Set of supplementary notes.

4.3. NEARLY FREE ELECTRON THEORY 63
If the lattice potential V were zero, then the sum in the second term of this equation
would vanish. We would be left, then with a set of independent simultaneous equations for
E, which have solutions E = h¯ 2/(2m)q 2 , if c q ≠ 0 and all the other c q−G ′ = 0, and generally
E = h¯ 2/(2m)(q − G ′ ) 2 for a particular c q−G ′ ≠ 0, when all the other coefficients apart from
that one are zero. We obtain a set of parabolic bands (Fig. 4.2) for the unperturbed solutions,
which will then hybridise, where the bands cross, when the lattice potential is non-zero.
To make the calculation more specific, we work out the actual dispersion for a one-dimensional
chain. We use a simplified atomic potential which just contains the leading Fourier components,
i.e
V (x) = 2V 2π/a cos 2πx
(4.14)
a
If V 2π/a is small, we should be able to treat it perturbatively, remembering to take care of
degeneracies. If V 2π/a = 0, we get the free electron eigenvalues
which are repeated, offset parabolas.
E (m)
0 (k) = h¯ 2
2m (k − 2πm/a)2 , m = ..., −2, −1, 0, 1, 2, ... (4.15)
Now suppose V 2π/a is turned on, but is very small. It will be important only for those
momenta at which two free electron states are nearly degenerate, for example, m=0,1 are
degenerate when k = π/a. Near that point, we can simplify the band structure to the 2x2
matrix (
)
2
h¯2
( )
2m k 2 − E V 2π/a
ck
V2π/a
∗ h¯2
(k − (4.16)
2π
2m a )2 − E c k−2π/a
The solution of the determinantal leads to a quadratic equation:
E ± (k) = h¯ 2 1
2m 2 (k2 + (k − 2π/a) 2 ) ± 1 √
( h¯ 2
2 2m k2 − h¯ 2
2m (k − 2π/a)2 ) 2 + 4V2π/a 2 (4.17)
Exactly at k = π/a, the energy levels are
E ± (π/a) = E 0 π/a ± |V 2π/a |, (4.18)
and if we choose the potential to be attractive V 2π/a < 0, the wavefunctions are (aside from
normalisation)
ψ − (π/a) = cos(πx/a) ,
ψ + (π/a) = sin(πx/a) . (4.19)
4.3.3 Example calculations for 3D metals
Fig. 4.3 illustrates the results of a three dimensional nearly free electron calculation. You can
see that
• Because of Bloch’s theorem, for every ∣ ∣ ψ
n
k+G
〉
there is an identical state |ψ
m
k 〉. Therefore,
E k has the same periodicity as the reciprocal lattice.