Computational Physics

6.2 Band structures and Bloch’s theorem 125

three instead of two atoms: then the atomic levels split into three different ones,
and so on. A solid consists of an infinite number of atoms moved close together
and therefore each atomic level splits into an infinite number, forming aband.Itis
our aim to calculate these bands.
We shall now prove the famous Bloch theorem which says that the eigenstates of
the Hamiltonian with a periodic potential are the same in the lattice cells located at
RiandRjup to a phase factor exp[iq·(Ri−Rj)]for some reciprocal vectorq.We
shall see that a consequence of this theorem is that the energy spectra are indeed
composed of bands.
We write the Schrödinger equation in reciprocal space. The potentialVis periodic
and it can therefore be expanded as a Fourier sum over reciprocal lattice vectorsK:

V(r)=

∑

K

eiK·rVK. (6.7)

An arbitrary wave functionψcan expanded as a Fourier series with wave vectors
qallowed by the periodic boundary conditions(6.6):^1

ψ(r)=

∑

q

eiq·rCq. (6.8)

Writingq=k+K, the Schrödinger equation reads (in atomic units)
[
1
2

(k+K)^2 −ε

]

Ck+K+

∑

K′

VK−K′Ck+K′=0. (6.9)

This equation holds for each vectorkin the first Brillouin zone: in the equation,
wave vectorsk+Kandk+K′are coupled by the term with the sum overK′,but
no coupling occurs betweenk+Kandk′+K′for differentkandk′. Therefore,
for eachkwe can, in principle, solve the eigenvalue equation(6.9)and obtain the
energy eigenvaluesεand eigenvectorsCkwith componentsCk+K, leading to wave
functions of the form (see (6.8))

ψk(r)=eik·r

(

∑

K

Ck+KeiK·r

)

. (6.10)

The eigenvalues form a discrete spectrum for eachk. The levels vary withkand
therefore give rise to energy bands. Equation (6.9) yields an infinite spectrum for
eachk. We might attach a labeln, running over the spectral levels, alongside the
labelk, to the energy levelε:ε=εnk.
We can rewrite(6.10)in a more transparent form. To this end we note that the
expression in brackets in this equation is a periodic function inr. Denoting this
periodic function byuk(r), we obtain

ψk(r)=eik·ruk(r). (6.11)

(^1) For an infinite solid, the sum overqbecomes an integral.