Computational Physics

272 Quantum molecular dynamics

molecule 9.3 An example: quantum molecular dynamics for the hydrogen

In this subsection we work out an application of the Car–Parrinello method to the
hydrogen molecule in some detail. Our example is based on the Hartree–Fock calcu-
lation of the hydrogen molecule considered inChapter 4, in particular Problem 4.9.
There are two spin-orbitals with opposite spin and the same orbital part. Therefore,
the wave function is completely specified by the form of this orbital. We use the
GTO basis set of Problem 4.9 with eight basis s-functionsχr, four on each atom.
The molecular dynamics method can be restricted to the electronic structure part
of the total energy, keeping the nuclear positions fixed. We do this first; later we
shall extend the method to include nuclear displacements.

9.3.1 The electronic structure

The energy can be written as

Etot= 2

∑

rs

CrhrsCs+

∑

rstu

CrCsCtCu〈rt|g|su〉+

1

X

. (9.26)

Note that there is no indexkas the two electrons occupy only one orbital. The Fock
matrixFis given by
Frs=hrs+

∑

tu

CtCu〈rt|g|su〉 (9.27)

(all sums over indicesr,s,t,urun over the basis states, so in our case from 1 to 8).
The normalisation condition for the orbital is
∑

rs

CrSrsCs=1. (9.28)

Therefore, the equation of motion for theCr(without friction) is given by
μ
4
C ̈r=−

∑

s

hrsCs−

∑

stu

CsCtCu〈rt|g|su〉−λ

∑

s

SrsCs

=−

∑

s

(Frs+λSrs)Cs. (9.29)

We shall use the Verlet algorithm for solving the equations of motion. In this
form, they read forμ=4:

Cr(t+h)= 2 Cr(t)−Cr(t−h)−h^2

∑

s

(Frs+λSrs)Cs(t). (9.30)

Suppose we know theCr(t)and theCr(t−h). The solution to the equation of
motion proceeds in two stages. First we calculate

C ̃r(t+h)= 2 Cr(t)−Cr(t−h)−h^2

∑

s

FrsCs(t). (9.31)