PHZ6426: Fall 2013
FINAL EXAM
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:
R∞
x2
π2
• 0 dx cosh
2 x = 12
R∞
2
• 0 dx exx−1 = π6
P1a 30
P1b 20
P2 50
P1
a. A monolayer of graphene is electron-doped (µ kB T > 0 but µ is still much smaller than the
bandwidth). The scattering mean free time is some function of energy: τ = τ (ε). Find the Seebeck
coefficient (thermoelectric power), defined as E 0 = S∇T , to lowest order in kB T /µ. Approximate
the electron spectrum by εk = ±h̄vF k near both the K and K 0 points, and assume that transport
can be described by the (linearized) Boltzmann equation in the relaxation-time approximation
−v · ∇T
∂f0
f − f0
ε − µ ∂f0
− e (v · E0 )
=−
,
T ∂ε
∂ε
τ (ε)
(1)
where f0 is the equilibrium Fermi function, and E0 = E + ∇µ/e is the effective electric field.
Solution
The Seebeck coefficient is measured under the open circuit conditions, when the electric current is absent.
Solve (1) for δf , substitute δf into an expression for the electric current
R
j = −2e (d2 k/2π)3 vδf , and require that j = 0. Since graphene is treated in the isotropic
approximation here, we obtain for S:
R
∂f0
ε−µ
2
τ
(ε)
−
dεg(ε)v
(ε)
T
∂ε
1
,
S=−
(2)
e R dεg(ε)v 2 (ε)τ (ε) − ∂f0
∂ε
where g(ε) is the density of states. For graphene g(ε) = |ε|/πh̄2 vF2 (per one K point) and
v(ε) = vF = const. Since graphene is electron-doped, we can neglect the contribution of the
valence band to S, which means the integrals in Eq. (2) run from 0 to ∞. The integral in
the denominator (proportional to the conductivity) can evaluated at T = 0 which amounts to
0
replacing − ∂f
∂ε = δ(ε − µ):
Z
∂f0
dεg(ε)v 2 (ε)τ (ε) −
= vF2 g(µ)τ (µ).
(3)
∂ε
In the numerator, shift the energy by µ and extend the lower limit of the integral over ≡ ε − µ
from −µ to −∞:
Z
Z ∞
1
∂f0
∂f0
2
dεg(ε)v (ε) (ε − µ) τ (ε) −
≈
(4)
d(µ + )τ (µ + ) −
∂ε
∂
πh̄2 −∞
0
Since ∂f
∂ is an even function of , we need to expand the product as (µ + )τ (µ + ) = εµτ (µ) +
2
[τ (µ) + µτ 0 (µ)] + . . .. The 2 is necessary in order to get a finite result for S. The resulting
integral is solved by introducing x ≡ /2kB T and using the first integral from Useful Math:
Z ∞
Z ∞
Z ∞
2
∂f0
2
x2
1
2 2π
2 2
d2 −
d
=
k
T
=
=
4k
T
.
(5)
B
B
∂
2kB T 0
3
cosh2 2kB
cosh2 x
−∞
0
T
Collecting everything together, we obtain the final result
kB π 2 kB T
τ 0 (µ)
1+µ
S=−
e 3 µ
τ (µ)
(6)
b. Show that S = 0 if τ (ε) ∝ |ε|−1 regardless of the sign and magnitude of µ. (To see this, you need
to consider the contributions from both the conduction- and valence bands.)1
0
(µ)
= 0 for τ (ε) = aε−1 ,
Solution The result can already be anticipated from the fact that 1 + µ ττ (µ)
where a = const. To see that S vanishes identically in this case, re-write the integral in the
numerator of Eq. (2) as a sum of the conduction- and valence-band contributions:
Z ∞
Z 0
Z
∂f0
∂f0
a
∂f0
2
dε(ε − µ) −
dεg(ε)v (ε) (ε − µ) τ (ε) −
=
+
dε(ε − µ) −
∂ε
∂ε
∂ε
πh̄2 0
−∞
Z ∞
a
∂f0
=0
(7)
=
dε(ε − µ) −
2
∂ε
πh̄ −∞
∂f0
∂ε
because
is an even function of ε − µ.
P3 In addition to the usual acoustic and optical modes, a free-standing membrane of graphene supports a
bending mode. This is a transverse mode with dispersion ω(q) = aq 2 , where a = const. Assume that
this dispersion law holds for 0 ≤ q ≤ qD . Find the contribution of the bending mode to the specific
heat in two limiting cases a) kB T h̄ωD and b) kB T h̄ωD , where ωD ≡ ω(qD ).
Solution The density of states of the bending mode is found in the usual way:
2πqdq
1
= g(ω)dω → g(ω) =
(2π)2
4πa
(8)
The thermal energy (per unit area)
Z
E=
ωD
dωg(ω)
0
1
=
4πa
−1
h̄ω
e
h̄ω
kB T
Z
ωD
dω
0
(kB T )2
=
4πah̄
−1
h̄ω
e
h̄ω
kB T
Z
0
h̄ωD
kB T
dx
ex
x
.
−1
(9)
In case a), the upper limit of the integral is 1, hence one can expand ex = 1 + x + . . .. This gives
E=
kB T ωD
∂E
kB ωD
q2
→C=
=
= kB D
4πa
∂T
4πa
4π
(10)
In case b), the upper limit can be extended to ∞, which (on using the second integral in Useful math)
gives
E=
1
2
(kB T )2 π 2
π kB
T
→C=
4πah̄ 6
12 ah̄
FYI: such a dependence of τ on ε is typical for scattering from short-range defects in graphene.
(11)