5.2 Lattice Vibrations
97
Fig. 5.2 Visualization of
transverse (‘T’) and
longitudinal (‘L’) waves in
a linear monoatomic chain
at different wavevectors
T
L
T
L
T
L
k
a
=1/2 /
k a
= /
a
k
a
=1/4 /
Mω
2 u n = C (2u n − u n−1 − u n+1 ) .
(5.3)
If, also, the solution is periodic in space, i.e. is a (one-dimensional) plane wave, i.e. u n (x, t) =
v 0 exp(i(kx −ωt)) with x = n a, we find from the periodic boundary condition exp(ik N a) = 1 and thus
k =
2 π
a
n
N
, n ∈ N .
(5.4)
It is important that, when k is altered by a reciprocal space vector, i.e. k
= k+2π n/a, the displacements
u n are unaffected. This property means that there are only N values for k that generate independent
solutions. These can be chosen as k = −π/a, . . . , π/a, so that k lies in the Brillouin zone of the lattice
(Fig. 5.3). Since the properties are periodic with the Brillouin zone, the k-values can be imagined being
on a circle as visualized in Fig. 5.3; the angle φ = k a run from 0 to 2π or from −π to +π as you like.
In the Brillouin zone there is a total number of N k-values, i.e. one for each lattice point. The
distance between adjacent k-values is
2π
N a
=
2π
L
,
(5.5)
L being the lateral extension of the system.
The displacements at the lattice points n and n + m are now related to each other via
u n+m = v 0 exp(ik (n + m) a)
(5.6)
= v 0 exp(ikna) exp(i k m a) = exp(i k m a) u n .