Now back to the Central equations (1 dim). These are the algebraic equations we get when we put
the Fourier- series in the Schrödinger equation:
(
~^2 k^2
2 m
−E
)
Ck+∑GUGCk−G= 0The central equations couple coefficientsk to other coefficients that differ by a reciprocal lattice
wavevectorG. So we choose akand already know~andm, we only do not know the energyE. The
points of the bravais lattice in reciprocal space areG 0 , 2 G 0 and so on. Thus we know theG 0 ’s so the
energies are the eigenvalues of a matrix. We make a matrix like this:
...
(~^2 (k− 2 m 2 G 0 ))^2 −E UG
0 U^2 G 0
U−G 0 (~^2 (k 2 −mG^0 ))^2 −E UG 0 U 2 G 0
U− 2 G 0 U−G 0 ~ 22 mk^2 −E UG 0 U 2 G 0
U− 2 G 0 U−G 0 (~^2 (k 2 +mG^0 ))^2 −E UG 0 U 2 G 0
U− 2 G 0 U−G 0 (~^2 (k+2 2 mG^0 ))^2 −E UG 0
...
..
.
Ck+2G 0
Ck+G 0
Ck
Ck−G 0
Ck− 2 G 0
..
.
= 0(51)The minimum of nearest neighbors in a 1 dimensional problem is one. You get a matrix (2 by 2),
which can be solved and plotted. What you get is the dispersion relationship with the energy overk
(it starts parabolic, then bends over to the Brillouin zone boundary, then there’s a gap, and then the
second band - see fig. 28).
Figure 28: Dispersion relationE(k)evaluated with Central EquationsYou can also expand the matrix and take more terms (two to the right and one to the left). Every-
thing is coupled to it’s two nearest neighbors (just a linear chain). For this 4 by 4 matrix you get four
energies (four bands). Thus by making the matrix bigger, higher energies can be calculated.