0198506961.pdf

4.3 The central-field approximation 65

S(r). This follows because the closed sub-shells within the core have
a spherical charge distribution, and therefore the interactions between
the different shells and between shells and the valence electron are also
spherically symmetric. In thiscentral-field approximationthe total po-
tential energy depends only on the radial coordinate:

VCF(r)=−

Ze^2 / 4 π 0

r

+S(r). (4.3)

In this approximation the Hamiltonian becomes

HCF=

∑N

i=1

{

−

^2

2 m

∇^2 i+VCF(ri)

}

. (4.4)

For this form of potential, the Schr ̈odinger equation forN electrons,
Hψ=Eatomψ,canbeseparatedintoN one-electron equations, i.e.
writing the total wavefunction as a product of single-electron wavefunc-
tions, namely
ψatom=ψ 1 ψ 2 ψ 3 ···ψN, (4.5)

leads toNequations of the form
{
−

^2

2 m

∇^21 +VCF(r 1 )

}

ψ 1 =E 1 ψ 1 , (4.6)

and similar for electronsi=2toN. This assumes that all the elec-
trons see the same potential, which is not as obvious as it may appear.
This symmetric wavefunction is useful to start with (cf. the treatment
of helium before including the effects of exchange symmetry); however,
we know that the overall wavefunction for electrons, including spin,
should be antisymmetric with respect to an interchange of the particle
labels. (Proper antisymmetric wavefunctions are used in the Hartree–
Fock method mentioned later in this chapter.) The total energy of the
system isEatom=E 1 +E 2 +...+EN. The Schr ̈odinger equations for
each electron (eqn 4.6) can be separated into parts to give wavefunctions
of the formψ 1 =R(r 1 )Yl 1 ,m 1 ψspin(1). Angular momentum is conserved
in a central field and the angular equation gives the standard orbital
angular momentum wavefunctions, as in hydrogen. In the radial equa-
tion, however, we haveVCF(r) rather than a potential proportional to
1 /rand so the equation forP(r)=rR(r)is
{
−

^2

2 m

d^2

dr^2

+VCF(r)+


^2 l(l+1)

2 mr^2

}

P(r)=EP(r). (4.7)

To solve this equation we need to know the form ofVCF(r) and compute
the wavefunctions numerically. However, we can learn a lot about the
behaviour of the system by thinking about the form of the solutions,
without actually getting bogged down in the technicalities of solving the
equations. At small distances the electrons experience the full nuclear
charge so that the central electric field is

E(r)→

Ze

4 π 0 r^2

̂r. (4.8)