PHZ6426: Fall 2012
FINAL EXAM: SOLUTIONS
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.
Useful math: the integral
R∞
0
x
dxxn (exe+1)2 is equal to ln 2 for n = 1 and to 9ζ(3)/2 for n = 3.
P1 30
P2 40
P3 30
P1 Find the electric conductivity of graphene at the Dirac point (the chemical potential µ = 0) at finite
temperature. Approximate the electron spectrum by εk = ±~vF k near both K and K 0 points. Assume that transport can be described by the (linearized) Boltzmann equation in the relaxation-time
approximation
∂f0
f − f0
=−
,
(1)
∂ε
τ
where f0 is the equilibrium Fermi function and τ does not depend on the electron’s energy.
Solution
In the Dirac approximation, graphene is isotropic and thus σxx = σyy , while σxy = σyx = 0. The
direction of the electric field can be arbitrary, let’s choose its along the x axis. Then
Z
Z
d2 k
d2 k
jx = −2e
v
f
=
−2e
vx (f − f0 ).
(2)
x
2
(2π)
(2π)2
−e (v · E)
Substituting the solution of (1) into
f − f0 = eEτ vx
∂f0
∂ε
into (2), we obtain for σxx
2
Z
d2 k 2
∂f0
v −
.
(2π)2 x
∂ε
Z
∂f0
d2 k 2
v −
.
(2π)2
∂ε
σxx = 2e τ
Since σxx = σyy = (σxx + σyy )/2,
2
σxx = e τ
For an isotropic spectrum, 2d2 k/(2π)2 = g(ε)dε, where g(ε) is the density of states. For each of the K
and K 0 points, 4πkdk/(2π)2 = g(ε)dε or g(ε) = |ε|/π~2 vF2 . Because integration over d2 k goes over K
and K 0 points, this adds a factor of 2. Now, the derivative of f0 is an even function, therefore one can
fold the integral over ε into the one from 0 to ∞. This adds another factor of 2. Finally,
Z
Z
2e2 τ ∞
∂f0
2e2 τ ∞
ε
eε/kB T
σxx = σyy =
dεε
−
=
dε
2
2
π~ 0
∂ε
π~ 0
kB T eε/kB T + 1 2
Z
2e2 τ kB T ∞
ex
2 ln 2 e2 kB T τ
=
dxx
=
,
(3)
2
2
π~
π ~ ~
(ex + 1)
0
where we used the n = 1 value of the integral in Useful math.
P2 In the same model of the electron spectrum as in P1 and also at µ = 0, find the thermal conductivity
of graphene. Assume that the temperature of the system varies slowly along the graphene plane and
that thermal transport can be described by the Boltzmann equation
v · ∇f = −
f − f0
.
τ
(4)
P
Hint 1: Define the heat conductivity tensor καβ by the relation jhα = β καβ ∂β T with α, β = x, y
R D
and jh = 2 d kv(εk − µ)f /(2π)D being the heat current.
Hint 2: Replace f in the left-hand side of Eq. (4) by f0 with spatially varying temperature.
Solution
~ as
Since the gradient of T is small, we can simplify ∇f
1
∇f ≈ ∇f0 = ∇
exp
ε
kB T (r)
=
+1
∇T xex
,
T ex + 1
where x = ε/kB T and T is the average temperature. Using the rotational symmetry of the spectrum,
we arrive at
Z ∞
3 2 Z ∞
3 2
eε/kB T
2kB
T
ex
9ζ(3) kB
T τ
ε2
3
2
=
−
dxx
=
−
.
dεg(ε)
κxx = κyy = −2vF τ
2
2
2
x
ε/k
T
kB T e B + 1
π~
(e + 1)
π
~2
0
0
(5)
It is more conventional to define
the
relation
between
the
heat
current
and
the
temperature
gradient
P
with a minus sign jhα = − β καβ ∂β T . With such a definition, καβ is a minus one times the result in
(5).
P3 Suppose that the electron spectrum for a tetragonal lattice can be described by the following equation
εk =
~2 ky2
~2 kx2
~2 kz2
+
+
2mxx
2mxx
2mzz
with mxx 6= mzz . Using the Onsager formula for the cyclotron mass, mc = (~2 /2π)∂A/∂ε with A being
the area of an isoenergetic contour perpendicular to the magnetic field, find mc for the case when the
magnetic field is a) along the z axis and b) in the x − y plane.
Solution
For B||z,
A = π(kx2 + ky2 ) = π
2mxx ε mxx
−
~2
mzz
and mc = mxx . For√B in the x − y plane,
√ the isonergetic countour is an ellipse with semi-major/minor
√
axes given by a = 2mxx ε/~ and b = 2mzz ε/~. The area A = πab, which gives mc = mxx mzz .