Computational Physics

12.2 The variational Monte Carlo method 379

ASis the Slater determinant (seeChapter 4) andφis a function which contains
the two-particle correlation effects. For identical bosons, all the minus-signs in the
determinant are replaced by pluses. The particular form we chose in the helium case
is a simple form of a class called Padé–Jastrow wave functions[7]. Inclusion of three
and four point correlations is obviously possible. We shall not go into the problem
of finding the best Slater determinants andφ-functions but restrict ourselves to
a short discussion of the requirements which we can derive for special particle
configurations – these are the ‘cusp conditions’: boundary conditions satisfied at
the points where the potential diverges. Near these points the kinetic and potential
energy contributions of the Hamiltonian are both very large, and they should cancel
out for a large part. This leads to large statistical fluctuations which are avoided
by respecting the cusp conditions. In the next section we shall see that these cusp
conditions are essential for trial wave functions used in the DMC method. We
have already dealt with a similar problem inChapter 2of this book, when we
found appropriate boundary conditions for the numerical solution of the radial
Schrödinger equation with a Lennard–Jones potential, which diverges strongly at
r=0. Now we consider singularities in the Coulomb potential.
In the helium atom, the potential diverges when one of the electrons approaches
the nucleus, or when the electrons are close to each other. The Schrödinger equation
can be solved analytically for these configurations since the Coulomb potential
dominates all other terms except the kinetic one. Suppose that one of the electrons,
labelledi, is very close to a nucleus (which we take at the origin) with chargeZ.In
that case the Schrödinger equation becomes approximately
[
−

1

2

∇i^2 −

Z

ri

]

ψ(r 1 ,...,rN)=0. (12.16)

Writing out the kinetic energy in spherical coordinates of particlei, we arrive at a
radial Schrödinger equation of the form (r=ri)

[

d^2

dr^2

+

2

r

d

dr

+

2 Z

r

−

l(l+ 1 )

r^2

]

R(r)=0. (12.17)

If, as is usually the case, the wave function is radially symmetric inriforrismall,
we have exclusively anl=0 contribution, and the two terms containing the factor
1 /rmust cancel (the first term does not contribute for a function which is regular
at the origin). ForR( 0 )=0 this leads to

1

R

dR

dr

=−Z, r=0; (12.18)

so thatR(r)=exp(−Zr).