solved to a two-by-two matrix, which means that we only take one fourier coefficient into account.
(
(k−G 0 )^2 U 0 k^2 U 1
(k−G 0 )^2 U 1 k^2 U 0
)
·(
ck−G
ck)
=ω^2(
ck−G
ck)Fork=G 2 it is
G^20
4(
U 0 U 1
U 1 U 0)
·(
ck+G
ck)
=ω^2(
ck+G
ck)Finding the eigenvalues of a two-by-two matrix is easy, we can do that analytically. As a result for
the eigenvalues we get the two frequencies
ω=G 0
2
√
U 0 −U 1 and ω=G 0
2
√
U 0 +U 1.So we can estimate how large the gap is. Further we know that very long wavelengths do not see the
modulation, so the speed of light in this region is the average speed. If we know the modulation and
where the brillouin zone boundary is, we cananalytically figure outwhat thesize of the gapis.
5.5 Density of States
From the dispersion relationship we can determine the density of states. This means how many states
are there with a particular frequency. When we have light travelling through a periodic medium, the
density of states is different than in vacuum. So some of those things calculated for vacuum, like the
radiation pressure or the specific heat, are now wrong because of the different density of states.
In vacuum the density of states in one dimension is just constant, because in the one-dimensional
k-space the allowed values in the periodic boundary conditions are just evenly spaced alongk. The
allowed wave that fits in a one-dimensional box of some lengthLis evenly spaced ink. So in this
region the allowed states is a linear function so it is evenly spaced inωtoo.
In a material we start out with a constant density of states. When we get close to the brillouin
zone boundary, it is still evenly spaced in terms ofk. But the density of states increases at the bril-
louin zone boundary and then drops to zero, because there are no propagating modes in this frequency
range (photons with these frequencies will get reflected back out). After the gap, the density of states
bunches again and then gets kind of linear (see fig. 9(b)).
In fig. 10 it looks a bit like a sin-function near the brillouin zone boundary, soω ≈ωmax|sinka 2 |.
So we getk=^2 asin−^1
(
ω
ωmax). The density of states in terms ofωis
D(ω) =D(k)dk
dω.
The density of states in terms ofkis a constant. The density of states inωis
D(ω)∝1
√
1 − ω
2
ω^2 max(41)
It looks like the plot in fig. 11(a).