−t 0 0 −t
−t −t 0 ··· 0
0 −t −t 0
0 0 −t ..
.
..
.
...
−t
−t 0 0 ··· −t
1
ei^2 πj/N
ei^4 πj/N
ei^6 πj/N
..
.
ei^2 π(N−1)j/N
= (−t·(ei^2 πj/N+ei2(N−1)πj/N))
1
ei^2 πj/N
ei^4 πj/N
ei^6 πj/N
..
.
ei^2 π(N−1)j/N
(54)
This kind of matrix has a known solution as showed in eqn. (54). It looks like plane waves. The first
term of the wavefunction is 1 , the next term has a phase factorj, which shows the position in the
chain. It is just an integer andNis the total number of atoms. The next elements always grow with
the same phase factor. If you multiply this vector by the matrix you get three terms and the vector
as you see below. So these terms are the eigenvalues of the matrix:
−t(
ei 2 πj
N +e
−i 2 πj
N)
=− 2 tcos(
2 πj
N)They can be translated fromjtokby^2 Nπj=ka. So the dispersion relationship is
E(k) =− 2 tcos (ka),which is shown in fig. 31.
Figure 31: Dispersion relation for the tight binding model