Solid State Theory FS 15
Solution 4 PD V. Geshkenbein
Problem 4.1 Brillouin zone of simple crystals
a) Triangular lattice:
The Bravais lattice is identical to the crystal lattice. It is generated by vectors
√
a1 = a(1, 0), a2 = a2 (1, 3). (1)
They are illustrated in the left part of Fig. 1.
The reciprocal lattice is generated by vectors b1,2 such that
ai · bj = 2πδij . (2)
This definition means that b1 ⊥ a2 and b2 ⊥ a1 . Direction of vectors b1,2 is found
quickly by drawing only. Length of the vectors is fixed by condition (2). Calculation
of the dot product reveals that
b1 = 2πa
1, − √1
3
, b2 = a4π
√ (0, 1)
3
(3)
√
so that |b1 | = |b2 | = 4π/ 3. Both vectors as well as the generated reciprocal lattice
are shown in the right part of Fig. 1. Construction of the first Brillouin zone is also
indicated therein.
a2
b2
a1
b1
Figure 1: Left: Lattice vectors of a triangular lattice. Right: The reciprocal lattice to
triangular lattice is a triangular lattice. To construct the Brillouin zone around a given
point we have to connect it to its nearest neighbours (solid lines). In the next step we
draw symmetry axes of these connecting segments (dashed lines). The first Brillouin zone
is the interior of the dashed lines, i.e. a hexagon in this case.
b) Honeycomb lattice:
A honeycomb lattice is not a Bravais lattice, i.e. there is no pair of vectors a1,2
such that n1 a1 + n2 a2 , n1,2 ∈ Z would correspond to the lattice sites. In fact, the
underlying Bravais lattice is triangular and contains two atoms per lattice site. The
lattice vectors are √ √
a
a1 = a 3, 0 , a2 = 2 3, 3 (4)
1
√
and are depicted in Fig. 2. Since this is just 3-multiple
√ of the lattice vectors in
part (a), the corresponding reciprocal vectors are 3-times smaller, so that equation
(2) would be preserved, i.e.
b1 = a2π
√
3
1, − √1
3
, b2 = 4π
3a
(0, 1) . (5)
The Brillouin zone of a honeycomb lattice has the same shape as for the triangular
lattice.
a2
a1
Figure 2: The honeycomb lattice has a triangular Bravais lattice with two sites per unit
cell.
b) Sodium chloride: The underlying Bravais lattice of this crystal is face-centered
cubic with two atoms per unit cell. Vectors that generate the Bravais lattice are
a1 = a(0, 1, 1), a2 = a(1, 0, 1), a3 = a(1, 1, 0) (6)
and are depicted in the left part of Fig. 3.
Reciprocal lattice is generated by vectors
a2 × a3 π
b1 = 2π = a
(−1, 1, 1) (7)
a1 · (a2 × a3 )
a3 × a1 π
b2 = 2π = a
(1, −1, 1) (8)
a2 · (a3 × a1 )
a1 × a2 π
b3 = 2π = a
(1, 1, −1) . (9)
a3 · (a1 × a2 )
These vectors generate a body-centered cubic lattice. Its basic building block is
depicted in the middle part of Fig. 3. The Brillouin zone has a complicated shape
depicted in the right part of the same figure.
Problem 4.2 Two-orbital tight-binding model in 2d
The Bloch-waves constitute a basis of the (quasi-2-dimensional) Hilbert space of the sys-
tem. Since the Wannier functions are the “Fourier-transforms” of the Bloch-waves, they,
too, span the whole Hilbert space. Thus, we can write
X
H= hwα (r − rj )|H|wα0 (r − rj0 )ic†αj cα0 j0 , (10)
α, α0 , j, j0
2
z
a1 a2
a3
y x
Figure 3: Left: Cubic crystal of NaCl crystal viewed along the body diagonal. The
dashed lines indicate the underlying face-centered cubic structure. Middle: Building
block of the reciprocal body-centered cubic lattice. Right: The corresponding Brillouin
zone.
which can be split in two terms as
X X
H= Hα + Hα,α0 . (11)
α α6=α0
a) We restrict to the nearest neighbour hopping. The intra-band elements of (10) are
X
Hα = εα c†αj cαj + (txα c†α(j+x̂) cαj + tyα c†α(j+ŷ) cαj + h.c.) (12)
j
with
εα =hwα (r)|H|wα (r)i, (13)
txα =hwα (r − ax̂)|H|wα (r)i, (14)
tyα =hwα (r − aŷ)|H|wα (r)i. (15)
Considering the overlap elements tx,y
α for both bands, we recognize that
txpx = typy , (16)
txpy = typx (17)
due to the symmetry properties of the square lattice and of the atomic orbitals.
Notice that the inter-band elements
hwα (r − ax̂)|H|wα0 (r)i and hwα (r − aŷ)|H|wα0 (r)i (18)
are zero to due symmetry reasons. For example, an element
hwpx (r − ax̂)|H|wpy (r)i (19)
is zero because |wpy (r)i is odd under reflection with respect to x-axis, while hwpx (r−
ax̂)| is even.
3
b) We approximate the Wannier functions (which are orthogonal to each other) by
atomic (hydrogen) states (which are not orthogonal to each other), and we choose
the orientation of the orbitals such that
(
positive, x > 0,
sign(wpx (r)) = (20)
negative, x < 0,
(
positive, y > 0,
sign(wpy (r)) = (21)
negative, y < 0.
Using
" #
X
Hj=0 |wα (r)i = (Hkin + V (r)) |wα (r)i = εα + V (r − rj ) |wα (r)i, (22)
j6=0
we find for the matrix elements
" #
X
txpx =hwpx (r − ax̂)| εpx + V (r − rj ) |wpx (r)i (23)
j6=0
" #
X
typx =hwpx (r − aŷ)| εpx + V (r − rj ) |wpx (r)i (24)
j6=0
" #
X
txpy =hwpy (r − ax̂)| εpy + V (r − rj ) |wpy (r)i (25)
j6=0
" #
X
typy =hwpy (r − aŷ)| εpy + V (r − rj ) |wpy (r)i. (26)
j6=0
Consider first the case of (23) and (26). The main contribution to these matrix
elements comes from the region between the two lattice sites where the two orbitals
have opposite sign. As εα < 0 and V (r) < 0, we obtain that txpx = typy > 0. On the
other hand for (24) and (25) the orbitals have the same sign, hence typx = txpy < 0.
Performing the Fourier transformation
1 X −ik·rj
cαj = √ e cαk (27)
N k
of the annihilation operators in the Hamiltonian, we obtain
X
Hα = εα,k c†αk cαk (28)
k
with
εpx ,k = ε + 2t1 cos(kx a) − 2t2 cos(ky a) (29)
εpy ,k = ε − 2t2 cos(kx a) + 2t1 cos(ky a) (30)
where ε = εpx = εpy , t1 = txpx , and t2 = −typx > 0. The band structure and the Fermi
surface for half-filling are visualized in Fig. 4.
4
1.0
0.5
k yΠ
0.0
-0.5
-1.0
kxΠ
-1.0 -0.5 0.0 0.5 1.0
1.0 1.0
0.5 0.5
k yΠ
k yΠ
0.0 0.0
-0.5 -0.5
-1.0 -1.0
kxΠ kxΠ
-1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0
Figure 4: The band structure in the absence of interband coupling is visualized by a 3d plot
and contour plot of the two bands. Also shown is the Fermi surface (t1 = 0.4, t2 = 0.1).
c) We now include the next-nearest neighbour hopping. This modifies part of the
Hamiltonian Hα , and it also leads to non-zero Hα,α0 . We first discuss Hα only, and
inter-band coupling will be discussed later.
The next-nearest neighbour contribution to intraband coupling is
" #
X
t̃px =hwpx (r ± ax̂ ± aŷ)| εpx + V (r − rj ) |wpx (r)i (31)
j6=0
" #
X
t̃py =hwpy (r ± aŷ ± aŷ)| εpy + V (r − rj ) |wpy (r)i. (32)
j6=0
Due to lattice symmetries, all four terms in (31) and (32) are equal and t̃px = t̃py ≡
t3 . The largest contribution comes from area where the two orbitals have opposite
sign, therefore t3 > 0.
Performing a Fourier transformation (27) leads to modified energy bands.
ε̃px ,k = ε + 2t1 cos(kx a) − 2t2 cos(ky a) + 4t3 cos(kx a) cos(ky a) (33)
ε̃py ,k = ε − 2t2 cos(kx a) + 2t1 cos(ky a) + 4t3 cos(kx a) cos(ky a). (34)
Now we consider the next-nearest neighbour contribution to the interband coupling
X † †
−
Hα,α0 = t+
αα0 cα(j+x̂+ŷ) cα0 j + tαα0 cα(j+x̂−ŷ) cα0 j + h.c. (35)
j
5
with
t±
αα0 =hwα (r − a(x̂ ± ŷ)|H|wα0 (r)i (36)
Due to symmetry properties and the analogue consideration as above, we obtain
− −
t+ +
px py = tpy px = −tpx py = −tpy px ≡ t4 > 0. Performing a Fourier transform of the
Hamiltonian, we obtain
X
Hα,α0 = −4t4 sin(kx a) sin(ky a)c†αk cα0 k . (37)
k
Defining gk = −4t3 sin(kx a) sin(ky a), the complete Hamiltonian can be written as
!T !
X c† ε̃px k gk
cp x k
px k
H= (38)
c†py k gk ε̃py k cpy k
k
such that to diagonalize the Hamiltonian, we have to find the Eigenvalues Ek± of
the matrix above determined by the equation
(ε̃px k − Ek± )(ε̃py k − Ek± ) − gk2 = 0. (39)
The calculation is straightforward and we obtain
± 1 q 2 2
Ek = ε̃px k + ε̃py k ± ε̃px k − ε̃py k − gk (40)
2
The resulting band structure and the Fermi surface for half-filling are plotted in
Fig. 5.
d) The pz orbitals are odd under reflection with respect to xy-plane, while both px and
py orbitals are even. The consequence is that coupling between pz and px,y orbitals
vanishes in the studied two-dimensional model. The pz -band thus can be studied
independently.
Within the plane the pz orbitals are effectively s-like. The pz -band structure can be
easily found to be
εpz k = εz − 2t5 cos(kx a) − 2t5 cos(ky a) − 4t6 cos(kx a) cos(ky a) (41)
where we defined εpz ≡ εz and
" #
X
t5 = − hwpz (r ± ax̂)| εpx + V (r − rj ) |wpz (r)i (42)
j6=0
" #
X
= − hwpz (r ± aŷ)| εpx + V (r − rj ) |wpz (r)i > 0 (43)
j6=0
" #
X
t6 = − hwpz (r ± ax̂ ± aŷ)| εpx + V (r − rj ) |wpz (r)i > 0. (44)
j6=0
6
1.0
0.5
k yΠ
0.0
-0.5
-1.0
kxΠ
-1.0 -0.5 0.0 0.5 1.0
1.0 1.0
0.5 0.5
k yΠ
k yΠ
0.0 0.0
-0.5 -0.5
-1.0 -1.0
kxΠ kxΠ
-1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0
Figure 5: The band structure in the presence of interband coupling is visualized by a
3d plot and contour plot of the two bands. Also shown is the Fermi surface (t1 = 0.4,
t2 = 0.1, t3 = 0.05, t4 = 0.1).
