Two points to note 1 the bandwidth is proportional to

Two points to note
1. The bandwidth is proportional to the hopping term.
2. It is surprisingly easy to generalize the result to 2 and 3 dimensions, because all we need to do is sum
over the nearest neighbours! If they are symmetrically located then the bandwidth would simply be
2 zt where z is the number of nearest neighbours or the co-ordination number of the lattice.
PROBLEM : Calculate the group velocity of a particle at the bottom of the band and at the corner
( k = ± π/a ). Show that there is a point of inflection (where the second derivative changes sign) somewhere
between k = 0 and k = ± π/a .
How does the Bloch function look?
Here’s a plot of how the functions look. The k = 0 wavefunction is shown for reference, because that
has the maximum resemblance with the ”atomic” wavefunctions. Here we assumed that the atomic
wavefunction is a gaussian. See Fig. 5.2.

5.1. DIATOMIC MOLECULE AND LINEAR CHAIN OF ATOMS 85
Figure 5.2: The dots are the atomic sites, the k = 0 wavefunction shows what the atomic states are like.
The other two show what the linear combination of those wavefunctions, as given by Bloch’s theorem ,
would look like. See eqn. 5.13.

86 CHAPTER 5. TIGHT BINDING OR LINEAR COMBINATION OF ATOMIC ORBITALS (LCAO)
Generalising to 2 and 3 dimensions: with 1 oribital per site
The generalisation is easy. To handle 2 and 3d lattices we need to write the wavefunction as
| ψ
k
) =
1
√
N
summationdisplay
R
e
i k . R
| φ
R
) (5.26)
( r | ψ
k
) =
1
√
N
summationdisplay
R
e
i k . R
φ ( r − R ) (5.27)
Where the sum runs over all direct lattice vectors R and | φ
R
) is the atomic state centered at R . While
taking the expectation value of energy we will group the series of terms into three and ignore the interaction
between sites which are not nearest neighbours or next-nearest-neighbours:
H = T + V
1
+ V
2
+ V
3
+ ... + V
N
(5.28)
E ( k ) = ( ψ
k
| H | ψ
k
) (5.29)
=
1
N
summationdisplay
R = R ′
( φ
R ′
| H | φ
R
) +
1
N
summationdisplay
R , R ′
nearest
neigh-
bours
e
i k . ( R − R
′
)
( φ
R ′
| H | φ
R
) +
a0
a0
a0
a0
a0
a0
a0
a0 a64
a64
a64
a64
a64
a64
a64
a64
1
N
summationdisplay
R , R ′
further
than
nearest
neigh-
bours
... (5.30)
≈ E
0
+
summationdisplay
nearest
neigh-
bours
e
i k . R
t
R
(5.31)
Since all sites are identical, it is sufficient to sum over the nearest neighbours of the site at R = 0.
PROBLEM : Consider a 2-d rectangular lattice with sides a and b .
1. Show that following eqn. 5.31 the bandstructure would be of the form
E ( k
x
,k
y
) = E
0
− 2 t
1
cos( ak
x
) − 2 t
2
cos( bk
y
) (5.32)
2. What is the reciprocal lattice? Draw the first Brillouin zone.
3. Plot the constant energy contours, assuming t
1
> t
2
> 0 and a < b . Why is this physically
reasonable?
4. Plot some constant energy contours. How do the contours look for small k ? How do the shapes
change at slightly larger k ? Do all constant energy contours close within the first Brillouin zone?
Similarly for 3d lattices like BCC with 8 nearest neighbours and FCC with 12 neighbours can be summed
up.