Concise Physical Chemistry

c17 JWBS043-Rogers September 13, 2010 11:28 Printer Name: Yet to Come

HELIUM 277

with a similar expression forE 2 except thatbreplacesain the first two terms on the

right

E 2 =

b^2

2

−Zb+

ab

(

a^2 + 3 ab+b^2

)

(a+b)^3

The parametersaandbin the Slater-type orbitals for electrons 1 and 2 are min-

imization parameters representing an effective nuclear charge as “experienced” by

each electron, partially shielded by the other electron from the full nuclear charge.

The SCF strategy is to minimizeE 1 using an arbitrary startingband to findaat

the minimum. Thisavalue is then used to findbat the minimumE 1. This value

then replaces the startingbvalue and a new minimization cycle produces a new

aandb, and so on until there is no progressive difference inE 1. The electrical

field experienced by the electrons is now self-consistent, hence it is aself-consistent

fieldSCF.

In this particular case, the calculations are completely symmetrical. Everything

we have said forawe can also say forb. At self-consistency,a=band we can

substituteaforbat any point in the iterative process, knowing that as we ap-

proach self-consistency for one, we approach the same self-consistent value for the

other.

A reasonable thing to do at the end of each iteration would be to calculate

the total energy of the atom as the sum of its two electronic energies EHe=

E 1 +E 2 , but in so doing, we would be calculating the interelectronic repulsion

ab(a^2 + 3 ab+b^2 )/(a+b)^3 twice, once as anr 12 repulsive energy and once as an

r 21 repulsion. Ther 21 repulsion should be dropped to avoid double counting, leaving

EHe=E 1 +E 2 =E 1 +

b^2

2

−Zb

as the correct energy of the helium atom.

Although the Hartree procedure for atoms reaches self-consistency and gives a

qualitative picture of the electron probability densities, the ionization energy is still in

error by a substantial amount (0.014 h∼=37 kJ mol−^1 on three iterations). Something

is missing. A problem arises with the Hartree product when we exchange one atomic

orbital with its adjacent neighbor

ψi(ri)ψj(rj)→ψj(rj)ψi(ri)

The Hartree product is the same after exchange as it was before exchange. This is

equivalent to saying thatψi(ri) andψj(rj) are identical in all respects, even though

we recall from elementary chemistry that no two electronic wave functions can be

exactly alike and that each electron is represented by a unique set offourquantum

numbersn,l,m,ands. In three-dimensional Cartesian space, it is not difficult to

suppose that there are three quantum numbersn,l,andmin the complete solution