PHZ6426: Fall 2011
FINAL EXAM
1a 30
1b 20
2a 30
2b 20
P1 The energy spectrum of electrons on a 2D square lattice is given by the tight-binding model
εk = −t [cos (kx a) + cos (ky a)] .
Assume that transport can be desribed by the (linearized) Boltzmann equation in the relaxation-time
approximation
−e (v · E)
∂f0
f − f0
=−
.
∂ε
τ
(a) Find the conductivity tensor (in zero magnetic field and at T = 0)
!
σxx σxy
σ̂ =
σyx σyy
at half-filling. [30 points]
2
σαβ = 2e τ
= 2e2 τ
Z
d2 k
2 vα vβ
(2π)
Z
d2 k
∂f0
−
∂ε
2 vα vβ δ (εk − εF )
(2π)
I
e2 τ
dl
=
vα vβ .
2π 2 h̄
|v|
From the third line in the equation above it is obvious that σxy = σyx = 0. Indeed, each of vx,y
is odd in kx,y , whereas εk is even. Therefore, each of the integrals over kx and ky vanish by
symmetry. Diagonal components
I
dl 2
e2 τ
v
σxx =
2π 2 h̄
|v| x
I
e2 τ
dl 2
σyy =
v
2π 2 h̄
|v| y
By symmetry of the square lattice,
σxx = σyy
e2 τ
= (1/2) (σxx + σyy ) =
4π 2 h̄
ta
|v| =
h̄
q
sin2 kx a + sin2 ky a
I
dl |v| .
√
On
of the FS, ky = π/a − kx → |v| = 2ta |sin kx a| and dl =
q one of the segments
p
√
2
(dkx )2 + (dky ) = dkx 1 + (dky /dkx )2 = 2dkx
σxx
√ √ e2 τ ta Z π/a
dkx sin (kx a)
= σyy = 4 × 2 2 2 2
4π h̄h̄ 0
2 e2 τ t
= 2
π h̄ h̄
Notice that the effective mass at the bottom of the band is related to t via
1
h̄2
= 2.
∗
m
ta
At half-filling, the number density n = 1/a2 . Therefore, the result above for σxx = σyy can be
written as
σxx = σyy =
2 ne2 τ
,
π 2 m∗
which is by a factor of 2/π 2 smaller than the prediction of the Drude theory for electrons with
mass m∗ .
(b) Suppose that the electric field is applied at angle θ to the x axis. Find the Joule heat generated
per unit time per unit area of the sample. [20 points]
jx = σxx E cos θ
jy = σyy E sin θ
Q = jx Ex + jy Ey = σxx E 2 cos2 θ + σyy E 2 sin2 θ = σxx E 2 =
2 e2 τ t 2
E .
π 2 h̄ h̄
P2 (a) Find the energy spectrum in the nearest-neighbor hopping approximation of the tight-binding
model for a simple hexagonal lattice. The hopping matrix element is t and the bond length
is a. [30 points]
Vectors connecting the central atom of the hexagon to its six nearest neighbors are
√ !
1
3
s1 = a
,
2 2
s2 = a (1, 0)
s3
s4
√ !
1
3
,−
= a
2
2
√ !
1
3
= a − ,−
2
2
s5 = a (−1, 0)
√ !
3
1
s6 = a − ,
2 2
Diagonalizing the Hamiltonian
Ĥ = −t
X
n.n.
c†Rn cRn + H.c.
, we obtain for the energy spectrum
√ √ √ 3
3
i k 1 +k
a
−i kx 21 +ky 23 a
i k 1 −k
i
εk = −t e x 2 y 2
+e
+ eikx a + e−ikx a + e x 2 y 2 + e
#
"
√ !
√ !
3
3
1
1
a + cos kx − ky
a + cos (kx a)
= −4t cos kx + ky
2
2
2
2
√ ky a
1
= −4t 2 cos kx a
3
cos
+ cos (kx a)
2
2
−kx 12 +ky
√
3
2
+ c.c.
(b) Find the approximate form of the energy spectrum and determine the tensor of effective mass
near the Γ point (kx = ky = 0) of the Brillouin zone [20 points]
1 2 2
3 2 2
kx2 a2
= const + 3ta2 (kx2 + ky2 )
εk = −4t 2 1 − kx a
1 − ky a + 1 −
8
8
2
1
1
6ta2
=
= 2
mxx
myy
h̄
1
= 0.
mxy