PHZ6426: Fall 2013
MIDTERM: 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.
1a 20
1b 15
1c 15
2a 30
2b 20
P1 The energy spectrum of charge carriers in graphene is described by the Dirac-like dispersion relation
(see Fig. 1):
εk = ±h̄v0 |k| .
Find the T dependence of the electronic part of the specific heat for several positions of the Fermi
energy with respect to the “Dirac point” (εk = 0):
a) εF = 0;
b) εF > 0;
c) εF < 0.
For cases b) and c), consider only the low-temperature limit: kB T εF . You may need to use the
fact that
Z ∞
x2
3
dx x
= ζ(3) ≈ 1.803,
e
+
1
2
0
where ζ(x) is the Riemann zeta-function.
Solution The grand-canonical free energy of an electron gas with εF = 0 is given by
X Z d2 k
(j)
−εk /kB T
F = −2kB T
,
2 ln 1 + e
(2π)
j=±
where j = +(−) corresponds to the conduction (valence) band. The density of states in graphene is
found from the Pauli principle:
2
2πkdk
(2π)
2
= g (ε) dε
1 k (ε)
1 |ε|
=
.
π |dε/dk|
πh̄2 v02
g (ε) =
Apparently, g is an even function of energy. Switching from an integration over k to that over ε, and
taking into account both the conduction and valence bands, we obtain
Z ∞
Z 0
−ε/kB T
ε/kB T
F = −kB T
dεg (ε) ln 1 + e
+
dεg (ε) ln 1 + e
0
−∞
Z
Z ∞
2 kB T ∞
dεε ln 1 + e−ε/kB T
= −2kB T
dεg (ε) ln 1 + e−ε/kB T = − 2 2
πh̄ v0 0
0
Integration by parts gives
F =−
1 1
πh̄2 v02
Z
∞
dε
0
ε2
eε/kB T + 1
3
=−
1 (kB T )
πh̄2 v02
Z
|0
∞
x2
dx x
,
e +1
{z
}
=(3/2)ζ(3)
where ζ (z) is the Riemann function.
CV = −T
3 2
∂2F
9ζ (3) kB
T
=
.
2
2
∂T
v02
πh̄
For cases b) and c), there is no need to re-do the calculation from the scratch. It suffices to recall that
CV for a Fermi gas with arbitrary dispersion and at kB T |εF | is given by
CV =
π2
π |εF | 2
2
g (εF ) kB
T = 2 2 kB
T.
3
3h̄ v0
εk
ky
kx
FIG. 1: Dirac energy spectrum.
P2
a) Identify the primitive unit cell of a two-dimensional triangular lattice. Find the basis vectors.
b) Construct the basis vectors of the reciprocal unit cell.
Solution
See Fig. 2. Denoting the side of an equilateral triangle as a, the basis vectors can be chosen as
√
a
3
a1 = ax̂; a2 = x̂ +
aŷ.
2
2
The reciprocal lattice vectors are defined by the condition
ai · bj = 2πδij
or, in components,
We then obtain b2x = 0, b2y = 2π/a2y
or
a1x b1x + a1y b1y
a2x b2x + a2y b2y
a1x b2x + a1y b2y
a2x b1x + a2y b1y
√
= 4π/a 3, b1x =
= 2π
= 2π
=0
= 0.
√
b1x = −2π/a 3
2π/a1x = 2π/a, b1y = − aa2x
2y
2π
2π
x̂ − √ ŷ
a
3a
4π
= √ ŷ.
a 3
b1 =
b2
(1)
The reciprocal primitive unit cell is a parallelogram obtained by rotating the unit cell in the coordinate
space by π/2 and by rescaling the sides by a factor of 2π/a.
y
a2
a1
FIG. 2: Primitive unit cell (shaded) and basis vectors.
x