Computational Physics

152 Solving the Schrödinger equation in periodic solids

Even if both|K|and|K′|are smaller thanKmax, their difference can attain lengths
up to 2Kmax!
A two-dimensional representation of the situation is depicted inFigure 6.14,
where the sphere of radiusKmaxis indicated as a dashed circle, calledC1. The
kinetic energy is evaluated only for values ofKwithin this circle. When evaluating
the kinetic energy, only the points lying insideC1 must be taken into account.
However, the Fourier transorm of the density, which satisfies periodicity inK-space,
has nonzero components for points inside the bigger circleC2.
Now let us again consider the contribution to the Hamiltonian of alocal potential.
The matrix elements are given by

VK,K′=

1

∫

e−i(K−K

′)·r

V(r)d^3 r≈

1

N^3

∑

r

e−i(K−K

′)·r

V(r)=V(K−K′).

(6.72)
The last expression on the right hand side is the discrete Fourier transform of the
potential in real space. We regularly must perform Fourier transforms, for which
we use the FFT algorithm (see Appendix A9). There exist packages containing FFT
routines, and we mention here the FFTW package (http:// http://www.fftw.org)..)
Often, these packages have their own type definitions for real and complex numbers,
which should then be used throughout your program.
The simplest possible nontrivial case is the one with seven wave vectors inC1:
one in the origin and two along each of the three Cartesian axes. These are used for
the Hamiltonian and the wave functions. The wave vectors for which the density is
evaluated run over a grid with a linear size at least four times as large as the cut-off
wave vectorKmax(seeFigure 6.14), which would suggest a unit cell with a side of
four grid points. We take the grid size one larger (i.e. a 5× 5 ×5 grid) in order
to avoid the coincidence of point pairs like(2, 0, 0)and(−2, 0, 0)for functions or
operators which are nonperiodic in reciprocal space (such as the pseudopotential
of an ion which is not located at a real-space grid point, see above).

6.7.4 Free particle in a box

We start by considering a free particle in the box. The Hamiltonian only contains
the kinetic term:

HK,K′=

K^2

2

δ(K−K′). (6.73)

For a box of sizeL×L×L, the seven vectorsKwe take into account are the null
vector and the vectors with size 2π/Lalong the positive and negative Cartesian
axes. The eigenvalues are therefore equal to 0 (with multiplicity 1) and 2π^2 /L^2
(with multiplicity 6). If we put four electrons in the box, the total energy is given by