via a combinations of
the dimensionless integrals
"
#
2
I5/2−s 2
I7/2−s
kB
−
L= 2
,
e
I3/2−s
I3/2−s
R∞
where Iα ≡ 0 dxxα e−x = Γ (α + 1) and Γ (x) is a Gamma function with a property that Γ (x + 1) = xΓ (x).
Using this property, we obtain
L=
kB
e
2 5
−s .
2
The result L = (3/2)(kB /e)2 derived in AM corresponds to s = 1 (when τ (ε) ∝ 1/ε), in which case the diffusion
coefficient D = (1/3)v 2 (ε)τ (ε) does not depend on the electron energy, and the phenomenological treatment
presented in AM is correct. For s 6= 1, the phenomenological picture breaks down because one need to perform
proper averaging over the energy distribution, as we did above.
P3 [20 points ]
• Cyclotron motion
Using the Onsager formula for the cyclotron mass
mc =
~2 ∂S
2π ∂ε
with S being the area of an isoenergetic contour, find mc for electrons (εF > 0) and holes (εF < 0) in graphene
described by the Dirac-like dispersion εk = ±v0 k.
Solution
mc =
~2 ∂S .
2π ∂ε ε=εF
For both electrons and holes,
S = πk 2 = π
mc =
ε2
~2 vF2
εF
.
vF2
For holes εF < 0 and thus mc < 0.
P4 [20 points ] Normal modes of a square lattice
Consider a square lattice of atoms of mass M , in which the nearest neighbors are coupled to each other by springs
with constants K. Derive the equation of motion for the mode in which atomic displacements are normal to the
crystal plane, and find the dispersion of this mode. Analyze the behavior of this mode in the continuum limit.
FIG. 2: Square lattice.
Solution
Label the adjacent lattice sites as shown in the figure. The potential energy of the nearest-neighbor interaction
is
K
U=
(un,m − un,m−1 )2 + (un,m − un,m+1 )2 + (un,m − un+1,m )2 + (un,m − un−1,m )2
2
The equation of motion for atom (n, m):
M
∂ 2 un,m
∂U
= −
2
∂t
∂nn,m
= K [4un,m − un,m−1 − un,m+1 − un,m+1 − un,m − un,m+1 ] .
(7)
Upon substituting
un,m = A exp i (naqx + maqy − ωt),
the equation of motion reduces to
2K
(2 − cos qx a − cos qy a) ,
M
which is just a trivial generalization of the 1D case. For qx a 1 and qy a 1, the dispersion reduces to an
isotropic acoustic form
ω2 =
ω2 =
Ka2 2
Ka2 2
(qx + qy2 ) =
q .
M
M
P5 25 points Elastic properties of layered systems
A popular model for describing elastic properties of layered systems (such as high Tc cuprates or graphite) is a
stack of planes coupled by Hooke’s forces. The planes are treated in the continuum approximation, which is OK
as long as the temperature is much smaller than the Debye temperature of the in-plane phonon modes. On the
other hand, since the planes are weakly coupled, the corresponding Debye temperature is small, and the model
allows to follow a transition between a quasi-discrete regime at higher T , when the inter-plane spacings are
still resolved, to a completely continuous regime at lower T , when the crystal becomes equivalent to anisotropic
elastic continuum.
Let zn (x, y, t) be a vertical displacement of plane #n at point x, y. The equation of motion for zn (x, y, t) reads
2
∂ 2 zn (x, y, t)
∂
∂2
2
= s||
+ 2 zn (x, y, t) − ω02 [2zn (x, y, t) − zn+1 (x, y, t) − zn−1 (x, y, t)] ,
(8)
∂t2
∂x2
∂y
where s|| is the in-plane sound velocity and ω0 characterizes the strength of the inter-plane force.
a) Find the spectrum of phonon modes in this model using a substitution
zn (x, y, t) = A exp i q|| · r|| + nqz c − ωt ,
(9)
where r|| = {x, y} and c is the inter-plane distance.
Solution
Substituting (9) into (8), we obtain
ωq2 = s2|| q||2 + ω02 2 − eiqz c − e−iqz c
or
ωq2 = s2|| q||2 + 2ω02 (1 − cos qz c) .
b) Find the density of states of phonon modes using the general formula
X Z d3 q
d(ω) =
δ(ω − ωλ (q)),
(2π)3
(10)
λ
where λ is the mode index. Hint: Notice that the integrals over qx and qz should run from −∞ to +∞,
because planes are considered as continuous, whereas the integral over qz should run over the first Brillouin
zone, i.e., from −π/c to π/c.
Solution For a single mode the density of states is given by
Z
d3 q
g(ω) =
δ(ω − ωq ).
(2π)3
Using an identity
δ(f (x)) =
X δ(x − xi )
i
|f 0 (xi )|
,
where xi are the roots of the equation f (xi ) = 0, we obtain
Z
Z
d3 q
d3 q
g(ω) =
δ(ω
−
ω
)
=
2ω
δ(ω 2 − ωq2 ).
q
3
(2π)
(2π)3
(for ω > 0). This step it is not necessary but it does make the subsequent integrals easier. Now, switch to
th cylindrical coordinates
Z
π/c
g(ω) = 2ω
−π/c
dqz
2π
∞
Z
0
dq|| q||
(2π)2
Z
0
2π
dφ 2
qz c δ ω − s2|| q||2 − 2ω02 sin2
.
2π
2
(11)
The integral over φ gives just 2π. Introduce a new variable x = q||2 which is a non-negative quantity. This
needs to be taken into account when resolving the δ function because it imposes constraints on the possible
values of qz . The root of the δ-function is at
x=
ω 2 − 2ω02 sin2
s2||
qz c
2
≥ 0.
(12)
If ω > 2ω0 , the RHS of (12) is non-negative. In this case, there is no constraint on qz . If ω < 2ω0 , the
RHS is non-negative only if ω 2 ≥ ω02 sin2 (qz c/2) or | sin(qz c/2)| ≤ ω/2ω0 or |qz | ≤ (2/c) arcsin(ω/2ω0 ).
Therefore, the answer for g(ω) depends on whether ω < 2ω0 or ω ≥ 2ω0 . For ω > 2ω0 ,
2ω
g(ω) =
2π 2
Z
π/c
Z
dqz
0
ω
=
.
2πcs2||
0
∞
qz c dxδ ω 2 − s2|| x − 2ω02 sin2
2
(13)
FIG. 3: g(ω) as a function of ω/2ω0 .
For ω < 2ω0 ,
ω
g(ω) =
2π 2
Z
2
c
arcsin
ω
2ω0
Z
∞
0
ω
ω
= 2 2 arcsin
.
π s|| c
2ω0
qz c dxδ ω 2 − s2|| q||2 − 2ω02 sin2
2
(14)
Comparing (13) and (14), we see that g(ω) is continuous at ω = 2ω0 but its derivative is discontinuous.
For ω ω0 , g(ω) ∝ ω 2 , which is a 3D-like behavior for an acoustic mode. For ω ω0 , g(ω) ∝ ω, which
is a 2D-like behavior. At higher energies, the wavelength of phonons is short, and they behave as a 2D
system. At lower energies, the wavelength is long, and phonons behave as a 3D system.
c) Find the asymptotic expressions for the lattice specific heat in two limits: a) T ~ω0 /kB and b)T ~ω0 /kB .
Solution
Internal energy
Z
Z
~ωq
~ω
d3 q
=
dωg(ω)
(15)
E(T ) =
~ω
(2π)3 exp ~ωq − 1
exp
−
1
kB T
kB T
At kB T ω0 , one can replace g(ω) by its 3D-like asymptotic form g(ω) = Aω2 valid at ω ω0 . In this
case, we recover the isotropic 3D result: E(T ) ∝ T 4 and hence CV ∝ T 3 . At kB T ω0 , one can replace
g(ω) by its 2D-like asymptotic form g(ω) = Aω valid at ω ω0 . In this case, we recover the isotropic 2D
result: E(T ) ∝ T 3 and hence CV ∝ T 2 . Therefore, CV (T ) starts off as T 3 at lower T but changes to T 2
at higher temperatures. At T higher than the Debye temperature for in-plane vibrations, the T 2 behavior
will be replaced by the classical Dulong-Petite law (CV = const) but this behavior of beyond the current
model in which planes are treated in the continuum approximation.
1
This result was obtained by Paul Drude, who implicitly assumed that s = 1(because in this case the diffusion coefficient
D = (1/3)v 2 τ does not depend on the electron energy) and also made a pure arithmetical mistake: his result had a factor of
3 instead of 3/2. In this form, the result was remarkably close to result for Fermi-Dirac statistics κ/σT = (π 2 /3)(kB /e)2 ≈
3.29(kB /e)2 , relevant to metals.