Advanced Solid State Physics

−ω^2 M 2 v=C(u(1 +eika)− 2 v)

These equations can be written in a matrix
(
ω^2 M 1 − 2 C C(1 +e−ika)
C(1 +eika) ω^2 M 1 − 2 C

)(

u

v

)

= 0

In order for this equation to be true, the determinant of the matrix has to vanish, which leads to the
following equation (once again the Euler identities were used):

M 1 M 2 ω^4 − 2 C(M 1 +M 2 )ω^2 + 2C^2 (1−cos(ka)) = 0

And forωthe following equation must hold:

ω^2 =C

(

1

M 1

+

1

M 2

)

±C

√(

1

M 1

+

1

M 2

) 2

−

4 sin^2 (ka)

M 1 M 2

= 0

As one can see in fig. 15(a), there are two solutions forω^2 and therefore also two branches if we

plot

√

2 C

(

1

M 1 +

1

M 2

)

, an optical and an acoustical branch (without loss of generality we assumed

M 1 > M 2 ).

(^00) pi/(2a)k pi/a
ω
(a)
(^0) ω
D(ω
)
(b)
Figure 15: a) Density of states for a 1-dimensional chain of two different types of atoms; b) Dispersion
relation of a 1-dimensional chain of two different types of atoms
The density of states for this chain can be seen in fig. 15(b). If both masses are the same, we have
the same case as in fig. 13(a), where the optical branch vanishes and the Brillouin zone boundaries
are atπa(a is the distance between two atoms). However, if the masses are only slightly different, the
Brillouin zone gets smaller ( 2 πa), because the lattice constant is now 2 ainstead ofa. The dispersion
relation now looks like as shown in fig. 16, because the branches from the neighboring zones reach in
and there is an optical branch once more.