150 Solving the Schrödinger equation in periodic solids
ourselves to cubic unit cells. Although all quantities are expanded in a basis of
plane waves with wave vectors on a grid in reciprocal space, we cannot always use
periodicity on this grid. Consider for example the potential of a nucleus or ion core,
located at positionr 0 in the unit cell. We expand this potential using a grid of wave
vectors
K=2 π
L(nx,ny,nz) (6.61)withnxrunning from 1 toN,L/Nbeing the resolution in real space. If we represent
the potential inrealspace on the real-space grid, its Fourier transform is periodic
inK-space, with a period 2πN/Lin each Cartesian direction. Usually we are given
the (Fourier transform of the) pseudopotential for an ion core located at the origin.
If the atom is actually located at some placer 0 , the Fourier transfrom acquires an
extra structure factor exp(−iK·r 0 ).Ifr 0 does not lie on the real-space grid, this
structure factor is nonperiodic in reciprocal space!
Another example of a nonperiodic operator in reciprocal space is the kinetic
energy, for which we know the Fourier transform of the operator in continuum
space:
T=^2 K^2
2 m(6.62)
(in atomic units, this reduces toK^2 /2). This expression is cut off at some maximum
wave vector beyond which the components of the orbitals are supposed to be very
small. Note that we do not use the periodic discrete form of the kinetic energy (with
Fourier transform 3−cos(Kxa)−cos(Kya)−cos(Kza);a=L/N); the form (6.61)
is a more accurate representation.
As a basis, we use Fourier waves eiK·r/
√
(remember that is volume of the
unit cell). An orbitalφ(j)is then expanded in these basis vectors as
φ(j)(r)=1
√
∑
Kc(Kj)eiK·r. (6.63)The coefficients c(Kj) come out of a diagonalisation routine and are usually
normalised according to ∑
K|c(Kj)|^2 =1. (6.64)Therefore we have:
∑
r|φ(j)(r)|^2 =1
∑
r∑
K,K′c(Kj)c(Kj)′∗
ei(K−K
′)·r
=N^3
. (6.65)
The density due to all orbitals is therefore given by
n(r)=∑
jfj|φ(j)(r)|^2 , (6.66)